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1 SPRING QUARTER - 2008 M 230: INTRODUCTION TO REGIONAL PLANNING: THE EVOLUTION OF REGIONAL PLANNING DOCTRINES Instructor: Edward W. Soja - Room 2242 - Wednesday, 2-4:50 - _______________________________________________________________________________ M230 is one of the few courses
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Publié le : mardi 27 mars 2012
Lecture(s) : 36
Source : damtp.cam.ac.uk
Nombre de pages : 3
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Part III  Lent Term 2005 Approximation Theory – Lecture 5
5 Bestapproximation inC[a, b] 5.1 Characterization 1Theorem 5.1 (Kolmogorov[1948])LetUbe a linear subspace ofC(K)element. Anp∈ Uis a best approximation tofC(K)if and only if max [f(x)p(x)]q(x)0q∈ U,(5.1) x∈Z ∗ ∗ whereZis the set of all points for which|f(x)p(x)|=kfpk. Proof.1) Suppose that condition (5.1) is fulfilled. Take anyp∈ Uand setq=pp. We see that there is a pointx0∈ Zsuch that[f(x0)p(x0)]q(x0)0Then
22 |f(x0)p(x0)|=|f(x0)p(x0) +q(x0)| 22 =|f(x0)p(x0)|+ 2[f(x0)p(x0)]q(x0) +|q(x0)| 22 ≥ |f(x0)p(x0)|=kfpk, i.e., anyp∈ Uapproximatesfnot better thanp. 2) Conversely, if (5.1) is not true, then there exists aqsuch that max [f(x)p(x)]q(x) =2ε x∈Z for someε >0. By continuity, there is an open subsetGofKsuch that G⊃ Z,[f(x)p(x)]q(x)<εonG. ∗ ∗ Let us see howfis approximated byp:=pλq. Setm=kqkandE(f) =kfpk. a) ForxG, we have 22 |f(x)p(x)|=|f(x)p(x) +λq(x)| 22 2 =|f(x)p(x)|+ 2λ[f(x)p(x)]q(x) +λ|q(x)| 2 22 < E(f)2λε+λ m . 2 22 If we takeλ < ε/m, then we obtain that|f(x)p(x)|< E(f)λεonG. b) On the complementF=K\G, we have|f(x)p(x)|< E(f). ButFis closed, hence there is also aδ >0such that|f(x)p(x)| ≤E(f)2δonF. Takingλ < δ/m, we obtain |f(x)p(x)| ≤ |f(x)p(x)|+λ|q(x)| ≤E(f)2δ+λm < E(f)δonF .¤ Example 5.2ForfC[a, b]andU=P0(the constant functions), the best approximation is given by 1 p(x) =(maxf+ minf). 2
1 Andrey Kolmogorov (or, Kolmogoroff), 19031987, Russian mathematician, made fundamental contributions to prob ability, topology, functional analysis, mechanics, etc., etc. In short: one of the greatest mathematician of the 20th century.
