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Math Help - Problem 21

  1. #16
    Newbie slobone's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    As f(x)>0 for all x>=0 all roots are negative. Let these be -b1, -b2, .. -bn,
    with all the b's >0.

    Then:

    f(x) =(x+b1)(x+b2) ..(x+bn)

    Also as the constant term is 1 we know that b1.b2. .. bn = 1.

    Now 2 + bk = 1 + 1 + bk >= 3 cuberoot(1.1.bk), by the Arithmetic-Geometric
    mean inequality.
    So (2+bk) >= 3cuberoot(bk)

    So for x=2:

    f(2) = (2+b1)(2+b2) ...(2+bn) >= 3^n cuberoot(b1.b2. .. bn) = 3^n

    RonL
    That was the thing I needed. I suspected it was true, but I didn't know it was a theorem. Good problem.
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  2. #17
    Grand Panjandrum
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    Quote Originally Posted by ecMathGeek View Post
    Was my proof wrong?
    Yes, because the inductive step does not hold as you have the product
    of the first k r's equal to 1, the k+1 st r must be 1. So this proof only works if
    all the roots are -1, but this is in general not true.

    RonL
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  3. #18
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Yes, because the inductive step does not hold as you have the product
    of the first k r's equal to 1, the k+1 st r must be 1. So this proof only works if
    all the roots are -1, but this is in general not true.

    RonL
    I wasn't arguing that the r_{x} roots must be 1 (that is that every root must be 1), but that if the roots for the n = k case are r_1, r_2, ..., r_k, where r_1, r_2, ..., r_k are real numbers, and that the first "k" roots of the n = k + 1 case are r_1, r_2, ... , r_k, where r_1, r_2, ..., r_k are the same values from the n = k case, then the r_{k + 1} root must be 1.

    [Edit] There is a problem with this logic that just occurred to me: The roots of n = k + 1 don't have to be the same as those of n = k.
    Last edited by ecMathGeek; May 10th 2007 at 11:58 AM.
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