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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 31 "Polinomios y Bases de Lag
range." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 256 60 "Como hemos visto la ba
se de  Lagrange es aquella que cumple " }{XPPEDIT 18 0 "L[i](P[k]) = d
elta[ik];" "6#/-&%\"LG6#%\"iG6#&%\"PG6#%\"kG&%&deltaG6#%#ikG" }{TEXT 
-1 4 " .  " }{TEXT 257 97 "Si nos centramos en el caso de la interpola
cion polinomial clasica y en una tabla de tres puntos " }{TEXT 280 25 
"(x0,f0), (x1,f1), (x2,f2)" }{TEXT 281 143 ", tendremos que los conoci
dos polinomios de Lagrange tienen la siguiente forma, que evaluamos di
rectamente en los puntos dados por la lista xd." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 21 "xd:=[x0=0,x1=1,x2=2];" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#xdG7%/%#x0G\"\"!/%#x1G\"\"\"/%#x2G\"\"#" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "l0:= x-> eval((x-x1)*(x-x2)/((x0-x1
)*(x0-x2)),xd);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#l0GR6#%\"xG6\"6$
%)operatorG%&arrowGF(-%%evalG6$*&*&,&9$\"\"\"%#x1G!\"\"F3,&F2F3%#x2GF5
F3\"\"\"*&,&%#x0GF3F4F5\"\"\",&F;F3F7F5\"\"\"!\"\"%#xdGF(F(F(" }}}
{PARA 11 "" 1 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 6 "l0(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,&%\"xG\"\"\"!\"\"F
'F',&F&F'!\"#F'F'#F'\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
50 "l1:= x-> eval((x-x0)*(x-x2)/((x1-x0)*(x1-x2)),xd);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%#l1GR6#%\"xG6\"6$%)operatorG%&arrowGF(-%%evalG6$
*&*&,&9$\"\"\"%#x0G!\"\"F3,&F2F3%#x2GF5F3\"\"\"*&,&%#x1GF3F4F5\"\"\",&
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1 0 6 "l1(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&%\"xG\"\"\",&F%F&
!\"#F&F&!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "l2:= x-> e
val((x-x0)*(x-x1)/((x2-x0)*(x2-x1)),xd);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#l2GR6#%\"xG6\"6$%)operatorG%&arrowGF(-%%evalG6$*&*&,&9$\"\"\"
%#x0G!\"\"F3,&F2F3%#x1GF5F3\"\"\"*&,&%#x2GF3F4F5\"\"\",&F;F3F7F5\"\"\"
!\"\"%#xdGF(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "l2(x);" 
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\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 260 36 "El aspecto
 de estos polinomios sera:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
34 "plot([l0(x),l1(x),l2(x)],x=0...2);" }}{PARA 13 "" 1 "" {INLPLOT "6
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}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 258 87 "En el caso en e
l que los valores de la funcion interpolada esten dados por la lista y
d:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "yd:=[f0=1,f1=2,f2=1];
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ydG7%/%#f0G\"\"\"/%#f1G\"\"#/%#
f2GF(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 259 47 "La soluc
ion del problema de interpolacion sera:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 50 "sol:= x-> eval(f0*l0(x) + f1*l1(x) + f2*l2(x),yd);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solGR6#%\"xG6\"6$%)operatorG%&arro
wGF(-%%evalG6$,(*&%#f0G\"\"\"-%#l0G6#9$F2F2*&%#f1GF2-%#l1GF5F2F2*&%#f2
GF2-%#l2GF5F2F2%#ydGF(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
7 "sol(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*&,&%\"xG\"\"\"!\"\"F'
F',&F&F'!\"#F'F'#F'\"\"#*&F&F'F)\"\"\"F**&F&F.F%F.F+" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }{TEXT 261 75 "La grafica hace patente como la
 solucion es la suma pesada con los valores " }{TEXT 262 5 "f(x) " }
{TEXT 263 30 "de los polinomios de Lagrange." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 41 "plot([l0(x),l1(x),l2(x),sol(x)],x=0...2);" }}
