Fourier expansion between two different intervals

[LASmith]

Homework Statement

f(x) = x+1 for -1,x<0
x-1 for 0<x<1
0 for x=0

expand it in an appropriate cosine or sine series

Homework Equations

f(x) = a₀/2 + \sum [a_ncos (n\pix/p) + b_n sin (n\pix/p)

a₀ = 1/p \intf(x).dx

a_n = 1/p \int f(x)cos (n\pix/p).dx

b_n = 1/p \int f(x)sin (n\pix/p).dx

The Attempt at a Solution

As there are two functions within this f(x), I am unsure of how to go ahead and do this.
I realize the overall function is odd, therefore would only need to expand the b_n part of the Fourier series, however the individual functions are not odd.
How would I go about setting this up?

 

[Herr Malus]

You'll need to break the integrals into two parts. The first will integrate x+1 from -1 to 0, the second will integrate x-1 from 0 to 1.

 

[HallsofIvy]

\int_{-1}^1 f(x)dx= \int_{-1}^0 f(x)dx+ \int_0^1 f(x)dx

 

[LASmith]

\int_{-1}^1 f(x)dx= \int_{-1}^0 f(x)dx+ \int_0^1 f(x)dx

Can this only be done the long winded way of calculating a₀, a_n and b_n for each equation, therefore doing 6 integrals? Or will some cancel for being odd/even functions, which I cannot see at first glance?

Also would the value of 'p' in my 'relevant equations' be -1 for x+1 and 1 for x-1?

 

[vela]

Can this only be done the long winded way of calculating a₀, a_n and b_n for each equation, therefore doing 6 integrals? Or will some cancel for being odd/even functions, which I cannot see at first glance?

In a sense, you always have to calculate a_n and b_n. It's just that if you have an even or odd function, you can sometimes use tricks to simplify or avoid lengthy calculations. When you have an odd integrand O(x) and a symmetric interval [-a,a], you know that\int_{-a}^a O(x)\,dx = 0(You should be able to prove this.) If you have an even integrand E(x), you can say\int_{-a}^a E(x)\,dx = 2\int_{0}^a E(x)\,dx(Again, you should be able to prove this.)

In this problem, you want to write down the integrals for a_n and b_n, take a look at the integrands, and see if either of the above apply. If they do, you can use them to simplify evaluating the integrals for a_n and b_n.

Also would the value of 'p' in my 'relevant equations' be -1 for x+1 and 1 for x-1?

You need to understand what the quantities that appear in a formula represent otherwise you're just guessing. So what does p represent in these formulas?

 

[LASmith]

You need to understand what the quantities that appear in a formula represent otherwise you're just guessing. So what does p represent in these formulas?

I thought p represented the value for which we integrate between, as all the examples so far have been where the limit is between p and -p, however, because these functions are between -1 and 0 & 0 and 1, I would presume it was the non-zero number, although I do not understand where the 1/p at the beginning of each integral comes from, so I guess that is where my miss-understanding occurs.

 

[vela]

The function f(x) defined on the interval [-1,1]. That's what determines what p is equal to. How f(x) is actually defined, in this case, piecewise, has nothing to do with p.

 

[LASmith]

The function f(x) defined on the interval [-1,1]. That's what determines what p is equal to. How f(x) is actually defined, in this case, piecewise, has nothing to do with p.

Therefore p will be 1 for both parts of the function, regardless of where the two parts of the function lie, between this interval?

 

[vela]

Yup. Take a₀ for example. You get
\begin{align*}
a_0 &= \frac{1}{p}\int_{-p}^p f(x)\,dx \\
&= \frac{1}{1}\int_{-1}^1 f(x)\,dx \\
&= \int_{-1}^1 f(x)\,dx
\end{align*}
because a single period is over [-1,1]. Now when you substitute in for f(x), you would split the integral up because of the two pieces to get
\begin{align*}
a_0 &= \int_{-1}^1 f(x)\,dx \\
&= \int_{-1}^0 f(x)\,dx + \int_{0}^1 f(x)\,dx \\
&= \int_{-1}^0 (x+1)\,dx + \int_{0}^1 (x-1)\,dx
\end{align*}
Note that this isn't necessarily the way you'd calculate a₀ since you're not taking advantage of the oddness of f(x). It was just to show that p doesn't depend on the fact that f(x) is made out of two or more pieces.