# Mathematical Analysis: Supplementary notes I

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 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 F1-F7 below hold true Note tht multipliction cn lso be denoted by, or simply by juxtposition (F1) Associtivity of ddition: (r + s) + t = (r + s) + t for ll r, s, t R; (F2)(i) Existence of dditive identity: There exists 0 R such tht r + 0 = r for ll r R; (F2)(ii) Existence of dditive inverse: Given r R there exists s R such tht r + s = 0 (we write s = r); (F3) Commuttivity of ddition: r + s = s + r for ll r, s R; (F4) Associtivity of multipliction: (rs)t = r(st) for ll r, s, t R; (F5)(i) Existence of multiplictive identity: There exists 1 R with 1 0 such tht r 1 = r for ll r R; (F5)(ii) Existence of multiplictive inverse: Given r R \ {0} there exists t R such tht rt = 1 (we write t = r 1 ); (F6) Commuttivity of multipliction: rs = sr for ll r, s R; (F7) Distributive Lw: r(s + t) = rs + rt for ll r, s, t R 1

2 1 SEQUENCES DEFINITION Let f : N R (or N 0 R) Then f is clled sequence If f(n) = n, then n is clled the nth term It is customry to write such sequence s { n } (or { 1, 2, }) DEFINITION A sequence { n } is sid to converge to it if for every ε > 0 there is n integer N such tht n < ε whenever n N (The number N my depend on ε; in prticulr, smller ε my (nd often does) require lrger N) In this cse we write n n = or n s n A sequence tht does not converge is sid to diverge This cn be further qulified (eg the sequence diverges to + ) LEAST UPPER BOUND AND GREATEST LOWER BOUND DEFINITION Let E R, E A number b R is clled n upper (resp lower) bound for E if x b (resp b x) for every x E In the cse tht such b exists, we sy tht E is bounded bove (resp below) If E hs both n upper bound nd lower bound, then we sy tht E is bounded DEFINITION By supremum (or lest upper bound) of E we men number s R such tht (i) s is n upper bound for E nd (ii) if b is n upper bound for E, then s b If E hs supremum s, we write s = sup E (The nottion s = lub E is used in some textbooks) The infimum (or gretest lower bound) of E is defined similrly It is lower bound for E which is lrger thn every other lower bound for E We write inf E (or glb E) Axiom 5 (Supremum principle) Every nonempty set of rel number tht is bounded bove hs supremum in R Every nonempty set of rel numbers tht is bounded below hs n infimum in R MONOTONE SEQUENCES DEFINITION A sequence { n } is sid to be : nondecresing if n n+1 ; nonincresing if n n+1 ; strictly incresing if n < n+1 ; 2

3 strictly decresing if n > n+1, for every n N All such sequences re sid to be monotone, in the ltter two cses strictly monotone EXAMPLE The number e For every n N, let Then (i) { n } is strictly incresing, (ii) {b n } is strictly decresing, (iii) n n = n b n n = ( n ) n, bn = ( n ) n+1 The common it of these two sequences is denoted by e There holds 2 < e < 4 LIMSUP nd LIMINF Even though sequence need not hve it, there re two importnt rel numbers which cn be ssocited with every sequence of rel numbers DEFINITION Let { n } R For every k N define We hve Let y k = inf{ n ; n N, n k} z k = sup{ n ; n N, n k} y 1 y 2 y 3 nd z 1 z 2 z 3 y = k y k = sup{y k ; k N} z = k z k = inf{z k ; k N} The numbers y nd z re clled the it inferior of { n } nd the it superior of { n }, respectively Note: y nd/or z my be + or We write y = inf n n = sup inf{ n ; n N, n k}, k z = sup n = inf sup{ n; n N, n k} n k A useful tool in the study of sequences is the notion of cluster point DEFINITION A point x is clled cluster point of the sequence {x n } if for every ε > 0 there re infinitely mny vlues of n with x n x < ε 3

4 Let {x n } be sequence in R tht is bounded bove The it superior of x n ( sup x n ) n is the gretest cluster point of {x n }; equivlently, it is the supremum of the set of cluster points If the sequence is not bounded bove, sup n x n = Similrly, if {x n } is bounded below, the it inferior of x n ( inf n set of cluster points If {x n } is not bounded below, inf n CAUCHY SEQUENCES x n = x n) is the infimum of the DEFINITION A sequence {x n } of rel numbers is clled Cuchy sequence if for every number ε > 0, there is n integer N (depending on ε) such tht x n x m < ε whenever n N nd m N 2 SERIES DEFINITION Let { n } R nd define s n = n k = n k=1 for ech n N The symbol n=1 n or is clled n infinite series hving nth term n nd nth prtil sum s n DEFINITION A series n=1 n is sid to converge to R if the sequence of prtil sums s n = n k=1 k converges to, nd if so we write Otherwise, we sy tht n=1 n diverges = n n=1 Cuchy criterion Let { n } R A series n=1 n converges if nd only if for every ε > 0 there exists n 0 N such tht q n < ε whenever q > p n 0 n=p+1 Geometric series Let R nd r R Then r n converges nd its sum is n=0 r < 1 If 0 nd r 1, then this series diverges By the formul for geometric progressions we hve (r 1): 1 r if s n = n k=0 r k = 1 rn+1 1 r 4

5 Assuming tht r < 1, n s n = If 0 nd r 1, then 1 r rn 0 nd this shows tht the series cnnot converge Comprison test Suppose tht 0 n b n for ll but finitely mny n N (1) If n b n converges, then n n converges (2) If n n diverges, then n b n diverges ABSOLUTE CONVERGENCE DEFINITION We sy tht n=1 n converges bsolutely if n=1 n converges Leibniz s lternting test Let { n } be nonincresing sequence of positive numbers such tht n n = 0 Then n=1 ( 1)n n is convergent Root test (Cuchy) Let n=1 n nd ρ = sup n (i) If ρ < 1, the series converges bsolutely, (ii) If ρ > 1, the series diverges (iii) If ρ = 1, then the test is inconclusive n n Rtio test (d Alembert) Let n=1 n be series with n 0 for every n (i) If sup n+1 n n < 1, then the series converges bsolutely (ii) If inf n+1 n n > 1, then the series diverges (iii) If n+1 n n = 1, then the test is inconclusive 3 LIMITS AND CONTINUITY DEFINITION Let X R nd R We cll it point (or ccumultion point) of X if every intervl ( δ +, + δ), δ > 0, contins t lest one point of X other thn ( need not be in X) All points of X tht re not it points of X re clled isolted points of X DEFINITION Let f : X R, X R nd let be it point of X We sy tht f converges to the it l s x pproches if for every ε > 0 there is δ > 0 such tht f(x) l < ε whenever 0 < x < δ nd x X Theorem 1 Let x f(x) = l nd x g(x) = m Then 5

6 (i) x (f(x) ± g(x)) = x f(x) ± x g(x) = l ± m, (ii) x αf(x) = α x f(x) = αl for every α R, (iii) if m 0, then x f(x) g(x) = x f(x) x g(x) = l m, (iv) x f(x)g(x) = x f(x) x g(x) = lm Theorem 2 (Squeeze principle) Let f : X R, g : X R, h : X R be such tht f(x) g(x) h(x) for ll x X If f(x) = h(x) = l, then g(x) = l x x x LIMITS AT INFINITY DEFINITION Let f : [, ) R We sy tht x f(x) = l if nd only if for every ε > 0 there exists M > 0 (think of M s needing to be lrge if ε is smll) such tht f(x) l < ε whenever x M We sy tht x f(x) = if nd only if for every N > 0 there exists M > 0 such tht f(x) N whenever x > M In similr mnner we define its t The squeeze principle remins vlid for its t ( ) EXAMPLE Let f(x) = sin x For every x R we hve 1+ x 1 Since x 1 + x x sin x 1 + x x = 0, we lso hve f(x) = 0 x CONTINUOUS FUNCTIONS DEFINITION Let f : (, b) R nd x 0 (, b) We sy tht f is continuous t x 0 if x x 0 f(x) = f(x 0 ) This mens: f is continuous t x 0 if for every ε > 0 there exists δ > 0 such tht f(x) f(x 0 ) < ε whenever x x 0 < δ The function f is continuous on (, b) if f is continuous t ech point of (, b) Theorem 3 Let f nd g be defined on (, b) nd continuous t the point x 0 (, b), nd let λ be constnt Then (i) λ f(x) is continuous t x 0, (ii) f(x) ± g(x) is continuous t x 0, 6

7 (iii) f(x) g(x) is continuous t x 0, (iv) f(x) g(x) is continuous t x 0, provided g(x 0 ) 0 Theorem 4 (composition of functions) Let x x0 g(x) = l nd let f be continuous t l: then there holds x x0 f((g(x)) = f(l) If g is continuous t x 0 nd f is continuous t g(x 0 ), then f g(x) = f(g(x)) is continuous t x 0 DEFINITION Let I be n intervl, f : I R We sy tht f is uniformly continuous on I if for every ε > 0 there is δ > 0 such tht x, y I nd x y < δ imply f(x) f(y) < ε Theorem 5 Let f : [, b] R be continuous Then f is uniformly continuous PROPERTIES OF CONTINUOUS FUNCTIONS Intermedite Vlue Theorem Let λ be constnt nd f : [, b] R continuous on [, b] If f() < λ < f(b) or f() > λ > f(b), then there is c (, b) such tht f(c) = λ DEFINITION We sy tht function f ttins its minimum (mximum) vlue on set Ω R t Ω if f(x) f() (f(x) f()) for ll x Ω We sy tht function f hs n extremum t on Ω if it ttins its mximum or its minimum vlue on Ω t Extreme vlue theorem A continuous function f on closed intervl [, b] ttins its mximum nd minimum vlues on [, b] 4 SEQUENCES AND SERIES OF FUNCTIONS An importnt wy to construct nontrivil functions is to obtin them s its of sequences or series of given functions DEFINITION (pointwise convergence) Let X R nd f n : X R for ech n N The sequence {f n } is sid to be pointwise convergent on X if there exists function f : X R such tht f(x) = n f n (x) for every x X In this cse, f is clled the pointwise it on X of the sequence {f n } DEFINITION (uniform convergence) The sequence {f n } is sid to be uniformly convergent on X if there exists function f : X R such tht for every ε > 0 there exists n N N such tht f n (x) f(x) < ε whenever n N nd x X The number N my depend on ε but not on x 7

8 Theorem 1 Let f n : [, b] R be continuous for ech n N If f n f uniformly on [, b], then f is continuous on [, b] Theorem 2 (Cuchy condition) Let X R nd f n : X R Then the following two ssertions re equivlent: (i) {f n } is uniformly convergent, (ii) for every ε > 0 there exists some N N such tht for every x X SERIES OF FUNCTIONS Let f n : X R, X R nd f n (x) f m (x) < ε whenever n, m N s n (x) = f 1 (x) + + f n (x) DEFINITION If {s n (x)} is pointwise convergent on X to s(x), then we sy tht the series f k (x) is pointwise convergent on X nd tht s is its pointwise sum on X If {s n (x)} k=1 is uniformly convergent on X to s(x), then we sy tht the series convergent on X nd tht s is its uniform sum on X f k (x) is uniformly Theorem 3 (Weierstrss M-test) Let X R nd f n : X R Suppose tht there exists sequence {M n } of nonnegtive numbers such tht M n < nd f n (x) M n for x X nd n N Then n=1 f n (x) is uniformly convergent n=1 Theorem 4 (Abel s test) Let A R nd φ n : A R be decresing sequence of functions, tht is, φ n+1 (x) φ n (x) for every x A Suppose tht there is constnt M such tht φ n (x) M for every x A nd every n If f n (x) converges uniformly on A, then so does φ n (x)f n (x) n=1 Theorem 5 (Dirichlet s test) Let s n (x) = n=1 k=1 n f k (x) for sequence f k : A R Assume tht k=1 there is constnt M such tht s n (x) M for every x A nd every n Let g n : A R be such tht g n 0 uniformly, g n 0 nd g n+1 (x) g n (x) Then f n (x)g n (x) converges uniformly on A n=1 8

