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Equal Coefficients Polynomials

  1. Jun 30, 2009 #1
    Here is something that always bugged me, and I think I have an explanation for it now, but I am wondering if it is correct. Alright, the problem to me was that back when I was in Diff-eq, to use undetermined coefficients with polynomials, we would always group together the terms on one side, and set the undetermined coefficients equal on the other side, as such:

    2(x^2)+3x+5=a(x^2)+bx+c

    would imply that a=2, b=3 and c=5

    However, this method always seemed a little uncertain to me, and when I would ask about why we could set the coefficients equal, I was always given a handwaving argument or "doesn't it seem reasonable?"
    Today, I came across an exercise in my linear algebra book that I think may justify this method; the exercise says "prove that if v1,...,vn are linearly independent vectors in the vector space U and each ai,bi belongs to the field F then a1*V1+...an*Vn=b1*V1+...bn*Vn [tex]\leftrightarrow[/tex] a1=b1,...,an=bn with

    The proof in the forward direction (all I will prove):
    Since a1*V1+...an*Vn=b1*V1+...bn*Vn[tex]
    subtract the right side of the equation to end up with
    (a1-b1)*V1+...(an-bn)*Vn=0

    and since these vectors are linearly independent, each ai-bi=0, so ai=bi for i=1,...,n

    Now, looking at the space Pn of polynomials of degree n, the set {1,x,(x^2),...,(x^n)} forms a basis, so each of these vectors must be linearly independent. That means that, as above, if we have

    2(x^2)+3x+5=a(x^2)+bx+c

    Since the set vectors {1,x,(x^2)} is linearly independent, one can set the coefficients equal.

    Am I correct about using this theorem to justify this method of setting coefficients equal? I know it seems trivial, but if this is correct, someone who doesn't really understand why this works could try this method on a set of linearly dependent vectors, for which it isn't necessarily true!
     
    Last edited: Jun 30, 2009
  2. jcsd
  3. Jun 30, 2009 #2
    Yes, the fact that two polynomials are equal if and only if the coefficients of the corresponding powers are equal follows from the fact that {1,x,...,x^n} is a basis.

    Notice that this proof does not involve considering the polynomials as functions. If we consider two polynomials P(x) = Q(x), then we can generate an infinite set of linear equations:

    P(1) - Q(1) = 0

    P(2) - Q(2) = 0

    .
    .
    .

    from here you can if you wish complete the proof of the same fact using the tools of linear algebra.
     
  4. Jun 30, 2009 #3
    I suppose the same argument goes for the trig functions sine and cosine, since they are linearly independent.
     
    Last edited: Jun 30, 2009
  5. Jun 30, 2009 #4

    HallsofIvy

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    Even more simply, given two polynomials [itex]P(x)= a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0[/itex] and [itex]Q(x)= b_mx^m+ b_{m-1}x^{m-1}+ \cdot\cdot\cdot+ b_1x+ b_0[/itex], with P(x)= Q(x) for all x, setting x= 0 shows that [itex]a_0= b_0[/itex]. But we must also have P'(x)= Q'(x) and setting x= 0 shows that [itex]a_1= b_1[/itex]. Each successive derivative, at x= 0, gives the equality of the next coefficients.
     
  6. Jun 30, 2009 #5

    statdad

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    Yet another proof. If

    [tex]
    a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 = b_n x^n + b_{n-1} x^{n-1} + \cdots + b_1 x + b_0
    [/tex]

    for all x, then

    [tex]
    (a_n - b_n) x^n + (a_{n-1} - b_{n-1})x^n + \cdots + (a_1 - b_1) x + (a_0 - b_0) = 0
    [/tex]

    for all x, so the second polynomial has more than [tex] n [/tex] zeros, and so must itself be the zero polynomial. This means [tex] a_i = b_i \, \, i = 1, \dots, n [/tex]
     
  7. Jun 30, 2009 #6
    Yes, that is in fact "linear independence". The textbook may not say that, however, if it does not have linear algebra as a prerequisite. More advanced DE textbooks will, of course, mention the concept of "linear independence".
     
  8. Jun 30, 2009 #7

    Hurkyl

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    I would like to add some things.

    There are a variety of ways to define polynomials -- in at least one of them, "equal polynomials" actually literally means "they have the same coefficients".

    Equality of the functions these polynomials represent is a slightly trickier issue. The result statdad mentioned is the typical approach.

    It turns out that in familiar situations, equality of polynomials and equality of the functions they represent turn out to be the same.




    In some more esoteric situations, two polynomials may be different despite representing the same polynomial function. If you're familiar with modular arithmetic, you can check that if we define the functions

    [tex]f(x) = x \pmod 2[/tex]
    [tex]g(x) = x^2 \pmod 2[/tex]

    then f=g, despite [itex]x \pmod 2[/itex] and [itex]x^2 \pmod 2[/itex] being different polynomials.
     
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