# Linear Methods of Applied Mathematics Differentiating Fourier series

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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]=  1

Recall that % refers to the previous calculation, %% to the one before that,

etc. The phrase /. k -> 3 tells Mathematica to 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  Pi

If 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```