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Unformatted text preview: Mathematics 250 Fall 2006 Selected Solutions, Assignment #6 Hubbard and Hubbard, Problem 3.3.5: Find the cardinality of the set of multiexponents of degree m in with n entries. (Equivalently, find the number of monomials of degree m in n variables. Equivalently, find the dimension of the vector spact P m ( R n of homogeneous polynomials of degree m in n variables.) Response: Let D ( m, n ) be the number of multiexponents of degree m with n entries. One way to compute D ( m, n ) would be by induction  find a formula that expresses D ( m, n ) in terms of D ( k, ` ) for “smaller”. pairs ( k, ` ). For example, a given multiexponent may have a zero for its last entry. Then it could be considered a multiexponent in only n 1 entries. It would still have dgree m . If the last entry is nonzero, then if we can subtract 1 from it, and we will get a multiexponent of degree one less. This shows that D ( m, n ) = D ( m 1 , n ) + D ( m, n 1). This gives a recursive formula for computing D ( m, n ). It may remind one of the recursive formula for the binomial coefficient n k , denoting the number of way to choose a set of k objects out of a collection of ` objects. Binomial coefficients satisfy the recursive relation ` k = ` 1 k + ` 1 k 1 . If we set D ( m, n ) = m + n m , then D ( m, n 1) = m + n 1 m and D ( m 1 , n ) = n + m 1 m 1 , so the recursion relation fits. However, the small degrees don’t work. We have D ( m, 1) = 1 for all m , while D ( m, 1) = m + 1 m = m + 1. However, if we adjust the proposal to D 00 ( m, n ) = m + n 1 m 1 , then we still get the right recursion, and now D 00 ( m, 1) = m m = 1. We can also check that D 00 (0 , n ) = n 1 = 1 = D (0 , n ). Since D ( m, n ) is completely determined by its recursion relation and the side conditions D (0 , n ) = 1 = D ( m, 1), we conclude that D ( m, n ) = D 00 ( m, n ) = m + n 1 m . Remark: This connection between the dimensions of spaces of homogeneous polynomials and binomial coefficients is quite striking, and might provoke one to wonder if there is a way to think about multiexponents in a way that makes the connection appear natural, rather than being the result of some clever manipulations.in a way that makes the connection appear natural, rather than being the result of some clever manipulations....
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 Fall '06
 RogerHowe
 Math, Exponents, Derivative, Sin Cos, sin cos r2

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