Question

(a) Expand \({\raise0.7ex\hbox{\(1\)} \!\mathord{\left/ {\vphantom {1 {\sqrt[4]{{1 + x}}}}}\right.\\}\!\lower0.7ex\hbox{\({\sqrt[4]{{1 + x}}}\)}}\)as a power series.

(b) Use part (a) to estimate \({\raise0.7ex\hbox{\(1\)} \!\mathord{\left/ {\vphantom {1 {\sqrt[4]{{1.1}}}}}\right.\\}\!\lower0.7ex\hbox{\({\sqrt[4]{{1.1}}}\)}}\) correct to three decimal places.

(c) Expand \({\raise0.7ex\hbox{\(1\)} \!\mathord{\left/ {\vphantom {1 {\sqrt[4]{{1 + x}}}}}\right.\\}\!\lower0.7ex\hbox{\({\sqrt[4]{{1 + x}}}\)}}\)as a power series.

Short answer

\(1 + \sum\limits_{n = 1}^\infty {{{( - 1)}^n}\frac{{1 \cdot 5 \cdot 9 \cdot 13......(4n - 3)}}{{{4^n} \cdot n!}}} {x^n}\)

Step-by-step solution

Step 1: Use binomial series

We can use the binomial series with \(k = - \frac{1}{4}\). Plug the values in then write out some series terms. Look for the patterns to rewrite the summation without binomial thing.

\(\begin{array}{l}{(1 + x)^k} = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}}k\\n\end{array}} \right){x^n}} \\ = 1 + kx + \frac{{k(k - 1)}}{{2!}}{x^2} + \frac{{k(k - 1)(k - 2)}}{{3!}}{x^3} + ..........,R = 1\\{(1 + x)^{{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 4}}\right.\\}\!\lower0.7ex\hbox{$4$}}}} = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}}{{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 4}}\right.\\}\!\lower0.7ex\hbox{$4$}}}\\n\end{array}} \right){x^n}} \end{array}\)\(\begin{array}{l} = 1 - \frac{1}{4}x + \frac{{{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 4}}\right.\\}\!\lower0.7ex\hbox{$4$}}\left( {{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 4}}\right.\\}\!\lower0.7ex\hbox{$4$}} - 1} \right)}}{{2!}}{x^2} + \frac{{{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 4}}\right.\\}\!\lower0.7ex\hbox{$4$}}\left( {{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 4}}\right.\\}\!\lower0.7ex\hbox{$4$}} - 1} \right)\left( {{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 4}}\right.\\}\!\lower0.7ex\hbox{$4$}} - 2} \right)}}{{3!}}{x^3} + \frac{{{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 4}}\right.\\}\!\lower0.7ex\hbox{$4$}}\left( {{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 4}}\right.\\}\!\lower0.7ex\hbox{$4$}} - 1} \right)\left( {{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 4}}\right.\\}\!\lower0.7ex\hbox{$4$}} - 2} \right)\left( {{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 4}}\right.\\}\!\lower0.7ex\hbox{$4$}} - 3} \right)}}{{4!}}{x^4} + ......\\ = 1 - \frac{1}{4}x + \frac{5}{{{4^2} \cdot 2!}}{x^2} - \frac{{5 \cdot 9}}{{{4^3} \cdot 3!}}{x^3} + \frac{{5 \cdot 9 \cdot 13}}{{{4^4} \cdot 4!}}{x^4} + .......\end{array}\)\( = 1 + \sum\limits_{n = 1}^\infty {{{( - 1)}^n}\frac{{1 \cdot 5 \cdot 9 \cdot 13......(4n - 3)}}{{{4^n} \cdot n!}}} {x^n}\)

Step 2: Convergence\(1 + \sum\limits_{n = 1}^\infty  {{{( - 1)}^n}\frac{{1 \cdot 5 \cdot 9 \cdot 13......(4n - 3)}}{{{4^n} \cdot n!}}} {x^n}\)Where the \(1 \cdot 5 \cdot 9 \cdot 13......(4n - 3)\) part means if there are any numbers between \(1\;\& \;(4n - 3)\), multiply \(1 \cdot 5 \cdot 9 \cdot13......up\;to(4n - 3)\)This converges when \(\left| x \right| < 1\)(b) Use part (a) to estimate \({\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 {\sqrt[4]{{1.1}}}}}\right.\\}\!\lower0.7ex\hbox{${\sqrt[4]{{1.1}}}$}}\)  correct to three decimal places.

\({\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1{\sqrt[4]{{1.1}}}}}\right.\\}\!\lower0.7ex\hbox{${\sqrt[4]{{1.1}}}$}} = 0.976\)

Step 1: Solution we have from part a

\(1 + \sum\limits_{n = 1}^\infty {{{( - 1)}^n}\frac{{1 \cdot 5 \cdot 9 \cdot 13......(4n - 3)}}{{{4^n} \cdot n!}}} {x^n}\)

Where the \(1 \cdot 5 \cdot 9 \cdot 13......(4n - 3)\) part means if there are any numbers between \(1\;\& \;(4n - 3)\), multiply \(1 \cdot 5 \cdot 9 \cdot 13......up\;to(4n - 3)\)

This converges when \(\left| x \right| < 1\)

Step 2: Use of series for estimation

To estimate \(\frac{1}{{\sqrt[4]{{1.1}}}} = \frac{1}{{\sqrt[4]{{1 + 0.1}}}}\)we use the series with \(x = 0.1\)

It is an alternating series, so we look for the first term that is \(\left| {{a_n}} \right| < 0.001\)

Then we can use the partial sum up to but not including that term.

\(n\)	\({a_n}\)
\(1\)	\( - 0.025\)
\(2\)	\(0.0015625\)
\(3\)	\( - 0.0001171875\)

So we only need up to\(n = 2\)

\(\frac{1}{{\sqrt[4]{{1.1}}}} \approx 1 + ( - 0.025) + (0.0015625) \approx 0.976\)