Q. 65 - 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 

$\frac{\sin x}{1-x^2}$

Step-by-Step Solution

Answer

The first four nonzero terms in the Maclaurin series for the function $f\left(x\right)=\frac{\sin x}{1-x^2}$ are as follows:

$x+\frac{5}{6}{x}^3+\frac{101}{120}{x}^5+\frac{4241}{5040}{x}^7$

The interval of convergence for the Maclaurin series of the given function is, $(-1,1)$.

Step 1. Given Information

Consider the function as follows:

$f\left(x\right)=\frac{\sin x}{1-x^2}$

Step 2. To find the first four nonzero terms of the Maclaurin series

To find the first four nonzero terms of the Maclaurin series for the product of functions mentioned in the above function and also the interval of convergence.

The Maclaurin series for the function $\sin x$ is,

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

Expand the above series in the following way:

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

The Maclaurin series for the function $\frac{1}{1-x}$ is

$\frac{1}{1-x}=\sum _{k=0}^{\infty }{x}^k$

Expand the above series in the following way:

$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$

So the Maclaurin series for the function $\frac{1}{1-x^2}$ is

$\frac{1}{1-x^2}=\sum _{k=0}^{\infty }{\left(x^2\right)}^k$

Step 3: Find the interval of convergence

Expand the above series in the following way:

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

Multiply the preceding two series together term by term to get first four nonzero terms in the Maclaurin series for the function $f\left(x\right)=\frac{\sin x}{1-x^2}$.

There will be no constant term, since the series for$\sin x$does not contains any constant terms, so after multiplying the series for $\sin x$ and $\frac{1}{1-x^2}$. we get the series having the smallest degree of $x$ is $1$.

Therefore, the coefficient of $x$ term is,

$1·1=1$

The coefficient of $x^3$ term is,

$$\begin{gathered} 1·1-\frac{1}{3!}·1=1-\frac{1}{6} \\ =\frac{5}{6} \end{gathered}$$

Also, the coefficient of $x^5$ term is,

$$\begin{gathered} 1·1-\frac{1}{3!}·1+\frac{1}{5!}·1=1-\frac{1}{6}+\frac{1}{120} \\ =\frac{120-20+1}{120} \\ =\frac{101}{120} \end{gathered}$$

The coefficient of $x^7$ term is,

$$\begin{gathered} 1·1-\frac{1}{3!}·1+\frac{1}{5!}·1-\frac{1}{7!}·1=1-\frac{1}{6}+\frac{1}{120}-\frac{1}{5040} \\ =\frac{4241}{5040} \end{gathered}$$

Therefore, the first four nonzero terms in the Maclaurin series for the function $f\left(x\right)=\frac{\sin x}{1-x^2}$ are as follows:

$x+\frac{5}{6}{x}^3+\frac{101}{120}{x}^5+\frac{4241}{5040}{x}^7$

The interval of convergence for the Maclaurin series of the given function is, $(-1,1)$.