ACM 105: Applied Real and Functional Analysis. Solutions to Homework # 2.


 Stephen Harmon
 2 years ago
 Views:
Transcription
1 ACM 05: Applied Rel nd Functionl Anlysis. Solutions to Homework # 2. Andy Greenberg, Alexei Novikov Problem. RiemnnLebesgue Theorem. Theorem (G.F.B. Riemnn, H.L. Lebesgue). If f is n integrble function on (, + ), then lim + 0 f(x) cos dx = 0 Proof. We first note tht if the intervl of integrtion were finite [, b] with f(x) on it, then / f(x) cos dx = cos dx = cos(y)dy = O() 0 s 0, since the integrl of the cosine is bounded. We shll reduce the generl cse to this simple cse. First off, due to the integrbility of f on (, + ) nd the following estimtes, f(x) cos ( x ) dx + + b / + f(x) cos f(x) cos ( x ) dx we cn lwys find finite intervl [, b], such tht + f(x) cos dx dx f(x) cos f(x) cos f (x) dx + dx < ε 3 + b dx f (x) dx (by bounding the integrls long the tils ). Furthermore, since f is integrble, there exists sequence of integrble simple functions f n (x), such tht implying tht for N lrge enough, f(x) dx = lim n f N (x) dx f n (x) dx f(x) dx < ε 3
2 We thus hve f(x) cos dx f N (x) cos dx dx f(x) f N (x) cos f(x) f N (x) dx < ε 3 Now, we cn lso find (see proof of Royden s Proposition 3.22 in Homework ) set A with µ([, b]\a) < δ, for some δ > 0, nd step function g(x), such tht g(x) = f N (x) on A with f N (x) dx g(x) dx < ε 3 But g(x) is step function, so there exists finite number of points x j, j = 0,..., m with = x 0 < x <... < x m = b, such tht g(x) const = c j on [x j, x j ]. Therefore g(x) cos dx = m x j j= x j c j cos dx = m ( ) x j c j sin 0 j= x j s 0. Since the bove mnipultions gurntee tht + f(x) cos dx g(x) cos dx < ε for n rbitrrily smll ε, we re done. Problem 2. Lx Equivlence Theorem: Solution Opertor. For the initil vlue problem du(t) dt = Lu(t), 0 t T u(0) = u 0, () nd the norm V in the spce V u(t), we denote S(t) the solution opertor: if u(t) is the solution of (), then u(t) = S(t)u 0, u 0 V 0. We hve lim t 0 (u(t + t) u(t) Lu(t)) t = 0 (2). From the continuous dependence property we hve sup S(t)(u 0 u 2 (u 0) C 0 0 u 2 V 0) nd 0 t T sup S(t)u 0 V C 0 u 0 V u 0 V 0 0 t T This implies tht S(t) C 0, so S(t) is bounded on V 0. Since V 0 is dense in V, the required extension of S(t) on the whole spce V exists nd is unique by the extension theorem for opertors. 2 V
3 b. (continuity of generlized solutions). Choose sequence {u 0,n } V 0 which converges to u 0 in V : u 0,n u 0 V 0 s n Fix t 0 [0, T ] nd let t [0, T ]. We write u(t) u(t 0 ) = S(t)u 0 S(t 0 )u 0 = S(t)(u 0 u 0,n ) + (S(t) S(t 0 ))u 0,n S(t 0 )(u 0 u 0,n ) In estimting the norm of this difference, we use the fct tht S(t) C 0 in the first nd the lst terms of the righthnd side: u(t) u(t 0 ) V 2C 0 u 0,n u 0 V + (S(t) S(t 0 ))u 0,n V Given n ε > 0 we choose sufficiently lrge n, such tht 2C 0 u 0,n u 0 V < ε 2 For this n we use the definition of the solution (2) to come up with δ > 0, such tht for t t 0 < δ, (S(t) S(t 0 ))u 0,n V < ε 2 Then for t [0, T ] with t t 0 < δ,we hve u(t) u(t 0 ) V < ε, so tht u(t) is continuous. c. (semigroup property). The solution to the problem () is u(t) = S(t)u 0. We hve u(s) = = S(s)u 0 nd S(t)u(s) is the solution of the differentil eqution on [s, T ] with the initil condition u(s) imposed t time s. By the uniqueness of solution, i.e., S(t)u(s) = u(t + s) S(t)S(s)u 0 = S(t + s)u 0, nd therefore S(t + s) = S(t)S(s), s required. Problem 3. Lx Equivlence Theorem: Proof. Recll the definitions. Definition. A difference method is oneprmeter fmily of opertors C( t) : V V, nd there exists t 0 > 0, such tht t (0, t 0 ] C( t) c (C is uniformly bounded). The pproximte solution is u t (m t) = C( t) m, m =, 2,... Definition 2. A difference method is consistent if V c V, dense subspce of V, such tht u 0 V c for the corresponding solution of the initil vlue problem (), we hve u(t + t) C( t)u(t) lim = 0 t 0 t uniformly on [0, T ]. Definition 3. A difference method is convergent if for ny fixed t [0, T ] nd ny u 0 V we hve lim t i 0 (C( t i) m i S(t))u 0 = 0 3
