It is easy to prove in a similar manner that \begin{aligned} \arctan x = \int_{0}^{x} \frac{dt}{1 + t^{2}} &= \int_{0}^{x}(1 – t^{2} + t^{4} – \dots)\, dt\\ &= x – \tfrac{1}{3} x^{3} + \tfrac{1}{5} x^{5} – \dots,\end{aligned} provided that $$-1 \leq x \leq 1$$. The only difference is that the proof is a little simpler; for, since $$\arctan x$$ is an odd function of $$x$$, we need only consider positive values of $$x$$. And the series is convergent when $$x = -1$$ as well as when $$x = 1$$. We leave the discussion to the reader. The value of $$\arctan x$$ which is represented by the series is of course that which lies between $$-\frac{1}{4}\pi$$ and $$\frac{1}{4}\pi$$ when $$-1 \leq x \leq 1$$, and which we saw in Ch. VII (Ex. LXIII. 3) to be the value represented by the integral. If $$x = 1$$, we obtain the formula $\tfrac{1}{4}\pi = 1 – \tfrac{1}{3} + \tfrac{1}{5} – \dots.$

Example XCI
1. $$\log \left(\dfrac{1}{1 – x}\right) = x + \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \dots$$ if $$-1 \leq x < 1$$.

2. $$\\operatorname{arg tanh} x = \frac{1}{2} \log\left(\dfrac{1 + x}{1 – x}\right) = x + \frac{1}{3} x^{3} + \frac{1}{5} x^{5} + \dots$$ if $$-1 < x < 1$$.

3. Prove that if $$x$$ is positive then $\log(1 + x) = \frac{x}{1 + x} + \tfrac{1}{2} \left(\frac{x}{1 + x}\right)^{2} + \tfrac{1}{3} \left(\frac{x}{1 + x}\right)^{3} + \dots.$

4. Obtain the series for $$\log(1 + x)$$ and $$\arctan x$$ by means of Taylor’s theorem.

[A difficulty presents itself in the discussion of the remainder in the first series when $$x$$ is negative, if Lagrange’s form $$R_{n} = (-1)^{n-1} x^{n}/\{n(1 + \theta x)^{n}\}$$ is used; Cauchy’s form, viz. $R_{n} = (-1)^{n-1} (1 – \theta)^{n-1} x^{n}/(1 + \theta x)^{n},$ should be used (cf. the corresponding discussion for the Binomial Series, Ex. LVI. 2 and § 163).

In the case of the second series we have \begin{aligned} D_{x}^{n} \arctan x &= D_{x}^{n-1} \{1/(1 + x^{2})\}\\ &= (-1)^{n-1} (n – 1)! (x^{2} + 1)^{-n/2} \sin \{n\arctan(1/x)\}\end{aligned} (Ex. XLV. 11), and there is no difficulty about the remainder, which is obviously not greater in absolute value than $$1/n$$.]

5. If $$y > 0$$ then $\log y = 2 \left\{\frac{y – 1}{y + 1} + \frac{1}{3} \left(\frac{y – 1}{y + 1}\right)^{3} + \frac{1}{5} \left(\frac{y – 1}{y + 1}\right)^{5} + \dots\right\}.$

[Use the identity $$y = \biggl(1 + \dfrac{y – 1}{y + 1}\biggr) \bigg/ \biggl(1 – \dfrac{y – 1}{y + 1}\biggr)$$. This series may be used to calculate $$\log 2$$, a purpose for which the series $$1 – \frac{1}{2} + \frac{1}{3} – \dots$$, owing to the slowness of its convergence, is practically useless. Put $$y = 2$$ and find $$\log 2$$ to $$3$$ places of decimals.]

6. Find $$\log 10$$ to $$3$$ places of decimals from the formula $\log 10 = 3\log 2 + \log(1 + \tfrac{1}{4}).$

7. Prove that $\log \left(\frac{x + 1}{x}\right) = 2\left\{\frac{1}{2x + 1} + \frac{1}{3(2x + 1)^{3}} + \frac{1}{5(2x + 1)^{5}} + \dots\right\}$ if $$x > 0$$, and that $\log \frac{(x – 1)^{2}(x + 2)}{(x + 1)^{2}(x – 2)} = 2\left\{\frac{2}{x^{3} – 3x} + \frac{1}{3}\left(\frac{2}{x^{3} – 3x}\right)^{3} + \frac{1}{5}\left(\frac{2}{x^{3} – 3x}\right)^{5} + \dots\right\}$ if $$x > 2$$. Given that $$\log 2 = .693\ 147\ 1\dots$$ and $$\log 3 = 1.098\ 612\ 3\dots$$, show, by putting $$x = 10$$ in the second formula, that $$\log 11 = 2.397\ 895\dots$$.

8. Show that if $$\log 2$$, $$\log 5$$, and $$\log 11$$ are known, then the formula $\log 13 = 3\log 11 + \log 5 – 9\log 2$ gives $$\log 13$$ with an error practically equal to $$.000\ 15$$.

9. Show that $\tfrac{1}{2} \log 2 = 7a + 5b + 3c,\quad \tfrac{1}{2} \log 3 = 11a + 8b + 5c,\quad \tfrac{1}{2} \log 5 = 16a + 12b + 7c,$ where $$a = \\operatorname{arg tanh}(1/31)$$, $$b = \\operatorname{arg tanh}(1/49)$$, $$c = \\operatorname{arg tanh}(1/161)$$.

[These formulae enable us to find $$\log 2$$, $$\log 3$$, and $$\log 5$$ rapidly and with any degree of accuracy.]

10. Show that $\tfrac{1}{4}\pi = \arctan(1/2) + \arctan(1/3) = 4\arctan(1/5) – \arctan(1/239),$ and calculate $$\pi$$ to $$6$$ places of decimals.

11. Show that the expansion of $$(1 + x)^{1+x}$$ in powers of $$x$$ begins with the terms $$1 + x + x^{2} + 1/2 x^{3}$$.

12. Show that $\log_{10} e – \sqrt{x(x + 1)} \log_{10}\left(\frac{1 + x}{x}\right) = \frac{\log_{10} e}{24x^{2}},$ approximately, for large values of $$x$$. Apply the formula, when $$x = 10$$, to obtain an approximate value of $$\log_{10} e$$, and estimate the accuracy of the result.

13. Show that $\frac{1}{1 – x} \log\left(\frac{1}{1 – x}\right) = x + \left(1 + \tfrac{1}{2}\right)x^{2} + \left(1 + \tfrac{1}{2} + \tfrac{1}{3}\right)x^{3} + \dots,$ if $$-1 < x < 1$$. [Use Ex. LXXXI. 2.]

14. Using the logarithmic series and the facts that $$\log_{10} 2.3758 = .375\ 809\ 9\dots$$ and $$\log_{10} e = .4343\dots$$, show that an approximate solution of the equation $$x = 100 \log_{10}x$$ is $$237.581\ 21$$.

15. Expand $$\log\cos x$$ and $$\log(\sin x/x)$$ in powers of $$x$$ as far as $$x^{4}$$, and verify that, to this order, $\log\sin x = \log x – \tfrac{1}{45} \log\cos x + \tfrac{64}{45}\log\cos \tfrac{1}{2}x.$

16. Show that $\int_{0}^{x} \frac{dt}{1 + t^{4}} = x – \tfrac{1}{5}x^{5} + \tfrac{1}{9}x^{9} – \dots$ if $$-1 \leq x \leq 1$$. Deduce that $1 – \tfrac{1}{5} + \tfrac{1}{9} – \dots = \{\pi + 2\log(\sqrt{2} + 1)\}/4\sqrt{2}.$

[Proceed as in § 214 and use the result of Ex. XLVIII. 7.]

17. Prove similarly that $\tfrac{1}{3} – \tfrac{1}{7} + \tfrac{1}{11} – \dots = \int_{0}^{1} \frac{t^{2}\, dt}{1 + t^{4}} = \{\pi – 2\log(\sqrt{2} + 1)\}/4\sqrt{2}.$

18. Prove generally that if $$a$$ and $$b$$ are positive integers then $\frac{1}{a} – \frac{1}{a + b} + \frac{1}{a + 2b} – \dots = \int_{0}^{1} \frac{t^{a-1}\, dt}{1 + t^{b}},$ and so that the sum of the series can be found. Calculate in this way the sums of $$1 – \frac{1}{4} + \frac{1}{7} – \dots$$ and $$\frac{1}{2} – \frac{1}{5} + \frac{1}{8} – \dots$$.