Final answer:
The Fourier series expansion of f(x) = 1 results in a0 = 2 and all other Fourier coefficients an and bn equal to zero. Consequently, the Fourier series for this function is simply f(x) = 1 itself, as the integrals of sine and cosine over the symmetric interval are zero.
Step-by-step explanation:
The Fourier series expansion of a constant function f(x) = 1 can be determined using the formulas for the Fourier coefficients. For a function defined on an interval [-L, L], the Fourier series is given by:
F(x) = a0/2 + ∑ (an cos(nπx/L) + bn sin(nπx/L))
Where the Fourier coefficients are calculated as follows:
- a0 = (1/L) ∫-LL f(x) dx
- an = (1/L) ∫-LL f(x) cos(nπx/L) dx
- bn = (1/L) ∫-LL f(x) sin(nπx/L) dx
For f(x) = 1, the a0 coefficient is 2, and all other an and bn coefficients are 0 since the integrals of sine and cosine over a symmetric interval are zero.
Therefore, the Fourier series for f(x) = 1 is simply:
F(x) = 1