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Maths - proof

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    I need to prove that (1 + x/n)^n tends to e^x as n tends to infinity. could someone give me a step by step proof?
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    (Original post by Shinjimonkey)
    I need to prove that (1 + x/n)^n tends to e^x as n tends to infinity. could someone give me a step by step proof?
    Do you know where to start? Have you tried anything?
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    (Original post by RDKGames)
    Do you know where to start? Have you tried anything?
    I am trying to:

    ln(1+x/n)^n
    = n ln (1+x/n)

    but I do not know whether this is right or not
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    (Original post by Shinjimonkey)
    I need to prove that (1 + x/n)^n tends to e^x as n tends to infinity. could someone give me a step by step proof?
    We know that:

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

    \displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3  }{3!}+...+\frac{x^r}{r!}+...

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

    Can you work from there??
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    (Original post by RDKGames)
    We know that:

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

    \displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3  }{3!}+...+\frac{x^r}{r!}+...

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

    Can you work from there??
    how do we know the first rule that you have given?
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    (Original post by Shinjimonkey)
    how do we know the first rule that you have given?
    Because \displaystyle e^{n \ln(1+\frac{x}{n})}= e^{\ln[(1+\frac{x}{n})^n]}=(1+ \frac{x}{n})^n

    Simple logarithm manipulation whereby e^{\ln(b)}=b
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    (Original post by RDKGames)
    Because \displaystyle e^{n \ln(1+\frac{x}{n})}= e^{\ln[(1+\frac{x}{n})^n]}=(1+ \frac{x}{n})^n

    Simple logarithm manipulation whereby e^{\ln(b)}=b
    I see... as n tends to infinity, x/n tends to 0, so ln(1+x/n) tends to 0, therefore
    n ln(1+x/n) = 0. then wouldn't (1+x/n)^n tend to 1?
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    (Original post by Shinjimonkey)
    I see... as n tends to infinity, x/n tends to 0, so ln(1+x/n) tends to 0
    Correct.

    (Original post by Shinjimonkey)
    therefore n ln(1+x/n) = 0. then wouldn't (1+x/n)^n tend to 1?
    Not quite. Using the expansion for \ln(1+x) you would find that:

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

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

    Therefore from the RHS you can see that as n goes to infinity, you are left with x and not a 0.
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    (Original post by RDKGames)
    Correct.



    Not quite. Using the expansion for \ln(1+x) you would find that:

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

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

    Therefore from the RHS you can see that as n goes to infinity, you are left with x and not a 0.
    I don't quite understand why ln(1+x/n) expands to that?
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    (Original post by Shinjimonkey)
    I don't quite understand why ln(1+x/n) expands to that?
    I hope you understand where the expansion for \displaystyle \ln(1+x) comes from.

    If you're okay with that, the transition from \displaystyle \ln(1+x) to \displaystyle n \ln(1+\frac{x}{n}) is quite straight forward as \displaystyle x \mapsto \frac{x}{n} so you replace every x with \displaystyle \frac{x}{n}. Then of course, the whole series is simply multiplied by n
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    (Original post by RDKGames)
    I hope you understand where the expansion for \displaystyle \ln(1+x) comes from.

    If you're okay with that, the transition from \displaystyle \ln(1+x) to \displaystyle n \ln(1+\frac{x}{n}) is quite straight forward as \displaystyle x \mapsto \frac{x}{n} so you replace every x with \displaystyle \frac{x}{n}. Then of course, the whole series is simply multiplied by n
    I do not quite understand why the expansion of ln(1+x) come from? Could you explain please? thank you for the help!
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    (Original post by RDKGames)
    I hope you understand where the expansion for \displaystyle \ln(1+x) comes from.

    If you're okay with that, the transition from \displaystyle \ln(1+x) to \displaystyle n \ln(1+\frac{x}{n}) is quite straight forward as \displaystyle x \mapsto \frac{x}{n} so you replace every x with \displaystyle \frac{x}{n}. Then of course, the whole series is simply multiplied by n
    I understand now. I just realised that ln(!+x) = integral of 1/(1+t) to the limits x to 0. Thanks for the help!
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    (Original post by Shinjimonkey)
    I do not quite understand why the expansion of ln(1+x) come from? Could you explain please? thank you for the help!
    It follows from Maclaurin's Series whereby any function f(x) can be rewritten as a sum of derivatives with increasing powers of x.

    \displaystyle f(x)=f(0)+f'(0)x+\frac{f''(0)}{2  !}x^2+...+\frac{f^r(0)}{r!}x^r+.  ..

    So let \displaystyle f(x)=\ln(1+x) and perform this. You will get out with an infinite summation.

    https://www.youtube.com/watch?v=ibGZapvnB-M
 
 
 
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