By Garrett P.

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Topics in Computational Algebra

The most goal of those lectures is first to in short survey the basic con­ nection among the illustration thought of the symmetric staff Sn and the speculation of symmetric services and moment to teach how combinatorial equipment that come up evidently within the conception of symmetric services result in effective algorithms to precise quite a few prod­ ucts of representations of Sn by way of sums of irreducible representations.

Additional info for Hartogs' Theorem: separate analyticity implies joint (2005)(en)(5s)

Example text

25: 0 1 ! 1 1 1 1 1 C = t2 − t1 , (b) det B (a) det @ t1 t2 t3 A = (t2 − t1 )(t3 − t1 )(t3 − t2 ), t1 t2 2 2 2 t1 t2 t3 1 0 1 1 1 1 C Bt B 1 t2 t3 t4 C C (c) det B B t2 t2 t2 t2 C = (t2 − t1 )(t3 − t1 )(t3 − t2 )(t4 − t1 )(t4 − t2 )(t4 − t3 ). 29. 22. (a) By direct substitution: pd − bq aq − pc pd − bq aq − pc ax + by = a +b = p, cx + dy = c +d = q. ad − bc ad − bc ad − bc ad − bc ! 2; 0 2 4 0 10 10 ! 1 5 7 1 4 −2 1 4 det det = , y= =− . (ii) x = −2 6 3 −2 12 3 12 6 (c) Proof by direct0substitution, expanding all the determinants.

V1 (x, y) v2 (x, y) ! w1 (x, y) w2 (x, y) ! Distributivity: (c + d) " ! v1 (x, y) v2 (x, y) ! # (c + d) v1 (x, y) (c + d) v2 (x, y) ! =c ! v1 (x, y) v2 (x, y) ! +d ! # ! v1 (x, y) c d v1 (x, y) v1 (x, y) c d = = (c d) . v2 (x, y) c d v2 (x, y) v2 (x, y) Unit for Scalar Multiplication: ! v1 (x, y) v1 (x, y) . = 1 v2 (x, y) v2 (x, y) ! , w1 (x, y) w2 (x, y) ! 9. We identify each sample value with the matrix entry mij = f (i h, j k). In this way, every sampled function corresponds to a uniquely determined m × n matrix and conversely.

B) span { ex , e3 x }; (c) 2. 12. Each is a solution, and the general solution u(x) = c1 + c2 cos x + c3 sin x is a linear combination of the three independent solutions. 13. (a) e2 x ; (b) cos 2 x, sin 2 x; (c) e3 x , 1; (d) e− x , e− 3 x ; (e) e− x/2 cos 23 x, √ √ √ √ √ x x x x e− x/2 sin 23 x; (f ) e5 x , 1, x; (g) ex/ 2 cos √ , ex/ 2 sin √ , e− x/ 2 cos √ , e− x/ 2 sin √ . 14. (a) If u1 and u2 are solutions, so is u = c1 u1 + c2 u2 since u + 4 u = c1 (u1 + 4 u1 ) + c2 (u2 + 4 u2 ) = 0, u(0) = c1 u1 (0) + c2 u2 (0) = 0, u(π) = c1 u1 (π) + c2 u2 (π) = 0.