Q. 61 - 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.

$e^x\sin x$

Step-by-Step Solution

Answer

The Maclaurin series is $x+x^2+\frac{1}{3}{x}^3-\frac{1}{30}{x}^5$

Step 1, Given Information

Consider the function $e^x\sin x$

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

The Maclaurin series for $e^x$is $e^x=\sum _{k=0}^{\infty }\frac{x^k}{k!}$

Let us expand the above series in the following way, 

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

The Maclaurin series for $\sin x$ is $\sin x=\sum _{k=0}^{\infty }\frac{{(-1)}^k}{(2k+1)!}{x}^{2k+1}$

Let us expand the above series in the following way,

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$

Step 3 : Calculation

Multiply the preceding two series together term by term to get the first four nonzero terms in the Maclaurin series of the given function $f\left(x\right)=e^x\sin x$

There will be no constant terms as terms in the series of $\sin x$ does not contain any constant term, so after multiplication the smallest degree of $x$ is $1$

The coefficient for the term $x$ is $1$

The coefficient for the term $x^2$ is $1$

The coefficient for the term $x^3$ is $1(-\frac{1}{3!})+\frac{1}{2}=-\frac{1}{6}+\frac{1}{2}=\frac{-1+3}{6}=\frac{1}{3}$

The coefficient fot the term $x^4$ is $1(-\frac{1}{3!})+\frac{1}{3!}=-\frac{1}{3!}+\frac{1}{3!}=0$

Step 4: Simplify to get final result

The coefficient for the term $x^5$ is $\frac{1}{5!}+\frac{1}{2}(-\frac{1}{3!})+\frac{1}{4!}=\frac{1}{120}-\frac{1}{12}+\frac{1}{24}=-\frac{1}{30}$

Thus, the first four nonzero terms i the Maclaurin series of the function $f\left(x\right)=e^x\sin x$ are as follows,

$x+x^2+\frac{1}{3}{x}^3-\frac{1}{30}{x}^5$

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