Editing Appendix/Mathematics/Hypergeometric (section)

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

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

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<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>:

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<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>:

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<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,
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<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>
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But, <math>c_\pm = A_\pm</math>.  So we conclude, as did {{ VdBorght70 }}, that,

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<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,

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<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>
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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>