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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "Consider the following ra
tional function. A rational function is a polynomial or a quotient of \+
polynomials." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "f := x^3*(-
1+2*x^2)/(1-x^2+x^4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&*&)%
\"xG\"\"$\"\"\",&!\"\"F**&\"\"#F*)F(F.F*F*F*F*,(F*F**$F/F*F,*$)F(\"\"%
F*F*F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "int(f,x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*$)%\"xG\"\"#\"\"\"F(*&#F(\"\"%F(-%#
lnG6#,(F(F(F$!\"\"*$)F&F+F(F(F(F(*&#F(F'F(*&-%%sqrtG6#\"\"$F(-%'arctan
G6#,$*&,&F0F(*&F'F(F%F(F(F(F6F(#F(F9F(F(F0" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 41 "How can Maple compute the above integral?" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 268 16 "Three questions:" }
}{PARA 0 "" 0 "" {TEXT -1 65 "1) How do calculus books teach integrati
on of rational functions?" }}{PARA 0 "" 0 "" {TEXT -1 79 "2) Why is th
e standard calculus method not very suitable for the above example?" }
}{PARA 0 "" 0 "" {TEXT -1 103 "3) How can this method be improved; wha
t is the method that computer algebra systems such as Maple use?" }}}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 51 "Calculus method for integrating r
ational functions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 
"" 0 "" {TEXT -1 49 "The Maple commands: int, Int, simplify and value.
" }}{PARA 0 "" 0 "" {TEXT -1 315 "In Maple the command \"int\" is the \+
command for integration, and \"Int\" is the \"inert version\" of the i
ntegration command. An inert command means essentially: \"just display
 the command but do not do the computation yet\". The inert version of
 integration, \"Int\", does no computation until you apply the command
 \"value\"." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "int(x^3,x);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$)%\"xG\"\"%\"\"\"#F(F'" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Int(x^3,x);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#-%$IntG6$*$)%\"xG\"\"$\"\"\"F(" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#-%$IntG6$*$)%\"xG\"\"$\"\"\"F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 238 "The percentage sign % refers to the most recently evaluated ex
pression. The simplify command tries to simplify expressions but it do
es not tell \"Int\" to finally do something. Only the command \"value
\" awakes the \"Int\" from its hybernation." }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 9 "value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$)%
\"xG\"\"%\"\"\"#F(F'" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 97 "We know from calculus the following formula, which i
s valid for every real number n, except n=-1." }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 26 "Int(x^n, x) = int(x^n, x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%$IntG6$)%\"xG%\"nGF(*&)F(,&F)\"\"\"F-F-F-F,!\"\"" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 369 "Note that as far as Maple is con
cerned, an unassigned variable such as \"n\" is not considered to be e
qual to any particular real number, nor is it equal to any other unass
igned variable. So we have to be very careful about general formulas i
n Maple because whenever an equation is true in general (but with an e
xceptions like n=-1) then Maple will just use that formula." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "For n=-1 we hav
e:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "Int(x^(-1),x) = int(x
^(-1),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(%\"xG
!\"\"F)-%#lnG6#F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "More general
ly, if c is some complex number we have the following " }{TEXT 272 21 
"two calculus formulas" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 51 "Int( (x-c)^n ,x) = int( (x-c)^n,x ); # when n <> -1" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$),&%\"xG\"\"\"%\"cG!\"\"%
\"nGF)*&)F(,&F-F*F*F*F*F0F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
34 "Int( 1/(x-c) ,x) = int(1/(x-c),x);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%$IntG6$*&\"\"\"F(,&%\"xGF(%\"cG!\"\"F,F*-%#lnG6#F)" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "The above " }{TEXT 269 21 "two cal
culus formulas" }{TEXT -1 47 " are enough to integrate any rational fu
nction!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
105 "Why? Because every rational function is a linear combination of  \+
functions of the form (x-c)^n with c in " }{TEXT 257 1 "C" }{TEXT -1 
10 " and n in " }{TEXT 258 1 "Z" }{TEXT -1 3 "  (" }{TEXT 259 1 "C" }
{TEXT -1 37 " is the field of complex numbers and " }{TEXT 260 1 "Z" }
{TEXT -1 26 " is the ring of integers)." }}}{SECT 1 {PARA 4 "" 0 "" 
{TEXT -1 28 "Rational functions in Maple." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "Let f be a rational function, for \+
example:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "f := x^6/((x-3)^
2*(x-2)^2) - (5*x^2+x-3)/((x-3)*x^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%\"fG,&*&*$)%\"xG\"\"'\"\"\"F+*&),&F)F+\"\"$!\"\"\"\"#F+),&F)F+F1F
0F1F+F0F+*&,(*$)F)F1F+\"\"&F)F+F/F0F+*&F.F+F7F+F0F0" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 159 "We can normalize f, writing f as the quotient \+
of two polynomials, so that we have only one numerator and one denomin
ator. This is done by the command \"normal\"." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 10 "normal(f);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&
