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Rational function

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the second degree polynomial P(x)
    such that P(0)=1,P'(0)=0,and
    [itex]\int[/itex]P(x)/{x3(x-1)2} dx

    is a rational function

    2. Relevant equations
    this chapter is about integration techniques,L'Hopital's Rule, and Improper Integral
    partial fraction,partial integration are learnt.

    3. The attempt at a solution
    Since P(x) is second degree polynomial,i let P(x)=Ax2+Bx+C
    P(0)=1,so 1=A(0)+B(0)+c
    C=1
    P'(0)=0,so 2Ax+B=0,
    B=0
    So I know P(x) is Ax2+1,but i dont know how to do next on.Partial fraction is seem too many variables and have only 1 clue{P(0)=1},partial integration is hard too.I think "Rational function" is the key of this question,but i know rational fraction no more than it can be express as a/b,how to solve this question ?
     
  2. jcsd
  3. Jun 26, 2011 #2

    Dick

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    Partial fractions is the correct answer. Try and express the integrand in terms of partial fractions and find the value of A that will make the terms that would lead to answers in the integral that are not rational functions like logs, zero.
     
    Last edited: Jun 26, 2011
  4. Jun 26, 2011 #3
    [itex]\frac{Ax^2 +1}{x^3(x-1)^2}[/itex]=[itex]\frac{B}{x}[/itex]+[itex]\frac{C}{x^2}[/itex]+[itex]\frac{D}{x^3}[/itex]+[itex]\frac{E}{x-1}[/itex]+[itex]\frac{F}{(x-1)^2}[/itex]

    It is clear that constant B and E have to be zero so that P(x) is rational.However.the value of A is different when solving the equation,after eliminating other variable and only one variable(B or E) is left.Is this my mistake??
    ps;the answer is -3x2+1,while my answer is 3(using B) and 9(using E)....
    Is there any software or website can solve such a question?It is torturous to do for such a long time and the final answer is wrong.......:cry:

    Below is my attempt:

    Ax2+1=Bx2(x-1)2+Cx(x-1)2+D(x-1)2+Ex3(x-1)+Fx3



    when x=1,Ax+1=F
    x=0, 1=D
    x=-1,
    A+1=4B-4C+4D+2E-F
    2A+2=4B-4C+4+2E (F=Ax+1,D=1)
    A+1=2B-2C+2+E
    A-1=2B-2C+E-----(1)

    x=2,
    1+4A=4B+2C+D+8E+8F
    4A+1=4B+2C+1+8E+8A+8 (F=Ax+1,D=1)
    -4A-8=4B+2C+8E
    -2A-4=2B+C+4E----(2)

    x=-2,
    4A+1=4B(9)+(-2C)+9D-8E(-3)-8F
    4A+1=36B-18C+9D+24E-8F
    4A+1=36B-18C+9+24E-8A-8
    -4A=36B-18C+24E
    -2A=18B-9C+12E-----(3)


    (2)*9==》-2A-4=2B+C+4E
    (1)-(2)*9,
    16A+36=18C-24E
    8A+18=-9C-12E--(4)

    (3)-(1)*9,
    -11A+9=9C+3E-----(5)

    (4)+(5),
    -3A+27=-9e
    -A+9=-3E
    when A=9,E=0

    Similarly,
    (2)-(1)*4,
    -2A=-2B+3C---(6)


    (3)-(2)*6,
    A+3=3B-3C---(7)

    (6)+(7)
    -A+3=B
    when A =3, B=0
     
  5. Jun 27, 2011 #4

    Dick

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    No problem with what you are trying to do and how you are trying to do it, but the process is long enough that it's easy to make a mistake. I use a software package called 'maxima'. Type 'partfrac((A*x^2+1)/(x^3*(x-1)^2),x)'. You'll see you should have gotten the same condition for setting B and E to zero. Namely A+3=0 giving you the books answer.
     
  6. Jun 27, 2011 #5

    SammyS

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    Ax2+1=Bx2(x-1)2+Cx(x-1)2+D(x-1)2+Ex3(x-1)+Fx3

    x=0 gives D=1 .

    Multiply out the above expression, and equate coefficients for the various powers of x.

    You can also use WolframAlpha to do the partial fraction expansion for x2/(x3(x-1)2) and 1/(x3(x-1)2) separately, then combine the results.
     
    Last edited: Jun 27, 2011
  7. Jun 27, 2011 #6

    Dick

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    haleycomet2 got if x=0, then D=1 in section (1). I didn't get past section (3), but everything looked ok up till then. Must be some mistake after, gotta confess being lazy not to check the rest.
     
  8. Jun 28, 2011 #7
    What a brilliant software!!I can use this to check my work in future.Thank you!!:rofl:
     
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