a) Sketch:

π−ππ−πtf(t)Open image in a new page

Graph of `f(t)`, a sawtooth curve.

b) First, we need to find the Fourier coefficients `a_0`, `a_n` and `a_b`.


`=1/piint_(-pi)^pit\ dt`




Next, we use a result from the Table of Integrals:

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

We have:

`a_n=1/pi int_(-pi)^pif(t)cos nt\ dt`

`=1/pi int_(-pi)^pit\ cos nt\ dt`

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

`=1/(pin^2)[(cos n pi+0)` `{:-(cos(-n pi)+0)]`

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


To find `b_n`, once again we use a result from our Table of Integrals

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

We have:

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

`=1/piint_(-pi)^pit\ sin nt\ dt`

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

`=1/(n^2pi)([0-n pi\ cos n pi]` `{:-[0+n pi\ cos(-n pi)])`

`=(n pi)/(n^2pi)(-2\ cos n pi)`

`=-2/ncos npi`



Now to put it all together for the Fourier Series for our function:

`f(t)=a_0/2` `+sum_(n=1)^oo(a_n\ cos nt` `{:+b_n\ sin nt)`

`=0/2` `+sum_(n=1)^oo((0)\ cos nt` `{:+2/n(-1)^(n+1)sin nt)`

`=sum_(n=1)^oo(2/n(-1)^(n+1)sin nt)`

`=2\ sin t` `-sin 2t` `+2/3sin 3t-1/2sin 4t+...`

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