Question

$\mathit{x}\sqrt{\mathbf{1}\mathbf{+}\mathbf{x}}$

Short answer

The series is $x\sqrt{1+x}=\sum _{n=2}^{\infty }\frac{(-1{)}^{n-1}(2n-3)!!}{\left(2n\right)!!}{x}^{n+1}$

The sum is given below.

$S_1=x$

$S_2=x+\frac{1}{2}{x}^2$

$S_3=x+\frac{1}{2}{x}^2-\frac{1}{8}{x}^3$

$S_4=x+\frac{1}{2}{x}^2-\frac{1}{8}{x}^3+\frac{1}{16}{x}^4$

The graphs are shown below.

Step-by-step solution

Given Information

The Maclaurin series.

Definition of the MaclaurinSeries.

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.

$(1+x{)}^{1/2}=1+\frac{1}{2}x-\frac{1}{8}{x}^2+\frac{1}{16}{x}^3-\frac{5}{128}{x}^4+\dots$

$x(1+x{)}^{1/2}=x+\frac{1}{2}{x}^2-\frac{1}{8}{x}^3+\frac{1}{16}{x}^4-\frac{5}{128}{x}^5+\dots$

The general series is given below

$x\sqrt{1+x}=x\sum _{n=0}^{\infty }\left(\begin{array}{c}\frac{1}{2}\\ n\end{array}\right)x^n$

$x\sqrt{1+x}=\sum _{n=2}^{\infty }\frac{(-1{)}^{n-1}(2n-3)!!}{\left(2n\right)!!}{x}^{n+1}$

The sum is given below.

localid="1657352249901" $S_1=x$

$S_2=x+\frac{1}{2}{x}^2$

$S_3=x+\frac{1}{2}{x}^2-\frac{1}{8}{x}^3$

$S_4=x+\frac{1}{2}{x}^2-\frac{1}{8}{x}^3+\frac{1}{16}{x}^4$

Plot the graphs.

The graph of the function with $S_1,S_2,S_3,S_4$is given below