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Partial fractions and singularity point

  1. Oct 21, 2011 #1

    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


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    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.
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