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Thread: proof of that the definition limit of "e" exists... please help!

  1. #1
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    proof of that the definition limit of "e" exists... please help!

    Hi,

    I'm reviewing some basic calculus concepts, and I have the following questions that will probably be very easy for you guys to answer, although I'm stuck since a long time on them.

    Given the usual limit that is used in the definition of "e": lim (n->infinity) (1+1/n), how do we prove that the limit exists and it is finite?

    Can we use the binomial theorem expansion for a finite n, and prove that the coefficients tend to be 1/k! for n that goes to infinity, so we get the Taylor expansion? I'd like to see a rigorous proof of this derivation of e's expression as a power series, we shouldn't forget to prove that the coefficients of the binomial expansion don't diverge (in fact, i guess they should converge) as n goes to infinity...

    Or without expanding it, can we prove the following statements:

    let a(n)=(1+1/n)^n

    1-prove that a(n) is strictly increasing
    2-prove that a(n) is limited above by a finite positive number L (for instance, we could try to prove it for L=3 or L=4): a(n)<L for all n.

    Final questions:

    A-if we manage to prove the two statements above, then did we successfully prove that the limit exists?
    B-is there a better way than this to prove that the limit exists?

    Thanks! And please excuse me for the length of the post...
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  2. #2
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    Using the notation $\displaystyle e^x=\exp(x)$ we show it.
    It is easy to use Napierís inequality to show that $\displaystyle \dfrac{1}{n+1}\le\ln\left(1+\frac{1}{n}\right)\le\ dfrac{1}{n}.$

    Raise e to those powers $\displaystyle \exp\left(\frac{1}{n+1}\right)\le\left(1+\frac{1}{ n}\right)\le\exp\left(\frac{1}{n}\right).$.

    Raise each of those to the nth power
    $\displaystyle \exp\left(\frac{n}{n+1}\right)\le\left(1+\frac{1}{ n}\right)^n\le e.$.

    It is squeezed between e & e.
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  3. #3
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    Thank you for answering so soon, the proof is very simple, but how do you define ln(x) without proving the limit of e first? I mean, wouldn't it be circular logic?

    We should find a way to prove it without using the natural logarithm...

    Alternatively, we could define ln(x) as the integral of 1/t from 1 to x, and then napier's inequality should be simple to infer, but we need to use a integral definition...

    there should be an elementary method to prove it...
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  4. #4
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    There are so many different ways to do this.

    Notice that for $\displaystyle n\ge 2$ we have $\displaystyle \[
    \left( {1 - \frac{1}{{n^2 }}} \right)^n \geqslant \left( {1 - \frac{1}
    {n}} \right)\; \Rightarrow \;\left( {1 + \frac{1}{n}} \right)^n \geqslant \left( {1 + \frac{1}{{n - 1}}} \right)^{n - 1} $.

    So the sequence $\displaystyle \left( {1 + \frac{1}{n}} \right)^n $ is increasing and bounded above. That means it converges.

    Call that limit $\displaystyle e$.

    EDIT: If you need to show it is bounded.
    $\displaystyle \left( {1 + \frac{1}
    {n}} \right)^n = \sum\limits_{k = 0}^n {C(n,k)\left( {\frac{1}
    {n}} \right)^k } < 1 + \left( {1 + \frac{1}
    {2} + \cdots + \frac{1}
    {{2^{n - 1} }}} \right) < 3$
    Last edited by Plato; Oct 1st 2010 at 11:27 AM.
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  5. #5
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    Ok thank you very much, I understood how you deduced the facts from those inequalities, I will work to prove them.

    You have been very kind, thanks again.
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  6. #6
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    Here is a couple of links from PlanetMath.
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    Great, thanks again!
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  8. #8
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    By definition, $\displaystyle e$ is the base of the exponential function that is its own derivative.

    Therefore $\displaystyle \frac{d}{dx}(e^x) = e^x$.

    If we wish to evaluate $\displaystyle e$, we need to use this definition.


