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Consider the roots of unity of $z^n = 1$, say $1, \omega, \ldots, \omega^{n-1}$ where $\omega = e^{i\frac{2\pi}n}$.

It is a well known result that $\sum_{k=0}^{n-1}\omega^k = 0$, but what if we want to consider the alternating sum? I'm interested in

Finding the value of $S = 1 - \omega +\omega^2-\ldots +(-1)^{n-1}\omega^{n-1}$

Keeping in mind that $1, \omega, \ldots, \omega^{n-1}$ are the vertex of a regular $n$-gon in the plane, it is easy to see that when $n$ is even, $S = 0$

The problem arises when $n$ is odd. Here's what I've done so far: Take $x=\frac{2\pi}n$, then

$$S = \sum_{k=0}^{n-1}(-1)^k\omega^k = \sum_{k=0}^{n-1}(-1)^k(\cos kx +i \sin kx)$$

Dealing with the real part, we notice for $k=1,2,\ldots,n-1$ that $$\cos (n-k)x = \cos (2\pi k -kx) = \cos kx$$ Since $n-k$ and $k$ have different parity (remember $n$ is odd), we can see that $$\sum_{k=0}^{n-1}(-1)^k(\cos kx) = 1 + \sum_{k=1}^{n-1}(-1)^k(\cos kx)=1$$

But I have no idea on how to deal with the imaginary part. Asking almighty Wolfram, I got that $$\sum_{k=0}^{n-1}(-1)^k(\sin k\phi)=\sec(\frac \phi2) \sin(\frac {(n-1)(\phi+\pi)}2)\sin(\frac{n(\phi+\pi)}2)$$ Hence $$\sum_{k=0}^{n-1}(-1)^k(\sin kx) = \sec(\frac\pi n)\sin(\frac{\pi(n-1)(2+n)}{2n})\sin(\frac{\pi(2+n)}2)$$

In summary, I've got two questions:

a) How do you deduce the $\sum_{k=0}^{n-1}(-1)^k(\sin k \phi )$ formula?

b) Is there an alternative way to solve the original question?

Thanks in advance

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    $\begingroup$ Multiply $S$ by $1+\omega$, distribute terms, and see what you get... $\endgroup$
    – shalop
    Mar 6, 2016 at 9:27

3 Answers 3

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As $(-1)^r\cdot w^r=(-w)^r,$

$$S_{n-1}=\sum_{r=0}^{n-1}(-w)^r=\dfrac{1-(-w)^n}{1-(-w)}$$

If $n$ is odd,

$$S_{n-1}=\dfrac{1-(-1)}{1+w}=\dfrac2{1+\cos\dfrac{2\pi}n+i\sin\dfrac{2\pi}n}=\dfrac{\cos\dfrac{\pi}n-i\sin\dfrac{\pi}n}{\cos\dfrac{\pi}n}$$

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  • $\begingroup$ Thanks. I feel quite dumb now. $\endgroup$
    – EA304GT
    Mar 6, 2016 at 22:44
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Starting from what you wrote : $$S =\sum_{k=0}^{n-1}(-1)^k\, \big(\cos (kx) +i \sin (kx)\big)=\sum_{k=0}^{n-1}(-1)^k\,e^{ikx}=\frac{1-(-1)^n e^{i n x}}{1+e^{i x}}$$ Now, manipulate the result to extract the real and imaginary parts; this leads, after using trigonometric identities, to $$S_1=\sum_{k=0}^{n-1}(-1)^k\, \cos (kx)=\sec \left(\frac{x}{2}\right) \sin \left(\frac{1}{2} n (x+\pi )\right) \cos \left(\frac{1}{2} (n-1) (x+\pi )\right) $$ $$S_2=\sum_{k=0}^{n-1}(-1)^k\, \sin(kx)=\sec \left(\frac{x}{2}\right)\sin \left(\frac{1}{2} n (x+\pi )\right) \sin \left(\frac{1}{2} (n-1) (x+\pi )\right) $$ Now, setting $x=\frac {2\pi} n$, we get $$S_1=\frac{1}{2} (1-\cos (\pi n))$$ $$S_2=-\sin \left(\frac{\pi n}{2}\right) \cos \left(\frac{\pi }{n}-\frac{\pi n}{2}\right) \sec \left(\frac{\pi }{n}\right)$$

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For a), I would rewrite $\sin$ using Euler formula and then use the sum of the terms of a geometric sequence, and then a lot of calculations... But :

For b), why not simply using the same tip for S directly ? $$S = \sum_{k=0}^{n-1}(-1)^k\omega^k = \sum_{k=0}^{n-1}(-1)^ke^{\frac {2ik\pi} n}=\frac {1-(-1)^{n}e^{2i \pi}}{1+e^{\frac{2i\pi} n}}=\frac {1-(-1)^n} {1+e^{\frac{2i\pi} n}} $$

We clearly see that $S=0$ for n even.

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  • $\begingroup$ $e^{\frac{2\pi\color{#C00000}{i}}n}$ $\endgroup$
    – robjohn
    Mar 6, 2016 at 9:47

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