,0*$)%\"xG\"\")\"\"\"F)*&\"\"&F))F'F+F)!\"\"*&\"#MF))F'\"\"%F)F)*&\"#q
F))F'\"\"$F)F-*&\"#BF))F'\"\"#F)F)*&\"#gF)F'F)F)\"#OF-F)*(),&F'F)F5F-F
9F)),&F'F)F9F-F9F)F8F)F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 
"f; # f is unchanged:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&*$)%\"xG
\"\"'\"\"\"F)*&),&F'F)\"\"$!\"\"\"\"#F)),&F'F)F/F.F/F)F.F)*&,(*$)F'F/F
)\"\"&F'F)F-F.F)*&F,F)F5F)F.F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
156 "Note that \"normal\" or any other Maple command does not change f
, it just returns the normalized f. If we want to change f we have to \+
use an assignment \":=\"." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
15 "f := normal(f);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&,0*$)%
\"xG\"\")\"\"\"F+*&\"\"&F+)F)F-F+!\"\"*&\"#MF+)F)\"\"%F+F+*&\"#qF+)F)
\"\"$F+F/*&\"#BF+)F)\"\"#F+F+*&\"#gF+F)F+F+\"#OF/F+*(),&F)F+F7F/F;F+),
&F)F+F;F/F;F+F:F+F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "nume
r(f); # the numerator" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,0*$)%\"xG\"
\")\"\"\"F(*&\"\"&F()F&F*F(!\"\"*&\"#MF()F&\"\"%F(F(*&\"#qF()F&\"\"$F(
F,*&\"#BF()F&\"\"#F(F(*&\"#gF(F&F(F(\"#OF," }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 27 "denom(f); # the denominator" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#*(),&%\"xG\"\"\"\"\"$!\"\"\"\"#F'),&F&F'F*F)F*F')F&F*F'
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 159 "The command normal does more
 than just bring everything under the same denominator. It also remove
s common roots of the numerator and denominator. For example:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "f:=( numer(f) * (x^2 - 2*x +
 1) ) / ( denom(f) * (x-1)^2 );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
\"fG*&*&,0*$)%\"xG\"\")\"\"\"F,*&\"\"&F,)F*F.F,!\"\"*&\"#MF,)F*\"\"%F,
F,*&\"#qF,)F*\"\"$F,F0*&\"#BF,)F*\"\"#F,F,*&\"#gF,F*F,F,\"#OF0F,,(*$F;
F,F,*&F<F,F*F,F0F,F,F,F,**),&F*F,F8F0F<F,),&F*F,F<F0F<F,F;F,),&F*F,F,F
0F<F,F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 198 "Now the numerator and
 denominator have a common root: namely x=1 is a common double root. T
he common roots are the roots of the \"gcd\", the \"greatest common di
visor\" of the numerator and denominator." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 9 "numer(f);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,0*$)
%\"xG\"\")\"\"\"F)*&\"\"&F))F'F+F)!\"\"*&\"#MF))F'\"\"%F)F)*&\"#qF))F'
\"\"$F)F-*&\"#BF))F'\"\"#F)F)*&\"#gF)F'F)F)\"#OF-F),(*$F8F)F)*&F9F)F'F
)F-F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "denom(f);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#**),&%\"xG\"\"\"\"\"$!\"\"\"\"#F'),&F&
F'F*F)F*F')F&F*F'),&F&F'F'F)F*F'" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "gcd( numer(f), denom(f) );" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#*$),&%\"xG\"\"\"F'!\"\"\"\"#F'" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 174 "The normal command divides both numerator and denomin
ator by the \"gcd\". So after you apply \"normal\", there can no longe
r be any common roots of the numerator and denominator." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "f:=normal(f);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"fG*&,0*$)%\"xG\"\")\"\"\"F+*&\"\"&F+)F)F-F+!\"\"*&
\"#MF+)F)\"\"%F+F+*&\"#qF+)F)\"\"$F+F/*&\"#BF+)F)\"\"#F+F+*&\"#gF+F)F+
F+\"#OF/F+*(),&F)F+F7F/F;F+),&F)F+F;F/F;F+F:F+F/" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 26 "gcd( numer(f), denom(f) );" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 181 "After
 you do \"normal\", the gcd of the numerator and denominator is always
 1, meaning that a complex number c can never be the root of the numer
ator and denominator at the same time." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 30 "The roots of numer(f) are the " }
{TEXT 263 11 "roots of f." }{TEXT -1 53 " The function f takes the val
ue zero at these points." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 30 "The roots of denom(f) are the " }{TEXT 264 11 "pol
es of f." }{TEXT -1 49 " The function f goes to infinity at these poin
ts." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "Th
e roots of f are:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(
numer(f));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6*-%'RootOfG6$,0*$)%#_ZG\"
\")\"\"\"F+*&\"\"&F+)F)F-F+!\"\"*&\"#MF+)F)\"\"%F+F+*&\"#qF+)F)\"\"$F+
F/*&\"#BF+)F)\"\"#F+F+*&\"#gF+F)F+F+\"#OF//%&indexGF+-F$6$F&/F@F;-F$6$
F&/F@F7-F$6$F&/F@F3-F$6$F&/F@F--F$6$F&/F@\"\"'-F$6$F&/F@\"\"(-F$6$F&/F
@F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 149 "The expression RootOf( po
lynomial, index = i) refers to the i'th root of that polynomial. It is
 a short way to denote a root of a polynomial exactly." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "The poles of f are:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(denom(f));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6(\"\"!F#\"\"#F$\"\"$F%" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 193 "We see that all three poles have multipl
icity 2 because they all appear 2 times. So the function f goes to inf
inity at these points \"just as fast\" as 1/x^2 goes to infinity when \+
x goes to zero." }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 31 "Series expan
sion of a function." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 81 "If R is some function that is convergent at x=0 we can \+
