The Numerical Solution of Singular Fredholm Integral Equations of the Second Kind
|
|
- Phebe Elliott
- 5 years ago
- Views:
Transcription
1 WDS' Proceedigs of Cotributed Papers, Part I, 57 64, 2. ISBN MATFYZPRESS The Numerical Solutio of Sigular Fredholm Itegral Equatios of the Secod Kid J. Rak Charles Uiversity, Faculty of Mathematics ad Physics, Prague, Czech Republic., Departmet of Numerical Mathematics Abstract. This paper deals with umerical solutio of a sigular itegral equatio of the secod kid with special sigular kerel fuctio. The umerical solutio i this paper is based o Nystrom method. The Nystrom method is based o approximatio of the itegral i equatio by umerical itegratio rule. Covergece of the umerical solutio is show. The paper is completed with umerical examples. Itroductio Itegral equatios of the secod kid are ecoutered i various pieces of sciece (plasticity, plasticity, heat ad mass trasfer, oscillatio theory, fluid dyamics, filtratio theory, electrostatics, electrodyamics,..). Oe reaso for usig itegral equatios istead of differetial equatios is that all coditios specifyig boudary value problem or iitial value problem ca be reduced to oe itegral equatio. May methods for umerical solutio were developed. Most of them are described i [5]. The mai idea is to trasform itegral equatio ito system of liear equatios. Oe way is to fid solutio that satisfies itegral equatio approximately. We will obtai collocatio ad Galerki methods. Other way is to approximate itegral by umerical quadrature of cubature. We will obtai Nystrom methods. Problem is whe the kerel fuctio i itegral equatio has sigularity. Assume that λ. We will describe umerical method for the itegral equatio of the followig form λy(x) k(x, t)y(t)dt = f(x) () k(x, t) is called the kerel fuctio. For existece ad uiqueess of solutio of () let s rewrite () ito operator form (λi K)y = f (2) where K : C[, ] C[, ] is liear operator defied by Ky(x) = k(x, t)y(t)dt (3) For this paper assume a special case of sigularity called diagoal. It meas that there exist fuctios h C([, ]) C([, ]) ad g C(, ] L (, ) such that the kerel fuctio is of the form k(x, t) = g( x t )h(x, t) (4) ad fuctio g has sigularity at x =. Operator K is assumed to be compact itegral operator o C([, ]). The existece ad uiqueess of solutio of () ad (2) is described by Fredholm alterative theorem. See [4] for details. From Fredholm alterative we also have that if the operator (λi K) exists it is bouded operator. Numerical solutio - Nystrom method Now we will show umerical method for solutio of () with kerel fuctio satisfyig (4). Let s have umerical itegratio rule v(x)dx ω,j v(x,j ) (5) where x,j are called the ode poits ad satisfy x, < x,2 <... < x, < x, 57
2 Asume that umerical itegratio rule (5) coverges for all cotiuous fuctios. Here the itegratio rule ca t be applied directly to () because the kerel fuctio of form (4) is sigular. We will oly use itegratio rules that satisfy for itervals I = [a, b) ad (a, b], where b a / the coditio where c < is positive costat ad {j;x,j I } ω,j c (6) ω,j (7) Coditios (6) ad (7) are satisfied Romberg itegratio rules ad usual compoud itegratio rules. We have to formulate more coditios of fuctio g i (4). Assume that fuctio g(x) is o-icreasig fuctio o (, ] ad for all x (, ], g(x). Let {δ } be sequece δ = /. Let s defie approximatio g of g { g(r), r [δ, ] g : g (r) = (8) g(δ ), r [, δ ] Assume, that ad g () as (9) δ g(r) g (r) dr () The it easy to see that g is also o-icreasig fuctio, g C[, ] ad g (x) g(x) () g () g m (), m The sigularity i kerel fuctio i () ca be weake by followig steps. First let s rewrite () ito the form as i [2]. [ ] λ k(x, t)dt y(x) k(x, t)[y(t) y(x)] dt = f(x) (2) The the kerel fuctio i the right itegral o the left side of (2) ca be approximated by fuctio k (x, t) = g ( x t )h(x, t) (3) Note that k (x, t) = k(x, t), x t δ (4) We obtai [ λ ] k(x, t)dt y(x) ad we ca use umerical itegratio rule. We obtai [ λ ] k(x, t)dt y(x) k (x, t)[y(t) y(x)] dt = f(x) ω,j k (x, x,j )[y(x,j ) y(x)] = f(x) (5) Now let s ru x through the ode poits x,i ad we obtai system of liear equatios λ + ω,j k (x,i, x,j ) k(x i,, t)dt y(x,i ) ω,j k (x,i, x,j )y(x,j ) = f(x,i ),j i,j i (6) The itegral i (6) ca be calculated aalytically or by some special quadrature scheme. This system of liear equatios give us umerical solutio at the ode poits. I the followig we will write ω j istead of ω,j ad x j istead of x,j. 58
