Editing Appendix/Ramblings/ConcentricEllipsoidalDaringAttack (section)

==New Approach==

===Setup===

The surface of an ellipsoid with semi-major axes (a, b, c) is defined by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{x}{a}\biggr)^2 + \biggl( \frac{y}{b}\biggr)^2 + \biggl( \frac{z}{c}\biggr)^2 \, .</math>
  </td>
</tr>
</table>
This is identical to our expression for <math>~\lambda_1</math> if we make the associations,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a = \lambda_1 \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; </td>
  <td align="center">
<math>~b = \frac{\lambda_1}{q} \ ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; </td>
  <td align="left">
<math>~c = \frac{\lambda_1}{p} \, .</math>
  </td>
</tr>
</table>

Now, given that <math>~\lambda_3</math> does not functionally depend on <math>~z</math>, let's consider that the choice of <math>~z</math> is tightly associated with the specification of the second coordinate, <math>~\lambda_2</math>.  Specifically, let's adopt the definition,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_2^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~1 - \biggl( \frac{z}{c}\biggr)^2 \, ,</math>
  </td>
</tr>
</table>
in which case, we see that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~z^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~c^2(1-\lambda_2^2) = \frac{\lambda_1^2(1-\lambda_2^2)}{p^2} \, ,</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{x}{a}\biggr)^2 + \biggl( \frac{y}{b}\biggr)^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \lambda_2^2 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x^2 + q^2 y^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \lambda_1^2 \lambda_2^2 \, .</math>
  </td>
</tr>
</table>

[Note that <font color="red">in the case of spherical coordinates</font> (q<sup>2</sup> = p<sup>2</sup> = 1), <math>~\lambda_1 \rightarrow r</math>, and this "second" coordinate, <math>~\lambda_2</math>, becomes <math>~\sin\theta</math>.]  Combining this last expression with the <math>~x - y</math> relationship that is provided by the definition of <math>~\lambda_3</math>, gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_1^2 \lambda_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{y^{2/q^2}}{\lambda_3^2} + q^2y^2 \, .</math>
  </td>
</tr>
</table>

In general, the exponent of <math>~2q^{-2}</math> that appears in the first term on the right-hand side of this expression prevents us from being able to analytically prescribe the function, <math>~y(\lambda_1, \lambda_2, \lambda_3)</math>.  But a solution is obtainable for selected values of <math>~q^2 > 1</math>.

===Examine the Case: q<sup>2</sup> = 2===

If we set <math>~q^2 = 2</math>, then this last combined expression becomes a quadratic equation for <math>~y</math>.  Specifically, we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ 0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2y^2 + \frac{y}{\lambda_3^2} - \lambda_1^2 \lambda_2^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \Rightarrow~~~ y</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4} \biggl\{
-\frac{1}{\lambda_3^2} \pm \biggl[ \frac{1}{\lambda_3^4} + 8 \lambda_1^2 \lambda_2^2 \biggr]^{1 / 2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^2} \biggl\{
\biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1
\biggr\} \, .
</math>
  </td>
</tr>
</table>
(Note that, for reasons of simplicity <font color="red">for the time being</font>, in this last expression we have retained only the "positive" solution.)  Again, calling upon the <math>~x - y</math> relationship that is provided through the definition of <math>~\lambda_3</math>, we find (when q<sup>2</sup> = 2),
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{y}{\lambda_3^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^4} \biggl\{
\biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pm 
\frac{1}{2\lambda_3^{2}} \biggl\{
\biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1
\biggr\}^{1 / 2} \, .
</math>
  </td>
</tr>
</table>


<table border="1" cellpadding="8" align="center" width="80%"><tr><td align="left">
<div align="center">Summary (q<sup>2</sup> = 2)</div>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~z(\lambda_1, \lambda_2, \lambda_3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\lambda_1(1-\lambda_2^2)^{1 / 2}}{p} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~y(\lambda_1, \lambda_2, \lambda_3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^2} \biggl\{
\biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1
\biggr\} 
=
\frac{(\Lambda - 1)}{4\lambda_3^2}\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~x(\lambda_1, \lambda_2, \lambda_3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\lambda_3^{2}} \biggl\{
\biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1
\biggr\}^{1 / 2} 
=
\frac{(\Lambda - 1)^{1 / 2}}{2\lambda_3^{2}}
\, .
</math>
  </td>
</tr>
</table>
</td></tr></table>

