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Thread: New member, induction problem

  1. #1
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    New member, induction problem

    I have just joined this forum. I studied maths 45+ years ago, and I have not used it since - so I have forgotten nearly everything! I decided to learn calculus again, but do not have access to tutors etc, which makes it hard when I want to check answers. Anyway ...

    Show that n! > n^3 if n is large enough by induction? I get stuck.
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by pastmyprime View Post
    I have just joined this forum. I studied maths 45+ years ago, and I have not used it since - so I have forgotten nearly everything! I decided to learn calculus again, but do not have access to tutors etc, which makes it hard when I want to check answers. Anyway ...

    Show that n! > n^3 if n is large enough by induction? I get stuck.
    Have you done any induction problems before, do you know/understand the basic ideas behind the proof by induction?

    Or is it just setting up the problem initially that you're struggling with?
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  3. #3
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    Quote Originally Posted by craig View Post
    Have you done any induction problems before, do you know/understand the basic ideas behind the proof by induction?
    Or is it just setting up the problem initially that you're struggling with?
    I should have been more precise. Yes, I understand induction and have solved other problems OK.

    First, the smallest n such that $\displaystyle n! > n^3$ is 6.

    Second, assuming it is true for n = k > 6 then show that

    $\displaystyle (k+1)! > (k+1)^3$

    We know that $\displaystyle (k+1)! = k!(k+1) > k^3(k+1)$

    At which point I get stuck!

    PS: My first attempt at using LaTex and I don't know why a formulae is moved down relative to the rest of the line that it is in.
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  4. #4
    Senior Member roninpro's Avatar
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    It looks like you're doing fine.

    I'd like to note that it suffices to show that $\displaystyle k^3(k+1)\geq (k+1)^3$ for $\displaystyle k\geq 6$. Can you do it?
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  5. #5
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    No, sorry, I can't prove it.
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  6. #6
    Senior Member roninpro's Avatar
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    Did you try using induction?
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  7. #7
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    Show $\displaystyle k^3(k+1) \geq (k+1)^3$ for $\displaystyle k \geq 6$

    Step 1 rearrange

    $\displaystyle k^3(k+1) \geq (k+1)^3 \Rightarrow k^4 + k^3 \geq k^3 + 3k^2 + 3k + 1$
    $\displaystyle \Rightarrow k^4 - (3k^2 + 3k + 1) \geq 0$

    Step 2 induction

    First it is true for $\displaystyle k = 6$ by substitution.

    Second, assume it is true for some $\displaystyle k > 6$. Then

    $\displaystyle (k+1)^4 = k^4 + 4k^3 + 6k^2 +4k + 1$
    $\displaystyle 3(k+1)^2 + 3(k+1) + 1 = 3k^2 + 9k + 7$

    So show $\displaystyle k^4 + 4k^3 + 3k^2 -5k - 6 \geq 0$ for $\displaystyle k > 6$

    Now for $\displaystyle k \geq 6 \ \ 3k^2 > 18k > 17k + 6$

    So

    $\displaystyle k^4 + 4k^3 + 3k^2 -5k - 6 > k^4 + 4k^3 + 17k +6 -5k - 6 = k^4 + 4k^3 + 12k > 0$ for all k > 6

    Is that OK?
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