3 Covergece of the umerical solutio Existece of umerical solutio i (6) is here doe by operator calculus. Let s defie operator K : C[, ] C[, ]. K y(x) = ω j k (x, x j )y(x j ) (7) We ca se that operator K : C[, ] C[, ] is for each bouded ad compact operator (it is cotiuous liear ad fiite raked operator - see [4] for details). For rewritig (5) ito operator form we eed to defie differet operator K : C[, ] C[, ]. K y(x) = ω j k (x, x j )[y(x j ) y(x)] + k(x, t)y(x)dt (8) The (5) ca be rewritte ito If we defie fuctio u we ca se that (λi K )y = f (9) K = K (K u Ku)I This operator is ot compact uless K u = Ku. First let s proof existece of (λi K ) for eough large. We will use followig theorem from []: Theorem Let X be a Baach space, let operators S, T be bouded o X ad let the operator S be compact. For give λ let s assume that λi T is bijectio o X (which meas (λi T ) exists, is bouded ad R(λI T ) = X). If (T S)S < λ (λi T ) (2) the (λi S) exists, is bouded ad (λi S) If (λi T )y = f ad (λ S)z = f, the + (λi T ) S λ (λi T ) (T S)S (2) If y z (λi S) T y Sy (22) (K K )K (23) the the assumptio (2) is satisfied ad theorem ca be used for T = K ad S = K. Usig theorem there exists such that for > (λi K ) exists, is bouded ad if (λi K)y = f ad (λi K )z = f the y z (λi K ) Ky K y We eed to proof (23) Let s defie aother operator K : C[, ] C[, ] Now where K y(x) = k (x, t)y(t)dt (24) (K K )K = (K K + K K )K (K K )K + K K K (25) e (x, ξ) = (K K )K max x [,] k (x, v)k (v, ξ)dv ω j e (x, x j ) ω j k (x, x j )k (x j, ξ) 59
4 This is the umerical itegratio error for cotiuous itegrad k (x,.)k (., ξ). I [] is show, that max e (x, ξ) x,ξ [,] ad the first term i (25) goes to zero. Fuctio h from (4) is cotiuous. The there exists costat M < M = max h(x, t) x,t [,] For the secod term i (25) hece (K K )y max x [,] [k(x, t) k (x, t)]y(t) dt y max x [,] k(x, t) k (x, t) dt = δ = y max k(x, t) k (x, t) dt M y 2 g(r) g (r) dr x [,] t [,], x t <δ K K 2M δ g(r) g (r) dr by assumptio (). Now we eed to boud K idepedetly of. We will use followig lemma Lemma 2 Let [a, b] be bouded iterval. Assume that v is cotiuous o-icreasig fuctio o [a, b] ad v(x) for every x [a, b]. Assume that umerical itegratio rule coverges for all cotiuous fuctios ad satisfies coditios (6), (7). The ω j v(x j ) c b v(a) + c v(t)dt (26) PROOF Let fuctio v be o-icreasig fuctio o [a, b] ad for all x [a, b], v(x). Let s defie umbers a j, = a + (j ) b a, b j, = a + j b a, j =,..., ad itervals I j, = [a j,, b j, ), j =,..., The ad Let s defie ω j v(x j ) = i= {j;x j I,i} I j, = [a, b) I j, I i, =, j i ω = ω j v(x j )+ ω v(b) a { ω if x = b if x < b v(a i ) i= where c is costat from (6). By other calculatios i= v(a i) + v(b) = v(a ) + v(a i) + v(b ) = v(a) i=2 From the defiitio of Rieamm itegral ad the proof is complete. i= v(b i) b a {j;x j I,i} + v(t)dt i= ω j +ω v(b) c ( i= ) v(a i) + v(b) v(b i) + v(b ) v(a) + i= v(b i) 6
5 Now let s choose some x [, ] ad y C[, ] such that y =. By lemma 2 K y(x) M ω j g( x j x ) 2Mc[ g () + By (9) g / ad is bouded for all large eough. g(x). By () g (t)dt g(t)dt g (t)dt] We assumed that g L (, ). so there exists a costat c such that K c. For proof of existece (I K ) ca be used followig theorem from [3]. Theorem 3 Let operators M, L be bouded liear operators o a Baach space X. Assume that (I M) exists as a bouded Liear operator o X ad The (I L) exists ad is bouded, ad = (I M) (L M)L < (I L) + (I M) L (I L) y (I M) y (I M) Ly My + (I M) y for y (I L)X. If L is compact the (I L)X = X Let s use theorem 3 for M = λ K ad L = λ K. We eed <. Note that for u the ad from (27) ad (23) K = K (K u Ku)I K K = K u Ku (27) (K K ) K (28) Sice (λi K) is bouded liear operator, (I M) exists ad is also bouded liear operator. From (28) there exists such that (I M) λ 2 (K K ) K < for Hece < for operators M = λ K ad L = λ K whe >. By theorem 3 operator (I λ K ) exists as bouded operator from (I λ K )C([, ]) to C([, ]). Hece operator (λi K ) exists. Eve from (27) for all sufficietly large is K K < λ ad by iverse theorem (see [4]) exists (λi K + K ) from C([, ]) to C([, ]). Operator (λi K ) ca be rewritte as λi K = [I K (λi K + K ) ](λi K + K ) Sice K (λi K + K ) is compact operator from C([, ]) to C([, ]) it follows that if (λi K ) exists the (λi K ) : C([, ]) C([, ]). Existece of (λi K ) was doe by theorem 3. So we have that (λi K ) : C([, ]) C([, ]). Now let s derive error estimate. Let y be solutio of (λi K)y = f ad y be solutio of (λi K )y = f. y y = (λi K ) f y = (λi K [ ) f (λi K ] )y = (λi K [ ) (λi K)y (λi K ] )y 6
6 hece y y (λi K ) K y Ky K y(x) Ky(x) = where ad ω j k (x, x j )[y(x j ) y(x)] E y(x) = E 2 y(x) = ω j k (x, x j )[y(x j ) y(x)] k (x, t)[y(t) y(x)] dt k(x, t)[y(t) y(x)] dt = E y(x) + E 2 y(x) (29) k (x, t)[y(t) y(x)] dt (3) k(x, t)[y(t) y(x)] dt (3) E is umerical itegratio error ad E 2 depeds o approximatio of sigular kerel fuctio k(x, t) by k (x, t). More estimates will be doe by followig example. Example Let s have a equatio y(t) y(x) dt = f(x) (32) x t γ where γ (, ) Here the fuctio g from (4) is g is oicreasig o (, ]. Let s defie δ = /. Approximatio (8) g(r) = r γ (33) g (r) = { r γ, r [δ, ] γ, r [, δ ] (34) We eed to verify compactess of K o o C[, ] Ky(x) = ad two assumptios (9) ad (). If we defie K y(x) = g ( x t )y(t)dt we ca se that the latter itegral operator K is compact for each. Ky(x) K y(x) ad g( x t ) g ( x t )y(t) dt 2 y K K y(t) dt (35) x t γ (r γ γ )dr = 2γ γ γ y 2γ (36) γ γ K is a sequece of compact operators o C[, ]. By (36) K K. Hece K is compact operator o C[, ]. This is a stadard result ad ca be foud i ay book of fuctioal aalysis (for example i [4].) We have eve proofed (). Sice g () = γ as is (9) also satisfied. Let s try compoud midpoit itegratio rule for (32) with γ = /2. ω j = /, x j = 2 + j, =, 2,..., ad f(x)dx ω j f(x j ) 62
7 Assume that v C 2 [, ]. The compoud midpoit rule satisfies (see [6] for details) v(x) ω j v(x j ) 24 2 max x [a,b] v (x) Hece For E 2 E 2 y(x) max x [,] E y(x) D 2,where D = v (x) 24 k(x, t) k (x, t) y(t) y(x) dt 2 y k(x, t) k (x, t) dt = = 2 y g( x t ) g ( x t ) dt = 4γ γ γ y If we add other assumptios o fuctio y(x) we ca have better results. Assume that The E 2 y(x) (37) y(t) y(x) A t x α for < α ad A < (38) k(x, t) k (x, t) y(t) y(x) dt A k(x, t) k (x, t) t x α dt t [,], t x < A α k(x, t) k (x, t) dt = 2 A γ α+ γ (39) To demostrate error behavior let s choose five differet fuctio f(x). The system of liear equatios (6) is solved by Gauss elimiatio. All itegratio is doe by software Maple [7]. The error developmet is described by ratio fuctio r where is umber of ode poits I first example let s choose r = y(x) y /2(x) y(x) y (x) f(x) = x ( 4 3 x x x x x) The exact solutio is y(x) = x ad satisfies (38) with α =. Followig two table shows umerical solutio y (x) ad error i the ode poits. Followig table shows error evolutio. 5 ode poits x y(x) y (x) y(x) y (x),,,966474,335926,3,3, ,743679,5,5,5,,7,7, ,743679,9,9, , odes y(x) y (x) r odes y(x) y (x) r 5, ,4444 3,84,234 3,2 6,384 3,2 2,6383 3,63 32,57 2,73 4,78 3,74 64,8 2,8 From (37) ad (39) the theoretical value of r should be approximately 2, 83. For other examples oly error evolutio tables will be show. Let s ow choose f(x) = x 2 ( 6 5 x2 x x2 x x x + 2 x) 5 The exact solutio is y(x) = x 2 ad also satisfies (38) with α =. So we are expectig r 2, 83. Followig table shows error evolutio. 63
8 odes y(x) y (x) r odes y(x) y (x) r 5, ,2399 3,, ,69 6,45 3,9 2,4254 3,89 32,36 3,5 4,3857 3,69 64,435 3,3 Let s ow choose f(x) such that exact solutio is y(x) = e x. (38) is also satisfied with α =. we expect r Error evolutio is show by followig table. odes y(x) y (x) r odes y(x) y (x) r 5, ,426 3,8, ,58 6,482 2,95 2,68 3,83 32,636 2,95 4, ,84 64,549 2,98 Now let s choose f(x) so that the exact solutio is y(x) = x. The solutio satisfies (38) with α = /2. The r is expected approximately 2. odes y(x) y (x) r odes y(x) y (x) r 5, ,564 2,8, ,2 6, ,7 2, ,7 32, ,5 4, ,28 64,5234 2,2 For last example let s choose f(x) so that the exact solutio is y(x) = 4 x. The solutio satisfies (38) with α = /4. r is expected approximately, 68. Coclusio odes y(x) y (x) r odes y(x) y (x) r 5, ,48252,87, ,8 6,227937,83 2, ,32 32,25873,8 4,77643,9 64,7393,79 From the tables we ca se that results correspods to theory. Other methods - projectio methods or product itegratio methods (see [5]) have better error behavior. But here oly diagoal members of the matrix are itegrals (this is ot true for projectio ad product itegratio methods). Others are easy to compute from the kerel fuctio. This is the advatage of this method whe usig some iterative scheme for solvig liear systems of equatios, because we ca reduce memory usage. Disadvatage is that matrixes are fully populated. Refereces [] Kedall E. Atkiso, Weimi Ha: Theoretical Numerical Aalysis, Spriger-Verlag, ISBN: , 2. [2] L. V. Katorovich, V. I. Krylov: Approximate Methods of Higher Aalysis, Itersciece, New York, 958. [3] P. M. Aseloe: Collectively compact operator approximatio theory, Pretice-Hall, Egelwood Cliffs, NJ, 97. [4] Jaroslav Lukes: Zapisky z fukcioali aalyzy (Karolium), ISBN [5] Kedall E. Atkiso: The Numerical Solutio of Itegral Equatios of the Secod Kid (Cambridge Uiversity Press), ISBN: , 997. [6] Prem K. Kythe, Michael R. Schaferkotter: Hadbook of Computatioal Methods for Itegratio (Chapma ad Hall/CRC), ISBN: , 24. [7] Maplesoft: Maple, 64
Sequences 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 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 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 informationA collocation method for singular integral equations with cosecant kernel via Semi-trigonometric interpolation
Iteratioal Joural of Mathematics Research. ISSN 0976-5840 Volume 9 Number 1 (017) pp. 45-51 Iteratioal Research Publicatio House http://www.irphouse.com A collocatio method for sigular itegral equatios
More informationn 3 ln n n ln n is convergent by p-series for p = 2 > 1. n2 Therefore we can apply Limit Comparison Test to determine lutely convergent.
06 微甲 0-04 06-0 班期中考解答和評分標準. ( poits) Determie whether the series is absolutely coverget, coditioally coverget, or diverget. Please state the tests which you use. (a) ( poits) (b) ( poits) (c) ( poits)
More informationMATH 10550, EXAM 3 SOLUTIONS
MATH 155, EXAM 3 SOLUTIONS 1. I fidig a approximate solutio to the equatio x 3 +x 4 = usig Newto s method with iitial approximatio x 1 = 1, what is x? Solutio. Recall that x +1 = x f(x ) f (x ). Hece,
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 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 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 informationIntegrable Functions. { f n } is called a determining sequence for f. If f is integrable with respect to, then f d does exist as a finite real number
MATH 532 Itegrable Fuctios Dr. Neal, WKU We ow shall defie what it meas for a measurable fuctio to be itegrable, show that all itegral properties of simple fuctios still hold, ad the give some coditios
More informationNumerical Conformal Mapping via a Fredholm Integral Equation using Fourier Method ABSTRACT INTRODUCTION
alaysia Joural of athematical Scieces 3(1): 83-93 (9) umerical Coformal appig via a Fredholm Itegral Equatio usig Fourier ethod 1 Ali Hassa ohamed urid ad Teh Yua Yig 1, Departmet of athematics, Faculty
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 informationNTMSCI 5, No. 1, (2017) 26
NTMSCI 5, No. 1, - (17) New Treds i Mathematical Scieces http://dx.doi.org/1.85/tmsci.17.1 The geeralized successive approximatio ad Padé approximats method for solvig a elasticity problem of based o the
More informationA NEW CLASS OF 2-STEP RATIONAL MULTISTEP METHODS
Jural Karya Asli Loreka Ahli Matematik Vol. No. (010) page 6-9. Jural Karya Asli Loreka Ahli Matematik A NEW CLASS OF -STEP RATIONAL MULTISTEP METHODS 1 Nazeeruddi Yaacob Teh Yua Yig Norma Alias 1 Departmet
More information1 Approximating Integrals using Taylor Polynomials
Seughee Ye Ma 8: Week 7 Nov Week 7 Summary This week, we will lear how we ca approximate itegrals usig Taylor series ad umerical methods. Topics Page Approximatig Itegrals usig Taylor Polyomials. Defiitios................................................
