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==Blaes (1985)== ===Equilibrium Configuration=== In our [[Apps/PapaloizouPringleTori#Solution|separate discussion of PP84]], we showed that the equilibrium structure of a PP-torus is defined by the enthalpy distribution, <div align="center"> <math> H = \frac{GM_\mathrm{pt}}{\varpi_0} \biggl[ (\chi^2 + \zeta^2)^{-1/2} - \frac{1}{2}\chi^{-2} - C_B^' \biggr] . </math> </div> Normalizing this expression by the enthalpy at the "center" — ''i.e.,'' at the pressure maximum — of the torus which, as we have [[Apps/PapaloizouPringleTori#Pressure_Maximum|already shown]], is <div align="center"> <math> H_0 = \frac{GM_\mathrm{pt}}{2\varpi_0} [1-2C_B^' ] \, </math> </div> gives, <div align="center"> <math> [1-2C_B^' ]\biggl(\frac{H}{H_0}\biggr) = 2(\chi^2 + \zeta^2)^{-1/2} - \chi^{-2} -1 + [1 - 2C_B^' ]. </math> </div> Now, in our review of [[Apps/PapaloizouPringleTori#Model_as_Described_by_Kojima|Kojima's (1986)]] work, we showed that the square of the Mach number at the "center" of the torus is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathfrak{M}_0^2 \equiv \frac{(v_\varphi|_0)^2}{(c_s|_0)^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{2(n+1)}{\gamma}\biggl[ \frac{1}{\chi_-} - 1 \biggr]^{-2}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2n [ 1- 2C_B^' ]^{-1} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~~ [1 - 2C_B^'] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{2n}{\mathfrak{M}_0^2} \, , </math> </td> </tr> </table> </div> where, in obtaining this last expression we have related the adiabatic exponent to the polytropic index via the relation, <math>~\gamma = (n+1)/n</math>. Instead of specifying the system's Mach number, [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes (1985)] defines the dimensionless parameter, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\beta^2 </math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{2n}{\mathfrak{M}_0^2} \, .</math> </td> </tr> </table> </div> Implementing this parameter swap, the equilibrium expression becomes, <div align="center"> <math> \beta^2 \biggl(\frac{H}{H_0}\biggr) = 2(\chi^2 + \zeta^2)^{-1/2} - \chi^{-2} -1 + \beta^2 \, , </math> </div> or, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{H}{H_0} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 - \frac{1}{\beta^2}\biggl[\chi^{-2} - 2(\chi^2 + \zeta^2)^{-1/2} + 1 \biggr] \, .</math> </td> </tr> </table> </div> Looking at Figure 1 of Blaes85 — see also the coordinate definitions given in his equation (2.1) — I conclude that, <div align="center"> <math>~\chi = 1 - x\cos\theta</math> and <math>\zeta = x\sin\theta \, .</math> </div> Hence, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{H}{H_0} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 - \frac{1}{\beta^2}\biggl\{ [1 - x\cos\theta]^{-2} - 2[(1 - x\cos\theta)^2 + x^2\sin^2\theta]^{-1/2} + 1 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 - \frac{1}{\beta^2}\biggl\{ [1 - x\cos\theta]^{-2} - 2[(1 - 2x\cos\theta + x^2\cos^2\theta) + x^2(1-\cos^2\theta)]^{-1/2} + 1 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 - \frac{1}{\beta^2}\biggl[ (1 - x\cos\theta)^{-2} - 2(1 - 2x\cos\theta + x^2)^{-1/2} + 1 \biggr] \, .</math> </td> </tr> </table> </div> This matches equation (2.2) of Blaes85, if we acknowledge that Blaes uses <math>~f</math> — instead of the parameter notation, <math>~\Theta_H</math>, found in [[SSC/Structure/Polytropes#Governing_Relations|our discussion of equilibrium polytropic configurations]] — to denote the normalized enthalpy; that is, <div align="center"> <math>~f_\mathrm{Blaes85} = \Theta_H \equiv \frac{H}{H_0} \, .</math> </div> This expression for the enthalpy throughout a Papaloizou-Pringle torus is valid for tori of arbitrary thickness <math>~(0 < \beta < 1)</math>. When considering only slim tori, [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes (1985)] points out that this expression can be written in terms of the following power series in <math>~x</math> (see his equation 1.3): <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Theta_H</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 - \frac{1}{\beta^2}\biggl[x^2 + x^3(3\cos\theta - \cos^3\theta) + \mathcal{O}(x^4) \biggr] \, .</math> </td> </tr> </table> </div> Blaes then adopts a related parameter that is constant on isobaric surfaces, namely, <div align="center"> <math>\eta^2 \equiv 1 - \Theta_H \, ,</math> </div> which is unity at the surface of the torus and which goes to zero at the cross-sectional center of the torus. Notice that <math>~\eta</math> tracks the "radial" coordinate that measures the distance from the center of the torus; in particular, keeping only the leading-order term in <math>~x</math>, there is a simple linear relationship between <math>~\eta</math> and <math>~x</math>, namely, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\eta</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~[1 - \Theta_H]^{1/2} \approx \frac{x}{\beta} \, .</math> </td> </tr> </table> </div> ===Cubic Equation Solution=== For later use, let's invert the cubic relation to obtain a more broadly applicable <math>~x(\eta)</math> function. Because we are only interested in radial profiles in the equatorial plane — that is, only for the values of <math>~\theta = 0</math> or <math>~\theta=\pi</math> — the relation to be inverted is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x^2 \pm 2 x^3</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~(\beta\eta)^2</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~~ x^3 \pm \tfrac{1}{2}x^2 \mp \tfrac{1}{2}(\beta\eta)^2</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~0 \, .</math> </td> </tr> </table> </div> <div align="center"> <table border="1" align="center" cellpadding="8"> <tr> <td align="center" colspan="6"><font size="+1"><b>Table 1: Example Parameter Values</b></font><p></p> determined by iterative solution for <math>~\beta = \tfrac{1}{10}</math></td> </tr> <tr> <td align="center" rowspan="2"><math>~\eta</math></td> <td align="center" rowspan="2"><math>~\Gamma^2 = 54\beta^2\eta^2</math></td> <td align="center" colspan="2">Inner solution <math>~(\theta = 0)</math> <p></p>[''Superior'' sign in cubic eq.]</td> <td align="center" colspan="2">Outer solution <math>~(\theta = \pi)</math> <p></p>[''Inferior'' sign in cubic eq.]</td> </tr> <tr> <td align="center"><math>^\dagger\biggl(\frac{x_\mathrm{root}}{\beta}\biggr)</math></td> <td align="center"><math>~6(S + T)</math></td> <td align="center"><math>^\dagger\biggl(\frac{x_\mathrm{root}}{\beta}\biggr)</math></td> <td align="center"><math>~6(S + T)</math></td> </tr> <tr> <td align="right">0.25</td> <td align="center">0.03375</td> <td align="right">0.244112</td> <td align="right">1.14647</td> <td align="right">0.256675</td> <td align="right">-0.84600</td> </tr> <tr> <td align="right">1.0</td> <td align="center">0.54</td> <td align="right">0.91909</td> <td align="right">1.55145</td> <td align="right">1.1378</td> <td align="right">-0.31732</td> </tr> <tr> <td align="left" colspan="6"> <sup>†</sup>Here, <math>~x_\mathrm{root}</math> has been determined via a brute-force, iterative technique. </td> </tr> </table> </div> Following [http://mathworld.wolfram.com/CubicFormula.html Wolfram's discussion of the cubic formula], we should view this expression as a specific case of the general formula, <div align="center"> <math>~x^3 + a_2x^2 + a_1x + a_0 = 0 \, ,</math> </div> in which case, as is detailed in equations (54) - (56) of [http://mathworld.wolfram.com/CubicFormula.html Wolfram's discussion of the cubic formula], the three roots of any cubic equation are: <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x_1</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -\frac{1}{3}a_2 + (S + T) \, , </math> </td> </tr> <tr> <td align="right"> <math>~x_2</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -\frac{1}{3}a_2 - \frac{1}{2} (S + T) + \frac{1}{2} \it{i} \sqrt{3} (S-T)\, , </math> </td> </tr> <tr> <td align="right"> <math>~x_3</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -\frac{1}{3}a_2 - \frac{1}{2} (S + T) - \frac{1}{2} \it{i} \sqrt{3} (S-T)\, , </math> </td> </tr> </table> </div> where, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~S</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~[R + \sqrt{D}]^{1/3} \, ,</math> </td> </tr> <tr> <td align="right"> <math>~T</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~[R - \sqrt{D}]^{1/3} \, ,</math> </td> </tr> <tr> <td align="right"> <math>~D</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~Q^3 + R^2 \, ,</math> </td> </tr> <tr> <td align="right"> <math>~Q</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{3a_1 - a_2^2}{3^2} \, ,</math> </td> </tr> <tr> <td align="right"> <math>~R</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{3^2a_2 a_1 - 3^3a_0 - 2a_2^3}{2\cdot 3^3} \, . </math> </td> </tr> </table> </div> ====Outer [inferior sign] Solution==== Focusing, first, on the ''inferior'' sign convention, which corresponds to the "outer" solution <math>~(\theta = \pi)</math>, we see that the coefficients that lead to our specific cubic equation are: <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~a_2</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \tfrac{1}{2} \, ,</math> </td> </tr> <tr> <td align="right"> <math>~a_1</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~0 \, ,</math> </td> </tr> <tr> <td align="right"> <math>~a_0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tfrac{1}{2}(\beta\eta)^2 \, .</math> </td> </tr> </table> </div> Applying Wolfram's definitions of the <math>~Q</math> and <math>~R</math> parameters to our particular problem gives, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~Q</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{3a_1 - a_2^2}{3^2} = -\biggl(\frac{a_2}{3}\biggr)^2 = - \frac{1}{2^2\cdot 3^2} \, ;</math> </td> </tr> <tr> <td align="right"> <math>~R</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{3^2a_2 a_1 - 3^3a_0 - 2a_2^3}{2\cdot 3^3} = \frac{1}{2\cdot 3^3} \biggl[ -\frac{ 3^3}{2}(\beta\eta)^2 + \frac{1}{2^2}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{1}{2^3\cdot 3^3} \biggl[ 1 - 2\cdot 3^3(\beta\eta)^2\biggr] \, . </math> </td> </tr> </table> </div> Defining the parameter, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Gamma^2</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ 2\cdot 3^3(\beta\eta)^2 \, ,</math> </td> </tr> </table> </div> we therefore have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(2\cdot 3)^6 D</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~( 1 - \Gamma^2 )^2-1 \, ,</math> </td> </tr> <tr> <td align="right"> <math>~(2\cdot 3)^3S^3</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~(2\cdot 3)^3 R + \sqrt{(2\cdot 3)^6D} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~(1-\Gamma^2) + \sqrt{( 1 - \Gamma^2 )^2-1}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~(1-\Gamma^2) + i\sqrt{1-( 1 - \Gamma^2 )^2} \, ;</math> </td> </tr> <tr> <td align="right"> <math>~(2\cdot 3)^3T^3</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~(1-\Gamma^2) - i\sqrt{1-( 1 - \Gamma^2 )^2} \, .</math> </td> </tr> </table> </div> <div align="center" id="CubeRootImaginary"> <table border="1" width="60%" cellpadding="8"> <tr><td align="left"> <font color="purple" size="+1"><b>ASIDE:</b></font> The cube root of an imaginary number … <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\ell^3</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~A \pm i \sqrt{1-A^2}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~r_\ell e^{i\theta_\ell} \, ,</math> </td> </tr> </table> where, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~r_\ell</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~( A^2 + 1-A^2 )^{1/2} = 1 \, ,</math> </td> </tr> </table> and, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\theta_\ell</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\pm \tan^{-1}\biggl( \frac{\sqrt{1-A^2}}{A} \biggr) = \pm \cos^{-1}A \, .</math> </td> </tr> </table> Now, according to [http://math.stackexchange.com/questions/8760/what-are-the-three-cube-roots-of-1 this online resource], the three roots <math>~(j=0,1,2)</math> of <math>~\ell^3</math> are, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="center"><math>~\ell_j = r_\ell^{1/3}e^{i(\theta_\ell + 2j\pi)/3)} \, ,</math></td> </tr> </table> </div> which, for our specific problem gives, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\ell_j</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~e^{i\theta_\pm/3} \cdot e^{i(2j\pi/3)} \, ,</math> </td> </tr> </table> </div> where the subscript on <math>~\theta</math> refers to the <math>~\pm</math> in our original expression for <math>~\ell</math>. <!-- <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\ell_j</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\cos[(2j\pi +\theta_\ell)/3] + i \sin[(2j\pi + \theta_\ell)/3] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \cos\biggl[ {\frac{1}{3}\biggl(2j\pi \pm \cos^{-1}A\biggr)} \biggr] + i \sin\biggl[ {\frac{1}{3}\biggl(2j\pi \pm \cos^{-1}A\biggr)} \biggr] </math> </td> </tr> </table> </div> --> </td></tr> </table> </div> In our particular case, after associating <math>~A \leftrightarrow (1-\Gamma^2)</math>, we can write, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ 2\cdot 3(S + T) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[(1-\Gamma^2) + i\sqrt{1-( 1 - \Gamma^2 )^2} \biggr]^{1/3} + \biggl[(1-\Gamma^2) - i\sqrt{1-( 1 - \Gamma^2 )^2} \biggr]^{1/3} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ e^{i\theta_+/3} \cdot e^{i(2j\pi/3)} + e^{i\theta_-/3} \cdot e^{i(2j\pi/3)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~e^{i(2j\pi/3)} \biggl\{ e^{i[\cos^{-1}(1-\Gamma^2)]/3} + e^{-i[\cos^{-1}(1-\Gamma^2)]/3} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~e^{i(2j\pi/3)} \biggl\{\cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] + i\sin\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~+ \cos\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] - i\sin\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2 e^{i(2j\pi/3)} \cdot \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, .