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How to show that $$ \sum_{n=2}^\infty \frac{\sin{(nx)}}{\log n} $$ not the Fourier series of any function?

I have shown that the series is convergent by Dirichlet test.

Let $a(n)=\frac{1}{\log n}$. What is $\sum (a(n))^2$, to apply Parseval's theorem?

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    $\begingroup$ See Katznelson chapter I, 4.2. Briefly, a sine series $\sum a_n \sin{(nx)}$ is not a Fourier series of an integrable function if $a_n \gt 0$ and $\sum_n \frac{a_n}{n} = \infty$. $\endgroup$ – t.b. Apr 9 '12 at 8:21
  • $\begingroup$ Do you mean $\sin \frac{nx}{ \log n}$ or $\frac{\sin nx}{\log n}$? $\endgroup$ – Rudy the Reindeer Apr 9 '12 at 8:26
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    $\begingroup$ @MattN: the former has no chance of being a Fourier series... $\endgroup$ – t.b. Apr 9 '12 at 8:29
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    $\begingroup$ @t.b.: google link to An Introduction to Harmonic Analysis 4.2 (perhaps... :-)) $\endgroup$ – Raymond Manzoni Apr 9 '12 at 8:35
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Since $\Bigl\lvert\sum\sin(n x)\Bigr\rvert\leq M$ for $M<\infty$ and $\Bigl\{\frac{1}{\log(n)}\Bigr\}\to 0$ monotonically, by Dirichlet's test, $\sum_{n=2}^{\infty}\frac{\sin(nx)}{\log(n)}$ converges.


If $\sum_{n=2}^{\infty}\frac{\sin(nx)}{\log(n)}$ is a Fourier series, by Parseval's theorem, there exist a Riemann integrable function $f$ such that $$ \int_{-\pi}^{\pi}\lvert f(x)\rvert^2dx=2\pi\sum_{n=2}^{\infty}\frac{1}{\log^2\lvert n\rvert} $$

Theorem:

Suppose $a_1\geq a_2\geq\cdots\geq 0$. Then the series $\sum a_n$ converges iff $\sum 2^ka_{2^k}$ converges. $$ \sum_{n=2}^{\infty}\frac{2^k}{k^2\log^2(2)}\geq\sum_{n=2}^{\infty}\frac{1}{k}=\infty $$


Therefore, by Cauchy's condensation test, the $\sum_{n=2}^{\infty}\frac{1}{\log^2\lvert n\rvert}$ does not converge which contradicts Parseval's theorem. Thus, $\sum_{n=2}^{\infty}\frac{\sin(nx)}{\log(n)}$ is not a Fourier series.

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Suppose that it is the Fourier series of $f(x)$ in $(-\pi,\pi)$, namely, $$ f(x)=\sum_{n=2}^\infty \frac{\sin(nx)}{\log n},\quad x\in(-\pi,\pi).$$ Then by Pareraval's Identity, $$\int_{-\pi}^\pi f(x) \, dx=\sum_{n=2}^\infty\frac{1}{\log^2n}=\infty.$$

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    $\begingroup$ This answer is incomplete. Parseval's identity applies when $\int_{-\pi}^\pi |f(x)|^2\,dx<\infty$. What about the case when that integral is infinite but $\int_{-\pi}^\pi |f(x)| \, dx<\infty$? Those functions also have Fourier series. $\endgroup$ – Michael Hardy Mar 11 '13 at 18:31
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    $\begingroup$ . . . also, did you intend $\int_{-\pi}^\pi |f(x)|^2\,dx$ where you wrote $\int_{-\pi}^\pi f(x)\,dx$? $\endgroup$ – Michael Hardy Mar 11 '13 at 18:33

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