**Problem:** For the following function find its Fourier series, its
sine Fourier series and its cosine Fourier series; for each series,
determine its sum.

**Solution:**
This seems to be a standard problem, so we use the
usual procedure.

**Fourier series:**

When we extend the given function periodically, its period will be
*T* = 2.

Thus we get

that is,

To find the sum of this series we use
Jordan's conditions. First
we draw the periodic extension of the given function *f* and then at all
points of discontinuity we place dots at levels that average left and right
limits there.

**Sine Fourier series:**

Actually, the Fourier series consists only of sines, so it is already a sine
series. Indeed, the periodic extension that we drew above was an odd
function, so this is automatic.

What would happen if we applied the standard procedure for obtaining sine Fourier series? We would eventually get the same answer.

We get the sine Fourier series by considering an extension of the given function
*f* to *L* = 2,*T* = 2*L* = 4*L*
in place of *T*, applied to the original interval, but with the new
*a*_{0} and
*a*_{k} are automatically zero and we only need to
evaluate *b*_{k}.

Thus we get

Note that the expression ^{k}*k* odd, so odd multiples actually disappear from this
series. For even *k*, this expression becomes equal to 2. Thus we can
rewrite this series just for even multiples by replacing *k* with
2*k*.

We obtained the same series as above. To determine its sum we would start by
drawing an odd periodic extension of *f*, but note that in the picture
above, the extension of *f* is already odd, so that picture still
applies.

**Cosine Fourier series:**

Now we apply the standard procedure to find the cosine Fourier series. We
get it by considering an extension of the given function *f* to
*L* = 2,*T* = 2*L* = 4*b*_{k} are automatically zero and we
need to evaluate *a*_{0} and *a*_{k}.

Thus we get

Note that the expression ^{k} - 1*k* even, so even multiples actually disappear from this
series. For odd *k*, this expression becomes equal to *k*
with 2*k* + 1.

To determine the sum of this series we start by drawing an even periodic
extension of *f*, then we should handle points of discontinuity, but
since there are none, we know that the cosine Fourier series converges to
this extension and this convergence is uniform on the real line.