4 where the two sequences: {m i }, of integers nd { t i }, of step sizes re such tht lim i m i t i = t. Definition 4. A difference method is stble if the opertors {C( t) m 0 < t t0, m t T } re uniformly bounded, i.e., there exists constnt M 0 > 0 such tht C( t) m V V M 0 m : m t T, t t 0 Theorem (P.D. Lx). Suppose the initil vlue problem () is wellposed. Then for consistent difference method, stbility is equivlent to cnvergence. Proof. (Stbility = Convergence). Consider the error C( t) m u 0 u(t) = m j= C( t) j [C( t)u((m j) t) u((m j) t)] + u(m t) u(t) We use the hint: first consider u 0 V c. Then since the method is stble, C( t) m u 0 u(t) M 0 m t sup C( t)u(t) u(t + t) t t + u(m t) u(t) The sup goes to 0 by consistency, while continuity implies u(m t) u(t) 0, so convergence is estblished in this cse. Now consider the generl cse when u V nd not necessrily in V c. Since V c is dense in V, we hve sequence {u 0,n } V 0 such tht u 0,n u 0 in V. Writing (recll the construction in Problem 2) C( t) m u 0 u(t) = C( t) m (u 0 u 0,n ) + [C( t) m S(t)]u 0,n S(t)(u 0 u 0,n ), we obtin, by tringle inequlity, C( t) m u 0 u(t) C( t) m (u 0 u 0,n ) + [C( t) m S(t)]u 0,n + S(t)(u 0 u 0,n ) Since the initil vlue problem () is wellposed nd the method is stble, C( t) m u 0 u(t) c u 0 u 0,n + [C( t) m S(t)]u 0,n Given ny ε > 0, we cn find sufficiently lrge n, so tht For this n, tke t sufficiently smll, so tht c u 0 u 0,n < ε 2 [C( t) m S(t)]u 0,n ε 2 t smll, m t t < t nd we hve convergence. (Convergence = Stbility). We gin use the hint nd invoke proof by contrdiction. Suppose the method is not stble. Then there re sequences { t k } nd {m k } such tht m k t k T nd lim C( t k) m k = k 4
5 Since t k t, we my ssume tht the sequence { t k } is convergent. (Techniclly, we only know tht, since { t k } is bounded, it hs convergent subsequence, but we cn ssume here, without loss of generlity, tht it is the whole sequence.) First ssume tht {m k } is bounded. Then sup k C( t k ) m k sup C( t k ) m k < k which is contrdiction. Thus necessrily m k nd t k 0 s k. By the convergence of the method, sup C( t k ) m k u 0 < k The principle of uniform boundedness then implies lim C( t k) m k < k u 0 V contrdicting the ssumption tht the method is not stble, so the theorem is proved completely. Problem 4. Het Eqution: Discretiztion. u t = u xx + f(t, x) in [0, π] [0, T ] u(0, t) = u(π, t) = 0 0 t T u(x, 0) = u 0 (x) 0 x π (3). Let V = C 0 [0, π] = {v C[0, π], v(0) = v(π) = 0} with the norm v C0 = mx 0 x π v(x) Since both continuity nd the vlues t the boundries re preserved under liner combintions, V is obviously liner spce. To prove completeness, consider Cuchy (fundmentl) sequence in V : for ny ε > 0 N = N(ε) such tht for ny n, m > N we hve v n (x) v m (x) = mx 0 x π v n(x) v m (x) < ε By the uniform convergence criterion, the condition bove implies tht the Cuchy sequence converges uniformly on [0, π] to function v(x). But uniform limit of continuous functions is continuous, so v C[0, π]. By continuity, since v n (0) = v n (π) = 0 n, it follows tht lso v(0) = v(π) = 0, so v V, nd V is complete liner normed spce, i.e., Bnch spce. Now tke V 0, the spce of liner combintions of sines. Obviously ny v V 0 is continuous nd vnishes t the boundries (0 nd π). Therefore, V 0 V. A wellknown result from Fourier nlysis sys tht ny continuous function on [0, π], vnishing t the boundries, cn be represented s Fourier sine series, v(x) = k sin(kx) Therefore for ny function v V we cn find sequence {v n } V 0, v n (x) = k sin(kx) so tht v n v. Thus the closure of V 0, V 0 V, so V 0 is dense in V. 5
6 b. Define C( t, r)v(x) = ( 2r)v(x) + r(v(x + x) + v(x x)) Here h t = t, h x = x = t/r. If x ± x / [0, π], then we tke the odd 2πperiodic extension of v. Obviously, C( t) : V V is liner opertor. Moreover, so C( t)v V = ( 2r)v(x) + r(v(x + x) + v(x x)) V ( 2r)v V + rv V + rv V = ( 2r + 2r) v V nd the fmily {C( t)} is uniformly bounded. C( t) 2r + 2r, (4) For consistency nlysis, tke V c = V 0. For ny initil condition u 0 (x) V c we then obtin smooth solution s Fourier sine series u(t, x) = b k e k2t sin(kx) + F k (t) sin(kx) Here u 0 (x) = n b k sin(kx) nd F k (t) re due to the source term. (It cn be shown tht F k looks like F k (t) = t 0 f k (τ)e k2 (τ t) dτ where f(t, x) = n f k (t) sin(kx).) Now use Tylor series expnsion in (x, t) with reminder to write C( t)u(x, t) u(x, t + t) = ( 