compute the Taylor series:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
16 "R:=sin(x)/(x-2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"RG*&-%$sin
G6#%\"xG\"\"\",&F)F*\"\"#!\"\"F-" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "series(R,x=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#+/%
\"xG#!\"\"\"\"#\"\"\"#F&\"\"%F'#F&\"#C\"\"$#F&\"#[F*#!\"(\"$![\"\"&-%
\"OG6#F(\"\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 468 "Here O(x^6) ref
ers to some function that, when you divide it by x^6, it still converg
es at x=0. So when x goes to zero that function O(x^6) goes to zero at
 least as fast as x^6. In a small neighborhood of x=0 this function O(
x^6) takes very small values. The terms given above (given up to x^5) \+
form a good approximation of the function near x=0. If we want to have
 a more accurate approximation we simply have to take more terms of th
e Taylor expansion, say 15 terms:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "series(R,x=0,15);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#
+A%\"xG#!\"\"\"\"#\"\"\"#F&\"\"%F'#F&\"#C\"\"$#F&\"#[F*#!\"(\"$![\"\"&
#F1\"$g*\"\"'#!$V\"\"&?.%\"\"(#F8\"&S1)\"\")#!%*G\"\"(?:X\"\"\"*#F?\"(
SI!H\"#5#!&\"*3(\"*+WL>$\"#6#FF\"*+)o'Q'\"#7#!(`Zw#\",+k;;)\\\"#8#FM\"
,+GLK'**\"#9-%\"OG6#F(\"#:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 521 "No
w O(x^15) refers to some unknown function that is very small when x is
 close to zero, because divided by x^15 it would still converge at x=0
. So the above terms (given up to x^14) give a very accurate approxima
tion of f near x=0. Near other points, say near x=5 the above series i
s not accurate at all, it would never be accurate no matter how many t
erms we would use because the radius of convergence is only 2. So the \+
above terms only give a good approximation at values for x that are we
ll below 2 in absolute value." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 96 "An approximation of R near x=5 in the for
m of a Taylor series expansion of R at x=5 is given by:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "series(R,x=5);" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#+1,&%\"xG\"\"\"\"\"&!\"\",$-%$sinG6#F'#F&\"\"$\"\"!,&-%
$cosGF,F-*&#F&\"\"*F&F*F&F(F&,&F*#!\"(\"#a*&#F&F5F&F1F&F(\"\"#,&F1#F(F
9*&#\"\"(\"$i\"F&F*F&F&F.,&F*#F(\"%W>*&#F&FBF&F1F&F&\"\"%,&F1#FA\"%?(*
*&#F&\"%KeF&F*F&F&F'-%\"OG6#F&\"\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 78 "How can Maple compute a series expansion of a function that has
 a pole at x=c?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 279 "That's not so hard, just multiply the function by a powe
r of x-c so that the function at some point converges at x=c, then com
pute the coefficients of the Taylor series by computing the value of t
he derivatives of that function at x=c, and at the end divide by that \+
power of x-c." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "R := 1/(co
s(x)-1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"RG*&\"\"\"F&,&-%$cosG6
#%\"xGF&F&!\"\"F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "series
(R,x=0,10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#+-%\"xG!\"#F%#!\"\"\"\"
'\"\"!#F'\"$?\"\"\"##F'\"%CI\"\"%-%\"OG6#\"\"\"F(" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 44 "How could we have calculated that ourselves?" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 217 "Easy eno
ugh, first get rid of the pole by multiplying with x^2. Then compute t
he coefficients as follows. Note that diff(f, x) is the first derivati
ve of f. And diff(f,x,x) = diff(f,x$2) is the second derivative, etc.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "S := R*x^2;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"SG*&*$)%\"xG\"\"#\"\"\"F*,&-%$cosG6#F(F*
F*!\"\"F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "c0 := limit(S,
x=0)/0!;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#c0G!\"#" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c1 := limit(diff(S,x),x=0)/1!;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#c1G\"\"!" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 32 "c2 := limit(diff(S,x,x),x=0)/2!;" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%#c2G#!\"\"\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "c3 := limit(diff(S,x$3),x=0)/3!;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#c3G\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
32 "c4 := limit(diff(S,x$4),x=0)/4!;" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%#c4G#!\"\"\"$?\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 439 "One fina
l remark on a series expansion. Whenever we compute only finitely many
 terms, the function is (in some open subset of the complex numbers) o
nly approximately equal to its expansion. The function only equals its
 expansion if we take all terms, which we can not do on a computer bec
ause there are infinitely many terms. However, if we want to study loc
al properties of a function such as poles, then we only need finitely \+
many terms." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 49 "Poles are very important for integration because:" }}{PARA 0 "
" 0 "" {TEXT -1 44 "If f is a rational function, and if F is an " }
{TEXT 265 14 "antiderivative" }{TEXT -1 16 " of f, that is, " }{TEXT 
266 5 "F'=f " }{TEXT -1 39 "or in Maple language: diff(F,x)=f. Then" }
}{PARA 0 "" 0 "" {TEXT 267 96 "Important principle:  f has a pole at x
=c if and only if the antiderivative F has a pole at x=c." }}{PARA 0 "
