f(x)=|x|

Let's look at f(x)=|x|. Remember, the functions must be periodic with period 2Pi, so what we really consider is the periodic extension f^(x)=|x| from -Pi to Pi, |x-2Pi| from Pi to 3Pi, etc., which is continuous.

Note: For this example, you will want to click once on the plots that you get and drag the little black boxes to increase the size of the graph.


  part4=Plot[Abs[x+2Pi],{x,-3Pi,-Pi},DisplayFunction->Identity];
  part5=Plot[Abs[x],{x,-Pi,Pi},DisplayFunction->Identity];
  part6=Plot[Abs[x-2Pi],{x,Pi,3Pi},DisplayFunction->Identity];
  absgraph=Show[part4,part5,part6,
  DisplayFunction->$DisplayFunction,AspectRatio->Automatic];

Mathematica does not like to integrate piecewise defined functions so we must break it down. We cannot use the innerproduct function if we do not integrate between -Pi and Pi.


  Clear[a0,a,b,Sn]
  a0=(1/(2Pi))(Integrate[-x,{x,-Pi,0}]+Integrate[x,{x,0,Pi}]);


  a=(1/Pi)Table[Integrate[-x u[x][[i]],{x,-Pi,0}]+
  Integrate[x u[x][[i]],{x,0,Pi}],{i,1,10}];


  b=(1/Pi)Table[Integrate[-x v[x][[i]],{x,-Pi,0}]+
  Integrate[x v[x][[i]],{x,0,Pi}],{i,1,10}];


  a0
  a

Can you see the pattern?

Can you see the pattern? ...


  b

Well?

Well? ...

Let's calculate a few partial sums.


  Sn=Table[a0+Sum[a[[k]]Cos[k x],{k,1,n}],{n,1,10}];

Let's see how we did.


  Clear[sn1,sn3,sn5,sn10]


  sn1=Plot[Sn[[1]],{x,-2 Pi,2 Pi},DisplayFunction->Identity];
  Show[absgraph,sn1,PlotLabel->
  FontForm["a0+a1Cos[x]",{"Helvetica-Bold",12}]];

Pretty good, even with only one term!


  sn3=Plot[Sn[[3]],{x,-2 Pi,2 Pi},DisplayFunction->Identity];
  Show[absgraph,sn3,PlotLabel->
  FontForm["a0+a1Cos[x]+a2Cos[2x]+a3Cos[3x]",{"Helvetica-Bold",12}]];


  sn5=Plot[Sn[[5]],{x,-2 Pi,2 Pi},DisplayFunction->Identity];
  Show[absgraph,sn5,PlotLabel->
  FontForm["a0+Sum(k=1 to 5) akCos[kx]",{"Helvetica-Bold",12}]];


  sn10=Plot[Sn[[10]],{x,-2 Pi,2 Pi},DisplayFunction->Identity];
  Show[absgraph,sn10,PlotLabel->
  FontForm["a0+Sum(k=1 to 10) akCos[kx]",{"Helvetica-Bold",12}]];

Pretty impressive!

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