Q. 43 - Calculus

Question

In Exercises 41–50, find Maclaurin series for the given pairs of functions, using these steps:

(a) Use substitution and/or multiplication and the appropriate Maclaurin series to find the Maclaurin series for the given function f .

(b) Use Theorem 8.12 and your answer from part (a) to find the Maclaurin series for the antiderivative $F=\int f$ that satisfies the specified initial condition

$\left(a\right) f\left(x\right)=e^{\raisebox{1ex}{$-x^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \left(b\right) F\left(0\right)=0$

Step-by-Step Solution

Answer

Part (a)    $e^{\raisebox{1ex}{$-x^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\sum _{k=0}^{\infty }\left(-\frac{1}{3}\right)\frac{1}{k!}{x}^{2k}$

Part (b)   $F\left(x\right)=\sum _{k=0}^{\infty }\left(-\frac{1}{3}\right)\frac{1}{k!}\left(\frac{x^{2k+1}}{2k+1}\right)$

Part (a) Step 1. Given information

Let us consider the given function $f\left(x\right)=e^{\raisebox{1ex}{$-x^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Part (a) Step 2. Use substitution and/or multiplication and the appropriate Maclaurin series to find the Maclaurin series for the given function f .

The maclaurin series for $g\left(x\right)=e^x$ is :

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

So,the maclaurin series for $e^{\raisebox{1ex}{$-x^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ can be founded by substituting $x$ by $-\raisebox{1ex}{$x^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

Thus,

$$\begin{gathered} e^{\raisebox{1ex}{$-x^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.e^x}=\sum _{k=0}^{\infty }\frac{1}{k!}{\left(\frac{-x^2}{3}\right)}^k \\ =\sum _{k=0}^{\infty }\left(-\frac{1}{3}\right)\frac{1}{k!}{x}^{2k} \end{gathered}$$

Part (b) Step 1. Given information

Let us consider the given function $F=\int f\left(x\right)$

Part (b) Step 2. Use Theorem 8.12 and your answer from part (a) to find the Maclaurin series for the antiderivative that satisfies the specified initial condition

Put the value of function$f\left(x\right)$

$$\begin{gathered} F\left(x\right)=\int \left(\sum _{k=0}^{\infty }\left(-\frac{1}{3}\right)\frac{1}{k!}{x}^{2k}\right)dx \\ =\sum _{k=0}^{\infty }\left(-\frac{1}{3}\right)\frac{1}{k!}\int x^{2k}dx \\ =\sum _{k=0}^{\infty }\left(-\frac{1}{3}\right)\frac{1}{k!}\left(\frac{x^{2k+1}}{2k+1}\right)+C \end{gathered}$$

Since,the initial condition is$F\left(0\right)=0$

This implies that:

 $C=0$

Therefore,

$F\left(x\right)=\sum _{k=0}^{\infty }\left(-\frac{1}{3}\right)\frac{1}{k!}\left(\frac{x^{2k+1}}{2k+1}\right)$