" 0 "" {TEXT -1 410 "This means that when we want to compute F := int(
f,x), then we already know all the poles of F. Not only that, we also \+
know the pole orders, because if f has a pole of order 5 then F must h
ave a pole of order 4 at x=c. And if f has a pole of order 1 then F mu
st have a logarithm. So this trivial principle already gives a lot of \+
information about the antiderivative F, which we will have to use to d
etermine F." }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 122 "Writing a ratio
nal function f as a linear combination of (x-c)^n and integrating thes
e terms by the two calculus formulas." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "f := normal( x^6/((x-3)^
2*(x-2)^2) - (5*x^2+x-3)/((x-3)*x^2) );" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"fG*&,0*$)%\"xG\"\")\"\"\"F+*&\"\"&F+)F)F-F+!\"\"*&\"#MF+)F)
\"\"%F+F+*&\"#qF+)F)\"\"$F+F/*&\"#BF+)F)\"\"#F+F+*&\"#gF+F)F+F+\"#OF/F
+*(),&F)F+F7F/F;F+),&F)F+F;F/F;F+F:F+F/" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 53 "We have seen that the poles of f are c=0, c= 2, c= 3." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "A rationa
l function f is a polynomial if and only if it has no poles c in " }
{TEXT 256 1 "C" }{TEXT -1 59 ", in other words: If for every complex n
umber c, the limit:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "Limi
t(f, x=c);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%&LimitG6$*&,0*$)%\"xG
\"\")\"\"\"F,*&\"\"&F,)F*F.F,!\"\"*&\"#MF,)F*\"\"%F,F,*&\"#qF,)F*\"\"$
F,F0*&\"#BF,)F*\"\"#F,F,*&\"#gF,F*F,F,\"#OF0F,*(),&F*F,F8F0F<F,),&F*F,
F<F0F<F,F;F,F0/F*%\"cG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 199 "exists
 and is finite then f is a polynomial. For this particular f this is n
ot true for c=0,2,3, and f is not a polynomial. Note that Limit is the
 \"inert\" = \"do nothing\" version of the command limit." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Limit(f, x=0) = limit(f,x=0);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%&LimitG6$*&,0*$)%\"xG\"\")\"\"\"F-*
&\"\"&F-)F+F/F-!\"\"*&\"#MF-)F+\"\"%F-F-*&\"#qF-)F+\"\"$F-F1*&\"#BF-)F
+\"\"#F-F-*&\"#gF-F+F-F-\"#OF1F-*(),&F+F-F9F1F=F-),&F+F-F=F1F=F-F<F-F1
/F+\"\"!,$%)infinityGF1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "
Limit(f,x=2) = limit(f,x=2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%&Li
mitG6$*&,0*$)%\"xG\"\")\"\"\"F-*&\"\"&F-)F+F/F-!\"\"*&\"#MF-)F+\"\"%F-
F-*&\"#qF-)F+\"\"$F-F1*&\"#BF-)F+\"\"#F-F-*&\"#gF-F+F-F-\"#OF1F-*(),&F
+F-F9F1F=F-),&F+F-F=F1F=F-F<F-F1/F+F=%)infinityG" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 28 "Limit(f,x=3) = limit(f,x=3);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#/-%&LimitG6$*&,0*$)%\"xG\"\")\"\"\"F-*&\"\"&F-)F+F
/F-!\"\"*&\"#MF-)F+\"\"%F-F-*&\"#qF-)F+\"\"$F-F1*&\"#BF-)F+\"\"#F-F-*&
\"#gF-F+F-F-\"#OF1F-*(),&F+F-F9F1F=F-),&F+F-F=F1F=F-F<F-F1/F+F9%)infin
ityG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "limit(f,x=7);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6##\"(,*Rd\"&+'>" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 318 "We see that at x is 0, 2 or 3, the function f goes \+
to plus or minus infinity, and at any other x (such as the randomly ch
osen x=7) it does not go to plus or minus infinity. If we want to know
 more precisely what f \"does\" at the points x=0, x=2 and x=3, we hav
e to compute a power series expansion of f at those points." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "series(f,x=0);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#+/%\"xG!\"\"!\"##\"\"&\"\"$\"\"!#F(\"\"*\"\"\"#F
(\"#F\"\"##F(\"#\")F)-%\"OG6#F-\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 21 "We see that f equals:" }}{PARA 0 "" 0 "" {TEXT -1 34 "   -x^(-2
) + 5/3 + 5/9*x  + O(x^2)" }}{PARA 0 "" 0 "" {TEXT -1 141 "where O(x^2
) refers to a function that takes small values when x is close to 0. A
 function that, when divided by x^2, still converges at x=0." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "In particular f
 equals:" }}{PARA 0 "" 0 "" {TEXT -1 45 "  -x^(-2) + a function that c
onverges at x=0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "series(
f,x=2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#+1,&%\"xG\"\"\"\"\"#!\"\"\"
#k!\"#\"$?$F(#\"%$G$\"\"%\"\"!#\"%4fF.F&#\"&&3N\"#;F'#\"&pL#\"\")\"\"$
-%\"OG6#F&F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "We see that f equ
als:" }}{PARA 0 "" 0 "" {TEXT -1 75 "  64*(x-2)^(-2) + 320*(x-2)^(-1) \+
plus some function that converges at x=2. " }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 14 "series(f,x=3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#+
1,&%\"xG\"\"\"\"\"$!\"\"\"$H(!\"#!\"&F(#\"%tV\"\"*\"\"!#!&i;\"\"#FF&#
\"&]Q\"F2\"\"##!'k*R\"\"$V#F'-%\"OG6#F&\"\"%" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 17 "So f also equals:" }}{PARA 0 "" 0 "" {TEXT -1 68 "729*(
x-3)^(-2) - 5*(x-3)^(-1) plus a function that converges at x=3." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "If we com
bine all of this:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 77 "S := -x^(-2) + 64*(x-2)^(-2) + 320*(x-2)^(-1) + 729*(x-3)
^(-2) - 5*(x-3)^(-1)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 17 "then we see that:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 55 "f = S plus a function g that converges at
 x=0, 2 and 3." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "S := -x^(
-2) + 64*(x-2)^(-2) + 320*(x-2)^(-1) + 729*(x-3)^(-2) - 5*(x-3)^(-1);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"SG,,*&\"\"\"F'*$)%\"xG\"\"#F'!
