Proving Trig Identity for Odd n | Step-by-Step Solution

[ehrenfest]

[SOLVED] trig identity

Homework Statement

Can someone help me prove that
$$\sum_{k=1}^{(n-1)/2}\cos(2 \pi k / n) = -1/2$$

where n is an odd number?

Homework Equations

The Attempt at a Solution

I don't know where to start. You can easily verify it is true for n=3. But after that things get complicated. Maybe induction... I would go to complex exponentials but I reduced a problem involving complex exponentials to this, so I don't really want to go back...

 

[chaoseverlasting]

Actually, you would have to use complex numbers. I suppose you would start with $$x=cos\frac{\pi}{2} +i sin\frac{\pi}{2}$$.

 

[ehrenfest]

And use deMoivre? I don't see how that will help.

 

[NateTG]

I think there's a nice geometric argument available.

Can you show that:
$$\sum_{k=1}^{(n)}\cos(\frac{2 \pi k }{n}) = 0$$
And that:
$$\sum_{k=1}^{\frac{n-1}{2}}\cos(\frac{2 \pi k}{n})=\sum_{k=\frac{n+1}{2}}^{n}\cos(\frac{2 \pi k}{n})$$

 

[ehrenfest]

I think there's a nice gemoetric argument available.

Can you show that:
$$\sum_{k=1}^{(n)}\cos(\frac{2 \pi k }{n}) = 0$$

I assume the parenthesis around the n don't mean anything.

And the answer is no.That was actually the original problem. I thought I reduced the problem, but maybe I just made it worse...

I might as well state the original problem: Find the electric field at the center of regular 13-gon.

 

[NateTG]

I assume the parenthesis around the n don't mean anything.

Yeah, it's late, and I'm always sloppy.

And the answer is no.That was actually the original problem. I thought I reduced the problem, but maybe I just made it worse...

I might as well state the original problem: Find the electric field at the center of regular 13-gon.

It seems that, assuming the 13-gon has a uniform static charge, the field at the center is zero by symetry. (Assume by contradiction that it's not. Then there is at least one line from the center to a vextex that is not parralel to the field at the center of the polyon. Rotate by $\pi$ using that line as an axis. This will produce the same charge geometry, but a different electric field, which is a contradiction.)

Alternatively: The terms of the sum correspond to the x-component of the segments of a path around the perimiter of a regular n-sided polygon. Since the perimiter is closed, their sum must be zero. (Consider the vector sum of$<\cos \frac{2 \pi k}{n}, \sin \frac{2 \pi k}{n}>$.)

 

[ehrenfest]

Hmm. Its not obvious to me why the vector sum of the vertices of a regular n-gon must be 0. Obviously, if you sum the vector connecting vertex 0 to vertex 1, the vector connecting vertex 1 to vertex 2, and so on, then you will get zero, but that is not the same thing.

It is pretty obvious when you look at the symmetry. But there has got to be a numerical way to show it!

 

[morphism]

Hmm. Its not obvious to me why the vector sum of the vertices of a regular n-gon must be 0. Obviously, if you sum the vector connecting vertex 0 to vertex 1, the vector connecting vertex 1 to vertex 2, and so on, then you will get zero, but that is not the same thing.

It is pretty obvious when you look at the symmetry. But there has got to be a numerical way to show it!

There's a slick way to do it with complex numbers: Consider the n nth roots of unity, generated by cos(2*pi*i/n) + i sin(2*pi*i/n). Since these n numbers are precisely the roots of the polynomial x^n - 1, their sum must be 0 (-1 * the coefficient of x^(n-1)). Now take the real part of this sum.

 

[Gib Z]

I would have also done it complex numbers, but I would have done it another way, its probably longer though;

Write $$\cos (\frac{2\pi k}{n}) = Re\left( exp(\frac{2\pi ik}{n} )\right)$$ and then interchange the summand operator with the Re operator (extracts only the real part of the number), then it becomes an ordinary geometric series.

