Results 1 to 4 of 4

Math Help - Equating Coefficients of Polynomials Proof

  1. #1
    Junior Member
    Joined
    Apr 2008
    Posts
    31

    Equating Coefficients of Polynomials Proof

    Hi,

    I intuitively understand Equating Coefficients of Polynomials but I was hoping to find a proof of the method. Does anyone know where I might find one?

    Cheers,

    Scott.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Apr 2008
    Posts
    31
    I think I have figured it out.

    Given say three arbitrary functions of x, f_1(x), f_2(x) and f_3(x), the constants a,b,c,i,j,k and the following equation:

    af_1(x)+bf_2(x)+cf_3(x)=if_1(x)+jf_2(x)+kf_3(x)

    Can be written:

    (a-i)f_1(x)+(b-j)f_2(x)+(c-k)f_3(x)=0

    If the functions are linearly independent then the only solution to the equation is when the coefficients of the functions are all zero. This leads to:

    a=i
    b=j
    c=k

    Which shows the coefficients of each side of the original equation are equated. The idea can be extended to any number of linearly independent functions.

    The "x" terms of a polynomial are linearly independent so the coefficients can be equated also.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,974
    Thanks
    1121
    Of course, to complete the proof, one would have to show that the different powers of x, 1, x, [itex]x^2[/itex], etc. are "independent". That means showing that a_1+ a_2x+ a_3x^2+ \cdot\cdot\cdot+ a_{n+1}x^n= 0 only if a_1= a_2= a_3= \cdot\cdot\cdot= a_{n+1}= 0.

    One way to do that is to take n different values for x to get n equations for the coefficients. If we take x= 0 we get a_1= 0 immediately. If we take x= 0 we get [tex]a_1+ a+2+ a_3+ \cdot\cdot\cdot+ a_{n+1}= 0[/itex], etc.

    A simpler, but more sophisticated, way would be to take derivatives: since a_1+ a_2x+ a_3x^2+ \cdot\cdot\cdot+ a_{n+1}x^n= 0 for all x, it is a constant so its derivative, a_2+ 2a_3x+ \cdot\cdot\cdot+ na_{n+1} is also 0 for all x. Taking x= 0, we now get a_2= 0. Of course, since a_2+ 2a_3x+ \cdot\cdot\cdot+ na_{n+1}= 0 for all x, it is a constant and its derivative is 0. Continuing through the nth derivative gives us 0 for all n coefficients.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2008
    Posts
    31
    The way I see the linear independence of the polynomial terms is to take any two of them, say ax^1 and bx^0. If they were linearly dependent then there would be some a and b not zero and:

    ax^1 + bx^0=0
    ax^1=-bx^0
    x=-b/a

    But x is not a constant which shows a contradiction with the original assumption that the terms are linearly dependent. The only solution is when a and b are zero.

    This can be applied to any two terms of a polynomial of any degree which implies that the terms of any polynomial are linearly independent.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. The act of equating
    Posted in the Algebra Forum
    Replies: 11
    Last Post: December 11th 2011, 11:37 PM
  2. Equating two SI units of time
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 30th 2011, 09:00 AM
  3. Replies: 10
    Last Post: February 12th 2011, 09:50 AM
  4. Finding Polynomials based on coefficients
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: July 5th 2009, 11:49 PM
  5. Replies: 3
    Last Post: November 25th 2008, 03:03 PM

Search Tags


/mathhelpforum @mathhelpforum