Wikipedia:Reference desk/Archives/Mathematics/2012 August 14

= August 14 =

Convergence Question
Consider the following complex-valued two-way sequences: $$S = ...,a_{-2},a_{-1},a_{0},a_{1},a_{2},...$$ with the property that $$\sum_{n \in \mathbb{Z}} |a_n|^2$$ converges. Suppose you have a sequence of such sequences $$S_1,S_2,S_3,...$$ which and converges to $$...,b_{-2},b_{-1},b_{0},b_{1},b_{2}...$$ Show that $$\sum_{n \in \mathbb{Z}} |b_n|^2$$ converges too. Widener (talk) 12:34, 14 August 2012 (UTC)
 * How are you defining convergence in the space of sequences? Rckrone (talk) 21:48, 14 August 2012 (UTC)
 * They're convergent sequences, so presumably you just identify them with their limits. Any other definition would seen strange to me. --Tango (talk) 22:49, 14 August 2012 (UTC)
 * I think it means pointwise convergence. So if $$S_i = ...,a_{-2}^i, a_{-1}^i, a_0^i, a_1^i, a_2^i,...$$, then

$$b_n = \lim_i a_n^i$$. Of course, in this interpretation, the stated result is false. Define $$a_n^i = 1$$ if $$|n| < i$$ and $$=0$$ otherwise.--121.73.35.181
 * Sorry. It does not mean pointwise convergence; it means convergence under the metric $$d(S_i,S_j) = \sum_{k \in \mathbb{Z}} |a_k^i-a_k^j|^2$$. --Widener (talk) 00:06, 15 August 2012 (UTC)(talk) 23:54, 14 August 2012 (UTC)
 * You should think of using the triangle inequality with $$b_n = a_n^i - (a_n^i - b_n)$$.  Sławomir Biały  (talk) 01:40, 15 August 2012 (UTC)
 * If it helps: a sequence $$a_n$$ such that $$\sum |a_n|^2$$ converges must converge to ____?  Sławomir Biały  (talk) 22:41, 14 August 2012 (UTC)
 * I misunderstood your question. This observation will not help.   Sławomir Biały  (talk) 01:37, 15 August 2012 (UTC)


 * So you are dealing with a convergent sequence $$S_i$$ in the Hilbert space $$\ell^2(\Z)$$. A convergent sequence is bounded, and the norm is continuous. So $$\|S\|_2=\lim_{i\to\infty}\|S_i\|_2 < +\infty $$ (which you can prove directly by the triangle inequality as suggested above). --pm a 09:39, 15 August 2012 (UTC)
 * Thanks ! 00:28, 16 August 2012 (UTC)Widener (talk)

Example of Limitation of Riemann integration
How do You show That $$\sum_{k=2}^\infty \frac{1}{\log(k)}\sin(kx)$$ is Not a Fourier series Of a Riemann integrable Function? Widener (talk) 14:36, 14 August 2012 (UTC)


 * Why don't you start by trying to explain why
 * $$ \frac{\pi^2}{3} + \sum_{k=1}^{\infty} \frac{4(-1)^k}{k^2} \, \cos(kx) \, $$
 * is a Fourier series of a Riemann integrable function? — Fly by Night  ( talk )  17:17, 14 August 2012 (UTC)
 * Is it because the sequence of Fourier coefficients is not in $$l_2(\mathbb{Z})$$? Widener (talk) 23:06, 14 August 2012 (UTC)
 * Yes. Sławomir Biały  (talk) 00:19, 15 August 2012 (UTC)