Finding sum of $$\sum^{1007}_{k=1}\left(\cos \left(\frac{k\pi}{2007}\right)\right)^{2014}$$

$\bf{Attempt:}$ With the help of $\displaystyle \cos x = \frac{e^{ix}+e^{-ix}}{2}$ and substitute $\displaystyle \frac{\pi}{2007} = \theta$

$$\sum^{1007}_{k=1}\left(\cos k\theta\right)^{2014} = \sum^{1007}_{k=1}\bigg(\frac{e^{ix}+e^{-ix}}{2}\bigg)^{2014}$$

Could some help me to solve it, thanks

  • $\begingroup$ what result do you exspect? $\endgroup$ – Dr. Sonnhard Graubner Nov 19 '17 at 11:50
  • $\begingroup$ Why not consider $\sum_{k=1}^{2014}$ instead? Also you can now use the binomial theorem. $\endgroup$ – Lord Shark the Unknown Nov 19 '17 at 11:56
  • $\begingroup$ Are you sure that there should be $2007$, not $1007$? $\endgroup$ – Alex Ravsky Nov 22 '17 at 8:45

I will provide a solution for the case in which the denominator is $1007$, as noted in the comments (in my opinion, the value of $2007$ is a typo).

You are on the right way. Applying $\displaystyle \cos x = \frac{e^{ix}+e^{-ix}}{2}\,\,\,$ and setting $z=e^{\frac{\pi i}{1007}}\,$, we get that the summation is equivalent to

$$ \sum^{1007}_{k=1}\bigg(\frac{z^{k}+z^{-k}}{2}\bigg)^{2014}\\ = \frac{1}{2^{2014}} \sum^{1007}_{k=1}\bigg(z^{k}+z^{-k}\bigg)^{2014} $$

Expanding this last term with use of the binomial theorem, we get

$$ \frac{1}{2^{2014}} \sum^{2014}_{i=0} \sum^{1007}_{k=1} \binom{2014}{i} \bigg[z^{k(2014-i)}z^{-ki} \bigg] \\ = \frac{1}{2^{2014}} \sum^{2014}_{i=0} \sum^{1007}_{k=1} \binom{2014}{i} (z^w)^k $$

where $i=0,1,2...2014 \,\,\,\,$ and $w= 2014-2i \,\,\,\,$. Taking into account that $z^{2014}=1\,\,$, now note that the quantity $ \sum^{1007}_{k=1} (z^w)^k \,\, $ is equal, for $z^w \neq 1\,\,$, to a sum of the roots of unity in the complex plane, homogeneously spaced around the unit circle. So it is different from to zero only when $z^w=1\,\,$. This occurs for $i=0\,\,$,$i=1007\,\,$, and $i=2014\,\,$. So, calculating the sums only for these values of $i$, we get

$$ \frac{1}{2^{2014}} \left[1017+1017 \binom{2014}{1017}+1017 \right] \\ =\frac{1007}{2^{2014}} \left[2+\binom{2014}{1017}\right] \approx 17.9013... $$

This numerical result for the initial sum reported in the OP is confirmed by WA here.


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.