

From Wikisource
Jump to: navigation, search
This page needs to be proofread.

(b) Using only the first three terms,
\sin 1 = 1 - \tfrac{1}{6} + \tfrac{1}{120} = .841666,
the absolute error is less than \tfrac{1}{7}; i.e. \le \tfrac{1}{5040} (= .000198), and the percentage of error is less than \tfrac{1}{40} of 1 per cent.[1]
Moreover, the exact value of sin 1 lies between .8333 and .841666, since for an alternating series S_n is alternately greater and less than lim_{n \to \infty} S_n.


(a) when all terms beyond the second are neglected;
(b) when all terms beyond the third are neglected.
1. \cos 1. 4. \arctan 1. 7. e^{-\tfrac{1}{2}}.
2. \sin 2. 5. e^{-2}. 8. \arctan 2.
3. \cos \tfrac{1}{2}. 6. \sin \tfrac{\pi}{3}. 9. \sin 15^\circ.

II. The computation of \pi by series.

From Ex. 8, §145, we have

\arcsin x = x + \frac{1 \cdot x^3}{2 \cdot 3} + \frac{1 \cdot 3 x^5}{2 \cdot 4 \cdot 5} + \frac{1 \cdot 3 \cdot 5 x^7}{2 \cdot 4 \cdot 6 \cdot 7} + \cdots.

Since this series converges for values of x between -1 and +1, we may let x = \tfrac{1}{2}, giving

\frac{\pi}{6} = \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{3} \left( \frac{1}{2} \right)^3 + \frac{1 \cdot 3}{2 \cdot 4} \cdot \frac{1}{5} \left( \frac{1}{2} \right)^5 + \cdots,


\pi = 3.1415\cdots.

Evidently we might have used the series of Ex 9, §145, instead. Both of these series converge rather slowly, but there are other series, found by more elaborate methods, by means of which the correct value of \pi to a large number of decimal places may be easily calculated.

III. The computation of logarithms by series.


From Ex. 6, §145, we have

(A) \log \left( 1 + x \right) = x = \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \cdots.
  1. Since .000198 \div .841666 = .00023.