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Can you please help me with this problem?

Thank you so much!

Show that absolute value of f '(x) is less or equal than pi/8, when f(x)=

Series 1

Using that

Series 2

By Weierstrass M test, the series is uniformly convergent for all x, in particular each term can be differentiated. Moreover, the derivative of the sum is equal to the sum of the derivatives.

Then f '(x)=

Series

By Comparison,

f '(x) is less or equal than

Series 4

However, I do not see how I can prove that absolute value of f '(x) is less or equal than pi/8.

In one previous part of the problem, the domain is restricted to [0, pi/2]. It is not clear if for this part, the domain is restricted to that interval. However, I would not know how to use that.

Did made a interesting comment. Then, if helps, please eliminate x=0 from the domain. Like the interval (0, 2pi)

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    The derivative at $0$ is $\frac{\pi^2}6>\frac\pi8$ hence the upper bound $\frac\pi8$ does not hold (at least, on any interval containing $0$). What is your source? – 2017-02-26
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    @did Thank you for your fast answer. It is not a book. Maybe it is a mistake. I will check tomorrow with my professor. He gave me the problem. Then, is my reasoning correct? New question: What if 0 is not included? For example in the interval (0, pi/2]. – 2017-02-26
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    @Beginner: still $\lim_{x\to 0^+}f'(x)=\frac{\pi^2}{6}>\frac{\pi}{8}$. But the given inequality holds in a neighbourhood of $\pi$, for instance. – 2017-02-26
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    @JackD'Aurizio Thank you Jack! It seems that I can sleep tonight... – 2017-02-26
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    @Did Thank you again! My professor told me that he made a mistake... Ha! – 2017-02-28
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    "Did made a interesting comment." Thanks. "Then, if helps, please eliminate x=0 from the domain. Like the interval (0, 2pi)" Does not help at all. By continuity at 0, still a counterexample. – 2017-02-28

1 Answers 1

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We may start from $$ f_1(x) = \sum_{n\geq 1}\frac{\sin(nx)}{n} $$ that is the Fourier series of the sawtooth wave, a $2\pi$-periodic function that equals $\frac{\pi-x}{2}$ on $(0,2\pi)$.
It follows that $$ f_2(x) = \sum_{n\geq 1}\frac{1-\cos(nx)}{n^2} = \frac{\pi^2}{6}-\sum_{n\geq 1}\frac{\cos(nx)}{n^2} $$ equals $\frac{\pi x}{2}-\frac{x^2}{4}$ on the interval $(0,2\pi)$. It follows that $$ \sum_{n\geq 1}\frac{\cos(nx)}{n^2} = \frac{\pi^2}{6}-\frac{\pi x}{2}+\frac{x^2}{4} $$ is bounded between $-\frac{\pi^2}{48}$ and $\frac{\pi^2}{6}$ (see Bernoulli polynomials) and such bounds are tight.

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    Thank you Jack for your help! Amazing answer! I am still trying to figure out what you did! Ha.... – 2017-02-26
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    Thank you again! My professor told me that he made a mistake... Ha! – 2017-02-28