Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Partial fractions

  1. Feb 7, 2005 #1
    Hi, me with my really old book again. This time , a novel way of turning expressions into partial fractions.
    It would be best if I show you the examples in the book :

    [tex] \frac{3x^2 +12x +11} {(x+1)(x+2)(x+3)} [/tex]

    To express this fraction in the form

    [tex] \frac{A} {x+1} + \frac{B} {x+2} + \frac{C} {x+3} [/tex]

    We have to find the values of A , B and C by solving three simultaneous equations ( by the usual method )

    The method proposed in the book :

    equate the denominator of [tex]A[/tex] to zero and thus get the 'paravartya'
    Then substitute this value is the original equation but without the factor which is A's denominator. The result is the value of A.

    so for A, the 'paravartya' is -1 ( from x+1 = 0 )
    substituting this in
    [tex] \frac {3x^2 +12x +11} {(x+2)(x+3)} [/tex]
    we get 1, so A = 1

    similarly for B, the 'paravartya' is -2 (from x+2 =0)
    substituting this in
    [tex] \frac{3x^2 +12x +11} {(x+1)(x+3)} [/tex]
    we get 1, so B = 1

    For C, the paravartya is -3 ( from x+3 =0)
    substituting this in
    [tex] \frac{3x^2 +12x +11} { (x+1)(x+2) }[/tex]
    we get 1, so C = 1

    So, the partial fraction is
    [tex] \frac{1} {x+1} + \frac{1} {x+2} + \frac{1} {x+3} [/tex]

    This makes integration so much easier!
  2. jcsd
  3. Feb 7, 2005 #2
    That's basically the usual method of 'creating' partial fractions. You should notice that after multiplying the new fractions out, A and B get multiplied by the denominator of C, B and C by that of A, and A and C by that of B. So when you use the "paravartya" of A, the coefficients of B and C become zero, since the "paravartya" sets the denominator of A to zero. Now all that is left is comparing the RHS with the LHS (in particular, the nominators) after using the value of the "paravartya":

    [tex]3(-1)^{2}+12(-1)+11 \equiv A(-1+2)(-1+3)[/tex]

    [tex]2 \equiv 2A \; \Rightarrow \; A=1[/tex]
  4. Feb 7, 2005 #3


    User Avatar
    Homework Helper

    That's Vedic math, isn't it ? :smile:
  5. Feb 8, 2005 #4
    yes it is :)
  6. Dec 29, 2007 #5
    You can also solve for the constants using a system of linear equations.

    [tex] \frac{3x^2 +12x +11} {(x+1)(x+2)(x+3)}=\frac{A}{(x+1)}+\frac{B}{(x+2)} +\frac{C}{(x+3)} [/tex]

    Multiply both sides by the common denominator

    [tex] 3x^2 +12x +11=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2) [/tex]

    [tex] 3x^2 +12x +11=A(x^2+5x+6)+B(x^2+4x+3)+C(x^2+3x+2) [/tex]

    [tex] 3x^2 +12x +11=Ax^2+5Ax+6A+Bx^2+4Bx+3B+Cx^2+3Cx+2C [/tex]

    [tex] 3x^2 +12x +11=(A+B+C)x^2+(5A+4B+3C)x+(6A+3B+2C) [/tex]

    The coefficients on either side of the equation must be equal, so

    [tex] 3= A+ B+ C [/tex]
    [tex] 12=5A+4B+3C [/tex]
    [tex] 11=6A+3B+2C [/tex]

    Solving this set of equations will get the same results (A=1, B=1, C=1)

    Source: http://www.wholikeshomework.com/tutorials/partialfractions.pdf
  7. Dec 29, 2007 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    It's not Vedic math at all, but known as the Heaviside method, I believe.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook