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Math Help - Stuff with limits and integrals.. the usual

  1. #1
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    Stuff with limits and integrals.. the usual

    First prove the inequality
    e^{-t}\Big(1-\frac{e}{2n}t^2\Big)\leq\Big(1-\frac{t}{n}\Big)^n\leq e^{-t}


    for some reason this has been giving me problems, any idea would be appreciated


    Then they ask the following but I am not sure how to justify what I am doing to show it:

    Show
    \lim_{n\rightarrow\infty}\int_0^{n}\Big(1-\frac{t}{n}\Big)^{n} t^{x-1}dt= \int_0^\infty e^{-t}t^{x-1}dt
    If 0<x\leq1 you must consider a special case neat t=0


    Then they ask

    Show:
    \int_0^\infty e^{-t}t^{x-1}dt = \lim_{n\rightarrow\infty} \frac{n^xn!}{x(x+1)(x+2)\cdot\cdot\cdot(x+n)} for x>0


    Any ideas are appreciated, thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    Let's start fron the Newton expansion...

    (1-\frac{t}{n})^{n}= \sum_{k=0}^{n} (-1)^{k}\cdot \binom{n}{k}\cdot (\frac{t}{n})^{k} (1)

    ... that can be rearranged as...

    (1-\frac{t}{n})^{n}= 1-t+\sum_{k=2}^{n}(-1)^{k}\cdot a_{n,k}\cdot \frac{t^{k}}{k!} (2)

    ... where...

    a_{n,k}= \prod_{i=1}^{k-1}(1-\frac{i}{n}) (3)

    It is easy to verify from (3) that is...

    a_{n,k}> 0 \forall n>k

    a_{n+1,k}>a_{n,k} \forall n>k

    \lim_{n\rightarrow \infty} a_{n,k}=1 \forall k (4)

    From the last of (4) and (2) we derive the well known result...

    \lim_{n\rightarrow \infty} (1-\frac{t}{n})^{n}= \lim_{n\rightarrow \infty} 1-t+\sum_{k=2}^{n}(-1)^{k}\cdot a_{n,k}\cdot \frac{t^{k}}{k!} = \sum_{k=0}^{\infty}(-1)^{k}\frac{t^{k}}{k!}= e^{-t} (5)

    Now we will use (2) for computing the difference between two consecutive terms of the sequence (1-\frac{t}{n})^{n} ...

    (1-\frac{t}{n+1})^{n+1} - (1-\frac{t}{n})^{n} = \sum_{k=2}^{n} (-1)^{k} \cdot (a_{n+1,k}-a_{n,k}) \cdot \frac{t^{k}}{k!} + (-1)^{n+1}\cdot a_{n+1,n+1}\cdot \frac{t^{n+1}}{(n+1)!} (6)

    Since the (6) is a series with alternate decreasing terms and the first is positive, the sequence (1-\frac{t}{n})^{n} is increasing and, taking into account (5), we conclude that...

    (1-\frac{t}{n})^{n}<e^{-t} \forall n (7)

    ... so that half of your first question is answered...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor chisigma's Avatar
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    Post with errors!... please delete!... sorry!...
    Last edited by chisigma; April 23rd 2009 at 08:21 AM.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Now we proceed and the first step is computing the inverse of (1-\frac{t}{n})^{n}...

    (1-\frac{t}{n})^{-n}= 1 + t + (1+\frac{1}{n})\cdot \frac{t^{2}}{2!} + (1+\frac{1}{n})\cdot(1+\frac{2}{n})\cdot \frac {t^{3}}{3!} +\dots

    \dots + (1+\frac{1}{n})\cdot(1+\frac{2}{n})\dots (1+\frac{k-1}{n})\cdot \frac {t^{k}}{k!} +\dots (1)

    ... and after the inverse of  e^{-t}\cdot (1-\frac{e}{2n}\cdot t^{2})...

    \frac{e^{t}}{(1-\frac{e}{2n}\cdot t^{2})} = e^{t}\cdot \{1 + (\frac{e}{2n})\cdot t^{2} + (\frac{e}{2n})^{2}\cdot t^{4} + \dots  (\frac{e}{2n})^{k}\cdot t^{2k} + \dots\} =

     = 1 + t + \{(\frac{e}{2n}) + \frac{1}{2!}\}\cdot t^{2} + \{(\frac{e}{2n}) + \frac{1}{3!}\}\cdot t^{3} + \{(\frac{e}{2n})\cdot \frac{1}{2!} + (\frac{e}{2n})^{2} + \frac{1}{4!}\}\cdot t^{4} + \dots (2)

    Now we observe in (1) and (2) the coefficients of the term t^{k}. For k=0 and k=1 they are equal, for k=2 and \forall n>0 is...

    \frac{e}{2n} + \frac{1}{2} > \frac{1}{2n} + \frac{1}{2} (3)

    ... so that for -t_{0}<t<t_{0} and \forall n>0 is...

    \frac{e^{t}}{(1-\frac{e}{2n}\cdot t^{2})} > (1-\frac{t}{n})^{-n} \rightarrow e^{-t}\cdot (1-\frac{e}{2n}\cdot t^{2}) < (1-\frac{t}{n})^{n} (4)

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 23rd 2009 at 07:17 PM.
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  5. #5
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    Opalg's Avatar
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    Quote Originally Posted by KZA459 View Post
    First prove the inequality
    e^{-t}\Big(1-\frac{e}{2n}t^2\Big)\leq\Big(1-\frac{t}{n}\Big)^n\leq e^{-t}
    I think that for both inequalities you will need to assume that t<n. Otherwise, they may well fail. (If t = 2n then the right-hand inequality will fail for n even, and the left-hand one will fail for n odd.)

    For the right-hand inequality, take logs. You then need to show that \ln\bigl(1-\tfrac tn\bigr)\leqslant-\tfrac tn. But the inequality \ln(1+x)\leqslant x holds for all x>1 (easy proof by calculus).

    The left-hand inequality looks more delicate. But the logarithmic approach might still work there.
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  7. #7
    MHF Contributor chisigma's Avatar
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    The double inequality...

    e^{-t}\cdot(1-\frac{e}{2n}\cdot t^{2}) < (1-\frac{t}{n})^{n} < e^{-t} (1)

    ... has been proved [even if with some difficulties ...] in posts #2 and #4. What does matter however is that (1) holds only for 0<t<t_{0} where t_{0} must be found and the reason is evident if you consider that...

    \lim_{t \rightarrow +\infty} e^{-t}\cdot(1-\frac{e}{2n}\cdot t^{2}) = \lim_{t \rightarrow +\infty} e^{-t}=0, \forall n>0

    ... and...

    \lim_{t \rightarrow +\infty} (1-\frac{t}{n})^{n} = + \infty, n even,  = -\infty, n odd

    In post #2 has been demonstrated that...

    \lim_{n \rightarrow \infty} (1-\frac{t}{n})^{n}= e^{-t}, \forall t (2)

    ... and that validates the expression...

    \lim_{n \rightarrow \infty} \int_{0}^{n} (1-\frac{t}{n})^{n} \cdot t^{x-1}\cdot dt= \int_{0}^{\infty} t^{x-1}\cdot e^{-t}\cdot dt (3)

    The last question will take to me a little more time ...

    Kind regards

    \chi \sigma
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  8. #8
    MHF Contributor chisigma's Avatar
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    In post #2 it has been found the expansion...

    (1-\frac{t}{n})^{n} = 1 - t +\sum_{k=2}^{n} (-1)^{k}\cdot a_{n,k}\cdot \frac{t^{k}}{k!} (1)

    ... where...

    a_{n,k}= \prod_{i=1}^{k-1}(1-\frac{i}{n}) (2)

    ... and also that is...

    \lim_{n\rightarrow \infty} a_{n,k}= 1 , \forall k (3)

    Using (1) we obtain...

    (1-\frac{t}{n})^{n}\cdot t^{x-1} = t^{x-1} - t^{x} + \sum_{k=2}^{n} (-1)^{k}\cdot a_{n,k}\cdot \frac{t^{x+k-1}}{k!} (4)

    ... so that...

    \int_{0}^{n}(1-\frac{t}{n})^{n}\cdot t^{x-1}\cdot dx = \frac{n^{x}}{x} - \frac{n^{x+1}}{x+1} + \sum_{k=2}^{n}(-1)^{k}\cdot a_{n,k}\cdot \frac{n^{x+k}}{(x+k)\cdot k!}=

    = n^{x}\cdot \{\frac {1}{x} - \frac{n}{x+1} + \sum_{k=2}^{n}(-1)^{k}\cdot a_{n,k}\cdot \frac{n^{k}}{(x+k)\cdot k!}\} (5)

    In post #4 it has been demonstrated that...

    \int_{0}^{\infty}t^{x-1}\cdot e^{-t}\cdot dt = \lim_{n\rightarrow \infty} \int_{0}^{n} (1-\frac{t}{n})^{n}\cdot t^{x-1}\cdot dt (6)

    ... so that, taking into account (3) and (5), is...

    \int_{0}^{\infty}t^{x-1}\cdot e^{-t}\cdot dt = \lim_{n\rightarrow \infty} n^{x}\cdot \sum_{k=0}^{n}(-1)^{k}\cdot \frac{n^{k}}{(x+k)\cdot k!} (7)

    Now we have to search if (7) can be written in different way...

    Kind regards

    \chi \sigma
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