Deformations isomonodromiques et varieties de Frobenius by Sabbah C.

By Sabbah C.

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If, Let (x0 , y0 ) be a critical point of f . Then y0 2 P(x0 ) 0. Thus, a critical furthermore, (x0 , y0 ) ∈ Cf , then y0 P (r) 0. point of f on Cf is of the form (r, 0), where P(r) Differentiating once more, we see that (r, 0) is a saddle point if P (r) > 0. We now look at these conditions on P from an algebraic point of view. P(r) 0 means that r is a root of P. By the factor theorem, 0 the root factor (x − r) divides P. 9 38 3. Rationality, Elliptic Curves, and Fermat’s Last Theorem divides P.

K + 3)! 1 1 1 + + + ··· k+1 (k + 1)(k + 2) (k + 1)(k + 2)(k + 3) < 1 1 1 + 2 + 3 + ··· 2 2 2 Clearly, k! k j 0 1 −1 1 − (1/2) 1. 1/j! e. But this implies that e is irrational. ¬ Indeed, assume that e a/b, a, b ∈ N. e. ¬ Corollary. eq is irrational for all 0 q ∈ Q. Proof. eqm . Now choose m to If eq is rational, then so is any power (eq )m be the denominator of q (written as a fraction) to get a contradiction to Theorem 1. Remark. Q is dense in R in the sense that, given any real number r, we can find a rational number arbitrarily close to r.

4 2− 3+1 B12 How well do A12 and B12 approximate 2π? Does this method surpass Archimedes’ approximation of π using 96-sided polygons? √ 13. Evaluate the expansion of tan−1 at 1/ 3, and obtain π 6 1 1 1 1 + 2 − 3 + ··· . ) 14. Define the rth Gregory number tr , r ∈ R, as the angle (in radians) in an uphill road that has slope 1/r. Equivalently, let tr tan−1 (1/r). ) ♦ Use Størmer’s observation11 that the argument of the complex number a + bi is ta/b (and additivity of the arguments in complex multiplication) to derive Euler’s formulas t1 t2 + t3 , t1 2t3 + t7 , t1 5t7 + 2t18 − 2t57 .

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