4
$\begingroup$

Problem:

Evaluate for $n=11$$$\begin{align} \sin^{4n}\left(\frac{\pi}{4n}\right) + \cos^{4n}\left( \frac{\pi}{4n}\right) = \frac{1}{4^{2n-1}} \left[ \sum_{r=0}^{n-1} \binom{4n}{2r} \cos\left(1 - \frac{r}{n} \right) \pi \, + \frac{1}{2} \binom{4n}{2n} \right]. \end{align} $$

Sorry for this odd question. I saw this formula here on MSE which is quite helpful for a question I need to solve ($\sin^{4n}\frac{\pi}{4n} + \cos^{4n} \frac{\pi}{4n})$ for given values of $n$. Unfortunately I don't know how to evaluate this formula manually. I was thus hoping to use Wolfram Alpha to evaluate this for different values of $n$ but was unable to enter it correctly. I would be truly grateful if somebody would kindly show me how to input this formula into Wolfram Alpha or solve it for $n=11$. Many thanks in advance.

$\endgroup$
1
  • $\begingroup$ In general computer based systems have limitations. This appears to be one, in the general sense. The best reduction is to use the database of cosine values functions.wolfram.com/ElementaryFunctions/Cos/03/02 and write out the components of the series. There are also two typos and the formula should read \begin{align} \sin^{4n}\left(\frac{\pi}{4n}\right) + \cos^{4n}\left( \frac{\pi}{4n}\right) = \frac{1}{4^{2n-1}} \left[ \sum_{r=0}^{n-1} \binom{4n}{2r} \cos\left(1 - \frac{r}{n} \right) \pi \, + \frac{1}{2} \binom{4n}{2n} \right]. \end{align} $\endgroup$
    – Leucippus
    Commented Jul 15, 2015 at 14:40

3 Answers 3

6
$\begingroup$

With some trivial manipulation it straighforward to check that we just have to compute: $$ \sum_{r=0}^{2n}\binom{4n}{2r}\cos\frac{4\pi r}{n}=\text{Re}\sum_{r=0}^{2n}\binom{4n}{2r}\exp\left(8r\cdot\frac{2\pi i}{4n}\right).\tag{1} $$ Since: $$ \sum_{r=0}^{2n}\binom{4n}{2r} z^{2r} = \frac{1}{2}\sum_{k=0}^{4n}\binom{4n}{k}\left(z^k+(-z)^k\right)=\frac{(1+z)^{4n}+(1-z)^{4n}}{2}\tag{2}$$ by taking $\omega=\exp\frac{2\pi i}{n}$ we have: $$ \sum_{r=0}^{2n}\binom{4n}{2r}\cos\frac{4\pi r}{n}=\text{Re}\left(\frac{(1+\omega)^{4n}+(1-\omega)^{4n}}{2}\right)\tag{3}$$ then, since $1\pm \omega = 2\cos\left(\frac{\pi}{n}\right)\exp\left(\pm\frac{\pi i}{n}\right)$, $$ \sum_{r=0}^{2n}\binom{4n}{2r}\cos\frac{4\pi r}{n} = \frac{1}{2}\left(2\cos\frac{\pi}{n}\right)^{4n}\left(\cos(4\pi)+\cos(-4\pi)\right)=\color{red}{\left(2\cos\frac{\pi}{n}\right)^{4n}}.\tag{4}$$

$\endgroup$
2
  • $\begingroup$ Thanks a lot for your efforts Sir. Sir I'm really sorry but there was a typo in the formula I had written. According to Leucippus Sir the formula should have been $$\begin{align} \sin^{4n}\left(\frac{\pi}{4n}\right) + \cos^{4n}\left( \frac{\pi}{4n}\right) = \frac{1}{4^{2n-1}} \left[ \sum_{r=0}^{n-1} \binom{4n}{2r} \cos\left(1 - \frac{r}{n} \right) \pi \, + \frac{1}{2} \binom{4n}{2n} \right]. \end{align}$$ Sir please would it be possible for you to evaluate this series with any $n>11$ as per your choice? $\endgroup$ Commented Jul 15, 2015 at 16:21
  • $\begingroup$ @MakeaDifference: I am just too lazy to modify my answer accounting for your typo. My method works also in that case, you just have to take a different root of unity $\omega$. $\endgroup$ Commented Jul 15, 2015 at 16:32
1
$\begingroup$

$\displaystyle e^{ix}=\cos x+i\sin x\implies e^{-ix}=\cos(-x)+i\sin(-x)=\cos x-i\sin x$

