Editing Appendix/Mathematics/Hypergeometric (section)

===General Value for b===
[[File:CommentButton02.png|right|100px|Comment by J. E. Tohline:  In Van der Borght (1970), the fourth argument of the hypergeometric function is ax<sup>2</sup> whereas the more general power law form, ax<sup>b</sup>, applies.]]In association with his equation (3), {{ VdBorght70full }} states that a displacement function of the form,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\xi</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
x^c F(\alpha, \beta, \gamma, ax^b) \, ,
</math>
  </td>
</tr>
</table>
provides a solution to the following 2<sup>nd</sup>-order ODE:
<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>
x^2(ax^b - 1) \cdot \frac{d^2\xi}{dx^2} + (apx^b + \lambda)x \cdot \frac{d\xi}{dx} + (ar x^b + s)\cdot \xi
\, .</math>
  </td>
</tr>
</table>

Is this ODE essentially the same as the above-defined ''hypergeometric equation''?  

A mapping between the two differential equations requires,
<!--
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>ax^2</math>
  </td>
  <td align="center">
<math>\leftrightarrow</math>
  </td>
  <td align="left">
<math>
z \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>\frac{\xi}{x^c}</math>
  </td>
  <td align="center">
<math>\leftrightarrow</math>
  </td>
  <td align="left">
<math>
u
\, ,</math>
  </td>
</tr>
</table>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\Rightarrow~~~ \frac{dx}{dz}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d}{dz}\biggl(\frac{z}{a}\biggr)^{1 / 2}
=
a^{-1 / 2} \cdot \frac{dz^{1 / 2}}{dz}
=
\frac{1}{2} a^{-1 / 2} z^{-1 / 2}
=
\frac{1}{2ax} 
\, ,</math>
  </td>
</tr>
</table>
-->

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

<tr>
  <td align="right">
<math>ax^b</math>
  </td>
  <td align="center">
<math>\leftrightarrow</math>
  </td>
  <td align="left">
<math>
z \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>\frac{\xi}{x^c}</math>
  </td>
  <td align="center">
<math>\leftrightarrow</math>
  </td>
  <td align="left">
<math>
u
\, ,</math>
  </td>
</tr>
</table>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\Rightarrow~~~ \frac{dx}{dz}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d}{dz}\biggl(\frac{z}{a}\biggr)^{1 / b}
=
a^{-1 / b} \cdot \frac{dz^{1 / b}}{dz}
=
\frac{1}{b} \cdot a^{-1 / b} z^{-1 + 1/b}
=
\frac{1}{bz} \cdot \biggl(\frac{z}{a}\biggr)^{1 / b}
=
\frac{x}{bz} 
=
\frac{x^{1-b}}{ab} 
\, ,</math>
  </td>
</tr>
</table>

in which case,
<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>
z(1-z) \frac{d^2u}{dz^2} + [\gamma - (\alpha + \beta + 1)z] \frac{du}{dz} - \alpha \beta u
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
ax^b(1-ax^b)\biggl\{ \frac{dx}{dz} \cdot \frac{d}{dx}\biggl[\frac{dx}{dz} \cdot \frac{d}{dx}\biggl( \xi x^{-c} \biggr)  \biggr]\biggr\}
+ [\gamma - (\alpha + \beta + 1)ax^b] \frac{dx}{dz} \cdot \frac{d}{dx}\cdot\biggl( \xi x^{-c} \biggr) 
- \alpha \beta \biggl( \xi x^{-c} \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
ax^b(1-ax^b)\biggl\{ \frac{x^{1-b}}{ab}\cdot \frac{d}{dx}\biggl[\frac{x^{1-b}}{ab} \cdot 
\biggl(x^{-c} \frac{d\xi}{dx} - c \xi x^{-1-c} \biggr)  \biggr]\biggr\}
+ [\gamma - (\alpha + \beta + 1)ax^b] \cdot \frac{x^{1-b}}{ab}\cdot \biggl(x^{-c} \frac{d\xi}{dx} - c \xi x^{-1-c} \biggr)  
- \alpha \beta \biggl( \xi x^{-c} \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
ax^b(1-ax^b)\biggl\{ \frac{x^{1-b}}{a^2b^2}\cdot 
\frac{d}{dx}\biggl[ x^{1-b-c} \frac{d\xi}{dx} - c \xi x^{-b-c} \biggr]\biggr\}
+ \frac{[\gamma - (\alpha + \beta + 1)ax^b]}{ab} \cdot \biggl(x^{1-b-c} \frac{d\xi}{dx} - c \xi x^{-b-c} \biggr)  
- \alpha \beta \biggl( \xi x^{-c} \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
ax^b(1-ax^b) \cdot \frac{x^{1-b}}{a^2b^2}\cdot 
\biggl[ 
x^{1-b-c} \frac{d^2\xi}{dx^2}
+
(1-b-c)x^{-b-c} \frac{d\xi}{dx} 
- 
c x^{-b-c}\frac{d\xi}{dx}
- 
c(-b-c) \xi x^{-1-b-c}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \frac{[\gamma - (\alpha + \beta + 1)ax^b]}{ab} \cdot \biggl[ x^{1-b-c} \frac{d\xi}{dx} - c \xi x^{-b-c} \biggr]  
- \biggl( \alpha \beta x^{-c} \biggr)\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(1-ax^b) \cdot \frac{x}{ab^2}\cdot 
\biggl[ x^{1-b-c} \frac{d^2\xi}{dx^2} \biggr]
+
(1-ax^b) \cdot \frac{x}{ab^2}\cdot 
\biggl[ (1-b-c)x^{-b-c} \frac{d\xi}{dx} - c x^{-b-c}\frac{d\xi}{dx} \biggr]
+
(1-ax^b) \cdot \frac{x}{ab^2}\cdot 
\biggl[ c(b+c) \xi x^{-1-b-c}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \frac{[\gamma - (\alpha + \beta + 1)ax^b]}{ab} \cdot \biggl[ x^{1-b-c} \frac{d\xi}{dx} \biggr]  
- \frac{[\gamma - (\alpha + \beta + 1)ax^b]}{ab} \cdot \biggl[ c \xi x^{-b-c} \biggr]  
- \biggl( \alpha \beta x^{-c} \biggr)\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{(1-ax^b)}{ab^2} \cdot x^2 
\biggl[ x^{-b-c} \frac{d^2\xi}{dx^2} \biggr]
+\biggl\{
\frac{(1-ax^b) }{ab^2}\cdot 
\biggl[ (1-b-c)x^{1-b-c} - c x^{1-b-c}\biggr]
+ \frac{[\gamma - (\alpha + \beta + 1)ax^b]}{ab} \cdot \biggl[ x^{1-b-c} \biggr] 
\biggr\} \frac{d\xi}{dx} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+\biggl\{
- \frac{c[\gamma - (\alpha + \beta + 1)ax^b]}{ab} \cdot \biggl[ x^{-b-c} \biggr]  
- \biggl( \alpha \beta x^{-b-c} \biggr)x^b
+
\frac{(1-ax^b)}{ab^2} \cdot 
\biggl[ c(b+c) x^{-b-c}\biggr]
\biggr\}\xi \, .
</math>
  </td>
</tr>
</table>
Multiplying through by <math>(-ab^2 x^{b+c})</math> gives,

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

<tr>
  <td align="right">
<math>(-ab^2 x^{b+c}) \times 0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
x^2(ax^b-1) \frac{d^2\xi}{dx^2} 
+\biggl\{
(ax^b-1) \cdot 
\biggl[ (1-b-c) - c \biggr]
- b[\gamma - (\alpha + \beta + 1)ax^b] 
\biggr\} x \cdot \frac{d\xi}{dx} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+\biggl\{
bc[\gamma - (\alpha + \beta + 1)ax^b]   
+ \biggl( \alpha \beta a b^2 \biggr)x^b
-
(1-ax^b) \cdot 
\biggl[ c(b+c) \biggr]
\biggr\}\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
x^2(ax^b-1) \frac{d^2\xi}{dx^2} 
+\biggl\{
-(1 + b\gamma -b-2c)
+
ax^b (1-b-2c)
+ b(\alpha + \beta + 1)ax^b 
\biggr\} x \cdot \frac{d\xi}{dx} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+\biggl\{
bc\gamma - c(b+c)
- bc(\alpha + \beta + 1)ax^b   
+ \alpha \beta a b^2 x^b
+ c(b+c)ax^b
\biggr\}\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
x^2(ax^b-1) \frac{d^2\xi}{dx^2} 
+ (apx^b + \lambda) x \cdot \frac{d\xi}{dx} 
+ (arx^b + s) \xi \, ,
</math>
  </td>
</tr>
</table>
which matches equation (3) of {{ VdBorght70 }} if the expressions for the four new scalar coefficients are,

