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By Mary Gray

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The most function of those lectures is first to in short survey the elemental con­ nection among the illustration concept of the symmetric staff Sn and the speculation of symmetric capabilities and moment to teach how combinatorial equipment that come up obviously within the idea of symmetric capabilities bring about effective algorithms to precise a number of prod­ ucts of representations of Sn when it comes to sums of irreducible representations.

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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.

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