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 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 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 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 informationComparison Study of Series Approximation. and Convergence between Chebyshev. and Legendre Series
Applied Mathematical Scieces, Vol. 7, 03, o. 6, 3-337 HIKARI Ltd, www.m-hikari.com http://d.doi.org/0.988/ams.03.3430 Compariso Study of Series Approimatio ad Covergece betwee Chebyshev ad Legedre Series
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 informationMath 113, Calculus II Winter 2007 Final Exam Solutions
Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this
More informationf(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim
Math 3, Sectio 2. (25 poits) Why we defie f(x) dx as we do. (a) Show that the improper itegral diverges. Hece the improper itegral x 2 + x 2 + b also diverges. Solutio: We compute x 2 + = lim b x 2 + =
More informationMa 4121: Introduction to Lebesgue Integration Solutions to Homework Assignment 5
Ma 42: Itroductio to Lebesgue Itegratio Solutios to Homework Assigmet 5 Prof. Wickerhauser Due Thursday, April th, 23 Please retur your solutios to the istructor by the ed of class o the due date. You
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 informationMATH 31B: MIDTERM 2 REVIEW
MATH 3B: MIDTERM REVIEW JOE HUGHES. Evaluate x (x ) (x 3).. Partial Fractios Solutio: The umerator has degree less tha the deomiator, so we ca use partial fractios. Write x (x ) (x 3) = A x + A (x ) +
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 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 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 informationTMA4205 Numerical Linear Algebra. The Poisson problem in R 2 : diagonalization methods
TMA4205 Numerical Liear Algebra The Poisso problem i R 2 : diagoalizatio methods September 3, 2007 c Eiar M Røquist Departmet of Mathematical Scieces NTNU, N-749 Trodheim, Norway All rights reserved A
More informationAN INVERSE STURM-LIOUVILLE PROBLEM WITH A GENERALIZED SYMMETRIC POTENTIAL
Electroic Joural of Differetial Equatios, Vol. 7 (7, No. 4, pp. 7. ISSN: 7-669. URL: http://ejde.math.txstate.edu or http://ejde.math.ut.edu AN INVERSE STURM-LIOUVILLE PROBLEM WITH A GENERALIZED SYMMETRIC
More informationReal Analysis Fall 2004 Take Home Test 1 SOLUTIONS. < ε. Hence lim
Real Aalysis Fall 004 Take Home Test SOLUTIONS. Use the defiitio of a limit to show that (a) lim si = 0 (b) Proof. Let ε > 0 be give. Defie N >, where N is a positive iteger. The for ε > N, si 0 < si
More informationMath 312 Lecture Notes One Dimensional Maps
Math 312 Lecture Notes Oe Dimesioal Maps Warre Weckesser Departmet of Mathematics Colgate Uiversity 21-23 February 25 A Example We begi with the simplest model of populatio growth. Suppose, for example,
More informationMath 113 Exam 3 Practice
Math Exam Practice Exam will cover.-.9. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for you
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 informationCS321. Numerical Analysis and Computing
CS Numerical Aalysis ad Computig Lecture Locatig Roots o Equatios Proessor Ju Zhag Departmet o Computer Sciece Uiversity o Ketucky Leigto KY 456-6 September 8 5 What is the Root May physical system ca
More informationSolutions to Homework 7
Solutios to Homework 7 Due Wedesday, August 4, 004. Chapter 4.1) 3, 4, 9, 0, 7, 30. Chapter 4.) 4, 9, 10, 11, 1. Chapter 4.1. Solutio to problem 3. The sum has the form a 1 a + a 3 with a k = 1/k. Sice
More informationHOMEWORK #10 SOLUTIONS
Math 33 - Aalysis I Sprig 29 HOMEWORK # SOLUTIONS () Prove that the fuctio f(x) = x 3 is (Riema) itegrable o [, ] ad show that x 3 dx = 4. (Without usig formulae for itegratio that you leart i previous
More informationProbability 2 - Notes 10. Lemma. If X is a random variable and g(x) 0 for all x in the support of f X, then P(g(X) 1) E[g(X)].