    $\displaystyle \frac{d}{dx}(e^x) = \lim_{h \to 0}\frac{e^{x + h} - e^x}{h}$

    $\displaystyle = \lim_{h \to 0}\frac{e^xe^h - e^x}{h}$

    $\displaystyle = \lim_{h \to 0}\frac{e^x(e^h - 1)}{h}$

    $\displaystyle = e^x\lim_{h \to 0}\frac{e^h - 1}{h}$.


    Since we know by definition that $\displaystyle \frac{d}{dx}(e^x) = e^x$, that means

    $\displaystyle e^x\lim_{h \to 0}\frac{e^h - 1}{h} = e^x$

    $\displaystyle \lim_{h \to 0}\frac{e^h - 1}{h}= 1$

    $\displaystyle \lim_{h \to 0}(e^h - 1) = \lim_{h \to 0}h$

    $\displaystyle \lim_{h \to 0}e^h = \lim_{h \to 0}(1 + h)$

    $\displaystyle \lim_{h \to 0}(e^h)^{\frac{1}{h}} = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$

    $\displaystyle e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$.


    Now if we let $\displaystyle h = \frac{1}{n}$, then as $\displaystyle h \to 0, n \to \infty$.

    Thus $\displaystyle e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$.
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  9. #9
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    Another useful way to define ln(x) is simply:
    $\displaystyle ln(x)= \int_1^x \frac{1}{t}dt$

    Since $\displaystyle \frac{1}{t}$ is defined for all non-zero t, this defines ln(x) for all positive x. It is clearly continuous and differentiable for all x. By the fundamental theorem of Calculus, its derivative is 1/x. Since that is positive for all positive x, ln(x) is an increasing function and so a one-to-one function.

    It is not difficulty to prove, by substitutions in the integral, that
    1) For any positive x, $\displaystyle ln(1/x)= - ln(x)$.
    2) For any positive x and y, $\displaystyle ln(xy)= ln(x)+ ln(y)$.
    3) For any positive x and y any real number, $\displaystyle ln(x^y)= y ln(x)$.

    For example, for the last one, $\displaystyle ln(x^y)$ is, by this definition, $\displaystyle \int_1^{x^y}\frac{1}{t}dt$.

    If $\displaystyle y\ne 0$, define $\displaystyle u= t^{1/y}$ so that $\displaystyle t= u^y$, $\displaystyle du= y u^{y- 1}du$ so that $\displaystyle \frac{1}{t}dt= \frac{1}{u^y}(y u^{y-1}du)= y \frac{1}{u}du$. When t= 1, $\displaystyle u= 1^{1/y}= 1$ and when $\displaystyle t= x^y$, $\displaystyle u= (x^y)^{1/y}= x$ so the integral becomes
    $\displaystyle ln(x^y)= \int_1^x y \frac{1}{u}du= y\int_1^x \frac{1}{u}du= y ln(x)$

    (If y= 0 then $\displaystyle x^y= x^0= 1$ and $\displaystyle ln(x^y)= ln(1)= \int_1^1 \frac{1}{t}dt= 0= 0 ln(x)= y ln(x)$ still.)

    It follows then that ln(x) is a one-to-one function from the positive real numbers onto the real numbers and so has an inverse function, exp(x), from the real numbers to the positive real numbers.

    Finally, if y= exp(x), then x= ln(y). If x is not 0, divide both sides by x to get $\displaystyle 1= \frac{1}{x}ln(y)= ln(y^{1/x})$. Going back to the exponential form, $\displaystyle y^{1/x}= exp(1)$ so $\displaystyle y= (exp(1))^x$.
    (If x= 0, then y= 1 because $\displaystyle ln(1)= \int_1^1 \frac{1}{t}dt= 0$. So we still have $\displaystyle exp(x)= exp(0)= 1= (exp(1))^0$.
    Last edited by HallsofIvy; Oct 2nd 2010 at 02:59 AM.
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