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Prove n^3 less than or equal to 3^n for n=1,2,

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that [tex]n^3 \leq 3^n[/tex] for [itex]n=1,2,...[/itex]


    2. Relevant equations
    Binomial theorem


    3. The attempt at a solution
    Check for P(1)= [tex]1^3 \leq 3^1[/tex]. So the base case holds, now assume P(k) is true and show that it implies P(k+1).

    By assumption [tex]3\cdot k^3 \leq 3\cdot 3^n[/tex]. And this is where I get stuck because obviously [tex](k+1)^3 \nleq 3k^3[/tex]
     
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  3. Oct 14, 2009 #2

    Office_Shredder

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    If that last part is obvious, you should be able to prove it. It'll be tough...

    Consider instead expanding (k+1) and finding an upper bound for that based on k3
     
  4. Oct 14, 2009 #3
    Well it's not "obvious" but plotting [tex]3k^3 - (k+1)^3[/tex] shows negative values for positive k. On the other hand, I'm not sure what you mean by finding an upper bound.

    Expanding gives [tex]k^3+3k^2 +3k+1[/tex]. Comparing that with [tex]k^3 + k^3 +k^3[/tex]. I could try a term by term comparison of [tex]3\cdot 3^k=3^k + 3^k +3 ^k[/tex]. The first terms of each side are ok because that's by assumption. But [tex]3k^2 > 3^k[/tex] so hmmm.
     
  5. Oct 14, 2009 #4

    Office_Shredder

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    If you can find a natural number k that breaks the inequality, you're done. You've proven the hypothesis to be false.

    By inductive hypothesis
    [tex]3k^2<k^3<3^k[/tex] as long as [itex] k \geq 3[/itex]

    So your base case is in fact going to be a couple of values of k, but that's not a real problem
     
  6. Oct 14, 2009 #5

    HallsofIvy

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    Do you realize that you are saying that the very statement you are trying to prove is "obviously" false? [itex]3k^3= k^{k+1}[/itex] so if [itex](k+1)^3\nleq 3k^3= 3^{k+1}[/itex] then [itex]n^2< n^3[/itex] is not true for all n.
     
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