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Math Help - Rewriting e as an infinite series

  1. #1
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    Rewriting e as an infinite series

    I'm given the function f(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!} and I'm supposed to prove that f(x)=e^x.
    I was thinking that I should use a different representation of e, namely:

    e^x=\lim_{n \rightarrow \infty} (1+\frac{x}{n})^n

    and use the binomial theorem to expand it as an infinite series, but I can't get it to work properly. If it's possible, then how should it be done?
    Otherwise, any hints on how I should prove it instead?
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    Re: Rewriting e as an infinite series

    Quote Originally Posted by scounged View Post
    I'm given the function f(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!} and I'm supposed to prove that f(x)=e^x.
    The answer to this depends upon the level of rigor required in proof.
    There is one well respected calculus textbook simply points out that f(x)=f'(x) in this case.
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    Re: Rewriting e as an infinite series

    Quote Originally Posted by Plato View Post
    There is one well respected calculus textbook simply points out that f(x)=f'(x) in this case.
    Does that mean that f(x)=e^x is the only function that equals its own derivative?
    If that's the case, then I think that's the way I'm intended to prove it (in a previous part of this assignment I'm supposed to prove that f(x)=f'(x).
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    Re: Rewriting e as an infinite series

    Quote Originally Posted by scounged View Post
    Does that mean that f(x)=e^x is the only function that equals its own derivative?
    If that's the case, then I think that's the way I'm intended to prove it (in a previous part of this assignment I'm supposed to prove that f(x)=f'(x).
    Well there is a trivial function, f(x)=0. But clearly that is not the case here.
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    Re: Rewriting e as an infinite series

    Then I think I'll ask my teacher tomorrow if proving f(x)=f'(x) is enough. Thanks for the help.
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    Re: Rewriting e as an infinite series

    Quote Originally Posted by scounged View Post
    I'm given the function f(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!} and I'm supposed to prove that f(x)=e^x.
    I was thinking that I should use a different representation of e, namely:

    e^x=\lim_{n \rightarrow \infty} (1+\frac{x}{n})^n

    and use the binomial theorem to expand it as an infinite series, but I can't get it to work properly. If it's possible, then how should it be done?
    Otherwise, any hints on how I should prove it instead?
    Instead of using the Binomial Theorem, expand \displaystyle \begin{align*} e^x \end{align*} as a MacLaurin series.

    \displaystyle \begin{align*} f(x) = \sum_{n = 0}^{\infty}\frac{f^{(n)}(0) \cdot x^n}{n!}  \end{align*}
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    Re: Rewriting e as an infinite series

    Does that mean that is the only function that equals its own derivative?
    Quote Originally Posted by Plato View Post
    Well there is a trivial function, f(x)=0. But clearly that is not the case here.
    Not quite. f(x)= Ae^x, for A any constant (which, of course, includes 1 and 0), has that property.
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    Re: Rewriting e as an infinite series

    Yeah, I found it in my formula collection. If I had known about MacLaurin series yesterday I don't think I would've created this thread. Anyway, thanks for the help.
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