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Math Help - Math Induction...PLEASE HELP!!!!!

  1. #1
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    Math Induction...PLEASE HELP!!!!!

    Use math induction to show that
    1) n!>= n^2 for n>=6

    2) n!>=n^3 for n>=6
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  2. #2
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    Quote Originally Posted by xomichelleybelly View Post
    Use math induction to show that
    1) n!>= n^2 for n>=6
    I will assume that you have done the base case and that we assume this is true for some k \ge 6

    So we need to show that:

    (k+1)! \ge (k+1)^2,

    well:

    (k+1)! = k!~ k

    and by supposition k! \ge k^2, so:

    (k+1)! = k! k \ge k^3

    So if we can prove that k^3 \ge (k+1)^2 we are done, but k^3 \ge (k+1)^2 is equvalent to k^3-k^2-2k-1 \ge 0. So to compete the proof of the induction step you need to show that:

    k^3-k^2-2k-1 = 0

    has no real root greater than or equal to 6 (this is sufficient as for k greater than its largest real root the cubic is positive).

    This you do by making the substitution x=k-6, then using Descartes rule of signs to show that this has no positive roots.

    (In fact the only real root of k^3-k^2-2k-1 = 0 is near k=2.1)

    RonL
    Last edited by CaptainBlack; March 16th 2008 at 01:48 AM.
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    Quote Originally Posted by xomichelleybelly View Post
    Use math induction to show that

    2) n!>=n^3 for n>=6
    Exactly the same method as for part 1, but the equation that you need to
    show has no root greater than 6 is:

    k^4-k^2-2k-1=0

    and exactly the same method will work with this.

    RonL
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  4. #4
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    Quote Originally Posted by xomichelleybelly View Post
    Use math induction to show that
    1) n!>= n^2 for n>=6

    2) n!>=n^3 for n>=6
    Halve your workload, chum: Start with 2).

    By proving 2) you've automatically proved 1).

    (And I reckon a lawyer could argue that induction has been used to prove both )
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