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Math Help - proof by induction

  1. #1
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    proof by induction

    cant seem to get the symbols on, sorry, if anyone would like to help i could email them the question and solution

    Quote Originally Posted by mitch_nufc View Post
    1 + kx + 1/2k(k 1)x^2 + x + kx^2 + 1/2k(k 1)x^3

    1 + kx + 1/2k(k 1)x^2 + x + kx^2, as 1/2k(k 1)x^3 0,

    1 + (k + 1)x + ( 1/2k(k 1) + k)x^2

    can anyone explain why we can simply ignor 1/2k(k-1)x^3? this is the inductive step of the proof to show:

    (1+x)^2 ≥ 1+nx +1/2n(n-1)x^2

    Mr F edit: Is the above what the question was?
    Last edited by mr fantastic; January 2nd 2009 at 07:44 PM.
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  2. #2
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    If I get well, this is same that:
    x + epsilon >= x, if epsilon >= 0.

    It seems me evident.
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  3. #3
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    Quote Originally Posted by mitch_nufc View Post
    1 + kx + 1/2k(k 1)x^2 + x + kx^2 + 1/2k(k 1)x^3

    1 + kx + 1/2k(k 1)x^2 + x + kx^2, as 1/2k(k 1)x^3 0,

    1 + (k + 1)x + ( 1/2k(k 1) + k)x^2

    can anyone explain why we can simply ignor 1/2k(k-1)x^3? this is the inductive step of the proof to show:

    (1+x)^2 ≥ 1+nx +1/2n(n-1)x^2

    I'm going to prove to you that your lecturers actions were indeed correct.

    Let C = (x+1)(x+1)^k

    Let  B = \frac{1}{2}k(k-1)x^3 .

    Let A = 1+kx+\frac{1}{2}k(k-1)x^2+x+kx^2

    The equation you have, up until the point you don't understand, is:

     C \geq A + B

    We know that  B \geq 0

    Therefore we can conclude that A+B is always bigger or equal to A. (If B = 0, then A+B =A, if B>0, then A+B>A):  A +B \geq A

    So we have:

     A+B \geq A

    And the original statement was:

     C \geq A+B

    So C is greater than or equal to A + B, and A + B is greater than or equal to A, we can put this into one single inequality:

     C \geq A+B \geq A

     \therefore C \geq A

    From this we can conclude that the first term (C) is the biggest in the inequality, and the last term (A) is the smallest, and that that term in the middle is between the two (A+B). What we can conclude from that is:

     \therefore C \geq A

    Or you can go about it more algebraically:

     C \geq A+B \geq A

     C-B \geq A \geq A-B

     C-B \geq A-B

     C \geq A

    Which confirms your lecturers actions.
    Last edited by Mush; January 2nd 2009 at 11:05 AM.
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