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{SECT 0 {PARA 18 "" 0 "" {TEXT -1 37 "Linear Methods of Applied Mathem
atics" }}{PARA 256 "" 0 "" {TEXT -1 44 "Examples of Calculations with \+
Fourier Series" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 ""
0 "" {TEXT -1 62 "Copyright 1998, 2000 by Evans M. Harrell II and Jame
s V. Herod" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 45 "Section 1: Calcula
tions with the complex form" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 253 " We illustrate the calculations with the c
omplex form of the Fourier sereis. This form is completely equivalent \+
to the real series, with both sines and cosines. It may be instructive
to set up the calculations with the real functions for comparison." }
}{PARA 0 "" 0 "" {TEXT -1 101 " The complex Fourier coefficients f
or a function fo on an interval (a,b) are given by the formula" }}
{PARA 0 "" 0 "" {TEXT -1 18 " " }{XPPEDIT 18 0 "c(k,f
,a,b)=int(f(x)*exp(-2*Pi*I*k*x/(b-a)),x=a..b)/(b-a)" "6#/-%\"cG6&%\"kG
%\"fG%\"aG%\"bG*&-%$intG6$*&-F(6#%\"xG\"\"\"-%$expG6#,$*.\"\"#F3%#PiGF
3%\"IGF3F'F3F2F3,&F*F3F)!\"\"F=F=F3/F2;F)F*F3,&F*F3F)F=F=" }{TEXT -1
1 "." }}{PARA 0 "" 0 "" {TEXT -1 20 "We try computations." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "c:=(k,f,a,b)->int(f(x)*exp(-2*Pi*I*
k*x/(b-a)),x=a..b)/(b-a);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
18 "f:=1;\nc(k,f,0,Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "
normal(\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "convert(\",t
rig);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "c(0,f,0,Pi);\nc(1,
f,0,Pi);\nc(2,f,0,Pi);\nc(3,f,0,Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 30 "
We try a second example. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 12 "f:=x->x-x^3;" }}}{PARA 0 "" 0 "" {TEXT -1 81 " We compute th
e general coefficient with the assumption that k is an integer." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "c(k,f,-1,1);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(\");" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 35 "assume(k,integer);\nconvert(\",trig);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "k:='k';" }}}{PARA 0 "" 0 ""
{TEXT -1 126 " It seems appropriate to draw a graph. It is best to
convert these exponential to real quantities -- to sines and cosines.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "convert(sum('c(pp,f,-1,1
)*exp(2*Pi*I*pp*x/2)','pp'=-4..4),trig):\napprox:=unapply(combine(\"),
x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "plot([f(x),approx(x)
],x=-1..1);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 41 "Section 2: Differ
entiating Fourier Series" }}{PARA 0 "" 0 "" {TEXT -1 4 " " }}{PARA
0 "" 0 "" {TEXT -1 101 " We perform an experiment examining what h
apens if we dirrerentiate. If, as in Section 1, we take" }}{PARA 0 ""
0 "" {TEXT -1 45 " " }
{XPPEDIT 18 0 "f(x) = x - x^3" "6#/-%\"fG6#%\"xG,&F'\"\"\"*$F'\"\"$!\"
\"" }}{PARA 0 "" 0 "" {TEXT -1 4 "then" }}{PARA 0 "" 0 "" {TEXT -1 44
" " }{XPPEDIT 18 0 "diff(f(
x),x)= 1-3*x^2" "6#/-%%diffG6$-%\"fG6#%\"xGF*,&\"\"\"\"\"\"*&\"\"$F-*$
F*\"\"#F-!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 148 "The se
ries can also be differentiated. We compare the graph of the different
iated series with the graph of f'(x) out side of the region from [-1,1
]." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "diff(approx(x),x):\nDs
eries:=unapply(\",x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "pl
ot([1-3*x^2,Dseries(x)],x=-Pi..Pi);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 255 " We seem to get the same r
esult whether we differentiate the function and then calculate the Fou
rier series or differentiate the Fourier series of the original functi
on. Ordinarily these are in fact the same. Let us take a closer look
at this issue." }}{PARA 0 "" 0 "" {TEXT -1 793 " In Chapter III, \+
we examined the nature of the convergence of series. Since we are goin
g to be using representations of functions by their associated Fourier
series to solve differential equations, we should pause a bit to ask \+
if the derivative of a Fourier series representing f is a series which
will represent the derivative of f. To remember that there is somethi
ng to be said here, recall that differentiation is a limiting process,
as is integration. Also, forming an infinite series from the finite s
ums is a limiting process. So, to ask that the derivative on an infini
te sum is the sum of the derivatives is to ask if the limits can be in
terchanged. To be sure that the student recalls that there could be a \+
problem with interchanging limits, think of the following type example
. " }}{PARA 0 "" 0 "" {TEXT -1 37 " Think of a sequence of function
s " }}{PARA 0 "" 0 "" {TEXT -1 8 " " }{XPPEDIT 18 0 "S[n](x);"
"6#-&%\"SG6#%\"nG6#%\"xG" }{TEXT -1 40 " = Heaviside(x-1/n) - Heavisid
e(x+1/n). " }}{PARA 0 "" 0 "" {TEXT -1 60 "Imagine the graph of the fi
rst three terms of this sequence." }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64
"plot([seq(Heaviside(x+1/n) - Heaviside(x-1/n),n=1..3)],x=-2..2);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 45 "Choose an n. Compute the limit as x -> 0 for " }{XPPEDIT 18 0 "
S[n](x);" "6#-&%\"SG6#%\"nG6#%\"xG" }{TEXT -1 80 " . This limit is one
. Thus, the limit as n-> infinity of the limit as x -> 0 of " }
{XPPEDIT 18 0 "S[n](x);" "6#-&%\"SG6#%\"nG6#%\"xG" }{TEXT -1 8 " is on
e." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "n:=5;\nlimit(Heaviside
(x+1/n) - Heaviside(x-1/n),x=0);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 140 "Now we do it the ot
her way. Pick an x different from zero. Consider the limit as n -> inf
inity. For example, take n so large that 1/n < |x|." }}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 52 "x:=1/10;\nn:=11;\nHeaviside(x+1/n) - Heavis
ide(x-1/n);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 155 "It follows that \+
the limit as x->0 of the limit as n-> infinity is zero. This is not th
e same answer we got when the limits were taken in the reverse order.
