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.

58. \(y = {e^x}ln(1 + x)\)

Short answer

The First three nonzero terms is \(x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}\)

Step-by-step solution

Step 1: The equation of the function

\(y = {e^x}\ln (1 + x)\)

Get the series for \({e^x}\) and for \(\ln (1 + x)\) from the table (or derive it again yourself if you like).

\({e^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} = 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \cdots = 1 + x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} + \cdots {\rm{ }}\)\(\ln (1 + x) = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}} \frac{{{x^n}}}{n} = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + \cdots \)

Step 2: Multiply two series

Multiply the two series together.

\({e^x}\ln (1 + x) \approx {\rm{ }}\left( {1 + x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right)\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4}} \right)\)\( \approx x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + {x^2} - \frac{{{x^3}}}{2} + \frac{{{x^4}}}{3} - \frac{{{x^5}}}{4}\)\( + \frac{{{x^3}}}{2} - \frac{{{x^4}}}{4} + \frac{{{x^5}}}{6} - \frac{{{x^6}}}{8} + \frac{{{x^4}}}{6} - \frac{{{x^5}}}{{12}} + \frac{{{x^6}}}{{18}} - \frac{{{x^7}}}{{24}}\)

Step 3: Final proof

We have lowest exponents of 1,2,3, so we can drop the terms with higher exponents.

\({e^x}\ln (1 + x){\rm{ }} \approx x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} + {x^2} - \frac{{{x^3}}}{2} + \frac{{{x^3}}}{2}\)\( \approx x - \frac{{{x^2}}}{2} + \frac{{2{x^2}}}{2} + \frac{{{x^3}}}{3}\)\( \approx x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}\)