So, I've done most of the problem to this point, but just cannot figure out the last piece. I may just be missing the math skills needed to complete the proof (differential equations).

Problem (from Rudin): If $\alpha$ is real and $-1 < x < 1$, prove Newton's binomial theorem;

$(1+x)^\alpha$ = $1 + \sum{\frac{\alpha (\alpha-1) \dots(\alpha - n + 1)}{n!}}$$x^n$

Where the sum goes from $n=1$ to infinity. The book suggests calling the right side $f(x)$, proving the series converges, proving $(1+x)f'(x) = \alpha f(x)$. I've done all this. All that is left is solving this differential equation, which I simply cannot figure out. Could anyone help with this last step? It would be appreciated!

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    $\begingroup$ Do you know the identity $\frac{\mathrm d}{\mathrm dx}\log(f(x))=\frac{f^\prime(x)}{f(x)}$? $\endgroup$ Apr 7, 2013 at 16:25
  • $\begingroup$ That was the one I needed! Totally slipped my mind. Thank you! $\endgroup$
    – user23658
    Apr 7, 2013 at 16:35

3 Answers 3


To settle this: there is the formula for logarithmic differentiation,

$$\frac{\mathrm d}{\mathrm dx}\log(f(x))=\frac{f^\prime(x)}{f(x)}$$

which means

$$\frac{\mathrm d}{\mathrm dx}\log(f(x))=\frac{\alpha}{1+x}$$

Integrate both sides, mind the boundary condition, and you should be able to recover your needed result.

  • $\begingroup$ Can showing $(1+x)f'(x)=\alpha f(x)$ be proven a quicker way? I essentially used a long chain of algebra. $\endgroup$
    – user23658
    Apr 7, 2013 at 17:45
  • $\begingroup$ Can you edit your question to summarize this "long chain" you speak of? $\endgroup$ Apr 7, 2013 at 17:51
  • $\begingroup$ I started by using the limit definition of f'(x) with lim of x+1 going to zero but this feels like an odd way to approach it. I actually think it may be completely wrong. I'm a non-math student trying some Rudin problems so you'll have to forgive me if I'm missing something elementary. I appreciate the help! $\endgroup$
    – user23658
    Apr 7, 2013 at 18:01
  • $\begingroup$ Ack. Are you already at the stage where you can justify when an infinite series can be differentiated term-by-term? (P.S. I'm a non-mathematician, too.) $\endgroup$ Apr 7, 2013 at 18:06
  • $\begingroup$ I've used that if f(x) = $\sum{cn x^n}$ (n=0 to inf) converges for x $\in (-1,1)$, (which I have shown), then $f'(x) = \sum{n cn x^{n-1}}$ (n=1 to inf), if that's what you're referring to. $\endgroup$
    – user23658
    Apr 7, 2013 at 18:16

For compactness, use Knuth's notation for falling factorial powers: $$ \alpha^{\underline{n}} = \alpha \cdot (\alpha - 1) \cdot \ldots \cdot (\alpha - n + 1) = \prod_{0 \le k \le n - 1} (\alpha - k) $$ Now use Maclaurin's formula: $$ \begin{align*} f(z) &= (1 + z)^\alpha \\ f^{(n)}(x) &= \alpha^{\underline{n}} (1 + z)^{\alpha - n} \end{align*} $$ This gives directly: $$ f(z) = \sum_{n \ge 0} \frac{f^{(n)}(0)}{n!} z^n = \sum_{n \ge 0} \frac{\alpha^{\underline{n}}}{n!} z^n $$ We can define: $$ \binom{\alpha}{n} = \frac{\alpha^{\underline{n}}}{n!} $$ to get the familiar: $$ (1 + z)^\alpha = \sum_{n \ge 0} \binom{\alpha}{n} z^n $$


Note that the differential equation $$f'-\frac{\alpha}{1+x}f=0$$ with initial conditions $f(0)=1$ has a unique solution.