</math> </td> </tr> </table> </div> Similarly, we can write, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ 2\cdot 3(S - T) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[(1-\Gamma^2) + i\sqrt{1-( 1 - \Gamma^2 )^2} \biggr]^{1/3} - \biggl[(1-\Gamma^2) - i\sqrt{1-( 1 - \Gamma^2 )^2} \biggr]^{1/3} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ e^{i\theta_+/3} \cdot e^{i(2j\pi/3)} - e^{i\theta_-/3} \cdot e^{i(2j\pi/3)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~e^{i(2j\pi/3)} \biggl\{ e^{i[\cos^{-1}(1-\Gamma^2)]/3} - e^{-i[\cos^{-1}(1-\Gamma^2)]/3} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~e^{i(2j\pi/3)} \biggl\{\cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] + i\sin\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~- \cos\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] + i\sin\biggl[ \tfrac{1}{3} \cos^{-1}(1-\Gamma^2)\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2i e^{i(2j\pi/3)} \cdot \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, . </math> </td> </tr> </table> </div> Focusing specifically on the "j=0" root, and setting <math>~a_2 = -\tfrac{1}{2}</math>, we therefore have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~6x_1-1</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 6(S + T) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, ;</math> </td> </tr> <tr> <td align="right"> <math>~6x_2-1</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{1}{2} \biggl\{ 2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] - i\sqrt{3} \cdot 2i \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -2\biggl\{ \frac{1}{2}\cdot \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] +\frac{\sqrt{3}}{2} \cdot \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) + \frac{2\pi}{3}\biggr] \biggr\} \, ; </math> </td> </tr> <tr> <td align="right"> <math>~6x_3-1</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{1}{2} \biggl\{ 2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] + i\sqrt{3} \cdot 2i \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -2\biggl\{ \frac{1}{2}\cdot \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] -\frac{\sqrt{3}}{2} \cdot \sin\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) - \frac{2\pi}{3}\biggr] \biggr\} \, . </math> </td> </tr> </table> </div> <div align="center"> <table border="1" align="center" cellpadding="8"> <tr> <td align="center" colspan="8"><font size="+1"><b>Table 1: Analytically Evaluated Roots</b></font><p></p> determined for <math>~\beta = \tfrac{1}{10}</math></td> </tr> <tr> <td align="center" rowspan="2"><math>~\eta</math></td> <td align="center" rowspan="2"><math>~\Gamma^2 = 54\beta^2\eta^2</math></td> <td align="center" colspan="3">Inner solution <math>~(\theta = 0)</math> <p></p>[''Superior'' sign in cubic eq.]</td> <td align="center" colspan="3">Outer solution <math>~(\theta = \pi)</math> <p></p>[''Inferior'' sign in cubic eq.]</td> </tr> <tr> <td align="center"><math>~x_1/\beta</math></td> <td align="center"><math>~x_2/\beta</math></td> <td align="center"><math>~x_3/\beta</math></td> <td align="center"><math>~x_1/\beta</math></td> <td align="center"><math>~x_2/\beta</math></td> <td align="center"><math>~x_3/\beta</math></td> </tr> <tr> <td align="right">0.25</td> <td align="center">0.03375</td> <td align="right">-4.98744</td> <td align="right" bgcolor="yellow">0.24411</td> <td align="right">-0.25667</td> <td align="right">4.98744</td> <td align="right">-0.24411</td> <td align="right" bgcolor="yellow">0.25667</td> </tr> <tr> <td align="right">1.0</td> <td align="center">0.54</td> <td align="right">-4.78128</td> <td align="right" bgcolor="yellow">0.91909</td> <td align="right">-1.1378</td> <td align="right">4.78128</td> <td align="right">-0.91909</td> <td align="right" bgcolor="yellow">1.1378</td> </tr> <tr> <td align="center" colspan="2"> </td> <td align="center" colspan="3">CONFIRMATION: In all cases, <p></p><math>~x^2 + 2x^3 = (\beta\eta)^2</math></td> <td align="center" colspan="3">CONFIRMATION: In all cases, <p></p><math>~x^2 - 2x^3 = (\beta\eta)^2</math></td> </tr> </table> </div> <!-- ************************* --> ====Inner [superior sign] Solution==== Next, examing the ''superior'' sign convention, which corresponds to the "inner" solution <math>~(\theta = 0)</math>, we see that the coefficients that lead to our specific cubic equation are: <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~a_2</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tfrac{1}{2} \, ,</math> </td> </tr> <tr> <td align="right"> <math>~a_1</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~0 \, ,</math> </td> </tr> <tr> <td align="right"> <math>~a_0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \tfrac{1}{2}(\beta\eta)^2 \, .</math> </td> </tr> </table> </div> Following the same set of steps that were followed in [[Appendix/Ramblings/PPTori#Outer_.5Binferior_sign.5D_Solution|determining the "outer" solution]], here we find: <math>~Q</math> remains the same; <math>~R</math> has the same magnitude, but changes sign; and, hence, <math>~D</math> remains the same. We therefore have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(2\cdot 3)^3S^3</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- (1-\Gamma^2) + i\sqrt{1-( 1 - \Gamma^2 )^2} \, ;</math> </td> </tr> <tr> <td align="right"> <math>~(2\cdot 3)^3T^3</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- (1-\Gamma^2) - i\sqrt{1-( 1 - \Gamma^2 )^2} \, ,</math> </td> </tr> </table> </div> which leads to the following expressions for the three "inner" roots: <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~6x_1+1</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- 2 \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2)\biggr] \, ;</math> </td> </tr> <tr> <td align="right"> <math>~6x_2+1</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) + \frac{2\pi}{3}\biggr] \biggr\} \, ; </math> </td> </tr> <tr> <td align="right"> <math>~6x_3+1</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - 2\biggl\{ \cos\biggl[ \tfrac{1}{3}\cos^{-1}(1-\Gamma^2) - \frac{2\pi}{3}\biggr] \biggr\} \, . </math> </td> </tr> </table> </div> ===Analytically Prescribed Eigenvector=== ====Our Notation==== As is explicitly defined in [[Appendix/Ramblings/AzimuthalDistortions#Figure1|Figure 1 of our accompanying detailed notes]], we have chosen to represent the spatial structure of an eigenfunction in the equatorial-plane of toroidal-like configurations via the expression, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl\{ f_m(\varpi)e^{-im\phi_m} \biggr\} \, .</math> </td> </tr> </table> </div> In general, we should assume that the function that delineates the radial dependence of the eigenfunction has both a real and an imaginary component, that is, we should assume that, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~f_m(\varpi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\mathcal{A}(\varpi) + i\mathcal{B}(\varpi) \, ,</math> </td> </tr> </table> </div> in which case the square of the modulus of the function is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~|f_m|^2 \equiv f_m \cdot f^*_m </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\mathcal{A}^2 + \mathcal{B}^2 \, .</math> </td> </tr> </table> </div> We can rewrite this complex function in the form, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~f_m(\varpi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~|f_m|e^{-i[\alpha(\varpi) + \pi/2]} \, ,</math> </td> </tr> </table> </div> if the angle, <math>~\alpha(\varpi)</math> is defined such that, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\sin\alpha = \frac{\mathcal{A}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}}</math> </td> <td align="center"> and </td> <td align="left"> <math>~\cos\alpha = \frac{\mathcal{B}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}}</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~~ \alpha</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\tan^{-1}\biggl(\frac{\mathcal{A}}{\mathcal{B}}\biggr) = \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] \, .</math> </td> </tr> </table> </div> Hence, the spatial structure of the eigenfunction can be rewritten as, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~|f_m(\varpi)|e^{-i[\alpha(\varpi) + \pi/2+ m\phi_m]} \, . </math> </td> </tr> </table> </div> From this representation we can see that, at each radial location, <math>~\varpi</math>, the phase angle(s) at which the fractional perturbation exhibits its maximum amplitude, <math>~|f_m|</math>, is identified by setting the exponent of the exponential to zero. That is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\phi_m = \phi_\mathrm{max}(\varpi)</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~-\frac{1}{m}\biggl[\alpha(\varpi) +\frac{\pi}{2}\biggr] = -\frac{1}{m}\biggl\{ \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] +\frac{\pi}{2} \biggr\} \, .</math> </td> </tr> </table> </div> An equatorial-plane plot of <math>~\phi_\mathrm{max}(\varpi)</math> should produce the "constant phase locus" referenced, for example, in recent papers from the [[Appendix/Ramblings/ToHadleyAndImamura#Summary_for_Hadley_.26_Imamura|Imamura & Hadley collaboration]]. <!-- SECOND ATTEMPT This is the form that has been adopted broadly by the numerical simulation community, as graphical displays of <math>~f_m(\varpi)</math> and <math>~\phi_m(\varpi)</math> have been used to study the structure of unstable eigenmodes — see, for example, our discussion of [[Appendix/Ramblings/ToHadleyAndImamura#Summary_for_Hadley_.26_Imamura|simulation results published by the Imamura & Hadley collaboration]]. Multiplying this expression through by its complex conjugate gives the square of the modulus of the function. <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl| \frac{\delta\rho}{\rho_0}\biggr|^2_\mathrm{spatial} \equiv \biggl[ \frac{\delta\rho}{\rho_0}\biggr] \cdot \biggl[ \frac{\delta\rho}{\rho_0}\biggr]^* </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~f^2_m(\varpi)e^{-im[\phi_m(\varpi)]} \cdot e^{+im[\phi_m(\varpi)]} = f^2_m(\varpi) \, .</math> </td> </tr> </table> </div> We see, therefore, that written in this manner, <math>~f_m</math> is the modulus of the eigenfunction. Alternatively, we could choose to omit explicit reference to an angular phase function and write the perturbation amplitude as a function with an imaginary as well as a real part, say, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\mathcal{A}(\varpi) - i\mathcal{B}(\varpi) \, .</math> </td> </tr> </table> </div> This is the form often used in research papers that seek to find analytic expressions for the structure of unstable eigenmodes, such as the works of [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes85] and [http://adsabs.harvard.edu/abs/1986MNRAS.221..339G GGN86]. A mapping from one representation to the other is accomplished by, first, constructing the modulus of the complex perturbation amplitude and equating it to <math>~f_m</math>, that is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~f_m(\varpi) = \sqrt{\biggl[ \frac{\delta\rho}{\rho_0}\biggr] \cdot \biggl[ \frac{\delta\rho}{\rho_0}\biggr]^* }</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\sqrt{\mathcal{A}^2(\varpi) + \mathcal{B}^2(\varpi)} \, .</math> </td> </tr> </table> </div> Second, the phase function is obtained via the relation, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~m\phi_m(\varpi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tan^{-1}\biggl[ \frac{\mathcal{B}(\varpi)}{\mathcal{A}(\varpi)} \biggr] \, ,</math> </td> </tr> </table> </div> where, more thoroughly it must be understood that, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\cos(m\phi_m) = \biggl[ \frac{\mathcal{A}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}} \biggr]</math> </td> <td align="center"> and </td> <td align="left"> <math>~\sin(m\phi_m) = \biggl[ \frac{\mathcal{B}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}} \biggr] \, .</math> </td> </tr> </table> </div> To demonstrate that this is the proper mapping for the phase function, let's plug the expression for <math>~f_m</math> along with the expressions for <math>~\cos(m\phi_m)</math> and <math>~\sin(m\phi_m)</math> into our first relation, that is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~f_m(\varpi) \{ \cos[m\phi_m(\varpi)] - i\sin[m\phi_m(\varpi)] \}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\sqrt{\mathcal{A}^2 + \mathcal{B}^2}\biggl\{ \biggl[ \frac{\mathcal{A}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}} \biggr] - i\biggl[ \frac{\mathcal{B}}{\sqrt{\mathcal{A}^2 + \mathcal{B}^2}} \biggr] \biggr\}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \mathcal{A} - i\mathcal{B} \, .</math> </td> </tr> </table> </div> Q.E.D. --> <!