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Math Help - proving the existence of "e"

  1. #1
    Junior Member
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    Unhappy proving the existence of "e"

    hello,

    i don't know how to do proving the existence of e ....

    okay, i know that it is by definition :

    \displaystyle  e = \lim_{n\to \infty} (1+\frac {1}{n})^n

    or

    \displaystyle e = \sum_{n=0} ^{\infty} \frac {1}{n!}

    I have no idea how to do it, can some one help me with these please


    P.S. I'm really really sorry if this shouldn't be posted here
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  2. #2
    Member Traveller's Avatar
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    e is usually defined as :

    \displaystyle  e = \lim_{n\to \infty} (1+\frac {1}{n})^n

    Its equivalence to the form \displaystyle e = \sum_{n=0} ^{\infty} \frac {1}{n!} can be proved by many methods including using the generalized binomial theorem or directly the Taylor series.

    http://en.wikipedia.org/wiki/Taylor_series
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  3. #3
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    thanks

    hm... it should not (i think) be shown existence of e using Taylor series, because this we will not do Taylor series for long time ....

    i do't have problem using "e" like it is defined... (lol solved a lot problems using that) but when it comes to prove it's existence i'm think i really don't get it... probably i'm having to much problem of figure it out theoretically, and understanding it to
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  4. #4
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    I will not give you details. Here is an outline.

    I is easy to show that \dfrac{1}{n+1}<\ln\left(1+\frac{1}{n}\right)<\dfra  c{1}{n} . (use Napier's inequality)

    From that we see e^{\frac{1}{n+1}}<\left(1+\frac{1}{n}\right)<e^\fr  ac{1}{n}

    Raise to n^{th} power e^{\frac{n}{n+1}}<\left(1+\frac{1}{n}\right)^n<e

    Use squeeze play.
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  5. #5
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    Quote Originally Posted by Plato View Post
    I will not give you details. Here is an outline.

    I is easy to show that \dfrac{1}{n+1}<\ln\left(1+\frac{1}{n}\right)<\dfra  c{1}{n} . (use Napier's inequality)

    From that we see e^{\frac{1}{n+1}}<\left(1+\frac{1}{n}\right)<e^\fr  ac{1}{n}

    Raise to n^{th} power e^{\frac{n}{n+1}}<\left(1+\frac{1}{n}\right)^n<e

    Use squeeze play.
    thank you very much
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  6. #6
    MHF Contributor chisigma's Avatar
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    If we accept the definition...

    \displaystyle \lim_{n \rightarrow \infty} (1+\frac{1}{n})^{n}= e (1)

    ... what You have to demonstrated it that the limit exists. From the Newton's expansion...

    \displaystyle a_{n}= (1+\frac{1}{n})^{n}= 1 + \frac{\binom{n}{1}}{n} + \frac{\binom{n}{2}}{n^{2}} + \frac{\binom{n}{3}}{n^{3}} + \dots + \frac{\binom{n}{n}}{n^{n}} =

    \displaystyle = 1 + 1 + \frac{(1-\frac{1}{n})}{2!} + \frac{(1-\frac{1}{n})\ (1-\frac{2}{n})}{3!} + \dots + \frac{(1-\frac{1}{n})\ (1-\frac{2}{n}) \dots (1-\frac{n-1}{n})}{n!} (2)

    ... it is evident that...

    a) the sequence a_{n} is increasing with n and for n>1 is a_{n} > 2...

    b) \displaystyle a_{n} < 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{n!} (3)

    Now is 3!= 3\ 2 > 2^{2}, 4!=4\ 3! > 2^{3} , \dots , n!= n\ (n-1)!> 2^{n-1} , so that...

    \displaystyle a_{n} < 1 + 1 + \frac{1}{2} + \frac{1}{2^{2}} + \dots + \frac{1}{2^{n-1}} = 1 + \frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}} = 1 + 2\ (1-\frac{1}{2^{n}}}) (4)

    ... and from (4) You derive that for all n is a_{n}< 3. The conclusion is that the limit (1) exists and is 2< e < 3. For the 'pratical' computation of e the (1) is very slow and is better the relation found by Isaac Netwon himself...

    \displaystyle e= \sum_{n=0}^{\infty} \frac{1}{n!} (5)

    Kind regards

    \chi \sigma
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  7. #7
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    If you prefer to use e= \displaystyle\sum_{n=0}^\infty \frac{1}{n!} as the definition of e, then proving that e exists is just proving that the series converges- and that is easy to do using the "ratio test".
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