Smooth counterexamples to strong unique continuation for a Beltrami system in C 2
|
|
- Branden Morrison
- 5 years ago
- Views:
Transcription
1 Smooth couterexamples to strog uique cotiuatio for a Beltrami system i C Adam Coffma Yifei Pa October, 0 Abstract We costruct a example of a smooth map C C which vaishes to ifiite order at the origi, ad such that the ratio of the orm of the z derivative to the orm of the z derivative also vaishes to ifiite order. This gives a couterexample to strog uique cotiuatio for a vector valued aalogue of the Beltrami equatio. Itroductio We will costruct a example of a smooth fuctio u : C C which has a isolated zero of ifiite order at the origi z k uz 0asz 0 for all k 0, ad where the ratio of orms of derivatives u z / u z is small, also vaishig to ifiite order at z = 0. This behavior is obviously differet from that of a map u with u z 0, which would be holomorphic ad could ot have a isolated zero of ifiite order. This vector valued case is also differet from the complex scalar case, where solutios u : C C of the well-kow Beltrami equatio u z = azu z, for small az, also caot vaish to ifiite order at a isolated zero [B, [CH, [AIM, [R. More precisely, we will show i Sectio 4 that i a eighborhood of the origi, uz is a solutio of a Beltrami-type system of differetial equatios, which is liear, elliptic, ad has cotiuous coefficiets very close to those of the Cauchy-Riema system, but does ot have the strog uique cotiuatio property. I Theorem 4., we verify the coefficiets are ot Lipschitz cotiuous. Couterexamples to the weak uique cotiuatio property for elliptic Beltrami systems have bee cosidered by [IVV. The costructio was motivated by a example of Rosay [R ad questios posed by [IS, who were cosiderig the uique cotiuatio problem for systems of equatios from almost complex geometry. MSC 00: 35A0 Primary; 30G0, 3W50, 35J46 Secodary
2 I Sectio, we develop a geeral framework for costructig smooth maps C C vaishig to ifiite order. I Sectio 3, we preset both Rosay s example ad our ew example. I Sectio 5 we state some ope questios. Geeral Setup. Aular cutoff fuctios Start with a real valued fuctio sx whichissmoothor, with s 0o [0, 4, s icreasig o [ 4, 3 4, s =, s =,s = 0, ad s o[3 4,. For r > 0 ad two parameters 0 <r<r ad 0 < Δr <r r, deote the aulus A r,δr = {z = x + iy C : r z r +Δr} cotaied i the disk D r, ad defie a family of fuctios χ r,δr : A r,δr R by the formula χ r,δr z =s z r Δr. At a particular poit z = x, ỹ A r,δr, [ x [χ r,δrx, y x,ỹ = x + y s r x Δr x = s +ỹ r Δr = s z r Δr x z Δr. x,ỹ x x +ỹ The y derivative is similar, ad the z, z derivatives are complex liear combiatios. I particular, = Δr z χ r,δrz = z r z χ r,δrz =s x + iy Δr z Δr z χ r,δrz = z χ r,δrz m 0 Δr for some costat m 0 > 0 ot depedig o r, r or Δr. For higher derivatives of χ r,δr, the followig Lemma is a simplified versio of the Faà di Bruo formula for derivatives of composites. Lemma.. For k 0, there exist polyomials p abc x,x,x 3, q abc x,x,x 3 idexed by a, b, c 0, a + b = k, c k, with costat complex coefficiets ot depedig o r, r, Δr, ors, so that a b z a z b χ r,δrz = k c=0 z r s c pabcz, z, Δr+ z q abc z, z, Δr Δr z k Δr k. Proof. The k = 0 case is trivial ad the k = case is stated above. We record
3 the secod derivatives: z χ r,δrz = z χ r,δrz z r = s Δr z z χ r,δrz = s z z r z 4 z Δr + s Δr 4 z 3 Δr, 3 z r z r Δr 4Δr + s Δr 4 z Δr. The proof for all larger k is by iductio o k; the calculatio is straightforward ad omitted here. It follows as a cosequece of the Lemma that there are positive costats m ab idexed by a, b 0, a + b = k, ad depedig o the choices of s ad r, but ot depedig o r, Δr sothat = max z A r,δr a b z a z b χ r,δrz a b z a z b χ r,δrz m ab z k Δr k m ab r k Δr k. I various cases, i particular k = as i, the r k cabeimprovedwitha smaller expoet, but it is good eough to use later i Lemma.3.. The basic costructio of the examples Let r be a real sequece decreasig with limit = 0. Deote Δr = r r +. Let A deote the closed aulus A = A r+,δr = {z C : r + z r }, so the uio is a disk: D r = A {0}. The aular cutoff fuctios ca be idexed by : χ = χ r+,δr : A R. For N, letp be a icreasig positive iteger sequece; set p0 = 0. Let F be a positive real [ valued sequece; set F 0 =. Defie a fuctio 0 u : D r C by u0 = ad o the aulus A 0, for eve : uz = [ u z u z, u z = F z p, 4 u z = χ zf z p + χ zf +z p+. 5 For odd, switch the formulas for u, u. 3
4 So far, for ay s, r, p, F, the fuctio u is smooth o D r \{0}, ad exteds holomorphically for z r. We also have that u ad u z have o-zero value at every poit of D r \{0}; for eve ad switchig idices if is odd: u z = F p z p. 6 z u.3 Smoothess at the origi A importat property of the examples u : D r C we wat to costruct is that they are smooth at ad ear the origi. It is ot eough to check oly that the compoets vaish to ifiite order. I geeral, as easily costructed examples would show, for fuctios f : R N R M, f ca vaish to ifiite order at the origi: f x lim x 0 x k = 0 for all whole umbers k, but eed ot be smooth. Our approach to provig smoothess of our examples will be to show u, u, ad all their higher partial derivatives approach 0; this implies vaishig to ifiite order, as i the followig Lemma. Lemma.. Give f : R \{ 0} R, suppose f is smooth ad for each j, k =0,,, 3,..., k+j fx, y lim x,y 0 x k y j =0. The extedig f so that f0, 0 = 0 defies a smooth fuctio o R that vaishes to ifiite order at the origi. Proof. f is cotiuous at 0 byhypothesisj = k =0. Toshowf is smooth, we oly eed to show every partial derivative of order l of f exists at 0, ad has value 0; the it follows that for k + j = l, k+j fx,y is cotiuous at 0. x k y j The proof is by iductio o l; suppose for ay o-commutative word x k y j x ka y j b with k + k a +j +...+j b = k +j = l, l fx, y x k y j x ka y j b exists at 0 ad has value 0. The, the x-derivative at the origi is with the y-derivative beig similar: l fx, y x x k y j x ka y j b = lim t 0 = lim t 0 0,0 l f0 + t, 0 x k y j x ka y j b t l x k y j ft, 0 0 t. l f0, 0 x k y j x ka y j b 4
