Need Help with Fourier Series: sinh t & 1+ltl

[danai_pa]

I don't solve Fourier series of
1) sinh t :-1<t<1.
2) 1+ltl : -p<t<p
anyone please suggest to me. Thank you very much

 

[Tide]

What have you done so far? I suggest looking up the definition of Fourier series.

 

[quicknote]

Sure, I'd be happy to help with Fourier series. First, let's review what a Fourier series is. It is a way to represent a periodic function as a sum of sine and cosine functions. This can be useful in solving differential equations and understanding the behavior of a system.

For the first function, sinh t, we can use the definition of the hyperbolic sine function: sinh t = (e^t - e^-t)/2. Since this function is only defined for -1<t<1, we can extend it to be a periodic function with period 2 by repeating it outside of this interval. This means that the Fourier series for sinh t will only have odd terms, since the function is odd.

To find the Fourier coefficients, we can use the formula c_n = (1/T) * ∫f(t)sin(nωt)dt, where T is the period and ω is the angular frequency (2π/T). Plugging in the function and integrating over one period, we get c_n = 2/(nπ) * (1 - (-1)^n)/n. This simplifies to c_n = 4/(nπ^2) for odd values of n, and c_n = 0 for even values of n.

For the second function, 1+ltl, we can represent it as a piecewise function: f(t) = 1-t for -1<t<0 and f(t) = 1+t for 0<t<1. Again, we can extend this function to be periodic with period 2. This function is even, so the Fourier series will only have cosine terms.

Using the same formula as before, we can find the coefficients c_n = (1/T) * ∫f(t)cos(nωt)dt. For the first interval, this simplifies to c_n = 2/(nπ)^2 * (1 - (-1)^n)/n^2. For the second interval, we get c_n = 2/(nπ)^2 * (1 - (-1)^n)/n^2. This simplifies to c_n = 4/(nπ)^2 * (1 - (-1)^n)/n^2.

I hope this helps with your Fourier series calculations. Remember to always check your work and make sure your answer makes sense in the context of the problem. Good luck!