Question

Question: 55-58 = Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

57. \(y = \frac{x}{{sinx}}\)

Short answer

The first three nonzero terms is \(1 + \frac{{{x^2}}}{6} + \frac{{7{x^4}}}{{360}}\)

Step-by-step solution

Step 1: To Find the Maclaurin series

\(y = \frac{x}{{\sin x}}\)

Get the series for \(\sin x\) from the table (or derive it again yourself if you like).

\(\sin x{\rm{ }} = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{2n + 1}}}}{{(2n + 1)!}} = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \frac{{{x^7}}}{{7!}} + \cdots \)

\( = x - \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{120}} - \frac{{{x^7}}}{{5040}} + \cdots \)

Step 2: Divide the first three terms

Divide \(x{\rm{ }}\)by the series for sine. Using the first \(3{\rm{ }}\)terms should be enough. (click on the image if the automatic stretching is making it too big)

\(\begin{array}{l}1 - \frac{{{x^3}}}{6} +\frac{{{x^5}}}{{120}}\mathop{\left){\vphantom{1\begin{array}{l}x\\\underline {x -\frac{{{x^3}}}{6} + \frac{{{x^5}}}{{120}}} \end{array}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}x\\\underline {x - \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{120}}} \end{array}}}}\limits^{\displaystyle\,\,\, {1 + \frac{{{x^2}}}{6} + \frac{{7{x^4}}}{{360}}}}\\\begin{array}{*{20}{c}}{}&{}&{\begin{array}{*{20}{c}}{}&{}&{\begin{array}{*{20}{c}}{}&\begin{array}{l}\frac{{{x^3}}}{6} - \frac{{{x^5}}}{{120}}\\\underline {\frac{{{x^3}}}{6} - \frac{{{x^5}}}{{36}} + ...} \\\begin{array}{*{20}{c}}{}&{}\end{array}\frac{{7{x^5}}}{{360}}...\\\begin{array}{*{20}{c}}{}&{}\end{array}\frac{{7{x^5}}}{{360}}...\end{array}\end{array}}\end{array}}\end{array}\end{array}\)

Step 3: Final Proof

Therefore, the first three nonzero terms of the function

\(\begin{array}{l}\frac{x}{{\sin x}} \approx \frac{x}{{x - \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{120}}}}\\ \approx 1 + \frac{{{x^2}}}{6} + \frac{{7{x^4}}}{{360}}\end{array}\)