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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 23 "Computing the residues." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "restart; F:=2*log(x^2-2*x-2)
-6*log(x^3-3)+5*log(x-7);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"FG,(-
%#lnG6#,(*$)%\"xG\"\"#\"\"\"\"\"\"F,!\"#F0F/F--F'6#,&*$)F,\"\"$F.F/!\"
$F/!\"'-F'6#,&F,F/!\"(F/\"\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 30 "f:=normal(diff(F,x),expanded);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%\"fG*&,.*$)%\"xG\"\"&\"\"\"!\"**$)F)\"\"%F+\"$?\"*$)F)\"\"#F+!$z#F
)\"$E\"*$)F)\"\"$F+!$)>!#a\"\"\"F+,0*$)F)\"\"'F+F;F'F,F6\"#6F1\"#FF-\"
#7F)!#O!#UF;!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "a:=num
er(f);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG,.*$)%\"xG\"\"&\"\"\"!
\"**$)F(\"\"%F*\"$?\"*$)F(\"\"#F*!$z#F(\"$E\"*$)F(\"\"$F*!$)>!#a\"\"\"
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "b:=denom(f);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"bG,0*$)%\"xG\"\"'\"\"\"\"\"\"*$)F(\"\"&F
*!\"**$)F(\"\"$F*\"#6*$)F(\"\"#F*\"#F*$)F(\"\"%F*\"#7F(!#O!#UF+" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "resultant(a-z*diff(b,x),b,x)
;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,0*$)%\"zG\"\"'\"\"\"\"1+k[*GUyF
\"*$)F&\"\"&F(\"2+wP01e+:\"*$)F&\"\"%F(!2+?f%oo_LQ*$)F&\"\"$F(!3+;K$eZ
xdR%*$)F&\"\"#F(\"3+K3s$z&=!o$F&\"4+O(oS$3&>kQ!4+![7e!py-_&\"\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "factor(%);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#,$*(,&%\"zG\"\"\"!\"&F'F'),&F&F'!\"#F'\"\"#\"\"\"),&
F&F'\"\"'F'\"\"$F-\"1+k[*GUyF\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
39 "The example shows that the residues of:" }}{PARA 0 "" 0 "" {TEXT 
-1 22 "a/b  (with gcd(a,b)=1)" }}{PARA 0 "" 0 "" {TEXT -1 71 "are equa
l to the roots (as a polynomial in z) of resultant(a-z*b',b,x);" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 106 "Why is t
his true? Well, suppose that alpha is a root of b, so alpha is a pole \+
of f=a/b. Then we can write:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 38 "f = a / ( (x-alpha)*quo(b,x-alpha,x) )" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "alpha:=7; # one of the roots
 of b" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&alphaG\"\"(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "a/( (x-alpha)*quo(b,x-alpha,x) ); #
 equals f" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,.*$)%\"xG\"\"&\"\"\"!
\"**$)F'\"\"%F)\"$?\"*$)F'\"\"#F)!$z#F'\"$E\"*$)F'\"\"$F)!$)>!#a\"\"\"
F)*&,&F'F9!\"(F9\"\"\",.F%F9F+!\"#F4F?F/!\"$F'\"\"'FAF9\"\"\"!\"\"" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 175 "And we see that for the residue,
 we need the coefficient of (x-alpha)^(-1) of this expression, which i
s the same as the coefficient of (x-alpha)^0 of the following expressi
on:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "a/quo(b,x-alpha,x); \+
# equals f*(x-alpha)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,.*$)%\"xG\"
\"&\"\"\"!\"**$)F'\"\"%F)\"$?\"*$)F'\"\"#F)!$z#F'\"$E\"*$)F'\"\"$F)!$)
>!#a\"\"\"F),.F%F9F+!\"#F4F;F/!\"$F'\"\"'F=F9!\"\"" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 100 "How do you compute the t^0 coefficient of this, \+
where t is the local parameter x=alpha. Easy enough:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 16 "subs(x=alpha,%);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "And ther
