ordinal number, because it is the only variable factor of the exponent + 2 kr -]. Thus 2 k = represents that circumference whose place is designated by k.

#viT is a sign of impossibility, because it espresses a quantity greater than a maximum, and less than a minimum; but

a
the area of each concentric ring is out of that circle which serves as
a nucleus, and it is the diameter of that circle which is a maximum.
The diameter, then, of the exterior circle of this ring is greater
than a maximum. The whole area of the ring which exceeds this
diameter is then proved imaginary, which shows that the sign
+ N - 1 belongs absolutely to it. It now remains to explain the

ts
sign 5. A ring contains two circumferences of circles; to wit, an
external and an internal. Now 2 - only expresses one; the sign

causes it to express two; which I thus prove: Let y y = a a

.(19) be the equation to a circle : if we take the value of y we shall have y = raa

...(20) Here the double sign indicates two ordinates of an equal length drawn from any particular point of the diameter on each side of it. The positive ordinates, designated by +, extend only to half the circle; and the negative ordinates designated by -, extend to the other half: in order to obtain the ordinates which extend to the whole circle, we must unite the two signs, as in I. Now when this sign is accompanied by ✓ -], it does not mean + or -,

, but + and - ; because the imaginary quantities always go in pairs, and they cannot be separated without an absurdity, as I will prove.

Thus let 1 1 1 be the tangent to the central circle. This tangent

XX

:

1 0 1

tud als

mer

is the greatest ordinate which can be drawn to the exterior circle withoui entering into the central circle 1 1 1: its middle point is at

the

1 1 1

the same time the smallest of those which can be drawn in the interior circle, since it is reduced to this point, ], in which the two

ordinates coincide, the two ordinates, 1 1 and 11 having then the

1

point 1, which is common to both, and are connected by that point. Thus they form a continued right line, which is expressed by + N - T. If we refer this expression to the interior circumference, we have IV-I = 0, which is not imaginary, because

1 then it is the sign - l, which ought to be considered as 0. If,

♡ on the contrary, we refer it to the extreme circumference, we have I s = tn

i VIT, which only ceases to be imaginary at the two points. I l, which coincide with this extreme

ses a

0 1

1

circumference. To apply this principle to the double sine of the expression = 2 ka n - 1, let us divide into two equal parts the part 1 1 of the diameter 1 l, which is intersected between the two circumferences of the ring which extends beyond the central circle ; and through the point of division let us draw the dotted concentric circle; the circumference of the dotted circle will be an arithmetical mean proportional between the two extreme circumferences of the ring. If we take this dotted circumference as a line of abscissa, it is clear it will cut all the sections of the diameters intercepted between the extreme circumference into two equal parts. Each of these half parts will be equal ordinates drawn on each side of the circumference of the dotted circle, this circumference being taken as a line of abscissa, and the two extreme circumferences will be the curves described by means of these ordinates. As all these ordinates are imaginary, they have only two real points, which are their two extremes: one of these two extremes is a point in the dotted circle, and the other is a point in one of the circles already described : these three circles are then composed only of insulated points, the points of the dotted circle are double, and those of the circle described are simple. = 2 + - 1 expresses the sum of the points of the dotted circle ; that is to say, + 27 V I is the sum of the points of the exterior circle, and - 25 N - l, the

W sum of the points of the interior circie. Resuming all this explanation, we find = 2 kN-1 is the sign of the description of. +

I two concentric circles forming a ring by assuming for a line of abscissa a third concentric circle whose circumference is an arithmetical proportional mean between the circumferences to be described, the same as I vaa – x x is the sign of the description of a simple circumference by taking its diameter for a line of the abscissa.

This granted, in order to resolve the equations (1), (2), (3), I begin by multiplying their second member by the second member of equation (12), which gives me Q + 10 = 16 + 27 V -1 = the area* of the 16th circle, et

= • (fig. 1)

.(22) 12 + ac s 17 e

4 2 πν

= the area of the 17th circle ..(23) ER + al = 18 e

#27 VI

= the area of the 18th circle..(24). I assume for the roots of these equations,

o the

o the when

that

cum

.

Ce

+

26

4

dis

= 17 et 2n =1 = 18e+arva et 21

21

.

Х

17 + 21

2 t

4

X

@=e+\2.2 -T

= area of the 13th quadrant. .(25) =et.27N -T

= area of the 17th quadrant (26)
***.2 V
= area of the 21st quadrant

:(27) "

To prove that these are the true roots of equations (22), (23), (24), I substitute these roots for a, b, c; then take the differential*, considering the sign as the differential sign. These substitutions

# give me p+99.27 vitet

(42).