5 Let gt = l x k y ft, 0 for t 0; the lim gt = 0 by hypothesis, ad j t 0 g t = l+ x k+ ft, 0. yj L Hôpital s Rule applies to the above limit: gt lim t 0 t g t l+ = lim = lim t 0 t 0 x k+ ft, 0 = 0. yj The property of vaishig to ifiite order follows from Taylor approximatio at the origi. For our examples u, we wat to choose r, p, adf,sothatu is smooth ad vaishes to ifiite order at 0. The followig criterio for smoothess will be verified for both the Examples i Sectio 3. Lemma.3. If Δr /r Δr +/r + is a bouded sequece ad, for each iteger k 0, lim F +p + k r p+ 4k Δr /r k =0, 7 the u is smooth ad vaishes to ifiite order at the origi. Proof. From Lemma. ad the costructio of u, it is eough to show, for ay o-egative itegers a, b, with a + b = k, that a b a b max u z A z z, max u z A z z both have limit 0 as. The followig estimates for the derivatives assume is eve, ad sufficietly large compared to k. The derivatives of u = F z p 4 are easy: term: z a b u z F pp p k +z p k F p k r p k 8 The derivatives of u z 5 are cosidered oe term at a time. For the first a z z = F z = F b χ zf z p a a +a =a a b χ z z p z a b χ z z z 5 z a z p
6 The sum is over the a terms with may repeated that result from applyig the product rule a times. By Lemma., b χ z z a z m ab z a+b Δr a+b, ad a z p z = p p a + z p a p a z p a. Let m k = max m ab. The a+bk a b χ zf z p z z F a max a a { m ab z a+b Δr a+b } max a a F a m k z k Δr k p a z p a F k m k Δr k p k z p 3k {p a z p a } k F p k r p 3k m k Δr k. 9 Similarly for the secod term of u z, a b χ zf +z p+ z z k F +p + k r p+ 3k m k Δr k. 0 So, the criteria for all the derivatives of u to vaish at the origi are that the expressios 8, 9, ad 0 must all have limit 0 as.thehypothesis 7 is equivalet to 0 0. Comparig 0 to 8 by shiftig the idex i 8 from to +, this scalar multiple of 0 is much larger: F +p + k r p+ 3k Δr k >F +p + k r p+ k +, so if 0 has limit 0, the so does also implies F +r p+ k 0, which is eough to show u vaishes to ifiite order: z k uz 0asz 0. 6
7 Shiftig the idex i 9 from to + gives the followig quatity, which is comparable to 0: k F +p + k r p+ 3k + m k Δr + k < k F +p + k r p+ 4k Δr /r k m k Δr /r k Δr + /r + k, ad uder the additioal hypothesis that also implies Comparig first derivatives Δr /r Δr +/r + is a bouded sequece, 7 We wat to choose F, p, adr so that u z u z is small, as z 0. For z A, eve ad switchig idices if is odd, expadig the derivative ad usig gives: u z = z m 0 F z p + m 0 F + z p+ Δr Δr = m 0 Δr z p [ F z p p+ + F + z p+ p+. Usig 6 ad itroducig a factor g > 0, for z A : [ u z m 0r g F r p p + + F +r p+ p. u z Δr p gf The first fractio i the product is what we wat to make small for large, depedig o p ad Δr. The secod fractio we would like to make bouded, r depedig o F ad r, ad a arbitrary fudge factor g. The role of g is to maagethesizeoff ad simplify the calculatio provig boudedess of the secod factor, possibly at the expese of affectig the rate at which the first factor approaches 0. 3 Examples Example 3.. Rosay s example [R has p =, r = +,ad Δr r =. The becomes: u z u z m 0g [ F + F gf 7
8 The choice, as i [R, F = /, satisfies the recursive formula F = gf, 4 with g =. This simplifies the secod factor of the RHS of 3, so it is easily see to be bouded. The coclusio is u z u C for z A, ad sice log z o A, u z u z C 5 log z for all z D \{0}. To check that u is smooth ad vaishig to ifiite order at the origi, it is eough to verify the coditio of Lemma.3; for each fixed k 0: +/ + k + + 4k lim k =0. The goal of the ext example is to improve upo the order of vaishig of the ratio 5. Example 3.. Cosider r = l+,sor = l.44, r = l3,... This radius shriks much more slowly tha i Example 3.. Sice lim l+ l+ / l + =, there are costats C, c > 0sothatforall: c l + < Δr l + = r l + < C l +. 6 Let p = ; the the iequality becomes: u z u z m 0 l +g c [ [ + [ F + F + l+ l+ +. gf This motivates, i aalogy with the previous Example, this choice of g ad a recursive formula for F as i 4: So F = F 0 =, ad for >, g = l +, F = l + [ F. F = l + [ l + [ l5 4 l4 = [ l + l + l5 l4. 8
9 The the followig sequece of ratios is bouded above because it is coverget as : F l + [ + F + l + [+ gf = F l +[ + l+3 l+ [ F l+ [+ l + l + [ F = + l +3 l + l + [+ C 3. Here we used the elemetary calculus lemma that l+ l is a bouded sequece. The estimate for the ratio of derivatives o A, for eve, becomes: For z A, so u z u z u z u z m 0 l + l + c C 3 = m 0l + C 3 c = C 4l + 7 < C 5l l + z l + l + l + z + exp z +, u z u z C 5 z exp z, for all z D r \{0}. It remais to check that u is smooth, ad vaishes to ifiite order. The hypothesis o Δr of Lemma.3 is satisfied usig 6, so 9
