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anonymous

  • 5 years ago

prove induction n^2 < n! , for n>=4 , if it is easier n! > n^2 i didnt get far, the basis case is true and (k+1)!= (k+1)k! > ...

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  1. watchmath
    • 5 years ago
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    I am sorry did you just delete your earlier question with my answer in it?

  2. anonymous
    • 5 years ago
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    oh im sorry

  3. anonymous
    • 5 years ago
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    this program is strange , i didnt see anything in it

  4. anonymous
    • 5 years ago
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    i apologize earlier, anything you have to say is helpful :)

  5. anonymous
    • 5 years ago
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    it doesnt have to be induction

  6. anonymous
    • 5 years ago
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    but i have a proof of my own, the basis case is true 4! > 4^2 (k+1)! = k!(k+1) > k^2 ( 2 + 1) = k^2( 1 + 1 + 1) = k^2 + k^2 + k^2 > k^2 + 2k + 1

  7. anonymous
    • 5 years ago
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    oh wait, that doesnt work

  8. anonymous
    • 5 years ago
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    not the most elegant proof, but it might work

  9. watchmath
    • 5 years ago
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    Ok, if you want we can use induction here we go. So we assume that k^2 < k! Notice that that k>1 is equivalent to 2k+1 < 3k It follows that k>= 4 we have 2k+1 < 3k < (k)! k Now using the assumption k^2 < k! we have k^2 +(2k+1) < k! +(2k+1) + k! +(k)! k=(k+1)! So (k+1)^2 < (k+1)!

  10. anonymous
    • 5 years ago
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    that looks, good,. one sec, what was your other proof . now im curious

  11. watchmath
    • 5 years ago
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    So you haven't read it?

  12. anonymous
    • 5 years ago
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    im reading it , one sec

  13. watchmath
    • 5 years ago
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    The statement \(n^2<n!\) is equivalent to \(n<(n-1)!\) Notice that for \(n\geq 4\) we have \((n-1)!\geq (n-2)(n-1)\) So it is enough to prove \((n-2)(n-1)>n\) The inequality is equivalent to \(n^2-3n+2>n\) \(n^2-4n+2>0\) \(n(n-4)+2>0\) which is clearly true for \(n\geq 4\).

  14. anonymous
    • 5 years ago
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    ok thank you

  15. watchmath
    • 5 years ago
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    Are you trying to use this for some statement about the series? if it is the case maybe it will be easier to prove the statement for some bigger n (not 4) to make it more obvious.

  16. anonymous
    • 5 years ago
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    thats clever, can you look over my proof the basis case is true 4! > 4^2 (k+1)! = k!(k+1) > k^2 ( 2 + 1) = k^2( 1 + 1 + 1) = k^2 + k^2 + k^2 > k^2 + 2k + 1

  17. anonymous
    • 5 years ago
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    not that i know of about serious

  18. anonymous
    • 5 years ago
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    because k>2 for k>=4

  19. anonymous
    • 5 years ago
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    oh but then i need to show k^2 > 2k ?

  20. watchmath
    • 5 years ago
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    correct, that is what I was going to say :)

  21. watchmath
    • 5 years ago
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    well actually you need 2k^2 > 2k+1

  22. watchmath
    • 5 years ago
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    Notice that \((k-1)^2 \geq 0\) So \(k^2-2k+1 \geq 0\) So \(k^2\geq 2k-1\) For \(k\geq 4, k^2 > 2\) So \(2k^2> k^2+2\geq (2k-1)+2 =2k +1\)

  23. anonymous
    • 5 years ago
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    yes

  24. anonymous
    • 5 years ago
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    can i ask a quick question. im helping a student who thinks that you can form a solid of revolution , bounded by y=x^2, y = x+2, and revolve it about the y axis

  25. watchmath
    • 5 years ago
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    Yes I think you can but there is some overlap between the solid form by the region on the right of the y-axis and the solid form by the region on the left side of the y-axis

  26. anonymous
    • 5 years ago
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    ty

  27. anonymous
    • 5 years ago
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    keeps freezing on me

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