Q. 42 - 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

$\left(a\right) f\left(x\right)=x \cos {\left(x\right)}^3 \left(b\right) F\left(0\right)=-1$

Step-by-Step Solution

Answer

Part (a) $x \cos {\left(x\right)}^3=x.\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}{x}^{6k}$

Part (b) $F\left(x\right)=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}\left(\frac{x^{6k+7}}{6k+7}\right)-1$

Part (a) Step 1. Given information

Let us consider the given function$f\left(x\right)=x \cos {\left(x\right)}^3$

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 $g\left(x\right)=\cos x$ is :

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

So,the maclaurin series for $\cos {\left(x\right)}^3$ can be founded by substituting $x$ by $x^3$

Thus,

$$\begin{gathered} \cos {\left(x\right)}^3=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}{\left(x^3\right)}^{2k} \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}{x}^{6k} \end{gathered}$$

So,the maclaurin series for $x \cos {\left(x\right)}^3$ is:

$x \cos {\left(x\right)}^3=x.\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}{x}^{6k}$

Part (b) Step 1. Given information

Let us consider the given function $F=\int f$

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

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

$$\begin{gathered} F\left(x\right)=\int \sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}{x}^{6k+1}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}\int x^{6k+1}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}\left(\frac{x^{6k+1+1}}{6k+1+1}\right)+C \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}\left(\frac{x^{6k+7}}{6k+7}\right)+C \end{gathered}$$

Since,the initial condition is$F\left(0\right)=-1$

This implies that:

 $C=-1$

Therefore,

$F\left(x\right)=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}\left(\frac{x^{6k+7}}{6k+7}\right)-1$