Trigonometria Britannica

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## Chapter Four

Concerning Trisection.

For if the Subtended Chord of any Arc shall be given, and the Subtended Chord of the third [part] being sought, it is permitted to use the same source for our inspiration1 as previously, nevertheless the discovery is with a little more labour.

Let the given Subtending Chord of 36 Degrees be 061803398875.
Three of the Subtended Chord of 12:0': Degrees being equal to the given Subtended Chord of 36:0': and the Cube of the Subtended Chord 12:0' [i.e. being added together]: by the second consequence of the preceding chapter. This equation will be expressed thus2:

3 = 061803398875 + 1 or 3 - 1 = 061803398875

1. Given therefore the Subtended Chord being divided by three, and between the division always being added the Cube of the Quotient from that place [in the expansion]; finally the Quotient found will be the Latus [i.e. root] sought, truly the Third of the Subtended Chord sought. Especially therefore being noted the positions of the points of the Cubes, by beginning from the place of unity, with two places put in between: as you see here.

The first place is the place of unity towards the left; surely the place in which the radius has been placed.

Then being commenced with the Division by the given Third, which in the place of unity is not possible or for the first time being taken away: therefore Zero being placed in the quotient, and being sought whether and how much will be able to be reckoned in the nearby place: now 3 being found [ i.e. the initial divisor]. Being divided into the nearby place twice, I place therefore 2 into the Quotient, and of this Quotient the Cube is 8. Being added to the dividend at its own place, surely in the following place for the Cube; and by taking away the total the Triple of the Quotient found as you see here.

Then I progress asking how many [times] 3 [goes] into 2, the next place, and as it is not possible to take [anything] away, I place Zero in the Quotient, and I progress to the further remainder being added, found in the next three subsequent places, as the more appropriate nearest gnomon3 Cube to that place being added; and now I seek how many [times] 3 [goes] into 26, which can be had eight times: but where with the Gnomon the Cube will have been increased in to its place, the given divisor 3 will be able to have been taken away nine times. And so that we will be able to judge more properly which number should be placed in the [next] place in the Quotient, with the product from the given divisor 3, and the product from the square of the Quotient first found being tripled (which not inconveniently will be called the Divisor of the Cube) shall be added in the same place to the Quotient : most conveniently it will make the divisor being taken away smaller from being too large; as by the remaining numbers (which we will be able to call the Correct Divisor) we will be able to be surer of the number in the Quotient that should be placed to stand. These divisors therefore being brought together4, because it will hardly be able to done except both the one and the other should be put in its own place.

By having found the Correct Divisor, I go on and ask how many times 288 [goes] into 2603.
And with finding the number in the Quotient should be placed 9.

This Remainder gives 0 for the next place of the Quotient, and if the divisors should be brought together, the correct divisor will be 2868 which will give the Quotient 5 for the place after the zero that should be written.

Another Example

2. Let the given Subtended Chord of 150:00': Degrees be 1931851652578. The third of the Subtended arc being sought, namely 50:00': The equation being expressed thus:

3 = 1931851652578 +1

The first place of the root should be taken as large as possible. As here 8, which very much exceeds a third of the number 19 to be divided. Where indeed the Cube, namely,512 will have been added to the dividend, in the second place [lit. in the Place of the Cube one past the first Place], they will make 2443. And from this number the triple of the quotient 24 will be taken away. But if I should put 9 in the Quotient, the Cube of the Quotient, being added to the Dividend would have yielded 2660, from which the triple of the Quotient 27 is not able to be taken away6.

The next place of the Quotient shall be found thus. The product from the given divisor 3, in the nearest place of the Quotient has been taken away, and the product from the Square of the root triplicated 19200, in the same place of the Quotient (which has followed the particular part of the gnomon of the Cube): therefore the third given divisor should be made smaller, by taking away of the triplicate of the Square of the root being found in the usual way, being collected together as you see above [Table 4-4], and that remainder is 10800, being situated in the place of the divisor, which we are able to call the corrected Divisor, which scarcely ever fails [me] in extracting the Quotient; therefore of the division of 43851 by 10800. The Quotient will be 4, which gives the Gnomen of the Cube being added 80704*.