-- USEFUL (BUT NOT FULLY CORRECT) MANIPULATION OF COMPLEX EIGENFUNCTION EXPRESSIONS ... Recognizing that the leading factor, <math>~f_m</math>, is, in general, composed of both a real part and an imaginary part, we can rewrite this expression as, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[\mathrm{Re}(f_m) + i\mathrm{Im}(f_m)\biggr] \biggl[\cos(m\phi_m) - i\sin(m\phi_m) \biggr]</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\mathrm{Re}(f_m) \cos(m\phi_m) + \mathrm{Im}(f_m)\sin(m\phi_m)\biggr] + i \biggl[- \mathrm{Re}(f_m)\sin(m\phi_m) + \mathrm{Im}(f_m)\cos(m\phi_m)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2} \biggl\{ \biggl[\frac{\mathrm{Re}(f_m) \cos(m\phi_m) + \mathrm{Im}(f_m)\sin(m\phi_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2} }\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + i \biggl[\frac{- \mathrm{Re}(f_m)\sin(m\phi_m) + \mathrm{Im}(f_m)\cos(m\phi_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2} }\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl| \frac{\delta\rho}{\rho_0}\biggr|_\mathrm{spatial} \biggl\{\sin(\beta_f + m\phi_m) + i \cos(\beta_f + m\phi_m)\biggr\} \, , </math> </td> </tr> </table> </div> where, the modulus of this function is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl| \frac{\delta\rho}{\rho_0}\biggr|_\mathrm{spatial} \equiv \sqrt{\biggl[ \frac{\delta\rho}{\rho_0}\biggr]\cdot\biggl[ \frac{\delta\rho}{\rho_0}\biggr]^*} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl\{ \biggl[\mathrm{Re}(f_m) \cos(m\phi_m) + \mathrm{Im}(f_m)\sin(m\phi_m)\biggr]^2 + \biggl[- \mathrm{Re}(f_m)\sin(m\phi_m) + \mathrm{Im}(f_m)\cos(m\phi_m)\biggr]^2 \biggr\}^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl\{ [\mathrm{Re}(f_m) \cos(m\phi_m)]^2 + 2\mathrm{Re}(f_m)\mathrm{Im}(f_m)\sin(m\phi_m)\cos(m\phi_m)+ [\mathrm{Im}(f_m)\sin(m\phi_m)]^2 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + [\mathrm{Re}(f_m)\sin(m\phi_m)]^2 - 2\mathrm{Re}(f_m)\mathrm{Im}(f_m)\sin(m\phi_m)\cos(m\phi_m) + [\mathrm{Im}(f_m)\cos(m\phi_m)]^2 \biggr\}^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \sqrt{ [\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2} \, , </math> </td> </tr> </table> </div> and where we have introduced a new angle, <math>~\beta_f</math>, such that, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\sin\beta_f = \frac{\mathrm{Re}(f_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2}}</math> </td> <td align="center"> and </td> <td align="left"> <math>~\cos\beta_f = \frac{\mathrm{Im}(f_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2}} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~~ \beta_f </math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\tan^{-1}\biggl[\frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] \, .</math> </td> </tr> </table> </div> Alternatively, we could choose to write, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ \frac{\delta\rho}{\rho_0}\biggr]_\mathrm{spatial} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl| \frac{\delta\rho}{\rho_0}\biggr|_\mathrm{spatial} \biggl\{ \cos(\alpha_f - m\phi_m) + i \sin(\alpha_f - m\phi_m)\biggr\} = \biggl| \frac{\delta\rho}{\rho_0}\biggr|_\mathrm{spatial} e^{i(\alpha_f -m\phi_m)} \, , </math> </td> </tr> </table> </div> where we have introduced a new angle, <math>~\alpha_f</math>, such that, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\cos\alpha_f = \frac{\mathrm{Re}(f_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2}}</math> </td> <td align="center"> and </td> <td align="left"> <math>~\sin\alpha_f = \frac{\mathrm{Im}(f_m)}{\sqrt{[\mathrm{Re}(f_m)]^2 + [\mathrm{Im}(f_m)]^2}} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~~ \alpha_f </math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\tan^{-1}\biggl[\frac{\mathrm{Im}(f_m)}{\mathrm{Re}(f_m)} \biggr] \, .</math> </td> </tr> </table> </div> Notice that, when written in this form, it is clear from taking the complex conjugate of the function that, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\alpha_f - m\phi_m</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~0 </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~~ m\phi_m</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tan^{-1}\biggl[\frac{\mathrm{Im}(f_m)}{\mathrm{Re}(f_m)} \biggr] \, .</math> </td> </tr> </table> </div> FINISHED EXTRACTION OF COMPLEX FUNCTION MANIPULATION --> ====General Formulation==== From my initial focused reading of the analysis presented by [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes (1985)], I conclude that, in the infinitely slender torus case, unstable modes in PP tori exhibit eigenvectors of the form, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\delta W}{W_0} \equiv \biggl[ \frac{W(\eta,\theta)}{C} - 1 \biggr]e^{im\Omega_p t}e^{-y_2 (\Omega_0 t)} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl\{ f_m(\eta,\theta)e^{-i[m\phi_m + k\theta]} \biggr\} \, ,</math> </td> </tr> </table> </div> <!-- [<font color="red"><b>NOTE:</b></font> Initially, I wrote "+ k" rather than "- k" in the exponent of the exponential term on the RHS of this expression; but experience shows that "- k" is required to achieve proper behavior of the "constant phase locus" plot, as displayed below.] --> where we have written the perturbation amplitude in a manner that conforms with the notation that we have used in [[Appendix/Ramblings/AzimuthalDistortions#Figure1|Figure 1 of a related, but more general discussion]]. As is summarized in §1.3 of Blaes (1985), for "thick" (but, actually, still quite thin) tori, "exactly one exponentially growing mode exists for each value of the azimuthal wavenumber <math>~m</math>," and its complex amplitude takes the following form (see his equation 1.10): <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~f_m(\eta,\theta)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\beta^2 m^2 \biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2} \pm 4i\biggl(\frac{3}{2n+2}\biggr)^{1/2} \eta\cos\theta\biggr] + \mathcal{O}(\beta^3) \, . </math> </td> </tr> </table> </div> Aside from an arbitrary leading scale factor, we should therefore find that the amplitude (modulus) of the enthalpy perturbation is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl|\frac{\delta W}{W_0} \biggr|</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\sqrt{[\mathrm{Re}(f_m)]^2+ [\mathrm{Im}(f_m)]^2} \, ;</math> </td> </tr> </table> </div> and the associated phase function is, <!-- <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~m\phi_m - k\theta</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tan^{-1} \biggl\{ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr\} \, .</math> </td> </tr> </table> </div> [<font color="red"><b>NOTE:</b></font> Initially, I expected the argument inside the arctan function to be the ratio, <math>~\mathrm{Im}(f_m)/\mathrm{Re}(f_m)</math>; but experience shows that the reciprocal of this ratio is required to achieve proper behavior of the "constant phase locus" plot, as displayed below.] --> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~m\phi_\mathrm{max}(\varpi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -\biggl\{ \tan^{-1}\biggl[ \frac{\mathrm{Re}(f_m)}{\mathrm{Im}(f_m)} \biggr] +\frac{\pi}{2} +k\theta \biggr\} \, .</math> </td> </tr> </table> </div> Now, keeping in mind that, for the time being, we are only interested in examining the shape of the unstable eigenvector in the ''equatorial plane'' of the torus, we can set, <div align="center"> <math>~\cos\theta ~~ \rightarrow ~~ \pm 1 \, .</math> </div> Hence, we have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{1}{\beta^4 m^4}\biggl|\frac{\delta W}{W_0} \biggr|^2</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[2\eta^2 - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2}\biggr]^2 + 16\biggl[\biggl(\frac{3}{2n+2}\biggr)^{1/2} \eta \biggr]^2 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{[2^2(n+1)^2]^2}\biggl[2^3(n+1)^2\eta^2 - 3(n+1)\eta^2 - (4n+1) \biggr]^2 + \frac{2^3 \cdot 3\eta^2}{(n+1)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{[2^2(n+1)^2]^2}\biggl[(4n+1) - (n+1)[2^3(n+1)-3]\eta^2 \biggr]^2 + \frac{2^3 \cdot 3\eta^2}{(n+1)} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~~ \biggl[\frac{2(n+1)}{\beta m} \biggr]^4 \biggl|\frac{\delta W}{W_0} \biggr|^2</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[(4n+1) - (n+1)[2^3(n+1)-3]\eta^2 \biggr]^2 + 2^7 \cdot 3(n+1)^3\eta^2 \, . </math> </td> </tr> </table> </div> Also, keeping in mind that, because of the <math>~\cos\theta</math> factor, the sign on the imaginary term flips its sign when switching from the "inner" region to the "outer" region of the torus, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="center" bgcolor="blue"> </td> <td align="right"> <math>~m\phi_\mathrm{max}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \biggl\{ \tan^{-1}\biggl[ \frac{(4n+1) - (n+1)[2^3(n+1)-3]\eta^2 }{2^{7/2}\cdot 3^{1/2}(n+1)^{3/2}\eta} \biggr]+\frac{\pi}{2}\biggr\}</math> </td> <td align="center"> over </td> <td align="left"> inner <math>~(\theta=0)</math> region of the torus; </td> </tr> <tr><td colspan="6" align="center">while</td></tr> <tr> <td align="center" bgcolor="green"> </td> <td align="right"> <math>~m\phi_\mathrm{max}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \biggl\{ \tan^{-1}\biggl[- \frac{(4n+1) - (n+1)[2^3(n+1)-3]\eta^2 }{2^{7/2}\cdot 3^{1/2}(n+1)^{3/2}\eta}\biggr\} +\frac{3\pi}{2} \biggr\}</math> </td> <td align="center"> over </td> <td align="left"> outer <math>~(\theta=\pi)</math> region of torus. </td> </tr> </table> </div> ====Incompressible Slim Tori==== If we specifically consider geometrically slim, incompressible tori — that is, if we set the polytropic index, <math>~n=0</math> — to lowest order the eigenfunction derived by [http://adsabs.harvard.edu/abs/1985MNRAS.216..553B Blaes (1985)] takes the form, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~f_m(\eta,\theta)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\beta^2 m^2 \biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4} - \frac{1}{4} \pm 4i\biggl(\frac{3}{2}\biggr)^{1/2} \eta\cos\theta\biggr] + \cancelto{0}{\mathcal{O}(\beta^3)} \, . </math> </td> </tr> </table> </div> ====Check Validity of Blaes85 Eigenvector==== =====Step 1===== Equation (2.6) of Blaes85 states that, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\beta^2 \eta^2 = [x^2 + x^3(3\cos\theta - \cos^3\theta) ]</math> </td> <td align="center"> <math>~\Rightarrow</math> </td> <td align="left"> <math>~ \beta^2(1 - \eta^2) = [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \, .</math> </td> </tr> </table> </div> This means that, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial \eta^2}{\partial x}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{\beta^2}\biggl[ 2x +3x^2(3\cos\theta - \cos^3\theta) \biggr] \, ; </math> </td> </tr> </table> </div> and, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial \eta^2}{\partial\theta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{x^3}{\beta^2}\biggl[ -3\sin\theta + 3\sin\theta \cos^2\theta \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3x^3 \sin\theta}{\beta^2}\biggl[\cos^2\theta -1\biggr] \, .</math> </td> </tr> </table> </div> Furthermore, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial \eta}{\partial x}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{2\beta}\biggl[x^2 + x^3(3\cos\theta - \cos^3\theta)\biggr]^{-1/2}\biggl[ 2x +3x^2(3\cos\theta - \cos^3\theta) \biggr] \, ; </math> </td> </tr> </table> </div> and, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial \eta}{\partial\theta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{2\beta}\biggl[x^2 + x^3(3\cos\theta - \cos^3\theta)\biggr]^{-1/2}\biggl[ -3\sin\theta + 3\sin\theta \cos^2\theta \biggr] \, .</math> </td> </tr> </table> </div> =====Step 2===== Equation (4.14) of Blaes85 states that, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\nu + m</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \pm im\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta \, ; </math> </td> </tr> </table> </div> and equation (4.13) of Blaes85 states that, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\delta W}{A_{00}}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 + \beta^2 m^2\biggl[ 2\eta^2 \cos^2\theta - \frac{3\eta^2}{4(n+1)} - \frac{(4n+1)}{4(n+1)^2} \pm 4i\biggl( \frac{3}{2n+2} \biggr)^{1/2} \eta\cos\theta \biggr] </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~~\frac{1}{m^2}\biggl[\frac{\delta W}{A_{00}}-1\biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\beta^2 \biggl\{- \frac{(4n+1)}{4(n+1)^2} + \eta^2\biggl[ 2\cos^2\theta - \frac{3}{4(n+1)}\biggr] \pm i\biggl( \frac{2^3\cdot 3}{n+1} \biggr)^{1/2} \eta\cos\theta \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{\beta^2}{2^2(n+1)^2} \biggl\{- (4n+1) + \eta^2 [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \pm i ~[ 2^7\cdot 3(n+1)^3 ]^{1/2} \eta\cos\theta \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~~\Lambda \equiv \frac{2^2(n+1)^2}{m^2}\biggl[\frac{\delta W}{A_{00}}-1\biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- (4n+1)\beta^2 + [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [x^2 + x^3(3\cos\theta - \cos^3\theta)] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- (4n+1)\beta^2 + (n+1)x^2[ 2^3(n+1)\cos^2\theta - 3] \cdot [1 + xb] ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math> </td> </tr> </table> </div> <!