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**Additional info for A Do-it-yourself Guide To Energy Star Home Sealing - Sealing Air Leaks And Adding Attic Insulation**

**Example text**

Let |z| = 1. We have |f (z) − Pn(m) (f )(z)| = |z| · |g(z) − Qn,m (f )(z)| = |g(z) − Qn,m (f )(z)|, (m) which by the uniform convergence in D1 of Pn (f ) to f and by the maximum modulus principle implies the uniform convergence in D1 of Qn,m (f )(z) to g(z). 5in bernstein Approximation by Complex Bernstein and Convolution Type Operators 34 Since g is continuous in D1 and |g(z)| > 0 for all z ∈ D1 , there exist an index n1 ∈ N and a > 0 depending on g, so that |Qn,m (f )(z)| > a > 0, for all z ∈ D1 and all n ≥ n0 .

E. for any 0 < β < α, there is n0 so that for all n ≥ n0 we have (m) Re z[Pn ] (f )(z) (m) Pn (f )(z) (m) > β > 0, for all z ∈ D1 . (m) Since Pn (f )(z) differs from Bn (f )(z) only by a constant, this proves the first part in (ii). For the second part, the proof is identical with the first part, with the only difference that instead of D1 , we reason for Dr . 5in bernstein Bernstein-Type Operators of One Complex Variable 35 uniformly in D1 . 2, there (m) exists n1 (m, f ) so that Bn (f )(z) be univalent in D1 for all n ≥ n1 (m, f ).

J=1 f (4) (z)[1 − 6z(1 − z)]z(1 − z) . 4! · 8 · n3 This identity follows from the identities Tn,1 (z) = 0 and + f (z) 8 T4n,2 (z) T2n,2 (z) 1 Tn,2 (z) − · = 0, +2· − · 2! 3 16n2 4n2 3 n2 8 T4n,3 (z) T2n,3 (z) 1 Tn,3 (z) f (z) − · = 0, +2· − · 3! 3 64n3 8n3 3 n3 8 T4n,4 (z) T2n,4 (z) 1 Tn,4 (z) f (4) (z) − · +2· − · 4 4! 5in Approximation by Complex Bernstein and Convolution Type Operators 44 =− f (4) (z)nz(1 − z)[1 − 6z(1 − z)] . 4! 3 for p = 2, but with explicit constants), will immediately give an explicit constant d for [4] the estimate |Ln (f )(z) − f (z)| ≤ d/n3 .