The next place of the Quotient is not being found by the corrected Dividend found above: but when the Square of the Root being found being increased, because the triple ought to have been taken away from the given Divisor 3, it is necessary the corrected Divisor being diminished. As with the root found 84, the Square 7056, of which the triple 21168: that ought to be taken away from the three, thus

3. However this same line 19318516525578 shall subtend two Arcs, the smaller one the semi-arc of 150:0': the remaining larger of 210:0': (since together being equal to the whole Periphery of the Circle) being sought a third of this Arc, indeed of 70:0' Degrees. The equation is the same as before:

3 = 1931851652578 +1

That Remainder of three [times] the root has exceeded the [sum of] the Cube of the same root and the given Subtending Chord. There are therefore for this remainder by three [times] the root always to be added on to the dividend, and the Gnomons of the Cubic being taken away. As you see here7.

In the next place in the Quotient has to be put 1, for if 2 is being put in this place, the Cubic Gnomon will be larger than it will be possible to be taken away from the remainder, it is allowed for the triple of the Quotient being added to the same.

The corrected divisor 63 can be taken away 5 times from the remainder 371, but the total will be more than the Gnomon that can be taken away; therefore let four be the figure placed in the quotient. The Gnomon to be taken away 150544.

The subsequent figures which you see here having been added on being found by the same way. And the Subtended Chord of 70:0': degrees sought will be 1.114715282702. And by this method we will extract the Subtended Chord of the Third parts if the Subtended Chord of the Triple shall have been given.

4. But if the whole Circle shall be added to the Arc, of which the Subtended Chord shall be given, we will be able to find the Subtended Chord of the third part of the total being put together from the Circle and the Arc given. As shall be given the Subtended Chord of 81:0': degrees, 1298896066660366.
The Sum of the Arc given and the Circle [is] 441 Degrees, 0': Of this sum the third part [is] 147 Degrees, 0:' of which I wish to find the Subtended Arc. Of this sought Subtended Chord the Cube being equal to [the sum of] three of the root of the same Cubic (or the Subtended Chord being sought tripled) and of the given Subt. Ch. 81:0': Thus the equation being expressed:

3 + 129889 =

Everything being set up as before8.

It is allowed by the proposition, three of the Quotient together with the given number ought to be equal to the Cube: nevertheless this will not come out until the whole Quotient shall have been found; which tripled and being added to the given Subtended Chord being equal to the whole Cube, with all the Gnomons to be taken together.

The corrected Divisor in this and preceding with the working always being increased from before being smaller as we see in this 783, 79443, 8024667, 803156928, 80319.
It is by this Divisor, the figure to be found being placed in the Quotient , which will give the Gnomon of the Cubic, etc. But with that Divisor being made small, the other places towards the left not being changed, we will be able to use for that Divisor, in the common way, by neglecting the Gnomon, and the Quotient will be for the same subsequent total figures, clearly the same which the other way of working would arrive at. 9 For:

Therefore , by the third consequence of the Third Chapter, for any given Subtended Chord of an Arc, we shall be able to find the Subtended Chord of the triple of the Arc, and (if the Arc triplicated should be more than the whole circumference) the Subtended Chord of the excess over the Circle. We will be able also to find the Subtended Chord of a Third of the given Arc, and the Subtended Chord of the sum of the total of the Circle and the Arc given.

Notes On Chapter Four

1 The phrase 'ab eodem fonte ...dimanet' Lit. 'may spread (abroad) or (in different directions) from the same source or spring', of which Briggs was obviously fond, as he uses it in the Arithmetica.

2 This is the first time Briggs has used the ' = ' sign; He has set up cubic equations before in the Arithmetica Logarithmica, Chapter 29, when dealing with the diagonals of the regular heptagon and nonagon. The number in the circle refers to the power of the variable, not given a letter by Briggs, but understood implicitly to be present. Thus, Briggs' cubics are :

3x = 0.61803398875 + x3, or 3x - x3 = 0.61803398875,

corresponding to Note 2 of Chapter 3, or 3x = A + x3 in general, where 0 < A < 2. It is convenient to write f(x) = x3 - 3x + A.

