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Logarythmic
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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?
Looks to me like -1/2 to +1/2 will do nicely.Logarythmic said:What limits should I use for integration?
Can you show an example?Logarythmic said:I don't get it. The integral is never zero with m different from n.
Factor an [tex]e^{\pi i (m-n)}[/tex] out of your I. Does that help?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?
OK. This is it.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]?
I think the factoring step I did will work for any interval. Give it a try.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 can't speak for everyone, but it works for me.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.
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.
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.
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.
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.
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.