Page:EB1911 - Volume 27.djvu/297

of the series ${\frac {1}{1^{m}}}+{\frac {1}{3^{m}}}+{\frac {1}{5^{m}}}+\cdots$, and $W^{m}$ that of the series Sums of Powers of Reciprocals of Natural Numbers. ${\frac {1}{1^{m}}}-{\frac {1}{3^{m}}}+{\frac {1}{5^{m}}}-{\frac {1}{7^{m}}}+\cdots$, we obtain by taking logarithms in the formulae (25) and (26)

$\log(x{\text{ cosec }}x)=U_{2}\left({\frac {x}{\pi }}\right)^{2}+{\frac {1}{2}}U_{4}\left({\frac {x}{\pi }}\right)^{4}+{\frac {1}{3}}U_{6}\left({\frac {x}{\pi }}\right)^{6}+\cdots$,

$\log(\sec x)=V_{2}\left({\frac {2x}{\pi }}\right)^{2}+{\frac {1}{2}}V_{4}\left({\frac {2x}{\pi }}\right)^{4}+{\frac {1}{3}}V_{6}\left({\frac {2x}{\pi }}\right)^{6}+\cdots$;

and differentiating these series we get

(31)

${\frac {1}{2}}\cot x={\frac {1}{2x}}-{\frac {U_{2}}{\pi ^{2}}}x-{\frac {U_{4}}{\pi ^{4}}}x^{3}-{\frac {U_{6}}{\pi ^{6}}}x^{5}-\cdots$,

(32)

${\frac {1}{2}}\tan x={\frac {V_{2}}{\pi ^{2}}}2^{2}x+{\frac {V_{4}}{\pi ^{4}}}2^{4}x^{3}+{\frac {V_{6}}{\pi ^{6}}}2^{6}x^{5}+\cdots$,

In (31) $x$ must lie between $\pm \pi$ and in (32) between $\pm {\tfrac {1}{2}}\pi$. Write equation (30) in the form

$\sec x=\sum (-1)^{n}{\frac {(2n+1)\pi }{({\overline {2n+1}}\cdot {\frac {1}{2}}\pi )^{2}-x^{2}}}$,

and expand each term of this series in powers of $x^{2}$, then we get

(33)

$\sec x={\frac {2^{2}W_{1}}{\pi }}+{\frac {2^{4}W_{3}x^{2}}{\pi ^{3}}}+{\frac {2^{6}W_{5}x^{4}}{\pi ^{5}}}$

where $x$ must lie between $\pm {\tfrac {1}{2}}\pi$. By comparing the series (31), (32), (33) with the expansions of $\cot x$, $\tan x$, $\sec x$ obtained otherwise, we can calculate the values of $U_{2},U_{4}\dots$ $V_{2},V_{4}\dots$ $W_{1},W_{3}\dots$. When $U_{n}$ has been found, $V_{n}$ may be obtained from the formula $2^{n}V^{n}=(2^{n}-1)U_{n}$.

For Lord Brounker's series of π, see Circle. It can be got at once Continued Factors for π. by putting $a=1$, $b=3$, $c=5$,$\dots$ in Euler's theorem $={\frac {1}{a}}-{\frac {1}{b}}+{\frac {1}{c}}-\cdots ={\frac {1}{a\;+}}\;{\frac {a^{2}}{b-a\;+}}\;{\frac {b^{2}}{c-b\;+}}\cdots$

Sylvester gave (Phil. Mag., 1869) the continued fraction

${\frac {\pi }{2}}=1+{\frac {1}{1\;+}}\;{\frac {1\,.\,2}{1\;+}}\;{\frac {2\,.\,3}{1\;+}}\;{\frac {3\,.\,4}{1\;+}}\cdots$;

which is equivalent to Wallis's formula for π. This fraction was originally given by Euler (Comm. Acad. Petropol. vol. xi.); it is also given by Stern (in Crelle's Journ. vol. x.).

