Question

What is the Taylor expansion of ${{e}^{{-1}/{x}\;}}$?

Answer 1

Before solving this question you should know about the Taylor series of a function and you can expand any function by this till an infinite sum of terms which are expressed in terms of the function’s derivatives at a single Point. Mainly in most of the functions the function and the sum of its Taylor Series are equal near this point.

Complete step by step answer:
In this question we have to find the Taylor expansion of ${{e}^{{-1}/{x}\;}}$.
Let $f\left( x \right)={{e}^{{-1}/{x}\;}}$
The Taylor series about the point $x=a$ is given by
$f\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)+\dfrac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+.......+\dfrac{{{f}^{n}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}+......$
As no pivot point for the Taylor expansion series has been provided so it will be as usually assume that $a=0$, which will provide is Maclaurin series:
$f\left( x \right)=f\left( 0 \right)+f'\left( 0 \right)x+\dfrac{f''\left( 0 \right)}{2!}{{x}^{2}}+\dfrac{{{f}^{3}}\left( 0 \right)}{3!}{{x}^{3}}+.......+\dfrac{{{f}^{n}}\left( 0 \right)}{n!}{{x}^{n}}+.....$
So, $f\left( x \right)$ has an essential singularity when $x=0$ and so we cannot form the Maclaurin Series.
So, technically here the question has ended. There is no such series.
Using the well known series for ${{e}^{x}}$ we can expand a series by substituting x for ${-1}/{x}\;$. This gives us the power series of increasing negative powers.
So, we can however form a Taylor series about any other pivot point so let’s do so about $x=1$ firstly we have,
$f\left( 1 \right)={{e}^{-1}}=\dfrac{1}{e}$
No the first derivative ${{f}^{-1}}\left( x \right)=\dfrac{{{e}^{-1/x}}}{{{x}^{2}}}$
$\therefore {{f}^{-1}}\left( 1 \right)=\dfrac{{{e}^{-1}}}{1}=\dfrac{1}{e}$
And the second derivative (by quotient rule): -

& f''\left( x \right)=\dfrac{{{x}^{2}}\left( \dfrac{{{e}^{-1/x}}}{{{x}^{2}}} \right)-\left( {{e}^{-1/x}} \right)\left( 2x \right)}{{{\left( {{x}^{2}} \right)}^{2}}} \\\ & f''\left( x \right)=\dfrac{{{e}^{-1/x}}\left( 1-2x \right)}{{{x}^{4}}} \\\ & \therefore f''\left( 1 \right)={-1}/{e}\; \\\ & .. \\\ & \,.. \\\ \end{aligned}$$ And so the Taylor series about $$x=1$$ is given by: $$\begin{aligned} & f\left( x \right)=\dfrac{1}{e}+\dfrac{1}{e}\left( x-1 \right)+\dfrac{\dfrac{-1}{e}}{2!}{{\left( x-1 \right)}^{2}}+\dfrac{{{f}^{3}}\left( 0 \right)}{3!}{{\left( x-1 \right)}^{3}}+..... \\\ & f\left( x \right)=\dfrac{1}{e}+\dfrac{x-1}{e}-\dfrac{{{\left( x-1 \right)}^{2}}}{2e}+....... \\\ \end{aligned}$$ **Note:** For solving the expansion if any function you should know about to calculate the Maclaurin series. Because if the Maclaurin series is satisfied or formed then we can easily form the Taylor series and if it does not then you have to substitute any other value regarding that.