1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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 and singularity point

  1. Oct 21, 2011 #1
    Hi,

    I have a question regarding partial fractions. One of my Math lecturers said that to find partial fraction, we are actually finding the singularity points. I don't understand what happens at a singularity point that allows us to decompose a function into the sum of two other functions. How does taking the limit of a function as it approaches a singularity point help us find the partial fractions. I've looked at graphs of functions to try and understand it, but I don't see it.

    I have been wondering this for a while now and some help would be very much appreciated!

    Thank you.
     
  2. jcsd
  3. Oct 22, 2011 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Gosh, those are a lot of big words for a simple concept! If I want to write, say,
    [tex]\frac{3x- 2}{(x- 1)(x- 2)}[/tex]
    as "partial fractions" I write
    [tex]\frac{3x- 2}{(x- 1)(x- 2)}= \frac{A}{x- 1}+ \frac{B}{x- 2}[/tex]
    and multiply both sides by (x-1)(x- 2) to get
    [tex]3x- 2= A(x- 2)+ B(x- 1)[/tex]

    There are now any number of ways of determining what A and B are:
    a) multiply out 3x- 2= (A+ B)x+ (-2A- B) and set like coefficients equal
    b) Choose any two values of x to get two equations for A and B
    b') Specifically choose x= 2, and x= 1 because they make the equations very simple: if x= 1, 3- 2= 1= -A and if x= 2, 6- 2= 4= B- the equations are separated.

    Now, if we want to be very pedantic, we could object that the original expression does not exist at x= 1 and x= 2 so we should not use those numbers. But that is the same as objecting that, say, [itex](x^2- 4)/(x- 2)= x+ 2[/itex] is not true for x= 2, where the left side is "undetermined" because both numerator and denominator are 0. Any good teacher should point that out- repeatedly! But then note that the definition of limit does not require the value of the function at the target point. To take the limit of [itex](x^2- 4)/(x- 2)[/itex] at x= 2, we can take the limit as x goes to 2.

    And finding the values of A and B, we are, essentially, taking the limit, we can, after multiplying by x- 1 and x- 2, set x= 1 and x= 2.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Partial fractions and singularity point
  1. Partial fractions (Replies: 13)

  2. Partial fraction (Replies: 3)

  3. Partial Fraction (Replies: 4)

  4. Partial Fractions (Replies: 10)

  5. Partial Fractions (Replies: 7)

Loading...