10 for fixed k, cosider the expressio: = < < F +p + k r p+ 4k Δr /r k [ l +3 l + l4 + k + 4k l+ k l+ l+ [ l +3 l + l4 + k k l + + 4k c l+ [ l +3 l + l4 + 3k l + k c k l. ++ 4k The last expressio has limit zero by the Ratio Test: l+4 + l+3 l4 c k l++ 4k + 3k l +3 k l+3 l4 c k l++ 4k +3k l + k = l +4+ l + + 4k + 3k l +3 k l + + 4k + 3k l + k < l +4++k + 3k l + +3+k 0, + 3k agai usig the boudedess of l+. l 4 A Beltrami-type system [ u Ay smooth map u : C C, u = u, satisfies the followig Beltrami-type system of first-order differetial equatios at poits where u z 0: [ [ [ u z u z [u u z = u z z u z u z u z + u z u z [ [ u z u = z u z u z u z u z u 9 z u z u z = Q z [ u z u z. u z u z For u costructedasisectio,otheaulia with eve, u z 0, so the first row of the matrix i 9 is [0 0, ad similarly the secod row is [0 0 for odd. Defie Q0tobethezeromatrix. 0
11 For a matrix Qz defied as i 9 by some fixed fuctio u, the operator L = z Qz is complex liear. If, o some eighborhood of z =0,the z Qz etries are defied ad small eough, the L is elliptic i the sese of [AIM Sectio 7.4. I the followig Theorem, we cosider Qz fortheexampleuz fromexample 3.. If we restrict u ad Q to z i some sufficietly small eighborhood of the origi, u will be a solutio of the elliptic equatio Lu =0. Theorem 4.. For u as i Example 3., let q ij z = etry i the matrix Qz from 9. q ij C D r \{0} C 0 D r ; uī z uj z u z deote the i, j q ij vaishes to ifiite order: z k q ij z 0 as z 0 for ay k 0; The partial derivatives exist at the origi: x q ij0 = y q ij0 = 0; For ay 0 <r<r, q does ot have the Lipschitz property o D r. Proof. The C claim follows from the smoothess of u o D r ad the ovaishig of u z for z 0. The sum q + q + q + q is exactly u z, which vaishes to u z ifiite order as z 0foru as i Example 3.. It follows that each q ij also vaishes to ifiite order, which implies Q ad the etries q ij are cotiuous at the origi, with the previously assiged values q ij 0 = 0. The flatess also implies the existece of all directioal derivatives at the origi of C = R ;for the x directio, x q q ij x, 0 q ij 0, 0 ij = lim =0. x 0 x,y=0,0 x The last claim takes up the rest of the Proof; the pla is to show there is a sequece of poits x C approachig 0 so that u zu z z u z is a ubouded z=x sequece. If q had a Lipschitz property o D r, i.e., q z q z K z z for some K ad all z, z, the its derivatives o D r \{0} would be bouded; the uboudedess of the derivative also directly shows q / C D r. A real variable aalogue of this behavior is the fuctio exp / x cosexp/ x, which vaishes to ifiite order but has ubouded first derivative as x 0. It is eough, ad simpler, to cosider oly which are eve ad sufficietly large. This will ivolve some estimates for derivatives that are more precise tha 9. We choose the sequece x = r + + Δr +0i A ; the by costructio of s ad χ, χ x = χ,adgives z x = χ z x = Δr.From3, χ z z x = χ z x = χ z x = x Δr
12 this is where we use the s = 0 assumptio, to simplify the calculatio. For r as i Example 3., l+ <x < l+,adδr = l+ satisfies 0 <c 6 l + < <C 6 l +. Δr I the followig expressio, u z u z z u z = u z z u z u z + u z u zz u z u z u z z u z u z + u z u z u z 4 l+ = u z z u z u z + u z u zz u z u z + u z u z u z u z u z u zz + u z z u z + u z u zz u z 4, 0 the terms with the largest magitude are the oes ivolvig the secod z- derivatives, u zz ad u zz. Evaluated at poits i the sequece x,theseterms idividually grow at least as fast as some costat multiple of. However, due to some cacellatios, the overall growth rate turs out to be less tha ; the Theorem will be proved by showig oe of the terms is ubouded ad the remaiig terms have a slower rate of growth. The first cacellatio is that the secod ad last terms i the umerator of the secod fractio i 0 are exactly opposites. This leaves two terms with zz-derivatives; usig the power rule o u = F z i the iterior of A gives: u zz = u z. z 0 becomes: u z zu z u zu zu u z + z u zz u z z u z 4 u zu zu z zu z u z 4. We will show that the last of the three terms i is the domiat oe. First, cosider the ratio: u z / u z = F χ z z + χ z + F + χ z z+ + χ + z + F / z. The followig calculatio for z A is similar to that for the estimate 7. [ u χ F z z z + χ z u z F z χ F +[ z z + + χ + z + + F z
13 From Example 3., recallig F =l + F, l+ z l+,ad χ z m0 Δr, it follows that there is some costat C 7 > sothat u z u z C 7 l +. 3 To estimate the deomiators of, u z u u = + z z u z +C 7 l + = u z C 8 l + u z. So we get a lower boud for the third term i, u z u z u z z u z u z 4 ad cosider 4 oe factor at a time. χ z F z χ z u z u = z u z u z = z=x Δr u u z z u z z C8 l +4 u, 4 z 4 F +z+ F z l + x l +3 x + l + c 6 l + l + l +3 l c 4l +. 6 This lower boud 6 is comparable to the upper boud 7, which is used i the ext step to fid a lower boud for u z u z. Note that the expressio has two terms which, usig the equality of χ χ z ad z whe evaluated at x, match two of the terms from u z i 5: u z u z u = z z=x u + z x l + x + l +3 + x + l + u z l + u z x l +3 + l + + l + c 9 l + C 4 C 0 l + c 7 l
14 This lower boud is comparable to the upper boud 3. The remaiig factor from 4 ivolves secod derivatives: = u z z / u z F χ z z z + χ z z χ F + z z z+ + χ z + z + / F z F χ F z z + χ z z z + F + χ F z z z + + χ z + z +. Evaluatig at x agai, for sufficietly large, u z z u z x l + Δr x l +3 x Δr x x x + x Δr Δr l + c6 l + l + C 6l + l + l +3 C 6 l + l l + + c l So, the term from 4 is bouded below by a product icludig factors from 6, 7, ad 8: u z u z u z z c 4 l + c 7 l + c l + 3 u z 4 C8 x l +4 c l
15 From the secod term i, we cosider the followig quatity: u zz u z z χ = F z z + χ z z +χ z +F + χ z z+ χ z + z + + χ + +z + F χ z z + χ z F + χ z z + + χ + z +. The cacellatio of the 4 quatities is the key step. The ratio u zz u z z / u z = F χ z z + χ z 4 +3z χ z χ +F + z χ z+ z +4 +3z + + χ + +z + F / z 5
16 has a upper boud o the x sequece: u zz u z z u z x l + x Δr x l +3 x + + l Δr x x + x Δr + x + + +x Δr C 6 l l + + l + l +3 + C 6 l l l + C 3 l The middle term from, evaluated at poits x, has factors bouded by 7, 7, ad 30: u z u z u z u zz u z /z u 4 x u z u z u zz u z /z = u z u u z x z 4 u z 4 C 4 l + u z 4 x x u zz u z/z u z x c 7 l + 4 C 3l + 3 C 4 l
17 The first term from ivolves the secod z-derivative: u z z χ z F z χ u z = z F +z + F z u z z u z x F F x Δr C 6l + χ z z + + l + F + F x l + l + + l + 3 C 5. The first term from also approaches 0 for large : u z zu z u z z z u u x z χ z z +. + l +3 x + l +3 l + + u x z l + C 5 u z x 3. The coclusio from 0 ad is: u zu z z u z c l + 3 l C 4 l + C +3 5 x > c 6 x 3. 5 Remarks ad Questios Remark 5.. The regularity of the coefficiets is a importat cosideratio i the aalysis of uique cotiuatio properties for some PDEs for example, [LNW for strog UCP, ad [IVV for weak UCP, which is why we preseted the detailed Proof of Theorem 4.. However, we do ot yet uderstad the sharpess of Example 3. ad Theorem 4. for this particular uique cotiuatio problem. The questio that aturally arises is: would improved regularity of Qz i additio to, or istead of, the flatess property imply a strog uique cotiuatio property, or, oppositely, is there some couterexample where Qz is smooth? Remark 5.. [R shows how Example 3. ca be modified so that the origi is ao-isolatedzeroofu; it is a matter of replacig quatities z N i 4, 5 by z N z a for a sequece a ad re-workig the cutoff fuctios χ. Our Example 3. ca be modified i a aalogous way but we have ot worked out all the details. 7
18 Remark 5.3. By a costructio aalogous to 9, the fuctio u from Example 3. also satisfies a real liear, elliptic equatio of the form u z = Q u z. Qz is ot the same as Qz but also has etries vaishig to ifiite order. Remark 5.4. Aother differetial iequality, cosidered by [R, is u z K u α u z, for 0 < α <. Our attempts to use the costructio of Sectio to fid smooth fuctios u satisfyig the iequality ad vaishig to ifiite order at a isolated zero have ot yet met ay success. [R proves a weak uique cotiuatio property for α =, but the strog property remais a ope questio. Refereces [AIM [B [CH [IS [IVV K. Astala, T. Iwaiec, adg. Marti, Elliptic Partial Differetial Equatios ad Quasicoformal Mappigs i the Plae, PMS48, Priceto Uiversity Press, Priceto, 009. MR j:30040, Zbl B. V. Bojarski, Geeralized solutios of a system of differetial equatios of the first order ad elliptic type with discotiuous coefficiets, Uiv. Jyväskylä Dept. of Math. ad Statistics Report Trasl. from Russia, Mat. Sb. N.S MR j:30096, Zbl R. Courat ad D. Hilbert, Methods of Mathematical Physics. Vol. II: Partial Differetial Equatios, Wiley, New York - Lodo, 96. MR #46, Zbl S. Ivashkovich ad V. Shevchishi, Local properties of J-complex curves i Lipschitz-cotiuous structures, Math. Z , MR 88746, Zbl T. Iwaiec, G. Verchota, ada. Vogel, The failure of rak-oe coectios, Arch.Ratio.Mech.Aal.63 00, MR j:580, Zbl [LNW C.-L. Li, G. Nakamura, adj.-n. Wag, Optimal three-ball iequalities ad quatitative uiqueess for the Lamésystem, Duke Math. J , MR j:3535, Zbl [R J.-P. Rosay, Uiqueess i rough almost complex structures, ad differetial iequalities. A.Ist.Fourier660 00, MR h:3033, Zbl
(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3
MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special
More informationb i u x i U a i j u x i u x j
M ath 5 2 7 Fall 2 0 0 9 L ecture 1 9 N ov. 1 6, 2 0 0 9 ) S ecod- Order Elliptic Equatios: Weak S olutios 1. Defiitios. I this ad the followig two lectures we will study the boudary value problem Here
More informationSequences and Series of Functions
Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges
More informationProduct measures, Tonelli s and Fubini s theorems For use in MAT3400/4400, autumn 2014 Nadia S. Larsen. Version of 13 October 2014.
Product measures, Toelli s ad Fubii s theorems For use i MAT3400/4400, autum 2014 Nadia S. Larse Versio of 13 October 2014. 1. Costructio of the product measure The purpose of these otes is to preset the
More informationFall 2013 MTH431/531 Real analysis Section Notes
Fall 013 MTH431/531 Real aalysis Sectio 8.1-8. Notes Yi Su 013.11.1 1. Defiitio of uiform covergece. We look at a sequece of fuctios f (x) ad study the coverget property. Notice we have two parameters
More informationComplex Analysis Spring 2001 Homework I Solution
Complex Aalysis Sprig 2001 Homework I Solutio 1. Coway, Chapter 1, sectio 3, problem 3. Describe the set of poits satisfyig the equatio z a z + a = 2c, where c > 0 ad a R. To begi, we see from the triagle
More information6.3 Testing Series With Positive Terms
6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial
More informationPRELIM PROBLEM SOLUTIONS
PRELIM PROBLEM SOLUTIONS THE GRAD STUDENTS + KEN Cotets. Complex Aalysis Practice Problems 2. 2. Real Aalysis Practice Problems 2. 4 3. Algebra Practice Problems 2. 8. Complex Aalysis Practice Problems
More informationIt is often useful to approximate complicated functions using simpler ones. We consider the task of approximating a function by a polynomial.
