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{SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 37
"Module 10: Calculus on Fourier Series" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 801 " \+
In the previous module, we examined the nature of the convergence of \+
series. Since we are going to be using representations of functions by
their associated Fourier series to solve differential equations, we s
hould pause a bit to ask if the derivative of a Fourier series represe
nting f is a series which will represent the derivative of f. To remem
ber that there is something to be said here, recall that differentiati
on is a limiting process, as is integration. Also, forming an infinite
series from the finite sums is a limiting process. So, to ask that th
e derivative on an infinite sum is the sum of the derivatives is to as
k if the limits can be interchanged. To be sure that the student recal
ls that there could be a problem with interchanging limits, think of t
he following type example. " }}{PARA 0 "" 0 "" {TEXT -1 37 " Think \+
of a sequence of functions " }}{PARA 0 "" 0 "" {TEXT -1 8 " " }
{XPPEDIT 18 0 "S[n](x);" "6#-&%\"SG6#%\"nG6#%\"xG" }{TEXT -1 40 " = He
aviside(x-1/n) - Heaviside(x+1/n). " }}{PARA 0 "" 0 "" {TEXT -1 60 "Im
agine the graph of the first 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 " . T
his 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 one." }}{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 \+
other way. Pick an x different from zero. Consider the limit as n -> i
nfinity. 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) - Heav
iside(x-1/n);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 155 "It fol
lows that the limit as x->0 of the limit as n-> infinity is zero. This
is not the same answer we got when the limits were taken in the rever
se order." }}{PARA 0 "" 0 "" {TEXT -1 256 " We see that the limits
cannot be interchanged in that example. It matters whether we take th
e limit as x -> 0 first, or as n -> infinity first. Perhaps this examp
le makes the statement believable that some care must be taken when in
terchanging limits." }}{PARA 0 "" 0 "" {TEXT -1 25 " Here is the t
heorem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260
8 "THEOREM." }{TEXT -1 163 " If f(x) is periodic, continuous, and sect
ionally smooth, then the differentiated Fourier series of f(x) converg
es to f '(x) at every point x where f ''(x) exists." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "Example: Consider the f
unction 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 cosine terms in its Fourier series. By now you know h
ow to compute the coefficients." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 130 "n:='n':\nf:=x->x-x^3;\nassume(n,integer);\nInt((x-x^3)*sin(n*
Pi*x),x=-1..1)=\n int((x-x^3)*sin(n*Pi*x),x=-1..1);\nn:=
'n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 110 " To get an understanding for how the series approx
imates f(x), we plot 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: w
hat happens if we differentiate?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 51 "diff(f(x),x);\ndiff(sum(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 understa
nding for the nature of the results of the previous theorem and, secon
d, make it believable. Do remember that outside the interval [-1, 1], \+
we have that the series converges to the periodic extension of f(x)."
}}{PARA 0 "" 0 "" {TEXT -1 28 " What about integration?" }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 8 "THEOREM:" }
{TEXT -1 114 " if f(x) is periodic and sectionally continuous, then th
e Fourier series for f(x) may be integrated term-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]*sin(n*Pi*x),n=1..3),x);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "plot([[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 to worry. The graphs di
ffer by a constant. But, you recall that there is always an arbitrary \+
constant, called \"plus C\", floating around when you take antiderivat
ives. 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:=unappl
y(%,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 " Let f(x) = x (x+1) (x-1) on the inter
val [-1, 1]. Graph f and five terms of its Fourier series. Graph f '(x
) and the derivative of five terms of its Fourier series. Graph the de
finite integrals of f and terms of the series." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}}{MARK "1 0" 37 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
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