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Fiv$\"1![HZws-u$F17$F^w$\"1U+7fR[pTF17$Fcw$\"1CPI!pKig%F17$Fhw$\"1byUs
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F'7$F,$\"10w%H:)G&3\"Fcs7$F3$\"1Q&o9c/k:\"Fcs7$F8$\"1i0J!oWHB\"Fcs7$F=
$\"1NI!Rb;jI\"Fcs7$FB$\"1>o?3#ycP\"Fcs7$FG$\"1wjkX@sO9Fcs7$FL$\"1so974
i'\\\"Fcs7$FQ$\"1;..Ft-b:Fcs7$FV$\"1RjPBKm4;Fcs7$Fen$\"1@zpV/8i;Fcs7$F
jn$\"1%=*><$3_q\"Fcs7$F_o$\"1'Hv)G+>]<Fcs7$Fdo$\"1u\\lN=h\"z\"Fcs7$Fio
$\"1wfcl!zz#=Fcs7$F^p$\"1n0bm\\)z&=Fcs7$Fcp$\"1Z#pj![#**)=Fcs7$Fhp$\"1
)o!H2J!Q\">Fcs7$F]q$\"1=\"f:f5z$>Fcs7$Fbq$\"1D+>*)y&f&>Fcs7$Fgq$\"11(e
&>@?s>Fcs7$F\\r$\"1-X'>b6U)>Fcs7$Far$\"1cE#f-XJ*>Fcs7$Ffr$\"1]2$y58\")
*>Fcs7$F[s$\"1$Hxa-*****>Fcs7$Fas$\"1T4'zsT\")*>Fcs7$Fgs$\"1V2R\"**eL*
>Fcs7$F\\t$\"13a4kN)[)>Fcs7$Fat$\"1&*[G(z<D(>Fcs7$Fft$\"17\\'=Cqo&>Fcs
7$F[u$\"1')*HEM)QQ>Fcs7$F`u$\"1'R&G(R;S\">Fcs7$Feu$\"1t7Fw1m))=Fcs7$Fj
u$\"1%*4(Q[wz&=Fcs7$F_v$\"1K%)fs3&p#=Fcs7$Fdv$\"1!>M#[t_*y\"Fcs7$Fiv$
\"1r2(*\\A(4v\"Fcs7$F^w$\"1\"*fU\\o?2<Fcs7$Fcw$\"1Afg.\\*4m\"Fcs7$Fhw$
\"1HWwr())*3;Fcs7$F]x$\"12l%>x$Rb:Fcs7$Fbx$\"1nw$>NEq\\\"Fcs7$Fgx$\"1+
s\"R=)eN9Fcs7$F\\y$\"1(GE$QD,w8Fcs7$Fay$\"19Ipzn0/8Fcs7$Ffy$\"1\"G#[wH
PO7Fcs7$F[z$\"1F+C9)[2;\"Fcs7$F`z$\"1zAF'=B]3\"FcsFa]m-Fhz6&FjzF(F(F[[
l-%+AXESLABELSG6$Q\"x6\"%!G-%%VIEWG6$;F(Fez%(DEFAULTG" 2 438 438 438 
2 0 1 0 2 9 0 4 2 1.000000 45.000000 45.000000 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 1 1 0 0 0 532 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 264 113 "Cada uno de los polinomios de Lagrange e
s la solucion de un problema de interpolacion donde todos los valores \+
de " }{TEXT 265 5 "f(x) " }{TEXT 266 150 "en los puntos dato son cero \+
excepto en uno de ellos. Veamoslo resolviendo los tres problemas de in
terpolacion que nos proveen con la base de Lagrange:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 25 "unassign('f0','f1','f2');" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "yd:=[f0=1,f1=0,f2=0];" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%#ydG7%/%#f0G\"\"\"/%#f1G\"\"!/%#f2GF+" }}}
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*l1(x) + f2*l2(x),yd);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solGR6#%
\"xG6\"6$%)operatorG%&arrowGF(-%%evalG6$,(*&%#f0G\"\"\"-%#l0G6#9$F2F2*
&%#f1GF2-%#l1GF5F2F2*&%#f2GF2-%#l2GF5F2F2%#ydGF(F(F(" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 7 "sol(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#,$*&,&%\"xG\"\"\"!\"\"F'F',&F&F'!\"#F'F'#F'\"\"#" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 35 "plot([l1(x),l2(x),sol(x)],x=0...2);" }}
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[q$\"18W&4OMjb\"F47$F`q$\"1Wlrq6_p7F47$Feq$\"1)ok&4ACE(*F,7$Fjq$\"1;\\
Fkp-sqF,7$F_r$\"1vQ`Xs[#[%F,7$Fdr$\"1t\"HE@jiE#F,7$Fir$\"1H*ooWo2%\\Ff
_l7$F_s$!1$p8sPqC1#F,7$Fds$!1Lr.d(oDu$F,7$Fis$!1fqV'*Ri\"R&F,7$F^t$!1`
ZCk$oZ\"pF,7$Fct$!11cC=ZOF#)F,7$Fht$!1v#**R%\\EI$*F,7$F]u$!1zp#R-Ii.\"
F47$Fbu$!1(p*ok2o66F47$Fgu$!1s\\N%z#=u6F47$F\\v$!1F)3418Z@\"F47$Fav$!1
\\4<m;]T7F47$Ffv$!1&=PWE&**\\7F47$F[w$!1d*zLX_:C\"F47$F`w$!1THOE<=;7F4
7$Few$!1W@2!R)[r6F47$Fjw$!1+#*RE)346\"F47$F_x$!1,]&Q^j6.\"F47$Fdx$!1GL
pTB3V$*F,7$Fix$!1cVJ;a1(H)F,7$F^y$!1;S)HpWW\"pF,7$Fcy$!1%p!y!\\E:^&F,7
$Fhy$!1q8]CvwUQF,7$F]z$!1AR6,&p$y?F,Faz-Fez6&FgzFhzFhzF(-%+AXESLABELSG
6$Q\"x6\"%!G-%%VIEWG6$;F(Fbz%(DEFAULTG" 2 438 438 438 2 0 1 0 2 9 0 4 
2 1.000000 45.000000 45.000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 
0 0 0 2167 -28348 0 0 0 0 0 0 }}}{PARA 11 "" 1 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "unassign('f0','f1','f2');" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "yd:=[f0=0,f1=1,f2=0];" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ydG7%/%#f0G\"\"!/%#f1G\"\"\"/%#f2GF
