Q. 31 - Calculus

Question

Use the Maclaurin series for $\frac{1}{1-x}$ to find power series representations for $\frac{1}{1+x}$ , $\ln (1+x)$ , $\int \ln (1+x)dx$ , and ${\int }_0^2\ln (1+x)dx$.

Step-by-Step Solution

Answer

The power series are

$$\begin{gathered} \frac{1}{1+x}=1-x+x^2-x^3+x^4 \\ \ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-... \\ \int \ln (1+x)=\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}+\frac{x^6}{30}-.... \\ {\int }_0^2\ln (1+x)dx=\frac{2^2}{2}-\frac{2^3}{6}+\frac{2^4}{12}-\frac{2^5}{20}+\frac{2^6}{30}-.... \end{gathered}$$

Step 1: Given information

We are given a power series of $\frac{1}{1-x}$

Step 2: Find the power series of 1 1 + x

The power series of $\frac{1}{1-x}$ can be given as

$\frac{1}{1-x}=1+x+x^2+x^3+x^4+.....$

Replacing x by -x we get,

$\frac{1}{1+x}=1-x+x^2-x^3+x^4-.....$

Step 3: Find the power series of ln ( 1 + x )

We have find the power series of $\frac{1}{1+x}$ which is

$\frac{1}{1+x}=1-x+x^2-x^3+x^4-....$

On integrating we get,

$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-....$

Again integrating the power series we get,

$\int \ln (1+x)=\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}+\frac{x^6}{30}-...$

And finally on integrating within the limits

$$\begin{gathered} {\int }_0^2\ln (1+x)dx={{[\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}+\frac{x^6}{30}-..]}^2}_0 \\ {\int }_0^2\ln (1+x)dx=[2-\frac{8}{6}+\frac{16}{12}-\frac{32}{20}+\frac{64}{30}-... \end{gathered}$$