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Spivak's Calculus: sum of derivatives of a polynomia

  1. Jun 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Let f be a polynomial function of degree n such that f(x) ≥ 0 for all x (note that n must be even). Prove that f + f' +f'' + f''' + ... + f^(n) ≥ 0.


    2. Relevant equations

    I believe that is all - the derivative of some term in the polynomial ax^n is anx^(n-1).

    3. The attempt at a solution

    I attempted induction on n (stepping up by two, not one, each time), actually writing out the final summed polynomial in terms of its coefficients, etc. but have made no real progress.

    Thank you for hints/solutions in advance!
    (if you have a suggestive hint, I actually would appreciate that MORE than a solution, but if you want to give a solution I'd still be happy).
     
  2. jcsd
  3. Jun 20, 2012 #2
    Never mind, I found the solution a while ago. For anyone curious, here it is:

    Note that the first coefficient of f must be positive, otherwise f(x) would be negative for large enough x.

    Let Q(x) = (f + f' + f'' +... +f^(n))(x). Note that the first coefficient of Q is the first coefficient of f, so Q is positive for large enough (in absolute value) x, so Q(x) is positive for all x such that x<A for some number A, and Q(x) is positive for all x such that x>B for some number B. Let x2>B .

    If Q(x) is positive for all x, then we're done. But assume false , i.e there is some interval that is a subset of (A,B) such that Q is negative on said interval, in particular say Q(n) < 0 for some n such that n<B.

    Then by the continuity of Q there is a z such that n<z<x2 such that Q(z) = 0, and z is the closest root to n, i.e Q is negative on (n,z). Then again by the continuity and differentiability of Q, there is a y[itex]\ni[/itex](n,z) such that Q'(y) = [itex]\displaystyle\frac{Q(z) - Q(n) = -Q(n)}{z-n}[/itex], so Q'(y) is positive. Because y[itex]\ni[/itex](n,z), Q(y) is negative. So Q(y) - Q'(y) is negative.

    But notice that Q(y) -Q'(y) = (f + f' + f'' +...)(y) - (f' + f'' + f''' +...)(y) = f(x). So the existence of an n such that Q(n) < 0 implies that f is negative at some y which is false.

    I really like this book!
     
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