e is the residue. So we have to evaluate f*(x-alpha) at x=alpha. Now t
he denominator b of f equals:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 27 "(x-alpha)*quo(b,x-alpha,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*
&,&%\"xG\"\"\"!\"(F&F&,.*$)F%\"\"&\"\"\"F&*$)F%\"\"%F,!\"#*$)F%\"\"$F,
F0*$)F%\"\"#F,!\"$F%\"\"'F8F&F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
26 "The derivative of that is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 17 "diff(%,x); # = b'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,0*$)%\"x
G\"\"&\"\"\"\"\"\"*$)F&\"\"%F(!\"#*$)F&\"\"$F(F-*$)F&\"\"#F(!\"$F&\"\"
'F5F)*&,&F&F)!\"(F)F),,F*F'F.!\")F1!\"'F&F;F5F)F)F)" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 33 "So b' = diff(b,x) is of the form:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "quo(b,x-alpha,x) + (x-alpha)*someth
ing;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,0*$)%\"xG\"\"&\"\"\"\"\"\"*$)
F&\"\"%F(!\"#*$)F&\"\"$F(F-*$)F&\"\"#F(!\"$F&\"\"'F5F)*&,&F&F)!\"(F)F)
%*somethingGF)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 143 "And if we su
bstitute x=alpha in that, then the (x-alpha*something becomes zero and
 we get the same as substituting x=alpha in quo(b,x-alpha,x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "subs(x=alpha, quo(b,x-alpha,
x) );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"&?7\"" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 94 "Therefore, substituting x=alpha in b' is the same \+
as substituting x=alpha in quo(b,x-alpha,x);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 25 "subs(x=alpha, diff(b,x));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"&?7\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "So inst
ead of:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "subs(x=alpha, a/
quo(b,x-alpha,x)); # = coeff of (x-alpha)^(-1) of a/b" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#\"\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "We \+
may just as well do the following because at x=alpha, the functions di
ff(b,x) and quo(b,x-alpha,x) take the same values." }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 28 "subs(x=alpha, a/diff(b,x) );" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#\"\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "No
w if r is a residue, so" }}{PARA 0 "" 0 "" {TEXT -1 31 " r = subs(x=al
pha, a/diff(b,x))" }}{PARA 0 "" 0 "" {TEXT -1 5 "then:" }}{PARA 0 "" 
0 "" {TEXT -1 18 " a - diff(b,x) * r" }}{PARA 0 "" 0 "" {TEXT -1 78 "i
s zero at x=alpha. So if z is a variable, then r is a root of the poly
nomial:" }}{PARA 0 "" 0 "" {TEXT -1 30 " subs(x=alpha,  a-diff(b,x)*z)
" }}{PARA 0 "" 0 "" {TEXT -1 30 "where z is used as a variable." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 127 "The prod
uct of all these a-diff(b,x)*z, this product taken over all roots alph
a of the polynomial b, that equals the resultant:" }}{PARA 0 "" 0 "" 
{TEXT -1 90 "resultant(a-diff(b,x)*z, b,x)  =  \"product(subs(x=i,a-di
ff(b,x)*z), i = [all roots of f])\"" }}{PARA 0 "" 0 "" {TEXT -1 74 "Th
e quotes indicate that the part \"all roots of f\" is not real Maple c
ode." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "Ex
ample:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "f := (2*x^3-3)/(x
^6+6*x^3-2*x^2+9);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&,&*$)%\"
xG\"\"$\"\"\"\"\"#!\"$\"\"\"F+,**$)F)\"\"'F+F.F'F2*$)F)F,F+!\"#\"\"*F.
