2.1. Newton's method for determining the branches of a plane curve 5 As our first approximation, we can use our first transformation to solve for y in terms of x and y\: y = Cl x5° + yi x6°. Now the second transformation gives us y = Cl x6° + c2xSo+^ + y2x6o+^. We can iterate this procedure to get the formal fractional series (2.3) y = cix6° + c2xSoJr^ + c 3 / 0 + ^ + ^ T + • • • . Theorem 2.1. There exists an io such that Si £ N for i io. Proof, ri = mult(/i(0,?/i)) are monotonically decreasing, and positive for all i, so it suffices to show that n = n+i implies (SjGN. Without loss of generality, we may assume that i = 0 and r*o = r\. /i(#i, yi) is given by the expression (2.1). Set g(t) = hfat) = Y, M*+*)'"• g(t) has degree r$. Since r\ = ro, we also have mult (#(£)) = ro- Thus g(t) = a0rotr°, and ] P a ^ =a 0 ro(^-ci) r o . In particular, since if has characteristic 0, the binomial theorem shows that (2.4) ev 0 _i ^ 0, where i is a natural number with i + So{ro — 1) = Soro- Thus 5 o G N. • We can thus find a natural number ra, which we can take to be the smallest possible, and a series p(t) = X * such that (2.3) becomes (2.5) y=p(X™). For n £ N, set Using induction, we can show that Pn{t) = J2bif' i = l mult(/(t m ,p n (t))-oo

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