Taylor Polyomials ad Taylor Series It is ofte useful to approximate complicated fuctios usig simpler oes We cosider the task of approximatig a fuctio by a polyomial If f is at least -times differetiable
More informationMAT1026 Calculus II Basic Convergence Tests for Series
MAT026 Calculus II Basic Covergece Tests for Series Egi MERMUT 202.03.08 Dokuz Eylül Uiversity Faculty of Sciece Departmet of Mathematics İzmir/TURKEY Cotets Mootoe Covergece Theorem 2 2 Series of Real
More informationInfinite Sequences and Series
Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet
More informationTopics. Homework Problems. MATH 301 Introduction to Analysis Chapter Four Sequences. 1. Definition of convergence of sequences.
MATH 301 Itroductio to Aalysis Chapter Four Sequeces Topics 1. Defiitio of covergece of sequeces. 2. Fidig ad provig the limit of sequeces. 3. Bouded covergece theorem: Theorem 4.1.8. 4. Theorems 4.1.13
More information1 Generating functions for balls in boxes
Math 566 Fall 05 Some otes o geeratig fuctios Give a sequece a 0, a, a,..., a,..., a geeratig fuctio some way of represetig the sequece as a fuctio. There are may ways to do this, with the most commo ways
More informationChapter 10: Power Series
Chapter : Power Series 57 Chapter Overview: Power Series The reaso series are part of a Calculus course is that there are fuctios which caot be itegrated. All power series, though, ca be itegrated because
More informationLecture Notes for Analysis Class
Lecture Notes for Aalysis Class Topological Spaces A topology for a set X is a collectio T of subsets of X such that: (a) X ad the empty set are i T (b) Uios of elemets of T are i T (c) Fiite itersectios
More informationChapter 6 Infinite Series
Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat
More informationAssignment 5: Solutions
McGill Uiversity Departmet of Mathematics ad Statistics MATH 54 Aalysis, Fall 05 Assigmet 5: Solutios. Let y be a ubouded sequece of positive umbers satisfyig y + > y for all N. Let x be aother sequece
More informationlim za n n = z lim a n n.
Lecture 6 Sequeces ad Series Defiitio 1 By a sequece i a set A, we mea a mappig f : N A. It is customary to deote a sequece f by {s } where, s := f(). A sequece {z } of (complex) umbers is said to be coverget
More informationChapter 8. Euler s Gamma function
Chapter 8 Euler s Gamma fuctio The Gamma fuctio plays a importat role i the fuctioal equatio for ζ(s that we will derive i the ext chapter. I the preset chapter we have collected some properties of the
More informationSequences and Limits
Chapter Sequeces ad Limits Let { a } be a sequece of real or complex umbers A ecessary ad sufficiet coditio for the sequece to coverge is that for ay ɛ > 0 there exists a iteger N > 0 such that a p a q
More informationLesson 10: Limits and Continuity
www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals
More informationDefinition 4.2. (a) A sequence {x n } in a Banach space X is a basis for X if. unique scalars a n (x) such that x = n. a n (x) x n. (4.
4. BASES I BAACH SPACES 39 4. BASES I BAACH SPACES Sice a Baach space X is a vector space, it must possess a Hamel, or vector space, basis, i.e., a subset {x γ } γ Γ whose fiite liear spa is all of X ad
More informationOptimally Sparse SVMs
A. Proof of Lemma 3. We here prove a lower boud o the umber of support vectors to achieve geeralizatio bouds of the form which we cosider. Importatly, this result holds ot oly for liear classifiers, but
More informationMATH301 Real Analysis (2008 Fall) Tutorial Note #7. k=1 f k (x) converges pointwise to S(x) on E if and
MATH01 Real Aalysis (2008 Fall) Tutorial Note #7 Sequece ad Series of fuctio 1: Poitwise Covergece ad Uiform Covergece Part I: Poitwise Covergece Defiitio of poitwise covergece: A sequece of fuctios f
More informationPart A, for both Section 200 and Section 501
Istructios Please write your solutios o your ow paper. These problems should be treated as essay questios. A problem that says give a example or determie requires a supportig explaatio. I all problems,
More information7 Sequences of real numbers
40 7 Sequeces of real umbers 7. Defiitios ad examples Defiitio 7... A sequece of real umbers is a real fuctio whose domai is the set N of atural umbers. Let s : N R be a sequece. The the values of s are
More informationRoger Apéry's proof that zeta(3) is irrational
Cliff Bott cliffbott@hotmail.com 11 October 2011 Roger Apéry's proof that zeta(3) is irratioal Roger Apéry developed a method for searchig for cotiued fractio represetatios of umbers that have a form such
More informationChapter 3. Strong convergence. 3.1 Definition of almost sure convergence
Chapter 3 Strog covergece As poited out i the Chapter 2, there are multiple ways to defie the otio of covergece of a sequece of radom variables. That chapter defied covergece i probability, covergece i
More informationChapter 7 Isoperimetric problem
Chapter 7 Isoperimetric problem Recall that the isoperimetric problem (see the itroductio its coectio with ido s proble) is oe of the most classical problem of a shape optimizatio. It ca be formulated
More informationUniversity of Colorado Denver Dept. Math. & Stat. Sciences Applied Analysis Preliminary Exam 13 January 2012, 10:00 am 2:00 pm. Good luck!
