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  1. #1
    Kärnfysikern's Avatar
    Kärnfysikern is offline Retired: AR-Hall of Famer
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    Maths help needed

    How do I prove that

    2^n > n^3 for n>10 or n=10??? I am suposed to use the induktion principle(this is a straight translation from swedish so might not be the correct english term).

    I have done it in steps.

    Step 1. 2^10 > 10^3. The difference is valid for n=10.

    Step 2. Assume that the difference is true for n=p where p=10 or p>10

    2^p > p^3 <=> 2^p - p^3 > 0

    Step 3

    2^(p+1) - (p+1)^3 = 2*2^p - (p^3 + p^2 + 3p + 1)=
    = 2*2^p - p^3 - p^2 - 3p -1 > 0

    How do I continue from there?? Or have I made a error in step 3?
    Should I replace 2*2^p with 2*p^3?? damn I hate this

  2. #2
    Calipso's Avatar
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    This thread scared me and reminded me I'm going to college some day.

  3. #3
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    2+2=4...... F-O-U-R


  4. #4
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    Quote Originally Posted by johan
    How do I prove that

    2^n > n^3 for n>10 or n=10??? I am suposed to use the induktion principle(this is a straight translation from swedish so might not be the correct english term).

    I have done it in steps.

    Step 1. 2^10 > 10^3. The difference is valid for n=10.

    Step 2. Assume that the difference is true for n=p where p=10 or p>10

    2^p > p^3 <=> 2^p - p^3 > 0

    Step 3

    2^(p+1) - (p+1)^3 = 2*2^p - (p^3 + p^2 + 3p + 1)=
    = 2*2^p - p^3 - p^2 - 3p -1 > 0

    How do I continue from there?? Or have I made a error in step 3?
    Should I replace 2*2^p with 2*p^3?? damn I hate this
    ok ok.........i think you are totally making this more complicated then it really is.......first off......does ^ mean multiplication?!?!?........you need to explain all the terms.........

    no clue what the induktion principle is................please explain

    what Math class are you taking?............

    "where p=10 or p>10"....quoting you.........now if P is 10..........then how is P greater then 10......???......10 equals 10.......

    im just trying to help you............if i have all the correct terms and know what you are trying to solve...........this problem will be a breeze for me........

  5. #5
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    2*2^p - p^3 - p^2 - 3p -1 > 0 should be your final answer

    Plug in 10 and you should have your proof. If F(P+1) is true, then so is F(P) where n=10, n>10.

    I think? Been awhile since I did these. Calc 1 right? Analytic?

  6. #6
    BDTR's Avatar
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    This post made me so happy i'll never have to take Calc again. God i hated college.

  7. #7
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    I can solve this, but have 2 questions. What level math is this and are you allowed to use L'Hoptial's rule?

  8. #8
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    Quote Originally Posted by bdtr
    This post made me so happy i'll never have to take Calc again. God i hated college.
    I am laughing at one of my friends who is taking Calc II, and his prof. won't let him use his graphing calculator... only a normal scientific one!!! I feel sorry for him, but glad my math is done!

  9. #9
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    Prove:
    2^(n+1) > (n+1)^3 forall n >= 10

    Inductive step
    2^(n+1) > (n+1)^3 =
    (2^n) + (2^n) > (n^3) + (3n^2 + 3n + 1)
    By the inductive hypothesis we know that 2^n > n^3 forall n > 10
    so it suffices to prove that n^3 > 3n^2 + 3n + 1

    Since we know n > 10 then we know that
    n^3 > 10n^2 and also
    n^3 > 3n^2 + 3n^2 + 3n^2
    and since 3n^2 > 3n > 1 we can conclude
    n^3 > 3n^2 + 3n + 1
    QED

  10. #10
    BDTR's Avatar
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    I would have never made it through Cal II without a graphing calculator.