For convenience, we have defined,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Lambda^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~1 + 8\lambda_1^2 \lambda_2^2 \lambda_3^4 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \lambda_1^2 \lambda_2^2 \lambda_3^4 </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{1}{8}\biggl( \Lambda^2 - 1\biggr) \, .</math>
  </td>
</tr>
</table>

<table border="1" width="90%" align="center" cellpadding="8">
<tr>
<td align="center" bgcolor="pink">
'''Test Example'''
</td>
</tr>
<tr>
<td align="left">
<math>~q^2 = 2, p^2=3.15, (x, y, z) = (0.4, 0.63581, 0.1)</math><p></p>
<math>~(\lambda_1, \lambda_2, \lambda_3) = (1, 0.98412, 1.99344)</math><p></p>
<math>~\ell_{3D} = 0.730058, ~~ \ell_q = 0.750164</math><p></p>
<math>~h_1 = 0.730058</math>
</td>
</tr>

<tr>
<td align="center">
<math>~\Lambda^2-1 = 122.34879 ~~~\Rightarrow ~~~ \Lambda = 11.10625</math><p></p>
</td>
</tr>

<tr><td align="left">
Do we get the correct values of <math>~(x, y, z)</math> &nbsp;?

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~z(\lambda_1, \lambda_2, \lambda_3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\lambda_1(1-\lambda_2^2)^{1 / 2}}{p} = 0.1000000 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~y(\lambda_1, \lambda_2, \lambda_3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(\Lambda - 1)}{4\lambda_3^2} = 0.635807\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~x(\lambda_1, \lambda_2, \lambda_3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(\Lambda - 1)^{1 / 2}}{2\lambda_3^{2}} = 0.400000
\, .
</math>
  </td>
</tr>
</table>
</td></tr>

<tr><td align="left">
Evaluate a few partial derivatives &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(1-\lambda_2^2)^{1 / 2}}{p} = 0.1\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial y}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^2}{\Lambda} \biggr]
= 0.693054
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2\lambda_1 \lambda_2^2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} 
= 0.218008
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ h_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[
\biggl(\frac{\partial x}{\partial \lambda_1}\biggr)^2
+ \biggl(\frac{\partial y}{\partial \lambda_1}\biggr)^2
+ \biggl(\frac{\partial z}{\partial \lambda_1}\biggr)^2
\biggr]^{1 / 2} 
=
0.733383
\, .
</math>
  </td>
</tr>
</table>
This matches the numerical value for <math>~h_1</math> as determined [[#h1evaluated|below]], but it does not match the numerical value obtained previously (0.730058) for <math>~h_1</math>.  The most likely piece that needs adjustment is the partial of "z" with respect to &lambda;<sub>1</sub>.  It needs to be &hellip;
<div align="center"><math>~\frac{\partial z}{\partial \lambda_1} = \biggl[ h_1^2 - \biggl( \frac{\partial x}{\partial \lambda_1} \biggr)^2 - \biggl( \frac{\partial y}{\partial \lambda_1} \biggr)^2 \biggr]^{1 / 2} = 0.071647</math>.</div>

Alternatively,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~h_1^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)
=
(0.730058)^2 \biggl[ \frac{p^2z}{\lambda_1} \biggr]
</math>
  </td>
</tr>
</table>

</td></tr></table>

Next, let's examine all nine partial derivatives, noting at the start that,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{\partial\Lambda}{\partial \lambda_1} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\Lambda}\biggl[16\lambda_1 \lambda_2^2 \lambda_3^4 \biggr] 
=
\frac{(\Lambda^2-1)}{\lambda_1 \Lambda} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial\Lambda}{\partial \lambda_2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\Lambda}\biggl[16\lambda_1^2 \lambda_2 \lambda_3^4 \biggr] 
=
\frac{(\Lambda^2-1)}{\lambda_2 \Lambda} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial\Lambda}{\partial \lambda_3} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\Lambda}\biggl[32\lambda_1^2 \lambda_2^2 \lambda_3^3 \biggr] 
=
\frac{2(\Lambda^2-1)}{\lambda_3 \Lambda} 
\, .
</math>
  </td>
</tr>
</table>

