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Prove a linear mapping of a polynomial function is a map

  1. Dec 3, 2015 #1
    1. The problem statement, all variables and given/known data

    The bane of all physicists... 'Proof' questions...

    So we have the mapping,

    Δ : P3→P3
    Δ[P(x)] = (x2-1) d2P/dx2 + x dP/dx

    And I need to prove that this is a linear mapping

    2. Relevant equations

    Linear maps must satisfy:
    Δ[P(x+y)] = Δ[P(x)] + Δ[P(y)]
    Δ[P(αx)] - αΔ[P(x)]

    3. The attempt at a solution

    I'm not sure what to do. I've tried working through the actual mapping, performing the differential operations on the polynomial:
    and on:
    then expanded the brackets to see if I could separate the x and y terms to show that it's possible to pull them apart and demonstrate equality with Δ[P(x)] + Δ[P(y)]
    But it doesn't work. My best bet is that I've done something wrong regarding the part where I need to do
    d2P/d(x+y)2. I've never really had to work in that way before. Maybe I've gone wrong completely.

    Any advice?

    Thanks! :oldsmile:
  2. jcsd
  3. Dec 3, 2015 #2


    Staff: Mentor

    The above isn't right.
    It should be Δ[P(x) + P(y)] = Δ[P(x)] + Δ[P(y)]
    IOW, the Δ map carries a sum of functions to the sum of the transformations of the functions.
  4. Dec 3, 2015 #3
    Bingo, you've got it. I remember now.

    Such a little thing - thanks for pointing it out!
  5. Dec 3, 2015 #4


    User Avatar
    Homework Helper

    Taking the differential on ## P(x) = Ax^3 + Bx^2 + Cx + D ## should give:
    ##\Delta P(x) = (x^2-1)(6Ax+2B)+x(3Ax^2+2Bx+C) = 9Ax^3 +4Bx^2+(C-6A)x-2B##
    As Mark pointed out, you want to compare two polynomials ##P_1(x) = ax^3 + bx^2 + cx + d## and ##P_2(x) = hx^3 + ix^2 + jx + k##, not polynomials of different variables.
    This should make sense, because a polynomial of greater than first order is known to not be a linear function.
  6. Dec 3, 2015 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Shouldn't that be ##\Delta [P_1(x) + P_2(x)] = \Delta P_1(x) + \Delta P_2(x) \: \forall x##?
  7. Dec 3, 2015 #6


    Staff: Mentor

    Now that you mention it, yes...
    Thanks for the correction.
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