By Roy R.

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29) that rn(0) satisfies (0) −n 2rn+1 (0) 2rn−1 − n = n2 . 6 Continued Fractions 11 (0) and a similar relation holds for rn−1 . 31), some calculation gives us (1) 2rn+1 − (n − 2) (1) 2rn−1 − (n + 2) = n2 . 32) Once again, rn(1) ≈ n. 32) to get, after simplification, (2) (2) (2) (2) 16sn−1 sn−1 − 2(n + 4)sn+1 − 2(n − 4)sn−1 − 4 = 0. 32), multiply this last equation by 4, set sn(2) = rn(2) /22 , and add n2 to both sides to get (2) 2rn+1 − (n − 4) (2) 2rn−1 − (n + 4) = n2 . 33) We then have rn(1) = n + 22 /rn(2) .

However, Brouncker was very proficient in languages as well as mathematics. He did all his surviving mathematical work in association with Wallis, with the exception of his series for ln 2. In addition to the continued fraction for π , he wrote a short piece on the rectification of the 3 semicubical parabola y = x 2 , probably after seeing William Neil’s work. He also gave a method for solving Fermat’s problem of finding integer solutions of x 2 − Ny 2 = 1 for a given positive integer N. This solution can also be described in terms of continued fractions, but when Wallis wrote up Brouncker’s method, he did not use that form.

1) where sum Sn(p) = 1p + 2p + · · · + np . The work of Archimedes and al-Haytham showed that Sn(p) could be expressed as a polynomial in n; the asymptotic value simply yields the term with the highest power. Because Madhava and his school were primarily interested in integration, and thus in the highest power, they failed to note the full significance of the polynomial itself. In the seventeenth century, Fermat was very interested in asymptotic values, since a he too wished to evaluate 0 x p dx.