<table border="1" align="center" width="60%" cellpadding="10">
<tr><td align="center">'''Required Mapping Expressions'''</td></tr>
<tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="left" width="25%">
1<sup>st</sup>:
  </td>
  <td align="right">
<math>p</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(1-b-2c) + b(\alpha + \beta + 1) = 1 - 2c + b(\alpha + \beta) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="left" width="25%">
2<sup>nd</sup>:
  </td>
  <td align="right">
<math>\lambda</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-(1 + b\gamma -b-2c) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="left" width="25%">
3<sup>rd</sup>:
  </td>
  <td align="right">
<math>s</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
bc\gamma - c(b+c) = c(b\gamma - b - c)\, ,
</math>
  </td>
</tr>

<tr>
  <td align="left" width="25%">
4<sup>th</sup>:
  </td>
  <td align="right">
<math>r</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- bc(\alpha + \beta + 1)   
+ \alpha \beta b^2 
+ c(b+c)
\, .
</math>
  </td>
</tr>
</table>
</td></tr></table>

If, for any given problem, we are given the values of these four scalar coefficients along with a choice of the exponent, <math>b</math>, that appears in the fourth argument of the hypergeometric series, we can determine values the other three arguments of the hypergeometric series &#8212; <math>\alpha, \beta, \gamma</math> &#8212; and the exponent, <math>c</math>.  In what follows we show how this is done.

====Determining the Value of the Exponent, <i>c</i>====

Equating <math>(b\gamma)</math> in the 2<sup>nd</sup> and 3<sup>rd</sup> of the ''required mapping expressions'', gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>-(1  -b-2c +\lambda)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{s}{c} + b + c
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ (1+\lambda)  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
c - \frac{s}{c}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 0  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
c^2 - c(1+\lambda)- s \, .
</math>
  </td>
</tr>
</table>
The pair of roots, <math>c_\pm</math>, of this quadratic equation are then obtained from the relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>2c_\pm  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(1+\lambda) \pm [
(1+\lambda)^2 + 4s
]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>

<span id="cplusminus">Note for further use below that,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>2(c_+ - c_-)  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\{(1+\lambda) + [(1+\lambda)^2 + 4s]^{1 / 2} \}
-
\{(1+\lambda) - [(1+\lambda)^2 + 4s]^{1 / 2} \}
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2[(1+\lambda)^2 + 4s]^{1 / 2}
\, .
</math>
  </td>
</tr>
</table>

</td></tr>
<tr>
  <td align="left">
Consistent with our derivation, {{ VdBorght70 }} states, <font color="darkgreen">"&hellip; if <math>A_1</math>, <math>A_2</math> are the solutions of <math>A^2 - (\lambda + 1)A-s = 0</math> &hellip; then <math>c = A_1</math> &hellip;"</font>
  </td>
</tr>

====Determining the Value of the Coefficient, <i>&gamma;</i>====

<tr><td align="left">
Combining our 3<sup>rd</sup> ''required mapping expression'' with the quadratic equation for <math>c_\pm</math> in such a way as to eliminate <math>s</math>, we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>cb(\gamma - 1) -c^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
c^2 - c(1+\lambda)
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ b(\gamma - 1) </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2c_\pm - (1+\lambda) \, .
</math>
  </td>
</tr>
</table>

