Example 1

Recall that

This converges to f(x) where f(x) is continuous and converges to
1/2 [f(x+)+f(x-)] where f(x) has a jump discontinuity.

Remember, with Legendre polynomials p(x)=1, the eigenfunctions are the Legendre polynomials, and the limits of integration are -1 and 1.

Recall from what we saw above that the denominator in the expression for cn is 2/(2n+1).

In[15]:=

  f[x_]:=Cos[Pi x/2]

In[16]:=

  fplot=Plot[f[x],{x,-3,3}];

In[17]:=

  Clear[c0,c,Sn]
  c0=1/2 Integrate[f[x]LegendreP[0,x],{x,-1,1}];
  Simplify[c0]
  c=Table[(2k+1)/2 Integrate[f[x]LegendreP[k,x],{x,-1,1}],{k,1,6}];
  Simplify[c]

Out[17]=

  2
  --
  Pi

Out[18]=

                  2                       2     4
      10 (-12 + Pi )     18 (1680 - 180 Pi  + Pi )
  {0, --------------, 0, -------------------------, 0, 
             3                        5
           Pi                       Pi
   
                          2         4     6
    26 (-665280 + 75600 Pi  - 840 Pi  + Pi )
    ----------------------------------------}
                        7
                      Pi

Why all the zeroes at odd k?

Why all the zeroes at odd k? ...

In[19]:=

  Sn=Table[c0 LegendreP[0,x]+Sum[c[[k]]LegendreP[k,x],{k,1,n}],{n,1,6}];

Let's see how we did.

In[20]:=

  sn1=Plot[Sn[[1]],{x,-3,3},DisplayFunction->Identity];
  Show[fplot,sn1,DisplayFunction->$DisplayFunction,
  PlotLabel->FontForm["c0P0+c1P1",{"Helvetica-Bold",12}]];

In[21]:=

  sn3=Plot[Sn[[3]],{x,-3,3},DisplayFunction->Identity];
  Show[fplot,sn3,PlotLabel->
  FontForm["c0P0+c1P1+c2P2+c3P3",{"Helvetica-Bold",12}]];

In[22]:=

  sn6=Plot[Sn[[6]],{x,-3,3},DisplayFunction->Identity];
  Show[fplot,sn6, PlotLabel->
  FontForm["Sum(n=0 to 6) cnPn",{"Helvetica-Bold",12}]];

Up to Legendre Polynomials