Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/257



Find three terms of the expansion in each of the following functions:

Show that $$\log x$$ cannot be expanded by Maclaurin's Theorem. Compute the values of the following functions by substituting directly in the equivalent power series, taking terms enough until the results agree with those given below.

$$e = 2.7182\cdots.$$ Solution. Let $$x = 1$$ in series of Ex. 1; then



$$\arctan \left( \tfrac{1}{5} \right) = 0.1973\cdots;$$ use series in Ex. 9. $$ \cos 1 = 0.5403\cdots;$$ use series in Ex. 2. $$ \cos 10^\circ = 0.9848\cdots;$$ use series in Ex. 2. $$\sin .1 = .0998\cdots;$$ use series $$x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots.$$
 * First term
 * $$= 1.00000$$
 * Second term
 * $$= 1.00000$$
 * Third term
 * $$= 0.50000$$
 * Fourth term
 * $$= 0.16667\cdots$$
 * (Dividing third term by 3.)
 * Fifth term
 * $$= 0.04167\cdots$$
 * (Dividing fourth term by 4.)
 * Sixth term
 * $$= 0.00833\cdots$$
 * (Dividing fifth term by 5.)
 * Seventh term
 * $$= 0.00139\cdots$$
 * (Dividing sixth term by 6.)
 * Eighth term
 * $$= 0.00019\cdots,$$ etc.
 * (Divding sevent term by 7.)
 * align="right"|Adding, $$e$$
 * $$= 2. 71825\cdots$$ Ans.
 * }
 * Seventh term
 * $$= 0.00139\cdots$$
 * (Dividing sixth term by 6.)
 * Eighth term
 * $$= 0.00019\cdots,$$ etc.
 * (Divding sevent term by 7.)
 * align="right"|Adding, $$e$$
 * $$= 2. 71825\cdots$$ Ans.
 * }
 * $$= 2. 71825\cdots$$ Ans.
 * }
 * }

$$ \arcsin 1 = 1.5708\cdots;$$ use series in Ex. 8. $$\sin \tfrac{\pi}{4} = 0.7071\cdots;$$ use series (B), &sect;145. $$\sin .5 = 0.4794\cdots;$$ use series (B), &sect;145.

$$e^2 = 1 + 2 + \tfrac{2^2}{2!} + \tfrac{2^3}{3!} + \cdots = 7.3891.$$

$$\sqrt{e} = 1 \frac{1}{2} + \frac{1}{2^2\left(2!\right)} + \frac{1}{2^3\left(3!\right)} + \cdots = 1.6487.$$

In more advanced treatises it is shown that, for values of $$x$$ within the interval of convergence, the sum of a power series is differentiable and that its derivative is obtained by differentiating the series term by term as in an ordinary sum. Thus from (B), &sect;145, Differentiating both sides,

Differentiating both sides, we get