By Rotman J.J.
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Example text
66) and so R(x, y) = A(x 1/q , x −β/q (y − φ(x 1/q ))). 69) where α = β − q and, since φ has degree β − h, where h ≥ 1, ψ(x) = −x β φ(1/x) is a polynomial of the form ψ(x) = cx h + · · · , where c = 0. This may be regarded as the inverse of the formula A(s, t) = P(s −q, φ(s)s −q + ts α ). 71) and, if q ≥ β − h, y remains bounded on this curve as x → 0. e. A(x, 0) is a bounded polynomial as x → ∞. This implies A(x, 0) is constant, and so P(x q, y) is constant on the curve. 3 Real analytic polynomials at infinity 19 and in particular β ≥q +2 and α > h.
31) k=0 The terms of the matrix are given by wi, j = b j−i wi, j = c j−i+n wi, j = 0 for i ≤ j ≤ n + i and 1 ≤ i ≤ n, for i − n ≤ j ≤ i and n + 1 ≤ i ≤ 2n, otherwise. 32) As recorded earlier these relations imply that the maximum degree of wi, j is given by degree of wi, j ≤ n− j +i −j + i (i ≤ j ≤ n + i and 1 ≤ i ≤ n), (i − n ≤ j ≤ i and n + 1 ≤ i ≤ 2n). 33) Otherwise degree of wi, j = −∞ (so trivially satisfies the above inequalities). 34) i=1 where the sum ranges over all permutations φ(i) of the numbers 1, 2, .
We compare two terms by considering for a suitable global subsequence the ratio of the absolute values of the two terms. If the limit of the ratio is finite and non-zero, we say the terms are comparable. If the limit is zero, then the upper term of the ratio is smaller than the lower term; and vice versa. In this way we can order the groups of comparable terms and pick out the largest grouping in this ordering. It 20 1 The algebra of polynomials is clear that the largest group must contain at least two terms, for if it contained only one, then this term would dominate the sum, and so the sum would tend to ∞.