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Examine the continuity and differentiability of functions:

a) $\displaystyle f(x)=\sum_{n=1}^{+\infty}\frac{\sin(nx)}{n^3}$

b) $\displaystyle f(x)=\sum_{n=1}^{+\infty}\arctan\left(\frac{x}{n^2} \right)$

in the case of differentiability explore the sign of f '(0).

So, we are dealing with function series I think. I tried a): Let $f_n(x)=\frac{\sin(nx)}{n^3}$. With Weierstrass M-test $|f_n|\le \frac{1}{n^3}$ so the series $\sum_{n=1}^{+\infty}f_n$ converges uniformly. The same applies for series of derivatives that is: $\sum_{n=1}^{+\infty}\frac{\cos(nx)}{n^2}$ so function $f$ is differentiable and so it is continuous. f'(0)>0

Is this argumentation ok?

I don't know how to approach b).

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    I only don't understand argumentation for b) does not converge uniformly on all $\mathbb{R}$.. can you make it clear for me? I know that $\arctan(x)\in(-\pi/2; \pi/2)$ but when $n$ is large then $x/n^2$ is close to $0$ so I think I can't limit the terms of this sum from the bottom..2012-04-12

1 Answers 1

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Your argument for part a) is correct.

For b), think about where the problem occurs when we try to do similar steps. We can't prove uniform convergence immediately, because $x$ could be unbounded. So instead, for now just consider $x\in (-a,a).$ Then since $|\arctan x| \leq |x| $ we have $\biggr|\sum_{n=1}^{\infty} \arctan \left( \frac{x}{n^2} \right) \biggr| \leq \sum_{n=1}^{\infty} \frac{a}{n^2} < \infty. $

Thus the sum converges uniformly in $(-a,a).$

Now, the sequence of term by term derivatives converges uniformly in $(-a,a)$ as well since (as sos440 mentioned above) $ \frac{d}{dx} \arctan \left( \frac{x}{n^2} \right) \leq \frac{1}{n^2}.$

Thus, we conclude the sum is differentiable for all $x\in (a,-a).$ Since $a>0$ was arbitrary, the sum is differentiable for all $x\in \mathbb{R},$ with derivative f'(x) = \sum_{n=1}^{\infty} \frac{1}{n^2 + (x/n)^2} . In particular, f'(0) = \pi^2/6 > 0.