STØRMER NUMBERS PAUL HEWITT For nearly two millennia Archimedes method was the best technique for approximating π. In the south Indian state of Kerala during the late medieval period, mathematicians began developing infinite series expressions for trigonometric functions. These led to much faster and more accurate methods for hand computation, especially when combined with the Hindu decimal numeration. Unfortunately none of this work became widely known in the rest of the world. At that time, the Chinese and Islamic empires were in decline, and Europe was not quite ready for such sophisticated mathematics. Eurasia would be swept by Mongol-Turkic armies, then plagues, then new generations of Mongol-Turkic conquerors. When the dust had settled, much of Eurasia was ruled by inward-looking religious zealots, not much interested in expanding the frontiers of math or science. They were all at a very high level of scientific advancement, due to the work of the freer and more open societies that preceded them, but they viewed innovation as a political threat. One exception was Europe. After the trauma of the late Middle Ages, parts of Europe were ready to throw off the hegemony of the Roman church. This only added to the upheaval. So, while much of Eurasia was settling into a comfortable, conservative lifestyle, Europeans were fomenting artistic, religious, political, and scientific revolutions. Like most revolutions, these were messy and bloody. But Europe started this period much the poorer and more backward, compared to its Eurasian brethren, and ended up by far the richer and more powerful. They would dominate the entire globe during what we now call the Colonial Era. In this climate science and mathematics flourished. This culminated in the invention of Calculus. One of the most successful applications of Calculus is methods of approximation. Europeans rediscovered series expansions for trigonometric and other functions. In particular they found better ways to approximate π, based on identities for the arctangent function. There are lots of such identities. In the 20th century, the Swedish mathematician Størmer developed a nice way to generate and study these identities. There is a brief account of Størmer numbers in the fun Book of Numbers, by Conway and Guy [1]. His method use the gaussian integers, which is a system of complex arithmetic which has many of the properties of ordinary whole-number arithmetic. You can find out more about gaussian integers in [1]. The application of gaussian integers to the problem of approximating the real number π nicely illustrates Hadamard s aphorism. Date: 4 November 2008. 1
2 PAUL HEWITT 1. Gregory s series Arctangent is the inverse of the tangent function. In other words, (1) α = arctan(y/x) if tan(α) = y/x. α x Of course, there is an ambiguity here. We resolve this by assuming that α lies in the interval ( π/2, π/2). If t = tan(α) then dt (2) dα = sec2 (α) = 1 + tan 2 (α) = 1 + t 2. Hence the Inverse Function Rule tells us that dα (3) dt = 1 dt/dα = 1 1 + t 2. To expand arctangent in a series we will first expand its derivative, and then integrate term by term. This was the point of view eventually adopted by Newton: to him calculus was about manipulation of (possibly infinite) series, since derivatives and integrals of these are so straightforward. To expand 1/(1+t 2 ) we use the geometric series. We start with the finite version: 1 r is a factor of 1 r n, since r is a root of the latter. When we divide we find that 1 r n (4) 1 r = 1 + r + r2 + + r n 1. You can check this by multiplying both sides by 1 r. To get the infinite series we simply let n : this works when r < 1, since then both sides of equation (4) converge: (5) 1 r n 1 r 1 1 r = 1 + r + r2 + as n. If we combine equations (3) and (5) we get Gregory s series 1 : (6) arctan(t) = 1 t 2 + t 4 = t t3 3 + t5 5 This series converges fairly rapidly when t is small. It just barely converges when t = 1, yielding the identity π (7) 4 = 1 1 3 + 1 5 1 7 + 1 9 This series, first noted by Leibniz, converges far too slowly to be useful for computation, but it is quite pretty. y 1 Gregory died before Newton and Leibniz did their work. His book Geometriae pars universalis was the first appearance of a proof of what we now call the Fundamental Theorem of Calculus. Read more about Gregory at MacTutor [3].
