# Thread: [SOLVED] Finding a power series

1. ## [SOLVED] Finding a power series

Hello

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 :

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

Thanks

2. By the Binomial Theorem: $\displaystyle \sum\limits_{n=0}^{\infty} {\binom{-\frac{1}{2}}{n}\cdot{(-4)^n}\cdot{x^n}} =\frac{1}{\sqrt{1-4x}}$

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

Remember that $\displaystyle \binom{u}{n}=\frac{u\cdot{...}\cdot{(u-n+1)}}{n!}$ $\displaystyle u \in \mathbb{R}/n \in \mathbb{N}$

3. Another way: $\displaystyle \int_{ - \pi }^\pi {\cos ^{2n} \left( x \right)dx} = \tfrac{{2\pi }} {{4^n }} \cdot \binom{2n}{n}$ for all naturals n

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

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

Then: $\displaystyle 2\pi \cdot \sum\limits_{n = 0}^\infty {\binom{2n}{n} \cdot z^n } =\int_{ - \pi }^\pi {\tfrac{{dx}} {{1 - 4 \cdot z \cdot \cos ^2 \left( x \right)}}}$

4. Another way. ($\displaystyle \Gamma$ is unit circle).

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

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

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

Now use residue theorem and this integral calculates to $\displaystyle \frac{1}{\sqrt{1-4x}}$.

5. Originally Posted by PaulRS

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

6. $\displaystyle \sum\limits_{n = 0}^\infty {\binom{2n}{n} \cdot x^n } = \sum\limits_{n = 0}^\infty {\tfrac{{\left( {\tfrac{{\left( {2n} \right)!}} {{n!}}} \right)}} {{n!}} \cdot x^n }$ let $\displaystyle a_n = \tfrac{{\left( {2n} \right)!}} {{n!}}$

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

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

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