!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "a,b := numer(f), d
enom(f);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6$%\"aG%\"bG6$,&*$)%\"xG
\"\"$\"\"\"\"\"#!\"$\"\"\",**$)F+\"\"'F-F0F)F4*$)F+F.F-!\"#\"\"*F0" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "R:=resultant(a-diff(b,x)*z,
b,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"RG,**$)%\"zG\"\"'\"\"\"!*
oz]r#*$)F(\"\"%F*\"*)[:=5*$)F(\"\"#F*!)Ops7\"'*GI&\"\"\"" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "R:=factor(R);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"RG,$*$),&*$)%\"zG\"\"#\"\"\"\"\")!\"\"\"\"\"\"\"$F-
!'*GI&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "residues:=\{solve
(R,z)\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)residuesG<$,$*$-%%sqrtG
6#\"\"#\"\"\"#!\"\"\"\"%,$F'#\"\"\"F/" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "r1,r2 := op(residues);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>6$%#r1G%#r2G6$,$*$-%%sqrtG6#\"\"#\"\"\"#!\"\"\"\"%,$F)#\"\"\"F1
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 501 "Now note that we have only 2
 distinct residues, but we have 6 poles. So there must be several dist
inct poles that have the same residue. How do we find all poles with r
esidue r1? Note that as usual, we don't really want to find those pole
s because they may be complicated algebraic numbers. We just want to f
ind a square-free polynomial such that the roots of that polynomial ar
e the poles with residue r1. If alpha is any complex number, then we h
ave a pole at x=alpha with residue r1 if and only if:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "alpha:='alpha';" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%&alphaGF$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 36 "subs(x=alpha, \{a-diff(b,x)*r1,  b\});" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#<$,**$)%&alphaG\"\"'\"\"\"\"\"\"*$)F'\"\"$F)F(*$)F'\"\"
#F)!\"#\"\"*F*,(F+F0!\"$F**&,(*$)F'\"\"&F)F(F.\"#=F'!\"%F*-%%sqrtG6#F0
F)#F*\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 325 "are both zero. Why
? Well, for one thing, b(alpha) should be zero, otherwise alpha wouldn
't even be a pole. But as we've seen before, at a pole x=alpha, the re
sidue r satisfied the equation a-diff(b,x)*r=0. So both of these equat
ions must hold. In other words, we're looking for a solution of the fo
llowing two equations in x:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
18 "a-diff(b,x)*r1, b;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$,(*$)%\"xG\"
\"$\"\"\"\"\"#!\"$\"\"\"*&,(*$)F&\"\"&F(\"\"'*$)F&F)F(\"#=F&!\"%F+-%%s
qrtG6#F)F(#F+\"\"%,**$)F&F1F(F+F$F1F2!\"#\"\"*F+" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 231 "So we have two polynomials, call them f1 and f2 f
or now, and we're looking for a polynomial f3 such that whenever alpha
 is a solution of f1 and f2 at the same time, then alpha is a root of \+
f3. It is now clear what we need: the gcd." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 7 "gcd(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*$)%\"x
G\"\"$\"\"\"\"\"\"*&-%%sqrtG6#\"\"#F(F&F)F)F'F)" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 83 "Just to be sure, lets check that if alpha is a root,
 then the residue really is r1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 19 "alpha:=RootOf(%,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&alph
aG-%'RootOfG6#,(*$)%#_ZG\"\"$\"\"\"\"\"\"*&-%%sqrtG6#\"\"#F-F+F.F.F,F.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "subs(x=alpha, a/diff(b,
x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&*$)-%'RootOfG6#,(*$)%#_ZG
\"\"$\"\"\"\"\"\"*&-%%sqrtG6#\"\"#F/F-F0F0F.F0F.F/F5!\"$F0F/,(*$)F'\"
\"&F/\"\"'*$)F'F5F/\"#=F'!\"%!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "evala(%); # OK, this equals r1." }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,$*$-%%sqrtG6#\"\"#\"\"\"#!\"\"\"\"%" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 56 "Now do this for all residues with the following
 command:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "A:=add( r*log(
gcd(a-diff(b,x)*r,b)), r=residues);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%\"AG,&*&-%%sqrtG6#\"\"#\"\"\"-%#lnG6#,(*$)%\"xG\"\"$F+\"\"\"*&F'F+F
2F4F4F3F4F4#!\"\"\"\"%*&F'F+-F-6#,(F0F4F5F7F3F4F4#F4F8" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 126 "Then it must be true that f-diff(F,x) ha
s no poles at all, so it must be a polynomial, so it should be easy to
 integrate that." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "f-diff(
A,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*&,&*$)%\"xG\"\"$\"\"\"\"\"
#!\"$\"\"\"F*,**$)F(\"\"'F*F-F&F1*$)F(F+F*!\"#\"\"*F-!\"\"F-*&*&-%%sqr
tG6#F+F*,&F2F)*$F9F*F-F-F*,(F&F-*&F9F*F(F-F-F)F-F6#F-\"\"%*&*&F9F*,&F2
