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I'm trying to understand a certain way of proving that

$$(1+x)^\alpha = \sum_{n=0}^{\infty}\binom{\alpha}{n}x^n$$

for $\alpha \in \mathbb{Q}$ and $|x|<1$.

Now, my notes tackle that proof by saying that for $\alpha, \beta \in \mathbb{R}$:

$$\sum_{n=0}^{\infty}\binom{\alpha}{n}z^n \cdot \sum_{n=0}^{\infty}\binom{\beta}{n}z^n = \sum_{n=0}^{\infty}c_nz^n$$

with $$c_n =\sum_{k=0}^{\infty}\binom{\alpha}{k}\binom{\beta}{n-k}=\binom{\alpha + \beta}{n} $$

and proceeding by proving that by induction.

Now, I can't even understand why proving the second notion would prove the initial theorem, so if anybody could explain or provide further material since I can't find anything about that particular way of proof, I'd be very thankful.

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    $\begingroup$ I suppose that, when you wrote $(1+x)^\alpha=\sum_{n=0}^{\infty}\binom{n}{k}x^n$, what you meant was $(1+x)^\alpha=\sum_{n=0}^{\infty}\binom\alpha nx^n$. $\endgroup$ Aug 28 '19 at 18:38
  • $\begingroup$ Note your original series should have $\binom{\alpha}{n}$ instead of $\binom{n}{k}$. The indices in your other series are also highly suspect. $\endgroup$
    – MPW
    Aug 28 '19 at 18:39
  • $\begingroup$ $f_a(x) = \sum_{n=0}^\infty {a \choose n}x^n$ from your formula you have $f_{ma}(x) f_a(x) = f_{(m+1)a}(x)$ and this is what you need to obtain $f_{u/v}(x)^v = f_u(x) = (1+x)^u$ (where $u,v\in \Bbb{Z}_{\ge 1}$ so that $f_u(x) = (1+x)^u$ is easy to show) $\endgroup$
    – reuns
    Aug 28 '19 at 18:43
  • $\begingroup$ Sorry, just realized that I had mistakes in my notes, changed them to fit the original script. $\endgroup$
    – psyph
    Aug 28 '19 at 19:03
  • $\begingroup$ This is one of lesser known proofs of binomial theorem. The proof with all details is available in my post: paramanands.blogspot.com/2016/07/… $\endgroup$ Sep 2 '19 at 15:31
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My guess is that what you wrote on the second paragraph is not meant to be a proof of the fact that $$(1+x)^\alpha=\sum_{n=0}^{\infty}\binom\alpha nx^n,\tag1$$but a proof of the fact that, if you define $(1+x)^\alpha$ by $(1)$, then$$(1+x)^\alpha(1+x)^\beta=(1+x)^{\alpha+\beta},\tag2$$since what it says there is precisely what is needed to prove that $(2)$ holds.

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  • $\begingroup$ I have just edited the question to match the original script, thanks for pointing out the mistakes. And you're right, (2) is there to prove a property that is later used to prove that $\alpha$ can be rational. $\endgroup$
    – psyph
    Aug 28 '19 at 19:17
  • $\begingroup$ I'm glad I could help. $\endgroup$ Aug 28 '19 at 19:23
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Here's an alternate proof, if you're interested.

One may show that $$p_n(\alpha)=\left[\frac{\partial}{\partial x}x^\alpha\right]_{x=1}=\prod_{r=1}^{n}(\alpha-r+1).$$ So then the Taylor series for $x^\alpha$ about $x=1$ is given by $$x^\alpha=\sum_{k\ge0}\frac{p_k(\alpha)}{k!}(x-1)^k.$$ We notice that $${n\choose k}=\prod_{i=1}^n\frac{n+1-i}{i}.$$ So we have $$(1+x)^\alpha=\sum_{k\ge0}{\alpha\choose k}x^k.$$

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