# Fourier series coefficients in a not centered interval

• I
• Phys pilot
In summary: The only difference is the period, but the coefficients are calculated the same way. So for the given example of $f(x)=x+\frac{1}{2}$ on [0,1], the coefficients would be:$$a_0=\frac{1}{2}\int_{0}^{1}(x+\frac{1}{2})dx= \frac{1}{2}(\frac{x^2}{2}+\frac{x}{2})\bigg|_{0}^{1}=\frac{3}{8}$$$$a_n=\frac{1}{2}\int_{0}^{1}(x+\frac{1}{2})\cos{n\pi x}dx=\frac{1 Phys pilot TL;DR Summary what are the coefficients for a non centered in zero interval? Hello, so for a Fourier series in the interval [-L,L] with L=L and T=2L the coefficients are given by$$a_0=\frac{1}{L}\int_{-L}^Lf(t)dta_n=\frac{1}{L}\int_{-L}^Lf(t)\cos{\frac{n\pi t}{L}}dtb_n=\frac{1}{L}\int_{-L}^Lf(t)\sin{\frac{n\pi t}{L}}dt$$But if we have an interval like [0,L] with L=L and T=2L (I'm not sure about that), how do we get the coefficients? like this?$$a_0=\frac{2}{L}\int_{0}^Lf(t)dta_n=\frac{2}{L}\int_{0}^Lf(t)\cos{\frac{n\pi t}{L}}dtb_n=\frac{2}{L}\int_{0}^Lf(t)\sin{\frac{n\pi t}{L}}dt$$Edit: Maybe this is the correct way. [0, 2L] so:$$a_0=\frac{1}{L}\int_{0}^{2L}f(t)dta_n=\frac{1}{L}\int_{0}^{2L}f(t)\cos{\frac{n\pi t}{L}}dtb_n=\frac{1}{L}\int_{0}^{2L}f(t)\sin{\frac{n\pi t}{L}}dt$$If that is rifgt, is this right for this example?$$f(x)=x+\frac{1}{2}\,\, \text{from 0 to 1/2 and from 1/2 to 1 is zero}$$so 2L=1, L=1/2$$a_0=2\int_{0}^{1}f(t)dta_n=2\int_{0}^{1}f(t)\cos{\frac{n\pi t}{L}}dtb_n=2\int_{0}^{1}f(t)\sin{\frac{n\pi t}{L}}dt

Last edited:
It shouldn't make a difference if you define your periodic function on $[-L,L]$ or $[0,2L]$.

## 1. How are Fourier series coefficients calculated for a non-centered interval?

The coefficients for a Fourier series in a non-centered interval can be calculated using the formula:
c_n = (1/L) ∫ f(x) e^(-i2πnx/L) dx,
where L is the length of the interval and f(x) is the periodic function.

## 2. Can Fourier series coefficients be calculated for any interval?

Yes, Fourier series coefficients can be calculated for any interval as long as the function is periodic on that interval.

## 3. How does the choice of interval affect the Fourier series coefficients?

The choice of interval can affect the Fourier series coefficients as it determines the period of the function and the range of values that the coefficients can take. A larger interval may result in a larger range of coefficients, while a smaller interval may result in a smaller range of coefficients.

## 4. Are there any specific techniques for calculating Fourier series coefficients in a non-centered interval?

There are various techniques for calculating Fourier series coefficients in a non-centered interval, such as the use of trigonometric identities, integration by parts, and complex analysis methods. The choice of technique may depend on the complexity of the function and the desired level of accuracy.

## 5. How do Fourier series coefficients in a non-centered interval differ from those in a centered interval?

The main difference between Fourier series coefficients in a non-centered interval and a centered interval is the range of values that the coefficients can take. In a non-centered interval, the coefficients may have a wider range of values, while in a centered interval, the coefficients are typically symmetric and have a smaller range of values.

Replies
2
Views
1K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
7
Views
499
Replies
4
Views
804
Replies
2
Views
870
Replies
4
Views
912
Replies
8
Views
2K