User:Egm6321.f10.team2.patterson/Homework 7

= Problem 8: Derivation of a Specialized Case of the Binomial Series =

Given
Given the following binomial series equation:


 * {| style="width:100%" border="0"

$$ \displaystyle \left( x +y \right)^r=\sum_{k=0}^{\infty}\dbinom{r}{k}x^{r-k}y^{k} $$ (eq 8.1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

where the binomial coefficient, "r choose k", is defined as follows:


 * {| style="width:100%" border="0"

$$ \displaystyle \dbinom{r}{k}=\frac{r(r-1)\cdots (r-k+1)}{k!} $$ (eq 8.2)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

and r is a real number, $$\displaystyle r \in \mathbb{R}$$

Derive the following equation:


 * {| style="width:100%" border="0"

$$ \displaystyle \left(1-x \right)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\alpha_{i}x^{i}
 * style="width:95%" |
 * style="width:95%" |

$$ (eq 8.3)
 * 
 * }

where


 * {| style="width:100%" border="0"

$$ \displaystyle \alpha_i = \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2i - 1)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot (2i)} $$ (eq 8.4)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Solution
In comparing the left hand side (LHS) of Eq 8.3 to the LHS of Eq 8.1, the following values for r, x, and y are obtained:


 * {| style="width:100%" border="0"

$$ \displaystyle r=-\frac{1}{2}, \qquad x=1, \qquad y=(-x) $$ (eq 8.5)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Plugging these values into the right hand side (RHS) of Eq. 8.1 yields the following:


 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\left( 1-x \right)^{-\frac{1}{2}}=\sum_{k=0}^{\infty}\dbinom{-\frac{1}{2}}{k}1^{\left(-\frac{1}{2}-k\right)}(-x)^{k}

$$ (eq 8.6)
 * 
 * }

Note that for every k

$$ \displaystyle 1^{\left(-\frac{1}{2}-k\right)}=1,\quad \forall k $$

Next, the binomial coefficient in Eq 8.6 is expanded by using Eq 8.2 and the values in Eq 8.5 to yield the following:


 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\left( 1-x \right)^{-\frac{1}{2}}=\sum_{k=0}^{\infty}\frac{-\frac{1}{2}(-\frac{1}{2}-1)(-\frac{1}{2}-2)\cdots (-\frac{1}{2}-k+1)}{k!} (-x)^{k}=\sum_{k=0}^{\infty}\frac{-\frac{1}{2}(-\frac{3}{2})(-\frac{5}{2})\cdots (-k+\frac{1}{2})}{k!} (-x)^{k}

$$ (eq 8.7)
 * 
 * }

Next, factor out -1/2 from each term in the numerator and change the summation index k to i as follows:


 * {| style="width:100%" border="0"

$$ \displaystyle \left( 1-x \right)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\frac{\left(-\frac{1}{2}\right)[1]\left(-\frac{1}{2}\right)[3]\left(-\frac{1}{2}\right)[5]\cdots \left(-\frac{1}{2}\right)[2i-1]}{i!} (-x)^{i}=\sum_{i=0}^{\infty}\left(\left(-\frac{1}{2}\right)(-x)\right)^i\frac{(1)(3)(5)\cdots (2i-1)}{i!} $$ (eq 8.8)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Eq 8.8 can be simplified as follows:


 * {| style="width:100%" border="0"

$$ \displaystyle \left( 1-x \right)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\frac{x^i}{2^i}\cdot \frac{(1)(3)(5)\cdots (2i-1)}{i!} $$ (eq 8.9)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Note that the terms in the denominator can be rewritten as:
 * {| style="width:100%" border="0"

$$ \displaystyle (2^i)(i!)=\underbrace{(2\cdot 2\cdot 2\cdot \ldots)}_{i\ times}(1\cdot 2\cdot 3\cdot \ldots \cdot i)=2\cdot 4\cdot 6\cdot \ldots \cdot 2i $$ (eq 8.10)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

The RHS of Eq 8.10 can replace the denominator of Eq 8.9 resulting in the equation found in Eq 8.3:


 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} \left( 1-x \right)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\cdot \underbrace{\frac{(1)(3)(5)\cdots (2i-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2i}}_{\alpha_i}x^i
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\end{align}$$ (eq 8.11)
 * 
 * }


 * The following link was consulted after this problem was solved in an attempt to provide the most clear and comprehensive presentation :

http://en.wikiversity.org/w/index.php?title=User:Egm6321.f09.Team1/HW5&oldid=510466#Problem_6:_Binomial_Series