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

In summary, In order to derive the orthogonality relations for functions e^{2 \pi inx} for n any integer, you use the definition of "orthogonality" and show that the inner product is \int f(x)g(x)dx is 0 as long as m \ne n.
  • #1
Logarythmic
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0
I know that the functions [tex]e^{2 \pi inx}[/tex] for [tex]n \in \mathbb{Z}[/tex] are a base in the space of functions whith period 1. How do I derive the orthogonality relations for these functions?
 
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  • #2
By using the definition of "orthogonality"! For functions in L2 (square integrable functions- [itex]\int f^2(x)dx[/itex] exists), the functions over which the Fourier series are defined, the inner product is [itex]\int f(x)g(x)dx[/itex]. To show that [itex]e^{2\pi inx}[/itex] for n any integer are orthogonal show that
[itex]\int e^{2\pi inx}e^{2\pi imx} dx[/itex] is 0 as long as [itex]m \ne n[/itex].
 
  • #3
What limits should I use for integration?
 
  • #4
I don't get it. The integral is never zero with m different from n. I guess I should use some weight function?
 
  • #5
Logarythmic said:
What limits should I use for integration?
Looks to me like -1/2 to +1/2 will do nicely.
Logarythmic said:
I don't get it. The integral is never zero with m different from n.
Can you show an example?
 
  • #6
[tex]\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 = [/tex]

[tex]= \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[/tex]

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

[tex]I = \frac{e^{2 \pi i (m-n)} - 1}{2 \pi i (m-n)}[/tex]

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?
 
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  • #7
Logarythmic said:
[tex]\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 = [/tex]

[tex]= \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[/tex]

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

[tex]I = \frac{e^{2 \pi i (m-n)} - 1}{2 \pi i (m-n)}[/tex]

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 [tex]e^{\pi i (m-n)}[/tex] out of your I. Does that help?
 
  • #8
Then I is zero if [tex]e^{\pi i (m-n)} = 0[/tex] or if [tex]-\pi i(m-n) = 2[/tex].
The first one gives [tex]m-n = \frac{2k+1}{2}[/tex] and the second one gives [tex]m - n = \frac{2i}{\pi}[/tex]?
 
  • #9
Logarythmic said:
Then I is zero if [tex]e^{\pi i (m-n)} = 0[/tex] or if [tex]-\pi i(m-n) = 2[/tex].
The first one gives [tex]m-n = \frac{2k+1}{2}[/tex] and the second one gives [tex]m - n = \frac{2i}{\pi}[/tex]?
OK. This is it.

[tex]I = \frac{e^{2 \pi i (m-n)} - 1}{2 \pi i (m-n)}[/tex]

[tex] 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] [/tex]

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.
 
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  • #10
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"?
 
  • #11
Logarythmic said:
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.
 
  • #12
So, if I use a general interval [a,b] with b-a=1, I get

[tex]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[/tex]

and everyone is happy.
 
  • #13
Logarythmic said:
So, if I use a general interval [a,b] with b-a=1, I get

[tex]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[/tex]

and everyone is happy.
I can't speak for everyone, but it works for me.
 
  • #14
Thanks for your help. =)
 

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

What are orthogonality relations of functions e^(2 pi i n x)?

The orthogonality relations of functions e^(2 pi i n x) refer to the relationship between different exponential functions with a complex argument. Specifically, these relations describe how these functions are orthogonal to each other, meaning that their inner product is equal to 0.

How are these orthogonality relations useful in mathematics?

These relations are useful in a variety of mathematical applications, including Fourier analysis, signal processing, and differential equations. They allow us to decompose a function into a series of orthogonal components, making it easier to analyze and manipulate mathematically.

What is the significance of the complex argument in these functions?

The complex argument in these functions allows for a more general representation of functions compared to real-valued arguments. It also introduces the concept of complex conjugates, which is essential in proving the orthogonality relations.

Do these orthogonality relations only apply to exponential functions?

No, these relations can also be extended to other types of functions, such as trigonometric functions and Legendre polynomials. However, exponential functions are commonly used due to their simpler mathematical properties.

How do these orthogonality relations relate to the concept of orthogonality in linear algebra?

In linear algebra, orthogonality refers to the perpendicularity of two vectors. Similarly, in the context of functions, the orthogonality relations describe how two functions are perpendicular or orthogonal to each other. This concept is also closely related to the inner product and the dot product in linear algebra.

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