Editing Appendix/Mathematics/Hypergeometric (section)

====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>
-->