Question

How do you find the Taylor expansion for ${e^{ - \dfrac{1}{x}}}$?

Answer 1

In order to calculate the Taylor series of the $f(x)$ which is centred at $x = a = 0$, but due to singularity in function consider the pivot point as $x = 1$, we first derive the series for ${e^{ - \dfrac{1}{x}}}$ by use the formula of Taylor series $f\left( x \right) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(1)}}{{n!}}{{(x - 1)}^n}}$, up to the value of nth term. Find the derivative of the function up to the order n using the property of $\dfrac{d}{{dx}}({e^X}) = {e^X}.\dfrac{d}{{dx}}\left( X \right)$ and chaining rule. Putting all the values back into the formula will give your required result.

Formula Used:
${T_n}(x) = \sum\limits_{i = 0}^n {\dfrac{{{f^{(i)}}(a)}}{{i!}}{{(x - a)}^i}}$
$\dfrac{d}{{dx}}({e^X}) = {e^X}.\dfrac{d}{{dx}}\left( X \right)$

Complete step by step solution:
We are given a function ${e^{ - \dfrac{1}{x}}}$. Let it be call $f(x)$
$f(x) = {e^{ - \dfrac{1}{x}}}$
First, We will start working on the Taylor series for $f(x) = {e^{ - \dfrac{1}{x}}}$
Let’s have a look into the Taylor formula which we are going to use,
${T_n}(x) = \sum\limits_{i = 0}^n {\dfrac{{{f^{(i)}}(a)}}{{i!}}{{(x - a)}^i}}$
If we closely look the formula, we conclude that each term in the formula requires:
1. A derivative of order (i). The $i$ above the f looks like an exponent but it is not. It is actually representing the order of the derivative.
2. Putting the value of $a$ in the derivatives.
3. Then Dividing every term by the factorial number $i!$
4. Finally Multiplying by $(x - a)$ raised to the power $i$ .
According to our question, we have not given the centred point around which we can expand our series, so we will expand our Taylor series around $x = a = 0$ .So, this is a Maclaurin Series.
Our formula becomes,

$$f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(0)}}{{n!}}{{(x - 0)}^i}} \\\ f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{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, (i.e. the Taylor series pivoted about $x = 0$ ).
So in such cases we can form Taylor series about some another pivot point , and in our case we will be taking this another pivot point as $x = 1$
Now our formula for series becomes.
${e^{ - \dfrac{1}{x}}} = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(1)}}{{n!}}{{(x - 1)}^n}}$-------------------(1)
This means, we have to work for the nth derivative as there is no range to the degree is given in question
Let’s find out some of the derivative up to order n
$f(x) = {e^{ - \dfrac{1}{x}}} \Rightarrow f(1) = {e^{ - \dfrac{1}{1}}} = {e^{ - 1}} = \dfrac{1}{e}$
Let’s find out the derivatives with respect to x , using the property of chain rule in derivative i.e. $\dfrac{d}{{dx}}({e^X}) = {e^X}.\dfrac{d}{{dx}}\left( X \right)$ .
1st derivative:
$f'(x) = {f^1}(x) = \dfrac{{{e^{ - \dfrac{1}{x}}}}}{{{x^2}}}\,\,\,\,\, \\\ \Rightarrow {f^1}(1) = \dfrac{{{e^{ - 1}}}}{1} = \dfrac{1}{e} \\\$
2nd derivative:

$$f''(x) = {f^2}(x) = \dfrac{{\left( {{x^2}} \right)\left( {\dfrac{{{e^{ - \dfrac{1}{x}}}}}{{{x^2}}}} \right) - \left( {{e^{ - \dfrac{1}{x}}}} \right)\left( {2x} \right)}}{{{{\left( {{x^2}} \right)}^2}}}\,\, = \dfrac{{\left( {{e^{ - \dfrac{1}{x}}}} \right)\left( {1 - 2x} \right)}}{{{x^4}}}\,\,\,\,\, \\\ \Rightarrow {f^2}(1) = - \dfrac{1}{e} \\\$$

Putting vales all these values together in our original formula i.e. equation (1.),we get

$${e^{ - \dfrac{1}{x}}} = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(1)}}{{n!}}{{(x - 1)}^n}} = \dfrac{1}{e} + \dfrac{1}{e}\left( {x - 1} \right) + \dfrac{{ - \dfrac{1}{e}}}{{2!}}{\left( {x - 1} \right)^2} + \dfrac{{{f^{\left( 3 \right)}}\left( a \right)}}{{3!}}{\left( {x - 1} \right)^3} + .... \\\ = \dfrac{1}{e} + \dfrac{1}{e}\left( {x - 1} \right) + \dfrac{{ - \dfrac{1}{e}}}{{2!}}{\left( {x - 1} \right)^2} + ....... \\\$$

**Therefore, the Taylor series for the ${e^{ - \dfrac{1}{x}}}$is equal to $\dfrac{1}{e} + \dfrac{1}{e}\left( {x - 1} \right) + \dfrac{{ - \dfrac{1}{e}}}{{2!}}{\left( {x - 1} \right)^2} + .......$.
**

Additional Information:
Factorial: The continued product of first n natural numbers is called the “n factorial “and denoted by $n!$.

Note:
1. Don’t forget to cross-check your answer at least once.
2. Meaning of Zero factorial is senseless to define it as the product of integers from 1 to zero. So, we define it as $0! = 1$ .
3. Factorials of any number can be calculated as $n! = n(n - 1)(n - 2)(n - 3).......(2)(1)$