Differentiating Fourier series

This is an evaluated

*In[1]:=*

(c) Copyright 1994,1995 by Evans M. Harrell, II. All rights reserved

Differentiating Fourier series. Let's begin by reminding *Mathematica*
about

some identities we will use:

*In[2]:=*

R1 = {Cos[Pi n_] -> (-1)^n, Sin[Pi n_] -> 0} Clear[c] Clear[f]

In this notebook we will use the complex form of the Fourier series. As you

recall, the complex series is completely equivalent to the full real series,
with

both sines and cosines. You may find it instructive to set the calculations

up with the real functions for comparison. The complex Fourier coefficients

for a function f on an interval (a,b) are given by the formula:

*In[3]:=*

c[k_,f_,a_,b_] := (1/(b-a)) Integrate[f[x] Exp[-2 Pi I k x/(b-a)], {x,a,b}]

In[4]:=f[x_] := 1

In[5]:=c[k,f,0,Pi]

Out[5]=-I I -2 I k Pi -- - E 2 2 -- + ------------ k k ----------------- Pi

In[6]:=% /. k -> 2

Out[6]=0

In[7]:=%% /. k -> 3

Out[7]=0

In[8]:=c[0,f,0,Pi]

Out[8]=1Recall that % refers to the previous calculation, %% to the one before that,

etc. The phrase /. k -> 3 tellsMathematicato evaluate the expression with

the additional rule that k = 3.

So, we appear to be up and running. A question to ponder: What would

have happened if I had written %%% /. k -> 0 in the line above?

Let's now calculate some more series. Suppose that

In[9]:=f[x_] := x - x^3

In[10]:=c[k,f,-1,1] /.{E^(I Pi k) -> (-1)^k,E^(-I Pi k) -> (-1)^k}

Out[10]=k 2 2 (-1) (6 - 6 I k Pi - 2 k Pi ) (-(-------------------------------) + 4 4 k Pi k 2 2 (-1) (6 + 6 I k Pi - 2 k Pi ) -------------------------------) / 2 4 4 k Pi

In[11]:=Simplify[%]

Out[11]=k 6 I (-1) --------- 3 3 k PiIf we want a plot, it will be best to use real quantities, so recall that this is

an odd function, with only sines in the usual Fourier series.

FSeries[x_, N_] = Sum[-12 (-1)^k Sin[ Pi k x ] /(k Pi)^3, {k, 1, N} ]

In[12]:=FSeries[x_, N_] = Sum[-12 (-1)^k Sin[ Pi k x ] /(k Pi)^3, {k, 1, N} ]

Out[12]=k -12 (-1) Sin[k Pi x] Sum[---------------------, {k, 1, N}] 3 3 k Pi

In[13]:=Plot[{f[x], FSeries[x,3]}, {x,-1,1}]

Out[13]=-Graphics-Wonderful! What happens if we differentiate? We get either the formula

1 - 3 x^2 or else the series

In[14]:=D[FSeries[x,N],x]

Out[14]=k -12 (-1) Cos[k Pi x] Sum[---------------------, {k, 1, N}] 2 2 k Pi

In[15]:=%/. N -> 4

Out[15]=12 Cos[Pi x] 3 Cos[2 Pi x] 4 Cos[3 Pi x] ------------ - ------------- + ------------- - 2 2 2 Pi Pi 3 Pi 3 Cos[4 Pi x] ------------- 2 4 Pi

In[16]:=Plot[{1 - 3 x^2, %}, {x, -1,1}]

Out[16]=-Graphics-Super! Just for fun, let's see the comparison outside the interval where we

cut off the polynom