As x is 'small' in this case, it is legitimate to neglect the cubed term initially, and take the first approximation x1 to be A/3; or x0 = 0. Thus begins one of the great tasks performed by Briggs in this book: the setting up of the iterative method we have come to know as Newton's Method, for the solution of polynomial equations. This eventually leads to an immaculate table of sines at (5/8) spacing to 22 decimal places [Table 13-2], his answer to Ptolemy's Table of Sines. This table would seem to be the basis for all other tables of sines thereafter.

3 The Latin or the English gnomon is the pin of the sundial, which casts the shadow. Briggs uses the term as the name of the little procedure he goes through to produce the correction to the numerator f(xn) following an iteration: we will retain this name, rather as one has a special name for a subroutine in a computer program, to which Briggs' gnomon is similar.

4 Briggs solves the above cubic f(x) by an iterative approach which is now called Newton's method, or the Newton-Raphson method, for finding an approximate root of f(x). The origins of the method seem to be obscure: The earliest use of such an algorithm that I have come across, by the Iranian Al-Kashi (d. 1429) in the paper Al-Kashi's Iteration Method for the Determination of sin 1 , by Asger Aaboe, Scripta Mathematica, XX, (1954), pp 24-29. Viete certainly used a method for extracting cube roots in an iterative manner, that may be found in a translation of some of Viete's Opus Restitutae Mathematicae Analyseos, seu Algebra Nova by T. Richard Witmer, The Analytic Art, (The Kent State U.P., 1983, p.311 onwards): this work suffers from a lack of commentary! Viete's method is commented on by Goldstine, A History of Numerical Analysis ... (Springer-Verlag, (1977), p.66), while Whiteside informs us that Newton was unaware of the existence of the Trigonometria Britannica! Newton, however, spent time mastering Viete's work, and these annotations are expanded on in The Mathematical papers of Isaac Newton, edited by D.T. Whiteside, Vol. 1, Ch. 2, p. 63 onwards. The business of cutting an arc into a number of equal smaller arcs and relating their chord lengths had been attended to by Viete and Anderson (a great uncle of James Gregory of telescope fame, then Professor of Mathematics in Paris), Ad Angulares Sectiones, to be found in the Opera Mathematica of Francois Viete, p.287 - 304; (Olms, 1970). For whatever reasons, the present exposition by Briggs has been almost completely ignored by commentators on the history of Newton's method. As it appears, Briggs amended the method of Viete so that it becomes identical with the modern Newton-Raphson method. Again, Briggs has done this in his own way, but arrives at equations the same as those of Viete and Anderson, which he does not acknowledge.

According to this method, if x1 is an approximation for the root then a better approximation x2 is given by: x2 = x1 - f(x1)/f'(x1); there are situations where this algorithm does not converge. However, in the present situation, there is no difficulty. It is appropriate to proceed through Briggs' calculations bearing this correction mechanism in mind:

1st iteration: x0 = 0, and x1 = x0 + (x03 - 3x0 + A)/(3 - 3x02) = A/3 = 0.2 (to 1 sig. fig.) ;

2nd iteration: x1 = 0.2, and x2 = x1 + (x13 - 3x1 + A)/(3 - 3x12) =

Note that Briggs gets to this stage in two moves; first he corrects the dividend, then he corrects the divisor. Now the calculations have become more time consuming, and Briggs has evolved a scheme for making the corrections to the remainder or dividend and the divisor. For argument's sake, consider the above iterations, and let x2 = x1 + , where = 0.009, found by inspection; then x3 = x2 + (x23 - 3x2 + A)/(3 - 3x22) = x2 + ((x1+ A)3 - 3(x1+ A) + A)/(3 - 3(x1+ A)2) = x2 + (f(x1) + g(x1) - 3A)/(f'(x1) - 6x1A - 3A2), where f(x1) has already been found, and the gnomon g(x1,) is defined by:
g(x1 , A) = 3x12A + 3x1A2 + A3, and 'points' to the next correction.
The gnomon for Table 4-2 corresponds to g(0.2,0.009) = 3(0.2)2(0.009) + 3(0.2)(0.009)2 + (0.009)3 =0.00108 + 0.0000486 + 0.000000729 = 0.0011293; hence, f(0.2) + g(0.2,0.009) - 3(0.009) = 0.026033988 + 0.0011293 - 0.027 = 0. 000163317750, as required for the dividend.

