Question

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

Answer 1

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 `-1/x`, the Maclaurin series for `e^(-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)+...`

Answer 2

See description:

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

`e^(-1/x) = 1/e + (x-1)/e - (x-1)^2/(2e) + ...`

Explanation:

Let `f(x)=e^(-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(x)` 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 `-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 CC`:

`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(1)=e^(-1) =1/e`

We need the first derivative:

`f'(x)=e^(-1/x)/x^2`
`:. f'(1)=e^(-1)/1 =1/e`

And the second derivative (using quotient rule):

`f''(x) = ( (x^2)(e^(-1/x)/x^2)-(e^(-1/x))(2x) ) / (x^2)^2`
`" " = ( e^(-1/x)(1-2x) ) / (x^4)`
`:. f''(1) = -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 + ...`
`" " = 1/e + (x-1)/e - (x-1)^2/(2e) + ...`