a/z . [phi](z+1)/[phi]z = (a/z+) (a/(z+1)+) (a/(z+2)+) (a/((z+3)+ ...))
Let z = 1/2; and let 4a = -x^2. Then (a/z)[phi](z+1) is -(x^2/2) ( 1 -
x^2/(2.3) + x^4/(2.3.4.5...)) or -(x/2) sin x. Again [phi]z is 1 - x^2/2 +
x^4/(2.3.4) or cos x: and the continued fraction is
(1/4)x^2/(1/2)+ (1/4)x^2/(3/2)+ (1/4)x^2/(5/2)+ ...
or -x/2 x/1+ -x^2/3+ -x^2/5+
...
{371} whence tan x = x/1+ -x^2/3+ -x^2/5+ -x^2/7+ ...
Or, as written in the usual way,
tan x = x
-------
1 - x^2
-------
3 - x^2
-------
5 - x^2
-------
7 - ...
This result may be proved in various ways: it may also be verified by
calculation. To do this, remember that if
a_1/b_1+ a_2/b_2+ a_3/b_3+ ... a_n/b_n = P_n/Q_n; then
P_1=a_1, P_2=b_2 P_1, P_3=b_3 P_2+a_3 P_1, P_4=b_4 P_3+a_4 P_2, etc.
Q_1=b_1, Q_2=b_2 Q_1+a_2, Q_3=b_3 Q_2+a_3 Q_1, Q_4=b_4 Q_3+a_4 Q_2, etc.
in the case before us we have
a_1=x, a_2=-x^2, a_3=-x^2, a_4=-x^2, a_5=-x^2, etc.
b_1=1, b_2=3, b_3=5, b_4=7, b_5=9, etc.
P_1=x Q_1=1
P_2=3x Q_2=3-x^2
P_3=15x-x^3 Q_3=15-6x^2
P_4=105x-10x^3 Q_4=105-45x^2+x^4
P_5=945x-105x^3+x^5 Q_5=945-420x^2+15x^4
P_6=10395x-1260x^3+21x^5 Q_6=10395-4725x^2+210x^4-x^6
We can use this algebraically, or arithmetically. If we divide P_n by Q_n,
we shall find a series agreeing with the known series for tan x, _as far
as_ n _terms_. That series is
x + x^3/3 + 2x^5/15 + 17x^7/315 + 62x^9/2835 + ...
{372} Take P_5, and divide it by Q_5 in the common way, and the first five
terms will be as here written. Now take _x_ = .1, which means that the
angle is to be one tenth of the actual unit, or, in degrees 5 deg..729578.
We find that when x = .1, P_6 = 1038.24021, Q_6 = 10347.770999; whence P_6
divided by Q_6 gives .1003346711. Now 5 deg..729578 is 5 deg.43'46-1/2";
and from the old tables of Rheticus[675]--no modern tables carry the
tangents so far--the tangent of this angle is .1003347670.
Now let x = (1/4)[pi]; in which case tan x = 1. If (1/4)[pi] be
commensurable with the unit, let it be (m/n), m and n being integers:
we know that (1/4)[pi] < 1. We have then
1=(m/n)/1- (m^2/n^2)/3- (m^2/n^2)/5- ... = m/n- m^2/3n- m^2/5n- m^2/7n-
...
Now it is clear that m^2/3n, m^2/5n, m^2/7n, etc. must at last become and
continue severally less than unity. The continued fraction is therefore
incommensurable, and cannot be unity. Consequently [pi]^2 cannot be
commensurable: that is, [pi] is an incommensurable quantity, and so also
is [pi]^2.
I thought I should end with a grave bit of appendix, deeply mathematical:
but paradox follows me wherever I go. The foregoing is--in my own
language--from Dr. (now Sir David) Brewster's[676] English edition of
Legendre's Geometry, (Edinburgh, 1824, 8vo.) translated by some one who is
not named. I picked up a notion, which others had at Cambridge in 1825,
that the translator was the late Mr. Galbraith,[677] then known at
Edinburgh as a writer and teacher.
{373} But it turns out that it was by a very different person, and one
destined to shine in quite another walk; it was a young man named Thomas
Carlyle.[678] He prefixed, from his own pen, a thoughtful and ingenious
essay on Proportion, as good a substitute for the fifth Book of Euclid as
could have been given in the space; and quite enough to show that he would
have been a distinguished teacher and thinker on first principles. But he
left the field immediately.
* * * * *
(The following is the passage referred to at Vol. II, page 54.)
Michael Stifelius[679] edited, in 1554, a second edition of the Algebra
(_Die Coss._), of Christopher Rudolff.[680] This is one of the earliest
works in which + and - are used.
Stifelius was a queer man. He has introduced into this very work of Rudolff
his own interpretation of the number of the Beast. He determined to fix the
character of Pope Leo: so he picked the numeral letters from LEODECIMVS,
and by taking in X from LEO X. and striking out M as standing for
_mysterium_, he hit the number exactly. This discovery completed his
conversion to Luther, and his determination to throw off his monastic vows.
Luther dealt with him as straight-forwardly as with Melanchthon about his
astrology: he accepted the conclusions, but told him to clear his mind of
all the premises about the Beast. Stifelius {374} did not take the advice,
and proceeded to settle the end of the world out of the prophet Daniel: he
fixed on October, 1533. The parishioners of some cure which he held, having
full faith, began to spend their savings in all kinds of good eating and
drinking; we may charitably hope this was not the way of preparing for the
event which their pastor pointed out. They succeeded in making themselves
as fit for Heaven as Lazarus, so far as beggary went: but when the time
came, and the world lasted on, they wanted to kill their deceiver, and
would have done so but for the interference of Luther. {375}
* * * * *