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up vote 2 down vote favorite

I know that $f(x)=\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n$ converges for $|x|<1$

What I then have to show is that $(1+x)f'(x)=\alpha f(x)$ for $|x|<1$ and that any such $f$ is of the form $c(1+x)^\alpha$ for some constant c, and to use that fact to establish the binomial series.

I tried taking $$(1+x)f'(x)=(1+x)\left ( \sum_{n=0}^{\infty}\binom{\alpha}{n} x^n \right )'=(1+x)\left ( \sum_{n=0}^{\infty}\frac{\alpha !}{n!(\alpha-n)!} x^n \right )' = (1+x) \sum_{n=0}^{\infty}\frac{n \alpha !}{n!(\alpha-n)!} x^{n-1}=\alpha\sum_{n=0}^{\infty}\frac{ (\alpha-1) !}{(n-1)!((\alpha-1)-(n-1))!} (x^{n-1}+x^n)$$

But I'm not sure if my approach was right, and I'm not sure how to deal with the 2 $x$ terms if it was right.

I have no idea where to start for the second part.

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up vote 3 down vote accepted

$$\begin{align*}(1+x)f'(x) &= (1+x) \sum_{n=1}^\infty \binom a n n x^{n-1} = (1+x) \sum_{n=0}^\infty \binom a {n+1} (n+1) x^n \\ &= \sum_{n=0}^\infty \binom a {n+1} (n+1) x^n + \sum_{n=1}^\infty \binom a {n} n x^n \\ &= a + \sum_{n=1}^\infty \binom a {n+1} (n+1) x^n + \binom a {n} n x^n = a + \sum_{n=1}^\infty a_n x^n \end{align*}$$ where $$\begin{align*} a_n = \binom a {n+1} (n+1) + \binom a {n} n &= \frac{a!}{(n+1)!(a-n-1)!}(n+1) + \frac{a!}{n!(a-n)!}n \\ &= \frac{a!a}{n!(a-n)!} \end{align*}$$ Hence $$(1+x)f'(x) = a + \sum_{n=1}^\infty a_n x^n = a + \sum_{n=1}^\infty \frac{a!a}{n!(a-n)!} x^n = af(x) $$ The second part can be solved using logarithms, as RRL pointed out.

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How does the a on the left of the summation get cancelled out? – George 1 hour ago
    
@George What do you mean by cancelled out? Where? – Henry W. 1 hour ago
    
For $a + \sum_{n=1}^\infty \frac{a!a}{n!(a-n)!} x^n$, where does the a on the left of the summation go? Since f(x) is just the summation part. – George 1 hour ago
    
@George You can try putting $n = 0$ on the right. It show that $a$ can be merged into the series seamlessly. – Henry W. 1 hour ago
    
@HenryW. Looks good. +1 – RRL 1 hour ago
up vote 2 down vote

Hint:

$$(1+x)f'(x) = \alpha f(x) \implies (\log f(x))' = \alpha(1+x)^{-1}.$$

Now integrate both sides to show $f(x) = c(1 + x)^\alpha.$

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