Question

$\frac{\mathbf{1}}{\mathbf{x}}\mathit{s}\mathit{i}\mathit{n}\mathit{x}$

Short answer

The series is $\frac{\sin \left(x\right)}{x}=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{(2n+1)!}$.

The sum is given below.

$S_1=1-\frac{x^2}{6}$

$S_2=1-\frac{x^2}{6}+\frac{x^4}{120}$

$S_3=1-\frac{x^2}{6}+\frac{x^4}{120}-\frac{x^6}{5040}$

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.

$\sin \left(x\right)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$

$\frac{\sin \left(x\right)}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\dots$

$\frac{\sin \left(x\right)}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}-\frac{x^6}{5040}+\dots$

The general series is given below.

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

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

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

The sum is given below.

$S_1=1-\frac{x^2}{6}$

$S_2=1-\frac{x^2}{6}+\frac{x^4}{120}$

$S_3=1-\frac{x^2}{6}+\frac{x^4}{120}-\frac{x^6}{5040}$

Plot the graphs.

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