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The Full Question:

Show that $(1-4x)^{-\frac{1}{2}}$ generates the sequence $\binom{2n}{n}$, $n\in \mathbb N$

My Research

How to show that $1 \over \sqrt{1 - 4x} $ generate $\sum_{n=0}^\infty \binom{2n}{n}x^n $

Show $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$

Both these questions seem similar, but I found they used a very different approach (more calculus based) to find their answers whereas my question requires me to use more algebra and definitions to get the answer.

My Work

1) Let $y = -4x$

2) We know the $r^{th}$ term of$(1-y)^{-1/2}$ will have the coefficient of $\binom{-1/2}{r}(-4^r)$

3) $\binom{-1/2}{r}(-4^r) = \frac{(-1/2)(-1/2-1)(-1/2-2)\cdots(-1/2-r+1)}{r!}(-4)^r$

4) If we factor out $(-1/2)$ from the whole thing, we have: $\frac{(1)(1+2)(1+4)\cdots(1+2r-2)}{r!}(2^r)=\frac{(2r-1)(2r-3)\cdots(5)(3)(1)}{r!}2^r$

5) If we multiply top and bottom by $r!$ we get:$\frac{(2r-1)(2r-3)\cdots(5)(3)(1)2^rr!}{r!r!}$

6) All that remains is to show the numerator to equal $(2r)!$ and I'm golden, however I don't really know how to do this.

My Question

Can anyone help me with step 6? I feel I'm very close but haven't quite been able to see it.

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Dunka
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1 Answers1

1

Note: Avoiding a small calculation error in OPs work in 4.) and considering a small trick in 6.) will close the gap. Here's a calculation:

\begin{align*} \binom{-\frac{1}{2}}{r} &=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)\cdot\ldots\cdot \left(-\frac{1}{2}-(r-1)\right)}{r!}\\ &=\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdot\ldots\cdot \left(-\frac{2r-1}{2}\right)}{r!}\\ &=\left(-\frac{1}{2}\right)^r\cdot\frac{(2r-1)(2r-3)\cdot\ldots\cdot5\cdot3\cdot1}{r!}\tag{1}\\ &=\left(-\frac{1}{2}\right)^r\cdot\frac{(2r)(2r-1)(2r-2)(2r-3)\cdot\ldots\cdot5\cdot4\cdot3\cdot2\cdot1} {r!(2r)(2r-2)(2r-4)\cdot\ldots\cdot4\cdot2}\tag{2}\\ &=\left(-\frac{1}{2}\right)^r\cdot\frac{(2r)!} {r!2^r\cdot r(r-1)\cdot\ldots\cdot2\cdot1}\\ &=\left(-\frac{1}{4}\right)^r\frac{(2r)!}{r!r!}\\ &=\left(-\frac{1}{4}\right)^r\binom{2r}{r}\\ \end{align*}

Comment:

  • In (1) we factor out $\left(-\frac{1}{2}\right)^r$. Please note that $2^r$ is part of the denominator (check corr. part (4) in your calculation)

  • In (2) we fill in even factors to complete the numerator to $(2r)!$ and we also write these factors in the denominator as compensation.

Clever notation: In (1) it's convenient to use the compact notation of double factorials.We observe $$(2r)!=(2r)!!(2r-1)!!=2^rr!(2r-1)!!$$ and using this notation in (1):

\begin{align*} \binom{-\frac{1}{2}}{r} &=\ldots\\ &=\left(-\frac{1}{2}\right)^r\frac{(2r-1)!!}{r!}\\ &=\left(-\frac{1}{2}\right)^r\frac{(2r)!}{r!(2r)!!}\\ &=\left(-\frac{1}{2}\right)^r\frac{(2r)!}{r!2^rr!}\\ &=\left(-\frac{1}{4}\right)^r\binom{2r}{r}\\ \end{align*}

Markus Scheuer
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