This can be seen by taking $f,f_0$ two solutions and looking at $g=\frac{f}{f_0}$. You'll get that $$g'=\frac{ff_0^\prime-f'f_0}{f_0^2}$$

But then $$g'=\frac{1}{f_0^2} \left(f \frac{\alpha}{1+x}f_0-\frac{\alpha}{1+x}f f_0\right)=0$$

It follows $g$ is contant, and since $f(0)=f_0(0)=1$, $g=1$ so $f_0=f$. Thus, all you need to show is that $$f=\sum_{n=0}^\infty {\alpha\choose n}x^n$$ converges over $|x|<1$,that so does its derivative and that $(1+x)f'=\alpha f$. This last equality follows from $$n{\alpha\choose n}=\alpha{{\alpha-1}\choose n-1}$$

A much more tedious alternative is possible: you can show that the remainder goes to $0$ for $|x|<1$. To this end, first consdier $0\leq x <1$. Then $$R_{m,0}(x)=\frac{f^{m+1}(t)}{(m+1)!}x^{m+1}$$ for some $t\in(0,1)$. But if $f=(1+x)^{\alpha}$ then $$f^{(m+1)}=\prod_{k=0}^m (\alpha-k)(1+x)^{\alpha-m-1}$$

so that $$R_{m,0}(x)={\alpha\choose m+1}(1+t)^{\alpha-m-1}x^{m+1}$$

But $$0<(1+t)^{\alpha-m-1}\leq 1$$ for $\alpha<m+1$ It follows $\lim_{m\to\infty} R_{m,0}=0$ for any $x\in[0,1)$. Now suppose $-1<x<0$. By Cauchy's remainder formula. $$R_{m,0}=\frac{f^{(m+1)}(t)}{m!}(x-t)^m x$$ for some $x<t\leq 0$. Yet again $$f^{(m+1)}=\prod_{k=0}^m (\alpha-k)(1+x)^{\alpha-m-1}$$ so that $$R_{m,0}(x)=(m+1){\alpha\choose m+1}\left(\frac{x-t}{1+t}\right)^m(1+t)^{\alpha-1}x$$

But $$\left| {\frac{{x - t}}{{1 + t}}} \right| = \left| x \right|\left| {\frac{{1 - t/x}}{{1 + t}}} \right|$$ and since $-1<x<t< 0$, $0\leq t/x$ and $0\leq t+1< 1$ so $$\frac{1}{1+t}\geq 1$$ and $$\left| x \right|\left| {\frac{{1 - t/x}}{{1 + t}}} \right| \leqslant \left| x \right|$$

On the other hand, $$|x(1+t)^{\alpha-1}|\leq M_{\alpha,x}$$ where $M_{\alpha,x}=\max(1,(1+x)^{\alpha-1})$. Indeed, if $\alpha \geq 1$ then $0<1+x<1+t<1$ so $0<(1+x)^{\alpha-1}<(1+t)^{\alpha-1}<1$ and if $\alpha<1$ then $1< (1+t)^{\alpha-1}<(1+x)^{\alpha-1}$

It follows that $$\displaylines{ \left| {{R_{m,0}}\left( x \right)} \right| = \left( {m + 1} \right){\alpha\choose m+1}{\left| {\frac{{x - t}}{{1 + t}}} \right|^m}\left| {{{\left( {1 + t} \right)}^{\alpha - 1}}x} \right| \cr \leqslant \left( {m + 1} \right){\alpha\choose m+1}{\left| x \right|^{m + 1}}{M_{\alpha ,x}} \cr = \alpha {\alpha-1\choose m}{\left| x \right|^m}{M_{\alpha ,x}} \cr} $$

and this last quantity goes to zero when $m\to\infty$.

All in all $${\left( {1 + x} \right)^\alpha } = \sum\limits_{n = 0}^\infty{\alpha\choose m} {{x^n}} $$ over $|x|<1$.

  • $\begingroup$ Do you have any suggestions for proving the original equality? $(1+x)f'(x) = \alpha f(x)$. I think my original thinking danced around a clear proof $\endgroup$
    – user23658
    Apr 7, 2013 at 17:48
  • $\begingroup$ @user23658 Yes. Use Pascal's Rule and the identity I remarked in the ODE based solution. $\endgroup$
    – Pedro
    Apr 7, 2013 at 17:54

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