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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 11 "Math 2270-1" }}{PARA 
257 "" 0 "" {TEXT -1 17 "Friday October 27" }}{PARA 258 "" 0 "" {TEXT 
-1 27 "Projecting onto polynomials" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 267 "Here's the inner product we will use.  W
e'll call it ``dot'' in honor of the dot product of R^n, to which this
 inner product is very similar indeed.  With the dot product we can th
erefore define the magnitude of vectors, as well as distance and angle
 between vectors." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 141 "dot:=(f,g)->int(f(x)*g(x),x=-1..1); #dot ta
kes a pair of\n      #functions and returns a real number by computing
\n      #a particular integral" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$d
otGR6$%\"fG%\"gG6\"6$%)operatorG%&arrowGF)-%$intG6$*&-9$6#%\"xG\"\"\"-
9%F3F5/F4;!\"\"F5F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "
mag:=f->sqrt(dot(f,f)); #computes the magnitude of\n     #a vector" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$magGR6#%\"fG6\"6$%)operatorG%&arrow
GF(-%%sqrtG6#-%$dotG6$9$F2F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 79 "dist:=(f,g)->mag(f-g);  #computes the ``distance'' be
tween\n     #two vectors.  " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%dist
GR6$%\"fG%\"gG6\"6$%)operatorG%&arrowGF)-%$magG6#,&9$\"\"\"9%!\"\"F)F)
F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "angle:=(f,g)->arccos(
dot(f,g)/(mag(f)*mag(g)));\n     #computes the ``angle'' between two v
ectors" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&angleGR6$%\"fG%\"gG6\"6$%
)operatorG%&arrowGF)-%'arccosG6#*&-%$dotG6$9$9%\"\"\"*&-%$magG6#F4F6-F
96#F5F6!\"\"F)F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 260 "Let's experiment \+
by computing some of these quantities for some polynomials.  Here's ou
r usual basis for the space of polynomials of degree at most three.  T
he restriction of these polynomials to the interval [-1..1] gives a su
bspace of our inner product space" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 40 "P0:=x->1;P1:=x->x;P2:=x->x^2;P3:=x->x^3;" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#P0G\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%#P1GR6#%\"xG6\"6$%)operatorG%&arrowGF(9$F(F(F(" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#P2GR6#%\"xG6\"6$%)operatorG%&arrowGF(*$)9$\"\"#\"\"
\"F(F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#P3GR6#%\"xG6\"6$%)opera
torG%&arrowGF(*$)9$\"\"$\"\"\"F(F(F(" }}}{PARA 0 "" 0 "" {TEXT -1 43 "
2a)  Compute the magnitudes of P0,P1,P2,P3." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 32 "mag(P0);mag(P1);mag(P2);mag(P3);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#*$-%%sqrtG6#\"\"#\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#,$*$-%%sqrtG6#\"\"'\"\"\"#F)\"\"$" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#,$*$-%%sqrtG6#\"#5\"\"\"#F)\"\"&" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#,$*$-%%sqrtG6#\"#9\"\"\"#F)\"\"(" }}}{PARA 0 "" 0 "" {TEXT -1 
195 "2b)  Which pairs of these four polynomials are perpendicular to e
ach other?  Is it true that if f is an even function and g is an odd f
unction, then f and g are perpendicular in this space?  Why?" }}{PARA 
259 "" 0 "" {TEXT -1 326 "Since the product of an even function with a
n odd function is an odd function, and since the integral of an odd fu
nction over an interval which is symmetric with respect to the origin \+
is zero, it IS true that even and odd functions are orthogonal to each
other.  So the following dots should be zero, indicating orthogonality
: " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "dot(P0,P1);dot(P0,P3);
dot(P1,P2);dot(P2,P3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 256 25 "The rest of the dots are:" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "dot(P0,P2);dot(P1,P3);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6##\"\"#\"\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
#\"\"#\"\"&" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 58 "so those \+
pairs of functions turn out to not be orthogonal." }}{PARA 0 "" 0 "" 
{TEXT -1 116 "2c)  Show that f(x)=1 and g(x)=x^2 - 1/3 are orthogonal.
