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Thread: 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 $\displaystyle k \ge 6$

    So we need to show that:

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

    well:

    $\displaystyle (k+1)! = k!~ k$

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

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

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

    $\displaystyle k^3-k^2-2k-1 = 0$

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

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

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

    RonL
    Last edited by CaptainBlack; Mar 16th 2008 at 12:48 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:

    $\displaystyle k^4-k^2-2k-1=0$

    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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