Even and Odd Functions
In[24]:=
Plot[Evaluate[Table[x^k,{k,0,10,2}]],{x,-3,3},
PlotRange->{-.5,10}, PlotLabel->
FontForm["x^k, k even",{"Helvetica-Bold",12}]];

In[25]:=
Plot[Evaluate[Table[x^k,{k,1,10,2}]],{x,-3,3},
PlotRange->{-10,10}, PlotLabel->
FontForm["x^k, k odd",{"Helvetica-Bold",12}]];

You probably can also recall from calculus that if f(x) has a Taylor series expansion about x0=0 containing only even/odd powers, then f(x) is an even/odd function.
In[26]:=
Series[Sin[x],{x,0,10}]
Out[26]=
3 5 7 9
x x x x 11
x - -- + --- - ---- + ------ + O[x]
6 120 5040 362880
In[27]:=
Series[Cos[x],{x,0,10}]
Out[27]=
2 4 6 8 10
x x x x x 11
1 - -- + -- - --- + ----- - ------- + O[x]
2 24 720 40320 3628800
By using the definitions of even and odd functions, you should convince
yourself of the following:
a. The sum of even (odd) functions is even (odd).
b. The product of two even or two odd functions is even.
c. The product of an even and an odd function is odd.
I'll prove the last one for you.
Let f be even and g be odd.
fg(-x)=f(-x)g(-x)=f(x)[-g(x)]=-f(x)g(x)=-fg(x)
You should also convince yourself (or recall from calculus) that the integral from -T to T of an odd function is zero and the integral from -T to T of an even function is twice the integral from 0 to T.
These two facts can greatly simplify the calculation of Fourier coeficients.