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Let $f(x)=\frac{1}{(1+x)^{1/2}}$. Supposed to find the Taylor series $Tf$ of $f$ around $0$, and show that it converges to $f$ on $[-1,1)$, (although I suspect there's a misprint in the book, and that this is supposed to be $(-1,1]$, since $f$ is not defined in $x=-1$).

If $f(x)=T_nf(x)+R_nf(x)$, where $T_nf(x)$ denotes the $n$th Taylor polynomial, then there's a result in my book which says, in short, that if the $(n+1)$-th derived $f^{(n+1)}$ of $f$ is such that there's a number $M$ with $|f^{(n+1)}(t)|\leq M$ for all $t\in(-1,1]$, then $|R_n f(x)|\leq \frac{M}{(n+1)!}|x|^{n+1}.$ My plan was to use this to show that $|R_n f(x)|\rightarrow 0$ as $n\rightarrow \infty$ for $x\in(-1,1]$. Can't seem to find any such $M$, however, and actually such an $M$ seems unlikely to me, seeing as, if I'm not mistaking, $|f^{(n+1)}(t)|=\frac{3*5*\cdots*(2n+1)}{2^{n+1}|1-t|^{\frac{2n+3}{2}}}.$ For doesn't this grow big when $t$ is close enough to $-1$?

I may however very well be on an entirely wrong track here...with the $R_n f$ argument and stuff.

Very thankful for any help.

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I think there is a misprint in the formula of $|f^{(n+1)}(t)|$. – Siminore May 5 '12 at 12:23

The proof uses a different strategy for the cases $0\le x<1$ and $-1<x<0$. Let $f(x)=(1+x)^\alpha$, $\alpha\in\mathbb{R}$, and let $$ \binom{\alpha}{n}=\frac{\alpha(\alpha-1)\dots(\alpha-n+1)}{n!}. $$ Then $$ \frac{f^{(n)}(x)}{n!}=\binom{\alpha}{n}(1+x)^{\alpha-n}. $$ The Taylor series for $f$ is $$ \sum_{n=0}^\infty\binom{\alpha}{n}x^n. $$ Using the ratio test we find that the radius of convergence of the series is $1$. This implies that for any $r\in(0,1)$ $$ \lim_{n\to\infty}\left|\binom{\alpha}{n}\right|r^n=0.\tag{1} $$ The remainder is $$ R_n(x)=f(x)-\sum_{k=0}^n\binom{\alpha}{k}x^k= \binom{\alpha}{n+1}(1+\theta)^{\alpha-n-1}x^{n+1}, $$ where $\theta$ is a real number between $0$ and $x$. We want to show that $\lim_{n\to\infty}R_n(x)=0$ for $x\in(-1,1]$

First case: $0\le x<1$. Then $$ |(1+\theta)^{\alpha-n-1}|\le\max(1,2^\alpha)\qquad\forall n\ge0. $$ The desired result follows from (1).

Second case: $-1/2<x<0$. Then $$ |(1+\theta)^{\alpha-n-1}|\le C_\alpha(1-|x|)^{-n-1}\qquad\forall n\ge0 $$ and $$ |(1+\theta)^{\alpha-n-1}x^{n+1}|\le C_\alpha\Bigl(\frac{|x|}{1-|x|}\Bigr)^n,\quad\frac{|x|}{1-|x|}<1. $$ The result follows again from (1)

Third case: $-1<x<1/2$. This seems to be more complicated. The integral form of the remainder might be useful. I'll keep looking into it.

Fourth case: $x=1$. This follows from Abel's Theorem.

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The binomial theorem states:$$ (1+x)^\alpha = \sum_{n=0}^\infty \begin{pmatrix}\alpha \\ n\end{pmatrix} x^n $$ Meaning that the Taylor coefficient $c_n$ in your case is: $$\begin{pmatrix}-1/2 \\ n\end{pmatrix}$$ So the ratio test gives the radius of convergence as: $$r= \lim_{n\rightarrow\infty}\left|\frac{c_{n+1}}{c_n}\right|=1$$

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This shows that the series converges, but it does not follow that it converges to the original function. – user29743 May 5 '12 at 13:37
Power series expansions are unique. – Siminore May 5 '12 at 14:33
I don't follow why this implies the result. The Taylor series for the bump function has radius of convergence infinity but does not converge to the bump function anywhere. How does the ratio test show that this power series converges to OP's function rather than, say, OP's function plus the bump function? – user29743 May 5 '12 at 14:56
The function $x \mapsto (1+x)^\alpha$ can be expanded in power series for $-1<x<1$. Or, equivalently, the power series $\sum_n \begin{pmatrix}\alpha \\ n \end{pmatrix} x^n$ converges to $(1+x)^\alpha$ for every $x \in (-1,1)$. – Siminore May 5 '12 at 15:44
I guess that the OP's approach leads nowhere. It seems that the most convenient approach is: (i) prove that the power series converges, and (ii) prove that the limit is indeed $f$. For (ii), see . Essentially, you can prove that the power series solves a Cauchy problem, and by uniqueness of the solution, the power series must coincide with $f$. – Siminore May 5 '12 at 15:58

This is a binomial expansion. See this reference. Every mathematician will probably solve this problem by mean of complex analytic functions. Actually, you can choose a continuous branch of $z \in \mathbb{C} \mapsto (1+z)^\alpha$, which turns out to be analytic for $|z|<1$. This is the argument by Ahlfors, Complex Analysis, page 180.

If you prefer a real-analysis approach, you can

  1. Prove that $\sum_n \begin{pmatrix} \alpha \\ n \end{pmatrix} x^n$ converges when $-1<x<1$;
  2. differentiate term by term this series and discover that it solves a differential equation with an initial condition. Then, by uniqueness of solution, the series is your $f$.

Important edit: I have found a complete and yet elementary proof. Please refer to this excerpt from Hijab, Introduction to calculus and analysis.

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Well, thanks a lot. This is a first year calculus exercise though, and neither complex analysis nor binomial coefficients involving non-natural numbers are part of the curriculum...far from it:) – Bart Patzer May 5 '12 at 17:28
I do not really know how to prove directly that your function can be expanded in power series. The rough estimate of the remainder seems to be hard. – Siminore May 6 '12 at 8:11
Please see my last edit for a reference with complete proof. – Siminore May 6 '12 at 8:50

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