 

[ehrenfest]

I would have also done it complex numbers, but I would have done it another way, its probably longer though;

Write $$\cos (\frac{2\pi k}{n}) = Re\left( exp(\frac{2\pi ik}{n} )\right)$$ and then interchange the summand operator with the Re operator (extracts only the real part of the number), then it becomes an ordinary geometric series.

You mean
$$Re\left( exp(\frac{2\pi i1}{n} )\right), Re\left( exp(\frac{2\pi i2}{n} )\right),Re\left( exp(\frac{2\pi i3}{n} )\right),...$$

with common ratio

$$Re\left( exp(\frac{2\pi i1}{n} )\right)$$

?

Is that really a geometric series?

 

[Gib Z]

Yes? Why not?

 

[ehrenfest]

Because Re(x)Re(y) is not equal to Re(xy).
Take for example x=y=i.

 

[Gib Z]

How does that relate to this?

 

[ehrenfest]

A geometric series has the form x,x^2,x^3,...

Letting $$x = Re\left( exp(\frac{2\pi i1}{n} )\right)$$ gives

$$Re\left( exp(\frac{2\pi i1}{n} )\right),Re\left( exp(\frac{2\pi i1}{n} )\right)Re\left( exp(\frac{2\pi i1}{n} )\right),Re\left( exp(\frac{2\pi i1}{n} )\right)Re\left( exp(\frac{2\pi i1}{n} )\right)Re\left( exp(\frac{2\pi i1}{n} )\right),...$$

which is not the same as

$$Re\left( exp(\frac{2\pi i1}{n} )\right), Re\left( exp(\frac{2\pi i2}{n} )\right),Re\left( exp(\frac{2\pi i3}{n} )\right),...$$

 

[Gib Z]

We are interchanging the summation order with the Re operator. I'm sure you agree that we can do that?

 

[NateTG]

Hmm. Its not obvious to me why the vector sum of the vertices of a regular n-gon must be 0. Obviously, if you sum the vector connecting vertex 0 to vertex 1, the vector connecting vertex 1 to vertex 2, and so on, then you will get zero, but that is not the same thing.

Each edge vector of the n-gon can be translated into a spoke vector by means of a fixed rotation and scaling. Since linear transforms are distributive over vectors, this means that the vector sum of the spokes is the same as this rotation and scaling applied to the vector sum of the edges, but we know that that's zero.

...

Gib Z is applying the Euler Identity:
$$e^{i \theta} =\cos \theta + i \sin \theta$$
to express the sum of the x components as:
$$\rm{Real} \sum_{k=1}^{n} \left({e^{i\frac{2 \pi k}{n}}\right)^k$$

 

[ehrenfest]

We are interchanging the summation order with the Re operator.

I don't know what this means.

Again, Re(x)Re(y) is not equal to Re(xy). Please write out the first few terms of the geometric series you are talking about and show the multiplication by the common factor. I am relatively sure that

$$Re\left( exp(\frac{2\pi i1}{n} )\right), Re\left( exp(\frac{2\pi i2}{n} )\right),Re\left( exp(\frac{2\pi i3}{n} )\right),...$$

is not a geometric series.

 

[NateTG]

Gib Z means that he's distributing 'Real Part' over the sum. (The real part of a sum is the sum of the real parts provided the sum converges.)

$$\sum_{k=1}^{n}\left({e^{i\frac{2 \pi }{n}}\right)^k=e^{i\frac{2 \pi} {n}} + e^{2 i \frac{2 pi}{n}} + e ^ {3 i \frac{2 pi}{n}}...$$

$$\rm{Real Part} \left(\sum_{k=1}^{n}\left({e^{i\frac{2 \pi }{n}}\right)^k\right) = \cos \frac{1(2 \pi)}{n} + \cos \frac{2(2\pi)}{n} + \cos \frac{3 (2 \pi)}{n}...$$

 

[ehrenfest]

I see what you two mean.

But please confirm the following statement:

$$Re\left( exp(\frac{2\pi i1}{n} )\right), Re\left( exp(\frac{2\pi i2}{n} )\right),Re\left( exp(\frac{2\pi i3}{n} )\right),...$$

IS NOT A GEOMETRIC SERIES.

 

[NateTG]

It's not, but it's close to one.