-- SEPARATE EVALUATION OF LAMBDA --> <div align="center"> <table border="1" cellpadding="8" align="center"> <tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained …'''</font></td></tr> <tr><td align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Lambda </math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ \frac{2^2(n+1)^2}{m^2}\biggl[\frac{\delta W}{A_{00}}-1\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \, , </math> </td> </tr> <tr><td colspan="3" align="left">where,</td></tr> <tr> <td align="right"> <math>~(\beta\eta)^2</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~x^2(1+xb) \, ,</math> </td> </tr> <tr> <td align="right"> <math>~b</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~3\cos\theta - \cos^3\theta \, .</math> </td> </tr> </table> </td></tr></table> </div> Note that, differentiating the left-hand-side with respect to either coordinate <math>~(x</math> or <math>~\theta)</math> gives, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial \Lambda}{\partial x_i} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{2^2(n+1)^2}{m^2}\cdot \frac{\partial }{\partial x_i} \biggl(\frac{\delta W}{A_{00}} \biggr) </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~~\frac{\partial }{\partial x_i} \biggl(\frac{\delta W}{A_{00}} \biggr) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[\frac{m}{2(n+1)}\biggr]^2 \cdot \frac{\partial \Lambda}{\partial x_i} \, .</math> </td> </tr> </table> </div> Given that <math>~\nu</math> has both real and imaginary parts, presumably, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\nu^2 \equiv \nu \cdot \nu^*</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl\{ -m \pm im\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta\biggr\} \biggl\{ -m \mp im\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta\biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~m^2 + m^2\biggl[\frac{3}{2(n+1)}\biggr]\beta^2 \, . </math> </td> </tr> </table> </div> For later reference, let's take the relevant partial derivatives of the function, <math>~\Lambda(x,\theta)</math>. Adopting the shorthand notation, <div align="center"> <math>~b \equiv (3\cos\theta-\cos^3\theta) \, ,</math> </div> we have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial \Lambda}{\partial x}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot \frac{\partial }{\partial x} \biggl\{ [x^2 + x^3b] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \frac{\partial }{\partial x} \biggl\{ [x^2 + x^3b ]^{1/2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [2x + 3x^2b] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot [1 + xb ]^{-1/2} \cdot [2 + 3xb ] </math> </td> </tr> </table> </div> <!-- SEPARATE EVALUATION OF \partial\Lambda/\partial x --> <div align="center"> <table border="1" cellpadding="8" align="center"> <tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained …'''</font></td></tr> <tr><td align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial\Lambda}{\partial x}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (n+1)[2^3(n+1)\cos^2\theta -3]x(2+3xb) ~~~\pm ~~i~\beta\cos\theta [2^5\cdot 3 (n+1)^3]^{1/2} \cdot \frac{x(2+3xb)}{(\beta\eta)} \, , </math> </td> </tr> <tr><td colspan="3" align="left">which is the same.</td></tr> </table> </td></tr></table> </div> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial^2 \Lambda}{\partial x^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot \frac{\partial }{\partial x}[2x + 3x^2b] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \frac{\partial }{\partial x}\biggl\{[1 + xb ]^{-1/2} \cdot [2 + 3xb ] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [2 + 6xb] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \biggl\{ -\tfrac{b}{2}[1 + xb ]^{-3/2} \cdot [2 + 3xb ] + [1 + xb ]^{-1/2} \cdot [3b ] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [1 + 3xb] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot \biggl\{ - [2b + 3xb^2 ] + [6b + 6xb^2 ] \biggr\} \frac{1}{2(1+xb)^{3/2}} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~2[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [1 + 3xb] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~\pm ~i~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot b[1 + xb ]^{-3/2} [4 + 3xb ] </math> </td> </tr> </table> </div> <!-- SEPARATE EVALUATION OF 2nd DERIVATIVE wrt "x" --> <div align="center"> <table border="1" cellpadding="8" align="center"> <tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained …'''</font></td></tr> <tr><td align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial^2\Lambda}{\partial x^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2(n+1)[2^3(n+1)\cos^2\theta -3](1+3xb) ~\pm~~i~\beta\cos\theta [2^3\cdot 3(n+1)^3]^{1/2} \biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr] \, , </math> </td> </tr> <tr><td colspan="3" align="left">which is the same.</td></tr> </table> </td></tr></table> </div> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial \Lambda}{\partial \theta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{\partial }{\partial \theta} \biggl\{ [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [x^2 + x^3(3\cos\theta - \cos^3\theta)] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~\pm ~i~\frac{\partial }{\partial \theta} \biggl\{ \beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{1/2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot x^3 \cdot [\sin\theta (-3 + 3\cos^2\theta)] + [x^2 + x^3(3\cos\theta - \cos^3\theta)] \cdot [ -2^4(n+1)^2 \sin\theta \cos\theta ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [x^2\cos^2\theta + x^3(3\cos^3\theta - \cos^5\theta) ]^{-1/2} \cdot (-\sin\theta) [2x^2\cos\theta + x^3(9\cos^2\theta - 5\cos^4\theta) ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -3 x^3 \sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] -2^4(n+1)^2 x^2\sin\theta \cos\theta [1 + x(3\cos\theta-\cos^3\theta) ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~\beta (-1) [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} \cdot x~\sin\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -3 x^3 \sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] -2^4(n+1)^2 x^2\sin\theta \cos\theta [1 + xb ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~\beta (-1) [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2} \cdot x~\sin\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] </math> </td> </tr> </table> </div> <!-- SEPARATE EVALUATION OF \partial\Lambda/\partial \theta --> <div align="center"> <table border="1" cellpadding="8" align="center"> <tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained …'''</font></td></tr> <tr><td align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial\Lambda}{\partial \theta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 9(n+1)x^3\sin^3\theta - 2^3\cdot 3(n+1)^2x^3\sin^3\theta \cos^2\theta -2^4 (n+1)^2 (\beta\eta)^2 \sin\theta\cos\theta </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~~i~(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2}\biggl\{ (\beta\eta)\sin\theta +\frac{3x^3}{2}\cdot\biggl[ \frac{\sin^3\theta \cos\theta}{(\beta\eta)} \biggr]\biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~(n+1)\sin\theta \biggl\{ -2^4 (n+1) (\beta\eta)^2 \cos\theta + 3x^3 \sin^2\theta \biggl[3 - 2^3(n+1)\cos^2\theta \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~~i~(-1)\beta \sin\theta [2^7\cdot 3 (n+1)^3 (\beta\eta)^2]^{1/2}\biggl\{ 1 +\frac{3x^3}{2}\cdot\biggl[ \frac{\sin^2\theta \cos\theta}{(\beta\eta)^2} \biggr]\biggr\} \, , </math> </td> </tr> <tr><td colspan="3" align="left">which is the same.</td></tr> </table> </td></tr></table> </div> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial^2 \Lambda}{\partial \theta^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -3 x^3 \frac{\partial }{\partial \theta} \cdot \biggl\{\sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \biggr\} -2^4(n+1)^2 x^2 \cdot \frac{\partial }{\partial \theta} \biggl\{\sin\theta \cos\theta [1 + x(3\cos\theta-\cos^3\theta) ] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~\beta (-1)x [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot \frac{\partial }{\partial \theta} \biggl\{ \sin\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -3 x^3 \cdot \biggl\{3\sin^2\theta \cos\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] - \sin^3\theta [ 2^3(n+1)^2 \cdot 2\sin\theta\cos\theta ] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -2^4(n+1)^2 x^2 \cdot \biggl\{ (\cos^2\theta - \sin^2\theta ) [1 + x(3\cos\theta-\cos^3\theta) ] - 3x\sin^4\theta \cos\theta \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~\beta (-1)x [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl\{ \cos\theta \cdot [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> - 3x\sin^2\theta [1 + x(3\cos\theta - \cos^3\theta) ]^{-1/2} \cdot (3 - 5\cos^2\theta) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \frac{3}{2} \cdot x\sin^4\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + x(3\cos\theta - \cos^3\theta) ]^{-3/2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -3 x^3 \cdot \biggl\{3\sin^2\theta \cos\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] - \sin^3\theta [ 2^3(n+1)^2 \cdot 2\sin\theta\cos\theta ] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -2^4(n+1)^2 x^2 \cdot \biggl\{ (1 - 2\sin^2\theta ) + x (3\cos\theta -\cos^3\theta - 6\sin^2\theta\cos\theta + 2\sin^2\theta\cos^3\theta - 3\sin^4\theta \cos\theta ) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~\beta (-1)x [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cdot \biggl\{ \cos\theta \cdot [2 + x(9\cos\theta - 5\cos^3\theta) ] \cdot [1 + xb ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> - 3x\sin^2\theta [1 + xb ] \cdot (3 - 5\cos^2\theta) + \frac{3}{2} \cdot x\sin^4\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-2^4(n+1)^2 x^2 (1 - 2\sin^2\theta ) - x^3 \cdot \biggl\{ [ 2^3\cdot 3^2(n+1)^2\sin^2\theta \cos^3\theta - 3^3(n+1)\sin^2\theta \cos\theta] - 2^4\cdot 3(n+1)^2 \sin^4\theta\cos\theta </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + 2^4(n+1)^2 b + 2^4(n+1)^2 \cdot( - 6\sin^2\theta\cos\theta + 2\sin^2\theta\cos^3\theta - 3\sin^4\theta \cos\theta )\biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~\beta (-1)x [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cdot \biggl\{ 2\cos\theta \cdot [2 + x(15\cos\theta - 7\cos^3\theta) + x^2 b(9\cos\theta - 5\cos^3\theta)] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> - 6x\sin^2\theta (1 + xb ) \cdot (3 - 5\cos^2\theta) + 3 \cdot x\sin^4\theta [2 + x(9\cos\theta - 5\cos^3\theta) ] \biggr\} </math> </td> </tr> </table> </div> <!-- SEPARATE EVALUATION OF 2nd DERIVATIVE wrt "theta" --> <div align="center"> <table border="1" cellpadding="8" align="center"> <tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained …'''</font></td></tr> <tr><td align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\partial^2\Lambda}{\partial \theta^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ x^2 \biggl\{2^4(n+1)^2(\sin^2\theta - \cos^2\theta) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + x^3\biggl\{-2^4\cdot 3 (n+1)^2\cos^3\theta + 2^4(n+1)^2\cos^5\theta + 3^2(n+1)(16n +19)\sin^2\theta \cos\theta -2^3\cdot 23 (n+1)^2\sin^2\theta \cos^3\theta \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~~i~(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2} \biggl\{ (\beta\eta)\cos\theta + \frac{3x^3\sin^2\theta}{2(\beta\eta)}(5\cos^2\theta -2) + \frac{3^2x^6\sin^6\theta\cos\theta}{2^2(\beta\eta)^3} \biggr\} \, . </math> </td> </tr> </table> </td></tr></table> </div> =====Step 3===== From our [[Apps/PapaloizouPringle84#isolatingBlaes85|accompanying discussion of the Blaes85 derivation]], we expect the following equality to hold (see his equations 4.1 and 4.2): <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\hat{L} (\delta W)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-2n(1-\Theta_H)(M\nu^2 + N\nu)(\delta W) \, ,</math> </td> </tr> </table> </div> where, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\hat{L} (\delta W)</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ \Theta_H x^2\cdot \frac{\partial^2(\delta W)}{\partial x^2} +\Theta_H \cdot \frac{\partial^2(\delta W)}{\partial\theta^2} + \biggl\{\Theta_H x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial\Theta_H}{\partial x} \biggr\} \cdot \frac{\partial (\delta W)}{\partial x} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \biggl[\frac{\Theta_Hx\sin\theta}{ (1-x\cos\theta) } + n\cdot \frac{\partial \Theta_H}{\partial\theta} \biggr] \cdot \frac{\partial (\delta W)}{\partial\theta} + \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2\Theta_H}{(1-x\cos\theta)^2} \biggr]\delta W \, , </math> </td> </tr> <tr> <td align="right"> <math>~M</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{x^2}{(1-\Theta_H)\beta^2} \, ,</math> </td> </tr> <tr> <td align="right"> <math>~N</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{2mx^2}{(1-\Theta_H)\beta^2(1-x\cos\theta)^2} \, .</math> </td> </tr> </table> </div> Immediately evaluating the right-hand-side (RHS) of the equality, we have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{RHS}{A_{00}}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-2n(1-\Theta_H)(M\nu^2 + N\nu)\cdot \biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{\delta W}{A_{00}}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-\frac{2nx^2}{\beta^2}\biggl[ \nu^2 + \frac{2m\nu}{(1-x\cos\theta)^2} \biggr] \cdot \biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{\delta W}{A_{00}}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-\frac{2nx^2}{\beta^2}\biggl[ \nu^2 + \frac{2m\nu}{(1-x\cos\theta)^2} \biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2 \biggr\} \, .