Orthogonality relations of functions e^(2 pi i n x)

[Logarythmic]

I know that the functions $$e^{2 \pi inx}$$ for $$n \in \mathbb{Z}$$ are a base in the space of functions whith period 1. How do I derive the orthogonality relations for these functions?

 

[HallsofIvy]

By using the definition of "orthogonality"! For functions in L² (square integrable functions- $\int f^2(x)dx$ exists), the functions over which the Fourier series are defined, the inner product is $\int f(x)g(x)dx$. To show that $e^{2\pi inx}$ for n any integer are orthogonal show that
$\int e^{2\pi inx}e^{2\pi imx} dx$ is 0 as long as $m \ne n$.

 

[Logarythmic]

What limits should I use for integration?

 

[Logarythmic]

I don't get it. The integral is never zero with m different from n. I guess I should use some weight function?

 

[OlderDan]

What limits should I use for integration?

Looks to me like -1/2 to +1/2 will do nicely.

I don't get it. The integral is never zero with m different from n.

Can you show an example?

 

[Logarythmic]

$$\int_{- \frac{1}{2}}^{\frac{1}{2}}f_n^*(x)f_m(x)dx = \int_{- \frac{1}{2}}^{\frac{1}{2}}e^{-2 \pi inx}e^{2 \pi imx}dx = \int_{- \frac{1}{2}}^{\frac{1}{2}}e^{2 \pi ix(m-n)}dx =$$

$$= \left[ \frac{e^{2 \pi ix(m-n)}}{2 \pi i(m-n)} \right]_{- \frac{1}{2}}^{\frac{1}{2}} = \frac{e^{\pi i (m-n)} - e^{- \pi i (m-n)}}{2 \pi i (m-n)} = \frac{1}{\pi (m-n)} \sin{\pi (m-n)} = 0$$

Because m and n are integers. But this is only a solution for [-1/2,1/2] ? What if the interval is [0,1]? The I get

$$I = \frac{e^{2 \pi i (m-n)} - 1}{2 \pi i (m-n)}$$

and this is only zero for m = n...

So this shows that the functions are orthogonal only on the interval [-1/2,1/2] and this is my solution. Am I right?

 

[OlderDan]

$$\int_{- \frac{1}{2}}^{\frac{1}{2}}f_n^*(x)f_m(x)dx = \int_{- \frac{1}{2}}^{\frac{1}{2}}e^{-2 \pi inx}e^{2 \pi imx}dx = \int_{- \frac{1}{2}}^{\frac{1}{2}}e^{2 \pi ix(m-n)}dx =$$

$$= \left[ \frac{e^{2 \pi ix(m-n)}}{2 \pi i(m-n)} \right]_{- \frac{1}{2}}^{\frac{1}{2}} = \frac{e^{\pi i (m-n)} - e^{- \pi i (m-n)}}{2 \pi i (m-n)} = \frac{1}{\pi (m-n)} \sin{\pi (m-n)} = 0$$

Because m and n are integers. But this is only a solution for [-1/2,1/2] ? What if the interval is [0,1]? The I get

$$I = \frac{e^{2 \pi i (m-n)} - 1}{2 \pi i (m-n)}$$

and this is only zero for m = n...

So this shows that the functions are orthogonal only on the interval [-1/2,1/2] and this is my solution. Am I right?

Factor an $$e^{\pi i (m-n)}$$ out of your I. Does that help?

 

[Logarythmic]

Then I is zero if $$e^{\pi i (m-n)} = 0$$ or if $$-\pi i(m-n) = 2$$.
The first one gives $$m-n = \frac{2k+1}{2}$$ and the second one gives $$m - n = \frac{2i}{\pi}$$?

 

[OlderDan]

Then I is zero if $$e^{\pi i (m-n)} = 0$$ or if $$-\pi i(m-n) = 2$$.
The first one gives $$m-n = \frac{2k+1}{2}$$ and the second one gives $$m - n = \frac{2i}{\pi}$$?

OK. This is it.

$$I = \frac{e^{2 \pi i (m-n)} - 1}{2 \pi i (m-n)}$$

$$I = e^{\pi i(m - n)} \left[ {\frac{{e^{\pi i(m - n)} - e^{ - \pi i(m - n)} }}{{2\pi i(m - n)}}} \right] = e^{\pi i(m - n)} \left[ {\frac{{\sin \pi (m - n)}}{{\pi (m - n)}}} \right]$$

Compare this to your result for the [-1/2, 1/2] limits. In the form you had it, you cannot say I is zero when m = n; it is indeterminate. In this form, you can see the sine term is zero, but (sinx)/x is 1 for x = 0. When m = n the exponential is 1 and (sinx)/x is 1.

 

[Logarythmic]

So how do I do this on a general interval [a,b] where b - a = 1? How should I interpret the problem "Derive the orthogonality relations for these functions"?

 

[OlderDan]

So how do I do this on a general interval [a,b] where b - a = 1? How should I interpret the problem "Derive the orthogonality relations for these functions"?

I think the factoring step I did will work for any interval. Give it a try.

 

[Logarythmic]

So, if I use a general interval [a,b] with b-a=1, I get

$$I = e^{\pi i (m-n)(2a+1)} \frac{\sin{\pi (m-n)}}{\pi (m-n)} = \left\{\begin{array}{cc}0,&\mbox{ if }m \ne n\\1, & \mbox{ if } m = n\end{array}\right$$

and everyone is happy.

 

[OlderDan]

So, if I use a general interval [a,b] with b-a=1, I get

$$I = e^{\pi i (m-n)(2a+1)} \frac{\sin{\pi (m-n)}}{\pi (m-n)} = \left\{\begin{array}{cc}0,&\mbox{ if }m \ne n\\1, & \mbox{ if } m = n\end{array}\right$$

and everyone is happy.

I can't speak for everyone, but it works for me.

 

[Logarythmic]

Thanks for your help. =)