Probability 2 - Notes 0 Some Useful Iequalities. Lemma. If X is a radom variable ad g(x 0 for all x i the support of f X, the P(g(X E[g(X]. Proof. (cotiuous case P(g(X Corollaries x:g(x f X (xdx x:g(x
More informationTaylor polynomial solution of difference equation with constant coefficients via time scales calculus
TMSCI 3, o 3, 129-135 (2015) 129 ew Treds i Mathematical Scieces http://wwwtmscicom Taylor polyomial solutio of differece equatio with costat coefficiets via time scales calculus Veysel Fuat Hatipoglu
More informationInverse Nodal Problems for Differential Equation on the Half-line
Australia Joural of Basic ad Applied Scieces, 3(4): 4498-4502, 2009 ISSN 1991-8178 Iverse Nodal Problems for Differetial Equatio o the Half-lie 1 2 3 A. Dabbaghia, A. Nematy ad Sh. Akbarpoor 1 Islamic
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 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 informationNumerical Method for Blasius Equation on an infinite Interval
Numerical Method for Blasius Equatio o a ifiite Iterval Alexader I. Zadori Omsk departmet of Sobolev Mathematics Istitute of Siberia Brach of Russia Academy of Scieces, Russia zadori@iitam.omsk.et.ru 1
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 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 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 informationChapter 4 : Laplace Transform
4. Itroductio Laplace trasform is a alterative to solve the differetial equatio by the complex frequecy domai ( s = σ + jω), istead of the usual time domai. The DE ca be easily trasformed ito a algebraic
More informationA Fixed Point Result Using a Function of 5-Variables
Joural of Physical Scieces, Vol., 2007, 57-6 Fixed Poit Result Usig a Fuctio of 5-Variables P. N. Dutta ad Biayak S. Choudhury Departmet of Mathematics Begal Egieerig ad Sciece Uiversity, Shibpur P.O.:
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 informationMATH 413 FINAL EXAM. f(x) f(y) M x y. x + 1 n
MATH 43 FINAL EXAM Math 43 fial exam, 3 May 28. The exam starts at 9: am ad you have 5 miutes. No textbooks or calculators may be used durig the exam. This exam is prited o both sides of the paper. Good
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 informationx x x 2x x N ( ) p NUMERICAL METHODS UNIT-I-SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS By Newton-Raphson formula
NUMERICAL METHODS UNIT-I-SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS. If g( is cotiuous i [a,b], te uder wat coditio te iterative (or iteratio metod = g( as a uique solutio i [a,b]? '( i [a,b].. Wat
More informationSection A assesses the Units Numerical Analysis 1 and 2 Section B assesses the Unit Mathematics for Applied Mathematics
X0/70 NATIONAL QUALIFICATIONS 005 MONDAY, MAY.00 PM 4.00 PM APPLIED MATHEMATICS ADVANCED HIGHER Numerical Aalysis Read carefully. Calculators may be used i this paper.. Cadidates should aswer all questios.
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 informationMDIV. Multiple divisor functions
MDIV. Multiple divisor fuctios The fuctios τ k For k, defie τ k ( to be the umber of (ordered factorisatios of ito k factors, i other words, the umber of ordered k-tuples (j, j 2,..., j k with j j 2...
More informationMATH4822E FOURIER ANALYSIS AND ITS APPLICATIONS
MATH48E FOURIER ANALYSIS AND ITS APPLICATIONS 7.. Cesàro summability. 7. Summability methods Arithmetic meas. The followig idea is due to the Italia geometer Eresto Cesàro (859-96). He shows that eve if
More informationME NUMERICAL METHODS Fall 2007
ME - 310 NUMERICAL METHODS Fall 2007 Group 02 Istructor: Prof. Dr. Eres Söylemez (Rm C205, email:eres@metu.edu.tr ) Class Hours ad Room: Moday 13:40-15:30 Rm: B101 Wedesday 12:40-13:30 Rm: B103 Course
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 informationTHE SOLUTION OF NONLINEAR EQUATIONS f( x ) = 0.
THE SOLUTION OF NONLINEAR EQUATIONS f( ) = 0. Noliear Equatio Solvers Bracketig. Graphical. Aalytical Ope Methods Bisectio False Positio (Regula-Falsi) Fied poit iteratio Newto Raphso Secat The root of
More informationEntire Functions That Share One Value with One or Two of Their Derivatives
JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS 223, 88 95 1998 ARTICLE NO. AY985959 Etire Fuctios That Share Oe Value with Oe or Two of Their Derivatives Gary G. Guderse* Departmet of Mathematics, Ui
More informationECE-S352 Introduction to Digital Signal Processing Lecture 3A Direct Solution of Difference Equations
ECE-S352 Itroductio to Digital Sigal Processig Lecture 3A Direct Solutio of Differece Equatios Discrete Time Systems Described by Differece Equatios Uit impulse (sample) respose h() of a DT system allows
More information1.3 Convergence Theorems of Fourier Series. k k k k. N N k 1. With this in mind, we state (without proof) the convergence of Fourier series.
