Q. 45 - 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

$\left(a\right) f\left(x\right)=\frac{1}{1+x^3} \left(b\right) F\left(0\right)=-5$

Step-by-Step Solution

Answer

Part (a)  $\frac{1}{1+x^3} =\underset{k=0}{\overset{\infty }{\sum {(-1)}^k}}{x}^{3k}$

Part (b)  $F\left(x\right)=\sum _{k=0}^{\infty }{\left(-1\right)}^k\left(\frac{x^{3k+1}}{3k+1}\right)-5$

Part (a) Step 1. Given information

Let us consider the given function $f\left(x\right)=\frac{1}{1+x^2}$

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 $g\left(x\right)=\frac{1}{1-x}$ is :

$\frac{1}{1-x}=\sum _{k=0}^{\infty }{x}^4$

So,the maclaurin series for $f\left(x\right)=\frac{1}{1+x^3}$ can be founded by substituting $x$ by $x^3$

Thus,

$$\begin{gathered} \frac{1}{1+x^3}=\sum _{k=0}^{\infty }{\left(-x^3\right)}^k \\ =\underset{k=0}{\overset{\infty }{\sum {(-1)}^k}}{x}^{3k} \end{gathered}$$

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 F = ∫ f antiderivative that satisfies the specified initial condition

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

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

Since,the initial condition is $F\left(0\right)=-5$

This implies that:

 $C=-5$

Therefore,

$F\left(x\right)=\sum _{k=0}^{\infty }{\left(-1\right)}^k\left(\frac{x^{3k+1}}{3k+1}\right)-5$