Posers and Puzzles

Posers and Puzzles

  1. DonationAcolyte
    Now With Added BA
    Loughborough
    Joined
    04 Jul '02
    Moves
    3790
    24 Mar '04 00:375 edits
    Suppose f and g are two infinitely differentiable functions on the reals, and suppose that, for some real A, k>0, n>0, and for all x>A:
    f^(n)(x) - g^(n)(x) > k (where ^(n) means nth derivative.)

    Show that there exists a number B so that for all x>B, f(x) - g(x) > 0. Deduce that, if f is e^qx for q positive, and g is a polynomial, there exists a B for which f>g for all x>B.

    Once again I ask mathematical types not to call out the answer until others have had a chance to think about it. This one's off the top of my head, so I apologise if people find it too easy or hard, or if I've got the problem wrong.

    Note to the non-mathematical: a derivative is a 'rate of change', and the derivative of an nth derivative is a (n+1)th derivative. So the second derivative of f is 'rate of change of rate of change of f', and so on.
  2. Standard memberroyalchicken
    CHAOS GHOST!!!
    Elsewhere
    Joined
    29 Nov '02
    Moves
    17317
    24 Mar '04 18:00
    Originally posted by Acolyte
    Suppose f and g are two infinitely differentiable functions on the reals, and suppose that, for some real A, k>0, n>0, and for all x>A:
    f^(n)(x) - g^(n)(x) > k (where ^(n) means nth derivative.)

    Show that there exists a number B so that for all x>B, f(x) - g(x) > 0. Deduce that, if f is e^qx for q positive, and g is a polynomial, there exists a B for w ...[text shortened]... ivative. So the second derivative of f is 'rate of change of rate of change of f', and so on.
    This is clever, but isn't it only of interest to 'mathematical types' or maybe Thomas Malthus 😉?

  3. DonationAcolyte
    Now With Added BA
    Loughborough
    Joined
    04 Jul '02
    Moves
    3790
    25 Mar '04 18:20
    Originally posted by Acolyte
    Suppose f and g are two infinitely differentiable functions on the reals, and suppose that, for some real A, k>0, n>0, and for all x>A:
    f^(n)(x) - g^(n)(x) > k (where ^(n) means nth derivative.)

    Show that there exists a number B so that for all x>B, f(x) - g(x) > 0. Deduce that, if f is e^qx for q positive, and g is a polynomial, there exists a B for w ...[text shortened]... ivative. So the second derivative of f is 'rate of change of rate of change of f', and so on.
    Just noticed that f and g need only be n times differentiable. There are a few variations on this theme, and I was messing around with them after posting, hence the edits.
  4. DonationAcolyte
    Now With Added BA
    Loughborough
    Joined
    04 Jul '02
    Moves
    3790
    31 Mar '04 16:51
    Hint (and I don't care who answers it now): the magic word is Induction.
  5. my head
    Joined
    03 Oct '03
    Moves
    671
    13 Apr '04 17:21
    Originally posted by Acolyte
    Suppose f and g are two infinitely differentiable functions on the reals, and suppose that, for some real A, k>0, n>0, and for all x>A:
    f^(n)(x) - g^(n)(x) > k (where ^(n) means nth derivative.)

    Show that there exists a number B so that for all x>B, f(x) - g(x) > 0. Deduce that, if f is e^qx for q positive, and g is a polynomial, there exists a B for w ...[text shortened]... ivative. So the second derivative of f is 'rate of change of rate of change of f', and so on.
    must...learn....cal..cu...lus....
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