.3 Covergece Theorems of Fourier Series I this sectio, we preset the covergece of Fourier series. A ifiite sum is, by defiitio, a limit of partial sums, that is, a cos( kx) b si( kx) lim a cos( kx) b si(
More informationMIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS
MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will
More informationZ ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew
Problem ( poits) Evaluate the itegrals Z p x 9 x We ca draw a right triagle labeled this way x p x 9 From this we ca read off x = sec, so = sec ta, ad p x 9 = R ta. Puttig those pieces ito the itegralrwe
More informationResearch Article A New Second-Order Iteration Method for Solving Nonlinear Equations
Abstract ad Applied Aalysis Volume 2013, Article ID 487062, 4 pages http://dx.doi.org/10.1155/2013/487062 Research Article A New Secod-Order Iteratio Method for Solvig Noliear Equatios Shi Mi Kag, 1 Arif
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 informationAnalytic Continuation
Aalytic Cotiuatio The stadard example of this is give by Example Let h (z) = 1 + z + z 2 + z 3 +... kow to coverge oly for z < 1. I fact h (z) = 1/ (1 z) for such z. Yet H (z) = 1/ (1 z) is defied for
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 informationAdditional Notes on Power Series
Additioal Notes o Power Series Mauela Girotti MATH 37-0 Advaced Calculus of oe variable Cotets Quick recall 2 Abel s Theorem 2 3 Differetiatio ad Itegratio of Power series 4 Quick recall We recall here
More information1 Lecture 2: Sequence, Series and power series (8/14/2012)
Summer Jump-Start Program for Aalysis, 202 Sog-Yig Li Lecture 2: Sequece, Series ad power series (8/4/202). More o sequeces Example.. Let {x } ad {y } be two bouded sequeces. Show lim sup (x + y ) lim
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 informationINVERSE THEOREMS OF APPROXIMATION THEORY IN L p,α (R + )
Electroic Joural of Mathematical Aalysis ad Applicatios, Vol. 3(2) July 2015, pp. 92-99. ISSN: 2090-729(olie) http://fcag-egypt.com/jourals/ejmaa/ INVERSE THEOREMS OF APPROXIMATION THEORY IN L p,α (R +
More informationInverse Matrix. A meaning that matrix B is an inverse of matrix A.
Iverse Matrix Two square matrices A ad B of dimesios are called iverses to oe aother if the followig holds, AB BA I (11) The otio is dual but we ofte write 1 B A meaig that matrix B is a iverse of matrix
More informationf(w) w z =R z a 0 a n a nz n Liouville s theorem, we see that Q is constant, which implies that P is constant, which is a contradiction.
Theorem 3.6.4. [Liouville s Theorem] Every bouded etire fuctio is costat. Proof. Let f be a etire fuctio. Suppose that there is M R such that M for ay z C. The for ay z C ad R > 0 f (z) f(w) 2πi (w z)
More informationChapter 4. Fourier Series
Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,
More informationTaylor expansion: Show that the TE of f(x)= sin(x) around. sin(x) = x - + 3! 5! L 7 & 8: MHD/ZAH
Taylor epasio: Let ƒ() be a ifiitely differetiable real fuctio. A ay poit i the eighbourhood of 0, the fuctio ƒ() ca be represeted by a power series of the followig form: X 0 f(a) f() f() ( ) f( ) ( )
More informationLecture 3: Convergence of Fourier Series
Lecture 3: Covergece of Fourier Series Himashu Tyagi Let f be a absolutely itegrable fuctio o T : [ π,π], i.e., f L (T). For,,... defie ˆf() f(θ)e i θ dθ. π T The series ˆf()e i θ is called the Fourier
More informationSeries: Infinite Sums
Series: Ifiite Sums Series are a way to mae sese of certai types of ifiitely log sums. We will eed to be able to do this if we are to attai our goal of approximatig trascedetal fuctios by usig ifiite degree
More informationNumerical Solution of the Two Point Boundary Value Problems By Using Wavelet Bases of Hermite Cubic Spline Wavelets
Australia Joural of Basic ad Applied Scieces, 5(): 98-5, ISSN 99-878 Numerical Solutio of the Two Poit Boudary Value Problems By Usig Wavelet Bases of Hermite Cubic Splie Wavelets Mehdi Yousefi, Hesam-Aldie
More informationJANE PROFESSOR WW Prob Lib1 Summer 2000
JANE PROFESSOR WW Prob Lib Summer 000 Sample WeBWorK problems. WeBWorK assigmet Series6CompTests due /6/06 at :00 AM..( pt) Test each of the followig series for covergece by either the Compariso Test or
More information*X203/701* X203/701. APPLIED MATHEMATICS ADVANCED HIGHER Numerical Analysis. Read carefully
X0/70 NATIONAL QUALIFICATIONS 006 MONDAY, MAY.00 PM.00 PM APPLIED MATHEMATICS ADVANCED HIGHER Numerical Aalysis Read carefully. Calculators may be used i this paper.. Cadidates should aswer all questios.
