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Prove that: $(1+x)^{\alpha}=\sum_{n=0}^{+\infty}{\alpha \choose n} x^n$ for $x\in[0;1), \alpha \in\mathbb{R}$ based on Taylor's theorem with Lagrange remainder.

I don't feel such proofs. Isn't it kind of.. obvious? That when we expand $(1+x)^{\alpha}$ into series we get right side? I really want to understand this.

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    Shouldn't the limit be $x\in(0;1]$?2012-04-24

1 Answers 1

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You have that if

$y=(1+x)^\alpha$

then

y'(0)=\alpha y''(0)=\alpha(\alpha-1) $\cdots=\cdots$ $y^{(n)}(0)=\alpha(\alpha-1)\cdots(\alpha-n+1)$

$y^{(n)}(0)=\frac{\alpha!}{(\alpha-n)!}$

You can prove this last generalization by induction on $n$. The last equation is written for the sake of "order", but if you're not comfortable with it, you can use the $\Gamma$ function to denote the factorials.

Then

$\mathcal{T}_k(y,0)=\sum\limits_{n=0}^k \frac{\alpha!}{(\alpha-n)!}\frac{x^n}{n!}$

Then using the binomial coefficient notation

$\mathcal{T}_k(y,0)=\sum\limits_{n=0}^k {\alpha \choose n}x^n$

What you need now is that the series converges for $|x|<1$.

In my opinion you can argue as follows:

Fix $x$. Then the series

$\lim\limits_{k \to \infty} \mathcal{T}_k(y,0)=\sum\limits_{n=0}^\infty {\alpha \choose n}x^n$

will converge when

$ \lim\limits_{n \to \infty} \left|\frac{{\alpha \choose n+1}}{{\alpha \choose n}}\frac{x^{n+1}}{x^n}\right|<1$

$ \lim\limits_{n \to \infty} \left|\frac{{n+1}}{n-\alpha}x\right|<1$

$ \left|x\right|<1$

ADD: Here's a part of an answer to a similar question:

Assume

${\left( {1 + x} \right)^{\alpha}} = \sum\limits_{k = 0}^\infty {{a_k}{x^k}} $

Since if two series $\eqalign{ & \sum {{a_k}{{\left( {x - a} \right)}^k}} \cr & \sum {{b_k}{{\left( {x - a} \right)}^k}} \cr} $

sum up to the same function then

${a_k} = {b_k} = \frac{{{f^{\left( k \right)}}\left( a \right)}}{{k!}}$

for every $k \leq 0$, we can assume:

$a_k = \dfrac{f^{(k)}(0)}{k!}$

Putting $y = {\left( {1 + x} \right)^{\alpha}}$ we get

y'(0) = \alpha y''(0) = \alpha(\alpha-1) y'''(0) = \alpha(\alpha-1)(\alpha-2) $y^{{IV}}(0) = \alpha(\alpha-1)(\alpha-2)(\alpha-3)$

We can prove in general that

$y^{(k)}= \alpha(\alpha-1)\cdots(\alpha-k+1)$

or put in terms of factorials

$y^{(k)}(0)= \frac{\alpha!}{(\alpha-k)!}$

This makes

$a_k = \frac{\alpha!}{k!(\alpha-k)!}$

which is what we wanted.

${\left( {1 + x} \right)^\alpha } = \sum\limits_{k = 0}^\infty {\alpha \choose k} {{x^k}} $

You can prove this in a more rigorous manner by differential equations:

  1. Set $f(x) = \displaystyle \sum\limits_{k = 0}^\infty {\alpha \choose k} {{x^k}}$ and prove the radius of convergence is 1.
  2. Show that $f(x)$ is the solution to the ODE y' - \frac{\alpha }{{x + 1}}y = 0 with initial condition $f(0)=1$.
  3. By the theorem that the solution to the linear equation

y'+P(x)y=R(x)

with initial conditions $f(a) = b$ is unique, you can prove the assertion. (prove that $y = {\left( {1 + x} \right)^{\alpha}}$ also satifies the equation and you're done.)

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    The first part doesn't mention that one needs to show that $\mathcal T$ actually converges to $f$ for $n \to \infty$. It might converge but not to $f$. Therefore one needs to show that the limit of the remainder is zero.2018-01-27