44 the assumed circumference be 5' then, according to the "Duhamelian theorem," dividing the circumference by the diameter, we have by can 44 14 44 $ cellation = 22 X 5 5 $ 14 7 true ratio of the circumference to the diameter of the given circle. 1 By Case 1, 1/2=the tangent of the given arc, and the radius is But there are 22 double triangles contained in the given polygon. 44 Squaring the radius and dividing, we have (2)2 = 2; then 7 ÷ 2 = 3.142857, or 34, which is the true ratio of the circum ference to the diameter of the given circle. But the radius is the √2, and the diameter is twice the square root of two; thus 1/2 x 2 = 2√2. 44 Then, by division, we have 1/2 21/2 = 34, the true ratio of the circumference to the diameter of the given circle. Again, by Case 2, the sine of the given arc is double triangle for double the number of sides. But there are 3114 sides to the given polygon. scribed polygon for double the number of sides. Now the tangent of the given arc is 1 and the unit of comparison 70' Therefore there are 30800 squares contained in the given polygon, Therefore there are 9800 squares contained in the square described 1 upon the radius, each of which is represented by 4900 Then 30800 ÷ 9800 = 3.142857, 3.142857, or 34, the true ratio of the circumference to the diameter of the given circle. But the area of the circle is said to be the rectangle under the radius; that is, the rectangle contained by the circumference and the radius. If it is meant for the circumference to be a straight line without regard to the circular figure, then the definition is true; but it is admitted by all former mathematicians that this straight line has never yet been found. Moreover, it is asserted that it is not likely that it ever will be found, because "innumerable attempts have been made to find a solution of this problem, but these attempts have been made in vain." If it is meant for the circumference of a circle to be the boundary of the circular figure known as the circle, then, in that case, it is not true" that the rectangle contained by the radius and the circumference is equal to the area of the given circle," as can very readily be shown. For, witness the following demonstration: Let the straight line fP, Figures 1 and 2, Plate 6, be the radius of the given circle √2, and let the straight lines PF, PE be the tan 1 2 7 For, by ARTICLE 6, √⁄2 × √2= area of the circumscribed triangle. 1 2 Then, dividing this area by the given radius, we have÷√2 = 72, which is said to be the arc of the given cirele, but which is, in fact, only the tangent of the given arc, a result too large, because the tangent is greater than its arc. Again, let the straight line fh, fm, be the radius of the given circle 12, and the straight lines gh ge, lm ln, be the sines of the given arc, each of which = 1 5 ; then will the triangles fsm, fsn, be the inscribed double triangles for double the number of sides. double the number of sides. Dividing this area by the given radius, we have which is said to be the arc of the given circle, but which, in fact, is only the sine of the given arc-a result too small, because the sine is less than its arc. Again, let the straight line fs be the given radius V 2, and let the straight lines sr, st, be the tangents of the given arc, each of which 1 Dividing this area by the given radius, we have 70' 701 2÷1/2 which is said to be the arc of the given circle, but which is, in fact, only the tangent of the given arc-a result too large, because the tangent is greater than its arc. Again, let the straight line fs be the given radius = √/2, and let the straight lines sr, st, be the sines of the given arc, each of which is triangle for double the number of sides; for X √2 2; then will the triangles fsr, fst, be the inscribed double 1 99 2 = the area of the inscribed double triangle for double the number of sides. 1 Then, dividing the area by the given radius, we have ÷√2 2 99 991 2, which, it is said, is the arc of the given circle, but which is, in fact, only the sine of the given arc-a result too small, because the sine is less than its arc. For the next step let us take the sine, radius, and tangent of Cases 3, 4, 5, 6, or 7, and if they do not furnish an example sufficiently small, let the result be carried as far as it is possible for human power and endurance to carry it; and if this again is too small, let us picture to our imagination an example small enough to satisfy the most mathematical exactitude; the relative ratio will still be the same, namely: dividing the area of the circumscribed triangle by the radius gives for the result the tangent, which is too large; and, dividing the area of the inscribed double triangle for double the number of sides gives for the result the sine, which is too small; but the inscribed and circumscribed double triangles, by all former methods, are either inscribed or circumscribed polygons, and whatever is true of one member of a class is true of the whole class; therefore the result obtained by any former method is either too great or too small; and if, as there is nothing to prevent the inscribed polygon from extending beyond the circle, nor to prevent the circumscribed polygon from coming within it, which is very possible, for the limit of the two polygons is the limit obtained; then the result obtained by all former methods must be too small; and, consequently, the rectangle under the radius will not give the true area of the given circle. Again, let the radius of the given circle be 1/2, and the tangent of 1 70 the given are be then, by ARTICLE 6, V2X 99 the secant of the given arc is 70; then, dividing the area of the cir cumscribed double triangle by the secant of the given arc, we have 1 701/2÷ 99 1 2 X 70 1 991/2 the sine of the given arc. But the area of the inscribed double triangle for double the number 1 of sides divided by the radius gives the same result, namely: 2: the sine of the given arc. 991 = Therefore the area of the circumscribed double triangle divided by the secant is equal to the area of the inscribed double triangle divided 1 by the radius, namely: 991/2, which is a result too small. But by the present method the sum of the tangents, which is the perimeter of the circumscribed polygon, vanishes together with that polygon. And the sum of the sines, which is the perimeter of the inscribed polygon, becomes the circumference of the circle at the same time that the inscribed polygon becomes the circle; therefore the secant vanishes and the cosine becomes the radius, which is the √2. Consequently (PROPOSITION 7, THEOREM), The true ratio of the circumference to the diameter of the circle is as 3.142857, or 34, is to 1; and, therefore, the last term of the ratio vanishes also. The following demonstration is taken from "Elements of Euclid," by James Thompson, LLD., pp. 124-5. Belfast: Symmes & McIntyre. London: Longman & Co., and Simpson & Co.: PROP. XIX. THEOR.—In numbers which are continual proportionals, the difference of the first and second is to the first, as the difference of the first and last is to the sum of all the terms except the last. If A, B, C, D, E be continual proportionals, A- B : C + D. For, since (hyp.) A: B:: B: C::C:D :: D: E, we have (Sup. 8) A : B :: A+B+C+D:B+C+D+ E. Hence (conv.) A: A-B:: A+B+C+ D: A-E; and (inver.) A — B: A:: A-E: A+B+C+D. It is evident that if A were the least term, and E the greatest, we should get in a similar manner, B -A: A:: E-A: A+B+C+D. Therefore, in numbers, etc., − & : 4 :: 4 : }. For = 75 =, and × }=}• Therefore the sum of the series + + 33 Cor. If the series be an infinite decreasing one, the last term will vanish, and, if S be put to denote the sum of the series, the analogy will become A - B A :: A: S; and this, if rA be put instead of B, and the first and second terms be divided by A, will be changed into 1-r: 1 :: A: S. The number r is called the common ratio, or common multiplier, of the series, as by multiplying any term by it the succeeding one is obtained. |