Could someone help me with this please?

Prove that ∑k=1n−1(n−k)cos⁡2kπn=−n2\displaystyle\sum_{k=1}^{n-1}(n-k)\cos\frac{2k\pi}{n}=-\frac{n}{2} where n≥3n\ge3 is an integer.

#Algebra #ComplexNumbers #Trigonometry #Proofs

Note by Omkar Kulkarni
6 years, 3 months ago

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Comments

Definitely, you can approach it via @Incredible Mind's Method but I would like to share another method.

We can proceed by using Euler's Formula eix=cos⁡x+isin⁡xe^{ix} = \cos x + i\sin x ⇒cos⁡x=eix+e−ix2\Rightarrow \cos x = \frac{e^{ix}+e^{-ix}}{2}

Thus, setting x=2kπnx= \frac{2k\pi}{n}, we can compute the sum as follows:

∑k=1n−1(n−k)cos⁡2kπn=∑k=1n−1(n−k)ei2kπn+e−i2kπn2=n2∑k=1n−1(ei2kπn+e−i2kπn)−12∑k=1n−1k(ei2kπn+e−i2kπn)=n2∑k=1n−1(ei2kπn+e−i2kπn)−12∑k=1n−1k(ei2kπn+e−i2kπn)\begin{aligned} \sum_{k=1}^{n-1} (n-k)\cos \frac{2k\pi}{n} & = & \sum_{k=1}^{n-1} (n-k) \frac{e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}}{2} \\ & = & \frac{n}{2} \sum_{k=1}^{n-1} \left(e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}\right) - \frac{1}{2} \sum_{k=1}^{n-1} k \left( e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}\right) \\ & = & \frac{n}{2} \sum_{k=1}^{n-1} \left(e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}\right) - \frac{1}{2} \sum_{k=1}^{n-1} k \left( e^{i\frac{2k\pi}{n}}+e^{-i\frac{2k\pi}{n}}\right) \\ \end{aligned} Now that above series (the first sum) is a simple Geometric Expansion and the second one can be obtained by differentiating the first one. (Hint : ∑r=1nxr=x(xn−1)x−1\sum_{r=1}^{n} x^r = \frac{x(x^n-1)}{x-1} and ∑r=1nrxr=x+(nx−n−1)xn+1(x−1)2\sum_{r=1}^{n} rx^r = \frac{x+(nx-n-1)x^{n+1}}{(x-1)^2})

Finally, it evaluates to ∑k=1n−1(n−k)cos⁡2kπn=n2(−2)−12(−n)=−n2\sum_{k=1}^{n-1} (n-k)\cos \frac{2k\pi}{n} = \frac{n}{2}\left(-2\right)-\frac{1}{2}\left(-n\right) = -\frac{n}{2} (I urge the reader to verify the above result themselves)

EDIT : \textbf{EDIT : } A general approach for finding the sum of series of type ∑k=1nkrk\sum_{k=1}^n kr^k (doesn't involves differentiation)

Let Sn=∑k=1nkrk=r+2r2+3r3+…+nrnS_n = \sum_{k=1}^n kr^k = r + 2r^2 + 3r^3 + \ldots + nr^n

Then observing closely

Sn=r+2r2+3r3+…+nrnrSn=r2+2r3+…+(n−1)rn+nrn+1\begin{aligned} S_n & = & r + 2r^2 + 3r^3 + \ldots + nr^n \\ rS_n & = & \qquad r^2 + 2r^3 + \ldots + (n-1)r^n + nr^{n+1} \end{aligned}

Subtracting the second series from second one, we get

(1−r)Sn=(r+r2+r3+…+rn)−nrn+1=r(1−rn)1−r−nrn+1(1-r)S_n = (r + r^2 + r^3 + \ldots + r^n) - nr^{n+1} = \frac{r(1-r^n)}{1-r}-nr^{n+1} ⇒Sn=r(1−rn)(1−r)2−nrn+11−r\Rightarrow S_n =\frac{r(1-r^n)}{(1-r)^2}-\frac{nr^{n+1}}{1-r}

I hope that helps :):)

Kishlaya Jaiswal - 6 years, 3 months ago

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Oh okay. Thanks!

Omkar Kulkarni - 6 years, 3 months ago

To distinguish the 2 approaches, you should explain that the second summation can be obtained by methods other than differentiation, especially since Omkar said he doesn't know differentiation.

This is a useful formula to be aware of, to sum scenarios like this.

Calvin Lin Staff - 6 years, 3 months ago

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ok sir, I'll edit my solution, right now. :):)

Kishlaya Jaiswal - 6 years, 3 months ago

split it .then 1st can be known by using complex numbers.2nd sum from differentiation of

S=sinx+sin2x+sin3x...sin(n-1)x.........use the sum formula .then diff. w.r.t x both sides.then sub x=2pi/n.

i cannot type because i dont know LATEX

incredible mind - 6 years, 3 months ago

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Oh okay, but I don't know about differentiation. Is there any other method? I think complex numbers come handy.

Omkar Kulkarni - 6 years, 3 months ago

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But there too you need to use differentiation.

Anyways, I'll definitely think of some another approach which doesn't involves differentiation.

Kishlaya Jaiswal - 6 years, 3 months ago
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