Question

Suppose ()f(t) is such that

∫−()=5,∫−ππf(t)dt=5, and ∫−()sin()=0,∫−ππf(t)sin⁡(nt)dt=0, for all n, and
∫−()cos()=0∫−ππf(t)cos⁡(nt)dt=0 for all n except =2n=2 where
∫−()cos(2)=10.∫−ππf(t)cos⁡(2t)dt=10.
What is ()f(t) for t between −−π and π.?

Answer 1

The function f(t) is expressed in terms of the Fourier series, with all sine components and all cosine components except for n=2 vanishing. The function is composed of a constant term and a single cosine term, making f(t) = α + β cos(2t), with α being the average value, and β derived from the given integral values.

Step-by-step explanation:

The question is seeking the function f(t) which satisfies the given set of integral conditions within the interval −π to π. These conditions suggest that f(t) could be expressed in terms of Fourier series, which is a way to represent a function as a sum of sines and cosines. The given integrals indicate that all sine components of f(t)'s Fourier series vanish and all cosine components vanish as well, except for the term with n=2. The integral involving cos(2t) being 10 suggests that there is a cosine term with a frequency of 2 and amplitude related to this value.

The integral conditions given are characteristic of Fourier coefficients, which are calculated over a period from −π to π in the case of f(t). The integral ∣−ππ f(t) dt = 5 delivers the average value or the a0 term in the Fourier series, which will be halved when considering the standard Fourier series formulation. Meanwhile, the nonzero integral ∣−ππ f(t) cos(2t) dt = 10 gives us the Fourier coefficient a2, which corresponds to the cosine term at n=2. Since all other coefficients for sines and cosines are zero, we can conclude that f(t) is a combination of a constant term and a single cosine term at n=2.

Therefore, the function f(t) is determined to be:

f(t) = α + β cos(2t),

where α is the average value (a0/2) and β relates to the amplitude of the cos(2t) term.