In the case of the divisor, the modified divisor is f'(0.2) - 6(0.2)(0.009) - 3(0.009)2 = 2.88 - 0.0108 - 0.000243 = 2.8689. Thus, there is no doubt about the method being used. These are the computations that occupy a lot of the space in the table, set out methodically with squares, cubes, etc, of the approximate root , and with binomial coefficients that multiply these to give the correction gnomon on addition which is carried across to the point of application for the next division.

5 There is a typographical error here, where addition is mentioned while it is actually subtraction, which has been corrected. Numerous numerical errors in the tables, where clearly wrong numbers have been inserted by the typesetter, have been corrected without comment throughout the translation.

6 Note that cubics of the form f(x) = x3 - 3x + A cut the vertical axis at A, and have maximum turning point at x = -1 and a minimum turning point at x = 1. Hence, if 0 < A < 2, then there are three real roots. We seek the smaller positive root in case 2 outlined by Briggs. Recalling that Newton's Method 'works' by finding a tangent line that cuts the axis closer to the root than the original estimate, then this is not always possible, especially near a turning point. In the present case, where A = 1.9318516525578, the roots can be shown to be -1.9923893961812; 0 .8452365234576; and 1.1471528727235, and the positive roots the two positive roots are close together. Now, x1 = A/3 = 0.6 is not a good starting point here according to Briggs, for:
x2 = 0.6 + 0.547/1.92 = 0.88; x3 = 0.88 + (-.0266)/.6768 = 0.84:
presumably he was unhappy with the approximations oscillating by first increasing and then decreasing. However, in this case the oscillation does settle down; he must have had experiences of cases where the oscillations either grew larger and larger, or they settled into a loop. Hence, he is unwilling to chance it, and opts for a region where the convergence is well-behaved.

7 Newton's method fails when the either the first or a successive approximation passes through a turning point, as f'(x) = 0 leads to a correction at infinity. Briggs is aware of this, though he has chosen the first approximation as x1 = 1, for he has little choice in this case. Having done so, he evaluates f(x1), where he has chosen f(x) = 3x - A - x3;

Thus, f(x1) = 3x1 - A - x13 = 3 - (1.9318516525578 + 1) = 0.068148347422 : see [Table 4-8]. Note that the function has been inverted, to give a positive dividend, which is maintained throughout. Briggs cannot yet use the predictive property of the method, and resorts to setting
x2 = x1 + , where = 0.1; for a negative value of would converge on the previous root. The new value of the dividend is constructed, see [Table 4-9]: ; and the divisor can now be amended:

Subsequently, a scheme similar to the former is implemented on f(x) as defined here:

x3 = x2 + (x23 - 3x2 + A)/(3 - 3x22) = x2 + ((x1+ A)3 - 3(x1+ A) + A)/(3 - 3(x1+ A)2) = x2 + (f(x1) + g(x1) - 3A)/(f'(x1) - 6x1A - 3A2)

The working in the tables has been put into an abbreviated form, with the final root shown extracted.

8 Given (2R)sin(3/2) + (2R)3sin3(/2) = 3(2R)sin(/2) , as from note 2, Chapter 3. Then with 2R = 1, then 3 = 810 + 3600 = 4410, and = 1470;
2 sin(3/2) + (2 sin (/2))3= 3(2 sin(/2)), or 2 sin(3 147/2) + p3 = 3p, where p = 2 sin(147/2) is to be found. However, 2 sin(441/2) < 0, and the magnitude is given to be 1.29889: hence the equation shown.

9 The method eventually gives several more correct places than the one being sought at the left: in this way, the scheme is self-correcting, a great bonus to anyone involved in numerical analysis by hand. We show more of this with the quintic equations, where spreadsheet analysis brings out this feature.

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Ian Bruce January 2003