Hi :) I have a question about Taylor series and convergence.

Question

My function is $f(x)= \sqrt{1+\sin x}$ around $x_0=0$ . Taylor series should look like (I guess so)

$f^{(2n)}(x)=\frac{(-1)^n}{2^{2n}}f(x) \quad\to\quad f^{(2n)}(0)=\frac{(-1)^n}{2^{2n}}$

$f^{(2n+1)}(x)=\frac{(-1)^n}{2^{2n+1}}f'(x)\quad\to\quad f^{(2n+1)}(0)=\frac{(-1)^n}{2^{2n+1}}$

I don't really understand how to show when it converges. I want to find out how will be x defined if exists a $\lim_{n\to\infty} T_n(x)$

Thank you.

Answer 1

For your derivatives you got $|f^{(k)}(0)|=2^{-k}$. This means that the term-wise absolute series of the Taylor expansion is $$ \sum_{k=0}^\infty \left|\frac{f^{(k)}(0)}{k!}x^k\right|=\sum_{k=0}^\infty \frac{|x/2|^k}{k!}=e^{|x|/2} $$ which is finite for all $x\in\Bbb R$. Thus the Taylor expansion of $f$ is everywhere absolutely convergent.

Answer 2

Note that your function $f(x)=\sqrt{1+\sin x}$ is the composit function of $\sqrt{1+x}$ and $\sin x$

The Taylor series for $\sqrt {1+x}$ converges for $|x|\le 1$ and we have $|\sin x|\le 1$ for all $x$

Thus your power series converges on $(-\infty,\infty)$