  Hint: compute the inner product of P0 with P2 - (1/3)*P0. " }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "dot(P0,P2-1/3*P0);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
{TEXT 258 3 "yep" }}{PARA 0 "" 0 "" {TEXT -1 63 "2d)  Find the distanc
e between f(x)=1+2*x+x^3 and g(x)=x+3*x^2." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 25 "dist(P0+2*P1+P3,P1+3*P2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,$*$-%%sqrtG6#\"%5B\"\"\"#\"\"%\"$0\"" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#$\"+L3&4$=!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 304 "Let's denote the subspace spanned by P0,P1,P2 with the l
etter W2.  Here's how to use Gram-Schmidt to find an orthonormal basis
.  It should look familiar!  We will first find an orthogonal basis, t
hen we will normalize it.  We'll use Q0,Q1,Q2 for the orthogonal basis
, and u0,u1,u2 for the normalized one." }}{PARA 11 "" 1 "" {TEXT -1 0 
"" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 115 "Q0:=P0;   #first ortho
gonal basis element\nproj0:=f->(dot(f,Q0)/mag(Q0)^2)*Q0;\n      #proje
ction onto the span of Q0" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#Q0G\"
\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&proj0GR6#%\"fG6\"6$%)operat
orG%&arrowGF(*&*&-%$dotG6$9$%#Q0G\"\"\"F2F3F3*$)-%$magG6#F2\"\"#F3!\"
\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 145 "Q1:=P1 - proj0
(P1);  #second orthogonal basis element\nproj1:=f->proj0(f) + (dot(f,Q
1)/mag(Q1)^2)*Q1;\n      #projection onto the span of Q0 and Q1" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#Q1G%#P1G" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&proj1GR6#%\"fG6\"6$%)operatorG%&arrowGF(,&-%&proj0G6
#9$\"\"\"*&*&-%$dotG6$F0%#Q1GF1F7F1F1*$)-%$magG6#F7\"\"#F1!\"\"F1F(F(F
(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Q1(x);  #check our wor
k" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%\"xG" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 97 "Q2:=P2-proj1(P2);  #third orthogonal basis element
\nproj2:=f->proj1(f) + (dot(f,Q2)/mag(Q2)^2)*Q2;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#Q2G,&%#P2G\"\"\"#F'\"\"$!\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&proj2GR6#%\"fG6\"6$%)operatorG%&arrowGF(,&-%&proj1G6
#9$\"\"\"*&*&-%$dotG6$F0%#Q2GF1F7F1F1*$)-%$magG6#F7\"\"#F1!\"\"F1F(F(F
(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Q2(x); #check our work
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*$)%\"xG\"\"#\"\"\"F(#F(\"\"$!\"
\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "dot(Q0,Q1);dot(Q0,Q2)
;dot(Q1,Q2); #these should all\n      #be zero!" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 100 "u0:=x->Q0(x)/mag(Q0);u1:=x->Q1(x)/mag(Q1);u2:=x->Q2(
x)/mag(Q2);\n     #normalize our orthogonal basis" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#u0GR6#%\"xG6\"6$%)operatorG%&arrowGF(*&-%#Q0G6#9$\"
\"\"-%$magG6#F.!\"\"F(F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#u1GR6
#%\"xG6\"6$%)operatorG%&arrowGF(*&-%#Q1G6#9$\"\"\"-%$magG6#F.!\"\"F(F(
F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#u2GR6#%\"xG6\"6$%)operatorG%&
arrowGF(*&-%#Q2G6#9$\"\"\"-%$magG6#F.!\"\"F(F(F(" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 32 "u0(x);u1(x);u2(x); #unit vectors" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#,$*$-%%sqrtG6#\"\"#\"\"\"#F)F(" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#,$*&%\"xG\"\"\"-%%sqrtG6#\"\"'F&#F&\"\"#" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#,$*&,&*$)%\"xG\"\"#\"\"\"F*#F*\"\"$!\"\"F*-%
%sqrtG6#\"#5F*#F,\"\"%" }}}{PARA 0 "" 0 "" {TEXT -1 240 "Now let's use
 proj2 to find the projection of the exponential function exp onto W2.
  It should be the nearest polynomial of degree 2 to the exp function.
  In particular it should be closer that the degree two Taylor approxi
mation 1+x+x^2/2:" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 14 "proj2(exp)(x);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#,*-%$expG6#\"\"\"#F'\"\"#*&#F'F)F'-F%6#!\"\"F'F.*(\"\"$F'F,F'%\"
xGF'F'*(#\"#X\"\")F',&F$#F)F0*&#\"#9F0F'F,F'F.F',&*$)F1F)F'F'#F'F0F.F'
F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#,($\"+*>SH'**!#5\"\"\"*&$\"+C$QO5\"!\"*F'%\"xGF
'F'*&$\"+%>:sO&F&F')F,\"\"#F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 21 "dist(proj2(exp),exp);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$-
%%sqrtG6#,(*&-%$expG6#\"\"\"F--F+6#!\"\"F-\"$W\"*&\"$=&F-)F.\"\"#F-F0*
&\"#5F-)F*F5F-F0F-#F-F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "e
valf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+_u[&z$!#6" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "dist(P0+P1+P2/2,exp);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#,$*$-%%sqrtG6#,,\"%!4$\"\"\"*&\"$]%F*)-%$expG6#F
*\"\"#F*F**&\"%+FF*F.F*!\"\"*&F,F*)-F/6#F4F1F*F4*&F3F*F7F*F*F*#F*\"#I
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#$\"+>mvz%*!#6" }}}{PARA 0 "" 0 "" {TEXT -1 81 "S
o both approximations were close, but the projection was about 2.5 tim
es closer." }}{PARA 0 "" 0 "" {TEXT -1 103 "Let's get a feel for these
 distances geometrically by plotting the three functions under conside
ration:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 273 "with(plots):\nac
tual:=plot(exp(x),x=-1..1,color=`black`):\nnear:=plot(1+x+x^2/2,x=-1..
1,color=`blue`):\nnearest:=plot(.9962940199+1.103638324*x+.5367215194*
x^2,x=-1..1,\n        color=`red`):\n     #I used the mouse to paste t
his in from up above\ndisplay(\{actual,near,nearest\});" }}{PARA 13 "
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