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Math Help - Another Induction

  1. #1
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    Another Induction

    Got another problem on induction, this time I don't even know who to start and what to do.

    What I want to do is to show that 3^n \geq n^3 is true for all
    n \in N.
    Here we define the natural set of numbers decluding 0.

    From here I don't know much about what to do but show it true for n=1, witch I of course have done.
    Anyone got any ideas?
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  2. #2
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    Quote Originally Posted by a4swe
    Got another problem on induction, this time I don't even know who to start and what to do.

    What I want to do is to show that 3^n \geq n^3 is true for all
    n \in N.
    Here we define the natural set of numbers decluding 0.

    From here I don't know much about what to do but show it true for n=1, witch I of course have done.
    Anyone got any ideas?
    Next time, please post thy questions in a new thread.
    ---
    We note that,
    3^0\geq 0^3
    3^1\geq 1^3
    Let S represent all N such that it is true.
    We know that 0,1\in \mathbb{N}.

    Assume, k\in S, i.e. that it is true for k.
    Then,
    3^k\geq k^3
    Multiply by 3,
    3^{k+1}\geq 3k^3

    Note that,
    2k^3\geq 3k^2+3k+1
    For k\geq 2
    Add k^3 to both sides,
    3k^3\geq k^3+3k^2+3k+1
    Gaze upon ye perfect cubical,
    3k^3\geq (k+1)^3
    Transitivity,
    3^{k+1}\geq (k+1)^3
    Thus,
    (k+1)\in S

    Thus, S contain all N, i.e. true for all integers.
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  3. #3
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    Next time, please post thy questions in a new thread.
    Excuse me, the others forums I am active on keep a high "one topic only" policy, witch means that all subjects about let's say induction goes under one topic. That was not the case here I have to and will check the rules and posting guide lines.
    Once again, I am sorry.

    We know that 0,1\in \mathbb{N}.
    Well, that wasn't true for the definition of N I used, but ok it doesn't interfere with the soulution in general, I just don't care for the 0.
    Thanks for the soultion, I will now go back to my pencils and papers.
    Good night to you!
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  4. #4
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    Quote Originally Posted by a4swe
    Excuse me, the others forums I am active on keep a high "one topic only" policy
    Maybe it was okay to do that, I am just a moderator I like to boss people around, it makes me feel as a superior being (which of course I am).
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  5. #5
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    Quote Originally Posted by a4swe
    Excuse me, the others forums I am active on keep a high "one topic only" policy, witch means that all subjects about let's say induction goes under one topic. That was not the case here I have to and will check the rules and posting guide lines.
    Once again, I am sorry.
    Frankly I don't care what other site's policies are. Posting a new
    question into an existing thread confuses the thread, so very
    quickly nobody can tell which question any new replies apply to.

    If I find it happening I will split the thread, and if people persist
    after having this explained to them they will be given negetive
    reputation feedback.

    RonL
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  6. #6
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    Frankly I don't care what other site's policies are.
    Off course you don't, I just wanted to explain (not to make excuses) my wrongfull actions, it won't happen again.
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  7. #7
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    More on topic then:

    Note that,
    <br />
2k^3\geq 3k^2+3k+1<br />
    For
    Yes, I have noted that and yes it's true.
    But how do I show it?
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  8. #8
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    Quote Originally Posted by a4swe
    More on topic then:



    Yes, I have noted that and yes it's true.
    But how do I show it?
    You may find it easier to show
    3k^3 \ge (k + 1)^3

    or equivalently

    3 \ge \frac{(k+1)^3}{k^3}

    This is easily done by considering different values of k and noting that \frac{(k+1)^3}{k^3} monotonically decreases to a value of 1 as k increases indefinitely.

    -Dan
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  9. #9
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    A lot easier, thanks.
    Off to bed now.
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  10. #10
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    Quote Originally Posted by a4swe
    Yes, I have noted that and yes it's true.
    But how do I show it?
    Use another induction argument
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  11. #11
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    Use another induction argument
    Was afraid so.
    Ok, we try that.
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