Question

. (a) Use the binomial series to expand \({\raise0.7ex\hbox{\(1\)} \!\mathord{\left/ {\vphantom {1 {\sqrt {1 + {x^2}} }}}\right.\\}\!\lower0.7ex\hbox{\({\sqrt {1 + {x^2}} }\)}}\)

(b) Use part (a) to find the Maclaurin series for \(si{n^{ - 1}}x\) Use the binomial series to expand \({\raise0.7ex\hbox{\(1\)} \!\mathord{\left/ {\vphantom {1 {\sqrt {1 + {x^2}} }}}\right.\\}\!\lower0.7ex\hbox{\({\sqrt {1 + {x^2}} }\)}}\)

Short answer

\(1 + \sum\limits_{n = 1}^\infty {\frac{{1 \cdot 3 \cdot 5 \cdot 7......(2n - 1)}}{{{2^n} \cdot n!}}} {x^{2n}}\)

Step-by-step solution

Step 1: Use binomial series

\(\frac{1}{{\sqrt {1 - x{}^2} }} = {(1 - {x^2})^{ - {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\\}\!\lower0.7ex\hbox{$2$}}}} = {(1 + ( - {x^2}))^{ - {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\\}\!\lower0.7ex\hbox{$2$}}}}\)

We can use the binomial series with \(k = - \frac{1}{2}\;\& \; - {x^2} \to x\). Plug the values in then write out some series terms. Look for the patterns to rewrite the summation without binomial thing.

\(\begin{array}{l}{(1 + x)^k} = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}}k\\n\end{array}} \right){x^n}} \\ = 1 + kx + \frac{{k(k - 1)}}{{2!}}{x^2} + \frac{{k(k - 1)(k - 2)}}{{3!}}{x^3} + ..........,R = 1\\{(1 + ( - x))^{{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\\}\!\lower0.7ex\hbox{$2$}}}} = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}}{{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\\}\!\lower0.7ex\hbox{$2$}}}\\n\end{array}} \right){{( - {x^2})}^n} = } \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}}{{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\\}\!\lower0.7ex\hbox{$2$}}}\\n\end{array}} \right){{( - 1)}^n}} {x^{2n}}\end{array}\)\(\begin{array}{l} = 1 + \frac{1}{2}{x^2} + \frac{{{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\\}\!\lower0.7ex\hbox{$2$}}\left( {{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\\}\!\lower0.7ex\hbox{$2$}} - 1} \right)}}{{2!}}{x^4} - \frac{{{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\\}\!\lower0.7ex\hbox{$2$}}\left( {{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\\}\!\lower0.7ex\hbox{$2$}} - 1} \right)\left( {{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\\}\!\lower0.7ex\hbox{$2$}} - 2} \right)}}{{3!}}{x^6} + \frac{{{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\\}\!\lower0.7ex\hbox{$2$}}\left( {{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\\}\!\lower0.7ex\hbox{$2$}} - 1} \right)\left( {{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\\}\!\lower0.7ex\hbox{$2$}} - 2} \right)\left( {{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\\}\!\lower0.7ex\hbox{$2$}} - 3} \right)}}{{4!}}{x^8} + ......\\ = 1 + \frac{1}{2}{x^2} + \frac{3}{{{2^2} \cdot 2!}}{x^4} + \frac{{5 \cdot 3}}{{{2^3} \cdot 3!}}{x^6} + \frac{{3 \cdot 5 \cdot 7}}{{{2^4} \cdot 4!}}{x^8} + .......\end{array}\)\( = 1 + \sum\limits_{n = 1}^\infty {\frac{{1 \cdot 3 \cdot 5 \cdot 7......(2n - 1)}}{{{2^n} \cdot n!}}} {x^{2n}}\)

Step 2: Explanation for the Final Answer \(1 + \sum\limits_{n = 1}^\infty  {\frac{{1 \cdot 3 \cdot 5 \cdot 7......(2n - 1)}}{{{2^n} \cdot n!}}} {x^{2n}}\)Where the \(1 \cdot 3 \cdot 5 \cdot 7......(2n - 1)\) part means if there are any numbers between \(1\;\& \;(2n - 1)\), multiply \(1 \times 3 \times 5 \times7......up\;to(2n - 1)\)Use part (a) to find Maclaurin series for \(si{n^{ - 1}}x\).

\({\sin ^1}x = x + \sum\limits_{n = 1}^\infty {\frac{{1 \cdot 3 \cdot 5 \cdot 7......(2n - 1)}}{{(2n - 1){2^n} \cdot n!}}} {x^{2n + 1}}\)

Step 1: Solution we have from part a

\(1 + \sum\limits_{n = 1}^\infty {\frac{{1 \cdot 3 \cdot 5 \cdot 7......(2n - 1)}}{{{2^n} \cdot n!}}} {x^{2n}}\)

Where the \(1 \cdot 3 \cdot 5 \cdot 7......(2n - 1)\) part means if there are any numbers between \(1\;\& \;(2n - 1)\), multiply \(1 \times 3 \times 5 \times 7......up\;to(2n - 1)\)

Step 2: Use of series for estimation

The derivative of \(si{n^{ - 1}}x\;is\;\frac{1}{{\sqrt {1 - {x^2}} }}\)so we can integrate the series from part (a) to find the series for \({\sin ^{ - 1}}x\)

\(\begin{array}{l}\int {\left( {1 + \sum\limits_{n = 1}^\infty {\frac{{1 \cdot 3 \cdot 5 \cdot 7......(2n - 1)}}{{{2^n} \cdot n!}}{x^{2n}}} } \right)} dx = C + x + \sum\limits_{n = 1}^\infty {\frac{{1 \cdot 3 \cdot 5 \cdot 7......(2n - 1)}}{{(2n + 1){2^n} \cdot n!}}{x^{2n + 1}}} \\x = 0\\{\sin ^{ - 1}}0 = C + (0) + \sum\limits_{n = 1}^\infty {\frac{{1 \cdot 3 \cdot 5 \cdot 7......(2n - 1)}}{{(2n + 1){2^n} \cdot n!}}{0^{2n + 1}}} \\0 = C + 0 + 0\\C = 0\end{array}\)

So the arcsin is

\({\sin ^1}x = x + \sum\limits_{n = 1}^\infty {\frac{{1 \cdot 3 \cdot 5 \cdot 7......(2n - 1)}}{{(2n - 1){2^n} \cdot n!}}} {x^{2n + 1}}\)