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Prove that $1 \over (1-4x)^2$ generates ${2n \choose n}$, $n\in N$

I cant find theformula for that type of generating function, i have already tried expanding the combination and trying to expand the generating function. Any Ideas?

user1176409
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diegocuehdz
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2 Answers2

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First of all, the generating function is not $\frac{1}{(1-4x)^2}$ but is $\frac{1}{\sqrt{1-4x}}$.
Now, using the well known identity $\Gamma(1/2 - n) = \frac{(-4)^n n!}{(2n)!}\sqrt{\pi}$, which can be easily proved via induction by noting $\Gamma(1/2) = \sqrt{\pi}$, we find that

$$\begin{align} \frac{1}{\sqrt{1-4x}} = (1-4x)^{-1/2} &= \sum_{k=0}^\infty{-1/2 \choose k}(-1)^k (-4)^k x^k \\&=\sum_{k=0}^\infty \frac{\Gamma(1/2)}{\Gamma(1/2 - k) k!}(-4)^k x^k \\&=\sum_{k=0}^\infty \frac{\sqrt{\pi}(2k)! }{\sqrt{\pi}(-4)^k (k!)^2}(-4)^k x^k \\&=\sum_{k=0}^\infty \frac{(2k)!}{(k!)^2} x^k \\&=\sum_{k=0}^\infty \frac{(2k)!}{(2k-k)!\,k!}x^k \\&=\sum_{k=0}^\infty {2k \choose k}x^k \end{align}$$

Brevan Ellefsen
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  • If i start at what i put is the same or equivelent as what you put? How do i get from what they asked me to how you solved it? – diegocuehdz Dec 12 '17 at 07:18
  • No. The Generating Function for $\frac{1}{(1-4x)^2}$ is not the same as the one for $\frac{1}{(1-4x)^{1/2}}$. The problem you were given is wrong – Brevan Ellefsen Dec 12 '17 at 07:37
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Well, clearly $(1-4x)^{-2}$ is not the generating function, and the reason is this: $$ (1+x)^{m} = \sum_{n=0}^\infty \binom mn x^n $$ Where $\binom mn$ now stands for the generalized binomial coefficient, with $m$ being real and $n$ being a positive integer. (At least in our case)

With $m=-2$, what would this look like? $$ (1+4x)^{-2} = \sum_{n=0}^\infty \binom{-2}{n} (4x)^{n} $$

So the coefficients in this case are $\binom{-2}{n} 4^n$, which expands to $$ \frac{-2 \times -3 \times ... \times (-2-n+1)}{n!} \times 4^n $$

Simplify this for yourself, it is nowhere close to the coefficients you want.

Therefore, your problem as stated is incorrect.

However with $m= -0.5$ we'd get $(1-4x)^{0.5} = \sum_{n=0}^\infty \binom{-0.5}{n}(4x)^{n}$. So, the coefficients come out as $\binom{-0.5}{n} 4^n$. But then: $$ \frac{-0.5 \times -1.5 \times ... \times (-0.5-n+1)}{n!} \times 4^n = \frac{-1 \times -3 \times ... \times (1-2n)}{n!} \times 2^n \\ = (-1)^{2n}\frac{(2n-1) \times (2n-3) \times ... \times 1}{n!} \times 2^n = \frac{1 \times (2 \times 1) \times 3 \times (2 \times 2) ... \times (2n-1) \times (2 \times n)}{n!n!} = \binom{2n}{n} $$

and hence the whole thing actually works out.

Sarvesh Ravichandran Iyer
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  • But the problem is wrong or you just need to work out from what they gave me to the -1/2?? – diegocuehdz Dec 12 '17 at 07:19
  • It must have been a printing mistake, the problem is wrong. It is not your fault, I presume you would have got it had it been printed correctly. I knew it because I had seen it before. No, you must be given $\frac {-1}2$ from the starting, otherwise this problem is wrong. – Sarvesh Ravichandran Iyer Dec 12 '17 at 07:21
  • Thanks a lot for the help and the encouragement. I need to tell someone that this is wrong so that it can be changed in the exercises that we have to hand in. It was easy once that you use -1/2 instead of -2, if not this is imposible. – diegocuehdz Dec 12 '17 at 07:31
  • You are welcome! Also, If you like, you may upvote and accept any answer(s) of your choice. This will help close the question, since it is answered. Also, +1 for a good question. – Sarvesh Ravichandran Iyer Dec 12 '17 at 08:05