Then, dividing this area by the given radius, we have ÷ √2= 1 2 7 12, 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 = 1/2, and the straight lines gh ge, lm In; be the sines of the given 1 arc, each of which =; then will the triangles fsm, fsn, be the inscribed double triangles for double the number of sides. 1 ×√/2=√/2= area of the double inscribed triangle for 5V/2 double the number of sides. 1 Dividing this area by the given radius, we have √2÷√Z 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 = √2, and let the straight lines sr, st, be the tangents of the given arc, each of which 1 1. Dividing this area by the given radius, we have 70V/2÷1/2= 70' 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 1 equal to 991/2; then will the triangles fsr, fst, be the inscribed double triangle for double the number of sides; for X √2 1 99 area of the inscribed double triangle for double the number of sides. 1 99 Then, dividing the area by the given radius, we have ÷ √2 2 99 √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 the given arc be 1 70 ; then, by ARTICLE 6, 1/2 × 1 1 70 70V/2. And the secant of the given arc is 70; then, dividing the area of the circumscribed double triangle by the secant of the given arc, we have 70 1 1 2÷÷ 70V/2 99. 1 70 781/2X 99 991/2 the sine of the given arc. But the area of the inscribed double triangle for double the number of sides divided by the radius gives the same result, namely: the sine of the given arc. 1 991/2 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: 992, 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: A:: A—E; 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. 1 1 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., 2:4:: 4 :. For, and X}=} Then÷= X 4 = 1. Therefore the sum of the series++ 313 + 2401 6 79 49 7 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. PART SECOND. THE SQUARE ROOT OF TWO, OR THE COMMON MEASURE OF THE SIDE AND DIAGONAL OF THE SQUARE; BEING A SHORT, EASY, AND CONVENIENT METHOD OF FINDING EITHER THE SIDE OR DIAGONAL OF THE SQUARE, WHEN THE OTHER IS KNOWN, BY COMMON MULTIPLICATION AND DIVISION; ALSO, THE SQUARE ROOT OF TWO, BY DIVISION ALONE, TO ONE HUNDRED AND FORTY-FOUR DECIMAL PLACES. (135) |