Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
    
How does the a on the left of the summation get cancelled out? – George Mar 28 at 3:25
    
@George What do you mean by cancelled out? Where? – Henry W. Mar 28 at 3:26
    
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 Mar 28 at 3:31
    
@George You can try putting $n = 0$ on the right. It show that $a$ can be merged into the series seamlessly. – Henry W. Mar 28 at 3:32
    
@HenryW. Looks good. +1 – RRL Mar 28 at 3:34

Hint:

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

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

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.