\"\"F,*&\"#kF'*$),&F*F'F+F,F+F'F,F'*&\"$?$F'F1F,F'*&\"$H(F'*$),&F*F'\"
\"$F,F+F'F,F'*&\"\"&F'F8F,F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "So
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "g := f - S;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"gG,.*&,0*$)%\"xG\"\")\"\"\"F,*&\"\"&F,)F
*F.F,!\"\"*&\"#MF,)F*\"\"%F,F,*&\"#qF,)F*\"\"$F,F0*&\"#BF,)F*\"\"#F,F,
*&\"#gF,F*F,F,\"#OF0F,*(),&F*F,F8F0F<F,),&F*F,F<F0F<F,F;F,F0F,*&F,F,*$
F;F,F0F,*&\"#kF,*$FCF,F0F0*&\"$?$F,FDF0F0*&\"$H(F,*$FAF,F0F0*&F.F,FBF0
F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 392 "Now g must be a function w
ith no poles at x=0, x=2, x=3. Then why do we still see x, x-2 and x-3
 in the denominator? Well, that is because whenever we add or multiply
 rational functions, we must apply \"normal\" to get the normalization
. This will cause the roots x=0, x=2, x=3 in the denominator to cancel
 against roots of the numerator, which causes these roots of the denom
inator to go away." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "g:=no
rmal(g);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG,(*$)%\"xG\"\"#\"\"
\"F**&\"#5F*F(F*F*\"#jF*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 
"f = S + g;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#/*&,0*$)%\"xG\"\")\"\"
\"F**&\"\"&F*)F(F,F*!\"\"*&\"#MF*)F(\"\"%F*F**&\"#qF*)F(\"\"$F*F.*&\"#
BF*)F(\"\"#F*F**&\"#gF*F(F*F*\"#OF.F**(),&F(F*F6F.F:F*),&F(F*F:F.F:F*F
9F*F.,2*&F*F**$F9F*F.F.*&\"#kF**$FAF*F.F**&\"$?$F*FBF.F**&\"$H(F**$F?F
*F.F**&F,F*F@F.F.*$F9F*F**&\"#5F*F(F*F*\"#jF*" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 87 "As you can see, we have now written f as a linear comb
ination of functions of the form:" }}{PARA 0 "" 0 "" {TEXT -1 10 "   (
x-c)^n" }}{PARA 0 "" 0 "" {TEXT -1 10 "with c in " }{TEXT 261 1 "C" }
{TEXT -1 10 " and n in " }{TEXT 262 2 "Z." }}{PARA 0 "" 0 "" {TEXT -1 
64 "In other words we have written f as a sum of things of the form:" 
}}{PARA 0 "" 0 "" {TEXT -1 46 " (some number)*(x-some number)^(some in
teger)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
124 "We can integrate each of these terms, x^2, or 1/(x-2) or 1/(x-3)^
2, etc., by the calculus formulas given before, and we get:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "S+g;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,2*&\"\"\"F%*$)%\"xG\"\"#F%!\"\"F**&\"#kF%*$),&F(F%F)F*
F)F%F*F%*&\"$?$F%F/F*F%*&\"$H(F%*$),&F(F%\"\"$F*F)F%F*F%*&\"\"&F%F6F*F
**$F'F%F%*&\"#5F%F(F%F%\"#jF%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 9 "int(%,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,2*&\"\"\"F%%\"xG!\"
\"F%*&\"#kF%,&F&F%\"\"#F'F'F'*&\"$?$F%-%#lnG6#F*F%F%*&\"$H(F%,&F&F%\"
\"$F'F'F'*&\"\"&F%-F/6#F3F%F'*&#F%F4F%)F&F4F%F%*&F6F%)F&F+F%F%*&\"#jF%
F&F%F%" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 55 "Exercise: Compute an \+
integral with the calculus method." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 183 "Compute the poles of the following ratio
nal function by applying Maple's \"solve\" on the denom(f). Then compu
te the series expansions of f at those poles. Then, like above, write \+
f as:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 11 "
f =  S + g," }}{PARA 0 "" 0 "" {TEXT -1 57 "f = a linear combination o
f some (x-c)^n for some c's in " }{TEXT 270 2 "C " }{TEXT -1 11 "and n
's in " }{TEXT 271 2 "Z," }{TEXT -1 22 " plus some polynomial." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 105 "After th
at, integrating f is easy by applying the two calculus formulas. Here \+
is the rational function f:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "f:=(x^7-5*x^5-2*x^2-4*x-1+4*x^3)/(x
^6+2*x^5-2*x^4-4*x^3+x^2+2*x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"
fG*&,.*$)%\"xG\"\"(\"\"\"F+*&\"\"&F+)F)F-F+!\"\"*&\"\"#F+)F)F1F+F/*&\"
\"%F+F)F+F/F+F/*&F4F+)F)\"\"$F+F+F+,.*$)F)\"\"'F+F+*&F1F+F.F+F+*&F1F+)
F)F4F+F/*&F4F+F6F+F/*$F2F+F+*&F1F+F)F+F+F/" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 494 "Open a new Maple worksheet, copy this rational function \+
to your worksheet, find the poles, compute the series expansions at th
e poles, write f as a sum of things that look like (some number)*(x-so
me number)^(some number) plus a polynomial. Then integrate each of the
 terms in the sum to find an antiderivative of f. Verify your answer b
y computing the derivative (use Maple!) of your result, normalize that
 derivative, and see if it is the same as f. E-mail your worksheet to \+
hoeij@math.fsu.edu" }}{PARA 0 "" 0 "" {TEXT -1 43 "I will grade the wo
rksheets in the weekend." }}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 35 "A \+
problem with the calculus method." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "f := x^3*(-1+2*x^2)/(1-x^2+x
^4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&*&)%\"xG\"\"$\"\"\",&!