2r)u(x, t) + r(u(x + x, t) + u(x x, t)) u(x, t + t) = ( 2r)u(x, t) + r(2u(x, t) + u xx (x, t)( x) 2 ) + + r 4! ( x)4 (u xxxx (x + θ x, t) + u xxxx (x θ 2 x, t)) u(x, t) u t (x, t) t 2 u tt(x, t + θ 3 t)( t) 2 = t [ u t (x, t) + r ( x)2 u xx (x, t) + t +r ( x)2 ( x) 2 t 24 (u xxxx(x + θ x, t) + u xxxx (x θ 2 x, t)) ] 2 u tt(x, t + θ 3 t) t Since x = t/r, (r/ t)( x) 2 =, nd the first two terms combine to f(x, t) t. For consistency, the remining terms must go to zero fster thn t. We hve r ( x)2 ( x) 2 t 24 (u xxxx(x + θ x, t) + u xxxx (x θ 2 x, t)) 2 u tt(x, t + θ 3 t) t = [ x x = t t 24 (u xxxx(x + θ x, t) + u xxxx (x θ 2 x, t)) ] 2 u tt(x, t + θ 3 t) Now it is indeed cler tht the h x /h t = x/ t c is suffiecient to bound the term in the squre brckets, so tht (C( t)u(x, t) u(x, t + t) f(x, t)) t c t 0 uniformly, i.e., the difference method bsed on C( t) is consistent. 6
7 c. We prove tht r /2 is the necessry nd sufficient condition for stbility nd then invoke Lx s theorem to get the sme result for convergence. We hve estblished erlier (4) tht C( t) 2r +2r. It immeditely follows tht if r /2, then C( t), so tht C( t) m, i.e., uniformly bounded, so the method is stble. For the other direction (necessity), we prove tht for r > 0 (4) is ctully n equlity, i.e., tht mx C( t)v V = = ( 2r + 2r) v mx V is chieved. We just need to present the function v mx t which this mximum is ttined. For r /2, the mximum is obviously ttined on v mx. For 0 < r < /2, let x j = (j/π) x be the sptil discretiztion. Define piecewise liner function by { 0, j = 0 or j = n v mx (x j ) = ( ) j, otherwise with stright line segments connecting the vertices. (This is no other thn the fmilir swtooth shpe.) Obviously, the supremum norm of this function is, nd since its vlues on the lttice re lternting + s nd  s, the norm C( t)v mx = ( 2r)(±)+ +2r( ) = 4r = ( 2r + 2r) v V. Thus the inequlity in (4) is in fct n equlity. It is now esy to estblish the required result. If the method is stble, then C( t) m is uniformly bounded. Obviously, if C( t) >, then C( t) m, so the uniform bound cn exist only if C( t), i.e., 2r + 2r r /2. Thus r /2 is necessry nd sufficient for stbility of the difference method. Since the method is consistent, by the Lx equivlence theorem it is lso necessry nd sufficient for convergence. Problem 5. Geometric Series Theorem.. Geometric Series Theorem. Let V be Bnch spce nd L : V V with L <. Then the opertor I L is bijection on V, its inverse is bounded opertor nd (I L) L. (5) Proof. Define the following sequence of opertors on V, M n = n k=0 L k, n 0. For p, n+p n+p M n+p M n = L k n+p L k L k Since L <, we hve k=n+ k=n+ M n+p M n L n+ L k=n+ so sup M n+p M n 0 s n, nd {M n } is Cuchy sequence in L(V ), the spce of p liner opertors on V. But this spce is complete, so there exists n opertor M : V V, such tht M n M 0 s n. Now, (I L)M n = M n LM n = 7 L k k=0 L k+ = I L n+ k=0
8 Similrly, M n (I L) = I L n+, so (I L)M n = M n (I L) = I L n+. Letting n, we get (I L)M = M(I L) = I proving tht I L is bijection nd M = (I L) = lim L k = L n n k=0 n=0 Furthermore, M n k=0 Tking the limit s n, we obtin L k = L n+ L L M = (I L) L proving (5) nd concluding the proof of the geometric series theorem. We cn puse for second here to recognize tht wht we hve just obtined is just nother generliztion of the geometric series formul: if 0 < q <, then the sum of the geometric series + q + q q n +... = q We lso know from complex nlysis tht if z C is lso inside the unit disc (i.e., z < ), then the series z n n=0 converges uniformly to the function ( z), which is bounded on the unit disc. Now we lso know tht the sme holds for liner opertors cting on Bnch spce! b. The liner integrl eqution of the second kind, λu(x) k(x, y)u(y) dy = f(x) (6) with λ 0, k(x, y) C([, b] [, b]) nd f C[, b], cn be written in the form (λi K)u = f, where K is the integrl opertor generted by the kernel k(, ). Equivlently, By the geometric series theorem, if (I L)u = λ f, L = λ K L = λ K < then the inverse opertor (I L) exists nd (I L) L 8