Adopting the ''superior'' sign, we find that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>b(\gamma - 1) </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
2c_+ - (1+\lambda)
=
[(1+\lambda)^2 + 4s]^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(c_+ - c_-) \, ,
</math>
  </td>
</tr>
</table>
where, in order to make this last step we have drawn from the relation derived [[#cplusminus|immediately above]].
</td></tr>
<tr>
  <td align="left">
Consistent with this derivation, {{ VdBorght70 }} states, <font color="darkgreen">"&hellip; <math>(1-\gamma)b = A_2 - A_1</math>  &hellip;"</font>
  </td>
</tr>

====Determining the Values of the Coefficients, <i>&alpha;</i> and  <i>&beta;</i>====

Combining the 1<sup>st</sup> and 4<sup>th</sup> ''required mapping expressions'' in such a way as to cancel terms involving <math>bc(\alpha + \beta)</math>, we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>(1 - p)c - 2c^2 + bc(\alpha + \beta)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
bc(\alpha + \beta)   
+ bc   
- \alpha \beta b^2 
- c(b+c) + r
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ (1 - p)c - c^2 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \alpha \beta b^2 
+ r
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ (b\alpha)(b\beta )  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- (1 - p)c + c^2
+ r \, .
</math>
  </td>
</tr>
</table>

Also, from the 1<sup>st</sup> ''required mapping expression'' alone we can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
(b\alpha) 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(p-1) + 2c - b\beta \, .</math>
  </td>
</tr>
</table>
Together, then, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>(b\beta ) \biggl[  
(p-1) + 2c - b\beta
\biggr]  
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- (1 - p)c + c^2
+ r
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 
0
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(b\beta )^2 - (b\beta)\underbrace{[  (p-1) + 2c ]}_{(b\alpha + b\beta)}  
+ \overbrace{[c^2 + (p - 1 )c + r]}^{(b\alpha)\cdot(b\beta)  }
\, .
</math>
  </td>
</tr>
</table>
The pair of roots, <math>(b\beta)_\pm</math>, of this quadratic equation are then obtained from the relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>2(b\beta)_\pm  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
[ (p-1) + 2c ] \pm \biggl\{
[ (p-1) + 2c ]^2 - 4[c^2 + (p - 1 )c + r]
\biggr\}^{1 / 2}
\, .
</math>
  </td>
</tr>
</table>

Finally, plugging this expression for <math>(b\beta_\pm)</math> into the 1<sup>st</sup> ''required mapping expression'' gives <math>(b\alpha_\pm).</math>

<!--
{{ VdBorght70 }} states that if, <font color="darkgreen">"&hellip; <math>B_1, B_2</math> are solutions of <math>B^2 - (p-1)B + r = 0</math>, then &hellip; <math>b\beta = A_1 + B_2.</math>"</font>  Let's see if our derivation leads to this same conclusion.  First note that the roots, <math>B_\pm</math>, of ''this'' Van der Borght quadratic equation are,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>2B_\pm  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(p-1) \pm [(p-1)^2 -4r]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
If we assign the ''inferior'' root with Van der Borght's notation, <math>B_2</math>, then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>2(A_1 + B_2) = 2(c_+ + B_-)  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\{ (1+\lambda) + [(1+\lambda)^2 + 4s]^{1 / 2} \}
+
\{(p-1) - [(p-1)^2 -4r]^{1 / 2} \}
</math>
  </td>
</tr>
</table>
-->

====Alternate Determination of <i>&alpha;</i> and  <i>&beta;</i> by Completing Squares====

<table border="1" cellpadding="10" width="80%" align="center">
<tr><td align="left">
Again, let's draw upon the {{ VdBorght70 }} statement that, <font color="darkgreen">"&hellip; if <math>A_1</math>, <math>A_2</math> are the solutions of <math>A^2 - (\lambda + 1)A-s = 0</math> &hellip; then <math>c = A_1</math> &hellip;"</font>


{{ VdBorght70 }} also states that if, <font color="darkgreen">"&hellip; <math>B_1, B_2</math> are solutions of <math>B^2 - (p-1)B + r = 0</math>, then &hellip; <math>b\alpha = A_1 + B_1</math> and <math>b\beta = A_1 + B_2.</math>"</font>  

