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Math Help - ln(e)=1 proof

  1. #1
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    ln(e)=1 proof

    Hi.

    Hereīs my (di)lemma.

    Show that

    ln(e)=1 proof-picture-1.png

    I was supposed to do that using the intermediate-value theorem but since I didīnt know how to, I ended up doing it like this instead:

    ln(e)=1 proof-picture-3.png

    The only problem is that I can not assume ln(e)=1 and I donīt know how to prove it. They are each others inverses but ... ?

    Any help?
    Attached Thumbnails Attached Thumbnails ln(e)=1 proof-picture-2.png  
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  2. #2
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    Do you know the mean value theorem for integrals?
    \left( {\exists t \in \left[ {1,1 + \frac{1}<br />
{n}} \right]} \right)\left[ {\int\limits_1^{1 + \frac{1}<br />
{n}} {\frac{{dx}}<br />
{x}}  = \frac{1}<br />
{t}[\left( {1 + \frac{1}<br />
{n}} \right) - \left( 1 \right)} ]\right]
    Last edited by Plato; November 20th 2010 at 02:54 PM. Reason: TeX Fix
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  3. #3
    MHF Contributor chisigma's Avatar
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    The demonstration of the 'identity'...

    \displaystyle e^{x}= \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n} = \lim _{n \rightarrow \infty} \sum_{k=0}^{n} \frac{x^{k}}{k!} (1)

    ... is not an impossible task. We can start from the binomial expansion...

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

    ... where...

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

    For better clarity we write the terms (1+\frac{1}{n})^{n} expliciting the powers of x till to 3...

    \displaystyle  (1+\frac{x}{1})^{1} = 1+x

    \displaystyle  (1+\frac{x}{2})^{2} = 1 + x + (1-\frac{1}{2})\ \frac{x^{2}}{2!}

    \displaystyle  (1+\frac{x}{3})^{3} = 1 + x + (1-\frac{1}{3})\ \frac{x^{2}}{2!} + (1-\frac{1}{3})\ (1-\frac{2}{3})\ \frac{x^{3}}{3!}

    ...

    \displaystyle  (1+\frac{x}{n})^{n} = 1 + x + (1-\frac{1}{n})\ \frac{x^{2}}{2!} + (1-\frac{1}{n})\ (1-\frac{2}{n})\ \frac{x^{3}}{3!} + ...

    ...

    Now we observe the coefficients of the x^{k} proceeding 'by colums'. For k=0 and k=1 they are 1 and don't change if n increases. For k=2 the coefficient tends to \frac{1}{2!} if n tends to infinity. For k=3 the coefficient tends to \frac{1}{3!} if n tends to infinity. In general is...

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

    ... so that...

    \displaystyle  \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n} = \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac{x^{k}}{k!} (4)

    Kind regards

    \chi \sigma
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Maybe...

    Using the inequalities:

    1. ln(1+x)<x (for all x!=0)
    2. ln(1+x)>x/(1+x) (for all x)

    and also the Sandwich rule.

    We will have:

    1/{1+(1/n)}=n*{1/n}/{1+(1/n)}< ln(1+1/n)+ln(1+1/n)+...+ln(1+1/n)=ln(1+1/n)^n= ln(1+1/n)+ln(1+1/n)+...+ln(1+1/n)<1/n+...+1/n=n*1/n=1

    When n tends to infinity both sides tends to 1.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Do you know the mean value theorem for integrals?
    \left( {\exists t \in \left[ {1,1 + \frac{1}<br />
{n}} \right]} \right)\left[ {\int\limits_1^{1 + \frac{1}<br />
{n}} {\frac{{dx}}<br />
{x}}  = \frac{1}<br />
{t}[\left( {1 + \frac{1}<br />
{n}} \right) - \left( 1 \right)} ]\right]
    Thank you. How about this?
    ln(e)=1 proof-picture-4.png

    I am also wondering about the visual representation. Is it true that as n gets larger the interval from (1 --> (1+ one n:th ) gets smaller. Therefore the integral (area under the curve) gets really small. But then again we multiply this with n which is a large number. So we get the area 1 if t=1.

    I canīt help but declaring how proud I am of this community. Three replies in no time..WOW!

