0
$\begingroup$

I was reading about real analytic functions and I understood that to show a function is real analytic on $\mathbb{R}$, we need to show that there is a power series centered at $x_0$ which converges to an open interval around $x_0$ $\forall x_0 \in \mathbb{R}$
The example I am working on is the function $$f(x) = \frac{1}{1+x^2}$$ I can show that it converges for expansion around 0 by expanding it as a geometric series and this converges in (-1,1) when centered at x=0. How do I show it for other x $\in \mathbb{R}$ ?
I think I remember a fact that if power series centered around x=0 converges in (-1,1) then power series centered at any point in this interval would also converge(can someone give verify if this is indeed true?). But that still leaves us with the rest of the number line.

$\endgroup$
  • $\begingroup$ Have you learned about Taylor Series? $\endgroup$ – JavaMan Nov 21 '18 at 5:38
  • $\begingroup$ Yes, I have. Yea, maybe we can use taylor series to find the expansion. How do we show it converges? Also, is my statement correct about convergence about a different point in the interval? $\endgroup$ – Red Floyd Nov 21 '18 at 6:22
  • 1
    $\begingroup$ To show a Taylor Series converges, you have to show that $R_n(x) = \sum_{k=n+1}^{\infty} \frac{f^{(k)}(a) (x-a)^k}{k!} \to 0$. By Lagrange's Thm $R_n(x) = \frac{f^{(n+1)}(c) (x-a)^{n+1}}{(n+1)!}$ for some $c$ between $a$ and $x$. Now, the $n$th derivative is something like $f(x) \leq C x^{-n-2}$. Hence, $f^{(n)}(x) \to 0$ as $n \to \infty$. In particular $f^{(n+1)}(c)$ is bounded. Finally, $(x-a)^{n+1}$ is polynomial, and factorials go to zero faster than any polynomial (this follows since $x^n/n! \to 0$ as $n \to \infty$), which is demonstrated by convergence of Taylor series for $e^x$. $\endgroup$ – JavaMan Nov 22 '18 at 6:21
-2
$\begingroup$

Try just centering a geometric series around a different point. For example, take $$\frac{1}{1-(-(x-x_0)^2))}$$

If this does have power series that converges, what would it look like? This would mean that you can get a radius 1 of converge around any point that you desire.

$\endgroup$
  • 1
    $\begingroup$ Please check your solution again. $\endgroup$ – Jacky Chong Nov 21 '18 at 6:39
  • $\begingroup$ unless I'm mistaken, we should get the series $\sum_{i=1}^{\infty} (-1)^i (x-x_0)^{2i}$ which converges whenever $|x-x_0|<1$? $\endgroup$ – Alex Nov 21 '18 at 6:43
  • 2
    $\begingroup$ I think the OP wants to consider $(1+x^2)^{-1}$. $\endgroup$ – Jacky Chong Nov 21 '18 at 6:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.