Compute $$\sum_{n=0}^\infty(-1)^n\frac{\cos^2({3^nx})}{3^n}$$

The problem is pretty simple, but it was hard for me to segregate into the partial fractions(I wanted to make a form of telescoping).

Hmmmm... My attempts were: $$\sum_{n\ge0}(-1)^n\frac{\cos^2({3^nx})}{3^n}=\sum_{n\ge0}(-1)^n\frac{1+\cos(2\cdot3^nx)}{2\cdot3^n}={1\over2}\sum_{n\ge0}\left(-{1\over3}\right)^n+\Re \sum_{n\ge0}\frac{(-1)^ne^{i\cdot2\cdot3^nx}}{2\cdot3^n}$$

From here, could you please suggest me the idea in order to continue the calculation? I still cannot solve the series $$\sum_{n\ge0}\frac{(-1)^ne^{i\cdot2\cdot3^nx}}{2\cdot3^n}$$ because there is another exponents in the exponents of the natural constant $e$. I'm also pleasure to have a hint in a different perspective. Thanks for your interest.

[EDIT_1] I surely think that there must be some typo on the given series - for example, mistyping $\pi$ as $x$ as SangchulLee and DougM mentioned through the comments, or the location of $n$(such as $3nx\rightarrow3^nx$). But I suddenly wanted to deeply focus on this series, and I just started to doubtful about the existence of closed-form of it. Furthermore, just for the curious of math, if there's no closed-form, I want to prove that.

[EDIT_2] It's also welcome to suggest another possible typo. I'm still waiting the various opinions, suggestions, ideas, and creative solutions of the series. Besides, I'm also wondering whether there is a typical method to prove that the given series has no closed-form.

[EDIT_3] Can we evaluate the series with exponents in the denominator?

I recommend to skim what I've discussed so far. You don't have to reply all the questions. Thanks for your interest one more time.

  • 2
    $\begingroup$ Are you sure that inside $\cos^2$ it is $3^n x$ and not $3nx$? $\endgroup$ – mathcounterexamples.net Mar 2 at 15:33
  • 1
    $\begingroup$ In my print, it said $3^n$. It would be more great if it were $3nx$, so sad :( $\endgroup$ – ToBY Mar 2 at 15:35
  • $\begingroup$ Why do you think it has a closed form? Did you try to graph it (using a computer)? $\endgroup$ – metamorphy Mar 2 at 17:19
  • $\begingroup$ Considering the Weierstrass function as well as lacunary functions, I am very skeptical that this series has a closed form. $\endgroup$ – Sangchul Lee Mar 3 at 4:35
  • $\begingroup$ @metamorphy Um, actually, I'm still questionable having closed-form. Wolfram just said the series converges, and didn't show the exact value(I don't have any mathematical graph program in PC). But, I'm somehow believing that there exists closed-form, since if there isn't, there's no reason why this problem is included in my print that teacher gave. $\endgroup$ – ToBY Mar 3 at 4:36

Comment extended to a "not an answer" answer per request.

There is another possible form of typo $$\sum_{n=0}^\infty (-1)^n \frac{\cos^{\color{red}{3}}(3^n x)}{3^n}$$ which sums to a closed form.

Start from the triple angle formula for cosine, $$\cos(3\theta) = 4\cos^3\theta - 3\cos\theta \quad\iff\quad\cos^3\theta = \frac34\left[\cos\theta + \frac{\cos(3\theta)}{3}\right]$$ We have $$\begin{align} (-1)^n\frac{\cos^3(3^n x)}{3^n} &= (-1)^n \frac34\left[\frac{\cos(3^n x)}{3^n} + \frac{\cos(3^{n+1}(x)}{3^{n+1}}\right]\\ &= \frac34\left[ (-1)^n \frac{\cos(3^n x)}{3^n} - (-1)^{n+1} \frac{\cos(3^{n+1} x)}{3^{n+1}} \right]\end{align} $$ This allows us to turn the sum into a telescoping sum. The end result is $$\sum_{n=0}^\infty (-1)^n \frac{\cos^3(3^n x)}{3^n} = \frac34 \times (-1)^0 \frac{\cos(3^0 x)}{3^0} = \frac34 \cos(x)$$

| cite | improve this answer | |
  • 1
    $\begingroup$ Using triple formula allows the series to do the telescoping. The result made by your keen observation is quite surprising to me. Thanks, sir. $\endgroup$ – ToBY Mar 4 at 13:58

Lets say $$f(x)=\sum_{n=0}^\infty(-1)^n\frac{\cos^2({3^nx})}{3^n}$$ Then $$f'(x)=-\sum_{n=0}^\infty(-1)^n\sin({2*3^nx})$$ Now $$\sin(t)=t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+...$$ with $t=2*3^nx$ $$\sin(2*3^nx)=2*3^nx-\frac{(2*3^nx)^3}{3!}+\frac{(2*3^nx)^5}{5!}-\frac{(2*3^nx)^7}{7!}+...=2*3^nx-\frac{3^{3n}(2x)^3}{3!}+\frac{3^{5n}(2x)^5}{5!}-\frac{3^{7n}(2x)^7}{7!}+...$$ As $$\sum_{n=0}^\infty(-1)^n3^{mn}=\frac{1}{3^m+1} $$ The above relation becomes: $$f'(x)=-\sum_{k=0}^\infty\frac{(-1)^k(2x)^{2k+1}}{(1+3^{2k+1})(2k+1)!}$$ Not sure if this function has a closed form in terms of elementary functions.

| cite | improve this answer | |
  • $\begingroup$ I agree that what I did is formal, but I want to show that there is no closed form. $\endgroup$ – Gevorg Hmayakyan Mar 4 at 3:53
  • $\begingroup$ I haven't thought about the derivative version. Thanks for that. Is there any typical method to show that there does not exist a closed-form? $\endgroup$ – ToBY Mar 4 at 11:54
  • $\begingroup$ Actually the closed forms of known elementary functions does not have such form. May be we are looking for larger scope? $\endgroup$ – Gevorg Hmayakyan Mar 4 at 19:18
  • $\begingroup$ Just to clarify the factor $1+3^{2k+1}$ is not common for any elementary function. Of course this is not proof. $\endgroup$ – Gevorg Hmayakyan Mar 4 at 20:18
  • $\begingroup$ I see. I'll then look through how we can clarify the denominator, especially the exponents. Thanks, sir. $\endgroup$ – ToBY Mar 5 at 2:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.