STØRMER NUMBERS 3 2. Machin s identity Leibniz series (7) is an example of an alternating series, which means simply that the signs alternate between positive and negative. Note that Gregory s series (6) is alternating when t > 0 but not so when t < 0. In calculus we learned that if the absolute value of the terms of an alternating series decrease monotonically to 0, then the series converges. We deduced this by making an estimate of the error incurred when we truncate the series after (say) n terms: the remainder of the terms sum to something less (in absolute value) than the next term. For example, if we add up 1000 terms of Leibniz series (7) then the error is something less than 1/2001 = 0.0004997. This just gives 3-place accuracy. To get faster convergence, and thereby more accuracy for less effort, we want to express π in terms of smaller angles, ones whose tangents are much closer to 0. If we let (8) α n = arctan(1/n) then what we are seeking is an identity for α 1 = π/4 in terms of α n where n > 1. One example is the 1-2-3 identity: (9) α 1 = α 2 + α 3. There is a nice picture proof in [1], but this can also be proved using the addition formula for tangent: (10) tan(α ± β) = tan(α) ± tan(β) 1 tan(α) tan(β). When we apply this to α + α we obtain the double-angle formula. Let s apply the addition formula to α 2 + α 3. tan(α 2 + α 3 ) = 1 2 + 1 3 1 1 2 1 3 = 1. We can combine the 1-2-3 identity with the Gregory series, to obtain an approximation of π that works fairly well in practice: (11) π 4 = α 1 = 1 2 1 3 2 3 + 1 5 2 5 1 7 2 7 + 1 9 2 9 + + 1 3 1 3 3 3 + 1 5 3 5 By the alternating series estimate, the computation above is enough to get 4-place accuracy. We will see a series for arctangent in the next section which works much better with the 1-2-3 identity. A better identity for use with the Gregory series is due to Machin: (12) α 1 = 4α 5 α 239. The nice thing about Machin s identity is that it is relatively easy to compute the Gregory series for α 5 in the decimal system, while the Gregory series for α 239 requires very few terms for a given level of accuracy.
4 PAUL HEWITT 3. Euler s identities Euler had the habit of taking up the early work in calculus and improving it far beyond the combined effort of everyone before him. Series approximations of π were no exception. He improved on Gregory s series (6) by computing the Maclaurin series expansion of arctan(1/ x 1): 1 (13) arctan( ) = 1 x 1 x k=0 C k 1 (4x) k, where C k = 1 2k + 1 ( ) 2k. k Euler s series (13) is not alternating, and so the error estimate above does not apply. Nevertheless Euler s series can be seen to converge quite rapidly when x is large. One way to do this is to compare the series to an appropriate geometric series. Why is Euler s series (13) an improvement? Euler noticed that 2 = 5 1 and 3 = 10 1, while 1/(4 5) k = 1/20 k and 1/(4 10) k = 1/40 k are easy to compute in the decimal system. The integers C k are probably easiest to compute using Pascal s triangle. (Certainly not by computing factorials!) The term 1/ x can be computed efficiently in any of a number of ways, among them Newton s iteration. Euler also found several new identities among the α n, including two more that work well with his series: (14) α 1 = 2α 3 + α 7 = 5α 7 + 2α 18 2α 57. You compute α 7, α 18, and α 57 by taking x = 50, 325, and 3250 (respectively) in Euler s series (13). In fact, the computations of the terms in the series for α 18 and α 57 can be done essentially simultaneously. The difference between dividing by (4 325) k = 1300 k and by (4 3250) k = 13000 k is simply a matter of shifting the decimal place. These insights highlight Euler s practical genius. But it turns out to be quite easy to generate identities like these. In the 20th century a Swedish mathematician named Størmer 2 found a systematic way to generate all possible identities of this type. His method uses the arithmetic of the complex plane. 4. Størmer s method The angle α m is the argument of the complex number m + i. In the complex plane, the sum of the arguments is the argument of the product. Hence the angle sum α m + α n corresponds to the product (m + i) (n + i). Numbers of the form a + bi are called gaussian integers whenever a and b are ordinary integers. Gauss discovered that this expanded number system enjoys a key property of the ordinary integers, namely that every number except 0, ±1, and ±i factors into primes in an essentially unique way. Of course, an ordinary prime might be a gaussian composite. For example, 2 = (1 + i) (1 i) and 13 = (3 + 2i) (3 2i). On the other hand, some ordinary primes are also gaussian primes. For example, 3 is both an ordinary and gaussian prime. Let s prove this claim. 2 Størmer is known mostly for his mathematical theory describing the aurora borealis (the northern lights). You can read more about him at MacTutor [2].