(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "sol:= x-> eval(f0*l0(x
) + f1*l1(x) + f2*l2(x),yd);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$sol
GR6#%\"xG6\"6$%)operatorG%&arrowGF(-%%evalG6$,(*&%#f0G\"\"\"-%#l0G6#9$
F2F2*&%#f1GF2-%#l1GF5F2F2*&%#f2GF2-%#l2GF5F2F2%#ydGF(F(F(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "sol(x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,$*&%\"xG\"\"\",&F%F&!\"#F&F&!\"\"" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 25 "unassign('f0','f1','f2');" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 21 "yd:=[f0=0,f1=0,f2=1];" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%#ydG7%/%#f0G\"\"!/%#f1GF(/%#f2G\"\"\"" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "sol:= x-> eval(f0*l0(x) + f1*l1(x) \+
+ f2*l2(x),yd);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solGR6#%\"xG6\"6
$%)operatorG%&arrowGF(-%%evalG6$,(*&%#f0G\"\"\"-%#l0G6#9$F2F2*&%#f1GF2
-%#l1GF5F2F2*&%#f2GF2-%#l2GF5F2F2%#ydGF(F(F(" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 7 "sol(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&%
\"xG\"\"\",&F%F&!\"\"F&F&#F&\"\"#" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 3 "" 0 "" {TEXT -1 47 "Ejercicio: Base de Lagrange en Dos
 Dimensiones." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 0 "" }{TEXT 267 58 "Imaginemos que ahora tenemos una funci
on de dos variables " }{TEXT 268 7 "f(x,y) " }{TEXT 269 63 "que deseam
os interpolar, clasicamente,  en una tabla de puntos." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }{TEXT 270 37 "vamos a tomar como puntos dato en X: \+
" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "xd:=[x0=
0,x1=1,x2=2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#xdG7%/%#x0G\"\"!/%
#x1G\"\"\"/%#x2G\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "y en  Y:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "yd:=[y0=4,y1=6];" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ydG7$/%#y0G\"\"%/%#y1G\"\"'" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 271 40 "La funcion que vam
os a interpolar sera: " }{XPPEDIT 18 0 "f(x,y) = y^x;" "6#/-%\"fG6$%\"
xG%\"yG)F(F'" }{TEXT -1 1 " " }{TEXT 272 19 " que en los puntos " }
{TEXT 273 16 "(x0,y0), (x0,y1)" }{TEXT 274 14 ", etc. valdra:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "fd:=[f00=1,f01=1,f10=4,f11=6
,f20=16,f21=36];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#fdG7(/%$f00G\"
\"!/%$f01G\"\"\"/%$f10G\"\"%/%$f11G\"\"'/%$f20G\"#;/%$f21G\"#O" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 275 114 "Se pide resolver \+
este problema encontrando la base de Lagrange utilizando como espacio \+
solucion el expandido por \{" }{TEXT 276 13 "1, x, y, xy, " }{XPPEDIT 
18 0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "yx^2;" "
6#*$%#yxG\"\"#" }{TEXT -1 0 "" }{TEXT 277 1 "\}" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 0 "" }{TEXT 278 90 "Como ayuda: centraos en el comportam
iento de los polinomios de Lagrange en una dimension. " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }{TEXT 279 67 "La definicion de una funcion en dos \+
variables en Maple se hace asi:" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 21 "f3d := (x,y) -> y^x ;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$f3dGR6$%\"xG%\"yG6\"6$%)operatorG%&arrowGF))9%9$F)F)
F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "plot3d(f3d(x,y),x=0..
2, y=4..6, axes=boxed);" }}{PARA 13 "" 1 "" {INLPLOT "6%-%%GRIDG6%;\"
\"!$\"\"#F';$\"\"%F'$\"\"'F'W(\\bm\":\":3FF00000000000003FF00000000000
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