Calculating Fourier Coefficient of 2π-Periodic Function

In summary, the conversation was about finding the Fourier coefficient of a 2\pi-periodic function, and the formula \hat f\left( n \right) = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {f\left( x \right)e^{ - inx} dx}. There was a discussion about changing the limits of the integral and whether or not it affects the value of the integral. Ultimately, it was proven that the limits can be changed without changing the value of the integral due to the periodicity of the integrand.
  • #1
Zaare
54
0
I'm supposed to show
[tex]
\hat f\left( n \right) = - \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {f\left( {x + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}} \right)e^{ - inx} dx}
[/tex]
where [tex]\hat f\left( n \right)[/tex] is the Fourier coefficient and [tex]f(x)[/tex] is a [tex]2\pi[/tex]-periodic and Riemann integrable on [tex][\pi,-\pi][/tex].

This is what I've done:
I use the formulae for the Fourier coefficient of a [tex]2\pi[/tex]-periodic function
[tex]
\hat f\left( n \right) = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {f\left( x \right)e^{ - inx} dx}
[/tex]
and a simple change of variable
[tex]
\hat f\left( n \right) = \frac{1}{{2\pi }}\int\limits_{ - \pi + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}}^{\pi + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}} \right)e^{ - in\left( {x + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}} \right)} dx} = \frac{1}{{2\pi }}\int\limits_{ - \pi + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}}^{\pi + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}} \right)e^{ - inx} e^{ - i\pi } dx} = - \frac{1}{{2\pi }}\int\limits_{ - \pi + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}}^{\pi + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}} \right)e^{ - inx} dx}
[/tex].

Everything seems to agree except for the limits of the integral.
1) If I have done some mistake, I'd appreciate it someon would point it out.
2) If I haven't done any mistakes, what's the reasoning behind this? Don't the limits matter as longs as they are [tex]2\pi[/tex] apart?
 
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  • #2
You'll have to prove that

[tex] \int_{-\pi+\frac{\pi}{n}}^{-\pi} f\left(x+\frac{\pi}{n}\right) \ dx =-\int_{\pi}^{\pi+\frac{\pi}{n}} f\left(x+\frac{\pi}{n}\right) \ dx [/tex]

Daniel.
 
  • #3
Ok, here's my reasoning.
The integrand

[tex]
f(x+\frac{\pi}{n})e^{-inx}
[/tex]

is [tex]2\pi[/tex]-periodic. Hence a change from [tex]x+\frac{\pi}{n}[/tex] to [tex]x+\frac{\pi}{n}+2\pi[/tex] leaves the integrand unchanged and I get

[tex]
\int\limits_{ - \pi }^{ - \pi + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}} \right)e^{ - inx} dx} = \int\limits_\pi ^{\pi + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/
{\vphantom {\pi n}} \right.
\kern-\nulldelimiterspace} n}} \right)e^{ - inx} dx}
[/tex]

which allows me to change the limits of the original integral to [tex]-\pi[/tex] and [tex]\pi[/tex] without changing its value.
Am I right?
 
Last edited:

Related to Calculating Fourier Coefficient of 2π-Periodic Function

What is a 2π-periodic function?

A 2π-periodic function is a mathematical function that repeats itself every 2π units. This means that for any value of x, the function will have the same value at x+2π, x+4π, x+6π, and so on.

What is the Fourier series of a 2π-periodic function?

The Fourier series of a 2π-periodic function is a way of representing the function as a sum of sine and cosine functions with different frequencies and amplitudes. This series allows us to break down a complex function into simpler terms to better understand its behavior.

How do you calculate the Fourier coefficients of a 2π-periodic function?

To calculate the Fourier coefficients of a 2π-periodic function, you need to use the Fourier series formula, which involves integrating the function over one period (from 0 to 2π) and multiplying it by sine or cosine functions with different frequencies. This process is repeated for each coefficient, and the resulting values determine the amplitudes of the sine and cosine terms in the Fourier series.

What is the significance of the Fourier coefficients of a 2π-periodic function?

The Fourier coefficients of a 2π-periodic function represent the amplitudes of the sine and cosine terms in the Fourier series. They provide important information about the behavior of the function, such as its periodicity, symmetry, and smoothness. By analyzing these coefficients, we can better understand the properties of the original function.

Can the Fourier coefficients of a 2π-periodic function be calculated numerically?

Yes, the Fourier coefficients of a 2π-periodic function can be calculated numerically using numerical integration methods. This involves approximating the integral in the Fourier series formula using numerical techniques, such as the trapezoidal rule or Simpson's rule. This approach can be useful for complex functions that cannot be integrated analytically.

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