5.2 Chebyshevalternation theorem 2Theorem 5.3 (Chebyshev[1854])A polynomialp∈ Pnis the best approximant tofC[a, b]if and only if there exist(n+ 2)pointsat1<∙ ∙ ∙< tn+2bsuch that if(ti)p(ti) = (1)γ,|γ|=kfpk,(5.2) i.e., if and only if the differencef(x)p(x)takes consecutively its maximal value with alternating signs at least(n+ 2)times. ∗ ∗ Proof.1) Assume that the differencefptakes the valuekfpkwith alternating signs in mn+ 1points. Set ∗ ∗∗ ∗ Z={x[a, b] :|f(x)p(x)|=kfpk},Z±={x[a, b] :f(x)p(x) =±kfpk}. Then there exists an ordered set ofmdisjoint intervals(Ki)which contains bothZandZ+and such that, on adjacent intervals, the points fromZbelongs toZandZ+alternatively. Select any pointszkbetween adjacent setsKi: K1< z1< K2<∙ ∙ ∙< zm1< Km, and take the polynomial Q m1 q(x() =xzi), m1n. i=1 Thisqis inPn, it alternates in sign on adjacentKis, and for thisq(or forq) we obtain [f(x)p(x)]q(x)<0onZ, so that, by Theorem 5.1,pis not a best approximant. 2) Ifpis a polynomial that satisfies (5.2), then, for anyq∈ Pn, the condition [f(ti)p(ti)]q(ti)<0, i= 1, . . . , n+2, would forceqto change its sign in(n+ 2)points, hence to haven+ 1zero, which is impossible. Thus max [f(x)p(x)]q(x)0,q∈ Pn, x(ti) so that, by Kolmogorov Theorem,pis a best approximant.¤ Theorem 5.4A polynomialp∈ Tnis the best approximant tofC(T)if and only if there exist(2n+2) pointsπ < t1<∙ ∙ ∙< t2n+2πsuch that if(ti)p(ti) = (1)γ,|γ|=kfpk. Proof.The proof is essentially the same. 1) We notice that, due to periodicity, the difference|fp| can take its maximal value with alternating signs only even number of times, say2m. Therefore, if2m2n, there are2mpointszksuch that K1< z1< K2<∙ ∙ ∙< z2m1< K2m< z2m< K1+ 2π . The polynomialq∈ Tm(that violates the Kolmogorov criteria) is then defined as m m Y Y xz2i1xz2i q(x= (sin sin) =ai+ cos(xbi). 2 2 i=1i=1 2) For sufficiency, we use the fact that anyq∈ Tncan have no more than2nzeros on the period.¤ Remark 5.5ForU=PnorTn, the number of points required in alternation theorem is equal to dim(U) + 1. 2 Pafnutij Chebyshev (or Tchebycheff, or Tschebyshev, or Chebyshov, or .. . ),18211894, great Russian mathematician, got his international fame by proving the Bertrand postulate: there is always a prime betweennand2n, almost proved lnx the prime number theorem showing thatlimx→∞π(x1) =if the limit exists. x
5.3 Exercises 5.1.For the complex valued continuous functions, Kolmogorov criterion takes the form ∗ ∗ pis a b.a. tofmax Re[f(x)p(x)]q(x)0q∈ U. x∈Z Check the “real” proof to find the changes. 5.2.Apply Kolmogorov criterion to prove the following statements:
3 22 (a)K= [1,1], f(x) =x ,U= span{1, x} ⇒p(x) = (1x); 2(b)K= [1,1], f(x, y) =xy,U= span{1, x, y} ⇒p0; 2 21 (c)K={|x|+|y| ≤1}, f(x, y) =x+y ,U= span{1, x, y, xy} ⇒p(x, y+ 2) =xy; 2 n(d)K={zC:|z| ≤1}, f(z) =z ,U=Pn1p0. For the cases (a) and (c), find another b.a. tof. 5.3.ForTn1, the space of trig.polynomials of degreen1, find the value and the polynomial of best approximation tof(x) =acosnx+bsinnx. 5.4.Let ∞ ∞ X X k f(x) =akcos 3x, ak>0, ak<. k=0k=0 m m+1 Prove that, for3n <3, the polynomial m X k tn(x) =s3(f, x) =akcos 3x m k=0 (which is a partial Fourier series tof) is the b.a. toffromTnand find the value ofEn(f). 5.5.Use the previous result to prove thelethargy theorem: For any sequenceεn&0, there exists anfC(T)such thatEn(f)εnalln. (Hint.Setak:=εk1εk.) 5.6.Prove de La VallePoussin theorem: Ifp∈ Pnis a polynomial such that i f(ti)p(ti) = (1)εi,sgnεi= const, atn+ 2consecutive points, thenEn(f)mini|εi|. Use this theorem to establish the sufficiency half of the alternation theorem. 5.7.LetfC[a, b], and letp, q∈ Pn+1polinomials of degreebe alg.n+ 1determined by the condition i p(ti) =f(ti), q(ti) = (1),whereat1<∙ ∙ ∙< tn+2b. Prove that the polynomialr:=pλq, whereλis chosen so thatrbelongs toPn, is the best approximation tofon(ti)and find the error. Remark.This is a practical method for constructing the b.a. on the set ofn+ 2points.
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