9 Since K = mx x b k(x, y) dy we thus hve tht, s long s K < λ, (λi K) exists nd Therefore, if only (λi K) λ K mx k(x, y) dy < λ x b for ny f C[, b], there is unique solution u C[, b] to the integrl eqution of the second kind (6). Problem 6. Different Bnch Spces with the Sme Norm.. Consider the spce l of bounded rel sequences with the (i.e., supremum) norm. All xioms of the norm re trivilly verified. Also, liner combintions preserve boundedness, so l is liner normed spce. Consider Cuchy sequence {x n } l : ε > 0 N = N(ε) such tht n, m > N sup x n,k x m,k < ε k By the uniform convergence criterion x n x s n, so for n sufficiently lrge, sup x n,k x k < ε k so sup k x k sup k x n,k + ε for n rbitrry ε > 0, so x is lso bounded, proving tht l is complete, nd therefore Bnch spce. b. Following the sme line of rguments s bove, we estblish tht c is normed liner spce, nd if {x n } c is Cuchy sequence, then x n x with sup x n,k x k < ε k 2 for n rbitrry ε > 0 nd ll n > N (ε). Since {x n } c, it consists of converging sequences, so for ll k > N 2 (ε), we hve x n,k C < ε/2. Therefore, for ll k > mx{n, N 2 }, we hve x k C ε so x c nd c is Bnch spce. The exct sme rgument pplies to c 0. c. First consider c 0, the spce of sequences converging to 0, with the supremum norm. We show tht the spce of bounded liner functionls on c 0 is isomorphic to l, the spce of bsolutely summble sequences (i.e., those with x k < ), with the norm f = 9 f k
10 To prove this, first note tht for ny element f = (f, f 2,..., f n,...) l, the formul f(x) = x k f k (7) defines functionl f on the spce c 0, which is clerly liner. Moreover, so f(x) x f k f f (8) Consider the following vectors in c 0 : e = (, 0,..., 0,...), e 2 = (0,,..., 0,...), etc., nd let x (n) f k = f k e k (if f k = 0, then we set f k / f k = 0). Then x (n) c 0 nd x (n). Moreover, f(x (n) ) = f k f k f(e k ) = f k f k = f s n. This implies f f lim x (n) f. Together with (8), the opposite n inequlity estblished erlier, this implies f = f. Therefore the mpping crrying f into f is normpreserving mpping from the spce of liner functionls on c 0 to l. It remins to prove tht every functionl hs unique representtion like (7), where f l. Let x = (x,..., x k,...) c 0. Then x = x k e k where the series on the right converges in c 0 to the element x since n x x k = sup x k 0 k>n s n. Since the functionl f is continuous, ( ) f(x) = f lim x k e k = lim n n f(x k e k ) = x k f(ek ) so f hs unique representtion of the form (7). We only need to estblish tht f(e k ) < (9) For ny n, let x (n) = f(e k ) f(e k ) e k 0
11 Then x (n) c 0 nd x (n), so f(e f(e k ) k ) = f(e k ) f(e k ) = f( x (n) ) f < Since n cn be mde rbitrrily lrge, this proves (9). Thus the spce of bounded liner functionls on c 0 is isomorphic to l : x c 0 f(x) = f k x k with f k = f(e k ) (e k defined bove) nd f k <. Now consider the more generl spce c of ll bounded convergent sequences. The obvious ide is to notice tht for ny x c, converging to x 0, x = x x 0 c 0, so we should be ble to use the result for functionls on c 0 for the cse of c. We mke this ide precise in the following wy. Consider the vectors e = (, 0, 0,..., 0,...), e 2 = (0,, 0,..., 0,...), etc., for ll k nd lso e 0 = (,,,...). Then for {e n } n 0 c, nd for ny liner functionl f on c we cn define f n = f(e n ), n = 0,,... We use these vlues to represent ll bounded liner functionls on c. We first prove tht the set x (n) = λ i e i, i=0 λ i Q is dense subset of c. Indeed, if = (, 2,..., k,...) c, then 0 : lim k = 0 <. k Then for ny ε > 0 we cn find sufficiently lrge n, so tht 0e 0 ( i 0 )e i < ε 2 i=0 since the sequence ( 0 ) k 0 s k. Now, since i R, there exist rtionl numbers λ i Q, such tht λ i ( i 0 ) < ε/2 i+, so tht nd therefore λ i e i ( i 0 )e i < ε 2 i= i= x (n) = 0e 0 λ i e i < ε i= so the x (n) s re indeed dense in c. Moreover, since λ i Q, {x (n) } is countble dense subset of c. For ny x {x (n) } nd ny liner functionl f we cn represent f(x) = f 0 λ 0 + λ i f i i= where f i = f(e i ). This functionl will be bounded s long s f i <. Since x (n) is dense in c, then by the HhnBnch theorem we cn extend the liner functionl f to