We have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(1-\lambda_2^2)^{1 / 2}}{p} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\lambda_1 \lambda_2}{p(1 - \lambda_2^2)^{1 / 2}} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0 \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial y}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^2} \cdot \frac{\partial \Lambda}{\partial \lambda_1}
=
\biggl[ \frac{(\Lambda^2-1)}{4\lambda_3^2\lambda_1 \Lambda} \biggr]
=
\biggl[ \frac{8\lambda_1^2 \lambda_2^2 \lambda_3^4}{4\lambda_3^2\lambda_1 \Lambda} \biggr]
=
\biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^2}{\Lambda} \biggr]
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial y}{\partial \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^2} \cdot \frac{\partial \Lambda}{\partial \lambda_2}
=
\biggl[ \frac{(\Lambda^2-1)}{4\lambda_3^2\lambda_2 \Lambda} \biggr]
=
\biggl[ \frac{8\lambda_1^2 \lambda_2^2 \lambda_3^4}{4\lambda_3^2\lambda_2 \Lambda} \biggr]
=
\biggl[ \frac{2\lambda_1^2 \lambda_2 \lambda_3^2}{\Lambda} \biggr]
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial y}{\partial \lambda_3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^2} \cdot \frac{\partial \Lambda}{\partial \lambda_3}
- \frac{(\Lambda - 1)}{2\lambda_3^3}
=
\frac{1}{4\lambda_3^2} \cdot \biggl[ \frac{2(\Lambda^2 - 1)}{\lambda_3\Lambda} \biggr]
- \frac{\Lambda(\Lambda - 1)}{2\lambda_3^3 \Lambda}
=
\biggl[ \frac{(\Lambda^2 - 1) - \Lambda(\Lambda-1)}{2\lambda_3^3\Lambda} \biggr]
=
\frac{(\Lambda - 1) }{2\lambda_3^3\Lambda}
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \cdot \frac{\partial \Lambda}{\partial \lambda_1}
=
\frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \biggl[  \frac{(\Lambda^2 - 1)}{\lambda_1 \Lambda} \biggr]
=
\frac{(\Lambda^2 - 1)}{4 \lambda_1 \lambda_3^{2} \Lambda (\Lambda-1)^{1 / 2}} 
=
\frac{2\lambda_1 \lambda_2^2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \cdot \frac{\partial \Lambda}{\partial \lambda_2}
=
\frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \biggl[  \frac{(\Lambda^2 - 1)}{\lambda_2 \Lambda} \biggr]
=
\frac{(\Lambda^2 - 1)}{4 \lambda_2 \lambda_3^{2} \Lambda (\Lambda-1)^{1 / 2}} 
=
\frac{2\lambda_1^2 \lambda_2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \cdot \frac{\partial \Lambda}{\partial \lambda_3}
- \frac{(\Lambda-1)^{1 / 2}}{\lambda_3^{3}}
=
\frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \biggl[  \frac{2(\Lambda^2 - 1)}{\lambda_3 \Lambda} \biggr]
- \frac{(\Lambda-1)^{1 / 2}}{\lambda_3^{3}}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(\Lambda^2 - 1) -2\Lambda (\Lambda - 1) }{2\lambda_3^{3} \Lambda (\Lambda-1)^{1 / 2}} 
=
- \frac{ (\Lambda - 1)^{3 / 2}  }{2\lambda_3^{3} \Lambda} 
\, .
</math>
  </td>
</tr>
</table>

What about the derived scale-factors?
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{\partial x}{\partial \lambda_1}\biggr)^2
+ \biggl(\frac{\partial y}{\partial \lambda_1}\biggr)^2
+ \biggl(\frac{\partial z}{\partial \lambda_1}\biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} \biggr]^2
+ \biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^2}{\Lambda}  \biggr]^2
+ \biggl[ \frac{(1-\lambda_2^2)^{1 / 2}}{p}  \biggr]^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{4\lambda_1^2 \lambda_2^4 \lambda_3^{4 }}{\Lambda^2 (\Lambda-1)}  \biggr]
+ \biggl[ \frac{4\lambda_1^2 \lambda_2^4 \lambda_3^4}{\Lambda^2}  \biggr]
+ \biggl[ \frac{(1-\lambda_2^2)}{p^2}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{p^2 \Lambda^2(\Lambda - 1)} \biggl[
4 p^2 \lambda_1^2 \lambda_2^4 \lambda_3^4 \Lambda 
+ (1-\lambda_2^2) \Lambda^2(\Lambda - 1)  
\biggr] \, .
</math>
  </td>
</tr>
</table>

