Q.31 - Calculus

Question

Find Maclaurin series for the given pairs of functions, using these steps:

(a) Use substitution in the appropriate Maclaurin series to find the Maclaurin series for the given function.

(b) Use Theorem $8.11$ and your answer from part (a) to find the Maclaurin series for the given function.

(c) Find the Maclaurin series for the function in (b), using multiplication and substitution with the appropriate Maclaurin series. Compare your answers from (b) and (c).

 (a) $\sin \left(x^2\right)$

(b) $x\cos \left(x^2\right)$

Step-by-Step Solution

Answer

Part(a)The Maclaurin series  of $\sin \left(x^2\right)$ is $\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{4k+2}$

Part(b)The Maclaurin series of $x\cos \left(x^2\right)$ is $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{4k+1}\end{array}$.

Part(c) The answer of the function using both theorem is same

Part (a) Step 1:The given information.

The functions are  $\sin \left(x^2\right)$ and $x\cos \left(x^2\right)$.

Part (a) Step 2:The Maclaurin series of given functions

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

Substitute the value of $x$ by $x^2$ ,we get the Maclaurin series of $\sin \left(x^3\right)$ will be:

$$\begin{gathered} \sin \left(x^2\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^2}{(2k+1)!}{\left(x^2\right)}^{2k+1} \\ =\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{4k+2} \end{gathered}$$

So,the Maclaurin series is  $\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{4k+2}$

Step 3:the Maclaurin series from part a

The derivate of above function is 

$\begin{array}{rcl}{f}^{\text{'}}\left(x\right)& =& \frac{d}{dx}\left(\sin \left(x^2\right)\right)\\ & =& 2x\cos \left(x^2\right)\end{array}$

In this way the Maclaurin series of above function is:

$\begin{array}{rcl}2x\cos \left(x^2\right)& =& \frac{d}{dx}\left(\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{4+2}\right)\\ & =& \sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}\frac{d}{dx}(x{)}^{4k+2}\\ & =& \sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}(4k+2)x^{4k+2-1}\end{array}$

So,the Maclaurin series of $\begin{array}{rcl}& & x\cos \left(x^2\right)\end{array}$ is:

$\begin{array}{rcl}x\cos \left(x^2\right)& =& \frac{1}{2}\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}2(2k+1)x^{4k+1}\\ & =& \sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{4k+1}\end{array}$

From part $\left(a\right)$ , the Maclaurin series of $x\cos \left(x^2\right)$ is  $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{4k+1}\end{array}$

Part (b) Step 4: Use of multiplication and substitution with the appropriate Maclaurin series

The Maclaurin series of $g\left(x\right)=\cos x$ is $\cos x=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{2k}$

Substitute the value of $x$ by $x^2$ ,we can find the Maclaurin series of $\cos \left(x^2\right)$ is 

$$\begin{gathered} \cos \left(x^2\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{\left(x^2\right)}^{2k} \\ =\sum _{i=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{4k} \\ =\sum _{i=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{4k+1} \end{gathered}$$

The Maclaurin series of $x\cos \left(x^2\right)$ is $\sum _{i=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{4k+1}$.So, both the value from $\left(b\right)$ and $\left(c\right)$ is same.