\"\"F**&\"\"#F*)F(F.F*F*F*F*,(F*F**$F/F*F,*$)F(\"\"%F*F*F," }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "denom(f);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,(\"\"\"F$*$)%\"xG\"\"#F$!\"\"*$)F'\"\"%F$F$" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "solve(%,x);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6&,$*$-%%sqrtG6#,&\"\"#\"\"\"*&^#F)F*-F&6#\"\"$F*F*F
*#!\"\"F),$F$#F*F),$*$-F&6#,&F)F**&^#!\"#F*F-F*F*F*F0,$F5F3" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "c1, c2, c3, c4 := %;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>6&%#c1G%#c2G%#c3G%#c4G6&,$*$-%%sqrtG6
#,&\"\"#\"\"\"*&^#F0F1-F-6#\"\"$F1F1F1#!\"\"F0,$F+#F1F0,$*$-F-6#,&F0F1
*&^#!\"#F1F4F1F1F1F7,$F<F:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Now
 c1, c2, c3, c4 are the 4 poles of f." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 3 "c1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$-%%sqrtG6#,
&\"\"#\"\"\"*&^#F)F*-F&6#\"\"$F*F*F*#!\"\"F)" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 17 "series(f,x=c1,2);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#+',&%\"xG\"\"\"*&#F&\"\"#F&-%%sqrtG6#,&F)F&*&^#F)F&-F+6#\"\"$F&F
&F&F&*&**^##!\"\"F)F&F*F&,&F(F&*&^#F(F&F0F&F&F&F0F&F&,&*$F*F&F&*(F)F&F
*F&F8F&F7F7F7-%\"OG6#F&\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 19 "convert(%,polynom);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&**^##!
\"\"\"\"#\"\"\"-%%sqrtG6#,&F(F)*&^#F(F)-F+6#\"\"$F)F)F),&#F)F(F)*&^#F4
F)F0F)F)F)F0F)F)*&,&*$F*F)F)*(F(F)F*F)F3F)F'F),&%\"xGF)*&F4F)F*F)F)F)F
'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 179 "Converting the series expan
sion into a polynomial is done so that we can select coefficients. We \+
only need the coefficient of the only term that has a negative exponen
t in (x-c1)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "r1 := coeff
(%,(x-c1)^(-1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r1G*&**^##!\"\"
\"\"#\"\"\"-%%sqrtG6#,&F*F+*&^#F*F+-F-6#\"\"$F+F+F+,&#F+F*F+*&^#F6F+F2
F+F+F+F2F+F+,&*$F,F+F+*(F*F+F,F+F5F+F)F)" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "r1:=simplify(r1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%#r1G,&#\"\"\"\"\"%F'*&^#F&F'-%%sqrtG6#\"\"$F'F'" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 94 "r2, r3, r4 := seq(simplify(coeff(convert(ser
ies(f,x=c||i,2),polynom),(x-c||i)^(-1))), i=2..4);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>6%%#r2G%#r3G%#r4G6%,&#\"\"\"\"\"%F+*&^#F*F+-%%sqrtG6
#\"\"$F+F+,&F*F+*&^##!\"\"F,F+F/F+F+F3" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 71 "S := r1*(x-c1)^(-1) + r2*(x-c2)^(-1) + r3*(x-c3)^(-1)
 + r4*(x-c4)^(-1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"SG,**&,&#\"
\"\"\"\"%F)*&^#F(F)-%%sqrtG6#\"\"$F)F)F),&%\"xGF)*&#F)\"\"#F)-F.6#,&F5
F)*&^#F5F)F-F)F)F)F)!\"\"F)*&F'F),&F2F)*&#F)F5F)*$F6F)F)F;F;F)*&,&F(F)
*&^##F;F*F)F-F)F)F),&F2F)*&F4F)-F.6#,&F5F)*&^#!\"#F)F-F)F)F)F)F;F)*&FB
F),&F2F)*&#F)F5F)*$FHF)F)F;F;F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 94 "g:=evala(f-S); # use evala instead of normal for normalizing w
hen there are algebraic numbers." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
\"gG,$%\"xG\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "f = g +
 S;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&*&)%\"xG\"\"$\"\"\",&!\"\"F)
*&\"\"#F))F'F-F)F)F)F),(F)F)*$F.F)F+*$)F'\"\"%F)F)F+,,F'F-*&,&#F)F3F)*
&^#F7F)-%%sqrtG6#F(F)F)F),&F'F)*&#F)F-F)-F;6#,&F-F)*&^#F-F)F:F)F)F)F)F
+F)*&F6F),&F'F)*&#F)F-F)*$F@F)F)F+F+F)*&,&F7F)*&^##F+F3F)F:F)F)F),&F'F
)*&F?F)-F;6#,&F-F)*&^#!\"#F)F:F)F)F)F)F+F)*&FKF),&F'F)*&#F)F-F)*$FQF)F
)F+F+F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "Now apply the calculus
 formulas to integrate g+S" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
