Editing
Appendix/Ramblings/PPToriPt1A
(section)
Jump to navigation
Jump to search
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
====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>
Summary:
Please note that all contributions to JETohlineWiki may be edited, altered, or removed by other contributors. If you do not want your writing to be edited mercilessly, then do not submit it here.
You are also promising us that you wrote this yourself, or copied it from a public domain or similar free resource (see
JETohlineWiki:Copyrights
for details).
Do not submit copyrighted work without permission!
Cancel
Editing help
(opens in new window)
Navigation menu
Personal tools
Not logged in
Talk
Contributions
Log in
Namespaces
Page
Discussion
English
Views
Read
Edit
View history
More
Search
Navigation
Main page
Tiled Menu
Table of Contents
Old (VisTrails) Cover
Appendices
Variables & Parameters
Key Equations
Special Functions
Permissions
Formats
References
lsuPhys
Ramblings
Uploaded Images
Originals
Recent changes
Random page
Help about MediaWiki
Tools
What links here
Related changes
Special pages
Page information