30. It may be shown by means of a transformation of the series for $\cos x$ and ${\frac {\sin x}{x}}$ that $\tan x={\frac {x}{1\;-}}\;{\frac {x^{2}}{3\;-}}\;{\frac {x^{2}}{5\;-}}\;{\frac {x^{3}}{7\;-}}\cdots$ Continued Fractions for Trigonometrical Functions. This may be also easily shown as follows. Let $y=\cos {\surd x}$, and let $y',y''\dots$ denote the differential coefficients of $y$ with regard to $x$, then by forming these we can show that $4xy''+2y'+y=0$, and thence by Leibnitz's theorem we have

$4xy^{(n+2)}+(4n+2)y^{(n+1)}+y^{(n)}=0$

Therefore ${\frac {y}{y'}}=-2-{\frac {4x}{y'/y''}}$, ${\frac {y^{n}}{y^{(n+1)}}}=-2(2n+1)-{\frac {4x}{y^{(n+1)}/y^{(n+2)}}}$;

hence $-2{\sqrt {x}}\cot {\sqrt {x}}=-2-{\frac {4x}{-\;6\;-}}\;{\frac {4x}{-\;10\;-}}\;{\frac {4x}{-\;14\;-}}\cdots$

Replacing $\surd x$ by $x$ we have $\tan x={\frac {x}{1\;-}}\;{\frac {x^{2}}{3\;-}}\;{\frac {x^{3}}{5\;-}}\cdots$

Euler gave the continued fraction

$\tan nx={\frac {n\tan x}{1\;-}}\;{\frac {(n^{2}-1)\tan ^{2}x}{3\;-}}\;{\frac {(n^{2}-4)\tan ^{2}x}{5\;-}}\;{\frac {(n^{2}-9)\tan ^{2}x}{7\;-}}\cdots$;

this was published in Mém. de l'acad. de St Pétersb. vol. vi. Glaisher has remarked (Mess. of Math. vols. iv.) that this may be derived by forming the differential equation

$(1-x^{2})y^{(m+2)}-(2m+1)xy^{(m+1)}+(n^{2}-m^{2})y^{(m)}=0$,

where $y=\cos {(n\arccos x)}$, then replacing $x$ by $\cos x$, and proceeding as in the former case. If we put $n=0$, this becomes

$x={\frac {\tan x}{1\;+}}\;{\frac {\tan ^{2}x}{3\;+}}\;{\frac {4\tan ^{2}x}{5\;+}}\;{\frac {9\tan ^{2}x}{7\;+}}\cdots$;

whence we have

$\arctan x={\frac {x}{1\;+}}\;{\frac {x^{2}}{3\;+}}\;{\frac {4x^{2}}{5\;+}}\;{\frac {9x^{2}}{7\;+}}\cdots +{\frac {n^{2}x^{2}}{2n+1\;+}}\cdots$

31. It is possible to make the investigation of the properties of the simple circular functions rest on a purely analytical basis other than the one indicated in § 22. The sine of $x$ would be Purely Analytical Treatment of Circular Functions. defined as a function such that, if $x=\int _{0}^{y}{\frac {dy}{\surd (1-y^{2})}}$, then $y=\sin x$; the quantity ${\frac {\pi }{2}}$ would be defined to be the complete integral $\int _{0}^{1}{\frac {dy}{\surd (1-y^{2})}}$. We should then have ${\frac {\pi }{2}}-x=\int _{y}^{1}{\frac {dy}{\surd (1-y^{2})}}$. Now change the variable in the integral to $z$, where $y^{2}+z^{2}=1$, we then have ${\frac {\pi }{2}}-x=\int _{0}^{z}{\frac {dy}{\surd (1-z^{2})}}$, and $z$ must be defined as the cosine of $x$, and is thus equal to $\sin({\frac {1}{2}}\pi -x)$, satisfying the equation $\sin ^{2}x+cos^{2}x=1$.

Next consider the differential equation

${\frac {dy}{\surd (1-y^{2})}}+{\frac {dz}{\surd (1-z^{2})}}=0$.

This is equivalent to

$d\{y\surd (1-z^{2})+z\surd (1-y^{2})\}=0$;

hence the integral is

$y\surd (1-z^{2})+z\surd (1-y^{2})=$ a constant.

The constant will be equal to the value $u$ of $y$ when $z=0$; whence

$y\surd (1-z^{2})+z\surd (1-y^{2})=u$.

The integral may also be obtained in the form

$yz-\surd (1-y^{2})\surd (1-z^{2})=\surd (1-u^{2})$.

