# Proving Newton's Binomial Theorem

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!}}$$x^n Where the sum goes from n=1 to infinity. The book suggests calling the right side f(x), proving the series converges, proving (1+x)f'(x) = \alpha f(x). I've done all this. All that is left is solving this differential equation, which I simply cannot figure out. Could anyone help with this last step? It would be appreciated! • Do you know the identity \frac{\mathrm d}{\mathrm dx}\log(f(x))=\frac{f^\prime(x)}{f(x)}? Apr 7, 2013 at 16:25 • That was the one I needed! Totally slipped my mind. Thank you! Apr 7, 2013 at 16:35 ## 3 Answers To settle this: there is the formula for logarithmic differentiation,$$\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}$$Integrate both sides, mind the boundary condition, and you should be able to recover your needed result. • Can showing (1+x)f'(x)=\alpha f(x) be proven a quicker way? I essentially used a long chain of algebra. Apr 7, 2013 at 17:45 • Can you edit your question to summarize this "long chain" you speak of? Apr 7, 2013 at 17:51 • I started by using the limit definition of f'(x) with lim of x+1 going to zero but this feels like an odd way to approach it. I actually think it may be completely wrong. I'm a non-math student trying some Rudin problems so you'll have to forgive me if I'm missing something elementary. I appreciate the help! Apr 7, 2013 at 18:01 • Ack. Are you already at the stage where you can justify when an infinite series can be differentiated term-by-term? (P.S. I'm a non-mathematician, too.) Apr 7, 2013 at 18:06 • I've used that if f(x) = \sum{cn x^n} (n=0 to inf) converges for x \in (-1,1), (which I have shown), then f'(x) = \sum{n cn x^{n-1}} (n=1 to inf), if that's what you're referring to. Apr 7, 2013 at 18:16 For compactness, use Knuth's notation for falling factorial powers:$$ \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$. • Do you have any suggestions for proving the original equality?$(1+x)f'(x) = \alpha f(x)\$. I think my original thinking danced around a clear proof Apr 7, 2013 at 17:48
• @user23658 Yes. Use Pascal's Rule and the identity I remarked in the ODE based solution.
– Pedro
Apr 7, 2013 at 17:54