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Partial fraction decom position is used to integrate rational functions, but it also has othe r important applications that you will see later. Among others, partia l fractions are used in engineering for finding inverse Laplace transf orms. Let's look at the following example:" }}{PARA 0 "" 0 "" {TEXT 237 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Int(1/(x^2-4),x ); value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&\"\"\"F',&* $)%\"xG\"\"#F'F'\"\"%!\"\"F.F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*& #\"\"\"\"\"%F&-%#lnG6#,&%\"xGF&\"\"#!\"\"F&F&*&F%F&-F)6#,&F,F&F-F&F&F. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 145 "How did Maple find the integral? Exactly the same w ay that you could find it by hand: by partial fraction decomposition o f the rational function " }{XPPEDIT 18 0 "1/(x^2-4);" "6#*&\"\"\"F$,&* $)%\"xG\"\"#F$F$\"\"%!\"\"F+" }{TEXT 237 11 " . Namely:" }}{PARA 206 "" 0 "" {XPPEDIT 18 0 "1/(x^2-4) = 1/(4*(x-2))-1/(4*(x+2));" "6#/*&\" \"\"F%,&*$)%\"xG\"\"#F%F%\"\"%!\"\"F,,&*&F%F%*&F+F%,&F)F%F*F,F%F,F%*&F %F%*&F+F%,&F)F%F*F%F%F,F," }{TEXT 200 2 " ." }}{PARA 207 "" 0 "" {TEXT 200 0 "" }}{PARA 0 "" 0 "" {TEXT 237 38 "How can such a decomposition be found?" }}{PARA 0 "" 0 "" {TEXT 204 9 "Example 1" }{TEXT 237 57 ". Find, by hand, the partial fraction decomposition of " }{XPPEDIT 18 0 "1/(x^2-4);" "6#*&\"\"\"F$,&*$)%\"xG\"\"#F$F$\"\"%!\"\"F+" }{TEXT 237 5 " . " }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 260 "If the numerator of a given rational function is of a lower \+ degree than its denominator, you start your work by factoring the deno minator. (Otherwise, as in Example 5 below, you perform the long divis ion first.) The denominator of the above function factors to" } {XPPEDIT 18 0 "(x-2)*(x+2);" "6#*&,&%\"xG\"\"\"\"\"#!\"\"F&,&F%F&F'F&F &" }{TEXT 237 136 ". According to an important theorem in algebra, thi s means that the rational function can be decomposed into the followin g simpler terms" }}{PARA 208 "" 0 "" {XPPEDIT 18 0 "1/(x^2-4) = A/(x-2 )+B/(x+2);" "6#/*&\"\"\"F%,&*$)%\"xG\"\"#F%F%\"\"%!\"\"F,,&*&%\"AGF%,& F)F%F*F,F,F%*&%\"BGF%,&F)F%F*F%F,F%" }{TEXT 200 2 " " }}{PARA 0 "" 0 "" {TEXT 237 70 "for some constants A, B. To find these constants we u se the method of " }{TEXT 205 25 "undetermined coefficients" }{TEXT 237 56 ". If the above equality is to hold then it must be that:" }} {PARA 209 "" 0 "" {XPPEDIT 18 0 "1/(x^2-4) = (A*(x+2)+B*(x-2))/((x-2)* (x+2));" "6#/*&\"\"\"F%,&*$)%\"xG\"\"#F%F%\"\"%!\"\"F,*&,&*&%\"AGF%,&F )F%F*F%F%F%*&%\"BGF%,&F)F%F*F,F%F%F%*&F4F%F1F%F," }{TEXT 200 2 " ." }} {PARA 0 "" 0 "" {TEXT 237 94 "Since the denominators on both sides are the same, the numerators also must be the same. Hence" }}{PARA 210 "" 0 "" {XPPEDIT 18 0 "1 = A*(x+2)+B*(x-2);" "6#/\"\"\",&*&%\"AGF$,&%\"x GF$\"\"#F$F$F$*&%\"BGF$,&F)F$F*!\"\"F$F$" }{TEXT 200 2 " ." }}{PARA 0 "" 0 "" {TEXT 237 53 "Rearranging the terms in the left hand side we o btain" }}{PARA 211 "" 0 "" {XPPEDIT 18 0 "1 = (A+B)*x+2*A-2*B;" "6#/\" \"\",(*&,&%\"AGF$%\"BGF$F$%\"xGF$F$*&\"\"#F$F(F$F$*&F,F$F)F$!\"\"" } {TEXT 200 1 "." }}{PARA 0 "" 0 "" {TEXT 237 163 "We know that two poly nomials are equal if and only if coefficients at the corresponding pow ers of x are equal. Hence, we obtain the following equations for A and B" }}{PARA 212 "" 0 "" {XPPEDIT 18 0 "A+B = 0, 2*A-2*B = 1;" "6$/,&% \"AG\"\"\"%\"BGF&\"\"!/,&*&\"\"#F&F%F&F&*&F,F&F'F&!\"\"F&" }{TEXT 200 2 " ," }}{PARA 0 "" 0 "" {TEXT 237 15 "which gives us " }{XPPEDIT 18 0 "A = 1/4, B = -1/4;" "6$/%\"AG*&\"\"\"F&\"\"%!\"\"/%\"BG,$F%F(" } {TEXT 237 10 " . Thus " }}{PARA 213 "" 0 "" {XPPEDIT 18 0 "1/(x^2-4) = 1/(4*(x-2))-1/(4*(x+2));" "6#/*&\"\"\"F%,&*$)%\"xG\"\"#F%F%\"\"%!\" \"F,,&*&F%F%*&F+F%,&F)F%F*F,F%F,F%*&F%F%*&F+F%,&F)F%F*F%F%F,F," }{TEXT 200 2 " ." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 112 "We have obtained the partial fraction decomposition, which, o f course allows us to find easily the integral of " }{XPPEDIT 18 0 "1 /(x^2-4);" "6#*&\"\"\"F$,&*$)%\"xG\"\"#F$F$\"\"%!\"\"F+" }{TEXT 237 2 " ." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 243 "It is important to know what kind of partial fraction decomposition t o expect for a given rational function. The key is to factor the denom inator. Then the general rules are as follows. If the factored denomin ator contains a first degree term " }{XPPEDIT 18 0 "a*x+b;" "6#,&*&%\" aG\"\"\"%\"xGF&F&%\"bGF&" }{TEXT 237 72 " of mulitiplicity 1, this ter m gives rise to a single partial fraction " }{XPPEDIT 18 0 "A/(a*x+b) ;" "6#*&%\"AG\"\"\",&*&%\"aGF%%\"xGF%F%%\"bGF%!\"\"" }{TEXT 237 75 " \+ for some constant A, as in Example 1. If the denominator contains a te rm " }{XPPEDIT 18 0 "(a*x+b)^k;" "6#),&*&%\"aG\"\"\"%\"xGF'F'%\"bGF'% \"kG" }{TEXT 237 72 " of multiplicity k, this term gives rise to possi bly k partial fractions" }}{PARA 214 "" 0 "" {XPPEDIT 18 0 "A[1]/(a*x+ b);" "6#*&&%\"AG6#\"\"\"F',&*&%\"aGF'%\"xGF'F'%\"bGF'!\"\"" }{TEXT 200 3 " , " }{XPPEDIT 18 0 "A[2]/((a*x+b)^2);" "6#*&&%\"AG6#\"\"#\"\" \"*$),&*&%\"aGF(%\"xGF(F(%\"bGF(F'F(!\"\"" }{TEXT 200 9 " , ..., " } {XPPEDIT 18 0 "A[k]/(a*x+b)^k;" "6#*&&%\"AG6#%\"kG\"\"\"),&*&%\"aGF(% \"xGF(F(%\"bGF(F'!\"\"" }{TEXT 200 2 " ," }}{PARA 0 "" 0 "" {TEXT 237 19 "for some constants " }{XPPEDIT 18 0 "A[1];" "6#&%\"AG6#\"\"\"" } {TEXT 237 2 ", " }{XPPEDIT 18 0 "A[2];" "6#&%\"AG6#\"\"#" }{TEXT 237 4 ",..." }{XPPEDIT 18 0 "A[k];" "6#&%\"AG6#%\"kG" }{TEXT 237 69 ". If \+ the factored denominator contains an irreducible quadratic term " } {XPPEDIT 18 0 "a*x^2+b*x+c;" "6#,(*&%\"aG\"\"\"*$)%\"xG\"\"#F&F&F&*&% \"bGF&F)F&F&%\"cGF&" }{TEXT 237 94 ", that is, a quadratic term which \+ does not have real roots, that term gives rise to a fraction" }}{PARA 215 "" 0 "" {XPPEDIT 18 0 "(B*x+C)/(a*x^2+b*x+c);" "6#*&,&*&%\"BG\"\" \"%\"xGF'F'%\"CGF'F',(*&%\"aGF'*$)F(\"\"#F'F'F'*&%\"bGF'F(F'F'%\"cGF'! \"\"" }{TEXT 200 2 " " }}{PARA 0 "" 0 "" {TEXT 237 19 "for some const ants " }{TEXT 206 6 "B, C. " }{TEXT 237 69 "If the factored denominato r contains an irreducible quadratic factor " }{XPPEDIT 18 0 "(a*x^2+b* x+c)^k;" "6#),(*&%\"aG\"\"\"*$)%\"xG\"\"#F'F'F'*&%\"bGF'F*F'F'%\"cGF'% \"kG" }{TEXT 237 76 " of multiplicity k, that term gives rise to possi bly k fractions of the form" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 216 "" 0 "" {XPPEDIT 18 0 "(B[1]*x+C[1])/(a*x^2+b*x+c);" "6#*&,&*&&%\" BG6#\"\"\"F)%\"xGF)F)&%\"CGF(F)F),(*&%\"aGF)*$)F*\"\"#F)F)F)*&%\"bGF)F *F)F)%\"cGF)!\"\"" }{TEXT 200 3 " ," }{XPPEDIT 18 0 "(B[2]*x+C[2])/(( a*x^2+b*x+c)^2);" "6#*&,&*&&%\"BG6#\"\"#\"\"\"%\"xGF*F*&%\"CGF(F*F**$) ,(*&%\"aGF**$)F+F)F*F*F**&%\"bGF*F+F*F*%\"cGF*F)F*!\"\"" }{TEXT 200 10 " , ..., " }{XPPEDIT 18 0 "(B[k]*x+C[k])/(a*x^2+b*x+c)^k;" "6#*&, &*&&%\"BG6#%\"kG\"\"\"%\"xGF*F*&%\"CGF(F*F*),(*&%\"aGF**$)F+\"\"#F*F*F **&%\"bGF*F+F*F*%\"cGF*F)!\"\"" }{TEXT 200 2 " ." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 207 32 "An important th eorem in algebra " }{TEXT 237 339 "says that every polynomial with rea l coefficients can be factored into first degree polynomials and irred ucible quadratic polynomials with real coefficients. Furthermore, any \+ rational function such that its denominator has the degree greater tha n its numerator, can be decomposed into partial fractions according to the rules stated above." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 " " 0 "" {TEXT 237 320 "The theorem says that decomposition into partial fractions is possible. Quite another matter is finding such decomposi tion for a given rational function. As you can quess, the first obstac le may be factoring the denominator, which often is a difficult task. \+ Maple can handle a variety of situations with simple commands." }} {PARA 0 "" 0 "" {TEXT 237 2 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 68 "The basic command for finding a partial fraction decomposition is \+ \"" }{TEXT 208 8 "convert(" }{TEXT 209 11 "expression," }{TEXT 210 8 " parfrac," }{TEXT 211 8 "variable" }{TEXT 212 1 ")" }{TEXT 237 15 "\". \+ For example" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "convert(1/(x^2-4),parfrac,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*(#\"\"\"\"\"%F&F&F&,&%\"xGF&\"\"#!\"\"F+F&*(F%F&F&F &,&F)F&F*F&F+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 4 "" 0 "" {TEXT 238 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 213 11 "Example 2. " }{TEXT 237 33 " Consider the rational expression" }}{PARA 217 "" 0 "" {XPPEDIT 18 0 "w = (6*x^4-5*x+2)/(x^5+5*x^3+4*x^4 +4*x^2+4*x);" "6#/%\"wG*&,(*&\"\"'\"\"\"*$)%\"xG\"\"%F)F)F)*&\"\"&F)F, F)!\"\"\"\"#F)F),,*$)F,F/F)F)*&F/F)*$)F,\"\"$F)F)F)*&F-F)F*F)F)*&F-F)* $)F,F1F)F)F)*&F-F)F,F)F)F0" }{TEXT 200 3 " ." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 123 "Factor the denominator of w . Find the partial fraction decomposition. Find the indefinite integra l of w. Compare the three." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 36 "First let's define the ex pression w." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "w:=(6*x^4-5*x+2)/(x^5+5*x^3+4*x^4+4*x^2+4*x);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"wG*&,(*&\"\"'\"\"\")%\"xG\"\"%F)F) *&\"\"&F)F+F)!\"\"\"\"#F)F),,*$)F+F.F)F)*&F.F))F+\"\"$F)F)*&F,F)F*F)F) *&F,F))F+F0F)F)*&F,F)F+F)F)F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 82 "Now we want to factor t he denominator of w. To save retyping we use the command \"" }{TEXT 214 5 "denom" }{TEXT 237 79 "\" which gives the denominator of a given expression. Similarly, the command \"" }{TEXT 215 5 "numer" }{TEXT 237 112 "\", which we shall use later in this worksheet, gives the num erator of a given expression. This is how they act." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "numer(w);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*&\"\"'\"\"\")%\"xG\"\"%F&F&*&\" \"&F&F(F&!\"\"\"\"#F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "den om(w);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&%\"xG\"\"\",,*$)F$\"\"%F%F %*&\"\"&F%)F$\"\"#F%F%*&F)F%)F$\"\"$F%F%*&F)F%F$F%F%F)F%F%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 103 "As you see, Maple couldn't help itself \+ and somewhat simplifed the output. Let's factor the denominator." }} {PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "factor(denom(w));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(%\"xG\"\" \",&*$)F$\"\"#F%F%F%F%F%),&F$F%F)F%F)F%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 68 "We see a fir st degree term x of multiplicity 1, a first degree term " }{XPPEDIT 18 0 "(x+2)^2;" "6#*$),&%\"xG\"\"\"\"\"#F'F(F'" }{TEXT 237 55 " of mult iplicity 2, and an irreducible quadratic term " }{XPPEDIT 18 0 "x^2+1; " "6#,&*$)%\"xG\"\"#\"\"\"F(F(F(" }{TEXT 237 89 ". The rules above te ll us what kind of partial fraction decomposition to expect. Indeed:" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "convert(w,parfrac,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,**(# \"\"\"\"\"#F&F&F&%\"xG!\"\"F&*(#\"#a\"\"&F&F&F&*$),&F(F&F'F&F'F&F)F)*( #\"$$G\"#]F&F&F&F0F)F&*(#F&\"#DF&,&\"#ZF&*&\"\"%F&F(F&F&F&,&*$)F(F'F&F &F&F&F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 80 "The decomposition h as the form that we expected. Now let's look at the integral." }} {PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Int(w,x); value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$ *&,(*&\"\"'\"\"\")%\"xG\"\"%F*F**&\"\"&F*F,F*!\"\"\"\"#F*F*,,*$)F,F/F* F**&F/F*)F,\"\"$F*F**&F-F*F+F*F**&F-F*)F,F1F*F**&F-F*F,F*F*F0F," }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,,*&#\"\"\"\"\"#F&-%#lnG6#%\"xGF&F&*(# \"#a\"\"&F&F&F&,&F+F&F'F&!\"\"F&*&#\"$$G\"#]F&-F)6#F0F&F&*&#F'\"#DF&-F )6#,&*$)F+F'F&F&F&F&F&F1*&#\"#ZF:F&-%'arctanGF*F&F1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 453 " Clearly, the integral was found via partial fractions. You can see ho w to each fraction there corresponds a term in the integral. The last \+ fraction gives rise to the two last terms in the integral. They were o btained by splitting the fraction into two parts: one with 4x in the n umerator, which, after a simple substitution, gives the logarithm, and the second part with 47 in the numerator, which, of course, leads to \+ arctan(x). Similarly, as long as " }{XPPEDIT 18 0 "a*x^2+b*x+c;" "6#,( *&%\"aG\"\"\"*$)%\"xG\"\"#F&F&F&*&%\"bGF&F)F&F&%\"cGF&" }{TEXT 237 64 " is a polynomial with no real roots, any expression of the form " } {XPPEDIT 18 0 "(B*x+C)/(a*x^2+b*x+c);" "6#*&,&*&%\"BG\"\"\"%\"xGF'F'% \"CGF'F',(*&%\"aGF'*$)F(\"\"#F'F'F'*&%\"bGF'F(F'F'%\"cGF'!\"\"" }{TEXT 237 156 " can be easily be integrated (by substitution and completin g the square), giving rise to a term containing a logarithm and a term containing an arctangent." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 216 9 "Remark 1." }{TEXT 237 184 " Observe that w is de fined as an expression in terms of x and not a function. We could have defined the corresponding rational function f(x). In that case the pr oper syntax would be \"" }{TEXT 217 24 "convert(f(x),parfrac,x);" } {TEXT 237 40 "\", and not \"convert(f,parfrac,x);\". " }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 218 9 "Example 3" }{TEXT 237 70 ". Define an expression m which gives a formula for a rational function" }} {PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "m:=(x^8+x^7+2)/(x^6-5*x^4-9*x^2+8*x^3+8*x-3);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%\"mG*&,(*$)%\"xG\"\")\"\"\"F+*$)F)\"\"(F+F+\"\"#F+F +,.*$)F)\"\"'F+F+*&\"\"&F+)F)\"\"%F+!\"\"*&\"\"*F+)F)F/F+F8*&F*F+)F)\" \"$F+F+*&F*F+F)F+F+F>F8F8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 118 "Find the integral of m. Fi nd the partial fraction decomposition of m. Factor the denominator of \+ m. Compare the three. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 0 "" }} {PARA 0 "" 0 "" {TEXT 237 44 "Let's ask Maple for the integral of m fi rst:" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Int(m,x); value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#-%$IntG6$*&,(*$)%\"xG\"\")\"\"\"F,*$)F*\"\"(F,F,\"\"#F,F,,.*$)F*\"\" 'F,F,*&\"\"&F,)F*\"\"%F,!\"\"*&\"\"*F,)F*F0F,F9*&F+F,)F*\"\"$F,F,*&F+F ,F*F,F,F?F9F9F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,4*&#\"\"\"\"\"$F&) %\"xGF'F&F&*&#F&\"\"#F&)F)F,F&F&*&\"\"&F&F)F&F&*&#\"$Z&\"#!)F&-%#lnG6# ,&F)F&F'F&F&!\"\"*(#F&\"\"%F&F&F&*$),&F)F&F&F8F,F&F8F8*(#F/F;F&F&F&F>F 8F8*&#\"#n\"#;F&-F56#F>F&F&*&#\"\"(\"#SF&-F56#,&*$F-F&F&F&F&F&F8*&#F& \"#?F&-%'arctanG6#F)F&F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 119 "How did Maple obtain the va lue of the integral? By a partial fraction decomposition of the expres sion m. Let's find it." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "convert(m,parfrac,x);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#,2*$)%\"xG\"\"#\"\"\"F(F&F(\"\"&F(*(#\"$Z&\"#!)F (F(F(,&F&F(\"\"$F(!\"\"F0*(#F(F'F(F(F(*$),&F&F(F(F0F/F(F0F(*(#F)\"\"%F (F(F(*$)F5F'F(F0F(*(#\"#n\"#;F(F(F(F5F0F(*(#F(\"#?F(,&F(F0*&\"\"(F(F&F (F(F(,&F$F(F(F(F0F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 369 "Note that the partial fraction decomposition \+ of m contains a polynomial term. That is because the numerator of m is of a higher degree than its denominator, in which case the long divis ion is performed first. The resulting polynomial is left alone, and th e rational function whose denominator has greater degree than the nume rator is decomposed into partial fractions. " }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 221 "It is clear how terms of the decomposition correspond to the terms in the integral. Can you quess \+ from the decomposition what are the factors of the denominator of m an d with what multiplicities? Pretty much. Let's check." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "factor( denom(m));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(,&%\"xG\"\"\"\"\"$F&F& ,&*$)F%\"\"#F&F&F&F&F&),&F%F&F&!\"\"F'F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 34 "This is, ind eed, what we expected." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 4 "" 0 "" {TEXT 238 43 "Partial Fraction Decomposition Step-by-S tep" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 630 "As you see Maple does for you partial fraction decomposition \+ which you will find very useful in the future. How does Maple do it? T his one time, in Example 5 below, we shall retrace Maple's reasoning a nd do the partial fraction decomposition of m step-by-step. As you may expect, the algebra is going to be brutal. Don't worry, however. Mapl e will do all the hard work!. It will give you a good introduction to handling rational functions and doing other algebraic chores in Maple . Before we recreate the partial fraction decomposition of m, let's lo ok at a simple example of how to do the long division of polynomials i n Maple. " }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 219 26 "Example 4 (Long Division)." }{TEXT 237 34 " You can easily che ck by hand that" }}{PARA 218 "" 0 "" {TEXT 200 1 " " }{XPPEDIT 18 0 "( x^3+1)/(x^2+2*x) = x-2+(4*x+1)/(x^2+2*x);" "6#/*&,&*$)%\"xG\"\"$\"\"\" F*F*F*F*,&*$)F(\"\"#F*F**&F.F*F(F*F*!\"\",(F(F*F.F0*&,&*&\"\"%F*F(F*F* F*F*F*F+F0F*" }}{PARA 0 "" 0 "" {TEXT 237 30 "Let's do the long divisi on of " }{XPPEDIT 18 0 "x^3+1;" "6#,&*$)%\"xG\"\"$\"\"\"F(F(F(" }{TEXT 237 4 " by " }{XPPEDIT 18 0 "x^2+2*x;" "6#,&*$)%\"xG\"\"#\"\"\"F(*&F' F(F&F(F(" }{TEXT 237 14 " using Maple. " }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 46 "Let's first define the po lynomials in question" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "p1:=x^3+1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p1G,&*$)%\"xG\"\"$\"\"\"F*F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "p2:=x^2+2*x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p2G,&*$)%\"xG\"\"#\"\"\"F**&F)F*F(F*F*" }}}{EXCHG {PARA 0 "" 0 " " {TEXT 237 47 "The two basic commands for long division are \"" } {TEXT 220 3 "quo" }{TEXT 237 9 "\" and \"" }{TEXT 221 3 "rem" }{TEXT 237 56 "\". One gives you the quotient, the other the remainder." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "quo(p1,p2,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&%\"xG\"\"\"\"\"#!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "rem(p1,p2,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# ,&\"\"\"F$*&\"\"%F$%\"xGF$F$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 54 " The final form of the quotient p1/p2 can be written as" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "quo(p1 ,p2,x)+rem(p1,p2,x)/p2;" }}{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%" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 222 65 "Example 5 (Finding Part ial Fraction Decomposition Step-by-Step). " }{TEXT 237 117 " Find the \+ partial fraction decomposition of the expression m, step-by-step, usin g Maple to help you with the algebra." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 99 "Since our rational expressio n, m, is high up in the worksheet, let's remind ourselves of what m i s" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "m;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,(*$)%\"xG\"\" )\"\"\"F)*$)F'\"\"(F)F)\"\"#F)F),.*$)F'\"\"'F)F)*&\"\"&F))F'\"\"%F)!\" \"*&\"\"*F))F'F-F)F6*&F(F))F'\"\"$F)F)*&F(F)F'F)F)F " 0 "" {MPLTEXT 1 0 25 "quo(numer(m),denom(m),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*$)%\"xG\"\"#\"\"\"F(F&F(\"\"&F(" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 25 "rem(numer(m),denom(m),x);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#,.\"#<\"\"\"*&\"#SF%)%\"xG\"\"#F%F%*&\"#EF%)F)\"\"%F% F%*&\"\"$F%)F)\"\"&F%!\"\"*&\"#RF%)F)F0F%F3*&\"#PF%F)F%F3" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 52 "This means that the expression m can be \+ rewritten as" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "quo(numer(m),denom(m),x)+rem(numer(m),denom(m),x )/denom(m);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,**$)%\"xG\"\"#\"\"\"F( F&F(\"\"&F(*&,.\"# " 0 "" {MPLTEXT 1 0 37 "k:=rem(numer(m),denom( m),x)/denom(m);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG*&,.\"#<\"\" \"*&\"#SF()%\"xG\"\"#F(F(*&\"#EF()F,\"\"%F(F(*&\"\"$F()F,\"\"&F(!\"\"* &\"#RF()F,F3F(F6*&\"#PF(F,F(F6F(,.*$)F,\"\"'F(F(*&F5F(F0F(F6*&\"\"*F(F +F(F6*&\"\")F(F9F(F(*&FDF(F,F(F(F3F6F6" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 89 "To find the partial fraction decomposition of k, we have to first factor the denominator." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "factor(denom(k));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(,&%\"xG\"\"\"\"\"$F&F&,&*$)F%\"\"#F&F&F&F& F&),&F%F&F&!\"\"F'F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 28 "We see o ne first order term " }{XPPEDIT 18 0 "x+3;" "6#,&%\"xG\"\"\"\"\"$F%" } {TEXT 237 41 " of multiplicity 1, one first order term " }{XPPEDIT 18 0 "x-1;" "6#,&%\"xG\"\"\"F%!\"\"" }{TEXT 237 55 " of multiplicity 3, a nd one irreducible quadratic term " }{XPPEDIT 18 0 "x^2+1;" "6#,&*$)% \"xG\"\"#\"\"\"F(F(F(" }{TEXT 237 111 ". This tells us what form the \+ partial fraction decomposition of k is going to take. Let's call it \" decompk\"." }}{PARA 0 "" 0 "" {TEXT 237 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "decompk:=A/(x+3)+B/(x-1)+C/((x-1)^2)+H/((x-1)^3)+ (G*x+M)/(x^2+1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(decompkG,,*&%\" AG\"\"\",&%\"xGF(\"\"$F(!\"\"F(*&%\"BGF(,&F*F(F(F,F,F(*&%\"CGF(*$)F/\" \"#F(F,F(*&%\"HGF(*$)F/F+F(F,F(*&,&*&%\"GGF(F*F(F(%\"MGF(F(,&*$)F*F4F( F(F(F(F,F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 364 "For some constants A,B,C,H, G, M the ex pression k will have the above form. (We do not use \"D\" as a constan t as D has an assigned meaning in Maple.) The next step is finding co nstants A,B,C,H,G,M. It can be done in several ways. For example, we c ould follow the method of Example 1. However, there is a simpler way o f finding the constants, and we use it below." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 219 "" 0 "" {TEXT 200 20 "Finding A,B,C,G,H, M." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 354 "We know that decompk has to be equal to k for all values of x , except x=1,-3, where the denominators are 0. We have six unknowns A, B,C,G,H,M, let's choose six values for x, say 0,2,3,4,5,6, substitute \+ them one-by-one into k and decompk, and set up the corresponding equat ions. Let's label these equations eq1, eq2... and so on. Do not confus e the sign \"" }{TEXT 223 2 ":=" }{TEXT 237 40 "\", which means \"def ined as\", with \"" }{TEXT 224 1 "=" }{TEXT 237 72 "\", which means \" equals\". Below we define six equations. For example," }{TEXT 225 26 "eq1:=subs(x=0,k=decompk); " }{TEXT 237 162 "means that eq1 is defined as the equation that you obtain by substituting x=0 into both sides o f the equation k=decompk, that is, into both sides of the equation:" } }{PARA 0 "" 0 "" {XPPEDIT 18 0 "(17+40*x^2+26*x^4-3*x^5-39*x^3-37*x)/( x^6-5*x^4-9*x^2+8*x^3+8*x-3) = A/(x+3)+B/(x-1)+C/((x-1)^2)+H/((x-1)^3) +(G*x+M)/(x^2+1);" "6#/*&,.\"#<\"\"\"*&\"#SF'*$)%\"xG\"\"#F'F'F'*&\"#E F'*$)F,\"\"%F'F'F'*&\"\"$F'*$)F,\"\"&F'F'!\"\"*&\"#RF'*$)F,F4F'F'F8*& \"#PF'F,F'F8F',.*$)F,\"\"'F'F'*&F7F'F0F'F8*&\"\"*F'F*F'F8*&\"\")F'F;F' F'*&FGF'F,F'F'F4F8F8,,*&%\"AGF',&F,F'F4F'F8F'*&%\"BGF',&F,F'F'F8F8F'*& %\"CGF'*$)FOF-F'F8F'*&%\"HGF'*$)FOF4F'F8F'*&,&*&%\"GGF'F,F'F'%\"MGF'F' ,&F*F'F'F'F8F'" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 26 "eq1:= subs(x=0,k=decompk);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq1G/#!#<\"\"$,,*&#\"\"\"F(F,%\"AGF,F,%\"BG!\"\"%\" CGF,%\"HGF/%\"MGF," }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 75 "As you see , Maple evaluated k and decompk at x=0 and set both values equal." }} {PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "eq2:=subs(x=2,k=decompk);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$ eq2G/#\"$6\"\"#D,.*&#\"\"\"\"\"&F,%\"AGF,F,%\"BGF,%\"CGF,%\"HGF,*&#\" \"#F-F,%\"GGF,F,*&F+F,%\"MGF,F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "eq3:=subs(x=3,k=decompk);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %$eq3G/#\"#f\"#[,.*&#\"\"\"\"\"'F,%\"AGF,F,*&#F,\"\"#F,%\"BGF,F,*&#F, \"\"%F,%\"CGF,F,*&#F,\"\")F,%\"HGF,F,*&#\"\"$\"#5F,%\"GGF,F,*&#F,F>F,% \"MGF,F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "eq4:=subs(x=4,k =decompk);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq4G/#\"%(f\"\"%8K,.* &#\"\"\"\"\"(F,%\"AGF,F,*&#F,\"\"$F,%\"BGF,F,*&#F,\"\"*F,%\"CGF,F,*&#F ,\"#FF,%\"HGF,F,*&#\"\"%\"#F,%\"MGF,F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "eq5:=subs(x=5,k=decompk);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq5G/#\"$x\"\"$K),.*&#\"\"\"\"\")F,%\"AGF ,F,*&#F,\"\"%F,%\"BGF,F,*&#F,\"#;F,%\"CGF,F,*&#F,\"#kF,%\"HGF,F,*&#\" \"&\"#EF,%\"GGF,F,*&#F,F>F,%\"MGF,F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "eq6:=subs(x=6,k=decompk);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq6G/#\"%zJ\"&D;%,.*&#\"\"\"\"\"*F,%\"AGF,F,*&#F,\" \"&F,%\"BGF,F,*&#F,\"#DF,%\"CGF,F,*&#F,\"$D\"F,%\"HGF,F,*&#\"\"'\"#PF, %\"GGF,F,*&#F,F>F,%\"MGF,F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 86 "N ow we shall solve the corresponding system of equations. We label our \+ solutions sols." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "sols:=solve(\{eq1,eq2,eq3,eq4,eq5,eq6\},\{A,B ,C,G,H,M\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%solsG<(/%\"HG#\"\" \"\"\"#/%\"BG#\"#n\"#;/%\"MG#F)\"#?