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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 54 "Linear Algebra, Infinite Dimens
ional Spaces, and Maple" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 23 "Jim Herod, Georgia Tech" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 11 "S
ection 5: " }{TEXT -1 11 "Convergence" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 144 "     Graphs of the first four functi
ons in a sequence that converges pointwise, but not uniformly are made
 with MAPLE in a rather intuitive way." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "f:=(n,x)->n*x*exp(-n*x);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "plot([seq(f(n,x),n=1..4
)],x=0..6);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 50 "     The space C[-1,1] of continuous functions in " }{XPPEDIT 
18 0 "L^2" "6#*$%\"LG\"\"#" }{TEXT -1 190 "[-1,1] is an example of an \+
innerproduct space that is not complete. When asked to show it is not \+
complete, a novice might draw a sequence of functions in a manner sugg
ested with this syntax:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "g:=(n,x)->piecewise(x<=-1/n,0,-1/n \+
< x and x <= 0,1+n*x,\n          0<x and x<=1/n,1-n*x,0);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "plot(\{g(2,x),g(3,x),g(4,x)\},x=-1.
.1);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 137 
"     The sequence g(1,x), g(2,x), g(3,x), ... does appear to be conve
rging pointwise. We ask: Is this sequence converging in the norm of " 
}{XPPEDIT 18 0 "L^2" "6#*$%\"LG\"\"#" }{TEXT -1 13 "?  The space " }
{XPPEDIT 18 0 "L^2" "6#*$%\"LG\"\"#" }{TEXT -1 153 " is more complicat
ed than suggested by the last graph. TO see this, extablish that the s
equence suggested by the picture converges to zero by seeing that" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 11 "        \+
   " }{XPPEDIT 18 0 "int((g(n,x)-0)^2,x=-1..1)" "6#-%$intG6$*$,&-%\"gG
6$%\"nG%\"xG\"\"\"\"\"!!\"\"\"\"#/F,;,$\"\"\"F/\"\"\"" }{TEXT -1 3 " =
 " }{XPPEDIT 18 0 "2*int((1-n*x)^2,x=0..1/n)" "6#*&\"\"#\"\"\"-%$intG6
$*$,&\"\"\"F%*&%\"nGF%%\"xGF%!\"\"\"\"#/F.;\"\"!*&\"\"\"F%F-F/F%" }
{TEXT -1 3 " = " }{XPPEDIT 18 0 "2/(3*n)" "6#*&\"\"#\"\"\"*&\"\"$F%%\"
nGF%!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "2*int((1-n*x)^2,x=0..1/n);" }}}
{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 35 "This int
egral goes to zero as n -> " }{XPPEDIT 18 0 "infinity" "6#%)infinityG
" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 129 "     How does one ma
ke an example of a sequence in C[-1, 1] that converges but has a limit
 not in C[-1, 1]? Here is a suggestion." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 48 "h:=x->piecewise( x <= 0,-Pi*(1+x)/2,Pi*(1-x)/2);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "lim:=x->sum(sin(p*Pi*x)/p,p=
1..n);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "n:=5;\nplot([h(x)
,lim(x)],x=-1..1,discont=true,color=BLACK);" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 139 "n:='n';\nint(h(x)^2,x=0..1)\n       -2*sum(int(Pi
*(1-x)*sin(p*Pi*x)/(2*p),x=0..1),p=1..n)\n       +sum(int(( sin(p*Pi*x
)/p)^2,x=0..1),p=1..n);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "
subs(sin(p*Pi)=0,%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "lim
it(%,n=infinity);" }}}{PARA 0 "" 0 "" {TEXT -1 80 "     The limit is z
ero, and we have that the sequence has limit not in C[-1, 1]." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT 258 11 "Section 6: " }{TEXT -1 43 "Orthogonality and C
losest Point Projections" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 314 "     When one thinks of the graph of cos(¹ x/2) o
n the interval [-1,1], one is struck by the resemblance of this graph \+
to that of a quadratic, turned down and translated up, over the same i
nterval. It is curious to think about how close one might approximate \+
this transcandental function with a quadratic function." }}{PARA 0 "" 
0 "" {TEXT -1 72 "     The first think to think of is the Taylor Polyn
omial of degree two." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 27 "taylor(cos(Pi*x/2),x=0, 3);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "P:=unapply(convert(%,polynom),x);" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "plot(\{cos(Pi*x/2),P(x)\}
,x=-1..1);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 256 "     An alternate quadratic approximation that is more apropri
ate to this section is to use the polynomials of Problem 6.12. These a
re called the Legendre polynomials. Maple knows these polyomials, and \+
others. Here are the first three Legendre polynomials." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "with(or
thopoly);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "P(0,x); P(1,x)
; P(2,x); " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 57 "To get the best quadratic approximation for cos(¹x/2) in " }
{XPPEDIT 18 0 "L^2" "6#*$%\"LG\"\"#" }{TEXT -1 55 "[-1,1], we compute \+
the coefficients as in Problem 6.10." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 176 "a0:=int(cos(Pi*x/2)*P(0,x
),x=-1..1)/int(P(0,x)^2,x=-1..1);\na1:=int(cos(Pi*x/2)*P(1,x),x=-1..1)
/int(P(1,x)^2,x=-1..1);\na2:=int(cos(Pi*x/2)*P(2,x),x=-1..1)/int(P(2,x
)^2,x=-1..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "plot(\{cos
(Pi*x/2),a0*P(0,x)+a1*P(1,x)+a2*P(2,x)\},x=-1..1);" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 152 "     To further emphas
ize the importance of orthogonal polynomials, we make one more illustr
ation. Choose a function f: for the illustration, we choose " }
{XPPEDIT 18 0 "f(x)=1/(1+25*x^2)" "6#/-%\"fG6#%\"xG*&\"\"\"\"\"\",&\"
\"\"F**&\"#DF**$F'\"\"#F*F*!\"\"" }{TEXT -1 379 ".  Take n=1 points ch
osen evenly on the interval [-1, 1] and choose the point pairs \{x[i],
 y[i]\}, where y[i] = f(x[i]). Then, take the interpolating polynomial
 of degree n that fits the resulting n+1 point=pairs exactly. Do you t
hink this polynomial sequence might converge uniformly to f? The f we \+
have chosen provides a classical example that illustrates how wrong th
is can be." }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "f:=x->1/(1+25*x^2);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "for i from 1 to 9 do\n   s[i]:=-1+(
i-1)*2/8:\n   fs[i]:=f(s[i]):\nod:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 60 "fintrp:=x->interp([seq(s[i],i=1..9)],[seq(fs[i],i=1..
9)],x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "plot(\{f(x),fint
rp(x)\},x=-1..1);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 113 "     The situation -- lack of closeness to f -- does not
 improve with more points; just modify the syntas to see." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{MARK "0 0" 54 }{VIEWOPTS 1 1 0 1 1 1803 }