Question

Find the Maclaurin series expansion for the given expression, $\dfrac{1}{{(1 - {x^2})}}$

Answer 1

These kinds of questions are memory-based questions. Revive the correct formula of the Maclaurin series. Solve individually to avoid any mistakes in the later steps of assembling the expansion. Put the appropriate values in the expansion and get the correct answer.

Complete step by step solution:
The formula for the Taylor series expansion is,
$$$$$\sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(a)}}{{n!}}{{(x - a)}^n}} where a=0 for the Maclaurin series expansion. If we expand the above formula then it would look like, $$f(x) = f(0) + \dfrac{{f'(0)}}{{1!}}x + \dfrac{{f''(0)}}{{2!}}{x^2} + \dfrac{{f'''(0)}}{{3!}}{x^3} + \dfrac{{f''''(0)}}{{4!}}{x^4} + .... - - - (i)$$ \Rightarrow$$f\left( x \right){\text{ }} = \dfrac{1}{{(1 - {x^2})}}$$\Rightarrowf'\left( x \right){\text{ }} = \dfrac{{ - 2}}{{{{(1 + x)}^3}}}$$ $\Rightarrow f''\left( x \right){\text{ }} = \dfrac{6}{{{{(1 + x)}^4}}} $\Rightarrow$$$f'''\left( x \right){\text{ }} = \dfrac{{ - 24}}{{{{(1 + x)}^5}}}
$\Rightarrow$$$f''''\left( x \right){\text{ }} = \dfrac{{120}}{{{{(1 + x)}^6}}}$$
Putting the required values in place of x,

{f\left( 0 \right) = 0} \\\ \Rightarrow {f'\left( 0 \right) = - 2} \\\ \Rightarrow {f''\left( 0 \right) = 6} \\\ \Rightarrow {f'''\left( 0 \right) = - 24} \\\ \Rightarrow {f''''\left( 0 \right) = 120} \end{array}$$ Assembling all the values collected, in equation (i),

f\left( x \right) = 1 + \dfrac{{ - 2}}{1}x + \dfrac{6}{{2!}}{x^2} + \dfrac{{ - 24}}{{3!}}{x^3} + \dfrac{{120}}{{4!}}{x^4} + .... \\
\Rightarrow f\left( x \right) = 1 + \dfrac{{ - 2}}{1}x + \dfrac{6}{{\left( {2 \times 1} \right)}}{x^2} + \dfrac{{ - 24}}{{\left( {3 \times 2 \times 1} \right)}}{x^3} + \dfrac{{120}}{{\left( {4 \times 3 \times 2 \times 1} \right)}}{x^4} + .... \\
\Rightarrow f\left( x \right) = 1 + \dfrac{{ - 2}}{1}x + \dfrac{6}{2}{x^2} + \dfrac{{ - 24}}{6}{x^3} + \dfrac{{120}}{{24}}{x^4} + .... \\
\Rightarrow f\left( x \right) = 1 - 2x + 3{x^2} - 4{x^3} + 5{x^4} + .... \\

The Maclaurin series expansion of $\dfrac{1}{{(1 - {x^2})}}$ is $$1 - 2x + 3{x^2} - 4{x^3} + 5{x^4} + ....$$$$$$ **Note:** A Maclaurin series is a Taylor series expansion of a function about 0. Maclaurin series are a kind of series expansion during which all terms are nonnegative integer powers of the variable. The Taylor series of a function is an infinite sum of terms that are expressed in terms of the function's derivatives at one point. For many common functions, the function and also the sum of its Taylor series are equal near this time.