Uiversity of Colorado Dever Dept. Math. & Stat. Scieces Applied Aalysis Prelimiary Exam 13 Jauary 01, 10:00 am :00 pm Name: The proctor will let you read the followig coditios before the exam begis, ad
More information62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +
62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of
More informationFeedback in Iterative Algorithms
Feedback i Iterative Algorithms Charles Byre (Charles Byre@uml.edu), Departmet of Mathematical Scieces, Uiversity of Massachusetts Lowell, Lowell, MA 01854 October 17, 2005 Abstract Whe the oegative system
More informationMath 2784 (or 2794W) University of Connecticut
ORDERS OF GROWTH PAT SMITH Math 2784 (or 2794W) Uiversity of Coecticut Date: Mar. 2, 22. ORDERS OF GROWTH. Itroductio Gaiig a ituitive feel for the relative growth of fuctios is importat if you really
More informationMATH 324 Summer 2006 Elementary Number Theory Solutions to Assignment 2 Due: Thursday July 27, 2006
MATH 34 Summer 006 Elemetary Number Theory Solutios to Assigmet Due: Thursday July 7, 006 Departmet of Mathematical ad Statistical Scieces Uiversity of Alberta Questio [p 74 #6] Show that o iteger of the
More informationMa 530 Introduction to Power Series
Ma 530 Itroductio to Power Series Please ote that there is material o power series at Visual Calculus. Some of this material was used as part of the presetatio of the topics that follow. What is a Power
More informationChapter 8. Euler s Gamma function
Chapter 8 Euler s Gamma fuctio The Gamma fuctio plays a importat role i the fuctioal equatio for ζ(s) that we will derive i the ext chapter. I the preset chapter we have collected some properties of the
More informationThe natural exponential function
The atural expoetial fuctio Attila Máté Brookly College of the City Uiversity of New York December, 205 Cotets The atural expoetial fuctio for real x. Beroulli s iequality.....................................2
More informationMath Solutions to homework 6
Math 175 - Solutios to homework 6 Cédric De Groote November 16, 2017 Problem 1 (8.11 i the book): Let K be a compact Hermitia operator o a Hilbert space H ad let the kerel of K be {0}. Show that there
More informationHomework 2. Show that if h is a bounded sesquilinear form on the Hilbert spaces X and Y, then h has the representation
omework 2 1 Let X ad Y be ilbert spaces over C The a sesquiliear form h o X Y is a mappig h : X Y C such that for all x 1, x 2, x X, y 1, y 2, y Y ad all scalars α, β C we have (a) h(x 1 + x 2, y) h(x
More informationANSWERS TO MIDTERM EXAM # 2
MATH 03, FALL 003 ANSWERS TO MIDTERM EXAM # PENN STATE UNIVERSITY Problem 1 (18 pts). State ad prove the Itermediate Value Theorem. Solutio See class otes or Theorem 5.6.1 from our textbook. Problem (18
More informationMath 341 Lecture #31 6.5: Power Series
Math 341 Lecture #31 6.5: Power Series We ow tur our attetio to a particular kid of series of fuctios, amely, power series, f(x = a x = a 0 + a 1 x + a 2 x 2 + where a R for all N. I terms of a series
More information5. Matrix exponentials and Von Neumann s theorem The matrix exponential. For an n n matrix X we define
5. Matrix expoetials ad Vo Neuma s theorem 5.1. The matrix expoetial. For a matrix X we defie e X = exp X = I + X + X2 2! +... = 0 X!. We assume that the etries are complex so that exp is well defied o
More informationM17 MAT25-21 HOMEWORK 5 SOLUTIONS
M17 MAT5-1 HOMEWORK 5 SOLUTIONS 1. To Had I Cauchy Codesatio Test. Exercise 1: Applicatio of the Cauchy Codesatio Test Use the Cauchy Codesatio Test to prove that 1 diverges. Solutio 1. Give the series
More informationConvergence of random variables. (telegram style notes) P.J.C. Spreij
Covergece of radom variables (telegram style otes).j.c. Spreij this versio: September 6, 2005 Itroductio As we kow, radom variables are by defiitio measurable fuctios o some uderlyig measurable space
More informationSolutions to home assignments (sketches)
Matematiska Istitutioe Peter Kumli 26th May 2004 TMA401 Fuctioal Aalysis MAN670 Applied Fuctioal Aalysis 4th quarter 2003/2004 All documet cocerig the course ca be foud o the course home page: http://www.math.chalmers.se/math/grudutb/cth/tma401/
More informationMath 299 Supplement: Real Analysis Nov 2013
Math 299 Supplemet: Real Aalysis Nov 203 Algebra Axioms. I Real Aalysis, we work withi the axiomatic system of real umbers: the set R alog with the additio ad multiplicatio operatios +,, ad the iequality
More informationMath 104: Homework 2 solutions
Math 04: Homework solutios. A (0, ): Sice this is a ope iterval, the miimum is udefied, ad sice the set is ot bouded above, the maximum is also udefied. if A 0 ad sup A. B { m + : m, N}: This set does
More information3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,
3. Z Trasform Referece: Etire Chapter 3 of text. Recall that the Fourier trasform (FT) of a DT sigal x [ ] is ω ( ) [ ] X e = j jω k = xe I order for the FT to exist i the fiite magitude sese, S = x [
More informationLecture 2. The Lovász Local Lemma
Staford Uiversity Sprig 208 Math 233A: No-costructive methods i combiatorics Istructor: Ja Vodrák Lecture date: Jauary 0, 208 Origial scribe: Apoorva Khare Lecture 2. The Lovász Local Lemma 2. Itroductio
More informationCHAPTER I: Vector Spaces
CHAPTER I: Vector Spaces Sectio 1: Itroductio ad Examples This first chapter is largely a review of topics you probably saw i your liear algebra course. So why cover it? (1) Not everyoe remembers everythig
More informationDifferent kinds of Mathematical Induction
Differet ids of Mathematical Iductio () Mathematical Iductio Give A N, [ A (a A a A)] A N () (First) Priciple of Mathematical Iductio Let P() be a propositio (ope setece), if we put A { : N p() is true}
More informationMAS111 Convergence and Continuity
MAS Covergece ad Cotiuity Key Objectives At the ed of the course, studets should kow the followig topics ad be able to apply the basic priciples ad theorems therei to solvig various problems cocerig covergece
More informationx a x a Lecture 2 Series (See Chapter 1 in Boas)
Lecture Series (See Chapter i Boas) A basic ad very powerful (if pedestria, recall we are lazy AD smart) way to solve ay differetial (or itegral) equatio is via a series expasio of the correspodig solutio
More informationENGI Series Page 6-01
ENGI 3425 6 Series Page 6-01 6. Series Cotets: 6.01 Sequeces; geeral term, limits, covergece 6.02 Series; summatio otatio, covergece, divergece test 6.03 Stadard Series; telescopig series, geometric series,
