Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/256



Using the series (B) found in the last example, calculate $$sin 1$$ correct to four decimal places.

Solution. Here $$x = 1$$ radian; that is, the angle is expressed in circular measure. Therefore, substituting $$x = 1$$ in (B) of the last example,

Summing up the positive and negative terms separately,

$$ \begin{array}{rclrcl} 1 &=& 1.00000\cdots  & \qquad \frac{1}{3!} &=& 0.16667\cdots \\ &=& 0.00833\cdots  & \qquad \frac{1}{7!} &=& 0.00019\cdots \\ \hline & & 1.00833\cdots  &                      & & 0.16686\cdots \\ \end{array} $$ Hence $$ \sin 1 = 1.00833 - 0.16686 = 0.84147\cdots,$$

which is correct to four decimal places, since the error made must be less than; $$\tfrac{1}{9!};$$ i.e. less than .000003. Obviously the value of $$\sin 1$$ may be calculated to any desired degree of accuracy by simply including a sufficient number of additional terms.

Verify the following expansions of functions into power series by Maclaurin's Series and determine for what values of the variable they are convergent: