Find the Fourier series and comment the result

[Ezequiel]

Homework Statement

Let f(t) be defined on [-π,π] and by

f(t) = {
π² - t² if t ≠ 1/2ⁿ, n $\in$ N
t² if t = 1/2ⁿ, n $\in$ N
}

Find its Fourier series F(t) and comment the result (type of series, type of convergence, F(t) = f(t)?, ...).

Homework Equations

The Attempt at a Solution

I don't even know where to start, so any help would be appreciated :)

 

[mighty2000]

Thank you for your question. I am happy to assist you with finding the Fourier series of f(t) and commenting on the results.

To begin, let's recall the definition of the Fourier series. The Fourier series of a periodic function f(t) with period 2π is given by:

F(t) = a0 + ∑[an cos(nt) + bn sin(nt)]

where

a0 = (1/2π) ∫f(t)dt
an = (1/π) ∫f(t)cos(nt)dt
bn = (1/π) ∫f(t)sin(nt)dt

Now, let's apply this definition to the given function f(t). We can see that f(t) is defined on the interval [-π,π] and it is periodic with period 2π. Therefore, we can write:

F(t) = a0 + ∑[an cos(nt) + bn sin(nt)] = a0 + ∑[an cos(nt) + bn sin(nt)] = a0 + ∑[an cos(nt) + bn sin(nt)]

where

a0 = (1/2π) ∫f(t)dt = (1/2π) ∫[-π,π]f(t)dt = (1/2π) ∫[-π,π] (π2 - t2)dt = (1/2π) (π2t - (t3/3)) |[-π,π] = 0

an = (1/π) ∫f(t)cos(nt)dt = (1/π) ∫[-π,π] f(t)cos(nt)dt = (1/π) ∫[-π,π] (π2 - t2)cos(nt)dt = (1/π) (π2sin(nt)/n - (t3sin(nt)/3)) |[-π,π] = 0

bn = (1/π) ∫f(t)sin(nt)dt = (1/π) ∫[-π,π] f(t)sin(nt)dt = (1/π) ∫[-π,π] (π2 - t2)sin(nt)dt = (1/π) (π2cos(nt)/n - (t3cos(nt)/3)) |[-π,