Q. 47 - 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

$\left(a\right) f\left(x\right)=x^2\sin \left(5x^2\right) \left(b\right) F\left(0\right)=1$

Step-by-Step Solution

Answer

Part (a)  $x^2\sin \left(5x^2\right)=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{5}^{2k+1}{x}^{4k+4}$

Part (b)  $F\left(x\right)=\sum _{k=0}^{\infty } \frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{5}^{2k+1}\frac{x^{4k+5}}{4k+5}+1$

Part (a) Step 1. Given information

Let us consider the given function $f\left(x\right)=x^2\sin \left(5x^2\right)$

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 $g\left(x\right)=\sin x$ is :

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

So,the maclaurin series for $\sin \left(5x^2\right)$ can be founded by substituting $x$ by $5x^2$

Thus,

$$\begin{gathered} \sin \left(5x^2\right)=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{\left(5x^2\right)}^{2k+1} \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}.5^{2k+1}{\left(x\right)}^{4k+2} \end{gathered}$$

Now the maclaurin series for $x^2\sin \left(5x^2\right)$ is:

$$\begin{gathered} x^{2 }\sin \left(5x^2\right)=\underset{k=0}{\overset{\infty }{x^{2 }\sum }}\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{5}^{2k+1}{\left(x\right)}^{4k+2} \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{5}^{2k+1}{x}^{4k+4} \end{gathered}$$

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 F = ∫ f that satisfies the specified initial condition

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

$$\begin{gathered} F\left(x\right)=\int \left(\sum _{k=0}^{\infty } \frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{5}^{2k+1}{x}^{4k+4}\right)dx \\ =\sum _{k=0}^{\infty } \frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{5}^{2k+1}\int x^{4k+4} dx \\ =\sum _{k=0}^{\infty } \frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{5}^{2k+1}\frac{x^{4k+5}}{4k+5}+C \end{gathered}$$

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

This implies that:

 $C=1$

Therefore,

$F\left(x\right)=\sum _{k=0}^{\infty } \frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{5}^{2k+1}\frac{x^{4k+5}}{4k+5}+1$