11 "int(g+S,x);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,4*$)%\"xG\"\"#\"\"
\"F(*&#F(\"\"%F(-%#lnG6#,&F&F(*&#F(F'F(-%%sqrtG6#,&F'F(*&^#F'F(-F36#\"
\"$F(F(F(F(F(F(*(^#F*F(F,F(F8F(F(*&F*F(-F-6#,&F&F(*&#F(F'F(*$F2F(F(!\"
\"F(F(*(F<F(F>F(F8F(F(*&F*F(-F-6#,&F&F(*&F1F(-F36#,&F'F(*&^#!\"#F(F8F(
F(F(F(F(F(*(^##FDF+F(FGF(F8F(F(*&F*F(-F-6#,&F&F(*&#F(F'F(*$FKF(F(FDF(F
(*(FRF(FUF(F8F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 251 "We see that
 we get a terrible answer because we are computing with all roots of d
enom(f), which introduces nasty expressions such as nested square root
s. For other examples it can even get much worse. How come Maple gets \+
a nicer (i.e. shorter) answer?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 9 "int(f,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*$)%\"xG\"\"#\"
\"\"F(*&#F(\"\"%F(-%#lnG6#,(F(F(F$!\"\"*$)F&F+F(F(F(F(*&#F(F'F(*&-%%sq
rtG6#\"\"$F(-%'arctanG6#,$*&,&F0F(*&F'F(F%F(F(F(F6F(#F(F9F(F(F0" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 170 "Apparantly Maple does not calcula
te the roots c1,c2,c3,c4 of denom(f), because that was where the all t
he square roots in our answer (with the calculus method) came from." }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 
"" {TEXT -1 57 "Some rational functions that should be easy to integra
te." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 56 "F := (1-x)/(x^10-x^7+2*x^6-2*x^8-3*x^4-x^3+3*x^2+3*x+
1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"FG*&,&\"\"\"F'%\"xG!\"\"F',
4*$)F(\"#5F'F'*$)F(\"\"(F'F)*&\"\"#F')F(\"\"'F'F'*&F2F')F(\"\")F'F)*&
\"\"$F')F(\"\"%F'F)*$)F(F9F'F)*&F9F')F(F2F'F'*&F9F'F(F'F'F'F'F)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "diff(F,x);" }}{PARA 12 "" 1 
"" {XPPMATH 20 "6#,&*&\"\"\"F%,4*$)%\"xG\"#5F%F%*$)F)\"\"(F%!\"\"*&\"
\"#F%)F)\"\"'F%F%*&F0F%)F)\"\")F%F.*&\"\"$F%)F)\"\"%F%F.*$)F)F7F%F.*&F
7F%)F)F0F%F%*&F7F%F)F%F%F%F%F.F.*&*&,&F%F%F)F.F%,2*$)F)\"\"*F%F**&F-F%
F1F%F.*&\"#7F%)F)\"\"&F%F%*&\"#;F%F,F%F.*&FHF%F;F%F.*&F7F%F=F%F.*&F2F%
F)F%F%F7F%F%F%*$)F&F0F%F.F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
23 "f:=normal(%, expanded);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*
&,2*$)%\"xG\"\"(\"\"\"\"\"**&\"\"#F+)F)\"\"$F+F+*&\"\"&F+)F)F2F+!\"\"*
&\"\")F+)F)F.F+F4*&F.F+F)F+F+\"\"%F+*&\"#5F+)F)\"\"'F+F4*&F,F+)F)F9F+F
+F+,BF4F+*&F2F+F)F+F4*&\"#8F+)F)F,F+F+*&F=F+)F)F6F+F+*$)F)F;F+F4*$)F)
\"#9F+F4*&F2F+)F)FCF+F+*&F0F+)F)\"#:F+F4*&F,F+)F)\"#6F+F4*$)F)\"#<F+F+
*&F2F+F?F+F+*&F*F+F/F+F4*&F;F+F7F+F4*&F6F+F(F+F4*&F0F+F<F+F4*&F,F+F3F+
F+F4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "d := denom(f);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG,B!\"\"\"\"\"*&\"\"&F'%\"xGF'F&*
&\"#8F')F*\"\"*F'F'*&\"\"'F')F*\"\")F'F'*$)F*\"#5F'F&*$)F*\"#9F'F&*&F)
F')F*F,F'F'*&\"\"$F')F*\"#:F'F&*&F.F')F*\"#6F'F&*$)F*\"#<F'F'*&F)F')F*
\"\"%F'F'*&\"\"(F')F*F<F'F&*&F5F')F*\"\"#F'F&*&F2F')F*FIF'F&*&F<F')F*F
0F'F&*&F.F')F*F)F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "set
_of_poles := \{solve(% , x)\};" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%-s
et_of_polesG<),&*$),&\"$3\"\"\"\"*&\"#7F+-%%sqrtG6#\"#pF+F+#F+\"\"$F+#
F+\"\"'*&\"\"#F+*$)F)#F+F3F+!\"\"F+,(F'#F;F-*&F+F+*$)F)#F+F3F+F;F;*(^#
#F;F7F+-F/6#F3F+,&F'F4*&F7F+*$)F)#F+F3F+F;F;F+F+,(F'F=F>F;*(^##F+F7F+F
EF+FGF+F+-%'RootOfG6$,(*$)%#_ZG\"\"%F+F+FVF+F+F+/%&indexGF3-FQ6$FS/FYF
7-FQ6$FS/FYF+-FQ6$FS/FYFW" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 201 "Thi
s set of roots of d=denom(f) has 7 elements, say c1,..,c7. The roots c
1,..,c7 of d are the poles of f, because f is normalized which implies
 that the roots of denom(f) are not roots of the numer(f)." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 478 "So there 7 poles
, c1,..,c7. We would rather not compute with these c1,..,c7 because th
ey are not-so-simple expressions. Computing with c1,..,c7 will only le