Let $\alpha =\int _{0}^{y}{\frac {dy}{\surd (1-y^{2})}}$, $\beta =\int _{0}^{z}{\frac {dz}{\surd (1-z^{2})}}$, $\gamma =\int _{0}^{u}{\frac {du}{\surd (1-u^{2})}}$; we have $\alpha +\beta =\gamma$, and $\sin \gamma =\sin \alpha \cos \beta +\cos \alpha \sin \beta$, $\cos \gamma =\cos \alpha \cos \beta -\sin \alpha \sin \beta$, the addition theorems. By means of the addition theorems and the values $\sin {{\frac {1}{2}}\pi }=1$, $\cos {{\frac {1}{2}}\pi }=0$ we can prove that $\sin {({\frac {1}{2}}\pi +x)}=\cos x$, $\cos {({\frac {1}{2}}\pi +x)}=-\sin x$; and thence, by another use of the addition theorems, that $\sin {(\pi +x)}=-\sin x\cos(\pi +x)=-\cos x$, from which the periodicity of the functions $\sin x$, $cosx$ follows:—

We have also $\int {\frac {dy}{\surd (1-y^{2})}}=-\iota \log _{e}\{\surd (1-y^{2})+\iota y\}$; whence $\log _{e}\{\surd (1-y^{2})+\iota y\}+\log _{e}\{\surd (1-z^{2})+\iota z\}=$ a constant. Therefore $\{\surd (1-y^{2})\}+\iota y\{\surd (1-z^{2})+\iota z\}=\surd (1-u^{2})+\iota u$, since $u=y$ when $z=0$; whence we have the equation

$(\cos \alpha +\iota \sin \alpha )(\cos \beta +\iota \sin \beta )=\cos(\alpha +\beta )+\iota \sin(\alpha +\beta )$,

from which De Moivre's theorem follows.

References.—Further information will be found in Hobson's Plane Trigonometry, and in Chrystal's Algebra, vol. ii. For further information on the history of the subject, see Braunmühl's Vorlesungen über Geschichte der Trigonometrie (Leipzig, 1900).  (E. W. H.) 

TRIGONON, a small triangular harp, occasionally used by the ancient Greeks and probably derived from Assyria or Egypt. The trigonon is thought to be either a variety of the sambuca or identical with it. A trigonon is represented on one of the Athenian red-figured vases from Cameiros in the island of Rhodes, dating from the 5th century B.C., which are preserved in the British Museum. The triangle is here an irregular one, consisting of a narrow base to which one end of the string was fixed, while the second side, forming a slightly obtuse angle with the base, consisted of a wide and slightly curved sound-board pierced with holes through which the other end of the strings passed, being either knotted or wound round pegs. The third side of the triangle was formed by the strings themselves, the front pillar, which in modern European harps plays such an important part, being always absent in these early Oriental instruments. A small harp of this kind having 20 strings was discovered at Thebes in 1823.  (K. S.) 

TRIKKALA (anc. Trika), a town of Greece, capital of the department of Trikkala, and the see of an archbishop, 38 m. W. of Larissa. In winter, when great numbers of Vlach herdsmen take up their quarters in the town, its population exceeds that of Larissa. It has the appearance of a Mussulman town on account of its mosques (only two of which are in use) and it is a centre of trade in wheat, maize, tobacco and cocoons. The town was in ancient times a celebrated seat of the worship of Aesculapius. Pop. (1889), 14,820; (1907) 17,809; of the department, 90,548.

TRILEMMA (Gr. τρεῖς, three, λῆμμα, something taken), in logic, an argument akin to the dilemma (q.v.), in which there are three possibilities. By getting rid of two, the third is proved, provided the original three exhaust the number. The terms “tetralemma” (four possibilities) and “polylemma” (many) have also been used.

TRILOBITES, extinct Arthropoda, formerly classified with the Crustacea, but of late years relegated to the Arachnida (q.v.), which occurred abundantly in seas of the Cambrian and Silurian periods, but disappeared entirely at the close of the Palaeozoic epoch. Both their origin and the causes which led to their extinction are quite unknown. Widely divergent forms make their appearance suddenly in the Cambrian period amongst the earliest known fossils; and the high perfection of structure to which they had at that time attained