Chapter 8.6, Problem 20E

To determine

To find: a power series representation for f , and graph f and several partial sums $S_n(x)$ on the same screen and what happens as n increases.

Answer to Problem 20E

  $f(x)=\ln 4+{\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}{x}^{2n}}{2^{2n}.n}}$ .

Explanation of Solution

Given information: Given function $f(x)=\ln (x^2+4).$

Formula used:

  $\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}\mathrm{.....}={\displaystyle \sum _{n=0}^{\infty }{(-1)}^{n-1}}\frac{x^n}{n}$ .

Calculation:

  $\begin{array}{c}f(x)=\ln (x^2+4).\\ f(x)=\ln (4+x^2)=\ln [4(1+\frac{x^2}{4})]=\ln 4+\ln ((1+\frac{x^2}{4}).\\ $ \text{Now}\text{let}\text{X}=\frac{x^2}{4}={(\frac{x}{2})}^2.$f(x)=\ln 4+\ln (1+X)\\ f(x)=\ln 4+{\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}{X}^n}{n}}\\ =\ln 4+{\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}{({(\frac{x}{2})}^2)}^n}{n}}\\ =\ln 4+{\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}{x}^{2n}}{2^{2n}.n}}\\ f(x)=\ln 4+{\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}{x}^{2n}}{2^{2n}.n}}.\end{array}$

Below is the graph for first six partial sums and the function f(x) using graphing utility.

  

As n increases the approximation becomes worse when, get further away from x =0.