/%\"GG#!\"(F3/%\"CG#\"\"&\"\"%/%\"A G#!$Z&\"#!)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 40 "We have obtained the desired constants. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 140 "Once we have found the constants A,B,C,G,H,M we can sub stitute them into decompk and obtain the partial fraction decompositi on of k. Since " }{XPPEDIT 18 0 "m = x^2+x+5+k;" "6#/%\"mG,**$)%\"xG\" \"#\"\"\"F*F(F*\"\"&F*%\"kGF*" }{TEXT 237 168 " , we obtain the partia l fraction decomposition of m by subsituting the constants into decomp k and adding the polynomial. Let's call the result of this operation d ecm. " }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "decm:=x^2+x+5+subs(sols,decompk);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%decmG,2*$)%\"xG\"\"#\"\"\"F*F(F*\"\"&F**(#\"$Z&\"#! )F*F*F*,&F(F*\"\"$F*!\"\"F2*(#\"#n\"#;F*F*F*,&F(F*F*F2F2F**(#F+\"\"%F* F*F**$)F7F)F*F2F**(#F*F)F*F*F**$)F7F1F*F2F**&,&*&#\"\"(\"#?F*F(F*F2#F* FFF*F*,&F&F*F*F*F2F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{EXCHG {PARA 0 "" 0 "" {TEXT 237 211 "We have obtained the partial fr action decomposition of the expression m \"by hand\" using Maple only \+ to help us with algebra. Let's recall the result that we obtained usin g the \"convert(m,parfrac,x);\" command." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "convert(m,parfrac,x) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,2*$)%\"xG\"\"#\"\"\"F(F&F(\"\"&F (*(#\"$Z&\"#!)F(F(F(,&F&F(\"\"$F(!\"\"F0*(#\"#n\"#;F(F(F(,&F&F(F(F0F0F (*(#F)\"\"%F(F(F(*$)F5F'F(F0F(*(#F(F'F(F(F(*$)F5F/F(F0F(*(#F(\"#?F(,&* &\"\"(F(F&F(F(F(F0F(,&F$F(F(F(F0F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{EXCHG {PARA 0 " " 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 145 "Clearly, in the future, we are going to use the \"convert(..,parfrac,...)\" command r ather than finding partial fraction decompositions by hand. " }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 4 "" 0 "" {TEXT 238 50 "What if Roots of the Denominator are not Rational?" }}{PARA 0 "" 0 "" {TEXT 237 0 " " }}{PARA 0 "" 0 "" {TEXT 237 14 "The command \"" }{TEXT 226 23 "conve rt(..,parfrac,..);" }{TEXT 237 39 "\", similarly as the related comman d \"" }{TEXT 227 6 "factor" }{TEXT 237 444 "\", work very reliably as \+ long as the denominator has rational roots. As long as all coefficient s in your expression are rationals in a non-decimal form, both command s are, in fact, telling Maple to find a factorization or a partial fr action decomposition using only rational numbers as constants. If you r polynomial or the denominator of your rational function happen to ha ve irrational roots, the commands may not work at all. For example: " }}{PARA 0 "" 0 "" {TEXT 237 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "convert(1/(x^2-3),parfrac,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&\"\"\"F$,&*$)%\"xG\"\"#F$F$\"\"$!\"\"F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "factor(x^2-3);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#,&*$)%\"xG\"\"#\"\"\"F(\"\"$!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 502 "This is not what we would like to see. Maple can deal with irrational roots in se veral ways. The best approach is to tell Maple to look for a factoriza tion or a partial fraction decomposition which allows arbitrary real n umbers, not just rationals. We can do it by specifying \"real\" under \+ the corresponding command. Maple will respond by returning a partial f raction decomposition or a factorization in which coefficients and roo ts are floating point approximations of their exact values. For exampl e" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "convert(1/(x^2-3),parfrac,x,real);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#,&*($\"+X8v')G!#5\"\"\"F(F(,&%\"xGF($\"+330K " 0 "" {MPLTEXT 1 0 19 "factor(x^2-3,real);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&%\" xG\"\"\"$\"+330K " 0 "" {MPLTEXT 1 0 15 "evalf(sqrt(3));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$ \"+330K " 0 "" {MPLTEXT 1 0 31 "convert(1/(x^2-3.0),parfrac,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*($\"+X8v')G!