2
$\begingroup$

I have recently discovered $$\sum_{k=1}^{\infty}\frac{\cos\left(k\alpha\right)}{k^{2}}-\sum_{k=1}^{\infty}\frac{\cos\left(k\beta\right)}{k^{2}}=\int_{\alpha}^{\beta}\tan^{-1}\left(\cot\left(\theta/2\right)\right)\,\mathrm{d}\theta$$ which seems to be a nice way to attack Basel problem. The integral nicely reduces to $$\int_{0}^{\beta}\frac{\pi}{2}-\frac{\theta}{2}\,\mathrm{d}\theta=\frac{2\pi\beta-\beta^{2}-2\pi\alpha+\alpha^{2}}{4}=\frac{1}{4}\left(\alpha-\beta\right)\left(\alpha+\beta-2\pi\right)$$ when $\alpha,\,\beta\in\left[0,\,2\pi\right]$. The question: how do I choose $\alpha$ and $\beta$ such that $\cos\left(k\alpha\right)-\cos\left(k\beta\right)=1$ for all $k\in\mathbb{N}$? Does this $k$ even exist?

$\endgroup$
  • $\begingroup$ I don't think this is possible. Intuitively, for any $\alpha\neq\beta$, there should be some large $k$ such that $k\beta\approx\beta\pmod{2\pi}$, but $k\alpha\not\approx\alpha\pmod{2\pi}$, and thus $\cos(k\alpha)-\cos(k\beta)\neq\cos(\alpha)-\cos(\beta).$ $\endgroup$ – Samuel Jan 13 '13 at 20:51
1
$\begingroup$

Sadly, there's no such $\alpha$ and $\beta$. We'll try and solve the more general equation - $\cos(k\alpha)-\cos(k\beta)$ is a non-zero constant, say $C$.

Assume there are such $\alpha, \beta$. Let $x=\cos \alpha, y = \cos \beta$.

Note that $\cos(2t) = 2\cos(t)^2 - 1, \cos(3t) = 4\cos(t)^{3} - 3\cos(t)$. Then the following 2 equalities hold: $$x-y = C$$ $$2x^2-2y^2 = C$$ Factoring $2x^2-2y^2$ as $2(x-y)(x+y)$ and plugging the first equation, we find that $x+y = 0.5$ This leads to $x = \frac{C+0.5}{2}, y=\frac{0.5 - C}{2}$. Now we choose $k=3$: $$4x^3 - 3x - (4y^3 - 3y) = C$$ The LHS factors as $(x-y)(4(x^2+xy+y^2)-3)=(x-y)(4(x+y)^2-4xy-3)$, which is $$C(4(0.5)^2-4(\frac{1}{16} - \frac{C^2}{4})-3)=C$$ Cancelling the $C$s, it becomes $$C^2 = 3.25$$ So $C\neq 1$!

Now we note that $\cos(4t) = 8(\cos(t)^4-\cos(t)^2)+1$, which gives the following equation: $$8(x^4-x^2-(y^4-y^2)) = 1$$ The LHS factors as $8(x^2-y^2)(x^2+y^2-1)=1$. We can plug $2x^2-2y^2 = C$ and this becomes $$x^2+y^2 = \frac{1}{4C}+1$$ But $x^2+y^2 = \frac{C^2}{2}+\frac{1}{8} = 1.75$, so it follows that $C=\frac{1}{3}$, a contradiction to $C^2 = 3.25$.

$\endgroup$
2
$\begingroup$

Although the $\alpha$ and $\beta$ you seek cannot exist, taking $\alpha=0$ and $\beta = \pi$ will allow you to attack the Basel problem. These values give: $$\sum_{k=1}^{\infty}\frac{2}{(2k-1)^2} = \frac{\pi^2}{4}$$

and then since $$\sum_{k=1}^{\infty}\frac{1}{k^2}= \sum_{m=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(2^m(2k-1))^2}=\sum_{n=0}^{\infty} \frac{1}{4^n}( \sum_{k=1}^{\infty}\frac{1}{(2k-1)^2})=\frac{4}{3}\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}$$ The first equality comes from the fact any integer can be written uniquely as an odd integer times a power of 2. The second is from bringing the powers of 2 outside the sum. The $\frac{4}{3}$ arises as the sum of the geometric series. The Basel problem then follows easily.

$\endgroup$
  • $\begingroup$ I guess this doesn't strictly answer your question, I thought I'd share it anyway. $\endgroup$ – user45861 Jan 13 '13 at 21:40
  • $\begingroup$ Sorry, but how do you go from $\sum_{k=1}^{\infty}\frac{1-(-1)^k}{k^2}$ to $\sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}$? I got $\sum_{k=1}^{\infty}\frac{2}{(2k-1)^2}$ $\endgroup$ – Ian Mateus Jan 13 '13 at 21:50
  • $\begingroup$ You're quite right. I managed to introduce a few errors. I'll edit them. $\endgroup$ – user45861 Jan 13 '13 at 21:55
  • $\begingroup$ I'd be happy if you clarify a little more what you did next, I couldn't understand it. $\endgroup$ – Ian Mateus Jan 13 '13 at 21:57
  • $\begingroup$ That should be a bit better. $\endgroup$ – user45861 Jan 13 '13 at 22:14

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.