# [SOLVED] Finding a power series

• May 28th 2008, 03:59 AM
Moo
[SOLVED] Finding a power series
Hello :D

This can look like a stupid question, but I just can't see the trick (or maybe because I'm up since 5:30am, I'm too tired to think correctly...)

Prove :

$\frac{1}{\sqrt{1-4z}}=\sum_{n \ge 0} {2n \choose n} z^n$

Thanks :)
• May 28th 2008, 04:36 AM
PaulRS
By the Binomial Theorem: $
\sum\limits_{n=0}^{\infty} {\binom{-\frac{1}{2}}{n}\cdot{(-4)^n}\cdot{x^n}} =\frac{1}{\sqrt{1-4x}}
$

Now prove that: $\binom{-\frac{1}{2}}{n}\cdot{(-4)^n}=\binom{2n}{n}$

Remember that $\binom{u}{n}=\frac{u\cdot{...}\cdot{(u-n+1)}}{n!}$ $
u \in \mathbb{R}/n \in \mathbb{N}
$
• May 28th 2008, 04:48 AM
PaulRS
Another way: $
\int_{ - \pi }^\pi {\cos ^{2n} \left( x \right)dx} = \tfrac{{2\pi }}
{{4^n }} \cdot \binom{2n}{n}
$
for all naturals n

Thus: $
\sum\limits_{n = 0}^\infty {\left( {\int_{ - \pi }^\pi {\cos ^{2n} \left( x \right)dx} } \right) \cdot \left( {4z} \right)^n } = 2\pi \cdot \sum\limits_{n = 0}^\infty {\binom{2n}{n} \cdot z^n }
$

$
\sum\limits_{n = 0}^\infty {\left( {\int_{ - \pi }^\pi {\cos ^{2n} \left( x \right)dx} } \right) \cdot 4^n \cdot z^n } = \int_{ - \pi }^\pi {\left[ {\sum\limits_{n = 0}^\infty {4^n \cdot z^n \cdot \cos ^{2n} \left( x \right)} } \right]dx}
$

Then: $2\pi \cdot \sum\limits_{n = 0}^\infty {\binom{2n}{n} \cdot z^n } =\int_{ - \pi }^\pi {\tfrac{{dx}}
{{1 - 4 \cdot z \cdot \cos ^2 \left( x \right)}}}
$
• May 28th 2008, 08:51 AM
ThePerfectHacker
Another way. ( $\Gamma$ is unit circle).

${{2n}\choose n} = \frac{1}{2\pi i}\oint_{\Gamma} \frac{(1+z)^{2n}}{z^{n+1}}dz$

Thus, $\sum_{n=0}^{\infty} {{2n}\choose n} x^n = \frac{1}{2\pi i}\oint_{\Gamma} \sum_{n=0}^{\infty} \frac{(1+z)^{2n}}{z^{n+1}} x^n dz$

Geometric series, $\frac{1}{2\pi i}\oint_{\Gamma} \frac{z}{z+x(1+z)^2}\cdot \frac{1}{z} dz = \frac{1}{2\pi i}\oint_{\Gamma} \frac{dz}{z+x(1+z)^2}$.

Now use residue theorem and this integral calculates to $\frac{1}{\sqrt{1-4x}}$.
• May 28th 2008, 09:22 AM
Moo
Quote:

Originally Posted by PaulRS

Now prove that: $\binom{-\frac{1}{2}}{n}\cdot{(-4)^n}=\binom{2n}{n}$

Hmm, that's what I was struggling with.. I think I've understood ^^ It's all a matter of denominator ~

Pooh

The other methods don't correspond to the general method expected by the paper, but it's quite interesting :)

Thanks guys
• June 20th 2008, 08:09 AM
PaulRS
$
\sum\limits_{n = 0}^\infty {\binom{2n}{n} \cdot x^n } = \sum\limits_{n = 0}^\infty {\tfrac{{\left( {\tfrac{{\left( {2n} \right)!}}
{{n!}}} \right)}}
{{n!}} \cdot x^n }
$
let $
a_n = \tfrac{{\left( {2n} \right)!}}
{{n!}}
$

Note that: $
a_{n + 1} = 2 \cdot \left( {2n + 1} \right) \cdot a_n
$

Thus: $
\left( {\sum\limits_{n = 0}^\infty {\tfrac{{a_n }}
{{n!}} \cdot x^n } } \right)^\prime = \sum\limits_{n = 0}^\infty {\tfrac{{2 \cdot \left( {2n + 1} \right) \cdot a_n }}
{{n!}} \cdot x^n }
$
let $
\sum\limits_{n = 0}^\infty {\tfrac{{a_n }}
{{n!}} \cdot x^n } = y
$
we have: $
y^\prime = 2 \cdot \sum\limits_{n = 0}^\infty {\tfrac{{\left( {2n + 1} \right) \cdot a_n }}
{{n!}} \cdot x^n } = 4 \cdot x \cdot y^\prime + 2y
$

Thus our solution satisfies the differential equation: $
\left( {4 \cdot x - 1} \right) \cdot y^\prime + 2y = 0
$
which is quite easy to solve (we define $y(0)=\lim_{x\to 0}y(x)=1$)