In 1671 James Gregory mentioned in correspondence that he had found the inverse tangent of x equals the area under the curve of 1/(1+t^2) between t=0 and t=x. He didn't include his derivation, but it's not too difficult to see how, with the ideas of slopes and quadratures of functions that Gregory and others were developing at the time, this result could be found. Beginning with sin(u) t = tan(u) = ------ cos(u) we know that the slope of t versus u (in modern notation) is the derivative dt -- = sin(u) (-1/cos(u)^2) (-sin(u)) + (1/cos(u)) cos(u) du sin(u)^2 + cos(u)^2 = --------------------- = tan(u)^2 + 1 cos(u)^2 Therefore, since t = tan(u), we have dt/du = t^2 + 1, and so du 1 -- = --------- dt 1 + t^2 To invert the function t = tan(u) we need to express t in terms of u, which we can do by multiplying the above equation by dt and integrating both sides / / 1 | du = | ------- dt / / 1 + t^2 which is Gregory's result, i.e., we have x / 1 arctan(x) = | ------- dt / 1 + t^2 t=0 Indeed, Leibniz derived this same result independently just a few years later (around 1674). Now, by simple division of polynomials he also knew that 1 ------- = 1 - t^2 + t^4 - t^6 + t^8 - ... 1 + t^2 Furthermore, although this was prior to development of calculus proper by Newton and Leibniz, several people (including Fermat, Cavalieri, Pascal, etc) had already noticed that the area under the curve t^n from t=0 to x is (x^(n+1))/(n+1). Thus Gregory had found an infinite series for the inverse tangent of x: x^3 x^5 x^7 arctan(x) = x - --- + --- - --- + .... 3 5 7 (Some evidence suggests that this series was actually known in India by about 1500, but the discoverer's identity is not known, nor do we know by what method the result was derived. See Rajagopal, On Medieval Kerala Mathematics, 1986.) A little later, thinking along these same lines, Leibniz noted that since arctan(1)=pi/4 we have the remarkable relation pi 1 1 1 1 1 ---- = 1 - --- + --- - --- + --- - --- + ... 4 3 5 7 9 11 It's interesting to consider what Leibniz (or Gregory) might have done with this result. Unburdened by any concerns about re-arranging the terms of conditionally convergent series, they might well have noticed that the sum could be expressed (at least formally) as a product of geometric series in inverse primes pi / 1 1 \ / 1 1 \ / 1 1 \ -- = ( 1 - - + - - ... )( 1 + - + -- + ... )( 1 - - + -- - ... )... 4 \ 3 9 / \ 5 25 / \ 7 49 / Infinite products were certainly not unknown at this time. For example, Wallis had already given the closely related infinite product pi 2 4 4 6 6 8 8 -- = - - - - - - - ... 4 3 3 5 5 7 7 9 Anyway, notice that each geometric series has either alternating signs or is strictly additive, accordingly as p is congruent to -1 or +1 (mod 4). Of course, each of these geometric sums converges, and is given by 1/(1 +- 1/p), so throwing cauchy.. er, caution, to the wind, we could express Leibniz's series as the infinite product pi / 1 \ / 1 \ / 1 \ / 1 \ / 1 \ -- = ( ------- )( ------- )( ------- )( -------- )( -------- ).. 4 \1 + 1/3/ \1 - 1/5/ \1 + 1/7/ \1 + 1/11/ \1 - 1/13/ where the product is taken over all odd primes p, and the sign in the denominator is + or - depending on whether p is congruent to -1 or +1 modulo 4. This isn't the same as Wallis's product, as can be seen from the first few factors pi 3 5 7 11 13 17 -- = - - - -- -- -- ... 4 4 4 8 12 12 16 Numerically this product appears to converge, albeit very slowly. In any case, considering that Leibniz could digest the notion that 1+2+4+8+.. = -1 based on the geometric series 1 + x + x^2 + x^3 +.. equaling 1/(1-x), it's hard for me to imagine him being squeamish about this product. Now here's the interesting part (to me, least). Suppose we ask what happens if we reverse the signs in our infinite product. In other words, what if we use plus signs for primes congruent to -1 (mod 4), and minus signs for primes congruent to +1 (mod 4)? This gives the infinite product / 1 \ / 1 \ / 1 \ / 1 \ / 1 \ ( ------- )( ------- )( ------- )( -------- )( -------- ).. \1 - 1/3/ \1 + 1/5/ \1 - 1/7/ \1 - 1/11/ \1 + 1/13/ which, if we expand it into a sum, is 1 1 1 1 1 1 1 1 1 1 1 + --- - --- + --- + --- + --- - --- - --- - --- + --- + --- +... 3 5 7 9 11 13 15 17 19 21 where the sign of 1/n is plus or minus accordingly as n has an even or odd number of prime divisors (counting multiplicities) congruent to 1 (mod 4). It's easy to form the conjecture based on numerical evidence that this series (and the infinite product) converges on pi/2, i.e., exactly twice the former series (and product). Now, I don't know of any trigonometric interpretation of this series, analagous to Gregory's arctangent expansion for Leibniz's series, but there might be one. Also, at the risk of introducing some actual math into the discussion, I might just mention Dirichlet's celebrated class number formula h pi inf / D \ 1 ------------ = SUM ( ) --- 2 sqrt(|D|) n=1 \ n / n where the quantity in parentheses is 0 if n has a common factor with 2D, and otherwise it is the Jacobi symbol. With D=-1 this gives Leibniz's series, showing that the class number h equals 1 for D=-1. On the other hand, with D=-5 the class number is 2, and Derichlet's formula gives pi 1 1 1 1 1 1 1 1 ------- = 1 + --- + --- + --- - --- - --- - --- - --- + --- + ... sqrt(5) 3 7 9 11 13 17 19 21 where the sign of 1/n is positive if n = 1,3,7,9 (mod 20), and negative if n=11,13,17,19 (mod 20). But I digress. Returning to our hypothetical Leibniz and his reversal of the signs in the infinite product that formally corresponds to his famous series for pi/4, notice what happens if we multiply his two results together. We immediately have pi^2 / 1 \ / 1 \ / 1 \ / 1 \ ---- = ( --------- )( --------- )( --------- )( ---------- )... 8 \1 - 1/3^2/ \1 - 1/5^2/ \1 - 1/7^2/ \1 - 1/11^2/ 1 1 1 1 1 1 1 = 1 + --- + --- + --- + --- + ---- + ---- + ---- + ... 3^2 5^2 7^2 9^2 11^2 13^2 15^2 which is the sum of the inverses of all the ODD square numbers. Furthermore, since it's obvious that the sum of the inverses of the EVEN squares is simply 1/4 times the sum of the inverses of ALL the squares, we have S(all) = S(odd) + S(even) = S(odd) + S(all)/4 and so S(all) = (4/3) S(odd). (Of course, this also follows simply from multiplying the product involving the odd primes by the factor 1/(1 - 1/2) = 4/3.) Therefore, our hypothetical (and rather reckless) Leibniz has found that the sum of the inverses of all the squares is pi^2 / 6, anticipating Euler and solving the problem that defeated not only the historical Leibniz (judging from his lack of response to Oldenberg's query) but also his disciples the Bernoullis, who after much effort were able to prove only that the sum of the inverse squares converges to a value less than 2, but not to determine the value. It remained for Euler to finally solve the problem, using means that were, if anything, even more reckless and haphazard than the reasoning above (although, to be fair, he did later tidy things up). For more on this topic, see Factoring Zeta.

Return to MathPages Main Menu