More informationMcGill University Math 354: Honors Analysis 3 Fall 2012 Solutions to selected problems
McGill Uiversity Math 354: Hoors Aalysis 3 Fall 212 Assigmet 3 Solutios to selected problems Problem 1. Lipschitz fuctios. Let Lip K be the set of all fuctios cotiuous fuctios o [, 1] satisfyig a Lipschitz
More informationWe are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n
Review of Power Series, Power Series Solutios A power series i x - a is a ifiite series of the form c (x a) =c +c (x a)+(x a) +... We also call this a power series cetered at a. Ex. (x+) is cetered at
More informationSequences. Notation. Convergence of a Sequence
Sequeces A sequece is essetially just a list. Defiitio (Sequece of Real Numbers). A sequece of real umbers is a fuctio Z (, ) R for some real umber. Do t let the descriptio of the domai cofuse you; it
More informationArchimedes - numbers for counting, otherwise lengths, areas, etc. Kepler - geometry for planetary motion
Topics i Aalysis 3460:589 Summer 007 Itroductio Ree descartes - aalysis (breaig dow) ad sythesis Sciece as models of ature : explaatory, parsimoious, predictive Most predictios require umerical values,
More informationSOLUTION SET VI FOR FALL [(n + 2)(n + 1)a n+2 a n 1 ]x n = 0,
4. Series Solutios of Differetial Equatios:Special Fuctios 4.. Illustrative examples.. 5. Obtai the geeral solutio of each of the followig differetial equatios i terms of Maclauri series: d y (a dx = xy,
More informationAdvanced Analysis. Min Yan Department of Mathematics Hong Kong University of Science and Technology
Advaced Aalysis Mi Ya Departmet of Mathematics Hog Kog Uiversity of Sciece ad Techology September 3, 009 Cotets Limit ad Cotiuity 7 Limit of Sequece 8 Defiitio 8 Property 3 3 Ifiity ad Ifiitesimal 8 4
More information4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3
Exam Problems (x. Give the series (, fid the values of x for which this power series coverges. Also =0 state clearly what the radius of covergece is. We start by settig up the Ratio Test: x ( x x ( x x
More informationPAPER : IIT-JAM 2010
MATHEMATICS-MA (CODE A) Q.-Q.5: Oly oe optio is correct for each questio. Each questio carries (+6) marks for correct aswer ad ( ) marks for icorrect aswer.. Which of the followig coditios does NOT esure
More informationTR/46 OCTOBER THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION A. TALBOT
TR/46 OCTOBER 974 THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION by A. TALBOT .. Itroductio. A problem i approximatio theory o which I have recetly worked [] required for its solutio a proof that the
More informationMath 61CM - Solutions to homework 3
Math 6CM - Solutios to homework 3 Cédric De Groote October 2 th, 208 Problem : Let F be a field, m 0 a fixed oegative iteger ad let V = {a 0 + a x + + a m x m a 0,, a m F} be the vector space cosistig
More informationBeurling Integers: Part 2
Beurlig Itegers: Part 2 Isomorphisms Devi Platt July 11, 2015 1 Prime Factorizatio Sequeces I the last article we itroduced the Beurlig geeralized itegers, which ca be represeted as a sequece of real umbers
More informationA PROOF OF THE TWIN PRIME CONJECTURE AND OTHER POSSIBLE APPLICATIONS
A PROOF OF THE TWI PRIME COJECTURE AD OTHER POSSIBLE APPLICATIOS by PAUL S. BRUCKMA 38 Frot Street, #3 aaimo, BC V9R B8 (Caada) e-mail : pbruckma@hotmail.com ABSTRACT : A elemetary proof of the Twi Prime
More informationCSE 1400 Applied Discrete Mathematics Number Theory and Proofs
CSE 1400 Applied Discrete Mathematics Number Theory ad Proofs Departmet of Computer Scieces College of Egieerig Florida Tech Sprig 01 Problems for Number Theory Backgroud Number theory is the brach of
More informationDupuy Complex Analysis Spring 2016 Homework 02
Dupuy Complex Aalysis Sprig 206 Homework 02. (CUNY, Fall 2005) Let D be the closed uit disc. Let g be a sequece of aalytic fuctios covergig uiformly to f o D. (a) Show that g coverges. Solutio We have
More informationMA131 - Analysis 1. Workbook 3 Sequences II
MA3 - Aalysis Workbook 3 Sequeces II Autum 2004 Cotets 2.8 Coverget Sequeces........................ 2.9 Algebra of Limits......................... 2 2.0 Further Useful Results........................
More informationNEW FAST CONVERGENT SEQUENCES OF EULER-MASCHERONI TYPE
UPB Sci Bull, Series A, Vol 79, Iss, 207 ISSN 22-7027 NEW FAST CONVERGENT SEQUENCES OF EULER-MASCHERONI TYPE Gabriel Bercu We itroduce two ew sequeces of Euler-Mascheroi type which have fast covergece
More informationSequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet
More informationPRACTICE FINAL/STUDY GUIDE SOLUTIONS
Last edited December 9, 03 at 4:33pm) Feel free to sed me ay feedback, icludig commets, typos, ad mathematical errors Problem Give the precise meaig of the followig statemets i) a f) L ii) a + f) L iii)
More informationDeterminants of order 2 and 3 were defined in Chapter 2 by the formulae (5.1)
5. Determiats 5.. Itroductio 5.2. Motivatio for the Choice of Axioms for a Determiat Fuctios 5.3. A Set of Axioms for a Determiat Fuctio 5.4. The Determiat of a Diagoal Matrix 5.5. The Determiat of a Upper
More informationA NOTE ON BOUNDARY BLOW-UP PROBLEM OF u = u p
A NOTE ON BOUNDARY BLOW-UP PROBLEM OF u = u p SEICK KIM Abstract. Assume that Ω is a bouded domai i R with 2. We study positive solutios to the problem, u = u p i Ω, u(x) as x Ω, where p > 1. Such solutios
More informationSeunghee Ye Ma 8: Week 5 Oct 28
Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value
More informationA sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as
More information(3) If you replace row i of A by its sum with a multiple of another row, then the determinant is unchanged! Expand across the i th row:
Math 50-004 Tue Feb 4 Cotiue with sectio 36 Determiats The effective way to compute determiats for larger-sized matrices without lots of zeroes is to ot use the defiitio, but rather to use the followig
More informationLecture 19: Convergence
Lecture 19: Covergece Asymptotic approach I statistical aalysis or iferece, a key to the success of fidig a good procedure is beig able to fid some momets ad/or distributios of various statistics. I may
More informationMath 210A Homework 1
Math 0A Homework Edward Burkard Exercise. a) State the defiitio of a aalytic fuctio. b) What are the relatioships betwee aalytic fuctios ad the Cauchy-Riema equatios? Solutio. a) A fuctio f : G C is called
More informationand each factor on the right is clearly greater than 1. which is a contradiction, so n must be prime.
