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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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Just observe that $$ \frac{d}{dx} \arctan \left( \frac{x}{n^2} \right) = \frac{1}{n^2 + (x/n)^2} \leq \frac{1}{n^2},$$ so that their sum converges uniformly by Weierstrass $M$-test. –  sos440 Apr 12 '12 at 13:57
    
@sos440 What theorem are you using exactly? I've wondered if the sequence of term by term derivatives converging uniformly automatically implied the same for the original sum (as you seem to be suggesting). The theorem I know requires checking both $f_n \to f$ and $f'_n \to g$ uniformly (if so, then we can conclude $f'=g.$) –  Ragib Zaman Apr 12 '12 at 14:00
    
We know that for a sequence of functions $f : (a, b) \to \mathbb{R}$ on a bounded interval $(a, b)$, uniform convergence of $f_n' \to g$ and pointwise convergence of $f_n \to f$ implies uniform convergence of $f_n \to f$ with $f' = g$. But your weaker form of this theorem is still applicable in this situation. Of course, b) does not converge uniformly on all of $\mathbb{R}$ since $f(x)$ is unbounded while its partial sum not. Rather, it converges uniformly on every closed bounded interval. –  sos440 Apr 12 '12 at 14:08
    
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.. –  xan Apr 12 '12 at 14:48

1 Answer 1

up vote 1 down vote accepted

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, $f'(0) = \pi^2/6 > 0.$

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thank you very much! –  xan Apr 12 '12 at 14:39
    
@xan You are welcome. –  Ragib Zaman Apr 12 '12 at 14:41

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