# What is the taylor expansion of e^(-1/x)?

Mar 14, 2017

Maclaurin Series

The Maclaurin series for ${e}^{x}$ is given by:

e^x=sum_(n=0)^oox^n/(n!)=1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+...

Replacing $x$ with $- \frac{1}{x}$, the Maclaurin series for ${e}^{- \frac{1}{x}}$ is:

e^(-1/x)=sum_(n=0)^oo(-1/x)^n/(n!)=sum_(n=0)^oo(-1)^n/(n!)x^-n

color(white)(e^(-1/x))=1-1/x+1/((2!)x^2)-1/((3!)x^3)+1/((4!)x^4)+...

Mar 15, 2017

See description:

The Taylor Series about $x = 1$ is given by:

${e}^{- \frac{1}{x}} = \frac{1}{e} + \frac{x - 1}{e} - {\left(x - 1\right)}^{2} / \left(2 e\right) + \ldots$

#### Explanation:

Let $f \left(x\right) = {e}^{- \frac{1}{x}}$

The Taylor Series about the pivot point $x = a$ is given by:

 f(x) = f(a) + f'(a)(x-a) + (f''(a))/(2!)(x-a)^2 + (f^((3))(a))/(3!)(x-a)^3 + ... + (f^((n))(a))/(n!)(x-a)^n + ...

As no pivot point for the Taylor Expansion Series has been provided it would be usual practice to assume that $a = 0$ which gives us the Maclaurin Series;

 f(x) = f(0) + f'(0)x + (f''(0))/(2!)x^2 + (f^((3))(0))/(3!)x^3 + ... + (f^((n))(0))/(n!)x^n + ...

However, $f \left(x\right)$ has an essential singularity when $x = 0$ and so we cannot form the Maclaurin series, (ie the Taylor series pivoted about $x = 0$).

Technically this is the end of the question - There is no such series.

Using the well know series for ${e}^{x}$ we can expand a series by substituting $x$ for $- \frac{1}{x}$. This gives us a power series of increasing negative powers, and is known as a Laurent Series (As Laurent series typically have complex arguments we use $z$ by convention rather than $x$ where $z \in \mathbb{C}$:

 e^(-1/z) = 1-1/z+1/(2!z^2)-1/(3!z^3)+1/(4!z^4)+...

We can however form a Taylor Series about another pivot point so lets do so about $x = 1$.

Firstly, we have:

$f \left(1\right) = {e}^{- 1} = \frac{1}{e}$

We need the first derivative:

$f ' \left(x\right) = {e}^{- \frac{1}{x}} / {x}^{2}$
$\therefore f ' \left(1\right) = {e}^{- 1} / 1 = \frac{1}{e}$

And the second derivative (using quotient rule):

$f ' ' \left(x\right) = \frac{\left({x}^{2}\right) \left({e}^{- \frac{1}{x}} / {x}^{2}\right) - \left({e}^{- \frac{1}{x}}\right) \left(2 x\right)}{{x}^{2}} ^ 2$
$\text{ } = \frac{{e}^{- \frac{1}{x}} \left(1 - 2 x\right)}{{x}^{4}}$
$\therefore f ' ' \left(1\right) = - \frac{1}{e}$

$\vdots$

And so the Taylor Series about $x = 1$ is given by:

 f(x) = 1/e + 1/e(x-1) + (-1/e)/(2!)(x-1)^2 + (f^((3))(a))/(3!)(x-1)^3 + ...
$\text{ } = \frac{1}{e} + \frac{x - 1}{e} - {\left(x - 1\right)}^{2} / \left(2 e\right) + \ldots$