STØRMER NUMBERS 5 Of course 3 factors as ( i) (3i), but this is akin to the factorization ( 1) ( 3) in the ordinary integers. The numbers ±1 and ±i are all units, meaning that they have multiplicative inverses. Whenever you have a unit u then every number x factors as (u 1 ) (ux). So factorizations involving units are not very interesting. There are only two directions on the real line, which we call positive and negative. Every real number x is (essentially) uniquely ±w, where w is positive. (There is one ambiguity at the boundary: 0 = +0. We won t fuss over this detail.) By contrast, there are infinitely many directions in the complex plane. Let s focus on the four cardinal directions: east, north, west, and south. 01arg = π/4 west 00 11 00 11 000 111 000 111 0000 1111 0000 1111 00000 11111 00000 11111 000000 111111 000000 111111 0000000 1111111 0000000 1111111 00000000 11111111 00000000 11111111 000000000 111111111 000000000 111111111 1111111111 1111111111 1111111111 east 1111111111 1111111111 000000000 111111111 000000000 111111111 00000000 11111111 00000000 11111111 0000000 1111111 0000000 1111111 000000 111111 000000 111111 00000 11111 00000 11111 0000 1111 0000 1111 000 111 000 111 00 11 00 11 01 north south arg = π/4 The eastern region 3 contains the numbers whose argument is between ±π/4. Multiplication by i rotates the regions π/2 counter-clockwise. Every complex number z is (essentially) uniquely i k w, where w is eastern. (Again, we will not fuss too much deciding which numbers on the boundary lines lie in which regions. A little ambiguity is sometimes a healthy thing.) Now we can state the Unique Factorization Property more precisely: Every eastern gaussian number factors into eastern gaussian primes, uniquely except for order. Since arguments add when complex numbers are multiplied, and since the argument of 3 is 0, the only possible eastern factorization of 3 is (ax + ayi) (bx byi) = ab(x 2 +y 2 ), where a, b, x, and y are all nonnegative integers. Since 3 is an ordinary prime and there is no way to write 3 as a sum of two squares of positive integers, it must be that y = 0, x = 1, and either a = 1 or b = 1. I want to expand a bit on the fact that 3 is not the sum of two ordinary integral squares. For a small number such as 3 this can be checked by considering all of the possibilities. What about for a large prime, such as 397427 or 917617? Is either of these a sum of two integral squares? Gauss used his expanded number system to answer this question completely: An ordinary prime cannot be expressed as the sum of two ordinary integral squares if and only if it leaves a remainder of 3 when divided by 4. 3 In [1] they refer to this region as positive, but I feel this is a bit dangerous. The problem is that you might be lured into the trap that every complex number is either positive or negative, or that the product of positive complex numbers is positive. Both of these are false, and so the word positive does not seem very apt.