MATH 324 Summer 200 Elemetary Number Theory Solutios to Assigmet 2 Due: Wedesday July 2, 200 Questio [p 74 #6] Show that o iteger of the form 3 + is a prime, other tha 2 = 3 + Solutio: If 3 + is a prime,
More information10.1 Sequences. n term. We will deal a. a n or a n n. ( 1) n ( 1) n 1 2 ( 1) a =, 0 0,,,,, ln n. n an 2. n term.
0. Sequeces A sequece is a list of umbers writte i a defiite order: a, a,, a, a is called the first term, a is the secod term, ad i geeral eclusively with ifiite sequeces ad so each term Notatio: the sequece
More informationLecture 3: August 31
36-705: Itermediate Statistics Fall 018 Lecturer: Siva Balakrisha Lecture 3: August 31 This lecture will be mostly a summary of other useful expoetial tail bouds We will ot prove ay of these i lecture,
More informationLecture 8: Convergence of transformations and law of large numbers
Lecture 8: Covergece of trasformatios ad law of large umbers Trasformatio ad covergece Trasformatio is a importat tool i statistics. If X coverges to X i some sese, we ofte eed to check whether g(x ) coverges
More informationA Negative Result. We consider the resolvent problem for the scalar Oseen equation
O Osee Resolvet Estimates: A Negative Result Paul Deurig Werer Varhor 2 Uiversité Lille 2 Uiversität Kassel Laboratoire de Mathématiques BP 699, 62228 Calais cédex Frace paul.deurig@lmpa.uiv-littoral.fr
More informationMath 113 Exam 3 Practice
Math Exam Practice Exam 4 will cover.-., 0. ad 0.. Note that eve though. was tested i exam, questios from that sectios may also be o this exam. For practice problems o., refer to the last review. This
More information1+x 1 + α+x. x = 2(α x2 ) 1+x
Math 2030 Homework 6 Solutios # [Problem 5] For coveiece we let α lim sup a ad β lim sup b. Without loss of geerality let us assume that α β. If α the by assumptio β < so i this case α + β. By Theorem
More informationCommon Coupled Fixed Point of Mappings Satisfying Rational Inequalities in Ordered Complex Valued Generalized Metric Spaces
IOSR Joural of Mathematics (IOSR-JM) e-issn: 78-578, p-issn:319-765x Volume 10, Issue 3 Ver II (May-Ju 014), PP 69-77 Commo Coupled Fixed Poit of Mappigs Satisfyig Ratioal Iequalities i Ordered Complex
More informationLinear Regression Demystified
Liear Regressio Demystified Liear regressio is a importat subject i statistics. I elemetary statistics courses, formulae related to liear regressio are ofte stated without derivatio. This ote iteds to
More informationCHAPTER 10 INFINITE SEQUENCES AND SERIES
CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece
More informationRiesz-Fischer Sequences and Lower Frame Bounds
Zeitschrift für Aalysis ud ihre Aweduge Joural for Aalysis ad its Applicatios Volume 1 (00), No., 305 314 Riesz-Fischer Sequeces ad Lower Frame Bouds P. Casazza, O. Christese, S. Li ad A. Lider Abstract.
More informationSolutions to Tutorial 5 (Week 6)
The Uiversity of Sydey School of Mathematics ad Statistics Solutios to Tutorial 5 (Wee 6 MATH2962: Real ad Complex Aalysis (Advaced Semester, 207 Web Page: http://www.maths.usyd.edu.au/u/ug/im/math2962/
More informationMath 155 (Lecture 3)
Math 55 (Lecture 3) September 8, I this lecture, we ll cosider the aswer to oe of the most basic coutig problems i combiatorics Questio How may ways are there to choose a -elemet subset of the set {,,,
More informationThe Growth of Functions. Theoretical Supplement
The Growth of Fuctios Theoretical Supplemet The Triagle Iequality The triagle iequality is a algebraic tool that is ofte useful i maipulatig absolute values of fuctios. The triagle iequality says that
More informationSome Tauberian theorems for weighted means of bounded double sequences
A. Ştiiţ. Uiv. Al. I. Cuza Iaşi. Mat. N.S. Tomul LXIII, 207, f. Some Tauberia theorems for weighted meas of bouded double sequeces Cemal Bele Received: 22.XII.202 / Revised: 24.VII.203/ Accepted: 3.VII.203
More informationSequence A sequence is a function whose domain of definition is the set of natural numbers.
Chapter Sequeces Course Title: Real Aalysis Course Code: MTH3 Course istructor: Dr Atiq ur Rehma Class: MSc-I Course URL: wwwmathcityorg/atiq/fa8-mth3 Sequeces form a importat compoet of Mathematical Aalysis
More informationRead carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book.
THE UNIVERSITY OF WARWICK FIRST YEAR EXAMINATION: Jauary 2009 Aalysis I Time Allowed:.5 hours Read carefully the istructios o the aswer book ad make sure that the particulars required are etered o each
More informationHoggatt and King [lo] defined a complete sequence of natural numbers
REPRESENTATIONS OF N AS A SUM OF DISTINCT ELEMENTS FROM SPECIAL SEQUENCES DAVID A. KLARNER, Uiversity of Alberta, Edmoto, Caada 1. INTRODUCTION Let a, I deote a sequece of atural umbers which satisfies
More informationRecurrence Relations
Recurrece Relatios Aalysis of recursive algorithms, such as: it factorial (it ) { if (==0) retur ; else retur ( * factorial(-)); } Let t be the umber of multiplicatios eeded to calculate factorial(). The
More informationINFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES I geeral, it is difficult to fid the exact sum of a series. We were able to accomplish this for geometric series ad the series /[(+)]. This is
More informationMathematical Foundations -1- Sets and Sequences. Sets and Sequences
Mathematical Foudatios -1- Sets ad Sequeces Sets ad Sequeces Methods of proof 2 Sets ad vectors 13 Plaes ad hyperplaes 18 Liearly idepedet vectors, vector spaces 2 Covex combiatios of vectors 21 eighborhoods,
More information