214. The series for the inverse tangent

1. $\log \left({\displaystyle \frac{1}{1–x}}\right)=x+\frac{1}{2}{x}^2+\frac{1}{3}{x}^3+\dots$ if $-1\le x<1$.

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

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

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 $$\begin{array}{rl}{D}_x^n\arctan x& =D_x^{n-1}\{1/(1+x^2)\}\\ & =(-1{)}^{n-1}(n–1)!(x^2+1{)}^{-n/2}\sin \{n\arctan (1/x)\}\end{array}$$ (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 $$\log y=2\{\frac{y–1}{y+1}+\frac{1}{3}{\left(\frac{y–1}{y+1}\right)}^3+\frac{1}{5}{\left(\frac{y–1}{y+1}\right)}^5+\dots \}.$$

[Use the identity $y=(1+{\displaystyle \frac{y–1}{y+1}})/(1–{\displaystyle \frac{y–1}{y+1}})$. 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 $$\log 10=3\log 2+\log (1+\frac{1}{4}).$$

7. Prove that $$\log \left(\frac{x+1}{x}\right)=2\{\frac{1}{2x+1}+\frac{1}{3(2x+1{)}^3}+\frac{1}{5(2x+1{)}^5}+\dots \}$$ if $x>0$, and that $$\log \frac{(x–1{)}^2(x+2)}{(x+1{)}^2(x–2)}=2\{\frac{2}{x^3–3x}+\frac{1}{3}{\left(\frac{2}{x^3–3x}\right)}^3+\frac{1}{5}{\left(\frac{2}{x^3–3x}\right)}^5+\dots \}$$ if $x>2$. Given that $\log 2=.6931471\dots$ and $\log 3=1.0986123\dots$, show, by putting $x=10$ in the second formula, that $\log 11=2.397895\dots$.

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 $.00015$.

9. Show that $$\frac{1}{2}\log 2=7a+5b+3c,\frac{1}{2}\log 3=11a+8b+5c,\frac{1}{2}\log 5=16a+12b+7c,$$ where $$\begin{gathered} a= \\ operatornameargtanh(1/31) \end{gathered}$$, $$\begin{gathered} b= \\ operatornameargtanh(1/49) \end{gathered}$$, $$\begin{gathered} c= \\ operatornameargtanh(1/161) \end{gathered}$$.

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

10. Show that $$\frac{1}{4}\pi =\arctan (1/2)+\arctan (1/3)=4\arctan (1/5)–\arctan (1/239),$$ 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 $1+x+x^2+1/2x^3$.

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

14. Using the logarithmic series and the facts that ${\log }_{10}2.3758=.3758099\dots$ and ${\log }_{10}e=.4343\dots$, show that an approximate solution of the equation $x=100{\log }_{10}x$ is $237.58121$.

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, $$\log \sin x=\log x–\frac{1}{45}\log \cos x+\frac{64}{45}\log \cos \frac{1}{2}x.$$

16. Show that $${\int }_0^x\frac{dt}{1+t^4}=x–\frac{1}{5}{x}^5+\frac{1}{9}{x}^9–\dots$$ if $-1\le x\le 1$. Deduce that $$1–\frac{1}{5}+\frac{1}{9}–\cdots =\{\pi +2\log (\sqrt{2}+1)\}/4\sqrt{2}.$$

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

17. Prove similarly that $$\frac{1}{3}–\frac{1}{7}+\frac{1}{11}–\cdots ={\int }_0^1\frac{t^{2}dt}{1+t^4}=\{\pi –2\log (\sqrt{2}+1)\}/4\sqrt{2}.$$

18. Prove generally that if $a$ and $b$ are positive integers then $$\frac{1}{a}–\frac{1}{a+b}+\frac{1}{a+2b}–\cdots ={\int }_0^1\frac{t^{a-1}dt}{1+t^b},$$ 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 $\frac{1}{2}–\frac{1}{5}+\frac{1}{8}–\dots$.