Question

Expansion of sin inverse x by maclaurin's series .

Please write full solution not only the answer because i already have it.

Answer 1

sin inverse X Mclaren theorem

Answer 2

Given,

$\[f\left( x \right) = {\sin ^{ - 1}}x\]$

To find,

Maclaurin's series of $\[f\left( x \right).\]$

Solution,

Know that a Maclaurin's series is expressed as follows

$\[f\left( x \right) = f\left( 0 \right) + {f^'}\left( 0 \right)x + \frac{{{f^{''}}\left( 0 \right)}}{{2!}}{x^2} + \frac{{{f^{\left( 3 \right)}}\left( 0 \right)}}{{3!}}{x^3} - - - - - + \frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n} + - - - \]$

Therefore,

$\[\begin{array}{l} \Rightarrow f\left( x \right) = {\sin ^{ - 1}}\left( 0 \right) + \frac{1}{{\sqrt {1 - {x^2}} }}\frac{{{x^2}}}{{2!}} + \frac{x}{{{{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}}}\frac{{{x^3}}}{{3!}} + - - - \\ \Rightarrow f\left( x \right) = x + \frac{{{x^3}}}{6}\end{array}\]$

Hence, the expansion of the series results as $\[x + \frac{{{x^3}}}{6}.\]$