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How does $\sum_{n=0}^{\infty}\frac{p^n}{n!} = e^p$? I know that it is a geometric sum but I can't seem to work it out.

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It is not a geometric series: the ratio of consecutive terms is not constant, because of the factorial in the denominator. – Brian M. Scott May 1 '12 at 21:51
It depends on where you start. If you know $\dfrac{d^n}{dx^n}e^x=e^x$ then you could take a Taylor series – Henry May 1 '12 at 21:52
And in some treatments the infinite series is taken to be the definition of $e^p$. – Brian M. Scott May 1 '12 at 21:52
Another way: $f(x) = \sum_{n=0}^\infty x^n/n!$ satisfies: $f$ is continuous, increasing, $f(1)=e$ and $f(x+y) = f(x)f(y)$. – GEdgar May 1 '12 at 21:56
"So I take it, it is a 'standard' definition for e?" - it's one of the ways to define it. Which brings us to the question of what definition is being used in your notes... – J. M. May 1 '12 at 21:59
up vote 10 down vote accepted

Richard, the sum is not a geometric series, since the ratio is not constant:

$$\frac{a_{n+1}}{a_n}=\frac p {n+1} $$ It is rather the series representation for the exponential function $y=e^x$.

We know from the theory of Taylor polynomials and series that if we have a function $f$ that is differentiable inifinitely many times, then under appropriate conditions we can define it's Taylor series around a point $x=a$ as

$$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$

The appropriate condition is that the error produced by the $n$th approximation, which assuming the integrability of $f^{(n)}$, can be proven to be

$$E_n(a,x) = \int_a^x\frac{(x-t)^n}{n!}f^{(n+1)}(t)dt$$

will tend to zero for $n\to \infty$. The error expression which looks quite strange is not too hard to derive. I transcribe from this question of mine:

$$f(x)=f(a)+f'(a)(x-a)+R_1(x)$$ so that $$R_1(x) = f(x)-f(a) - f'(a) (x-a)$$ $${R_1}(x) = \int\limits_a^x {f'\left( t \right)dt} - \int\limits_a^x {f'(a)dt} $$ $${R_1}(x) = \int\limits_a^x {f'\left( t \right) - f'\left( a \right)dt} $$ So now we integrate by parts with $$f'\left( t \right) - f'\left( a \right) = u$$ $$t - x = v$$ to get $${R_1}(x) = \int\limits_a^x {\left( {x - t} \right)f''\left( t \right)dt} $$ We can similarily do this with $R_2(x)$, since $${R_2}(x) = {R_1}(x) - f''\left( a \right)\frac{{{{\left( {x - a} \right)}^2}}}{{2!}}$$ $${R_2}(x) = \int\limits_a^x {\left( {x - t} \right)f''\left( t \right)dt} - \int\limits_a^x {\left( {x - t} \right)f''\left( a \right)dt} $$ $${R_2}(x) = \int\limits_a^x {\left( {x - t} \right)\left( {f''\left( t \right) - f''\left( a \right)} \right)dt} $$ So again integrating by parts gives $${R_2}(x) = \int\limits_a^x {\frac{{{{\left( {x - t} \right)}^2}}}{{2!}}f'''\left( t \right)dt} $$

In the case of the exponential function, we can prove that the error indeed goes to zero, from where we can represent it around $x=a$ as

$$\exp(x)=\sum_{n=0}^{\infty} \frac{e^a }{n!}(x-a)^n$$

since it is the case $f^{(n)}(a) = e^a$ for the exponential function.

It is important to note some authors define $$\exp(x):=\sum_{n=0}^{\infty} \frac{x^n }{n!}$$

and derive then the properties of the function (note this is the series around $x=0$).

NOTE The error term in $(1)$ is used whenever $f^{(n)}$ is integrable. Else, we need other formulas, such as Cauchy's or Lagrange's.

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Thanks Peter, most appreciated. Really clear answer and gave me a better understanding! – Richard May 3 '12 at 14:31

A very polished development of this series expression for the exponential function is given in the opening section of W. Rudin's Real and Complex Analysis. Rudin's approach is to take the power series as defining a holomorphic function (since the radius of convergence is infinite) and briefly prove that the function equals its derivative. Moreover if one specifies the value at the origin $f(0)=1$, then this characterizes the exponential function. The usual properties of the exponential function can also be deduced from the series, e.g. $e^x e^{-x} = 1$, from which it follows that $e^x$ is never zero.

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Thanks for your reference hardmath. – Richard May 3 '12 at 14:30

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