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In class, our professor was very adamant that the following simplification is intuitive: \begin{align*} \sum^{\infty}_{n=0}x^n\binom{2n}{n}=\frac{1}{\sqrt{1-4x}} \end{align*}

I can get from the RHS to the LHS comfortably by using the identity that \begin{align*} \binom{-1/2}{n} &=(-1)^n 2^{-2n}\binom{2n}{n}. \end{align*}

However, I have no idea how to get from the LHS to the RHS without using any special identities (including the one above) or Cauchy's integral formula (I haven't learned about it yet, but someone suggested that this was a valid approach as well). Our professor insisted that we should be able to get the RHS result solely from massaging Taylor's theorem. I really want to make sure I can get these results on my own so I can try exploring other generating functions on my own, but it definitely doesn't feel intuitive in the slightest.

Anne Bauval
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Hot Tamale
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    You can use generalized binomial formula as done here: https://math.stackexchange.com/a/1064267/305862 – Jean Marie Sep 19 '22 at 06:47
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    I'm a pretty experienced mathematician and it wouldn't seem intuitive to me either. – Greg Martin Sep 19 '22 at 07:05
  • Just follow your professor's hint. Compute the successive derivatives of $f(x)=(1-4x)^{-1/2}.$ – Anne Bauval Sep 19 '22 at 08:20
  • You have to know the usual power series such as $(1+u)^\alpha$ otherwise you won't be able to notice them. If you expand the factorial you will see that the series is close to the power series of $(1+x)^{-\frac{1}{2}}$ then just do the change of variable $u=-4x$ to get the exact expansion. – Lelouch Sep 19 '22 at 09:25
  • @Lelouch I think the OP does not know the power series of $(1+x)^{-\frac12}$, otherwise he wouldn't have asked this question. – Anne Bauval Sep 19 '22 at 11:35
  • Well I dont see any other way to have the intuition then – Lelouch Sep 19 '22 at 11:36
  • Note that $\displaystyle \frac{1}{2\pi}\int_0^{2\pi}\sin^{2n}(x)dx=\frac{1}{4^n}\binom{2n}{n}$, because $\displaystyle\frac{1}{2\pi}\int_0^{2\pi}\sin^{2n}(x)dx=\frac{(-1)^n}{2\pi,4^n}\int_0^{2\pi}\big(e^{it}-e^{-it}\big)^{2n}dt$ and $\int_0^{2\pi}e^{ikt}dt=2\pi$, if $,k=0$; otherwise zero. Therefore, our sum is $\displaystyle S(x)=\frac{1}{2\pi}\sum_{n=0}^\infty(4x)^n\int_0^{2\pi}\sin^{2n}(t)dt$. Making summation first, $\displaystyle S(x)=\frac{1}{2\pi}\int_0^{2\pi}\frac{dt}{1-4x\sin^2t}\stackrel{s=\tan t}{=},\frac{2}{\pi}\int_0^{\infty}\frac{ds}{1+(1-4x)s^2}=\frac{1}{\sqrt{1-4x}}$ – Svyatoslav Sep 19 '22 at 17:25

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Just follow your professor's hint: let $f(x)=(1-4x)^{-1/2},$ compute the first derivatives, and then guess and prove (by induction) that $$f^{(n)}(x)=\frac{(2n)!}{n!}(1-4x)^{-\frac{2n+1}2}.$$ Hence $$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty\binom{2n}nx^n$$ as a formal power series, but also as an ordinary power series, with radius of convergence $\frac14.$

Another approach is to start from the LHS: let $a_n=\binom{2n}n,$ then $a_{n+1}=\left(4-\frac2{n+1}\right)a_n.$ Multiplying both sides by $x^{n+1}$ and summing up, you find that the series $f(x):=\sum a_nx^n$ satisfies $$f(x)-1=4xf(x)-2\int_0^xf(t)dt.$$ Solving the corresponding differential equation $(1-4x)f'(x)=2f(x),$ you discover the claimed RHS. For more details, see T. Koshy, Catalan Numbers with Applications, p. 26-28

Anne Bauval
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    I think the differential equation approach is pretty interesting since the function appears without knowing anything about usual power series. – Lelouch Sep 19 '22 at 12:18
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    Yup, exactly what professor intended with their hint. The solution with the least amount of knowledge necessary to fully understand it and come up with it immediately after the hint. – Will Sherwood Sep 19 '22 at 12:30
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If you write down a few terms: \begin{align*} f(x)=\sum^{\infty}_{n=0}x^n\binom{2n}{n}=1+2x+6x^2+20x^3+70x^4+... \end{align*}

Then, try to play around with it, like square it. $$ [f(x)]^2=1+(2+2)x+(6+4+6)x^2+(20+12+12+20)x^3+(70+40+36+40+70)x^4+...\\ =1+4x+16x^2+64x^3+256x^4+... $$

Luckily you find a G.P. Then $$ [f(x)]^2=\frac{1}{1-4x} $$ But in general, you need to know the series form of many functions in order to guess the answer form. For binomial coefficient type coefficient, maybe square is a good try.

Abel Wong
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$$ {1\over 1-4x}=1+4x+16x^2+64x^3+... $$

$$ \left(1+2x+6x^2+...{2n\choose n}x^2+...\right)^2 =\sum_{n=0}^\infty\left(\sum_{i=0}^n{2i\choose i}{2(n-i)\choose n-i}\right)x^n\\ =\sum_{n=0}^\infty 4^nx^n $$ The identity $$ \sum_{i=0}^n{2i\choose i}{2(n-i)\choose n-i}=\left(-{4}\right)^n\sum_{i=0}^n{-1/2\choose i}{-1/2\choose n-i}=\\(-4)^n{-1\choose n}=4^n $$ follows from Vandemonde's identity.

(Credit: I found this compact proof here. There are also many, many other proofs on math.stackexchange.net.)

Suzu Hirose
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