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Math Help - The "e" limit

  1. #1
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    The "e" limit

    e=\lim_{x\to\infty}\bigg(1+\frac1x\bigg)^x

    What's the proof of this?
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    Super Member Rebesques's Avatar
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    Well, this could easily be a definition


    But just using Taylor series for log, we get:

    \lim_{x\rightarrow\infty}(1+\frac{1}{x})^x=\lim_{x  \rightarrow\infty}{\rm e}^{x\log(1+\frac{1}{x})}=\lim_{x\rightarrow\infty  }{\rm e}^{x\left(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}+...\right)}=\lim_{x\  rightarrow\infty}{\rm e}^{1-\frac{1}{2x}+\frac{1}{3x^2}+...}

    and by continuity of the exponential function, this is equal to

    {\rm e}^{\lim_{x\rightarrow\infty}\left(1-\frac{1}{2x}+\frac{1}{3x^2}+...\right)}={\rm e}
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  3. #3
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    Ohhh... and without Taylor Series?
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    Quote Originally Posted by Krizalid View Post
    ... and without Taylor Series?
    Can you prove this inequality 0 < a < b\quad  \Rightarrow \quad \frac{1}{b} < \frac{{\ln (b) - \ln (a)}}{{b - a}} < \frac{1}{a}?

    Let a = 1\quad \& \quad b = 1 + \frac{1}{x} using the inequality we get
    \frac{1}{{x + 1}} < \ln \left( {1 + \frac{1}{x}} \right) < \frac{1}{x}.

    Using the property of exponents:
    e^{\left( {\frac{1}{{x + 1}}} \right)}  < \left( {1 + \frac{1}{x}} \right) < e^{\left( {\frac{1}{x}} \right)} \quad  \Rightarrow e^{\left( {\frac{x}{{x + 1}}} \right)}  < \left( {1 + \frac{1}{x}} \right)^x  < e\quad .

    Now the limit is apparent.
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    e=\lim_{x\to\infty}\bigg(1+\frac1x\bigg)^x

    What's the proof of this?
    One thing that you have to think about when faced with this question is:

    What definition of e do you have to work with?

    RonL
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  6. #6
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    Well... the base of natural logarithms.

    Can you show a proof of what I'm askin'?
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    Quote Originally Posted by Krizalid View Post
    Well... the base of natural logarithms.

    Can you show a proof of what I'm askin'?
    Well I would consider:

    <br />
f(x)=\lim_{n \to \infty} \left( 1+\frac{x}{n}\right)^n<br />

    and show that this satisfies the initial value problem:

    <br />
\frac{df}{dx}=f(x),\ f(0)=1<br />

    RonL
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  8. #8
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    Quote Originally Posted by CaptainBlack View Post
    Well I would consider:

    <br />
f(x)=\lim_{n \to \infty} \left( 1+\frac{x}{n}\right)^n<br />

    and show that this satisfies the initial value problem:

    <br />
\frac{df}{dx}=f(x),\ f(0)=1<br />
    Aha... how'd be the proof?
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  9. #9
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    Quote Originally Posted by CaptainBlack View Post
    Well I would consider:

    <br />
f(x)=\lim_{n \to \infty} \left( 1+\frac{x}{n}\right)^n<br />

    and show that this satisfies the initial value problem:

    <br />
\frac{df}{dx}=f(x),\ f(0)=1<br />

    RonL
    Quote Originally Posted by Krizalid View Post
    Aha... how'd be the proof?
    Put:

    <br />
f_n(x)= \left( 1+\frac{x}{n}\right)^n<br />

    then:

    <br />
\frac{ df_n}{d x}=n (1+x/n)^{n-1} \frac{1}{n}=f_n(x)/(1+x/n)<br />


    Now I will assume that you can interchange the two limits involved to get:

    <br />
\frac{df}{dx}=\lim_{n \to \infty} \frac{df_n}{dx}= \frac{\lim_{n \to \infty}f_n(x)}{\lim_{n \to \infty}(1+x/n)}=f(x)<br />

    Now as f_n(0)=1 \ \forall n,\ \ f(0)=1.

    So we have f(x) is the solution of the initial value problem:

    f'(x)=f(x), \ \ f(0)=1

    which is all we need.

    RonL
    Last edited by CaptainBlack; September 17th 2007 at 07:16 PM.
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  10. #10
    Super Member Rebesques's Avatar
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    What definition of e do you have to work with?
    Well... the base of natural logarithms.
    If you ever write a book, don't use this definition, it's going to be hell to derive basic properties
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  11. #11
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    Krizalid, the stadard way e^x is defined is first by defining,
    \ln x = \int_1^x \frac{dt}{t} \mbox{ for }x>0.
    And let e^x = \ln ^{-1} x.
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  12. #12
    Super Member Rebesques's Avatar
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    There's actually many ways to define e.

    The one I like most is e=\lim(1+1/n)^n, which comes natural historically (from medieval banking, that is )

    Another relatively easy way is by the (all familiar) series e=\sum_n 1/n!, but this is troublesome when it comes to deriving the basic properties of e^x.

    The most delicate way, is by describing the function \exp:\mathbb{R}\rightarrow\mathbb{R} to be the unique solution of the functional equation \exp(x+y)=\exp(x)\exp(y) to satisfy \exp(0)=1. Then e=\exp(1).


    Ok, I know you are bored by now, so I stop here ()
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    Quote Originally Posted by Rebesques View Post
    There's actually many ways to define e.
    But they are not as easy to use for analytical arguments as the \ln x one. If you were writing an analysis book I am sure you will use \ln x approach because it is by far the simplest to derive all the identities and properties for e^x.

    The most delicate way, is by describing the function \exp:\mathbb{R}\rightarrow\mathbb{R} to be the unique solution of the functional equation \exp(x+y)=\exp(x)\exp(y) to satisfy \exp(0)=1. Then e=\exp(1).
    What about \exp (x) = 1?
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  14. #14
    Super Member Rebesques's Avatar
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    But they are not as easy to use for analytical arguments as the...
    I' ll disagree here, the lim(1+1/n)^n works fine to that extent.




    What about... ?
    You just verified that there were three conditions altogether in that definiton, I just thought hard but couldn't remember if there was another one when making that post
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  15. #15
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    Here's a proof which involves integrals (But it's just a little integral.)

    Let

    \varphi = \lim_{x \to \infty } \left( {1 + \frac{1}<br />
{x}} \right)^x .

    Since the logarithm is continuous on its domain, we can interchange the function and taking limits.

    \ln \varphi = \ln \left[ {\lim_{x \to \infty } \left( {1 + \frac{1}<br />
{x}} \right)^x } \right] = \lim_{x \to \infty } x\ln \left( {1 + \frac{1}<br />
{x}} \right).

    Make the substitution u=\frac1x,

    \ln \varphi =\lim_{u\to0}\frac1u\ln(1+u).

    Since \ln (1 + u) = \int_1^{1 + u} {\frac{1}<br />
{\tau }\,d\tau } ,\,1 \le\tau\le 1 + u,

    \frac{1}<br />
{{1 + u}} \le \frac{1}<br />
{\tau } \le 1\,\therefore \,\frac{u}<br />
{{1 + u}} \le \int_1^{1 + u} {\frac{1}<br />
{\tau}\,d\tau} \le u.

    So \frac1{1+u}\le\frac1u\ln(1+u)\le1.

    Take the limit when u\to0, then by the Squeeze Theorem we can conclude that \lim_{u\to0}\frac1u\ln(1+u)=1.

    Finally \ln\varphi=1\,\therefore\,\varphi=e.
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