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 informationON CONVERGENCE OF BASIC HYPERGEOMETRIC SERIES. 1. Introduction Basic hypergeometric series (cf. [GR]) with the base q is defined by
ON CONVERGENCE OF BASIC HYPERGEOMETRIC SERIES TOSHIO OSHIMA Abstract. We examie the covergece of q-hypergeometric series whe q =. We give a coditio so that the radius of the covergece is positive ad get
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Fall 2013 Lecture 21 11/27/2013
MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Fall 2013 Lecture 21 11/27/2013 Fuctioal Law of Large Numbers. Costructio of the Wieer Measure Cotet. 1. Additioal techical results o weak covergece
More informationMost text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t
Itroductio to Differetial Equatios Defiitios ad Termiolog Differetial Equatio: A equatio cotaiig the derivatives of oe or more depedet variables, with respect to oe or more idepedet variables, is said
More informationA 2nTH ORDER LINEAR DIFFERENCE EQUATION
A 2TH ORDER LINEAR DIFFERENCE EQUATION Doug Aderso Departmet of Mathematics ad Computer Sciece, Cocordia College Moorhead, MN 56562, USA ABSTRACT: We give a formulatio of geeralized zeros ad (, )-discojugacy
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 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 informationCHAPTER 5. Theory and Solution Using Matrix Techniques
A SERIES OF CLASS NOTES FOR 2005-2006 TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS DE CLASS NOTES 3 A COLLECTION OF HANDOUTS ON SYSTEMS OF ORDINARY DIFFERENTIAL
More informationPart I: Covers Sequence through Series Comparison Tests
Part I: Covers Sequece through Series Compariso Tests. Give a example of each of the followig: (a) A geometric sequece: (b) A alteratig sequece: (c) A sequece that is bouded, but ot coverget: (d) A sequece
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 informationB Supplemental Notes 2 Hypergeometric, Binomial, Poisson and Multinomial Random Variables and Borel Sets
B671-672 Supplemetal otes 2 Hypergeometric, Biomial, Poisso ad Multiomial Radom Variables ad Borel Sets 1 Biomial Approximatio to the Hypergeometric Recall that the Hypergeometric istributio is fx = x
More informationPAijpam.eu ON DERIVATION OF RATIONAL SOLUTIONS OF BABBAGE S FUNCTIONAL EQUATION
Iteratioal Joural of Pure ad Applied Mathematics Volume 94 No. 204, 9-20 ISSN: 3-8080 (prited versio); ISSN: 34-3395 (o-lie versio) url: http://www.ijpam.eu doi: http://dx.doi.org/0.2732/ijpam.v94i.2 PAijpam.eu
More informationBernoulli, Ramanujan, Toeplitz e le matrici triangolari
Due Giori di Algebra Lieare Numerica www.dima.uige.it/ dibeede/gg/home.html Geova, 6 7 Febbraio Beroulli, Ramauja, Toeplitz e le matrici triagolari Carmie Di Fiore, Fracesco Tudisco, Paolo Zellii Speaker:
More informations = and t = with C ij = A i B j F. (i) Note that cs = M and so ca i µ(a i ) I E (cs) = = c a i µ(a i ) = ci E (s). (ii) Note that s + t = M and so
3 From the otes we see that the parts of Theorem 4. that cocer us are: Let s ad t be two simple o-egative F-measurable fuctios o X, F, µ ad E, F F. The i I E cs ci E s for all c R, ii I E s + t I E s +
More informationCS537. Numerical Analysis and Computing
CS57 Numerical Aalysis ad Computig Lecture Locatig Roots o Equatios Proessor Ju Zhag Departmet o Computer Sciece Uiversity o Ketucky Leigto KY 456-6 Jauary 9 9 What is the Root May physical system ca be
More informationMAT 271 Project: Partial Fractions for certain rational functions
MAT 7 Project: Partial Fractios for certai ratioal fuctios Prerequisite kowledge: partial fractios from MAT 7, a very good commad of factorig ad complex umbers from Precalculus. To complete this project,
More informationThe value of Banach limits on a certain sequence of all rational numbers in the interval (0,1) Bao Qi Feng
The value of Baach limits o a certai sequece of all ratioal umbers i the iterval 0, Bao Qi Feg Departmet of Mathematical Scieces, Ket State Uiversity, Tuscarawas, 330 Uiversity Dr. NE, New Philadelphia,
More informationNew Version of the Rayleigh Schrödinger Perturbation Theory: Examples
New Versio of the Rayleigh Schrödiger Perturbatio Theory: Examples MILOŠ KALHOUS, 1 L. SKÁLA, 1 J. ZAMASTIL, 1 J. ČÍŽEK 2 1 Charles Uiversity, Faculty of Mathematics Physics, Ke Karlovu 3, 12116 Prague
More information