Algebraic Approximation: A Guide to Past and Current - download pdf or read online

By Jorge Bustamante

ISBN-10: 3034801939

ISBN-13: 9783034801935

This booklet comprises an exposition of numerous effects similar with direct and communicate theorems within the concept of approximation through algebraic polynomials in a finite period. additionally, a few proof referring to trigonometric approximation which are beneficial for motivation and comparisons are integrated. the choice of papers which are referenced and mentioned record a few tendencies in polynomial approximation from the Fifties to the current day.

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Extra info for Algebraic Approximation: A Guide to Past and Current Solutions

Example text

27) was improved by Garkavi [133]. Set (k) Cn,r (f ) = inf max Tn ∈Tn 1≤k≤r f (k) − Tn En∗ (f (k) ) and Cn,r = sup f ∈W r (1,[0,2π]) Cn,r (f ). Garkavi proved that Cn,r = 4 (ln(p + 1)) + O(ln ln ln p)), π2 where p = min{n, r} and f (r) − Tn(r) π Cn,r En (f (r) ). 2 ≤ nr f − T n + 1 + One of the first results on simultaneous approximation is due to Gelfond in 1955. 2 (Gelfond, [141]). If f ∈ C m [a, b], for n ≥ n0 , there exists Pn ∈ Pn such that f (k) − Pn(k) ≤ C 1 nm−k ω f (m) , 1 n (0 ≤ k ≤ m).

There exists a constant R with the following property: let a ≥ 0 be a real number, r ≥ 1 an integer and ω a modulus of continuity. If a polynomial Pn satisfies the inequality | Pn (x) |≤ (Δn (x))r ω(Δn (x)) + a, x ∈ [−1, 1], then | Pn (x) |≤ R (Δn (x))r−1 ω(Δn (x)) + a(Δn (x))−1 , x ∈ [−1, 1]. 8. Simultaneous approximation 39 The last result was proved by Lebed [225] in the case a = 0. Other proofs were given in [107] and [379] (p. 219–226). According to Teliakovskii [371], for a > 0 the proof can be obtained with arguments similar to the one used in [379].

The End Points Effect Then | f (x) − pn (x) | ≤| f (x) − Fh (x) | + | Fh (x) − pn (x) | ≤ ω(f , h)+ | Fh (x) − pn (x) | . 35) There exist polynomials Qm such that | Fh (x) − Qm (x) |≤ C Δm (x) ω(Fh , Δm (x)). 36) Now, we use the representation ∞ Fh (x) − pn (x) = [Qns (x) − Qns−1 (x)] + Qn (x) − pn (x). 37) s=1 In this case we have | Qns (x) − Qns−1 (x) |≤ C Δns (x) ω (Fh , Δns (x)) . 9 (with a = 0) we obtain | Qns (x) − Qns−1 (x) |≤ C1 ω (Fh , Δns (x)) . 34) to estimate the difference Qn − pn . In fact | Qn (x) − pn (x) | ≤| Qn (x) − Fh (x) | + | Fh (x) − f (x) | + | f (x) − pn (x) | 1 ≤ C Δn (x) ω(Fh , Δn (x)) + hω(f , h) + AΔn (x)ω(f , Δn (x)) 2 1 ≤ C2 Δn (x) ω(f , Δn (x)) + hω(f , h).

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Algebraic Approximation: A Guide to Past and Current Solutions by Jorge Bustamante

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