Q. 63 - Calculus

Question

Use appropriate Maclaurin series to find the first four nonzero

terms in the Maclaurin series for the product functions in

Exercises 61–66. Also, give the interval of convergence for the

series.

$e^x \ln (1+x)$

Step-by-Step Solution

Answer

The answer is $x^3+x^4+\frac{1}{2}{x}^5-\frac{1}{3}{x}^6$

Step 1. Given Information

Consider the function as follows:

$f\left(x\right)=e^x\ln (1+x^3)$

The objective is to find the first four nonzero terms of the Maclaurin series for the product of the functions mentioned above and also the interval of convergence. 

Step 2 : Find the interval of convergence for the series.

The Maclaurin series for the function $e^x$ is,

$e^x=\sum _{k=0}^{\infty }\frac{x^k}{k!}$

Let us expand the above series in the following way:

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$

The Maclaurin series for the function $\ln (1+x)$ is,

$\ln (1+x)=\sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}{x}^k$

So, the Maclaurin series for the function $\ln \left(1+x^3\right)$ is,

$\begin{array}{rcl}\ln \left(1+x^3\right)& =& \sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}{\left(x^3\right)}^k\\ & =& \sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}{x}^3\end{array}$

Expanding the above series in the following way

$\ln \left(1+x^3\right)=x^3-\frac{1}{2}{x}^6+\frac{1}{3}{x}^4-\cdots$

Step 3: Simpification

Multiply the preceding two series together term by term to get first four nonzero terms in the Maclaurin series for the function $f\left(x\right)=e^x\ln \left(1+x^3\right)$

There will be no constant term, since the series for $f\left(x\right)=e^x\ln \left(1+x^3\right)$ does not contains any constant terms, so after multiplying the series for $e^x$ and $\ln \left(1+x^3\right)$, the series having the smallest degree of $x$ is $3$ .

Therefore, the coefficient of $x^3$ term is,

$1·1=1$

The coefficient of $x^4$ term is,

$1.1=1$

The coefficient of $x^5$ term is,

$\frac{1}{2!}·1=\frac{1}{2}$

Step 4: Calculation

Also, the coefficient of $x^6$ term is,

$\begin{array}{rcl}\frac{1}{3!}·1+1·\left(-\frac{1}{2}\right)& =& \frac{1}{6}-\frac{1}{2}\\ & =& -\frac{1}{3}\end{array}$

Therefore, the first four nonzero terms in the Maclaurin series for the function $f\left(x\right)=e^x\ln \left(1+x^3\right)$ are as follows:

$x^3+x^4+\frac{1}{2}{x}^5-\frac{1}{3}{x}^6$

The interval of convergence for the Maclaurin series of the given function is, $\mathrm{ℝ}.$