1
$\begingroup$

I was playing with the error function:

$$\text{erf}(x)=\int_{0}^x e^{-t^2}dt$$

In that process I expanded the integrand $e^{-t^2}$ and integrated term by term and found out that you can express the $\text{erf}(x)$ function in terms of an infinite series:

$$\text{erf}(x)=\frac{2}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{n!(2n+1)}$$

What really surprise me is the fact that this limit exists!(It seems unbounded, but it is not):

$$\lim_{x\rightarrow\infty}\text{erf}(x)= \lim_{x\rightarrow\infty}\ \frac{2}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{n!(2n+1)}=1 \tag{1}$$

You can argue that the error series converges for all finite $x$.

$$\lim_{n\rightarrow\infty} \Bigg| \frac{\frac{(-1)^{n+1}x^{2(n+1)+1}}{(n+1)!(2(n+1)+1)}}{\frac{(-1)^{n}x^{2n+1}}{n!(2n+1)}} \Bigg|$$

$$=\lim_{n\rightarrow\infty} \Bigg| \frac{x^2(2n+1))}{(2n+3)(n+1)} \Bigg|=0$$

This also means that the radius of convergence $|x^2|$ is the whole complex plane

Update

I want to know if you can evaluate the limit (1) in a different way ?, than from getting the error integral and integrate.

$\endgroup$
  • $\begingroup$ The series converges for each value of $x$ in the radius of convergence. But the limit of the series as $x\to \infty$ does not exist. $\endgroup$ – Mark Viola Sep 23 '15 at 23:35
  • $\begingroup$ I don't understand. Here the radius of convergence is the whole complex plane right ? $\endgroup$ – Keith Sep 23 '15 at 23:45
  • $\begingroup$ @Keith: it is the same for $\cos x=\sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n)!}$, but if you consider a single term of the series and let $x\to +\infty$ you obviously do not have a bounded contribute. The process of exchanging series/integrals/limits requires some care. $\endgroup$ – Jack D'Aurizio Sep 23 '15 at 23:52
  • $\begingroup$ I understand, I corrected the mistake and changed main content of the question $\endgroup$ – Keith Sep 23 '15 at 23:56
0
$\begingroup$

You're trying to compute $\lim_{x\to\infty}\text{erf}(x)$ using the series expansion: $$\text{erf}(x)=\frac{2}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{n!(2n+1)}$$ and finding that the terms of the series don't converge as $x\to\infty$. This is not a contradiction, because you've simply established that $$\lim_{x\to\infty}\lim_{m\to\infty}\sum_{n=0}^{m}\frac{(-1)^{n}x^{2n+1}}{n!(2n+1)} \ne\lim_{m\to\infty}\lim_{x\to\infty}\sum_{n=0}^{m}\frac{(-1)^{n}x^{2n+1}}{n!(2n+1)}\ . $$ In other words, interchanging the two limits doesn't yield the same result. The interchange of limits is not legal (i.e., the interchange won't give the same result) unless additional conditions are satisfied, such as uniform convergence.

See Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \dfrac{\sqrt \pi}{2}$ for alternative ways to evaluate this integral.

$\endgroup$

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.