F)F=!\"\"F-F*,(F&F-F?FEF)F-F6#FEFA" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 10 "normal(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "So the polynomial we end up wit
h is 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
639 "Lets recall what we have done. Previously we have seen that if f \+
is a rational function, and if the highest pole order is n, and if n>1
, then we can calculate a rational function A such that f-diff(A,x) ha
s pole orders <n. This could be repeated until we have pole order 1. W
e've seen above, that we can get rid of those by computing a function \+
A = add( r * log(...) , r=residues) where the ... is the polynomial wh
ose roots are the poles with residue r, (this is a square-free polynom
ial). Then we get that diff(A,x) has poles of order 1 with the same re
sidues as f, so f-diff(A,x) can not have any poles anymore and so it i
s a polynomial" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 153 "In the example the polynomial was 0. Could we have predi
cted in advance what that polynomial was going to be? Suppose the poly
nomial were something like:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
27 "n:=5; add(c[i]*x^i,i=0..n);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
\"nG\"\"&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,.&%\"cG6#\"\"!\"\"\"*&&F
%6#F(F(%\"xGF(F(*&&F%6#\"\"#F()F,F0\"\"\"F(*&&F%6#\"\"$F()F,F6F2F(*&&F
%6#\"\"%F()F,F;F2F(*&&F%6#\"\"&F()F,F@F2F(" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 122 "so if that was what was left of f after substracting dif
f(A,x), so that polynomial would then still have to be integrated." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 219 "Look aga
in at all the functions diff(A,x) that we substracted from f. For all \+
of them, you can see that they are all functions that converge to 0 wh
en x -> infinity. So what was it that did not change after all of this
:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 "*) i
f f was convergent for x->infinity, then the value of f did not change
." }}{PARA 0 "" 0 "" {TEXT -1 19 "Or, more generally:" }}{PARA 0 "" 0 
"" {TEXT -1 388 "Take the unique polynomial q such that f-q converges \+
to zero when x->infinity. Then, after we replaced f by diff(A,x) a num
ber of times to get rid of the poles of f, the resulting f still has t
hat same property, that f-q converges to zero when x->infinity. But at
 this point, f has become a polynomial, and the only way that the poly
nomial f-q can go to zero when x->infinity is when f=q." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 132 "Conclusion: the po
lynomial that you end up with at the end is the unique polynomial q su
ch that f-q converges to 0 when x->infinity." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 95 "In our exaple we had an f that \+
converges to 0 when x->infinity, so q was 0. In general we have:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "f:=(x^5+x-3)/(2*x^3-x^2);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&,(*$)%\"xG\"\"&\"\"\"\"\"\"F)
F,!\"$F,F+,&*$)F)\"\"$F+\"\"#*$)F)F2F+!\"\"!\"\"" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 23 "a,b:=numer(f),denom(f);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>6$%\"aG%\"bG6$,(*$)%\"xG\"\"&\"\"\"\"\"\"F+F.!\"$F.*&)
F+\"\"#F-,&F+F2!\"\"F.F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 
"q:=quo(a,b,x,'r');" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"qG,(*$)%\"x
G\"\"#\"\"\"#\"\"\"F)F(#F,\"\"%#F,\"\")F," }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 17 "r; # = rem(a,b,x)" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#,(!\"$\"\"\"%\"xGF%*$)F&\"\"#\"\"\"#F%\"\")" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 11 "new_f:=r/b;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%&new_fG*&,(!\"$\"\"\"%\"xGF(*$)F)\"\"#\"\"\"#F(\"\")F-*&)F)\"\"#F-,&F
)F,!\"\"F(\"\"\"!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "f \+
= new_f + q;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&,(*$)%\"xG\"\"&\"\"
\"\"\"\"F(F+!\"$F+F*,&*$)F(\"\"$F*\"\"#*$)F(F1F*!\"\"!\"\",**&,(F,F+F(
F+F2#F+\"\")F**&)F(\"\"#F*,&F(F1F4F+\"\"\"F5F+F2#F+F1F(#F+\"\"%F9F+" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "normal(lhs(%)-rhs(%)); # c
heck if that equation is true:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"
!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "Since a=q*b+r we have f=a/b=
q+r/b." }}{PARA 0 "" 0 "" {TEXT -1 264 "Now r/b is the part of f that \+
converges to 0 when x->infinity because remember from polynomial divis
ion that degree(r,x)<degree(b,x). This q is called the polynomial part
. You will see that if you apply the whole process on f, then you'll e
nd up with polynomial q." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "
f;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,(*$)%\"xG\"\"&\"\"\"\"\"\"F'F
*!\"$F*F),&*$)F'\"\"$F)\"\"#*$)F'F0F)!\"\"!\"\"" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 23 "a,b:=numer(f),denom(f);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>6$%\"aG%\"bG6$,(*$)%\"xG\"\"&\"\"\"\"\"\"F+F.!\"$F.*&)
F+\"\"#F-,&F+F2!\"\"F.F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 
"sqrfree(b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$\"\"\"7$7$%\"xG\"\"#
7$,&F'F(!\"\"F$F$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "d:=x;" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG%\"xG" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 11 "A1:=c0/d^1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#A1G*&%#c0G\"\"\"%\"xG!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 15 "a1:=diff(A1,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#a1G,$*&%#c
0G\"\"\"*$)%\"xG\"\"#F(!\"\"!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 19 "normal((f-a1)*d^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#*&,,*$)%\"xG\"\"&\"\"\"\"\"\"F'F*!\"$F**&%#c0GF*F'F*\"\"#F-!\"\"F),&
F'F.F/F*!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "rem(numer(
%),d,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&%#c0G!\"\"!\"$\"\"\"" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(\{%\},\{c0\});" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#<#/%#c0G!\"$" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 15 "A1:=subs(%,A1);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%#A1G,$*&\"\"\"F'%\"xG!\"\"!\"$" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "f1:=normal(f-diff(A1,x));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#f1G*&,&*$)%\"xG\"\"%\"\"\"\"\"\"!\"&F,F+*&F)\"\"\",&
F)\"\"#!\"\"F,\"\"\"!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
25 "a,b:=numer(f1),denom(f1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6$%
\"aG%\"bG6$,&*$)%\"xG\"\"%\"\"\"\"\"\"!\"&F.*&F+F.,&F+\"\"#!\"\"F.F." 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "factor(resultant(a-diff(b
,x)*z,b,x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,&!\"&\"\"\"%\"zGF
'F',&\"#zF'F(\"#;F'!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 
"residues:=\{solve(%,z)\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)resid
uesG<$\"\"&#!#z\"#;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "A:=a
dd(r*log(gcd(a-diff(b,x)*r,b)),r=residues);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"AG,&-%#lnG6#%\"xG\"\"&-F'6#,&F)\"\"\"#!\"\"\"\"#F.#
!#z\"#;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "f2:=normal(f1-di
ff(A,x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f2G,(*$)%\"xG\"\"#\"\"
\"#\"\"\"F)F(#F,\"\"%#F,\"\")F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 16 "q; # is the same" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*$)%\"xG
\"\"#\"\"\"#\"\"\"F'F&#F*\"\"%#F*\"\")F*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 10 "int(f2,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*$)%
\"xG\"\"$\"\"\"#\"\"\"\"\"'*$)F&\"\"#F(#F*\"\")F&F/" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 16 "Int(f,x)=A1+A+%;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%$IntG6$*&,(*$)%\"xG\"\"&\"\"\"\"\"\"F+F.!\"$F.F-,&*$
)F+\"\"$F-\"\"#*$)F+F4F-!\"\"!\"\"F+,.*&F-F-F+F8F/-%#lnG6#F+F,-F<6#,&F
+F.#F7F4F.#!#z\"#;F1#F.\"\"'F5#F.\"\")F+FG" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 6 "Check:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "rhs(%)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,.*&\"\"\"F%%\"xG!\"\"!\"$-%#lnG6
#F&\"\"&-F*6#,&F&\"\"\"#!\"\"\"\"#F0#!#z\"#;*$)F&\"\"$F%#F0\"\"'*$)F&F
3F%#F0\"\")F&F>" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "normal(f
 - diff(%,x)); # should be 0." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "OK." }}{PARA 0 "" 0 "" {TEXT 
-1 148 "What one usually does in the integration algorithm is to simpl
y start by computing the left-over polynomial at the end (which equals
 q) by computing" }}{PARA 0 "" 0 "" {TEXT -1 23 "     q:=quo(a,b,x,'r'
);" }}{PARA 0 "" 0 "" {TEXT -1 69 "This command gives you q, and puts \+
the remainder into the variable r." }}{PARA 0 "" 0 "" {TEXT -1 56 "The
n just integrate r/b, and add int(q,x) to the result." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "We have now covered th
e lecture notes until page 12." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "f:=(5*x^13-2079*
x+120)/(x^4-3)^3*x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&*&,(*$)
%\"xG\"#8\"\"\"\"\"&F*!%z?\"$?\"\"\"\"F0F*F0F,*$),&*$)F*\"\"%F,F0!\"$F
0\"\"$F,!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 258 11 "Assignment:" }
{TEXT -1 83 " integrate this function with the method explained in thi
s and previous worksheets." }}}}{MARK "12 0 0" 80 }{VIEWOPTS 1 1 0 1 
1 1803 }