</math> </td> </tr> </table> </div> And the similarly modified LHS is: <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{LHS}{A_{00}}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \Theta_H x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +\Theta_H \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} + \biggl\{\Theta_H x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial\Theta_H}{\partial x} \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \biggl[\frac{\Theta_Hx\sin\theta}{ (1-x\cos\theta) } + n\cdot \frac{\partial \Theta_H}{\partial\theta} \biggr] \cdot \frac{\partial\Lambda}{\partial\theta} + \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2\Theta_H}{(1-x\cos\theta)^2} \biggr] \cdot \biggl[\frac{2(n+1)}{m}\biggr]^2 \frac{\delta W}{A_{00}} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (1-\eta^2) x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +(1-\eta^2) \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} + \biggl\{(1-\eta^2) x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] + nx^2 \cdot \frac{\partial (1-\eta^2) }{\partial x} \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \biggl[\frac{(1-\eta^2) x\sin\theta}{ (1-x\cos\theta) } + n\cdot \frac{\partial (1-\eta^2) }{\partial\theta} \biggr] \cdot \frac{\partial\Lambda}{\partial\theta} + \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2(1-\eta^2) }{(1-x\cos\theta)^2} \biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (1-\eta^2) x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +(1-\eta^2) \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} + \biggl\{(1-\eta^2) x \biggl[\frac{1-2x \cos\theta}{ 1-x\cos\theta}\biggr] - nx^2 \cdot \frac{\partial \eta^2 }{\partial x} \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \biggl[\frac{(1-\eta^2) x\sin\theta}{ (1-x\cos\theta) } - n\cdot \frac{\partial \eta^2 }{\partial\theta} \biggr] \cdot \frac{\partial\Lambda}{\partial\theta} + \biggl[ \frac{2n x^2 m^2}{\beta^2(1-x\cos\theta)^4} - \frac{m^2 x^2(1-\eta^2) }{(1-x\cos\theta)^2} \biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2 \biggr\} </math> </td> </tr> </table> </div> Now multiply both sides by … <div align="center"> <math>~\beta^2 m^2 (1-x\cos\theta)^4 \, .</math> </div> We have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~m^2\mathcal{L}_{RHS} \equiv \beta^2 (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{RHS}{A_{00}}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- 2n m^2 x^2(1-x\cos\theta)^4 \biggl[ \nu^2 + \frac{2m\nu}{(1-x\cos\theta)^2} \biggr] \cdot \biggl\{\Lambda + \biggl[ \frac{2(n+1)}{m} \biggr]^2 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- 2n x^2 (1-x\cos\theta)^2 \biggl[ (1-x\cos\theta)^2 \nu^2 + (2m\nu) \biggr] \cdot \{m^2 \Lambda + [ 2(n+1) ]^2 \} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- 2n x^2 (1-x\cos\theta)^2 \{m^2 \Lambda + [ 2(n+1) ]^2 \} \biggl\{ m^2 (1-x\cos\theta)^2 \biggl[ 1+\frac{3\beta^2}{2(n+1)} \biggr] + 2m^2 \biggl[ -1 ~\pm~ i~\biggl[ \frac{3\beta^2}{2(n+1)} \biggr]^{1/2} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{n m^2 x^2}{(n+1)} \cdot (1-x\cos\theta)^2\{m^2 \Lambda + [ 2(n+1) ]^2 \} \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - 4 (n+1) \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\} \, . </math> </td> </tr> </table> </div> And, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~m^2\mathcal{L}_{LHS} \equiv \beta^2 (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{LHS}{A_{00}}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \beta^2 m^2 (1-x\cos\theta)^4 (1-\eta^2) x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +\beta^2 m^2 (1-x\cos\theta)^4 (1-\eta^2) \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \beta^2 m^2 (1-x\cos\theta)^3 \biggl[ (1-\eta^2) x (1-2x \cos\theta ) - nx^2 (1-x\cos\theta)\cdot \frac{\partial \eta^2 }{\partial x} \biggr] \cdot \frac{\partial \Lambda }{\partial x} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \beta^2 m^2 (1-x\cos\theta)^3 \biggl[(1-\eta^2) x\sin\theta - n (1-x\cos\theta)\cdot \frac{\partial \eta^2 }{\partial\theta} \biggr] \cdot \frac{\partial\Lambda}{\partial\theta} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + m^2 \biggl[ 2n x^2 - (1-x\cos\theta)^2 \beta^2 x^2(1-\eta^2) \biggr] \cdot \biggl\{m^2 \Lambda + \biggl[ 2(n+1) \biggr]^2 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ m^2 (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ]\cdot \frac{\partial^2 \Lambda}{\partial\theta^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + m^2 (1-x\cos\theta)^3 \biggl\{ [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x (1-2x \cos\theta ) - nx^2 (1-x\cos\theta)\cdot [ 2x +3x^2(3\cos\theta - \cos^3\theta)] \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + m^2 (1-x\cos\theta)^3 \biggl\{[ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x\sin\theta - 3n (1-x\cos\theta)\cdot ( \cos^2\theta -1 )x^3 \sin\theta \biggr\} \cdot \frac{\partial\Lambda}{\partial\theta} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + m^2 \biggl\{ 2n x^2 - (1-x\cos\theta)^2 x^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\} \cdot \biggl[m^2 \Lambda + 2^2(n+1)^2 \biggr] \, . </math> </td> </tr> </table> </div> =====Step 4===== Let's divide both sides by <math>~m^2</math> and swap a couple of terms between the sides in order to group, on the right, terms with no explicit mention of <math>~\Lambda</math>. <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathcal{L}_{RHS} </math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ \frac{\beta^2}{m^2} (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{RHS}{A_{00}} ~\pm~\mathrm{swaps} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{n x^2}{(n+1)} \cdot (1-x\cos\theta)^2\{[ 2(n+1) ]^2 \} \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ] - 4 (n+1) \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \biggl\{ 2n x^2 - (1-x\cos\theta)^2 x^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\} \cdot \biggl[2^2(n+1)^2 \biggr] </math> </td> </tr> </table> </div> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Rightarrow ~~~~ -~\frac{\mathcal{L}_{RHS} }{2^2(n+1)x^2} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ n \cdot (1-x\cos\theta)^2 \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ] - 4 (n+1) \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (n+1)\biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2n(n+1) - (1-x\cos\theta)^2 (n+1)[ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + n(1-x\cos\theta)^4 [ 2(n+1) + 3\beta^2 ] - 4 n(n+1) (1-x\cos\theta)^2 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~i~ (1-x\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2n(n+1) - (n+1)(1-x\cos\theta)^2[ 4n + \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] + n(1-x\cos\theta)^4 [ 2(n+1) + 3\beta^2 ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~i~ (1-x\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} \, . </math> </td> </tr> </table> </div> <!-- SEPARATE DEFINITION OF RHS_3 --> <div align="center"> <table border="1" cellpadding="8" align="center"> <tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained …'''</font></td></tr> <tr><td align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathrm{RHS}_3</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \mathrm{RHS}_1 ~~\pm~~ \mathrm{swaps} \biggr\} = \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \frac{1}{A_{00}}\biggl[ \frac{2(n+1)}{m} \biggr]^2\mathrm{RHS}_0 ~~\pm~~ \mathrm{swaps} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ 2^2(n+1)^2 x^2 \cdot \mathcal{A} \, , </math> </td> </tr> <tr><td colspan="3" align="left">where,</td></tr> <tr> <td align="right"> <math>~\mathcal{A}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ 2n + (1-x\cos\theta)^2\biggl[ 4n \biggl( \frac{\nu}{m}\biggr) - \beta^2 + x^2(1+xb)\biggr] + 2n (1-x\cos\theta)^4 \biggl(\frac{\nu}{m}\biggr)^2 \, , </math> </td> </tr> <tr> <td align="right"> <math>~\frac{\nu}{m}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~1~~ \pm~~ i~\biggl[\frac{3}{2(n+1)}\biggr]^{1/2} \beta \, . </math> </td> </tr> </table> </td></tr> <tr><td align="center"> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr><td colspan="3" align="left"> <font color="red" size="+1"><b> Case A: </b></font> <math>~\biggl(\frac{\nu}{m}\biggr)^2 = \frac{\nu \cdot \nu^*}{m^2} = 1+\frac{3\beta^2}{2(n+1)} ~~~\Rightarrow</math> </td></tr> <tr> <td align="right"> <math>~(n+1)\cdot \mathcal{A}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2n (n+1) + (n+1)(1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4 [2n(n+1) + 3n\beta^2 ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~~i~(1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} \, . </math> </td> </tr> </table> </div> </td></tr> <tr><td align="center"> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr><td colspan="3" align="left"> <font color="red" size="+1"><b> Case B: </b></font> <math>~\biggl(\frac{\nu}{m}\biggr)^2 = \frac{\nu }{m} \cdot \frac{\nu }{m} = 1 - \frac{3\beta^2}{2(n+1)} ~~\pm~i~(-1)\biggl[\frac{2\cdot 3\beta^2}{(n+1)}\biggr]^{1/2} ~~~\Rightarrow</math> </td></tr> <tr> <td align="right"> <math>~(n+1)\cdot \mathcal{A}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2n (n+1) + (n+1)(1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2 ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~~i~x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} \, . </math> </td> </tr> </table> </div> </td></tr></table> </div> And, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathcal{L}_{LHS} </math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ \frac{\beta^2}{m^2} (1-x\cos\theta)^4 \biggl[2(n+1)\biggr]^2 \frac{LHS}{A_{00}} ~\mp~\mathrm{swaps} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ]\cdot \frac{\partial^2 \Lambda}{\partial\theta^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (1-x\cos\theta)^3 \biggl\{ [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x (1-2x \cos\theta ) - nx^2 (1-x\cos\theta)\cdot [ 2x +3x^2(3\cos\theta - \cos^3\theta)] \biggr\} \cdot \frac{\partial \Lambda }{\partial x} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (1-x\cos\theta)^3 \biggl\{[ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] x\sin\theta - 3n (1-x\cos\theta)\cdot ( \cos^2\theta -1 )x^3 \sin\theta \biggr\} \cdot \frac{\partial\Lambda}{\partial\theta} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \biggl\{ 2n x^2 - (1-x\cos\theta)^2 x^2 [ \beta^2 - x^2 - x^3(3\cos\theta - \cos^3\theta) ] \biggr\} \cdot \biggl[m^2 \Lambda \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \frac{n x^2}{(n+1)} \cdot (1-x\cos\theta)^2\{m^2 \Lambda \} \biggl\{ (1-x\cos\theta)^2 [ 2(n+1) + 3\beta^2 ] - 4 (n+1) \pm i~ [2^3\cdot 3 (n+1)\beta^2 ]^{1/2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 - x^3b] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (1-x\cos\theta)^3 \biggl\{ x( \beta^2 - x^2 - x^3b ) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr] - n (1-x\cos\theta)\cdot x^3 \biggl[ ( 2 +3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 - x^3b ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot m^2 x^2\Lambda </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2 \Lambda </math> </td> </tr> </table> </div> <!-- SEPARATE DEFINITION OF LHS_3 --> <div align="center"> <table border="1" cellpadding="8" align="center"> <tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained …'''</font></td></tr> <tr><td align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathrm{LHS}_3</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \mathrm{LHS}_1 ~~\mp~~ \mathrm{swaps} \biggr\} = \beta^2 (1-x\cos\theta)^4 \cdot \biggl\{ \frac{1}{A_{00}}\biggl[ \frac{2(n+1)}{m} \biggr]^2\mathrm{LHS}_0 ~~\mp~~ \mathrm{swaps} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \beta^2(1-\eta^2) (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta)\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -~n x^3 (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] + m^2 x^2 \Lambda \cdot \mathcal{A} \, ,</math> </td> </tr> <tr><td colspan="3" align="left">where, as above in the definition of <math>~\mathrm{RHS_3} \, ,</math></td></tr> <tr> <td align="right"> <math>~\mathcal{A}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ 2n + (1-x\cos\theta)^2\biggl[ 4n \biggl( \frac{\nu}{m}\biggr) - \beta^2 + x^2(1+xb)\biggr] + 2n (1-x\cos\theta)^4 \biggl(\frac{\nu}{m}\biggr)^2 \, . </math> </td> </tr> </table> </td></tr></table> </div> The remaining question is, does <math>~\mathcal{L}_{LHS} = \mathcal{L}_{RHS}</math> — at least to lowest order(s) in <math>~x</math> — after the Blaes85 expression for the eigenfunction, <math>~\Lambda</math> (and its derivatives), is inserted into the LHS expression? =====Step 5===== Now let's evaluate the LHS terms, keeping only leading-orders in <math>~x</math> before plugging derivatives of <math>~\Lambda</math> into each term. For example, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ x^2 \biggl\{2[ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [1 + \cancelto{0}{3xb}] \pm ~i~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot b[1 + \cancelto{0}{xb} ]^{-3/2} [4 + \cancelto{0}{9xb} ] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~m^2\cdot \biggl\{ -2^4(n+1)^2 x^2 (1 - 2\sin^2\theta ) - \cancelto{0}{x^3 }\cdot \biggl[ [ 2^3\cdot 3^2(n+1)^2\sin^2\theta \cos^3\theta - 3^3(n+1)\sin^2\theta \cos\theta] - 2^4\cdot 3(n+1)^2 \sin^4\theta\cos\theta </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + 2^4(n+1)^2 b + 2^4(n+1)^2 \cdot( - 6\sin^2\theta\cos\theta + 2\sin^2\theta\cos^3\theta - 3\sin^4\theta \cos\theta )\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~\beta (-1)x [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cdot \biggl[ 2\cos\theta \cdot [2 + x(15\cos\theta - 7\cos^3\theta) + \cancelto{0}{x^2 b}(9\cos\theta - 5\cos^3\theta)] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> - 6x\sin^2\theta (1 + \cancelto{0}{xb }) \cdot (3 - 5\cos^2\theta) + 3 \cdot x\sin^4\theta [2 + \cancelto{0}{x}(9\cos\theta - 5\cos^3\theta) ] \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ x^2 \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2} [2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta] \biggr] </math> </td> </tr> </table> </div> Next, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (1-2x \cos\theta )\cdot \biggl\{ [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] \cdot [2x + \cancelto{0}{3x^2b}] ~~\pm ~~i~\beta [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot [1 + xb ]^{-1/2} \cdot [2 + \cancelto{0}{3xb }] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \sin\theta \cdot \biggl\{ -2^4(n+1)^2 \cancelto{0}{x^2}\sin\theta \cos\theta [1 + xb ] -3 \cancelto{0}{x^3} \sin^3\theta [ 2^3(n+1)^2\cos^2\theta - 3(n+1)] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ ~~\pm ~~i~\beta (-1) [ 2^5\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2} \cdot x~\sin\theta [2 + \cancelto{0}{x}(9\cos\theta - 5\cos^3\theta) ] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ (1-2x \cos\theta )\cdot \biggl[ 2x(n+1) [ 2^3 (n+1)\cos^2\theta - 3] ~~\pm ~~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ ~~\pm ~~i~\beta x (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2} \cdot \sin^2\theta </math> </td> </tr> </table> </div> <!