1. Not finding help here? Sign up for a free 30min 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 Question

  1. Sep 6, 2007 #1
    f(x) is a polynomial. A product of n distinct factors


    Prove that


    This I can do by writing f(x)=(x-a)g(x) where g(a)<>0. Then splitting




    for some h(x) so


    and differentiating f(x), f'(a)=g(a) so


    and I repeat that for each factor so I get the sum required.

    The next question I can't answer:

    Show that


    is 0 when r=0,1,...,n-2 and is 1 when r=n-1

    I've tried expanding f(x), differentiating and putting x equal to the roots and summing over all the roots so I see the result works but I'm having no success in proving it.
    Could someone help with a hint or two?
    Last edited by a moderator: Sep 6, 2007
  2. jcsd
  3. Sep 6, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    What do you get when n = 2, say?
  4. Sep 6, 2007 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    You might consider using logarithmic differentiation.

    If [itex]f(x)= a(x-x_1)(x- x_2)\cdot\cdot\cdot(x- x_n)[/itex]
    then [itex]ln(f(x)) ln a+ ln(x-x_1)+ \cdot\cdot\cdot+ ln(x-x_n)[/itex]
    what do you get when you differentiate that?
  5. Sep 7, 2007 #4
    Yes I see that the result works for n=2 and even 3. But I do not see how that helps starting off a proof by Induction.
  6. Sep 7, 2007 #5
    Yes I tried this too but I still cannot see how I get the [tex](a_{i})^r[/tex] terms unless I do something like what I tried in my first post i.e. use an expanded form of f(x) and differentiate as the highest order term is of order n-1.
  7. Sep 7, 2007 #6


    User Avatar
    Homework Helper

    How does 1/f(x) look if you use partial fractions... get a few terms... you'll see a pattern...

    Then examine what f'(ai) looks like.

    Relate the above 2 ideas...
  8. Sep 8, 2007 #7
    Thanks very much for the help.
    I see it now: Consider a1^r/f'(a1) as the polynomial x^r/[(x-a2)(x-a3)...(x-an)] and split this into partial fractions just like 1/f(x) above and you find a1^r/f'(a1) is just he negative of the sum of all other such terms with the other n-1 factors of f(x).
    This works for r<n-1. For the r=n-1 case I differentiated the expanded f(x), divided by f'(x), and then substituted in a1,a2,...,an and added. From above all terms with r<n-1 are zero so we are left with n=n*Sum(a^(n-1)/f'(a)).

    Thanks again for the hints.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Partial Fractions Question