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Derivative in an abstract polynomial ring

  1. Apr 15, 2008 #1
    "Derivative" in an abstract polynomial ring

    1. The problem statement, all variables and given/known data
    Let R be any ring and define D:R[X]-->R[X] by setting [itex]D[\sum a_nX^n]=\sum na_nX^{n-1}[/itex].

    a) Check that, if [itex]f(X)=\sum a_nX^n[/itex] and [itex]g(X)=\sum b_nX^n[/itex], then D[f+g]=D[f]+D[g]
    b) Check that [itex]D[a_ib_jX^{i+j}]=D[a_iX^i]b_jX^j+a_iX^iD[b_jX^j][/itex]
    c) Now using the additivity in a), show that D[f(X)g(X)]=D[f(X)]g(X)+f(X)D[g(X)] for all polynomials f and g in R.

    2. Relevant equations

    3. The attempt at a solution
    I see the clear parallel to polynomial derivatives, but it's not defined in terms of epsilons and deltas - it's just a "black box" function that really has no meaning. But in any case, I've been able to get parts (a) and (b) without any problems. But I'm having trouble with (c).

    I've chosen to write the product of [itex]f(X)=\sum_i a_iX^i[/itex] and [itex]g(X)=\sum_j b_jX^j[/itex] as just [itex]\sum_{i,j}a_ib_jX^{i+j}[/itex] so that it's similar to the result of part (b). Then
    [tex]D[f(X)g(X)]=D[\sum_{i,j}a_ib_jX^{i+j}]=\sum_{i,j} D[a_ib_jX^{i+j}]=\sum_{i,j} \left(D[a_iX^i]b_jX^j+a_iX^iD[b_jX^j]\right)[/tex]

    So does this work? I tried it another way by checking what's in the nth position on each side of the equation and showing them to be equal, but got bogged down in keeping track of indexes and stuff. My algebra professor said there was an easier way - is this it?

    Thanks for your help in advance.
  2. jcsd
  3. Apr 15, 2008 #2


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    Homework Helper

    About all you really did there is restate, without really quite proving, part b). But if you actually did do part a) and b), then you are a pretty good shape. a) shows that the Leibniz rule (a common name for what you are trying to prove) is linear over sums. b) shows that it's true for products of monomials. That makes c) easy. f(X)*g(X) is really just a sum of products of monomials if you expand it out.
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