Question

${\mathit{x}}^{\mathbf{2}}\mathit{l}\mathit{n}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{x}\mathbf{)}$

Short answer

The series is$x^2\ln (1-x)=-\sum _{n=1}^{\infty }\frac{x^{n+2}}{n}$

The sum is given below.

$S_1=-x^3$

$S_2=-x^3-\frac{-x^4}{2}$

$S_3=-x^3-\frac{-x^4}{2}-\frac{-x^5}{3}$

The graphs are shown below.

Step-by-step solution

Given Information

The Maclaurin series.

Definition of the Maclaurin Series.

A Maclaurin series is a function with an expansion series that gives the sum of the function's derivatives.

Find the series and sum.

The Maclaurin series is given below.

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

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

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

The general series is given below $x^2\ln (1-x)=-x^3-\frac{x^4}{2}-\frac{x^5}{3}-\dots -\frac{x^{n+2}}{n}$

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

The sum is given below.

$S_1=-x^3$

$S_2=-x^3-\frac{-x^4}{2}$

$S_3=-x^3-\frac{-x^4}{2}-\frac{-x^5}{3}$

Plot the graphs.

The graph of the function with $S_1,S_2,S_3$is given below.