#5\"\"\"F(F(,&%\"xGF($\"+330K " 0 "" {MPLTEXT 1 0 37 " convert(1/(x^2+2),parfrac,x,complex);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*($\"+2R`NN!#5\"\"\"%\"IGF(,&%\"xGF(*&$\"+iN@99!\"*F(F)F(F(!\"\"F (*(F%F(F)F(,&F+F0F,F(F0F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 1 "(" }{TEXT 231 2 " I" }{TEXT 237 43 " stands, of course for the imaginary unit)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 220 "" 0 "" {TEXT 200 17 "Homework Problems" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 232 11 "Problem 1. " }{TEXT 237 24 "Consider the polynomials" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }} {PARA 0 "" 0 "" {TEXT 237 5 " " }{XPPEDIT 18 0 "w1 = 7*x^5+4*x^4-2 *x^3+5*x^2-x+1;" "6#/%#w1G,.*&\"\"(\"\"\"*$)%\"xG\"\"&F(F(F(*&\"\"%F(* $)F+F.F(F(F(*&\"\"#F(*$)F+\"\"$F(F(!\"\"*&F,F(*$)F+F2F(F(F(F+F6F(F(" } {TEXT 237 7 " , " }{XPPEDIT 18 0 "w2 = -3*x^3-3*x-6;" "6#/%#w2G,(* &\"\"$\"\"\"*$)%\"xGF'F(F(!\"\"*&F'F(F+F(F,\"\"'F," }{TEXT 237 2 " ." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 26 "Perfo rm the long division " }{XPPEDIT 18 0 "w1/w2;" "6#*&%#w1G\"\"\"%#w2G! \"\"" }{TEXT 237 64 " using Maple. Have Maple display the final form o f the quotient " }{XPPEDIT 18 0 "w1/w2;" "6#*&%#w1G\"\"\"%#w2G!\"\"" } {TEXT 237 2 " ." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 233 12 "Problem 2. " }{TEXT 237 43 "For each the following rati onal expressions" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 4 " " }{XPPEDIT 18 0 "r1 = (4*x^3-1)/(x^5-3*x^4+3*x^3-x^2 );" "6#/%#r1G*&,&*&\"\"%\"\"\"*$)%\"xG\"\"$F)F)F)F)!\"\"F),**$)F,\"\"& F)F)*&F-F)*$)F,F(F)F)F.*&F-F)F*F)F)*$)F,\"\"#F)F.F." }{TEXT 237 8 " \+ , " }{XPPEDIT 18 0 "r2 = (x^2+1)/(x^4-4*x^2+x^3+2*x-12);" "6#/%#r2G *&,&*$)%\"xG\"\"#\"\"\"F+F+F+F+,,*$)F)\"\"%F+F+*&F/F+F'F+!\"\"*$)F)\" \"$F+F+*&F*F+F)F+F+\"#7F1F1" }{TEXT 237 3 " ," }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 33 "(a) Find the indefinite integral." }}{PARA 0 "" 0 "" {TEXT 237 77 "(b) Find the partial fract ion decomposition (using the simple Maple command, " }{TEXT 234 3 "not " }{TEXT 237 15 " step-by-step!)" }}{PARA 0 "" 0 "" {TEXT 237 27 "(c) \+ Factor the denominator." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 46 "Do you see the relationship between the three?" }} {PARA 0 "" 0 "" {TEXT 237 1 " " }}{PARA 0 "" 0 "" {TEXT 235 11 "Proble m 3. " }{TEXT 237 4 " For" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 221 "" 0 "" {XPPEDIT 18 0 "w = (3*x^2+x-2)/(x^3+3*x^2-4);" "6#/%\"wG*& ,(*&\"\"$\"\"\"*$)%\"xG\"\"#F)F)F)F,F)F-!\"\"F),(*$)F,F(F)F)F'F)\"\"%F .F." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 "" {TEXT 237 298 "find its partial fraction decomposition step-by-step, as in Example 5 . Then find the decomposition using the \"convert(..parfrac..)\" comma nd and compare the results. (Hint: Since the degree of the numerator i s less than the degree of the denominator, you do not have to perform \+ the long division.)" }}{PARA 0 "" 0 "" {TEXT 237 0 "" }}{PARA 0 "" 0 " " {TEXT 236 12 "Problem 4. " }{TEXT 237 36 "(a) Try to factor the po lynomial " }{XPPEDIT 18 0 "p = -4+10*x+6*x^2-14*x^3;" "6#/%\"pG,*\" \"%!\"\"*&\"#5\"\"\"%\"xGF*F**&\"\"'F**$)F+\"\"#F*F*F**&\"#9F**$)F+\" \"$F*F*F'" }{TEXT 237 53 " using the simple \"factor\" command. What happens?" }}{PARA 0 "" 0 "" {TEXT 237 114 " (b) U se the appropriate command to obtain a floating point factorization of the polynomial p." }}{PARA 0 "" 0 "" {TEXT 237 56 " \+ (c) Consider the rational function " }}{PARA 222 "" 0 "" {XPPEDIT 18 0 "z = (x^2-2)/(-4+10*x+6*x^2-14*x^3);" "6#/%\"zG*&,&*$)%\"xG\"\"# \"\"\"F+F*!\"\"F+,*\"\"%F,*&\"#5F+F)F+F+*&\"\"'F+F'F+F+*&\"#9F+*$)F)\" \"$F+F+F,F," }{TEXT 200 6 " ." }}{PARA 0 "" 0 "" {TEXT 237 0 "" }} {PARA 0 "" 0 "" {TEXT 237 182 "Try finding its partial fraction decomp osition with the simple \"convert(...parfrac,x);\" command. What happe ns? Use the appropriate command to obtain a floating point decompositi on." }}{PARA 223 "" 0 "" {TEXT 200 0 "" }}{PARA 224 "" 0 "" {TEXT 200 77 "MTH 142 Maple Worksheets written by B. Kaskosz and L. Pakula, Copy right 1999." }}}{EXCHG {PARA 225 "" 0 "" {TEXT 200 26 "Last modified A ugust 1999." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }