Q. 46 - 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  that satisfies the specified initial condition

$\left(a\right) f\left(x\right)= \ln (4+x^2) \left(b\right) F\left(0\right)=-2$

Step-by-Step Solution

Answer

Part (a)  $\ln (4+x^2)=\ln 4+\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k}{k.2^{2k}}.x^{2k}$

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

Part (a) Step 1. Given information

Let us consider the given function $f\left(x\right)=\ln (4+x^2)$

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 $g\left(x\right)=\ln (1+x)$ is :

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

So,the maclaurin series for $f\left(x\right)=\ln (4+x^2)$ Let us write the function as

$\ln (4+x^2)=\ln \left\{4\left[1+{\left(\frac{x}{2}\right)}^2\right]\right\}$

since, $\ln (a.b)=\ln a+\ln b$

Therefore,,

$\ln (4+x^2)=\ln 4+\ln \left[1+{\left(\frac{x}{2}\right)}^2\right]$

Now substitute x by  ${\left(\frac{x}{2}\right)}^2$in the maclaurin series of $g\left(x\right)=\ln (1+x)$ to find the maclaurin series of $\ln \left[1+{\left(\frac{x}{2}\right)}^2\right]$

Thus,

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

Inmplies that 

$\ln (4+x^2)=\ln 4+\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k}{k.2^{2k}}.x^{2k}$

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$f\left(x\right)$

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

Since,the initial condition is $F\left(0\right)=-2$

This implies that:

 $C=-2$

Therefore,

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