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Math Help - Follow the bouncing limit

  1. #1
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    Follow the bouncing limit

    Hi everyone, I'm running into a snag on this problem and am asking for assistance. Sorry that this is a jpg, I have no idea how to use the Math Formula options.

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  2. #2
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    Quote Originally Posted by HeirToPendragon View Post
    Hi everyone, I'm running into a snag on this problem and am asking for assistance. Sorry that this is a jpg, I have no idea how to use the Math Formula options.

    See this.
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  3. #3
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    While it does appear that this is what I'm looking for:

    <br />
n = (a_n+1)^n \geq  1 + na_n + \tfrac{1}{2}n(n-1)a_n^2 \geq \tfrac{1}{2}n(n-1)a_n^2<br />

    Can you tell me how you got the middle equation? The problem I'm having with this current book is that it does exactly what you did here. It shows me a resolution without showing any of the scratch.
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  4. #4
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    Quote Originally Posted by HeirToPendragon View Post
    <br />
n = (a_n+1)^n \geq  1 + na_n + \tfrac{1}{2}n(n-1)a_n^2 \geq \tfrac{1}{2}n(n-1)a_n^2<br />
    (Here n\geq 2).

    This follows by binomial theorem: (a+b)^n = \sum_{k=0}^n {n\choose k}a^kb^{n-k}

    If a,b\geq 0 then (a+b)^n = \sum_{k=0}^2 {n\choose k} a^k b^{n-k} + \sum_{k=2}^n {n\choose k}a^kb^{n-k} \geq \sum_{k=0}^2 {n\choose k}a^k b^{n-k} =  a^n + nab^{n-2}+\tfrac{1}{2}n(n-1)a^2b^{n-2}.
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  5. #5
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    I don't think I've ever even seen that theorem before, which means I probably don't have access to using it

    *sigh*
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  6. #6
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    Quote Originally Posted by HeirToPendragon View Post
    I don't think I've ever even seen that theorem before, which means I probably don't have access to using it
    This is the binomial theorem that you should have learned in high-school.
    If you are an analysis class, surly you must have learned it.
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  7. #7
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    It's possible. High School was 4 years ago and I didn't exactly have the best teachings (like, the sin and cos graphs were not part of my knowledge until I hit college and had to teach myself trig).

    But I'll take your word for it and use that.
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