Finding $[x^n]\frac{1}{\sqrt{1-x}}$

Question

I'm trying to find $[x^n]\frac{1}{\sqrt{1-x}}$ by using a simpler method than described here. By reading a paper I figured out that if we have a generating function as follows:

$$G_{a, b}(x) = \frac{1}{\sqrt{1−2ax+(a^2−4b)x^2}}$$

with $a$ and $b$ nonnegative integers, then:

$$[x^n]G_{a, b}(x) = \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k}\binom{2k}{k}a^{n-2k}b^k$$

If I make $a = \frac{1}{2}$ and $b = 1$ I can have what I'm looking for:

$$[x^n]\frac{1}{\sqrt{1-x}} = \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k}\binom{2k}{k}0.5^{n-2k}$$

The point is I need a closed-form and by using this strategy it seems to be very complicated to calculate, at least for me.

Does anyone could give me a simpler method?

Answer 1

Hint:

Note that $$\frac{1}{\sqrt{1-4y}}=\sum_{r=0}^{\infty}\binom{2r}{r}y^r$$

(Which is just the generating function of Central Binomial coefficients )

Thus substitute $\displaystyle y=\frac{x}{4}$ to get

$$\frac{1}{\sqrt{1-x}}=\sum_{r=0}^{\infty}\binom{2r}{r}\left(\frac{x}{4}\right)^r$$

Answer 2

If $f(x)=1/\sqrt{1-x}=(1-x)^{-1/2}$, then $f'(x)=(1/2)(1-x)^{-3/2}$ and so $$f(x)=2(1-x)f'(x).$$ Write $$f(x)=a_0+a_1x+a_2x^2+\cdots.$$ Then $$f'(x)=a_1+2a_2x+3a_3x^2+\cdots$$ and $$(1-x)f'(x)=a_1+(2a_2-a_1)x+(3a_3-2a_2)x^2+\cdots.$$ Comparing $x^n$ coefficients, $$a_n=2(n+1)a_{n+1}-2na_n$$ so that $$a_{n+1}=\frac{2n+1}{2(n+1)}a_n.$$ Since $a_0=1$, this recurrence gives $$a_n=\frac{1\times3\times\cdots\times(2n-1)}{2^nn!}=\frac{(2n)!}{2^nn!(2\times 4\times\cdots\times 2n)}=\frac{(2n)!}{4^nn!^2}=\frac1{4^n}\binom{2n}n.$$