" }}{PARA 0 "" 0 "" {TEXT -1 240 " We see that the limits cannot b
e interchanged. It matters whether we take the limit as x -> 0 first, \+
or as n -> infinity first. Perhaps this example makes the statement be
lievable that some care must be taken when interchanging limits." }}
{PARA 0 "" 0 "" {TEXT -1 25 " Here is the theorem." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 8 "THEOREM." }{TEXT -1
163 " If f(x) is periodic, continuous, and sectionally smooth, then th
e differentiated Fourier series of f(x) converges to f '(x) at every p
oint s where f ''(x) exists." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 38 "Example: Consider the function f(x) = " }
{XPPEDIT 18 0 "x-x^3;" "6#,&%\"xG\"\"\"*$F$\"\"$!\"\"" }{TEXT -1 152 "
on the interval [-1, 1]. This is an odd function. There will be no co
sine terms in its Fourier series. By now you know how to compute the c
oefficients." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "f:=x->x-x^3;
\nassume(n,integer);\nint((x-x^3)*sin(n*Pi*x),x=-1..1);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 110 " \+
To get an understanding for how the series approximates f(x), we p
lot three terms and compare the graphs." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 67 "for n from 1 to 3 do\n b[n]:=int((x-x^3)*sin(n*Pi*x
),x=-1..1);\nod;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "n:='n':
\nplot([[x,f(x),x=-1..1],[x,sum(b[n]*sin(n*Pi*x),n=1..3),x=-2..2]],x=-
2..2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "The graph of f(x) and \+
the truncated series match closely. " }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 41 "We ask: what happens if we differentiat
e?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "diff(f(x),x);\ndiff(su
m(b[n]*sin(n*Pi*x),n=1..3),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 86 "plot([[x,diff(f(x),x),x=-1..1],\n [x,diff(sum(b[n]*sin(n*Pi*
x),n=1..3),x),x=-2..2]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0
"" }}}{PARA 0 "" 0 "" {TEXT -1 270 "Perhaps this experimental evidence
gives, first, some understanding for the nature of the results of the
previous theorem and, second, make it beliveable. Do remember that ou
tside the interval [-1, 1], we have that the series converges to the p
eriodic extension of f(x)." }}{PARA 0 "" 0 "" {TEXT -1 28 " What a
bout integration?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT 259 8 "THEOREM:" }{TEXT -1 114 " if f(x) is periodic and section
ally continuous, then the Fourier series for f(x) may be integrated te
rm-by-term. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
258 8 "Example:" }{TEXT -1 29 " We take the antiderivatives." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "int(f(x),x);\nint(sum(b[n]*s
in(n*Pi*x),n=1..3),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "p
lot([[x,int(f(x),x),x=-1..1],\n [x,int(sum(b[n]*sin(n*Pi*x),n=1..3),
x),x=-2..2]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 254 "These two graphs are not the same! Not t
o worry. The graphs differ by a constant. But, you recall that there i
s always an arbitrary constant, called \"plus C\", floating around whe
n you take antiderivatives. Computing definite integrals should remove
this " }{TEXT 256 4 "faux" }{TEXT -1 9 " problem." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 35 "int(f(x),x=0..t):\nAf:=unapply(%,t);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "int(sum(b[n]*sin(n*Pi*x),n=1
..3),x=0..t):\nAs:=unapply(%,t);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 44 "plot([[t,Af(t),t=-1..1],[t,As(t),t=-2..2]]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 11 "Assignment:" }{TEXT -1 224 " L
et f(x) = x (x+1) (x-1) on the interval [-1, 1]. Graph f and five term
s of its Fourier series. Graph f '(x) and the derivative of five terms
of its Fourier series. Graph the definite integrals of f and terms of
the series." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 37 "Section 3. Integr
ating Fourier Series" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 76 " In this Section, we integrate a Fourier Series and
compare the results." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 12 "f:=x->x-x^3;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 95 "convert(sum('c(pp,f,-1,1)*exp(2*Pi*I*pp*x/2)','pp'=
-4..4),trig):\napprox:=unapply(combine(\"),x);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 12 "int(f(x),x);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 17 "int(approx(x),x);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 21 "plot([\"\",\"],x=-1..1);" }}}{PARA 0 "" 0 "" {TEXT
-1 248 " The two graphs do not match! But, how they differ shows w
hy they are different. It is the constant of integration for each. For
this example, the integral of the series seems to be the integral of \+
the function, up to a constant of integration." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "3 0 0" 62 }{VIEWOPTS 1 1 0 1 1
1803 }