!
.25v-= 16e+27 V-1
NT

(28)
( 13 + 21)
****.27vFitet
.2 VT

(29) et 27 vi +

(30) By taking the differential, I obtain (7.2 - 1) e

+ 20. 27 V 1 +(?2V et 17421)

= + 27vIx16 et 2 the area of the 16th ring

...(31) +(4.27 - 1)e+44.2* V-1 + (18+ 2127 V-)

() 251 - 1 e

2 πM-1

= + 2 * ♡ Txinan eu 25 v 1 area of the 17th ring

.(32) +(4:27 -1) +2.24 V-1 +(13:"? 2 7 V 1) N=

2 - )
(18 #1) 20 VFI

N
= + 2π W

4 2 π W

1 x 18 e area of the 18th ring

(33) The experimental quantities of equation (31) are reducible to

(34)
(Vid. equation 13)
1

(35)
1

(36) 17

(37) The equation (33) becomes e + 21 a v l....

(38) + 15 = M - 1 1

(39)

..(25

..(26

)

..(27

) 2), (2),

rential stitutions

By substituting these values of the exponential in the equations (31), (32), (33), they take the following forms :F 99.25 -17 4.27 V-1= + 2 + N - 1 x 16..(40)

=

) 74.25 V-174.27 -1= £ 27 - 1x 17..(41)

1 7 4.2* V-TF 4.25 -T= £ 277 - 1 18..(42)

. -TF90.2 Now 2 + expresses the circumference, of which one half is on the positive side, and the other half on the negative side: in respect to the same diameter, - 2 expresses the same things taken in contrary directions ; consequently + 2 + and + 2 + are two different expressions of the same circumference, when the diameters are the same; they then represent the same rings; and the last three equations give by reduction the following :

4

+ π W

..(31

)

../322

a = 4

- 1) TO

I +NI x 16

.(43) 34 x 34 1 ) = £27 Vrix = £27 N

+ x 17...

.(44) = + +2+ N - 1 x 4 = +27 - 1 N

N 1 18.

•(45) These last equations are evidently identical, and the problem proposed is now completely resolved by the roots (25), (26), (27), which, expressed arithmetically, are

! 31

.(46) = 45

.(47) (= * = 57

.(48) Unity being the area of any circle, which is an essential remark; for if unity were an abstract number, this solution would be absurd. It is only true when unity is expressed by e + 2 * v -1 of any ring or central circle.)

Remark. Wallis, who devoted much time and attention to the proposed problem, resolved the equations (1), (2), (3), by approximation. (Wallis's Algebra, Chap. 62.) His roots are, a = 2,525,513,986,744,158

.(49) b = 2,969,152,768,619,848

.(50) c= 3,240,580,681,617,174

.(51) Comparing these roots with the roots (46), (47), (48), it is easily perceived that they have not the least relation to each other. In order to prove how nearly the roots (49), (50), (51), verify the

(33)

e to

(= the area

(52)

come so.

60 Solution of a Problem of Col. Silas Titus. (JAN.
equations (1), (2), (3), Wallis has substituted them in the place of
a, b, c, by which he obtained the following result :-

16 = a + bc= 16,000,000,000,000,000
17 = 1 + ac= 17,000,000,000,000,000
=

.(53)
18 = 0 + ab = 17,999,999,999,999,997
2

.....(54) However near Wallis's approximation may be, the equations which he obtained are not for that reason the less absurd, nor are they the less

Equations, whose two members are not equal and never can le

In order that the two members should be equal, it is necessary that the inequality of the units compensate the inequality of the numbers, as in the equation 21. = 40s.; this compensation takes

.
place if each member and its unit are in an inverse ratio to each
other. This is the case in my solution, where the height of each
ring, multiplied by the mean circumference between the two extreme
circumferences, give for the product the constant area of the central
circle, as I thus demonstrate, for we have for each ring the follow-
ing proportion:
* The sum of the radii of the extreme circumferences of the rings

: To the tangent drawn to the smallest of these circumferences
:: This tangent, (which is = the radius of the central circle,)
: The difference of the radii of the extreme circumferences of the

rings. (This difference being the height of the ring, 185.) From whence we have the following theorem.

In a series of concentric rings, each of whose areas are equal lo the central circle. The rectangle formed by the sum of the radii of the extreme circumferences of any one of the rings and its height, is equal to the square of the central circles. (56.) Now the area of each ring is equal the area of a trapezium, which has for its base the height of the ring, and for the mean height half the sum of the extreme circumferences. But this half is equal to a mean circumference between the two extreme circumferences. We can then transform theorem (56) into the following.

In a series of concentric rings whose ureas are equal to the area of the central circle, the rectangle formed of the height of each ring and the circumference, which is a mean between the two extreme circumferences, is equal the area of the central circle, which is the proposition I had to demonstrate.

General Corollary. The preceding resolution of equations (1), (2), (3), gives a complete solution of Gauss's problem, viz. “ To divide a circumference into 17 equal parts.” The dotted isosceles triangle 011i (fig. 1) whose summit is the centre O, has for its base the continued line 111; it is this base whose extremities are 1 1, which divides the

17 16 17

Th