Let's see if we draw these same conclusions.  
</td></tr>
</table>

First, ''Complete the square'' in the quadratic equation for <math>B^2</math>:

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

<tr>
  <td align="right">
<math>B^2 - (p-1)B  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-r
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ B^2 - (p-1)B + \biggl[\frac{(p-1)}{2}\biggr]^2 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{(p-1)}{2}\biggr]^2 - r
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl[ B - \frac{(p-1)}{2}\biggr]^2 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{(p-1)}{2}\biggr]^2 - r
</math>
  </td>
</tr>
</table>

Second, ''complete the square'' in the quadratic equation for <math>A^2</math> &#8212; which also completes the square for <math>c^2</math>:

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

<tr>
  <td align="right">
<math>A^2 - (\lambda + 1)A  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
s
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ A^2 - (\lambda + 1)A  + \biggl[\frac{(\lambda + 1)}{2}\biggr]^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
s + \biggl[\frac{(\lambda + 1)}{2}\biggr]^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl[ A - \frac{(\lambda + 1)}{2} \biggr]^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
s + \biggl[\frac{(\lambda + 1)}{2}\biggr]^2
</math>
  </td>
</tr>
</table>

Third, ''complete the square'' in the quadratic equation for <math>(b\beta)^2</math>:

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

<tr>
  <td align="right">
<math> 
(b\beta )^2 - (b\beta)[  (p-1) + 2c ]  
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- [c^2 + (p - 1 )c + r]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math> 
\Rightarrow ~~~ (b\beta )^2 - (b\beta)[  (p-1) + 2c ] + \biggl[ \frac{(p-1)+2c}{2} \biggr]^2 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{(p-1)+2c}{2} \biggr]^2 - [c^2 + (p - 1 )c + r]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math> 
\Rightarrow ~~~ \biggl[ (b\beta ) - \frac{(p-1)+2c}{2} \biggr]^2 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{(p-1)+2c}{2} \biggr]^2 - \biggl[c^2 + (p - 1 )c + \frac{(p-1)^2}{2^2}\biggr] 
- r + \frac{(p-1)^2}{2^2}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- r + \frac{(p-1)^2}{2^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ B - \frac{(p-1)}{2}\biggr]^2 \, .
</math>
  </td>
</tr>
</table>
Taking the ''positive'' root of both sides of this expression, we find that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math> 
(b\beta ) - \frac{(p-1)}{2} - c_+  
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
B_+ - \frac{(p-1)}{2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math> 
\Rightarrow ~~~ (b\beta )  
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
B_+ + c_+ \, . 
</math>
  </td>
</tr>
</table>
But, <math>c_\pm = A_\pm</math>.  So we conclude, as did {{ VdBorght70 }}, that,

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

<tr>
  <td align="right">
<math> 
(b\beta )  
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
B_+ + A_+ \, . 
</math>
  </td>
</tr>
</table>

Alternatively, taking the ''negative'' root of the RHS of this expression, we find that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math> 
(b\beta ) - \frac{(p-1)}{2} - c_+  
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- B_- + \frac{(p-1)}{2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math> 
\Rightarrow ~~~ (b\beta )   
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{(p-1)}{2} + c_+ - B_- + \frac{(p-1)}{2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
c_+ - B_- + (p-1) \, . 
</math>
  </td>
</tr>
</table>

Also, given that,

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

<tr>
  <td align="right">
<math>
1 - 2c + b(\alpha + \beta) 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>p</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ (b\alpha) 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>p - 1 + 2c - (b\beta)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(p - 1) + 2c - \biggl[ c_+ - B_- + (p-1)  \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>2c - c_+ + B_- \, .</math>
  </td>
</tr>
</table>
As long as we assume that <math>c = c_+</math> in this expression, we also obtain the {{ VdBorght70 }} expression for <math>(b\alpha)</math>, namely,

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

<tr>
  <td align="right">
<math> 
(b\alpha )  
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
B_- + A_+ \, . 
</math>
  </td>
</tr>
</table>