-- EVALUATE 1st CROSS-TERM --> <div align="center"> <table border="1" cellpadding="8" align="center"> <tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained …'''</font></td></tr> <tr><td align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ x(n+1)[-6 + 2^4(n+1)\cos^2\theta] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - x^2(n+1)\cos\theta \{ [ 15 + 2^4(n+1) ] -\cos^2\theta[9 + 2^3\cdot 7 (n+1)] +2^3\cdot 3(n+1)\cos^4\theta \} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +x^3(n+1) \{9 - 2^2\cdot 3^2(1+2n)\cos^2\theta - [9 + 32(n+1)]\cos^4\theta +2^3(n+1)\cos^6\theta \} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~\pm~~i~\beta~\biggl[ \frac{2^5\cdot 3 (n+1)^3}{1+x(3\cos\theta-\cos^3\theta)} \biggr]^{1/2} \biggl\{ 2\cos\theta - x[2 - 7\cos^2\theta + 3\cos^4\theta ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~- x^2 \cos\theta [ 9 +4\cos^2\theta -\cos^4\theta ] \biggr\}</math> </td> </tr> </table> </td></tr></table> </div> Also, from above, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Lambda </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- (4n+1)\beta^2 + (n+1)\cancelto{0}{x^2}[ 2^3(n+1)\cos^2\theta - 3] \cdot [1 + xb] ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~- (4n+1)\beta^2 ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math> </td> </tr> </table> </div> <!-- EVALUATE 2nd CROSS-TERM --> <div align="center"> <table border="1" cellpadding="8" align="center"> <tr><td align="center" bgcolor="purple"><font color="white">'''Through a separate white-board derivation I have obtained …'''</font></td></tr> <tr><td align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ (2+3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ x\cdot 2^2(n+1)[2^3(n+1)\cos^2\theta -3] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + x^2\cdot 2^2\cdot 3(n+1)\cos\theta \{ [2^2n -5] +\cos^2\theta[2^4n + 19] - 2^2(n+1)\cos^4\theta \} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +x^3 \cdot 3^2(n+1)\cos^2\theta \{ -3^3 + 2\cdot 3^2\cos^2\theta[2^2n+5] - 99\cdot \cos^4\theta + 2^3(n+1)\cos^6\theta \} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +x^3 (n+1)\sin^4\theta \{ -3^3 + 3^2\cos^2\theta [ 2^3\cdot 3 n + 27] -2^3\cdot 3\cdot 5(n+1)\cos^4\theta \} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~\pm~~i~\beta~ \biggl[ \frac{2^5\cdot 3(n+1)^3}{1+xb} \biggr]^{1/2} \biggl\{ 4\cos\theta + 6x(2b\cos\theta + \sin^4\theta) + 3x^2(3b^2\cos\theta +2b\sin^4\theta + 3\sin^6\theta\cos\theta) \biggr\} </math> </td> </tr> </table> </td></tr></table> </div> Taken together, then, we have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathcal{L}_{LHS} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 - \cancelto{0}{x^3b}] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (1-x\cos\theta)^3 \biggl\{ x( \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3b} ) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr] - n (1-x\cos\theta)\cdot \cancelto{0}{x^3} \biggl[ ( 2 +3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 - \cancelto{0}{x^3}b ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot m^2 x^2\Lambda </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2 \Lambda </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl\{ x^2\cdot \frac{\partial^2 \Lambda}{\partial x^2} +m^2 \cdot \frac{\partial^2 \Lambda}{\partial\theta^2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (1-x\cos\theta)^3 \biggl\{ x( \beta^2) \biggl[ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot m^2 x^2\Lambda </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2 \Lambda </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl\{ x^2 \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2} [2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta] \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + x\beta^2(1-x\cos\theta)^3 \biggl\{ (1-2x \cos\theta )\cdot \biggl[ 2x(n+1)[ 2^3 (n+1)\cos^2\theta - 3] ~~\pm ~~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ ~~\pm ~~i~\beta x (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-1/2} \cdot \sin^2\theta \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + m^2 x^2\biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot \biggl\{- (4n+1)\beta^2 ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2}\biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm i~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} x^2 \cdot (1-x\cos\theta)^2 \cdot \biggl\{ - (4n+1)\beta^2 ~\pm ~i~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} \biggr\} </math> </td> </tr> </table> </div> Let's further simplify: <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathcal{L}_{LHS} </math> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ x^2\biggl\{ (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-x\cos\theta)^2 \cdot \beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \beta^2(1-x\cos\theta)^3 (1-2x \cos\theta )\cdot \biggl[ 2(n+1)[ 2^3 (n+1)\cos^2\theta - 3] \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - 2n(4n+1)\beta^2 m^2 + (4n+1)\beta^2 m^2 [4n + \beta^2 - x^2 ] (1-x\cos\theta)^2 - (4n+1)\beta^2 m^2 n (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ ~~\pm ~~i x\beta^3 (1-x\cos\theta)^3 (1-2x \cos\theta )\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + xb )^{-1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~m^2\beta (-1)x [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + xb ]^{-3/2} \cos\theta (1-x\cos\theta)^4 [ \beta^2 - x^2 ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm ~i~\beta m^2 x^2\biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} x\cos\theta [1 + x b ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> ~\pm ~i~\beta x^2 [ 2^3\cdot 3(n+1)^3 ]^{1/2} (1-x\cos\theta)^4 [ \beta^2 - x^2 ] \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + xb )^{-3/2} [2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta] \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ ~~\pm ~~i~\beta x^2 (-1) [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot \beta^2(1-x\cos\theta)^3 [1 + xb ]^{-1/2} \cdot \sin^2\theta </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm i~x^2(-1) (4n+1)\beta^2 \biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-x\cos\theta)^2 </math> </td> </tr> </table> </div> =====Step 6===== Hence, to lowest order we want to compare the following two expressions: <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~~ -~\frac{\mathcal{L}_{RHS} }{2^2(n+1)x^2} </math> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ 2n(n+1) - (n+1)(1-\cancelto{0}{x}\cos\theta)^2[ 4n + \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3}(3\cos\theta - \cos^3\theta) ] + n(1-\cancelto{0}{x}\cos\theta)^4 [ 2(n+1) + 3\beta^2 ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~i~ (1-\cancelto{0}{x}\cos\theta)^2 [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ 2n(n+1) - (n+1)[ 4n + \beta^2 ] + n [ 2(n+1) + 3\beta^2 ] \pm~i~ [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~3n\beta^2 + (n+1)\biggl[ 2n - 4n - \beta^2 + 2n \biggr] \pm~i~ [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~\beta^2 (2n-1) ~\pm~i~ [2^3\cdot 3 n^2(n+1)\beta^2 ]^{1/2} </math> </td> </tr> </table> </div> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathrm{Re}\biggl[\frac{\mathcal{L}_{LHS}}{x^2}\biggr] </math> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x^2} ] \biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -~\biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-x\cos\theta)^2 \cdot \beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cancelto{0}{x}\cos\theta [1 + x b ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \beta^2(1-\cancelto{0}{x}\cos\theta)^3 (1-\cancelto{0}{2x} \cos\theta )\cdot \biggl[ 2(n+1)[ 2^3 (n+1)\cos^2\theta - 3] \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - 2n(4n+1)\beta^2 m^2 + (4n+1)\beta^2 m^2 [4n + \beta^2 - \cancelto{0}{x^2} ] (1-\cancelto{0}{x}\cos\theta)^2 - (4n+1)\beta^2 m^2 n (1-\cancelto{0}{x}\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ \beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) -2^4 m^2 (n+1)^2 (1 - 2\sin^2\theta ) \biggr] + \beta^2\biggl[ 2^4 (n+1)^2\cos^2\theta -6(n+1)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~+ \beta^2\biggl\{- 2n(4n+1) m^2+ (4n+1)m^2 [4n ] - 2(4n+1) m^2 n \biggr\} + \beta^4\biggl\{(4n+1) m^2 - (4n+1) m^2 n \biggl[ \frac{3}{(n+1)} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ \beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) \biggr] + \beta^2\biggl[ 2^4 (n+1)^2\cos^2\theta -6(n+1)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + m^2\beta^2\biggl[ -2^4(n+1)^2 (1 - 2\sin^2\theta ) \biggr] + m^2\beta^2\biggl\{- 2n(4n+1) + 4n(4n+1) - 2n(4n+1) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (4n+1) m^2\beta^4\biggl[1 - \frac{3n}{(n+1)} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ 2\beta^2\biggl[ 2^4(n+1)^2\cos^2\theta - 6(n+1) \biggr] + m^2\beta^2\biggl[ 2^4(n+1)^2 (1 - 2\cos^2\theta ) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (4n+1) m^2\beta^4\biggl[1 - \frac{3n}{(n+1)} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ (1-m^2)2^5\beta^2(n+1)^2\cos^2\theta + 2^2(n+1)\beta^2\biggl[ 4m^2(n+1) -3\biggr] + (4n+1) m^2\beta^4\biggl[1 - \frac{3n}{(n+1)} \biggr] \, . </math> </td> </tr> </table> </div> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\pm~\mathrm{Im}\biggl[\mathcal{L}_{LHS} \biggr]</math> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ ~x\biggl\{\beta^3 (1-\cancelto{0}{x}\cos\theta)^3 (1-\cancelto{0}{2x} \cos\theta )\cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta \cdot (1 + \cancelto{0}{xb} )^{-1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> -~m^2\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot [1 + \cancelto{0}{xb} ]^{-3/2} \cos\theta (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x}^2 ] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +~x^2 \biggl\{ \beta m^2 \biggl\{ 2n - (1-x\cos\theta)^2 [ \beta^2 - x^2 ] + (1-x\cos\theta)^4 \biggl[ 2 + \frac{3\beta^2}{(n+1)} \biggr] n - 4 n (1-x\cos\theta)^2 \biggr\} \cdot [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cancelto{0}{x}\cos\theta [1 + x b ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> +~\beta [ 2^3\cdot 3(n+1)^3 ]^{1/2} (1-\cancelto{0}{x}\cos\theta)^4 [ \beta^2 - \cancelto{0}{x^2} ] \cdot \biggl[(12\cos^2\theta - 4\cos^4\theta)~~ - ~~m^2\cdot (1 + \cancelto{0}{xb} )^{-3/2} [2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta] \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -~\beta [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cdot \beta^2(1-\cancelto{0}{x}\cos\theta)^3 [1 + \cancelto{0}{xb} ]^{-1/2} \cdot \sin^2\theta </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -~ (4n+1)\beta^2 \biggl[\frac{2^3\cdot 3 n^2 m^4 \beta^2}{(n+1)} \biggr]^{1/2} \cdot (1-\cancelto{0}{x}\cos\theta)^2 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ ~x(1-m^2)\cdot \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> +~~x^2 \beta^3 [ 2^3\cdot 3(n+1)^3 ]^{1/2}\biggl\{(12\cos^2\theta - 4\cos^4\theta) - 4\sin^2\theta </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~- m^2 \biggl[2\cos^2\theta(15 - 7\cos^2\theta) - 6\sin^2\theta (3 - 5\cos^2\theta) + 6 \sin^4\theta -~ \frac{n (4n+1)}{(n+1)^2} \biggr]\biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ ~x(1-m^2)\cdot \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> +~~x^2 \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2}\biggl\{4\cos^2\theta - \cos^4\theta -1 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~- \frac{m^2}{2} \biggl[\cos^2\theta(15 - 7\cos^2\theta) -12\sin^4\theta +6 \sin^2\theta -~ \frac{n (4n+1)}{2(n+1)^2} \biggr]\biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ ~x(1-m^2)\cdot \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2} \cos\theta </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> +~~x^2 \beta^3 [ 2^7\cdot 3(n+1)^3 ]^{1/2}\biggl\{3-(1+\sin^2\theta)^2 - \frac{m^2}{2} \biggl[33\cos^2\theta-19 \cos^4\theta -6 -~ \frac{n (4n+1)}{2(n+1)^2} \biggr]\biggr\} </math> </td> </tr> </table> </div> =====Examples===== Evaluate various expressions using the parameter set: <math>~(n, \theta, x) = (1, \tfrac{\pi}{3}, \tfrac{1}{4})</math> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~b</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3}{2} - \frac{1}{8} = \frac{11}{8} </math> </td> <td align="right"> 1.375000000 </td> </tr> <tr> <td align="right"> <math>~(\beta\eta)^2</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl(\frac{1}{2^2}\biggr)^2\biggl[ 1 + \frac{11}{2^5} \biggr] = \frac{2^5 + 11}{2^9} = \frac{43}{2^9} </math> </td> <td align="right"> 0.083984375 </td> </tr> <tr> <td align="right"> <math>~\mathrm{Re}(\Lambda)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -5\beta^2 + \frac{43}{2^9} \biggl[ 2^3 - 6 \biggr] = -5\beta^2 + \frac{43}{2^8} </math> </td> <td align="right"> <math>~- 5\beta^2</math> + 0.167968750 </td> </tr> <tr> <td align="right"> <math>~\mathrm{Im}(\Lambda)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\beta}{2}\biggl[ 2^{10}\cdot 3 \cdot \frac{43}{2^9} \biggr]^{1/2} = \beta \biggl[ \frac{3\cdot 43}{2} \biggr]^{1/2} </math> </td> <td align="right"> 8.031189202 <math>~\beta</math> </td> </tr> <tr> <td align="right"> <math>~\mathrm{Re}\biggl( \frac{\partial\Lambda}{\partial x} \biggr)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ 2^3 - 6 \biggr]\frac{1}{2^2}\biggl( 2 + \frac{3\cdot 11}{2^2\cdot 2^3} \biggr) = \biggl( 1 + \frac{33}{2^6} \biggr) </math> </td> <td align="right"> 1.515625000 </td> </tr> <tr> <td align="right"> <math>~\mathrm{Im}\biggl( \frac{\partial\Lambda}{\partial x} \biggr)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\beta}{2} \biggl[ \frac{2^9\cdot 2^8\cdot 3}{43} \biggr]^{1/2} \biggl[ \frac{1}{2^2}\cdot \biggl(2 + \frac{3\cdot 11}{2^2\cdot 2^3}\biggr) \biggr] = \beta \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr) </math> </td> <td align="right"> 36.23373732 <math>~\beta</math> </td> </tr> <tr> <td align="right"> <math>~\mathrm{Re}\biggl( \frac{\partial\Lambda}{\partial \theta} \biggr)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2\biggl(\frac{3}{4}\biggr)^{1/2} \biggl\{-2^4 \cdot \frac{43}{2^9} +\frac{3}{2^6}\biggl(\frac{3}{4}\biggr)\biggl[3-4\biggr] \biggr\} = ~-~\frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9) </math> </td> <td align="right"> -2.388335684 </td> </tr> <tr> <td align="right"> <math>~\mathrm{Im}\biggl( \frac{\partial\Lambda}{\partial \theta} \biggr)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (-1) \beta~\biggl(\frac{3}{4}\biggr)^{1/2} \cdot \biggl[ \frac{2^{10} \cdot 3\cdot 43}{2^9} \biggr]^{1/2} \biggl\{1 + \frac{3\cdot 2^9}{2^7 \cdot 43} \biggl(\frac{3}{2^3}\biggr)\biggr\} = (-1) \beta~\biggl(\frac{3}{4}\biggr)^{1/2} \cdot [ 2\cdot 3\cdot 43 ]^{1/2} \biggl\{1 + \frac{3^2}{2\cdot 43} \biggr\} </math> </td> <td align="right"> (-1) × 15.36617018 <math>~\beta</math> </td> </tr> </table> </div> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathrm{Re}\biggl( \frac{\partial^2\Lambda}{\partial x^2}\biggr)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2^2[2^2-3]\biggl[1 + \frac{3}{2^2}\cdot \frac{11}{2^3} \biggr] = 2^2 + \frac{33}{2^3} = \frac{65}{8} </math> </td> <td align="right"> 8.125000000 </td> </tr> <tr> <td align="right"> <math>~\mathrm{Im}\biggl( \frac{\partial^2\Lambda}{\partial x^2}\biggr)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\beta~ 2^2\cdot \sqrt{3} \biggl[\frac{11}{2^3} \biggl(2^2+\frac{3}{2^2}\cdot \frac{11}{2^3} \biggr)\biggr] \biggl[ \frac{2^4}{43}\biggr]^{3/2} =\biggl[ \frac{2^3\cdot 3}{43^3}\biggr]^{1/2} \biggl[11\cdot (2^7+33)\biggr] \beta </math> </td> <td align="right"> 30.76957507<math>~\beta</math> </td> </tr> <tr> <td align="right"> <math>~\mathrm{Re}\biggl(\frac{\partial^2\Lambda}{\partial \theta^2}\biggr)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{2^4} \biggl\{2^6 \biggl(\frac{3}{2^2} - \frac{1}{2^2} \biggr) \biggr\} + \frac{1}{2^6}\biggl\{-2^3\cdot 3 + 2 + \frac{3^3\cdot 5\cdot 7}{2^2} -3\cdot 23 \biggr\} = 2 + \frac{1}{2^8}\biggl\{2^3 + 3^3\cdot 5\cdot 7 - 2^2\cdot 3\cdot 31 \biggr\} </math> </td> <td align="right"> 4.269531250 </td> </tr> <tr> <td align="right"> <math>~\mathrm{Im}\biggl( \frac{\partial^2\Lambda}{\partial \theta^2}\biggr)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~(-1)\beta [2^{10}\cdot 3]^{1/2} \biggl( \frac{43}{2^9} \biggr)^{1/2} \biggl\{ \frac{1}{2} + \frac{3}{2}\cdot \frac{2^9}{43} \cdot \frac{1}{2^6}\cdot \frac{3}{2^2} \biggl(\frac{5}{2^2} -2 \biggr) + \biggl( \frac{3}{2}\cdot \frac{1}{2^6} \cdot \frac{2^9}{43}\biggr)^2 \biggl( \frac{3}{2^2} \biggr)^3 \frac{1}{2} \biggr\} </math> </td> <td align="right"> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~(-1)\beta\biggl( \frac{3\cdot 43}{2} \biggr)^{1/2} \biggl\{ 1 - \frac{3^3}{2\cdot 43} + \frac{3^5}{(2\cdot 43)^2} \biggr\} </math> </td> <td align="right"> (-1) × 5.773638858 <math>~\beta</math> </td> </tr> </table> </div> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \mathrm{Re}\biggl\{ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl( 1 - \frac{1}{2^2} \biggr)\biggl( 1 + \frac{33}{2^6} \biggr) -~\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9) </math> </td> <td align="right"> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ ~\frac{(2\cdot 3\cdot 97)-3\cdot (2^3\cdot 43 + 9)}{2^9} </math> </td> <td align="right"> -0.931640625 </td> </tr> <tr> <td align="right"> <math>~ \mathrm{Im}\biggl\{ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl( 1 - \frac{1}{2^2} \biggr)\beta \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr) +~\frac{\sqrt{3}}{2}\cdot (-1) \beta~\frac{\sqrt{3}}{2}\cdot[ 2\cdot 3\cdot 43 ]^{1/2} \biggl\{1 + \frac{3^2}{2\cdot 43} \biggr\} </math> </td> <td align="right"> </td> </tr> <!-- <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\beta \biggl\{ \biggl[ \frac{3^3}{2^{3}\cdot 43} \biggr]^{1/2} (2^{6} + 3\cdot 11 ) -~\biggl[ \frac{3^4}{2^5\cdot 43} \biggr]^{1/2} ( 2\cdot 43 + 3^2 ) \biggr\} </math> </td> <td align="right"> NEW: 13.86780926 <math>~\beta</math> </td> </tr> --> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\beta \biggl[ \frac{3^3}{2^5\cdot 43} \biggr]^{1/2} [2 (2^6 + 33) - (2\cdot 43 + 3^2) ] </math> </td> <td align="right"> 13.86780926 <math>~\beta</math> </td> </tr> <tr> <td align="right"> <math>~ \mathrm{Re}\biggl\{ (2+3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl( 2 + \frac{3}{2^2}\cdot \frac{11}{2^3} \biggr)\biggl( 1 + \frac{33}{2^6} \biggr) +~3\cdot\biggl(\frac{3}{2^2}\biggr)^{3/2}\cdot \frac{\sqrt{3}}{2^8} \cdot (2^3\cdot 43 + 9) </math> </td> <td align="right"> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{97^2}{2^{11}} +~\frac{3^3}{2^{11}}\cdot (2^3\cdot 43 + 9) </math> </td> <td align="right"> 9.248046874 </td> </tr> <tr> <td align="right"> <math>~ \mathrm{Im}\biggl\{ (2+3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr\}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\beta \biggl\{ \biggl( 2 + \frac{3}{2^2}\cdot \frac{11}{2^3} \biggr) \biggl[ \frac{2^{13}\cdot 3}{43} \biggr]^{1/2} \biggl(1 + \frac{3\cdot 11}{2^{6}}\biggr) +~3\cdot\biggl(\frac{3}{2^2}\biggr)^{3/2}\cdot \biggl(\frac{3}{4}\biggr)^{1/2} \cdot [ 2\cdot 3\cdot 43 ]^{1/2} \biggl[1 + \frac{3^2}{2\cdot 43} \biggr]\biggr\} </math> </td> <td align="right"> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\beta \biggl\{ \biggl( \frac{3}{2^9\cdot 43} \biggr)^{1/2} \biggl[ ( 2^6 + 33)^2 +~3^3(2\cdot 43 +3^2 ) \biggr]\biggr\} </math> </td> <td align="right"> 139.7753772 </td> </tr> </table> </div> =====Step 7===== Let's begin by slightly redefining the LHS and RHS collections of terms. <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathrm{RHS}_4</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\mathrm{RHS}_3 - m^2 x^2 \Lambda \cdot \mathcal{A} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~x^2 ~ [ 2^2(n+1)^2 - m^2 \Lambda ] \cdot \mathcal{A} \, , </math> </td> </tr> <tr> <td align="right"> <math>~\mathrm{LHS}_4</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\mathrm{LHS}_3 - m^2 x^2 \Lambda \cdot \mathcal{A} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \beta^2(1-\eta^2) (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta)\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -~n x^3 (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] </math> </td> </tr> </table> <!-- Early attempt at lowest order **************** ======To Lowest Order====== Now, let's gather together only the lowest order components of all the LHS terms. <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathrm{LHS}_4</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ [ \beta^2 - \cancelto{0}{x^2} - \cancelto{0}{x^3}(3\cos\theta - \cos^3\theta) ] (1-\cancelto{0}{x}\cos\theta)^3 \biggl\{ (1-\cancelto{0}{x}\cos\theta) \biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -~n \cancelto{0}{x^3} (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ \beta^2~\biggl\{ \biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~~\frac{1}{\beta^2} \cdot \mathrm{LHS}_4</math> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ x^2 \biggl\{ 2(n+1)[2^3(n+1)\cos^2\theta -3](1+3\cancelto{0}{x}b) ~~\pm~~i~\beta\cos\theta [2^3\cdot 3(n+1)^3]^{1/2} \biggl[ \frac{b(4+3\cancelto{0}{x}b)}{(1+\cancelto{0}{x}b)^{3/2}} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + x^2 \biggl\{2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\} ~~\pm~~i~(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2} \biggl\{ \cancelto{x}{(\beta\eta)}\cos\theta \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + x\biggl\{ x(n+1)[-6 + 2^4(n+1)\cos^2\theta] ~~\pm~~i~\beta~\biggl[ \frac{2^5\cdot 3 (n+1)^3}{1+\cancelto{0}{x}(3\cos\theta-\cos^3\theta)} \biggr]^{1/2} \biggl[ 2\cos\theta - \cancelto{0}{x}(2 - 7\cos^2\theta + 3\cos^4\theta )\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ x^2 \biggl\{ 2(n+1)[2^3(n+1)\cos^2\theta -3] + 2^4(n+1)^2(\sin^2\theta - \cos^2\theta) + (n+1)[-6 + 2^4(n+1)\cos^2\theta] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> \pm~~i~\biggl\{(-1)\beta [2^7\cdot 3 (n+1)^3 ]^{1/2} \cdot x\cos\theta +\beta x [ 2^5\cdot 3 (n+1)^3 ]^{1/2} \cdot 2\cos\theta \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ x^2 \cdot 2(n+1)\biggl\{ [2^3(n+1)\cos^2\theta -3] + 2^3(n+1)(\sin^2\theta - \cos^2\theta) + [-3 + 2^3(n+1)\cos^2\theta] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> \pm~~i~x\beta \cos\theta\biggl\{ [ 2^5\cdot 3 (n+1)^3 ]^{1/2} \cdot 2 - [2^7\cdot 3 (n+1)^3 ]^{1/2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ x^2 \cdot 2(n+1)\biggl\{2^3(n+1)-6\biggr\} \pm~~i~x\beta \cos\theta\biggl\{ ~0~ \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ x^2 \cdot 4(n+1)(4n+1) \, . </math> </td> </tr> </table> Contrast this with, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathrm{RHS}_4</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~x^2 ~ [m^2 \Lambda - 2^2(n+1)^2 ] \cdot \mathcal{A} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~x^2 ~ \biggl\{ m^2 \biggl[ -(4n+1)\beta^2 ~~\pm~i~\cancelto{0}{x}\beta \cos\theta [2^7\cdot 3(n+1)^3]^{1/2}\biggr] - 2^2(n+1)^2 \biggr\} \cdot \mathcal{A} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~x^2 ~ \biggl\{ m^2 \biggl[ -(4n+1)\beta^2 \biggr] - 2^2(n+1)^2 \biggr\} \cdot \mathcal{A} </math> </td> </tr> </table> Now, to lowest order [<font color="red">Case B</font>], <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(n+1)\cdot \mathcal{A}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2n (n+1) + (n+1)(1-\cancelto{0}{x}\cos\theta)^2 [ \cancelto{0}{ x^2}(1+xb) - \beta^2 - 4n ] + (1-\cancelto{0}{x}\cos\theta)^4 [2n(n+1) - 3n\beta^2 ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~~i~\cancelto{0}{x}\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ 2n (n+1) + (n+1)[ - \beta^2 - 4n ]+ [2n(n+1) - 3n\beta^2 ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~- \beta^2(4n+1) </math> </td> </tr> </table> Hence, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{1}{\beta^2}\cdot \mathrm{RHS}_4</math> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~x^2 ~ \biggl\{ m^2 \biggl[ -(4n+1)\cancelto{0}{\beta^2} \biggr] - 2^2(n+1)^2 \biggr\} \cdot \biggl[ \frac{-(4n+1)}{(n+1)} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~x^2 ~ \biggl\{ 2^2(n+1) \biggr\} \cdot (4n+1) \, , </math> </td> </tr> </table> and we see that, to lowest order, the two sides do match. Notice that the mode number, <math>~m</math>, drops out in this lowest order approximation. ***************** --> ======Next Lowest Order====== Let's begin with the RHS (<font color="red">Case B</font>). <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(n+1)\cdot \mathcal{A}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2n (n+1) + (n+1)(1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2 ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~~i~x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2n (n+1) + (n+1)[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) ] [ x^2(1+xb) - \beta^2 - 4n ] + [1-4x\cos\theta + 6x^2\cos^2\theta + \mathcal{O}(x^3) ] [2n(n+1) - 3n\beta^2 ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~~i~x\cos\theta [1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) ] [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~x^0\biggl\{2n (n+1) -4n(n+1) + 2n(n+1) \biggr\} + x^1\biggl\{8n(n+1)\cos\theta - 8n(n+1)\cos\theta\biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + x^2\biggl\{-4n(n+1)\cos^2\theta + (n+1)\biggl[ 1 - \biggl(\frac{\beta}{x}\biggr)^2 \biggr] + 12n(n+1)\cos^2\theta - 3n \biggl(\frac{\beta}{x}\biggr)^2 \biggr\} + \mathcal{O}(x^3) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~~i~x^2\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~x^0\biggl\{0 \biggr\} + x^1\biggl\{0\biggr\} + x^2\biggl\{(n+1)[8n\cos^2\theta + 1] - (4n+1)\biggl(\frac{\beta}{x}\biggr)^2 \biggr\} + \mathcal{O}(x^3) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~~i~x^2\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3) \biggr] </math> </td> </tr> </table> </div> where, <table border="0" align="center" cellpadding="8"> <tr><td align="center"> <math>b_0 \equiv [ 2^7\cdot 3 (n+1)^3 \cos^2\theta ]^{1/2} \, .