ad to more complicated expressions. It shouldn't be necessary to compu
te with these complicated expressions c1,..,c7, after all, the integra
l int(f,x) = F is a simple expression (see top of this section). There
 has to be some quick computation with expressions not involving these
 complicated c1,..,c7 that will lead to F." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 100 "Main idea: Try to compute just
 with polynomials, instead of with the roots c1,..,c7 of a polynomial.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "Some \+
questions arise:" }}{PARA 0 "" 0 "" {TEXT -1 88 "1) How do we compute \+
the pole orders f if we don't even compute with the poles c1,..,c7?" }
}{PARA 0 "" 0 "" {TEXT -1 263 "2) Don't we need a series expansion at \+
each pole, and doesn't that make it necessary to compute with whatever
 expression it takes to write down that pole? (if we use the calculus \+
method then yes, then we do need to compute with that expression c1, o
r c2, .., c7)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 65 "Square-free decomposition (for more details see other wor
ksheet)." }}{PARA 0 "" 0 "" {TEXT -1 45 "If d = d1^1 * d2^2 * d3^3 * d
4^4 * ... * dn^n" }}{PARA 0 "" 0 "" {TEXT -1 4 "then" }}{PARA 0 "" 0 "
" {TEXT -1 48 "gcd(d, diff(d,x)) = d2^1 * d3^2 * ... * dn^(n-1)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "step1 := gcd(d, diff(d,x)); \+
# d2^1*d3^2*d4^3*..." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&step1G,4*$)
%\"xG\"#5\"\"\"F**$)F(\"\"(F*!\"\"*&\"\"#F*)F(\"\"'F*F**&F0F*)F(\"\")F
*F.*&\"\"$F*)F(\"\"%F*F.*$)F(F7F*F.*&F7F*)F(F0F*F**&F7F*F(F*F*F*F*" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "step2 := gcd(%, diff(%,x));
 # d3^1*d4^2*d5^3*..." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&step2G,(*$
)%\"xG\"\"$\"\"\"F*F(!\"\"F*F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 34 "gcd(%, diff(%,x)); # d4^1*d5^2*..." }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 145 "So ther
e is no d4 or d5 or.. So the highest order of any root of d is 3. The \+
roots of f that have order 3 are precisely the roots of d3 which is:" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "d3:=step2;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%#d3G,(*$)%\"xG\"\"$\"\"\"F*F(!\"\"F*F+" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "Now the polynomial given in step1 \+
equals d2*d3^2, so d2 is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
23 "d2:=normal(step1/d3^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#d2G,
(*$)%\"xG\"\"%\"\"\"F*F(F*F*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 
"So we find that d = (some number)*d2^2*d3^3, and that (some number) e
quals:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "normal(d/d2^2/d3^
3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 257 "What we did above, we can let Maple do that by the co
mmand sqrfree(d,x). Then it computes that (some number) which is 1 in \+
this example. It also computes d1 (absent, there are no roots of multi
plicity 1), d2 and d3, and indicates the multiplicities as well." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "sqrfree(d,x);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#7$\"\"\"7$7$,(*$)%\"xG\"\"$F$F$F*!\"\"F$F,F+7$,(
*$)F*\"\"%F$F$F*F$F$F$\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "Th
e above means d = 1 * (x^3-x-1)^3 * (x^4+x+1)^2." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 161 "As you can see we have d
etermined the order of the roots of d, in other words we have determin
ed the pole orders (2 and 3) of f, without computing with c1,..,c7." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "TO BE CO
NTINUED." }}}}}{MARK "6" 0 }{VIEWOPTS 1 1 0 3 2 1804 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }