Lecture 1. Functional series. Pointwise and uniform convergence.


 Hester Morgan
 1 years ago
 Views:
Transcription
1 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 defined s the series ( k cos kx + b k sin kx), k=1 where the coefficients k, b k re defined s k = 1 π π π f(x) cos kx dx, b k = 1 π with k = 0, 1,... (note tht this mens tht b 0 = 0). π π f(x) sin kx dx, This is n exmple of functionl series, which is series whose terms re functions:. As usul with series, we define the bove infinite sum s limit: = lim N N, providing the limit exists. Note tht different vlues of x will, in generl, give different limits, if they exist. In this lecture we shll look t functionl series, nd functionl sequences, nd we shll consider first the question of convergence. To del with this, we consider two types of convergence: pointwise convergence nd uniform convergence. There re three min results: the first one is tht uniform convergence of sequence of continuous functions gives us continuous function s limit. The second min result is Weierstrss Mjornt Theorem, which gives condition tht gurntees tht functionl series converges to continuous function. The third result is tht integrls of sequence of functions which converges uniformly to limit function f(x) lso converge with the limit being the integrl of f(x). These results re not only good for your mentl helth, they re lso importnt tools in our lter discussion of Fourier series, nd tht is the reson for looking t them. 1
2 2 Power series. A power series in the vrible x is series of the form x + 2 x x 3 + = k x k, where the coefficients 0, 1, 2,... re rel or complex numbers. We re (in principle) llowed to put the vrible x equl to ny number we wish. For instnce, with the power series k x k we my put x = 2 nd we obtin the numericl series Or, with the power series k 2 k = (note tht in this cse ll the coefficients re equl to 1) we obtin x k 2 k = if we put x = 2. However, we hve problem: how do we know tht the numericl series we obtin by putting x = 2 in the power series k x k is convergent? We cn lwys put x = 2 nd then investigte the convergence of the numericl series, but this is rther inefficient wy of deciding if prticulr vlue of x gives us convergent series. So we come to the following question: Given the power series k x k, with given choice of coefficients 0, 1, 2,..., wht vlues of x give us convergent series, nd which vlues give divergent series? There re two very useful results which help us in exmining this question. The first one is the following: 2
3 Theorem 2.1 For power series there re three possibilities: k x k 1. The power series 2. The power series k x k diverges for ll x 0 k x k converges for ll vlues of x 3. There is positive number R such tht k x k converges for ll vlues of x with x < R nd diverges for ll vlues of x with x > R. At first sight, this looks like very useless result, becuse it doesn t nswer the question of which vlues of x re llowed. However, it is very useful result: it tells us wht sort of behviour we cn expect, nd wht to look for in power series. In prticulr, it tells us wht is the decisive fctor in our subsequent investigtion: we need to find the number R, which is clled the rdius of convergence of the power series. One word of wrning: the theorem tells us tht, when we hve found the rdius of convergence R, then the series converges for x < R nd diverges whenever x > R, but it doesn t sy nything bout the cse when x = R, tht is, when x = R nd x = R. The theorem is silent on this mtter. In fct, we must investigte the cses x = ±R seprtely (tht is, we put in x = R or x = R nd we investigte the convergence of the series tht rise). This is smll price to py when we know wht R is. So, given our theorem, how do we go bout clculting R? One result is the following: Theorem 2.2 Given the power series exist: Then the following is true: k x k, suppose tht one of the following limits K = lim k+1 k k, K = lim k k. k 1. If K = 0 then the power series k x k converges for ll vlues of x; 2. If K > 0, then the rdius of convergence R of the power series R = 1 K. k x k is 3
4 If either of the limits lim k fils to exist, then the power series k+1 k, lim k k k. k x k diverges for ll vlues of x 0. This theorem is proved using the following result (the proof is given in Mts Neymrk s Kompendium om konvergens): Theorem 2.3 Suppose limits exists k is numericl series nd suppose tht one of the following K = lim k+1 k k, K = lim k k. k If 0 K < 1 then the series converges bsolutely. If K > 1 then the series diverges. If K = 1 then convergence or divergence of the series must be investigted using some other method. 3 Pointwise Convergence. Consider the power series x k = 1 1 x for x < 1. This just sys tht for ech x ] 1, 1[ the power series in the lefthnd side converges to the number 1/(1 x). If we put then we cn rephrse this s f n (x) = n x k, f(x) = 1 1 x, f n (x) f(x) s n for ech x ] 1, 1[. We give this type of convergence nme: pointwise convergence. Note tht we hve first defined sequence of functions f n by putting for ech n = 0, 1, 2,.... f n (x) = n x k 4
5 Definition 3.1 (Pointwise convergence.) Suppose {f n (x) : n = 0, 1, 2,... } is sequence of functions defined on n intervl I. We sy tht f n (x) converges pointwise to the function f(x) on the intervl I if f n (x) f(x), s n, for ech x I. We cll the function f(x) the limit function. Exmple 3.1 f n (x) = x 1 n. Then f n(x) converges pointwise to x for ech x R: f n (x) x = 1 n 0 s n. Exmple 3.2 f n (x) = e nx on [1, 3]. For ech x [1, 3] we hve nx s n nd therefore f n (x) 0 s n for ech x [1, 3]. Thus f n (x) converges pointwise to f(x) = 0 for ech x [1, 3]. Exmple 3.3 f n (x) = e nx on [0, 3]. For ech 0 < x 3 we hve nx s n nd therefore f n (x) 0 s n for ech 0 < x 3. However, t x = 0 we hve f n (0) = 1 for ll n. Thus f n (x) converges pointwise to the function f(x) defined by f(0) = 1, f(x) = 0 for ech 0 < x 3. This is not continuous function, despite the fct tht ech function f n (x) is continuous. The lst exmple shows wht cn hppen with pointwise convergence: the limit function my fil to be continuous, even though ll functions in the sequence re continuous. Exmple 3.4 Let the sequence f n be defined s n 2 x f n (x) = (nx + 1) 3 Then f n (0) = 0 nd for ech fixed x > 0 x [0, [. n 2 x f n (x) = (nx + 1) 3 n 2 x = n 3 (x + 1 n )3 = 1 x n (x s n. )3 n So tht f n (x) f(x) = 0 pointwise on [0, [. Then for x 0 we hve f n(x) = n2 (1 2nx) (nx + 1) 4 nd we see tht for x > 0 we hve f n(x) 0 s n wheres f n(0) = n 2. Here we see tht f n f only on for x > 0. This shows tht differentibility is not lwys respected by pointwise convergence. 5
6 The lst two exmples then led us to pose the question: wht extr condition (other thn just pointwise convergence) cn gurntee tht the limit function is lso continuous or differentible? The nswer to this is given by the concept of uniform convergence. 4 Uniform convergence We define for relvlued (or complexvlued) function f on nonempty set I the supremum norm of f on the set I: f I = sup f(x). x I Note tht if f is bounded function on I then sup f(x) = sup{ f(x) : x I} x I exists, by the soclled supremum xiom. Observe tht f(x) f I for ll x I, nd tht f(x) tkes on vlues which re rbitrrily ner f I. In prticulr f I = the lrgest vlue of f(x) whenever such vlue exists (such s when I is closed, bounded intervl nd f(x) is continuous function on I). The supremum norm hs the following properties for functions f nd g on set I: f I 0 nd f I = 0 f(x) = 0 for ll x I cf = c f I for ny constnt c f + g I f I + g I (tringle inequlity) f J f I when J is subset of I. The proof of these properties is left s n exercise for the interested reder. Now we come to the definition of uniform convergence: Definition 4.1 A sequence of functions f n (x) defined on n set I is sid to converge uniformly to f(x) on I if We write this s f n f I 0 s n. or s lim f n = f n uniformly on I 6
7 f n f uniformly on I s n. Uniform convergence implies pointwise convergence, however there re sequences which converge pointwise but not uniformly. Indeed we hve so tht f n (x) f(x) sup f n (x) f(x) = f n f I, x I f n f uniformly on I s n = f n (x) f(x) 0 for ech x I = f n f pointwise on I. We record this s result: Lemm 4.1 If the sequence of functions f n (x) converges uniformly to f(x) on the intervl I, then f n (x) converges pointwise to f(x). This Lemm sys tht the limit function obtined through uniform convergence (if this occurs) is the sme s the limit function obtined from pointwise convergence. Or: if f n (x) converges to f(x) uniformly, then it must converge to f(x) pointwise. This then tells us how to go bout testing for uniform convergence: first, obtin the pointwise limit f(x) nd then see if we hve uniform convergence to f(x). Exmple 4.1 f n (x) = e nx on [1, 3]. We hve seen bove tht f n (x) converges pointwise to f(x) = 0 for ech x [1, 3]. Then we hve f n (x) f(x) = f n (x) nd we then hve f n f = sup f n (x) x [1,3] = sup e nx x [1,3] = sup e nx x [1,3] = e n 0 s n. Thus we hve uniform convergence in this cse. Note tht the lst step follows from the observtion tht e nx is strictly decresing for x 0 with n 0, so tht e n e nx for ll x 1. 7
8 Exmple 4.2 f n (x) = xe nx on I = [0, [. Here the intervl is unbounded. First we look t pointwise convergence: f n (0) = 0 nd for x > 0 we hve tht f n (x) 0 s n. Thus f n (x) 0 pointwise on I. We now need to investigte uniform convergence. Since the limit function f(x) = 0 we hve f n f = sup xe nx x [0, [ = sup xe nx x [0, [ becuse f n (x) 0 for x 0. Now, we hve f n(x) = (1 nx)e nx for x > 0 (observe tht you should never differentite on closed intervls), nd we see tht f n(x) = 0 when x = 1/n. Further, f n(x) > 0 for 0 < x < 1/n, nd f n(x) < 0 for x > 1/n, so we conclude tht f n (x) hs mximum t x = 1/n nd hence f n f = sup xe nx x [0, [ = f n ( 1 n ) = 1 ne 0 s n. So we see tht the sequence of functions f n (x) = xe nx converges uniformly to 0 on the intervl I = [0, [. 5 Uniform convergence nd continuity. We now come to two importnt results. The first is the following. Theorem 5.1 Suppose f n (x) is sequence of continuous functions on n intervl I nd suppose lso tht f n (x) converges uniformly to f(x) on the intervl I. Then the limit function f(x) is lso continuous. Proof: First note tht for x, I we my write f(x) f() = [f(x) f n (x)] + [f n (x) f n ()] + [f n () f()] from which we obtin (using the tringle inequlity) f(x) f() f(x) f n (x) + f n (x) f n () + f n () f() (5.1) Then note tht, becuse sup f(x) f n (x) 0 s n we hve for ny given choice x I of ɛ > 0 nturl number N such tht sup f(x) f n (x) < ɛ x I 3 8
9 for ll n N. We lso hve f() f n () < ɛ 3 since f() f n () sup f(x) f n (x) < ɛ x I 3. Using this in eqution (5.1), we hve, for given ɛ > 0, nturl number N so tht f(x) f() 2ɛ 3 + f n(x) f n () for n N. Fix the choice of n N, sy n = N. We now use continuity of the f n (x): for ech ɛ > 0 there exists δ > 0 such tht f n (x) f n () < ɛ 3 whenever x < δ. Consequently, for x < δ we hve f(x) f() 2ɛ 3 + f n(x) f n () < 2ɛ 3 + ɛ 3 = ɛ. This mens tht for ny ɛ > 0 there exists δ > 0 such tht f(x) f() < ɛ whenever x < δ nd this mens tht f(x) f() s x. Tht is, f(x) is continuous t ech I. Remrk: This result is very useful s quick test for the bsence of uniform convergence: if (i) f n (x), n = 0, 1, 2,... is sequence of continuous functions (on some intervl); (ii) f n (x) converges pointwise to f(x); (iii) f(x) is not continuous; (iv) Then f n (x) does not converge uniformly to f(x). Exmple 5.1 f n (x) = e nx on [0, 3] is sequence of continuous functions, converging pointwise to f(x) defined by { 1 for x = 0 f(x) = 0 for 0 < x 3, which is not continuous, nd so, by Theorem 4.1, the sequence does not converge uniformly to f(x). The second result we mention is the following, which is of gret use in integrting series: 9
10 Theorem 5.2 Suppose tht f n (x) is sequence of continuous functions which converges uniformly to continuous function f(x) on bounded intervl [, b]. Then we hve lim n f n (x)dx = Proof: The proof is quite simple: Thus: f n (x)dx lim f n(x)dx = n b f(x)dx = (f n (x) f(x))dx = f n f f n (x) f(x) dx f n f dx 1 dx f(x)dx. = f n f (b ) 0 s n. f n (x)dx f(x)dx s n if f n f uniformly on I, which is wht we wnted to prove. Remrk 5.1 Theorem 5.2 is proved here for continuous functions so tht the integrls exist. It is however possible to replce the word continuous by the word integrble, nd the theorem is still true. We now give result bout uniform convergence nd differentibility: it tells us under which conditions the limit function f(x) is differentible whenever the functions of the sequence f n (x) re differentible. Theorem 5.3 Suppose tht {f n (x); n = 0, 1, 2,... } is sequence of functions on n intervl I nd stisfying the following conditions: (i) f n (x) is differentible on I for ech n = 0, 1, 2,... (ii) f n (x)converges pointwise to f(x) on I (iii) f n(x) is continuous for ech n nd f n g converges uniformly on I where g(x) is continuous function on I. Then the limit function f(x) is differentible nd f (x) = g(x). 10
11 This result is very useful, s we shll see, in exmining the differentibility of functionl series. Proof: Becuse of differentibility we hve (for, x I) f n (x) = f n () + x f n(t)dt Furthermore, since f n(t) g(t) uniformly on I, we know from Theorem 5.2 tht x f n(t)dt x g(t)dt when n. Also, f n (x) f(x) pointwise s n. From this it follows tht f(x) = f() + x g(t)dt on letting n. Since g(x) is continuous (it is the uniform limit of sequence of continuous functions, so it is continuous by Theorem 5.1), the integrl exists nd is primitive function of g(x). Differentiting this lst eqution, we obtin This concludes the proof. f (x) = g(x). 6 Applictions to functionl series. Definition 6.1 A functionl series is series where ech term of the series is function on n intervl I. We cn lso define pointwise convergence for functionl series: Definition 6.2 The functionl series is pointwise convergent for ech x I if the limit exists for ech x I. = lim N N 11
12 Thus, we lwys define sequence of prtil sums S N (x) given s so tht S N (x) = N S 0 (x) = u 0 (x), S 1 (x) = u 0 (x) + u 1 (x), S 2 (x) = u 0 (x) + u 1 (x) + u 2 (x),... nd if lim S N(x) N exists for x then we sy tht the series = lim N S N(x) converges t x. It converges pointwise on the intervl I if lim S N(x) N exists for ech x I. With these definitions, we deduce from Theorem 4.1 tht if the functions re ll continuous on I nd if the sequence of prtil sums S N (x) converges uniformly to S(x) on I, then S(x) is continuous. However, we would like n efficient wy of deciding if functionl series converges uniformly to (unique) limit. It is not t ll esy to pply the definition of uniform convergence to n infinite sum of functions, so nother method is desirble. The pproprite result is Weierstrss Mjornt Theorem: Theorem 6.1 Suppose tht the functionl series is defined on n intervl I nd tht there is sequence of positive constnts M k so tht for ll x I. If M k, k = 0, 1, 2,... converges, then converges uniformly on I. M k 12
13 Proof: If the conditions re fullfilled then we immeditely hve, from the Comprison Theorems for Positive Series, tht, for ech x I, the series is convergent, so tht is bsolutely convergent, nd therefore convergent. This mens tht is pointwise convergent on I, nd we denote the limit by S(x). We now show tht the prtil sums S N (x) = N converges uniformly to S(x) on I under the conditions of the theorem. We hve S(x) S N (x) = k=n+1 (ll we do is subtrct the first N terms from the series). Then it follows tht S(x) S N (x) k=n+1 k=n+1 for ech x I, since M k for ech x I ccording to our ssumption. Then S S N I k=n+1 We lso know (by ssumption) tht M k converges, so we must hve tht k=n+1 M k 0 s N. Consequently, M k. M k nd our result is proved. S S N I 0 s N, Corollry 6.1 If (i) the functionl series S(x) = converges uniformly on intervl I, 13
14 (ii) is continuous function on I for ech k = 0, 1, 2,..., then S(x) is continuous on I. Proof: Becuse finite sum of continuous functions is gin continuous function, it follows tht the prtil sums S N (x) = N re continuous functions for N = 0, 1, 2,.... Then by Theorem 5.1, we hve tht S(x) = lim N S N (x) is continuous function. Exmple 6.1 Tke the functionl series We hve k=1 = sin kx k 2 sin kx k 2. = sin kx k 2 1 k 2 since sin t 1 for ll rel t. We know (stndrd positive series) tht 1 converges (series of the form 1/k α converge for α > 1 nd diverge for α 1). Hence, by Weierstrss Mjornt Theorem, k=1 1 k 2 sin kx k 2 converges uniformly for ll x, nd by Corollry 5.1 this series is continuous function of x for ll x R. Remrk 6.1 One dvntge of Weierstrss Mjornt Theorem is tht we do not hve to clculte the vlue of the series t ech x I in order to decide if we hve uniform convergence. However, drwbck is tht the conditions of the theorem re only sufficient to estblish uniform convergence, they re not bsolutely necessry for uniform convergence. In the finl section of these lecture notes we give necessry nd sufficient condition for uniform convergence. Remrk 6.2 In our sttement of Weierstrss Mjornt Theorem, we hve not sid nything bout how to find the constnts M k. Usully we tke M k = sup, x I but this is not strictly necessry: ny sequence (of constnts) will do provided tht M k converges. 14
15 Another result of interest is the following: Theorem 6.2 If (i) the functionl series converges uniformly on the intervl I (ii) is continuous on I for ech k = 0, 1, 2,..., then x ( ) u k (t) dt = ( x ) u k (t)dt for ll, x I. In other words, if the series of continuous functions converges uniformly on I, then the integrl of the sum is the sum of the integrls of the functions, just s in the cse of finite sum. Proof: See Kompendium om Konvergens. We cn lso sy something bout the differentibility of the series, using Theorem 5.3 In this cse, s in the previous two theorems, we replce f n (x) by S N (t) nd f(x) by S(t). Thus, we wnt the following: S N (x) S(x) pointwise on I S N(x) G(x) uniformly on I S N (x) is continuously differentible for ech N nd then we my conclude tht S(x) is continuously differentible with S (x) = G(x). All we need is to formulte these requirements nd result s follows: Theorem 6.3 Suppose tht stisfies the following conditions: converges pointwise on I u k(x) converges uniformly on I is continuously differentible for ech k Then is continuously differentible nd ( ) d = dx Proof: See Kompendium om Konvergens. 15 u k(x).
16 7 APPENDIX. 7.1 Supremum nd Infimum: recpitultion. Definition 7.1 Let A R. Then the supremum of A, denoted by sup A, is defined s the smllest number R with the property tht x for ll x A. In mthemticl shorthnd we hve sup A = min{ R : x for ll x A}. Similrly, the infimum of A, denoted by inf A, is defined s the lrgest number b R with the property tht x b for ll x A. In mthemticl shorthnd we hve inf A = mx{b R : x b for ll x A}. Remrk 7.1 Note tht in these definitions neither the supremum nor the infimum need belong to the set A. Exmple 7.1 (i) A = [ 1, 3]. Here we hve sup A = 3, inf A = 1, nd both these belong to A. (ii) A =] 1, 3]. Here sup A = 3, inf A = 1, but only inf A belongs to A. (iii) A =] 1, 3[. Here sup A = 3, inf A = 1, nd both re not in A. (iv) A = [ 1, [. Here inf A = 1 wheres sup A does not exist. Definition 7.2 Let f : R R be function. Then the supremum of f(x) over A is defined s the smllest number R with the property tht f(x) for ll x A. In mthemticl shorthnd we hve sup f(x) = min{ R : f(x) for ll x A}. x A Similrly, the infimum of f(x) over A is defined s the lrgest number b R with the property tht f(x) b for ll x A. In mthemticl shorthnd we hve inf = mx{b R : f(x) b for ll x A}. x A Exmple 7.2 (i) f(x) = x 3, A = [ 1, 3] Then, since f(x) is strictly incresing function, we hve sup f(x) = 27, x A inf f(x) = 1. x A 16
17 (ii) f(x) = x 2, A = [ 1, 3] Then note tht f(x) = x 2 is not strictly incresing on this intervl: it is decresing on [ 1, 0] nd then strictly incresing on [0, 3]. So we hve nd sup f(x) = 1, x [ 1,0] inf f(x) = 0 x [ 1,0] sup f(x) = 9, x [0,3] Combining these two observtions, we find tht inf f(x) = 0. x [0,3] sup f(x) = 9, x [ 1,3] inf = 0. x [ 1,3] (iii) f(x) = rctn x, A = R. Here we hve strictly incresing function, nd we hve sup f(x) = π x R 2, inf f(x) = π x R 2. It is tempting to tke the lrgest vlue of function on n intervl s the supremum, nd the lest vlue for the infimum. The lst exmple shows tht the neither the supremum nor the infimum need be ttinble vlues of function. However, we hve the following simple but useful result: Lemm 7.1 Suppose tht f(x) is relvlued continuous function on the closed, bounded intervl [, b]. Then sup f(x) = mx{f(x) : x [, b]}, x [,b] inf f(x) = min{f(x) : x [, b]}. x [,b] Tht is, the supremum of continuous function over closed, bounded intervl is equl to its lrgest vlue over tht intervl, nd the infimum is the lest vlue of the function over the intervl. Proof: Since f(x) is continuous nd the intervl is closed, then f(x) hs lrgest vlue nd lest vlue on the intervl: there exist x 1, x 2 [, b] so tht f(x 1 ) f(x) f(x 2 ) for ll x [, b], nd we now see tht nd the result is proved. sup f(x) = f(x 2 ), x [,b] inf f(x) = f(x 1), x [,b] 17
UNIFORM 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 informationThe 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 informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
More informationChapter 6. Infinite series
Chpter 6 Infinite series We briefly review this chpter in order to study series of functions in chpter 7. We cover from the beginning to Theorem 6.7 in the text excluding Theorem 6.6 nd Rbbe s test (Theorem
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
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 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 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 informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
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 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 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 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 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 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 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 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 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 informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
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 informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
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 informationReview of basic calculus
Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below
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 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 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 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 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 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 information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
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 informationFUNDAMENTALS OF REAL ANALYSIS by. III.1. Measurable functions. f 1 (
FUNDAMNTALS OF RAL ANALYSIS by Doğn Çömez III. MASURABL FUNCTIONS AND LBSGU INTGRAL III.. Mesurble functions Hving the Lebesgue mesure define, in this chpter, we will identify the collection of functions
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationCalculus II: Integrations and Series
Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]
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 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 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 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 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 informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
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 informationCalculus III Review Sheet
Clculus III Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing
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 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 informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationc n φ n (x), 0 < x < L, (1) n=1
SECTION : Fourier Series. MATH4. In section 4, we will study method clled Seprtion of Vribles for finding exct solutions to certin clss of prtil differentil equtions (PDEs. To do this, it will be necessry
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 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 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 informationMA Handout 2: Notation and Background Concepts from Analysis
MA350059 Hndout 2: Nottion nd Bckground Concepts from Anlysis This hndout summrises some nottion we will use nd lso gives recp of some concepts from other units (MA20023: PDEs nd CM, MA20218: Anlysis 2A,
More informationMAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL
MAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL DR. RITU AGARWAL MALVIYA NATIONAL INSTITUTE OF TECHNOLOGY, JAIPUR, INDIA302017 Tble of Contents Contents Tble of Contents 1 1. Introduction 1 2. Prtition
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationContinuous Random Variables
STAT/MATH 395 A  PROBABILITY II UW Winter Qurter 217 Néhémy Lim Continuous Rndom Vribles Nottion. The indictor function of set S is relvlued function defined by : { 1 if x S 1 S (x) if x S Suppose tht
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
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 informationEntrance Exam, Real Analysis September 1, 2009 Solve exactly 6 out of the 8 problems. Compute the following and justify your computation: lim
1. Let n be positive integers. ntrnce xm, Rel Anlysis September 1, 29 Solve exctly 6 out of the 8 problems. Sketch the grph of the function f(x): f(x) = lim e x2n. Compute the following nd justify your
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 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 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 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 informationRecitation 3: More Applications of the Derivative
Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech
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 informationSYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus
SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is
More informationAdvanced Calculus I (Math 4209) Martin Bohner
Advnced Clculus I (Mth 4209) Spring 2018 Lecture Notes Mrtin Bohner Version from My 4, 2018 Author ddress: Deprtment of Mthemtics nd Sttistics, Missouri University of Science nd Technology, Roll, Missouri
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 information11 An introduction to Riemann Integration
11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in
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 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 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 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 informationAnalysis III. Ben Green. Mathematical Institute, Oxford address:
Anlysis III Ben Green Mthemticl Institute, Oxford Emil ddress: ben.green@mths.ox.c.uk 2000 Mthemtics Subject Clssifiction. Primry Contents Prefce 1 Chpter 1. Step functions nd the Riemnn integrl 3 1.1.
More informationBest Approximation in the 2norm
Jim Lmbers MAT 77 Fll Semester 111 Lecture 1 Notes These notes correspond to Sections 9. nd 9.3 in the text. Best Approximtion in the norm Suppose tht we wish to obtin function f n (x) tht is liner combintion
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 informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More information20 MATHEMATICS POLYNOMIALS
0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
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 informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More informationa n = 1 58 a n+1 1 = 57a n + 1 a n = 56(a n 1) 57 so 0 a n+1 1, and the required result is true, by induction.
MAS221(21617) Exm Solutions 1. (i) A is () bounded bove if there exists K R so tht K for ll A ; (b) it is bounded below if there exists L R so tht L for ll A. e.g. the set { n; n N} is bounded bove (by
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 informationMathematical Analysis: Supplementary notes I
Mthemticl Anlysis: Supplementry notes I 0 FIELDS The rel numbers, R, form field This mens tht we hve set, here R, nd two binry opertions ddition, + : R R R, nd multipliction, : R R R, for which the xioms
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 information1 Probability Density Functions
Lis Yn CS 9 Continuous Distributions Lecture Notes #9 July 6, 28 Bsed on chpter by Chris Piech So fr, ll rndom vribles we hve seen hve been discrete. In ll the cses we hve seen in CS 9, this ment tht our
More information1.9 C 2 inner variations
46 CHAPTER 1. INDIRECT METHODS 1.9 C 2 inner vritions So fr, we hve restricted ttention to liner vritions. These re vritions of the form vx; ǫ = ux + ǫφx where φ is in some liner perturbtion clss P, for
More informationReview on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones.
Mth 20B Integrl Clculus Lecture Review on Integrtion (Secs. 5.  5.3) Remrks on the course. Slide Review: Sec. 5.5.3 Origins of Clculus. Riemnn Sums. New functions from old ones. A mthemticl description
More informationAppendix to Notes 8 (a)
Appendix to Notes 8 () 13 Comprison of the Riemnn nd Lebesgue integrls. Recll Let f : [, b] R be bounded. Let D be prtition of [, b] such tht Let D = { = x 0 < x 1
More informationDuality # Second iteration for HW problem. Recall our LP example problem we have been working on, in equality form, is given below.
Dulity #. Second itertion for HW problem Recll our LP emple problem we hve been working on, in equlity form, is given below.,,,, 8 m F which, when written in slightly different form, is 8 F Recll tht we
More informationChapter 6. Riemann Integral
Introduction to Riemnn integrl Chpter 6. Riemnn Integrl WonKwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl
More informationLECTURE. INTEGRATION AND ANTIDERIVATIVE.
ANALYSIS FOR HIGH SCHOOL TEACHERS LECTURE. INTEGRATION AND ANTIDERIVATIVE. ROTHSCHILD CAESARIA COURSE, 2015/6 1. Integrtion Historiclly, it ws the problem of computing res nd volumes, tht triggered development
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 information13.4. Integration by Parts. Introduction. Prerequisites. Learning Outcomes
Integrtion by Prts 13.4 Introduction Integrtion by Prts is technique for integrting products of functions. In this Section you will lern to recognise when it is pproprite to use the technique nd hve the
More informationSection 7.1 Integration by Substitution
Section 7. Integrtion by Substitution Evlute ech of the following integrls. Keep in mind tht using substitution my not work on some problems. For one of the definite integrls, it is not possible to find
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More informationNumerical Integration
Chpter 5 Numericl Integrtion Numericl integrtion is the study of how the numericl vlue of n integrl cn be found. Methods of function pproximtion discussed in Chpter??, i.e., function pproximtion vi the
More information