Is it a Fourier series of a function?

Real Analysis
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Tolaso J Kos
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Is it a Fourier series of a function?

#1

Post by Tolaso J Kos »

Is the series \( \displaystyle \sum_{n=1}^{\infty} \frac{\cos n x}{\ln (n+2)} \) a Fourier series of some \( 2\pi \) periodical function?
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Riemann
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Re: Is it a Fourier series of a function?

#2

Post by Riemann »

The answer would be no due to Parsheval's identity. Indeed:

$$\sum_{n=1}^{\infty}a_n^2 =\frac{1}{\pi}\int_{-\pi}^{\pi}f^2(x)\,{\rm d}x \Leftrightarrow \sum_{n=1}^{\infty}\frac{1}{\ln^2 (n+1)}=\frac{1}{\pi}\int_{-\pi}^{\pi}f^2(x)\,{\rm d}x$$

which is a contradiction, since the series $\sum \limits_{n=1}^{\infty} \frac{1}{\ln^2 (n+1)}$ is known to diverge.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
S.F.Papadopoulos
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Re: Is it a Fourier series of a function?

#3

Post by S.F.Papadopoulos »

The sequence is convex.The series is Fourier series of Lebesgue integrable function.
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Re: Is it a Fourier series of a function?

#4

Post by Riemann »

S.F.Papadopoulos wrote:The sequence is convex.
I get the thing with the Fourier series and Lebesque but I do not get what do you mean by "the sequence is convex".
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Re: Is it a Fourier series of a function?

#5

Post by S.F.Papadopoulos »

If $2a_{n+1}<a_n+a_{n+2}$ for all natural $n$ the sequence is strictly convex.
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