Mathematical Analysis

# Goursat E.'s A course in mathematical analysis. - part.1 A complex PDF

By Goursat E.

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Additional resources for A course in mathematical analysis. - part.1 A complex variable

Sample text

H(m, n). n may be From the results of the previous sections we know that each p ∈ Hm n−1+m n written as a linear combination of dim Hm = n−1 ridge monomials (a · x)m n with certain given distinct directions. Thus h(m, n) ≤ dim Hm . We can also ask this same question with respect to all polynomials of degree m in n variables. , i g(m,n) pi (ai · x), p(x) = i=1 and directions a , i = 1, . . , g(m, n)? 19 it follows that h(m, n) ≤ g(m, n) ≤ n dim Hm . But this upper bound is not sharp. We will, in fact, prove that Π1m , h(m, n) ≤ g(m, n) ≤ i n−2+m n ).

N} we must have k > j . 3, we have that the ridge monomials {(k · x)m : k ∈ Ωm } are linearly independent. Since n the cardinality of Ωm is exactly dim Hm , we have that these ridge monomials m n . 10 By a linear transformation this same result holds if we replace Ωm by n ki bi : k ∈ Zn+ , |k| = m , Ωm = i=1 where the b are any n linearly independent vectors in Rn . 8 by using induction and, for example, choosing c = e1 . 1 Homogeneous Polynomials 43 if and only if i = j. If we let Gj (x) denote the linear polynomial that vanishes on Gj , then the m qj (x) = Gj (x), j = 1, .

P(x) := |k|≤m n+1 n+1 Thus p ∈ Hm . Similarly, if p ∈ Hm , then p(x) := p(x, 1) ∈ Πnm . n+1 results on Hm can be translated to results on Πnm and vice versa. Here are some results valid for Πnm obtained from the results of the As such, previous section and homogenization. 4 we obtain the following. 20 Let ai , i = 1, . . , r, be vectors in Rn . Then for every choice of real values {αi }ri=1 there exists a p ∈ Πnm satisfying p(ai ) = αi , i = 1, . . , r, if and only if the ridge polynomials {((ai · x) + 1)m }ri=1 are linearly independent.