Question

How do you find the Taylor series of $f\left( x \right) = \dfrac{1}{x}$?

Answer 1

In the given question, we have been given a function. This function is the inverse function, i.e., whatever value we put as the argument into the function, we get the reciprocal of the argument as the answer. We have to find the Taylor series of this reciprocal function. To solve this question, we write the general formula of the Taylor series, write the power series expansion formula for the function near to the given function, replace the value in the formula from so as to get to the function in the given question, and then just substitute the value to solve the answer.

Formula Used:
The formula of Taylor series for a function $f\left( x \right)$ about $x = a$ is,
$f\left( x \right) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}{{\left( {x - a} \right)}^n}}$

Complete step by step answer:
We know the standard Taylor series formula for $\dfrac{1}{{1 - x}}$, which is
$\dfrac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {{x^n}}$
Now, to get to the given question, we substitute the value of this Taylor series formula from $x$ to $1 - x$,
$\dfrac{1}{x} = \sum\limits_{n = 0}^\infty {{{\left( {1 - x} \right)}^n}}$
or $\dfrac{1}{x} = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\left( {x - 1} \right)}^n}}$

Note:
In the given question, we had to find the Taylor series of the reciprocal function. We write down the known and the unknown so that we have in front of us what we want to find and hence, we can steer in that direction. To solve this question, we wrote the formula for the series, wrote the power series expansion formula for the function near to the given function, replaced the value in the formula from $x$ to $\left( {1 - x} \right)$ so as to get to the function in the given question, and then just substituted the value to solve the answer.