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Math Help - Prove the series (x^n)/(n!) is e^x

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    Prove the series (x^n)/(n!) is e^x

    Problem: Prove the series (x^n)/(n!) is e^x

    I'm not really sure where to start here. We have just learned power series in class, but not specifically Taylor series yet, so I can use any of those techniques yet. I know that the derivative of the series remains the same, but I don't know how to apply this to prove it is e^x.
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  2. #2
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    Quote Originally Posted by uberbandgeek6 View Post
    Problem: Prove the series (x^n)/(n!) is e^x

    I'm not really sure where to start here. We have just learned power series in class, but not specifically Taylor series yet, so I can use any of those techniques yet. I know that the derivative of the series remains the same, but I don't know how to apply this to prove it is e^x.

    I think the solution is already in your message, but obviously you haven't yet studied this or perhaps you forgot it: there is one unique function which is infinitely differentiable everywhere, which equals 1 at zero and which equals its own derivative, and this is the exponential function e^x...
    Perhaps there's another way to solve your problem, but I can't remember it right now.

    Tonio
    Last edited by tonio; November 16th 2009 at 05:15 AM.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Two expressions of exponential function are...

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

    e^{x} = \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac{x^{k}}{k!} (2)

    Now we prove that (1) and (2) are equivalent. We start from the binomial expansion...

    (1+\frac{x}{n})^{n} = \sum_{k=0}^{n} \binom{n}{k}\cdot (\frac{x}{n})^{k} = 1 + x + \sum_{k=2}^{n} a_{n,k}\cdot \frac{x^{k}}{k!} (3)

    ... where...

    a_{n,k} = \prod_{i=1}^{k-1} (1-\frac{i}{n}) (4)

    In order to well understand the next step we write the sequence of polynomials taking in evidence the powers of x from 0 to 3...

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

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

    (1+\frac{x}{3})^{3} = 1+x + (1-\frac{1}{3})\cdot \frac{x^{2}}{2!} + (1-\frac{1}{3})\cdot (1-\frac{2}{3})\cdot \frac{x^{3}}{3!}

    \dots

    (1+\frac{x}{n})^{n} = 1+x + (1-\frac{1}{n})\cdot \frac{x^{2}}{2!} + (1-\frac{1}{n})\cdot (1-\frac{2}{n})\cdot \frac{x^{3}}{3!} + \dots

    \dots (5)

    We now observe the coefficients of the x^{k} 'by colums'. For k=0 and k=1 they are 1 and they don't change by increasing n. For k=2 the coefficient is  \frac{1-\frac{1}{n}}{2!} and for n \rightarrow \infty it tends to \frac{1}{2!}. For k=3 the coefficient is  \frac{(1-\frac{1}{n})\cdot (1-\frac{2}{n})}{3!} and for n \rightarrow \infty it tends to \frac{1}{3!}. In general is \lim_{n \rightarrow \infty} a_{n,k}=1 so that for  n \rightarrow \infty the coefficient of x^{k} tends to \frac{1}{k!}. The conlusion is that...

    \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n} = \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac{x^{k}}{k!} (6)

    Kind regards

    \chi \sigma
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    Wow, what a mouthful. I cant say I really understand it well, but I'm really impressed. Thanks!
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