12 bounded liner functionl on the whole spce c. Since this extension is unique, it hs to look like f(x) = f(x, x 2,..., x k,...) = f 0 x 0 + f i x i where, s bove, f i = f(e i ). In other words, we hve proved tht c = c 0 R, so tht the spce of bounded liner functionls on c is isomorphic to l R, just s we hd expected. References [] Royden H.L., Rel Anlysis, 3rd ed., PrenticeHll, 988. [2] Kolmogorov A.N., Fomin S.V., Introductory Rel Anlysis, Dover, 975. [3] Atkinson K., Hn W., Theoreticl Numericl Anlysis: Functionl Anlysis Frmework, Springer, 200. [4] Zorich V.A., Mthemticl Anlysis (Clculus), 2nd ed., MCNMO, 998 (in Russin). i= 2
The Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationMATH 174A: PROBLEM SET 5. Suggested Solution
MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous relvlued function on I), nd let L 1 (I) denote the completion
More informationMATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1
MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further
More informationSOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set
SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 Nottion: N {, 2, 3,...}. (Tht is, N.. Let (X, M be mesurble spce with σfinite positive mesure µ. Prove tht there is finite positive mesure ν on (X, M such
More information2 Fundamentals of Functional Analysis
Fchgruppe Angewndte Anlysis und Numerik Dr. Mrtin Gutting 22. October 2015 2 Fundmentls of Functionl Anlysis This short introduction to the bsics of functionl nlysis shll give n overview of the results
More informationProblem Set 4: Solutions Math 201A: Fall 2016
Problem Set 4: s Mth 20A: Fll 206 Problem. Let f : X Y be onetoone, onto mp between metric spces X, Y. () If f is continuous nd X is compct, prove tht f is homeomorphism. Does this result remin true
More informationUNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE
UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence
More informationProperties of the Riemann Integral
Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2
More informationMath 61CM  Solutions to homework 9
Mth 61CM  Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More informationNotes on length and conformal metrics
Notes on length nd conforml metrics We recll how to mesure the Eucliden distnce of n rc in the plne. Let α : [, b] R 2 be smooth (C ) rc. Tht is α(t) (x(t), y(t)) where x(t) nd y(t) re smooth rel vlued
More informationA HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES. 1. Introduction
Ttr Mt. Mth. Publ. 44 (29), 159 168 DOI: 1.2478/v1127956z t m Mthemticl Publictions A HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES Miloslv Duchoň Peter Mličký ABSTRACT. We present Helly
More informationBest Approximation. Chapter The General Case
Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given
More informationReview of Riemann Integral
1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.
More informationRiemann is the Mann! (But Lebesgue may besgue to differ.)
Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationAMATH 731: Applied Functional Analysis Fall Additional notes on Fréchet derivatives
AMATH 731: Applied Functionl Anlysis Fll 214 Additionl notes on Fréchet derivtives (To ccompny Section 3.1 of the AMATH 731 Course Notes) Let X,Y be normed liner spces. The Fréchet derivtive of n opertor
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationf(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all
3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the
More informationMath 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
More informationProf. Girardi, Math 703, Fall 2012 Homework Solutions: 1 8. Homework 1. in R, prove that. c k. sup. k n. sup. c k R = inf
Knpp, Chpter, Section, # 4, p. 78 Homework For ny two sequences { n } nd {b n} in R, prove tht lim sup ( n + b n ) lim sup n + lim sup b n, () provided the two terms on the right side re not + nd in some
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More information1. On some properties of definite integrals. We prove
This short collection of notes is intended to complement the textbook Anlisi Mtemtic 2 by Crl Mdern, published by Città Studi Editore, [M]. We refer to [M] for nottion nd the logicl stremline of the rguments.
More informationPresentation Problems 5
Presenttion Problems 5 21355 A For these problems, ssume ll sets re subsets of R unless otherwise specified. 1. Let P nd Q be prtitions of [, b] such tht P Q. Then U(f, P ) U(f, Q) nd L(f, P ) L(f, Q).
