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- 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

$$a_0=\frac{1}{L}\int_{-L}^Lf(t)dt$$

$$a_n=\frac{1}{L}\int_{-L}^Lf(t)\cos{\frac{n\pi t}{L}}dt$$

$$b_n=\frac{1}{L}\int_{-L}^Lf(t)\sin{\frac{n\pi t}{L}}dt$$

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?

$$a_0=\frac{2}{L}\int_{0}^Lf(t)dt$$

$$a_n=\frac{2}{L}\int_{0}^Lf(t)\cos{\frac{n\pi t}{L}}dt$$

$$b_n=\frac{2}{L}\int_{0}^Lf(t)\sin{\frac{n\pi t}{L}}dt$$

Edit: Maybe this is the correct way. [0, 2L]

so:

$$a_0=\frac{1}{L}\int_{0}^{2L}f(t)dt$$

$$a_n=\frac{1}{L}\int_{0}^{2L}f(t)\cos{\frac{n\pi t}{L}}dt$$

$$b_n=\frac{1}{L}\int_{0}^{2L}f(t)\sin{\frac{n\pi t}{L}}dt$$

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

$$a_0=2\int_{0}^{1}f(t)dt$$

$$a_n=2\int_{0}^{1}f(t)\cos{\frac{n\pi t}{L}}dt$$

$$b_n=2\int_{0}^{1}f(t)\sin{\frac{n\pi t}{L}}dt$$

$$a_0=\frac{1}{L}\int_{-L}^Lf(t)dt$$

$$a_n=\frac{1}{L}\int_{-L}^Lf(t)\cos{\frac{n\pi t}{L}}dt$$

$$b_n=\frac{1}{L}\int_{-L}^Lf(t)\sin{\frac{n\pi t}{L}}dt$$

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?

$$a_0=\frac{2}{L}\int_{0}^Lf(t)dt$$

$$a_n=\frac{2}{L}\int_{0}^Lf(t)\cos{\frac{n\pi t}{L}}dt$$

$$b_n=\frac{2}{L}\int_{0}^Lf(t)\sin{\frac{n\pi t}{L}}dt$$

Edit: Maybe this is the correct way. [0, 2L]

so:

$$a_0=\frac{1}{L}\int_{0}^{2L}f(t)dt$$

$$a_n=\frac{1}{L}\int_{0}^{2L}f(t)\cos{\frac{n\pi t}{L}}dt$$

$$b_n=\frac{1}{L}\int_{0}^{2L}f(t)\sin{\frac{n\pi t}{L}}dt$$

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

$$a_0=2\int_{0}^{1}f(t)dt$$

$$a_n=2\int_{0}^{1}f(t)\cos{\frac{n\pi t}{L}}dt$$

$$b_n=2\int_{0}^{1}f(t)\sin{\frac{n\pi t}{L}}dt$$

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