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Thread: 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?
    $\displaystyle \left( {\exists t \in \left[ {1,1 + \frac{1}
    {n}} \right]} \right)\left[ {\int\limits_1^{1 + \frac{1}
    {n}} {\frac{{dx}}
    {x}} = \frac{1}
    {t}[\left( {1 + \frac{1}
    {n}} \right) - \left( 1 \right)} ]\right]$
    Last edited by Plato; Nov 20th 2010 at 01:54 PM. Reason: TeX Fix
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  3. #3
    MHF Contributor chisigma's Avatar
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    The demonstration of the 'identity'...

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

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

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

    ... where...

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

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

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

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

    $\displaystyle \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!} $

    $\displaystyle ...$

    $\displaystyle \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!} + ... $

    $\displaystyle ... $

    Now we observe the coefficients of the $\displaystyle x^{k}$ 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 $\displaystyle \frac{1}{2!}$ if n tends to infinity. For k=3 the coefficient tends to $\displaystyle \frac{1}{3!}$ if n tends to infinity. In general is...

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

    ... so that...

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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?
    $\displaystyle \left( {\exists t \in \left[ {1,1 + \frac{1}
    {n}} \right]} \right)\left[ {\int\limits_1^{1 + \frac{1}
    {n}} {\frac{{dx}}
    {x}} = \frac{1}
    {t}[\left( {1 + \frac{1}
    {n}} \right) - \left( 1 \right)} ]\right]$
    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 $\displaystyle y =\lim_{n\to\infty}(1+1/n)^{n}$


    $\displaystyle \ln(y)= \lim_{n\to\infty}n\ln \left(1+\frac{1}{n}\right)\\
    = \lim_{n\to\infty} \frac{\ln \left(1+\frac{1}{n}\right)}{\frac{1}{n}}

    $

    Now apply L'hopital to obtain: $\displaystyle \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 $\displaystyle \ln y= 1$, i.e. $\displaystyle y= e$

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


    $\displaystyle \ln(y)= \lim_{n\to\infty}n\ln \left(1+\frac{1}{n}\right)\\
    = \lim_{n\to\infty} \frac{\ln \left(1+\frac{1}{n}\right)}{\frac{1}{n}}

    $

    Now apply L'hopital to obtain: $\displaystyle \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 $\displaystyle \ln y= 1$, i.e. $\displaystyle y= e$

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

    Another way is to define $\displaystyle ln(x)= \int_1^x \frac{dt}{t}$ and then define $\displaystyle e^x= exp(x)$ 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 "$\displaystyle a^x$" for all real numbers x and all positive numbers a, define "e" as the number such that $\displaystyle \displaytype\lim_{x\to 0}\frac{e^x- 1}{x}= 1$, then define ln(x) as the inverse function to $\displaystyle e^x$.

    Another way is to define $\displaystyle ln(x)= \int_1^x \frac{dt}{t}$ and then define $\displaystyle e^x= exp(x)$ 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 $\displaystyle \displaystyle \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n} = e^{x}$ , a 'theorem' that isn't immediate consequence of defintions of the exponential function and its inverse function...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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 $\displaystyle \displaystyle \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n} = e^{x}$
    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 $\displaystyle \displaystyle \lim _{n \to \infty } n\int_1^{1 + \frac{1}{n}} {\frac{{dt}}{t}} = 1$.
    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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