More information1 1D heat and wave equations on a finite interval
1 1D het nd wve equtions on finite intervl In this section we consider generl method of seprtion of vribles nd its pplictions to solving het eqution nd wve eqution on finite intervl ( 1, 2. Since by trnsltion
More informationSTUDY GUIDE FOR BASIC EXAM
STUDY GUIDE FOR BASIC EXAM BRYON ARAGAM This is prtil list of theorems tht frequently show up on the bsic exm. In mny cses, you my be sked to directly prove one of these theorems or these vrints. There
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationVariational Techniques for SturmLiouville Eigenvalue Problems
Vritionl Techniques for SturmLiouville Eigenvlue Problems Vlerie Cormni Deprtment of Mthemtics nd Sttistics University of Nebrsk, Lincoln Lincoln, NE 68588 Emil: vcormni@mth.unl.edu Rolf Ryhm Deprtment
More informationCzechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction
Czechoslovk Mthemticl Journl, 55 (130) (2005), 933 940 ESTIMATES OF THE REMAINDER IN TAYLOR S THEOREM USING THE HENSTOCKKURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When relvlued
More informationAMATH 731: Applied Functional Analysis Fall Some basics of integral equations
AMATH 731: Applied Functionl Anlysis Fll 2009 1 Introduction Some bsics of integrl equtions An integrl eqution is n eqution in which the unknown function u(t) ppers under n integrl sign, e.g., K(t, s)u(s)
More informationMath Solutions to homework 1
Mth 75  Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationNumerical Analysis: Trapezoidal and Simpson s Rule
nd Simpson s Mthemticl question we re interested in numericlly nswering How to we evlute I = f (x) dx? Clculus tells us tht if F(x) is the ntiderivtive of function f (x) on the intervl [, b], then I =
More informationConvergence of Fourier Series and Fejer s Theorem. Lee Ricketson
Convergence of Fourier Series nd Fejer s Theorem Lee Ricketson My, 006 Abstrct This pper will ddress the Fourier Series of functions with rbitrry period. We will derive forms of the Dirichlet nd Fejer
More informationA PROOF OF THE FUNDAMENTAL THEOREM OF CALCULUS USING HAUSDORFF MEASURES
INROADS Rel Anlysis Exchnge Vol. 26(1), 2000/2001, pp. 381 390 Constntin Volintiru, Deprtment of Mthemtics, University of Buchrest, Buchrest, Romni. emil: cosv@mt.cs.unibuc.ro A PROOF OF THE FUNDAMENTAL
More informationg i fφdx dx = x i i=1 is a Hilbert space. We shall, henceforth, abuse notation and write g i f(x) = f
1. Appliction of functionl nlysis to PEs 1.1. Introduction. In this section we give little introduction to prtil differentil equtions. In prticulr we consider the problem u(x) = f(x) x, u(x) = x (1) where
More informationMath 554 Integration
Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we
More informationThe HenstockKurzweil integral
fculteit Wiskunde en Ntuurwetenschppen The HenstockKurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft
More informationAdvanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015
Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n
More informationAnalytical Methods Exam: Preparatory Exercises
Anlyticl Methods Exm: Preprtory Exercises Question. Wht does it men tht (X, F, µ) is mesure spce? Show tht µ is monotone, tht is: if E F re mesurble sets then µ(e) µ(f). Question. Discuss if ech of the
More informationLecture 19: Continuous Least Squares Approximation
Lecture 19: Continuous Lest Squres Approximtion 33 Continuous lest squres pproximtion We begn 31 with the problem of pproximting some f C[, b] with polynomil p P n t the discrete points x, x 1,, x m for
More informationFunctional Analysis I Solutions to Exercises. James C. Robinson
Functionl Anlysis I Solutions to Exercises Jmes C. Robinson Contents 1 Exmples I pge 1 2 Exmples II 5 3 Exmples III 9 4 Exmples IV 15 iii 1 Exmples I 1. Suppose tht v α j e j nd v m β k f k. with α j,
More informationLecture 3. Limits of Functions and Continuity
Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live
More informationAM1 Mathematical Analysis 1 Oct Feb Exercises Lecture 3. sin(x + h) sin x h cos(x + h) cos x h
AM Mthemticl Anlysis Oct. Feb. Dte: October Exercises Lecture Exercise.. If h, prove the following identities hold for ll x: sin(x + h) sin x h cos(x + h) cos x h = sin γ γ = sin γ γ cos(x + γ) (.) sin(x
More informationThe Banach algebra of functions of bounded variation and the pointwise Helly selection theorem
The Bnch lgebr of functions of bounded vrition nd the pointwise Helly selection theorem Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics, University of Toronto Jnury, 015 1 BV [, b] Let < b. For f
More informationHilbert Spaces. Chapter Inner product spaces
Chpter 4 Hilbert Spces 4.1 Inner product spces In the following we will discuss both complex nd rel vector spces. With L denoting either R or C we recll tht vector spce over L is set E equipped with ddition,
More informationMath 270A: Numerical Linear Algebra
Mth 70A: Numericl Liner Algebr Instructor: Michel Holst Fll Qurter 014 Homework Assignment #3 Due Give to TA t lest few dys before finl if you wnt feedbck. Exercise 3.1. (The Bsic Liner Method for Liner
More informationHomework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.
Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of
More informationDEFINITION The inner product of two functions f 1 and f 2 on an interval [a, b] is the number. ( f 1, f 2 ) b DEFINITION 11.1.
