Question

What is the Taylor series expansion of $f\left( x \right) = \dfrac{1}{{{x^2}}}$ at $a = 1$?

Answer 1

Hint: Here in this question, we have to find the Taylor’s series expansion of a given function. By using the formula of Taylor's series $f(x) = \dfrac{{{f^0}(a)}}{{0!}}{(x - a)^0} + \dfrac{{f'(a)}}{{1!}}{(x - a)^1} + \dfrac{{f''(a)}}{{2!}}{(x - a)^2} + \dfrac{{f'''(a)}}{{3!}}{(x - a)^3} + \dfrac{{f''''(a)}}{{4!}}{(x - a)^4}...$ Substituting the value of a to the formula then simplifies by using the standard differentiated formula to get the required solution.

Complete step by step answer:
We have to determine the nth Taylor’s series by using the Taylor’s series expansion. Here the function is $f\left( x \right) = \dfrac{1}{{{x^2}}}$. The formula for the Taylor’s series expansion is given by $f(x) = f(a) + f'(a)(x - a) + \dfrac{{f''(a)}}{{2!}}{(x - a)^2} + \dfrac{{f'''(a)}}{{3!}}{(x - a)^3} + ...$ the value of a is 1. The general form of Taylor’s series expansion is written as,
$f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(a)}}{{n!}}} {(x - a)^n}$

Now consider the function $f\left( x \right) = \dfrac{1}{{{x^2}}}$. Let we find the derivatives of the function
${f^0}\left( x \right) = f\left( x \right) = \dfrac{1}{{{x^2}}} = {x^{ - 2}}$
The first derivative of the function is $f'(x) = - \dfrac{2}{{{x^3}}}$, The first derivative can be written as $f'(x) = - 2{x^{ - 3}}$
The second derivative of the function is $f''(x) = \dfrac{6}{{{x^4}}} = 6{x^{ - 4}}$,
The third derivative of the function is $f'''(x) = - \dfrac{{24}}{{{x^5}}} = - 24{x^{ - 5}}$,
The fourth derivative of the function is $f''''(x) = \dfrac{{120}}{{{x^6}}} = 120{x^{ - 6}}$,
Likewise the ${n^{th}}$ derivative is given by
${f^{(n)}}(x) = {( - 1)^{n - 1}}(n - 1)!{x^{ - \left( {n + 2} \right)}}$

Let us consider the Taylor series
$f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(a)}}{{n!}}} {(x - a)^n} = \dfrac{{{f^0}(a)}}{{0!}}{(x - a)^0} + \dfrac{{f'(a)}}{{1!}}{(x - a)^1} + \dfrac{{f''(a)}}{{2!}}{(x - a)^2} + \dfrac{{f'''(a)}}{{3!}}{(x - a)^3} + \dfrac{{f''''(a)}}{{4!}}{(x - a)^4}...$
                                                                                                                                                          -------(1)
$a$ is a constant we choose. Given $a = 1$ for simplicity. So, $x \to a$, except for the term ${\left( {x - a} \right)^n}$, which remains as ${\left( {x - 1} \right)^n}$.
The Taylor series or equation (1) becomes
$f(1) = \dfrac{{{f^0}(1)}}{{0!}}{(x - 1)^0} + \dfrac{{f'(1)}}{{1!}}{(x - 1)^1} + \dfrac{{f''(1)}}{{2!}}{(x - 1)^2} + \dfrac{{f'''(1)}}{{3!}}{(x - 1)^3} + \dfrac{{f''''(a)}}{{4!}}{(x - 1)^4}...$

Now substitute the values of derivatives.
$f(1) = \dfrac{{\dfrac{1}{{{1^2}}}}}{1}(1) + \dfrac{{\left( { - \dfrac{2}{{{1^3}}}} \right)}}{1}(x - 1) + \dfrac{{\left( {\dfrac{6}{{{1^4}}}} \right)}}{2}{(x - 1)^2} + \dfrac{{\left( { - \dfrac{{24}}{{{1^5}}}} \right)}}{6}{(x - 1)^3} + \dfrac{{\left( {\dfrac{{120}}{{{1^6}}}} \right)}}{{24}}{(x - 1)^4}...$
On simplification, we get
$f(1) = 1 + \left( { - 2} \right)(x - 1) + \dfrac{6}{2}{(x - 1)^2} + \dfrac{{ - 24}}{6}{(x - 1)^3} + \dfrac{{120}}{{24}}{(x - 1)^4}...$
$\therefore \,\,\,f(1) = 1 - 2(x - 1) + 3{(x - 1)^2} - 4{(x - 1)^3} + 5{(x - 1)^4}...$

Hence, the Taylor series expansion of $f\left( x \right) = \dfrac{1}{{{x^2}}}$ at $a = 1$ is $1 - 2(x - 1) + 3{(x - 1)^2} - 4{(x - 1)^3} + 5{(x - 1)^4}...$

Note: In Taylor’s series expansion we have the derivatives, so we have to apply the differentiation to the function. If the function is a logarithmic function then we use standard differentiation formula and then we determine the solution. While solving the above function we use simple arithmetic operations.