First, we need to define the function after observing the graph:

`f(t)={(-3, ,-pi {:<=:}t <0),(3, ,0 {:<=:}t <pi) :}`

We can see from the graph that it is periodic, with period `2pi`.

So `f(t) = f(t + 2π)`.

Also, `L=pi`.

We can also see that it is an odd function, so we know `a_0= 0` and `a_n= 0`. So we will only need to find bn.

Since `L=pi`, the necessary formulae become:

`b_n=1/piint_(-pi)^pif(t) sin nt\ dt`

`f(t)=sum_(n=1)^oob_n sin nt`

Now

`b_n=1/pi int_(-pi)^pif(t)sin nt\ dt`

`=1/pi(int_(-pi)^0 -3\ sin nt\ dt` `{:+int_0^pi 3\ sin nt\ dt)`

`=3/pi([(cos nt)/n]_(-pi)^0+[-(cos nt)/n]_0^pi)`

`=3/(pi n)(cos 0-cos(-pi n)` `-cos n pi` `{:+cos 0)`

`=3/(pi n)(2-cos pi n-cos pi n)`

`=6/(pi n)(1-cos pi n)`

`=12/(pi n)\ (n\ "odd") or`

`=0\ (n\ "even")`

We could write this as: `b_n=12/(pi(2n-1))\ \ n=1,2,3...` (Substitute `n=1,2,3...` to see how this works.)

So the Fourier series for our odd function is given by:

`f(t)=sum_(n=1)^oo b_n\ sin nt `

`=sum_(n=1)^oo 12/(pi(2n-1))sin(2n-1)t`

`=12/pisum_(n=1)^oo (sin(2n-1)t)/((2n-1)) `

NOTE: Since bn is non-zero for n odd, we must also have odd multiples of t within the sine expression (the even ones are multiplied by `0`, so will be `0`).

Checking, we graph the first 5 terms. It should closely resemble the square wave we started with.

`12/pi sum_(n=1)^5 (sin(2n-1)t)/((2n-1))`

`=12/pi(sin t` `+1/3 sin 3t` `+1/5 sin 5t` `+1/7 sin 7t` `{:+1/9 sin 9t)`

π−π-2π1234-1-2-3-4tf(t)Open image in a new page

Graph of `f(t)`, the Fourier series approximation of a square wave.

We see that the graph of the first 5 terms is certainly approaching the shape of the graph that was in the question. We can be confident we have the correct answer.

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