398 CHAPTER 11 ORTHOGONAL FUNCTIONS AND FOURIER SERIES 11.1 ORTHOGONAL FUNCTIONS REVIEW MATERIAL The notions of generlized vectors nd vector spces cn e found in ny liner lger text. INTRODUCTION The concepts
More informationRudin s Principles of Mathematical Analysis: Solutions to Selected Exercises. Sam Blinstein UCLA Department of Mathematics
Rudin s Principles of Mthemticl Anlysis: Solutions to Selected Exercises Sm Blinstein UCLA Deprtment of Mthemtics Mrch 29, 2008 Contents Chpter : The Rel nd Complex Number Systems 2 Chpter 2: Bsic Topology
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationPhysics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More informationSummary: Method of Separation of Variables
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 1 Summry: Method of Seprtion of Vribles 1. Seprtion of Vribles in Crtesin Coordintes 2. Fourier Series Suggested Reding: Griffiths: Chpter 3, Section
More informationA Convergence Theorem for the Improper Riemann Integral of Banach Spacevalued Functions
Interntionl Journl of Mthemticl Anlysis Vol. 8, 2014, no. 50, 24512460 HIKARI Ltd, www.mhikri.com http://dx.doi.org/10.12988/ijm.2014.49294 A Convergence Theorem for the Improper Riemnn Integrl of Bnch
More informationChapter 28. Fourier Series An Eigenvalue Problem.
Chpter 28 Fourier Series Every time I close my eyes The noise inside me mplifies I cn t escpe I relive every moment of the dy Every misstep I hve mde Finds wy it cn invde My every thought And this is why
More informationFourier series. Preliminary material on inner products. Suppose V is vector space over C and (, )
Fourier series. Preliminry mteril on inner products. Suppose V is vector spce over C nd (, ) is Hermitin inner product on V. This mens, by definition, tht (, ) : V V C nd tht the following four conditions
More informationCMDA 4604: Intermediate Topics in Mathematical Modeling Lecture 19: Interpolation and Quadrature
CMDA 4604: Intermedite Topics in Mthemticl Modeling Lecture 19: Interpoltion nd Qudrture In this lecture we mke brief diversion into the res of interpoltion nd qudrture. Given function f C[, b], we sy
More informationLecture 1: Introduction to integration theory and bounded variation
Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You
More informationII. Integration and Cauchy s Theorem
MTH6111 Complex Anlysis 200910 Lecture Notes c Shun Bullett QMUL 2009 II. Integrtion nd Cuchy s Theorem 1. Pths nd integrtion Wrning Different uthors hve different definitions for terms like pth nd curve.
More informationarxiv: v1 [math.ca] 11 Jul 2011
rxiv:1107.1996v1 [mth.ca] 11 Jul 2011 Existence nd computtion of Riemnn Stieltjes integrls through Riemnn integrls July, 2011 Rodrigo López Pouso Deprtmento de Análise Mtemátic Fcultde de Mtemátics, Universidde
More information0.1 Properties of regulated functions and their Integrals.
MA244 Anlysis III Solutions. Sheet 2. NB. THESE ARE SKELETON SOLUTIONS, USE WISELY!. Properties of regulted functions nd their Integrls.. (Q.) Pick ny ɛ >. As f, g re regulted, there exist φ, ψ S[, b]:
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More informationAnalysis III. Lecturer: Prof. Oleg Zaboronski, Typesetting: David Williams. Term 1, 2015
Anlysis III Lecturer: Prof. Oleg Zboronski, Typesetting: Dvid Willims Term, 25 Contents I Integrtion 2 The Integrl for Step Functions 2. Definition of the Integrl for Step Functions..........................
More informationLecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)
Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of
More informationFor a continuous function f : [a; b]! R we wish to define the Riemann integral
Supplementry Notes for MM509 Topology II 2. The Riemnn Integrl Andrew Swnn For continuous function f : [; b]! R we wish to define the Riemnn integrl R b f (x) dx nd estblish some of its properties. This
More informationA product convergence theorem for Henstock Kurzweil integrals
A product convergence theorem for Henstock Kurzweil integrls Prsr Mohnty Erik Tlvil 1 Deprtment of Mthemticl nd Sttisticl Sciences University of Albert Edmonton AB Cnd T6G 2G1 pmohnty@mth.ulbert.c etlvil@mth.ulbert.c
More informationNOTES AND PROBLEMS: INTEGRATION THEORY
NOTES AND PROBLEMS: INTEGRATION THEORY SAMEER CHAVAN Abstrct. These re the lecture notes prepred for prticipnts of AFSI to be conducted t Kumun University, Almor from 1st to 27th December, 2014. Contents
More informationChapter 4. Lebesgue Integration
4.2. Lebesgue Integrtion 1 Chpter 4. Lebesgue Integrtion Section 4.2. Lebesgue Integrtion Note. Simple functions ply the sme role to Lebesgue integrls s step functions ply to Riemnn integrtion. Definition.