</math> </td></tr> </table> Hence, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\mathrm{RHS}_4}{x^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ 2^2(n+1)^2\cdot \mathcal{A} + m^2 \cdot \mathcal{A} \biggl\{ -(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ 2^2(n+1)^2\cdot \mathcal{A} + m^2 \cdot \mathcal{A} (n+1)\biggl\{ -x^2\biggl(\frac{4n+1}{n+1}\biggr)\cdot \biggl(\frac{\beta}{x}\biggr)^2 + x^2(1+\cancelto{0}{x}b)[2^3(n+1)\cos^2\theta - 3] ~~~\pm ~~i~x^2 \biggl(\frac{\beta}{x}\biggr) \frac{b_0}{(n+1)} (1+\cancelto{0}{x}b)^{1/2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~\mathcal{A} (n+1)\biggl\{-~ 2^2(n+1) + m^2 \cancelto{0}{x^2} \biggl[ [2^3(n+1)\cos^2\theta - 3] -\biggl(\frac{4n+1}{n+1}\biggr)\cdot \biggl(\frac{\beta}{x}\biggr)^2 ~~~\pm ~~i~ \biggl(\frac{\beta}{x}\biggr) \frac{b_0}{(n+1)} \biggr] \biggl\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~-2^2(n+1)x^2\biggl\{(n+1)[8n\cos^2\theta + 1] - (4n+1)\biggl(\frac{\beta}{x}\biggr)^2 ~~~\pm~i~\biggl(\frac{\beta}{x}\biggr) \frac{nb_0}{2^2(n+1)} \biggl[1-2\cancelto{0}{x}\cos\theta + \cancelto{0}{x^2}\cos^2\theta + \cancelto{0}{\mathcal{O}}(x^3) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~~\frac{\mathrm{RHS}_4}{x^4}</math> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>2^2(n+1)(4n+1)\biggl(\frac{\beta}{x}\biggr)^2 ~-~2^2(n+1)^2[8n\cos^2\theta + 1] ~~\pm~i~(-1)\biggl(\frac{\beta}{x}\biggr) nb_0 \, . </math> </td> </tr> </table> This should be compared to, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\mathrm{LHS}_4}{x^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 - xb \biggr] (1-x\cos\theta)^3 \biggl\{ (1-x\cos\theta) \biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -~n x (1-x\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \, . </math> </td> </tr> </table> Now, from above, we can write, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~x^2\biggl\{ 2(n+1)[2^3(n+1)\cos^2\theta -3](1+3\cancelto{0}{x}b) ~\pm~~i~\beta \biggl(\frac{b_0}{2^2}\biggr)\biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr] \biggr\} + x^2 \biggl\{2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \cancelto{0}{x^3}\biggl\{-2^4\cdot 3 (n+1)^2\cos^3\theta + 2^4(n+1)^2\cos^5\theta + 3^2(n+1)(16n +19)\sin^2\theta \cos\theta -2^3\cdot 23 (n+1)^2\sin^2\theta \cos^3\theta \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \pm~~i~(-1)\beta b_0 ~x(1+\cancelto{0}{x}b)^{1/2}\biggl\{ 1 + \frac{3\cancelto{0}{x}\sin^2\theta (5\cos^2\theta -2)}{2(1+xb)\cos\theta } + \frac{3^2\cancelto{0}{x^2}\sin^6\theta}{2^2(1+xb)^2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~x^2\biggl\{ 2(n+1)[2^3(n+1)\cos^2\theta -3] + 2^4(n+1)^2(\sin^2\theta - \cos^2\theta)\biggr\} ~\pm~~i~x \biggl\{ \cancelto{0}{x\beta} \biggl(\frac{b_0}{2^2}\biggr)\biggl[ \frac{b(4+3xb)}{(1+xb)^{3/2}} \biggr]- \beta b_0\biggr\}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~2(n+1)x^2\biggl\{ 2^3(n+1)\sin^2\theta -3 \biggr\} ~\pm~~i~x^2\biggl\{ - \biggl(\frac{\beta}{x}\biggr) b_0\biggr\} \, .</math> </td> </tr> </table> </div> Also, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ x\biggl[ (1-2x \cos\theta )\cdot \frac{\partial \Lambda }{\partial x} + \sin\theta \cdot \frac{\partial\Lambda}{\partial\theta}\biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ x^2(n+1)[-6 + 2^4(n+1)\cos^2\theta] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \cancelto{0}{x^3}(n+1)\cos\theta \{ [ 15 + 2^4(n+1) ] -\cos^2\theta[9 + 2^3\cdot 7 (n+1)] +2^3\cdot 3(n+1)\cos^4\theta \} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +\cancelto{0}{x^4}(n+1) \{9 - 2^2\cdot 3^2(1+2n)\cos^2\theta - [9 + 32(n+1)]\cos^4\theta +2^3(n+1)\cos^6\theta \} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)~\biggl[ \frac{2^5\cdot 3 (n+1)^3}{1+\cancelto{0}{x}(3\cos\theta-\cos^3\theta)} \biggr]^{1/2} \biggl\{ 2\cos\theta - \cancelto{0}{x}[2 - 7\cos^2\theta + 3\cos^4\theta ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~- \cancelto{0}{x^2} \cos\theta [ 9 +4\cos^2\theta -\cos^4\theta ] \biggr\}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ 2(n+1)x^2[2^3(n+1)\cos^2\theta -3] ~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)b_0 \, . </math> </td> </tr> </table> Finally, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ nx\biggl[ (2+3xb )\cdot \frac{\partial \Lambda }{\partial x} - 3\sin^3\theta \cdot \frac{\partial\Lambda}{\partial\theta} \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + n \cancelto{0}{x^3}\cdot 2^2\cdot 3(n+1)\cos\theta \{ [2^2n -5] +\cos^2\theta[2^4n + 19] - 2^2(n+1)\cos^4\theta \} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +n \cancelto{0}{x^4} \cdot 3^2(n+1)\cos^2\theta \{ -3^3 + 2\cdot 3^2\cos^2\theta[2^2n+5] - 99\cdot \cos^4\theta + 2^3(n+1)\cos^6\theta \} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + n\cancelto{0}{x^4} (n+1)\sin^4\theta \{ -3^3 + 3^2\cos^2\theta [ 2^3\cdot 3 n + 27] -2^3\cdot 3\cdot 5(n+1)\cos^4\theta \} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~\pm~~i~nx^2\biggl(\frac{\beta}{x}\biggr)~ \biggl[ \frac{2^5\cdot 3(n+1)^3}{1+\cancelto{0}{x}b} \biggr]^{1/2} \biggl\{ 4\cos\theta + 6\cancelto{0}{x}(2b\cos\theta + \sin^4\theta) + 3\cancelto{0}{x^2}(3b^2\cos\theta +2b\sin^4\theta + 3\sin^6\theta\cos\theta) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] ~~\pm~~i~2nx^2\biggl(\frac{\beta}{x}\biggr)b_0 \, . </math> </td> </tr> </table> Inserting these three approximate expressions into the LHS_4 ensemble gives, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\mathrm{LHS}_4}{x^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 - \cancelto{0}{xb} \biggr] (1-\cancelto{0}{x}\cos\theta)^3 \biggl\{ (1-\cancelto{0}{x}\cos\theta) \biggl[x^2 \frac{\partial^2\Lambda}{\partial x^2} + \frac{\partial^2\Lambda}{\partial \theta^2} \biggr] + x\biggl[(1-2x\cos\theta)\frac{\partial\Lambda}{\partial x} + \sin\theta \frac{\partial\Lambda}{\partial\theta} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -~n x (1-\cancelto{0}{x}\cos\theta)^4\biggl[(2+3xb) \frac{\partial\Lambda}{\partial x} -3\sin^3\theta \frac{\partial\Lambda}{\partial\theta} \biggr] </math> </td> </tr> <tr> <td align="right"> <math>~</math> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ \biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr] \biggl\{ ~2(n+1)x^2\biggl[ 2^3(n+1)\sin^2\theta -3 \biggr] ~\pm~~i~x^2\biggl[ - \biggl(\frac{\beta}{x}\biggr) b_0\biggr] </math> </td> </tr> <tr> <td align="right"> <math>~</math> </td> <td align="center"> </td> <td align="left"> <math>~ + 2(n+1)x^2[2^3(n+1)\cos^2\theta -3] ~\pm~~i~x^2 \biggl(\frac{\beta}{x}\biggr)b_0 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -~x^2\cdot 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] ~~\pm~~i~2nx^2\biggl(\frac{\beta}{x}\biggr)b_0 </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~\frac{\mathrm{LHS}_4}{x^4}</math> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ \biggl[ \biggl(\frac{\beta}{x}\biggr)^2 - 1 \biggr] \biggl\{ ~2(n+1)\biggl[ 2^3(n+1)\sin^2\theta -3 \biggr] + 2(n+1)[2^3(n+1)\cos^2\theta -3] \biggr\} -~ 2^2n(n+1)[2^3(n+1)\cos^2\theta -3] </math> </td> </tr> <tr> <td align="right"> <math>~</math> </td> <td align="center"> </td> <td align="left"> <math>~ ~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ 2^2(n+1) \biggl(\frac{\beta}{x}\biggr)^2 [2^2(n+1) - 3 ] -(n+1) \biggl\{ [2^4(n+1) -12] +~ 2^2n[2^3(n+1)\cos^2\theta -3] \biggr\} ~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ 2^2(n+1) \biggl(\frac{\beta}{x}\biggr)^2 [4n+1 ] - 2^2(n+1)^2 [ 1+~ 2^3n\cos^2\theta ] ~\pm~~i~2n\biggl(\frac{\beta}{x}\biggr)b_0 \, . </math> </td> </tr> </table> ======Assessment====== The good news is that the real part of the <math>~\mathrm{LHS}_4</math> expression exactly matches the real part of the <math>~\mathrm{RHS}_4</math> expression. But the imaginary differ by a factor of 2. So, let's repeat the steps leading to the imaginary parts. '''<font color="red" size="+1">Case B:</font>''' <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathrm{Im}\biggl[ \frac{\mathrm{RHS_4}}{x^2} \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~2^2(n+1)\cdot \mathrm{Im}[(n+1)\mathcal{A}] +\frac{m^2}{(n+1)}\biggl\{ \mathrm{Im}[(n+1)\mathcal{A}]\cdot \mathrm{Re}[\Lambda] + \mathrm{Re}[(n+1)\mathcal{A}]\cdot \mathrm{Im}[\Lambda] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~2^2(n+1)\cdot x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +\frac{m^2}{(n+1)}\biggl\{ x\cos\theta (1-x\cos\theta)^2 [ 2^3\cdot 3 n^2\beta^2(n+1) ]^{1/2} \biggr\} \cdot \biggl\{ -(4n+1)\beta^2 + (\beta\eta)^2(n+1)[2^3(n+1)\cos^2\theta - 3] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +\frac{m^2}{(n+1)}\biggl\{ 2n (n+1) + (n+1)(1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4 [2n(n+1) - 3n\beta^2 ] \biggr\} \cdot \biggl\{\beta\cos\theta [2^7\cdot 3 (n+1)^3]^{1/2} (\beta\eta) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~x^2 \biggl(\frac{\beta}{x}\biggr) \biggl\{ (1-x\cos\theta)^2 nb_0 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +m^2 x^4\biggl(\frac{\beta}{x}\biggr)\biggl\{ (1-x\cos\theta)^2 \biggl[\frac{n b_0}{2^2(n+1)^2} \biggr] \biggr\} \cdot \biggl\{ -\biggl[ \frac{4n+1}{n+1}\biggr] \biggl(\frac{\beta}{x}\biggr)^2 + (1+xb)[2^3(n+1)\cos^2\theta - 3] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + m^2 x^2\biggl(\frac{\beta}{x}\biggr) \biggl\{ 2n + (1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4\cdot \biggl[2n - \frac{3n\beta^2}{(n+1)} \biggr] \biggr\} \cdot \biggl\{(1+xb)^{1/2} \biggr\} </math> </td> </tr> </table> </div> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Rightarrow~~~\mathrm{Im}\biggl[ \frac{\mathrm{RHS_4}}{x^4} \biggr]\biggl(\frac{x}{\beta}\biggr) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~(1-x\cos\theta)^2 nb_0 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + m^2 \biggl\{ 2n + (1-x\cos\theta)^2 [ x^2(1+xb) - \beta^2 - 4n ] + (1-x\cos\theta)^4\cdot \biggl[2n - \frac{3n\beta^2}{(n+1)} \biggr] \biggr\} \cdot (1+xb)^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +m^2 \biggl\{ -\biggl[ \frac{4n+1}{n+1}\biggr] \beta^2 + x^2(1+xb)[2^3(n+1)\cos^2\theta - 3] \biggr\} \cdot (1-x\cos\theta)^2 \biggl[\frac{n b_0}{2^2(n+1)^2} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~(1-x\cos\theta)^2 nb_0 + m^2 \biggl\{ 2n - 4n (1-x\cos\theta)^2 + 2n(1-x\cos\theta)^4 \biggr\} \cdot (1+xb)^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + m^2 (1-x\cos\theta)^2 \biggl\{ x^2(1+xb)^{3/2} - \beta^2\cdot (1+xb)^{1/2} \biggl[ 1 + \frac{3n(1-x\cos\theta)^2}{(n+1)} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +m^2 (1-x\cos\theta)^2 \biggl\{ x^2(1+xb)[2^3(n+1)\cos^2\theta - 3]\cdot\biggl[\frac{n b_0}{2^2(n+1)^2} \biggr] -\biggl[ \frac{nb_0(4n+1)}{2^2(n+1)^3}\biggr] \beta^2 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~(1-x\cos\theta)^2 nb_0 + m^2 \biggl\{ 2n - 4n [1-2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3)] + 2n[ 1-4x\cos\theta + 6x^2\cos^2\theta + \mathcal{O}(x^3)] \biggr\} \cdot (1+xb)^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + m^2 x^2 (1-x\cos\theta)^2 \biggl\{ 2^2(n+1)^2(1+xb)^{1/2} + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0 \biggr\} \cdot \frac{(1+xb)}{2^2(n+1)^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - m^2 \beta^2 (1-x\cos\theta)^2 \biggl\{ nb_0(4n+1) + 2^2(n+1)^2(1+xb)^{1/2} [ (n+1) + 3n(1-x\cos\theta)^2 ] \biggr\} \cdot \frac{1}{2^2(n+1)^3} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~ nb_0 [1 -2x\cos\theta + x^2\cos^2\theta + \mathcal{O}(x^3)] + m^2 \biggl\{ 8n x^2\cos^2\theta + \mathcal{O}(x^3)\biggr\} \cdot (1+\cancelto{0}{x}b)^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + m^2 x^2 (1-\cancelto{0}{x}\cos\theta)^2 \biggl\{ 2^2(n+1)^2(1+\cancelto{0}{x}b)^{1/2} + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0 \biggr\} \cdot \frac{(1+\cancelto{0}{x}b)}{2^2(n+1)^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - m^2 \beta^2 (1-\cancelto{0}{x}\cos\theta)^2 \biggl\{ nb_0(4n+1) + 2^2(n+1)^2(1+\cancelto{0}{x}b)^{1/2} [ (n+1) + 3n(1-\cancelto{0}{x}\cos\theta)^2 ] \biggr\} \cdot \frac{1}{2^2(n+1)^3} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~-~ nb_0 [1 -2x\cos\theta ] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~-nb_0x^2\cos^2\theta + m^2 x^2 \biggl\{2^5n(n+1)^2 \cos^2\theta + 2^2(n+1)^2 + [2^3(n+1)\cos^2\theta - 3]\cdot n b_0 \biggr\} \cdot \frac{1}{2^2(n+1)^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - m^2 \beta^2 \biggl\{ nb_0(4n+1) + 2^2(n+1)^2 [ (n+1) + 3n ] \biggr\} \cdot \frac{1}{2^2(n+1)^3} </math> </td> </tr> </table> </div>
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