Fourier Analysis: Finding Coefficients for f(t)=1

[Whitebread]

Homework Statement

I need to find the Fourier coefficients for a function f(t)=1 projected onto trigonometric polynomials of infinite order

Homework Equations

Equation finding the first coefficient, the constant term:

The Attempt at a Solution

So I feel quite stupid because this should be a very simple integral. The integral of a monotonic function over an interval that is symmetric about the origin SHOULD be equal to zero.

It seems my book has just integrated from 0 to pi for first Fourier coefficient. I'm here pulling my hair out trying to figure out why/how this is correct because the two integrals certainly are not equivalent. Can someone help me out here?

 

[gabbagabbahey]

Shouldn't that equation be:

a_0=\frac{1}{\pi} \int_{-\pi}^{\pi} f(t)dt

which gives a_0=0? ...Or were you trying to say that the answer key gives a_0=\frac{1}{\sqrt{2}}?...Are you sure that you are looking for the first term a_0 and not the first non-zero term?

 

[Whitebread]

See, that's what I think it should be (and yes I'm looking for the first term a(sub 0)). But the answer key gives the answer I posted above. I'm beginning to think its wrong. Although, going by what we think the answer is, the transform of the function f(t)=1 would just be 0, which doesn't make sense.

 

[Whitebread]

I can post the entire answer key answer if you would like to see it

 

[gabbagabbahey]

Are you looking for the Fourier Transform of f(t), or the Fourier Series representation? Assuming you are looking for the latter; just because the first term in the series is zero, doesn't mean the entire series is zero.

 

[Whitebread]

I'm looking for the transform, not the series.

 

[gabbagabbahey]

Well the Fourier transform is \hat{f}(\omega)= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt which is not an infinite series with coefficients: this is why I assumed you were looking for the Fourier series ...your original question makes no sense to me if you are looking for the transform and not the series!

I'm pretty sure your looking for the series, not the transform...are you sure f(t)=1? If I were a betting man, I'd bet dollars to dimes that the question actually has f(t)=1(t), where 1(t) is the unit (Heaviside) step function, namely:

1(t)= \left\{ \begin{array}{rl} 0, & t<0 \\ 1, & t \geq 0

In which case, the equation

a_0=\frac{1}{\sqrt{2} \pi} \int_{-\pi}^{\pi} f(t)dt

should give you the correct series coefficient.