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More information7.2 Riemann Integrable Functions
7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous
More informationDiscrete Leastsquares Approximations
Discrete Lestsqures Approximtions Given set of dt points (x, y ), (x, y ),, (x m, y m ), norml nd useful prctice in mny pplictions in sttistics, engineering nd other pplied sciences is to construct curve
More informationRegulated functions and the regulated integral
Regulted functions nd the regulted integrl Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics University of Toronto April 3 2014 1 Regulted functions nd step functions Let = [ b] nd let X be normed
More informationA BRIEF INTRODUCTION TO UNIFORM CONVERGENCE. In the study of Fourier series, several questions arise naturally, such as: c n e int
A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE HANS RINGSTRÖM. Questions nd exmples In the study of Fourier series, severl questions rise nturlly, such s: () (2) re there conditions on c n, n Z, which ensure
More informationPhil Wertheimer UMD Math Qualifying Exam Solutions Analysis  January, 2015
Problem 1 Let m denote the Lebesgue mesure restricted to the compct intervl [, b]. () Prove tht function f defined on the compct intervl [, b] is Lipschitz if nd only if there is constct c nd function
More informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
More informationInternational Jour. of Diff. Eq. and Appl., 3, N1, (2001),
Interntionl Jour. of Diff. Eq. nd Appl., 3, N1, (2001), 3137. 1 New proof of Weyl s theorem A.G. Rmm Mthemtics Deprtment, Knss Stte University, Mnhttn, KS 665062602, USA rmm@mth.ksu.edu http://www.mth.ksu.edu/
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More information1 Sets Functions and Relations Mathematical Induction Equivalence of Sets and Countability The Real Numbers...
Contents 1 Sets 1 1.1 Functions nd Reltions....................... 3 1.2 Mthemticl Induction....................... 5 1.3 Equivlence of Sets nd Countbility................ 6 1.4 The Rel Numbers..........................
More informationChapter 3. Vector Spaces
3.4 Liner Trnsformtions 1 Chpter 3. Vector Spces 3.4 Liner Trnsformtions Note. We hve lredy studied liner trnsformtions from R n into R m. Now we look t liner trnsformtions from one generl vector spce
More informationMATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals.
MATH 409 Advnced Clculus I Lecture 19: Riemnn sums. Properties of integrls. Drboux sums Let P = {x 0,x 1,...,x n } be prtition of n intervl [,b], where x 0 = < x 1 < < x n = b. Let f : [,b] R be bounded
More informationThe Wave Equation I. MA 436 Kurt Bryan
1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string
More information1. GaussJacobi quadrature and Legendre polynomials. p(t)w(t)dt, p {p(x 0 ),...p(x n )} p(t)w(t)dt = w k p(x k ),
1. GussJcobi qudrture nd Legendre polynomils Simpson s rule for evluting n integrl f(t)dt gives the correct nswer with error of bout O(n 4 ) (with constnt tht depends on f, in prticulr, it depends on
More informationMath 120 Answers for Homework 13
Mth 12 Answers for Homework 13 1. In this problem we will use the fct tht if m f(x M on n intervl [, b] (nd if f is integrble on [, b] then (* m(b f dx M(b. ( The function f(x = 1 + x 3 is n incresing
More informationEuler, Ioachimescu and the trapezium rule. G.J.O. Jameson (Math. Gazette 96 (2012), )
Euler, Iochimescu nd the trpezium rule G.J.O. Jmeson (Mth. Gzette 96 (0), 36 4) The following results were estblished in recent Gzette rticle [, Theorems, 3, 4]. Given > 0 nd 0 < s
More informationarxiv:math/ v2 [math.ho] 16 Dec 2003
rxiv:mth/0312293v2 [mth.ho] 16 Dec 2003 Clssicl Lebesgue Integrtion Theorems for the Riemnn Integrl Josh Isrlowitz 244 Ridge Rd. Rutherford, NJ 07070 jbi2@njit.edu Februry 1, 2008 Abstrct In this pper,
More informationConvex Sets and Functions
B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line
More informationCalculus of Variations: The Direct Approach
Clculus of Vritions: The Direct Approch Lecture by Andrejs Treibergs, Notes by Bryn Wilson June 7, 2010 The originl lecture slides re vilble online t: http://www.mth.uth.edu/~treiberg/directmethodslides.pdf
More information1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.
Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the
More informationIMPORTANT THEOREMS CHEAT SHEET
IMPORTANT THEOREMS CHEAT SHEET BY DOUGLAS DANE Howdy, I m Bronson s dog Dougls. Bronson is still complining bout the textbook so I thought if I kept list of the importnt results for you, he might stop.
More informationCalculus of Variations
Clculus of Vritions Com S 477/577 Notes) YnBin Ji Dec 4, 2017 1 Introduction A functionl ssigns rel number to ech function or curve) in some clss. One might sy tht functionl is function of nother function
More informationOrthogonal Polynomials and LeastSquares Approximations to Functions
Chpter Orthogonl Polynomils nd LestSqures Approximtions to Functions **4/5/3 ET. Discrete LestSqures Approximtions Given set of dt points (x,y ), (x,y ),..., (x m,y m ), norml nd useful prctice in mny
More information5.4, 6.1, 6.2 Handout. As we ve discussed, the integral is in some way the opposite of taking a derivative. The exact relationship
5.4, 6.1, 6.2 Hnout As we ve iscusse, the integrl is in some wy the opposite of tking erivtive. The exct reltionship is given by the Funmentl Theorem of Clculus: The Funmentl Theorem of Clculus: If f is
More informationMath& 152 Section Integration by Parts
Mth& 5 Section 7.  Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More information