Written in terms of Cartesian coordinates, this becomes,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{ 8 \lambda_1^2 \lambda_2^2 \lambda_3^4 (\lambda_2^2 ) }{2 \Lambda (\Lambda - 1)} 
+ \frac{(1-\lambda_2^2)}{p^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{ (\Lambda+1)\lambda_2^2  }{2 \Lambda } 
+ \frac{z^2}{\lambda_1^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{\lambda_1^2} \biggl[
\frac{ (\Lambda+1)\lambda_2^2 \lambda_1^2  }{2 \Lambda } 
+ z^2
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{\lambda_1^2} \biggl[
\frac{ (\Lambda+1)(x^2 + 2y^2) }{2 \Lambda } 
+ z^2
\biggr] \, .
</math>
  </td>
</tr>
</table>
Note that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Lambda -1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4x^2\lambda_3^4 = 4x^2 \biggl( \frac{y^2}{x^4} \biggr) = 4\biggl( \frac{y^2}{x^2} \biggr) </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \Lambda </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{x^2 + 4y^2}{x^2} \, .</math>
  </td>
</tr>
</table>

<span id="h1evaluated">Hence, the scale factor becomes,</span>

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{2 \lambda_1^2} \biggl[
(x^2 + 2y^2)  
+ \frac{ x^2(x^2 + 2y^2) }{(x^2 + 4y^2) } 
+ 2z^2
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{2 \lambda_1^2(x^2 + 4y^2) } \biggl[
(x^2 + 2y^2)  (x^2 + 4y^2) 
+ x^2(x^2 + 2y^2) 
+ 2z^2(x^2 + 4y^2) 
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{2 \lambda_1^2(x^2 + 4y^2) } \biggl[
(2x^4 + 8x^2y^2 +8y^4)
+ 2z^2(x^2 + 4y^2) 
\biggr] 
= \frac{1.911525}{3.554} = 0.537852
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ h_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
0.733384
\, .
</math>
  </td>
</tr>
</table>


----


Compare this expression with the one derived earlier, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_1^2 \biggr|_{q^2 = 2} = \biggl[\lambda_1^2 \ell_{3D}^2 \biggr]_{q^2 = 2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(x^2 + 2y^2 + p^2z^2)}{x^2 + 4y^2 + p^4z^2} \, .
</math>
  </td>
</tr>
</table>
Well &hellip; first we recognize that, when q<sup>2</sup> = 2,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_1^2 = x^2 + 2y^2 + p^2z^2 \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="center">
<math>~\lambda_2^2 = \frac{\lambda_1^2 - p^2 z^2}{\lambda_1^2}
=
\frac{x^2 + 2y^2}{x^2 + 2y^2 + p^2z^2} \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="left">
<math>~\lambda_3^2 = \frac{y}{x^2} \, .</math>
  </td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(\Lambda^2 - 1) = 8\lambda_1^2 \lambda_2^2 \lambda_3^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
8(x^2 + 2y^2 + p^2z^2)\biggl[ \frac{x^2 + 2y^2}{x^2 + 2y^2 + p^2z^2} \biggr]\frac{y^2}{x^4}
=
\biggl[ \frac{8y^2(x^2 + 2y^2)}{x^4 } \biggr] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \Lambda^2  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 + \biggl[ \frac{8y^2(x^2 + 2y^2)}{x^4 } \biggr]
=
\frac{1}{x^4}\biggl[x^4 + 8x^2y^2 + 16y^4  \biggr]
=
\frac{1}{x^4}\biggl[x^2 + 4y^4  \biggr]^2 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ (\Lambda+1)  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(x^2 + 4y^4)}{x^2} + 1
=
\frac{2x^2 + 4y^4}{x^2} \, ,
</math>
  </td>
</tr>
</table>
which means,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{8\lambda_1^2 \Lambda^2} \biggl\{
4\lambda_1^2 \lambda_2^2 \lambda_3 (\Lambda+1)
+ 4\lambda_1^2 \lambda_2^2 (\Lambda^2 - 1)
+ 8z^2\Lambda^2
\biggr\}
</math>
  </td>
</tr>
</table>