    Quote Originally Posted by navydevildoc
    I am laughing at one of my friends who is taking Calc II, and his prof. won't let him use his graphing calculator... only a normal scientific one!!! I feel sorry for him, but glad my math is done!

  11. #11
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    Calculus is the biggest pain in the ass class i have had thus far..besides Organic Chemistry.

  12. #12
    Full Intensity's Avatar
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    Quote Originally Posted by MMC78
    Prove:
    2^(n+1) > (n+1)^3 forall n >= 10

    Inductive step
    2^(n+1) > (n+1)^3 =
    (2^n) + (2^n) > (n^3) + (3n^2 + 3n + 1)
    By the inductive hypothesis we know that 2^n > n^3 forall n > 10
    so it suffices to prove that n^3 > 3n^2 + 3n + 1

    Since we know n > 10 then we know that
    n^3 > 10n^2 and also
    n^3 > 3n^2 + 3n^2 + 3n^2
    and since 3n^2 > 3n > 1 we can conclude
    n^3 > 3n^2 + 3n + 1
    QED
    that looks spot on to me cause for induction you first have to prove it correct for all "n" then for all "n+1" Fuck do the calc days bring back bad bad memories

  13. #13
    Dude-Man's Avatar
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    The worst part is the calcs keep going. I took calc 4 (differential equations) last semester. Hardest class i've ever taken.

  14. #14
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    calc I & II isn't too difficult....the intense algebra in it is what always threw me. but without a graphing calc....ouch.

  15. #15
    mass junkie's Avatar
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    Thanks for reminding me how much I suck at Math

  16. #16
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    www.math.com has tones of calculators thats the only way I passed my math class

  17. #17
    Kärnfysikern's Avatar
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    Thanks everyone.

    Yupp Billy its analytic

    Thanks for the correct answere MMC78

    Danielle, ^ menas...uhhh dont know the english word for it to be honest

    but x^2=x*x and x^3=x*x*x and so on. What is the correct english word for it?? It sux that we read the maths on swedish I would prefer it to be on english

    And I said "where p=10 or p>10" because the "equals or greater then" isymbol isnt aviable on the keyboard so I had to type it out like that instead.

    Cant figure out how to explain the inductive hypothesis on english either

    Now I know where to turn when I get stuck in maths expect those kinds of threads alot this weekend because I have a test on monday

    Thanks again!

  18. #18
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    No problem. BTW the english word for the ^ operator is exponentiation. 2^n is usually written with the n as a superscript, but it's common when writing on a computer to use ^.

  19. #19
    Kärnfysikern's Avatar
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    Quote Originally Posted by MMC78
    No problem. BTW the english word for the ^ operator is exponentiation. 2^n is usually written with the n as a superscript, but it's common when writing on a computer to use ^.
    btw just noticed a thing

    2^(n+1) > (n+1)^3 =
    (2^n) + (2^n) > (n^3) + (3n^2 + 3n + 1)

    shouldnt that be (n^3)*(3n^2+3n+1)??

    This is how my teacher wanted me to solve it

    2^(p+1)-(p+1)^3 = 2*2^p - (p^3+3p^2+3p+1) > 2*p^3 - (p^3+3p^2+3p+1)=p^3-3p^2-3p-1
    The above is > or = with
    10p^2-3p^2-3p-1=7p^2-3p-1 , p>10 or p=10
    That is > or = with
    70p-3p-1 , p>10 or p=10
    That in turn is > or = with
    670-1 , p>10 or p=10
    And then we have

    2^(p+1)-(p+1)^3 > 660 > 0

    So the difference according to the induktion is valid for all n>10 n= 10

    Dont know what way to solve it is the best one

  20. #20
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    [I](2^n) + (2^n) > (n^3) + (3n^2 + 3n + 1)
    shouldnt that be (n^3)*(3n^2+3n+1)??/[I]

    No, that's ok. I just grouped the addition with parenthesis to make the next step clearer. I don't know why your teacher didn't like this proof. It's basically a textbook soultion for a class I taught 2 years ago.

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