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I want to prove that $\sum e^{-nx + \cos(nx)}$ is defined and continuous on the given interval of $(a, \infty)$ where $a >0$.

Then, how exactly do I show it is defined? It just seems trivial, since the exponential function, the linear function and cosine function are all "Defined" on the reals.

Further, to show it is continuous, I proceed as follows:

$$\sum e^{-nx + \cos(nx)}$$

Then, since each of the functions $f_n$ are continuous, by the algebraic continuity theorem, the partial sums will also be continuous, and so our sum is continuous. $$(\text{CAVEAT})$$ Before I can make the above bolded claim, I must first show the series is uniformly convergent, correct??

So, if that is the case, we use the M-test. We must find an $M_n > 0$ such that:

$$|f_n(x)| \leq M_n$$ holds.

$$|f_n(x)| = |e^{-nx + \cos(nx)}| = e^{-nx}e^{\cos(nx)}$$ since $\max \cos(nx) = 1$ $$\implies e^{-nx}e^{\cos(nx)} \leq e^{-nx}e$$

We look for these properties on the interval $(a, \infty)$

Thus

$$|f_n(x)| \leq e^{-ax}e = M_n$$

How can we conclude that $\sum M_n$ converges, however? Is this a trivial fact, since the exponential decreases so fast? I question this last step, because $\frac{1}{x} \to 0$ but the sum does not converge.

Once I have established continuity, how can I establish it is differentiable with a continuous derivative?

Any help is very much appreciated!

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For the uniform convergence we have $$\sum_{n\geq1}e^{-nx+\cos\left(nx\right)}\leq\sum_{n\geq1}e^{-nx}\leq\sum_{n\geq1}e^{-na}=\frac{1}{e^{a}-1} $$ then we have uniform convergence by M-test. For the differentiation, we have to prove that $$f'\left(x\right)=\sum_{n\geq1}f_{n}'\left(x\right) $$ and so the uniform convergence of $\sum_{n\geq1}f_{n}'\left(x\right) $. Note that we have $$\left|\sum_{n\geq1}e^{-nx+\cos\left(nx\right)}\left(-n-n\sin\left(nx\right)\right)\right|\leq2\sum_{n\geq1}ne^{-na}=\frac{2e^{a}}{\left(e^{a}-1\right)^{2}} $$ then your claim.

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  • $\begingroup$ What about to show that it is continuous? Also, I am not quite sure what you have done for the differentiation. you are saying to show the uniform convergence of the series $\sum f'_n(x)$ and this proves that $f'(x) = \sum f'_n(x)$ ? $\endgroup$ – piman314 May 28 '15 at 8:55
  • $\begingroup$ @piman314 $f_{n}'\left(x\right)$ are continuous and and the sum is uniform convergent. About the derivative, I used this theorem dpmms.cam.ac.uk/~agk22/uniform.pdf. In other words, we need the convergence at one point plus uniform convergence of the series of derivatives. $\endgroup$ – Marco Cantarini May 28 '15 at 9:34

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