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 \(\frac{1}{2} – \frac{1}{5} + \frac{1}{8} – \dots\).


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