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Here is a standard identity:

$$\sum_{i=0}^{\infty}\frac{a^i}{i!}=e^a$$

Why does it hold true?

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  • $\begingroup$ Do you see the relationship between this and the Taylor series for the exponential function? $\endgroup$ – yoknapatawpha Feb 7 '14 at 1:18
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    $\begingroup$ For two reasons: (a) because it's the definition of $e^a$, or (b) because the left hand-side is the Taylor series of $e^a$ and this function is analytic, hence agreeing with its Taylor series. $\endgroup$ – Agustí Roig Feb 7 '14 at 1:19
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    $\begingroup$ Perhaps this link might prove useful. $\endgroup$ – Lucian Feb 7 '14 at 1:19
  • $\begingroup$ You argue by stages, depending of your definition of $e$. If $e=\sum 1/i!$, you can prove by simply multiplying the series repeatedly that the identity holds for $a$ rational. You can see the identity for $a$ irrational either as the definition of $e^a$, or as a consequence of continuity. $\endgroup$ – Andrés E. Caicedo Feb 7 '14 at 1:29
  • $\begingroup$ Ok, I got it! aahh mathematics are awesome :) Thank you all! $\endgroup$ – Remi.b Feb 7 '14 at 1:35
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A brief answer: Let's consider the exponential function $e^x$. The definition of $e$ is that $\frac{d}{dx}e^x = e^x$. Now let's assume that $e^x$ can be written as an infinite sum of the form $\sum_{i=0}^{\infty}a_ix^i$. Using the sum rule for derivatives, we have $\sum_{i=0}^{\infty}a_ix^i = \sum_{i=0}^{\infty}\frac{d}{dx}a_ix^i = \sum_{i=1}^{\infty}ia_{i}x^{i-1}$. Therefore, $a_i = \frac{a_{i-1}}{i}$, so starting with the base case of $i=0$, $a_i = \frac{1}{i!}$.


(Warning: variables change from the previous section)

Another way to look at this is to consider the more standard definition of $e$, which is $\lim_{n\to\infty}(1 + \frac{1}{n})^n$. Therefore, $e^a$ can be written as $\lim_{n\to\infty}(1 + \frac{1}{n})^{an}$. Using the binomial theorem, the expression expands to $$\lim_{n\to\infty}(1 + \frac{an}{n} + \frac{an(an - 1)}{2n^2}+ \frac{an(an-1)(an-2)}{6n^3} + \ldots)$$ The lesser powers in the numerators drop out, so the expression becomes $$1 + a + \frac{a^2}{2} + \frac{a^3}{6} + \ldots$$which is $\sum_{i=0}^{\infty}\frac{a^i}{i!}$ in sum notation.

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    $\begingroup$ How do you define $e^x$? What do you mean by "the definition is that $(e^x)'=e^x$"? $\endgroup$ – Andrés E. Caicedo Feb 7 '14 at 1:41
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    $\begingroup$ Why can we assume that $e^x$ is given by a power series? $\endgroup$ – Andrés E. Caicedo Feb 7 '14 at 1:42
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    $\begingroup$ One way to define the exponential function is to define it as the unique solution of $f'=f, f(0)=1$. If you want to use properties of holomorphic functions, then this already implies that $e^x$ can be expressed as power series. Otherwise, one could have a look at its Taylor series and spend some work on its convergence. $\endgroup$ – Roland Feb 7 '14 at 1:46
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The Taylor series for $e^x$ is

$$ \sum_{k = 0}^{\infty} \frac{x^k}{k!}. $$

Letting $x = a$, we obtain

$$ e^a = \sum_{k = 0}^{\infty} \frac{a^k}{k!} $$

as desired.

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  • $\begingroup$ I think you might have missed the point $\endgroup$ – user27182 Feb 10 '14 at 0:05

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