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Thread: Inductive Proof

  1. #1
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    Inductive Proof

    For which integers is 3^n > n^3?

    it seems to be for all but 3, but I am stuck on the proof.
    Assume 3^n > n^3
    n>3
    Then
    3^(n+1) > 3*n^3
    Now what?
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  2. #2
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    Re: Inductive Proof

    Quote Originally Posted by veronicak5678 View Post
    For which integers is 3^n > n^3?
    it seems to be for all but 3, but I am stuck on the proof.
    Assume 3^n > n^3 if n>3
    I will help you, but will not do this for you.
    $\displaystyle \text{If }n\ge 4\text{ then }3n^2+3n+1\le 48+12+1=61<3^n~.$

    So that means that $\displaystyle (n+1)^3\le n^3+3^n~.$

    You must find out how that helps you. Hint $\displaystyle 2<3$.
    Last edited by Plato; Aug 2nd 2011 at 04:48 PM.
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  3. #3
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    Re: Inductive Proof

    Quote Originally Posted by veronicak5678 View Post
    For which integers is 3^n > n^3?

    it seems to be for all but 3, but I am stuck on the proof.
    Assume 3^n > n^3
    n>3
    Then
    3^(n+1) > 3*n^3
    Now what?
    Another way to consider...

    You want to show

    $\displaystyle 3^{n+1}>(n+1)^3$

    Note that $\displaystyle 3^1>1^3,\;\;\;3^2>2^3,\;\;\;3^3=3^3$

    Hence n>3 for the proof...

    If $\displaystyle 3^n>n^3$, then $\displaystyle 3^{n+1}>3n^3$

    If we can show that $\displaystyle 3n^3>(n+1)^3$, for n>3, the proof is complete.

    Since we know n>3, then

    $\displaystyle 3n^3>n^3+3n^2+3n+1\;\;?$

    $\displaystyle 2n^3>3n^2+3n+1\;\;?$

    Now use the fact that if n>3, then $\displaystyle 2n^3>2(3)n^2$ and continue
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  4. #4
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    Re: Inductive Proof

    Thanks for the answer, but I don't follow it.

    What do you mean by the question marks?
    We know n > 3, so 3n^3 > 3*81
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    Re: Inductive Proof

    Quote Originally Posted by veronicak5678 View Post
    Thanks for the answer, but I don't follow it.

    What do you mean by the question marks?
    We know n > 3, so 3n^3 > 3*81
    The purpose of the question marks is...

    We are asking if $\displaystyle 3n^3>n^3+3n^2+3n+1,\;\;if\;\; n>3$

    Because if that's true, then $\displaystyle 3^{n+1}>(n+1)^3$

    Proof By Induction requires that you show $\displaystyle 3^{n+1}>(n+1)^3\;\;\;if\;\;\;3^n>n^3$

    Hence we are asking if

    $\displaystyle 2n^3\;\;is\;greater\;than\;\;\;3n^2+3n+1,\;\;n>3$

    Now if n>3, then $\displaystyle 2n^3>2(3)n^2$

    and you can finish by showing than $\displaystyle 6n^2>3n^2+3n+1$
    using the same reasoning,
    since $\displaystyle 2n^3>6n^2,\;\;\;n>3$
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Re: Inductive Proof

    Quote Originally Posted by veronicak5678 View Post
    For which integers is 3^n > n^3?

    it seems to be for all but 3, but I am stuck on the proof.
    Assume 3^n > n^3
    n>3
    Then
    3^(n+1) > 3*n^3
    Now what?
    Just to point out - you also need to consider the negative numbers, which won't be covered by your induction. That said, that aren't too hard if you thing about it correctly...(Where does $\displaystyle 3^{-n}$ lie? What about $\displaystyle (-n)^3$?)
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  7. #7
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    Re: Inductive Proof

    Got it. Thanks very much, everyone!
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