$\displaystyle e^{ix}+e^{-ix}=2\cos x,e^{ix}-e^{-ix}=2i\sin x$

$\displaystyle \left(2i\sin x\right)^{4n}+\left(2\cos x\right)^{4n}$ $\displaystyle =(e^{ix}+e^{-ix})^{4n}+(e^{ix}-e^{-ix})^{4n}$ $\displaystyle =2\sum_{r=0}^{2n}\binom{4n}{2r}(e^{ix})^{4n-2r}(e^{-ix})^{2r}$

$\displaystyle\implies16^n\left(\sin^{4n}x+\cos^{4n}x\right)$ $\displaystyle =2\sum_{r=0}^{2n}\binom{4n}{2r}e^{i4x(n-r)}$ $\displaystyle 2^{4n-1}\left(\sin^{4n}x+\cos^{4n}x\right)=\binom{4n}{2n}+\sum_{r=0}^{n-1}\left(\binom{4n}{2r}e^{i4x(n-r)}+\binom{4n}{4n-2r}e^{-i4x(n-r)}\right)$

As $\binom NR=\binom N{N-R},$

$\displaystyle\implies2^{4n-1}\left(\sin^{4n}x+\cos^{4n}x\right)$ $\displaystyle =\binom{4n}{2n}+\sum_{r=0}^n\binom{4n}{2r}\{e^{i4x(n-r)}+e^{-i4x(n-r)}\}$

$\displaystyle\implies2^{4n-1}\left[\sin^{4n}x+\cos^{4n}x\right]=\binom{4n}{2n}+\sum_{r=0}^{n-1}\binom{4n}{2r}2\cos\{4(n-r)x\}$

Set $x=\dfrac\pi{4n}$

Can you take it home from here?

$\endgroup$
4
  • $\begingroup$ Nomoshkar Sir. Thanks a lot Sir. Sir in the last result$$\displaystyle\implies16^n\left[\sin^{4n}x+\cos^{4n}x\right]=2\left[\binom {4n}{2n} 2+\sum_{r=0}^{n-1}\binom{4n}{2r}2\cos\{(n-r)4x\}\right]$$$\left[\sin^{4n}x+\cos^{4n}x\right]$ refers to the Greatest Integer Function right Sir? Also Sir please would you show me how to evaluate $2\left[\binom{4n}{2n}2+\sum_{r=0}^{n-1}\binom{4n}{2r}2\cos\{(n-r)4x\}\right]?$ I haven't yet learnt how to compute such series and thus resort to Wolfram Alpha for the computation. Unfortunately Sir I cannot understand how to input this series into Wolfram Alpha. $\endgroup$ Commented Jul 15, 2015 at 17:44
  • $\begingroup$ Sir please would you evaluate the formula for any $n\ge 11?$ I would feel really really grateful if you would kindly do this for me. $\endgroup$ Commented Jul 15, 2015 at 17:45
  • $\begingroup$ @MakeaDifference, Please find the updated version. This holds true for any positive integer $n$. So, I'm not sure about specialty of $n\ge11$ $\endgroup$ Commented Jul 16, 2015 at 5:44
  • $\begingroup$ Sorry Sir but I'm still unable to evaluate the sum $$\sum_{r=0}^{n-1}\binom{4n}{2r}2\cos\{4(n-r)x\}$$ for $x=\dfrac\pi{4n}$ and $n=11.$ I've never learnt how to evaluate these types of series. I would be really grateful Sir if you would please evaluate the above series for me for $n=11?$ Lastly Sir by $$\binom{4n}{2n}$$ do you mean $$\dfrac{44!}{22!22!}?$$ for $n=11?$ $\endgroup$ Commented Jul 16, 2015 at 10:32
0
$\begingroup$

if you meant $\cos(1-4n/2r)*\pi$, then using Maple I got

enter image description here

it seems for me that you cannot simplify the answer.

If you meant $\cos[(1-4n/2r)*\pi]$, then here is the result

enter image description here

$\endgroup$
2
  • 3
    $\begingroup$ This is not an answer. Everyone is able to use a computer algebra system, but the point is to provide a manageable closed form for the given sum: it is not difficult to compute it through standard human-manipulations. $\endgroup$ Commented Jul 15, 2015 at 14:59
  • $\begingroup$ try it first then do it by hand : ) I agree that we should not abuse computer alot! $\endgroup$ Commented Jul 15, 2015 at 15:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .