Dr. Katz DEq Homework Solutions 102

# Dr. Katz DEq Homework Solutions 102 - isnot aninteger Q.E.D...

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102 N.M. Katz: the points x = 0, 1, 2, and hence (~ + fl x) dx dR(x) (6.8.6.8) R(x) R(x) would be holomorphic at one of the points x = 0, 1, 2. In view of (6.8.6.5), this is absurd, hence R(x) is a polynomial of degree at least three. Because R(x) is a polynomial of degree at least three, it follows that (~ + fl x) dx (6.8.6.9) R(x) is holomorphic at x = oo, hence has zero residue there. Clearly, the residue at x--- oo of [ a/n bin e/n \ E ~-+~-~_1+~-~_2] dx=Ea/n, dX+fb/n'x d(X-x_ll) (6.8.6.10) d(x - 2) +~c/n x-2 is just -Ea/n-fb/n-gc/n, while the residue at x= oo of dR(x) (6.8.6.11) R(x) is -degree (R (x)). By (6.8.6.5), we must have (6.8.6.12) - degree R (x) + E a/n + ~ b/n + ~ c/n = 0 which is absurd' as f ( a+b+c)n
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Unformatted text preview: isnot aninteger. Q.E.D. 6.9. Conclusion of the Proof of 6.2 (6.9.0) Let a, b, c, n be integers, n> 1, and suppose that the hyper- geometric equation with parameters a/n, b/n, c/n has two "rood p" solutions for almost all p (i. e., suppose a/n, b/n, c/n verify (6.2.6)). We must prove that E(a/n, b/n, c/n) has a full set of algebraic solutions (cf. (5.4.2.1)), in order to conclude the proof of 6.2. (6.9.1) As we saw in 6.3, this is the case if any of the exponent differences (6.9.1.1) 1 - c/n, c/n - a/n - b/n, a/n - b/n is an integer. (6.9.2) We thus assume that none of the exponent differences is an integer; then 6.6.0 implies that none of the numbers (6.9.2.1) a/n, b/n, c/n - a/n, e/n - b/n...
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## This note was uploaded on 12/21/2011 for the course MAP 4341 taught by Professor Normankatz during the Fall '11 term at UNF.

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