6 PAUL HEWITT Thus, 397427 is not a sum of two ordinary integral squares but 917617 is. This theorem is part of a much larger theory, called quadratic reciprocity. This theory answers many questions about quadratic diophantine equations. It was regarded as the pinnacle of number theory at the beginning of the 20th century. Much of the effort of 20th century number theorists was directed towards generalizing Quadratic Reciprocity to higher-degree equations. We will resist the temptation to discuss Gauss theory further. Instead we will show how to use gaussian integers to find all additive relations between the Gregory numbers α n. Rather than state a precise and elaborate theory, we will simply work out a couple of examples in detail. A further example can be found in [1]. The one thing to keep in mind as we work thru the examples is that a gaussian factor of a + bi corresponds to an angle summand of arctan(b/a). Thus an ordinary integral factor corresponds to an angle summand of 0. That is, ordinary integral factors can be ignored when we interpret a gaussian product as an angle sum. Finally, note that α n = α n this is the argument of n i. Let s start with a very simple example: decompose α 23/15 = arctan(15/23) into a sum of angles ±α n, where n is an ordinary integer. We look for factors of 23+15i of the form n ± i. The trick is to find the smallest n which will cancel the largest prime factor of 23 2 + 15 2 = 754 = 2 13 29. For this we need the smallest n such that 29 divides n 2 + 1. It turns out that there is always an n not exceeding 29/2. We could simply check all of the integers up to 14, but we outline a more systematic way to do this, which is usually slightly faster as well. By Gauss theorem, 29 is a sum of two squares. A quick check of the numbers less than 29/2 = 3. reveals that 29 = 5 2 + 2 2. The extended euclidean algorithm, applied to the pair 5, 2 tells us that 1 = 1 5 2 2. Now if we set n = 2 5+1 2 = 12 then 29 divides n 2 + 1 = 145. Is there a smaller n? In this case, no. In general at this step the smallest n can be found first by finding the remainder r of n 29, and then taking the smaller value of r and 29 r. In this case, r = n = 12, and so there is no smaller value. Next we compute (23 + 15i) (12 ± i), and look for a factor of 29: (23 + 15i) (12 + i) = 261 + 203i = 29 (9 + 7i) (23 + 15i) (12 i) = 291 + 157i The first of these equations tells us that (15) α 23/15 + α 12 = α 9/7. Now repeat: 9 2 + 7 2 = 130 = 2 5 13 13 = 3 2 + 2 2 1 = 1 3 1 2 5 = 1 3 + 1 2 (9 + 7i) (5 + i) = 38 + 44i (9 + 7i) (5 i) = 52 + 26i = 26 (2 + i)
STØRMER NUMBERS 7 This last line tells us that (16) α 9/7 α 5 = α 2. If we combine the results (15) and (16) we find that (17) α 23/15 = α 2 + α 5 α 12. It turns out that none of the α n on the right-hand side of equation (17) can be decomposed further. This is because, for each of these n, the largest prime factor of n 2 + 1 is already at least 2n: 2 2 + 1 = 5, 5 2 + 1 = 2 13, 12 2 + 1 = 5 29. In fact, we chose these n for precisely this property. Størmer proved that if n is an ordinary integer then α n can be decomposed precisely when the largest prime factor of n 2 +1 is smaller than 2n. Conway and Guy refer to numbers which cannot be decomposed as Størmer numbers in their book [1]. For the second example we decompose α 57. 57 2 + 1 2 = 3026 = 2 17 89 89 = 8 2 + 5 2 1 = 2 8 3 5 34 = 3 8 + 2 5 (55 + i) (34 + i) = 1869 + 89i = 89 (21 + i) (55 + i) (34 i) = 52 + 26i This next-to-last line tells us that (18) α 55 + α 34 = α 21. Since 21 2 + 1 = 442 = 2 13 17, 21 is not a Størmer number, and so α 21 can be further decomposed: This last line tells us that 17 = 4 2 + 1 2 (21 + i) (4 + i) = 83 + 25i (21 + i) (4 i) = 85 17i = 17 (5 i) (19) α 21 α 4 = α 5. When we combine the results (18) and (19) we find that (20) α 55 = α 4 α 5 α 34. Exercises (1) Prove Machin s identity (12) in the following three steps. (a) Compute tan(2α 5 ), using the double-angle formula. (b) Compute tan(4α 5 ), using part (a) and the double-angle formula. (c) Use part (b) and the addition formula to show that tan(4α 5 α 239 ) = 1. (2) Derive Machin s identity (12) by applying Størmer s method to decompose α 239. (3) Give a geometric proof that α 1 = α 2 + α 3 based on the diagram on p 241 of [1].
8 PAUL HEWITT (4) For each of the following approximations, determine how many terms are needed to achieve 10-place accuracy. (a) The 1-2-3 identity (9). (b) Machin s identity (12). (c) Euler s identities (14). (5) Use Machin s identity (12) with Gregory s series (6) to compute π to 10 decimal places by hand. (6) Use the 1-2-3 identity (9) together with Euler s series (13) to compute π to 10 decimal places by hand. References [1] J Conway and R Guy, The Book of Numbers. Copernicus: 1996. [2] MacTutor biography of Størmer: http://www-history.mcs.st-and.ac.uk/biographies/stormer.html. [3] MacTutor biography of Gregory: http://www-history.mcs.st-and.ac.uk/biographies/gregory.html. University of Toledo