1
$\begingroup$

Consider $x \mapsto x \sin(\pi x), x\in[-1,1]$.

The task is to comput the Fourier series of this function and determine the pointwise limit of the Fourier series

$$c_0+\sum_{k=1}^{\infty}\Big(a_k\cos(k\pi x)+b_k \sin(k \pi x)\Big)$$

My solution is $c_0=\frac{1}{\pi}, a_1=\frac{-1}{2\pi}$ and $a_k=\frac{2(-1)^{k+1}}{(k^2-1)\pi}$ for $k>1$ so the Fourier series is:

$$\frac{1}{\pi}-\frac{\cos(\pi x)}{2\pi}+\sum_{k=2}^{\infty}\frac{2\cos(k\pi x)}{\pi(k^2-1)}(-1)^{k+1}$$

How can I now determine the pointwise limit of this series?

$\endgroup$
0
$\begingroup$

We need to check the $3$ Dirichlet conditions for this function.

  1. $$\int\limits_{-1}^1 \left| f(x)\right | dx < \infty $$ since $f$ is continuous. So $f$ is absolutely integrable.

  2. $$\int_{-1}^1 |f'(x)|dx < \infty $$ since $f'$ is continuous, hence $f$ is of bounded variation.

  3. Of course, $f$ is continuous everywhere on $[-1, 1]$.

Hence the Dirichlet conditions hold, and we can say that the Fourier series of $f$ converge to $f$ at every point of continuity of $f$, hence at every point where it's defined, since $f$ is continuous.

$\endgroup$
0
$\begingroup$

If i) $f$ is continuous on $[-1,1]$ and ii) the Fourier series of $f$ converges uniformly on that interval, then the Fourier series of $f$ converges uniformly (hence pointwise) on $[-1,1]$ to $f.$ This is an elementary result.

In our problem we have $f(x) = x\sin (\pi x),$ so i) is satisfied, and ii) the Fourier series of $f$ converges uniformly by the Weierstrass M test. Thus the given series converges uniformly to $x\sin (\pi x)$ on $[-1,1].$

$\endgroup$

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.