    (I took a screenshot of my maple screen but would like to convert it to latex so I could paste it in here. I canīt get it to work like I used to. Any ideas?)
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  6. #6
    Senior Member Dinkydoe's Avatar
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    Or you can use L'hopital, if you're allowed to use that, with a little trick:

    Let y =\lim_{n\to\infty}(1+1/n)^{n}


    \ln(y)= \lim_{n\to\infty}n\ln \left(1+\frac{1}{n}\right)\\<br />
= \lim_{n\to\infty} \frac{\ln \left(1+\frac{1}{n}\right)}{\frac{1}{n}}<br /> <br />

    Now apply L'hopital to obtain: \ln y = \lim_{n\to\infty}\frac{\left\frac{1}{1+1/n}\cdot -\frac{1}{n^2}}{-\frac{1}{n^2}} = \lim_{n\to\infty}\frac{1}{1+\frac{1}{n}} = 1

    Thus \ln y= 1, i.e. y= e

    Edit: Nevermind this post. You're supposed to do this with intermediate value theorem. (I expect you even know this proof!)
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  7. #7
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    Thumbs up

    Quote Originally Posted by Dinkydoe View Post
    Or you can use L'hopital, if you're allowed to use that, with a little trick:

    Let y =\lim_{n\to\infty}(1+1/n)^{n}


    \ln(y)= \lim_{n\to\infty}n\ln \left(1+\frac{1}{n}\right)\\<br />
= \lim_{n\to\infty} \frac{\ln \left(1+\frac{1}{n}\right)}{\frac{1}{n}}<br /> <br />

    Now apply L'hopital to obtain: \ln y = \lim_{n\to\infty}\frac{\left\frac{1}{1+1/n}\cdot -\frac{1}{n^2}}{-\frac{1}{n^2}} = \lim_{n\to\infty}\frac{1}{1+\frac{1}{n}} = 1

    Thus \ln y= 1, i.e. y= e

    Edit: Nevermind this post. You're supposed to do this with intermediate value theorem. (I expect you even know this proof!)
    That was great. I just love when there are some many ways to solve a problem in math.
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  8. #8
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    Quite frankly, before I saw "using the intermediate value theorem" I was going to ask "what are your definitions for e and ln?"

    One way of defining e is to define " a^x" for all real numbers x and all positive numbers a, define "e" as the number such that \displaytype\lim_{x\to 0}\frac{e^x- 1}{x}= 1, then define ln(x) as the inverse function to e^x.

    Another way is to define ln(x)= \int_1^x \frac{dt}{t} and then define e^x= exp(x) as its inverse function.

    In both of those cases, ln(e)= 1 follows immediately from the definition of "inverse function".
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  9. #9
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Quite frankly, before I saw "using the intermediate value theorem" I was going to ask "what are your definitions for e and ln?"

    One way of defining e is to define " a^x" for all real numbers x and all positive numbers a, define "e" as the number such that \displaytype\lim_{x\to 0}\frac{e^x- 1}{x}= 1, then define ln(x) as the inverse function to e^x.

    Another way is to define ln(x)= \int_1^x \frac{dt}{t} and then define e^x= exp(x) as its inverse function.

    In both of those cases, ln(e)= 1 follows immediately from the definition of "inverse function".
    Al that is, of course, perfectly correct!... in my opinion the true wilsburschmit's question, apart the questionable presentation, was the demonstration that \displaystyle \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n} = e^{x} , a 'theorem' that isn't immediate consequence of defintions of the exponential function and its inverse function...

    Kind regards

    \chi \sigma
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  10. #10
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    Quote Originally Posted by chisigma View Post
    in my opinion the true wilsburschmit's question, apart the questionable presentation, was the demonstration that \displaystyle \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n} = e^{x}
    Did any of you look in the OP at picture1?
    That contains the original question that wilbursmith wanted to show.
    Here it is: Prove that  \displaystyle \lim _{n \to \infty } n\int_1^{1 + \frac{1}{n}} {\frac{{dt}}{t}}  = 1.
    That question is tailor-made for the mean value theorem for integrals.
    Instead of mean value theorem, I think intermediate value theorem was written.
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  11. #11
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    Absolutely correct, mean value of course! Mixed up the two when translating.
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