a) Sketch:

(This is the same function we saw in the above Illustration example.)

π−ππtf(t)Open image in a new page

Graph of `f(t)`, a triangular waveform.

b) Since the function is even, we have bn = 0.

In this example, `L=pi`.

We have:

`a_0=2/Lint_0^Lf(t)dt`

`=2/piint_0^pit\ dt`

`=2/pi[t^2/2]_0^pi`

`=2/pi(pi^2)/2`

`=pi`

To find an, we use a result from before (see Table of Common Integrals):

`intt\ cos nt\ dt` `=1/n^2(cos nt+nt\ sin nt)`

We have:

`a_n=2/Lint_0^Lf(t)cos{:(n pi t)/L:}dt`

`=2/piint_0^pi t\ cos nt\ dt`

`=2/pi[1/n^2(cos nt+nt\ sin nt)]_0^pi`

`=2/(pi n^2)[(cos n pi+0)-(cos 0+0)]`

`=2/(pi n^2)[(cos n pi-1)]`

`=2/(pi n^2)[(-1)^n-1]`

When n is odd, the last line gives us `-4/(pin^2`.

When n is even, the last line equals `0`.

For the series, we need to generate odd values for n. We need to use `(2n - 1)` for `n = 1, 2, 3,...`.

So we have:

`f(t)=a_0/2+sum_(n=1)^oo a_n cos{:(n pi t)/L:}`

`=pi/2-4/pi sum_(n=1)^oo(cos(2n-1)t)/((2n-1)^2)`

`=pi/2` `-4/pi(cos t+1/9cos 3t` `{:+1/25cos 5t+...)`

Check: We graph different terms of the above expression to make sure our answer is correct. Here is the first term involving the cosine expression:

π−ππtf(t)Open image in a new page

Graph of `f(t)~~pi/2-4/pi cos(t)`.

Now add another term:

π−ππtf(t)Open image in a new page

Graph of `f(t)` `~~pi/2-4/pi (cos(t)` `{: + 1/9 cos(3t))`.

π−ππtf(t)Open image in a new page

Graph of `f(t)` `~~pi/2-4/pi (cos(t)` ` + 1/9 cos(3t)` `{:+1/25 cos(5t))`.

We can see our series is rapidly converging to our original triangular function.