9 5 DIFFERENTIATION OF FUNCTIONS OF ONE VARIABLE DEFINITION Let function f be defined on some open intervl contining x 0 R We sy tht f is differentible t x 0 if f (x 0 ) = h 0 f(x 0 + h) f(x 0 ) h exists We cll f (x 0 ) the derivtive of f t x 0 Rewriting this condition s f(x) f(x 0 ) f (x 0 )(x x 0 ) = 0, x x 0 x x 0 we see tht the stright line y = f(x 0 ) + f (x 0 )(x x 0 ), clled the tngent line to the grph of f t x 0, is good pproximtion to f ner x 0, nd rewriting it s [ ] f(x0 + x) f(x 0 ) f (x 0 ) = 0, x 0 x we see tht f (x 0 ), being the it of slopes of the secnt lines, cn be interpreted s the slope of the tngent line to the grph of f t (x 0, f(x 0 )) Proposition 1 If f is differentible t x 0, then f is continuous t x 0 Theorem 2 Let f, g : I R be defined on n open intervl I nd differentible t x I Let α R Then the functions αf, f + g, f g nd f (provided g 0) re differentible t g x Moreover, (i) (αf) (x) = αf (x), (ii) (f + g) (x) = f (x) + g (x), (iii) (f g) (x) = f(x)g (x) + g(x)f (x), (iv) ( ) f(x) g(x) = g(x)f (x) f(x)g (x) g(x) 2 Theorem 3 (Chin rule) Let f : I J nd g : J R, where I nd J re open intervls Suppose tht f is differentible t c I nd tht g is differentible t f(c) Then the composite function g f : I R defined by g f(x) = g(f(x)) is differentible t c nd (g f) (c) = g (f(c))f (c) LOCAL EXTREMA 9

10 DEFINITION Let f : E R, E R We sy tht f hs locl mximum (locl minimum) t c if there exists neighbourhood U of c such tht f(x) f(c) ( f(x) f(c)) for every x U If f hs either locl mximum or locl minimum t c, we sy tht f hs locl extremum t c The next theorem gives necessry, but not sufficient, condition tht locl extremum exists t given point Theorem 4 Let < c < b nd f : [, b] R be given If f hs locl extremum t c nd f (c) exists, then f (c) = 0 Remrks () The restriction tht c is not n endpoint of [, b] is necessry For instnce, the function f(x) = x on [0, 1] hs locl minimum t 0 nd locl mximum t 1, f +(0) = nd f (1) = 1 2 (b) The function f(x) = x 3 extremum t 0 on ( 1, 1) stisfies f (0) = 0 but does not hve locl (c) The theorem ssures us tht if we re seeking ll locl extrem of differentible function on n open intervl, then we need only consider, s cndidtes, those c for which f (c) = 0 (d) If f(x) = x for x R, then f hs locl minimum t c = 0, but f (0) does not exist MEAN VALUE THEOREMS Theorem 5 (Rolle s theorem) Suppose tht f : [, b] R is continuous on [, b], differentible on (, b) nd f() = f(b) Then there exists number ξ (, b) such tht f (ξ) = 0 Theorem 6 (Lgrnge) Let f : [, b] R be continuous on [, b], nd differentible on (, b) Then there exists point c (, b) such tht f(b) f() = f (c)(b ) Theorem 7 Suppose tht f is continuous on [, b] nd differentible on (, b) (i) If f (x) 0 for every x (, b), then f is nondecresing on [, b] (ii) If f (x) 0 for every x (, b), then f is nonincresing on [, b] (iii) If f (x) > 0 for every x (, b), then f is strictly incresing on [, b] (iv) If f (x) < 0 for every x (, b), then f is strictly decresing on [, b] (v) If f (x) = 0 for every x (, b), then f is constnt on [, b] Theorem 8 Suppose tht f is continuous on [, b] nd is twice differentible on (, b), nd tht x 0 (, b) (i) If f (x 0 ) = 0 nd f (x 0 ) > 0, then x 0 is strict locl minimum of f (ii) If f (x 0 ) = 0 nd f (x 0 ) < 0, then x 0 is strict locl mximum of f 10

11 6 INTEGRALS OF FUNCTIONS OF ONE VARIABLE DEFINITION Let f : [, b] R be bounded function We prtition [, b], which mens we choose n integer n nd points x 0, x 1,, x n 1, x n in such wy tht = x 0 < x 1 < < x n 1 < x n = b Denote such prtition by P, tht is, let P = {x 0, x 1,, x n } Then form two sums U(f, P ) = n i=1 M i (x i x i 1 ), where M i = sup f(x) x [x i 1,x i ] nd L(f, P ) = n i=1 m i (x i x i 1 ), where m i = inf x [x i 1,x i ] f(x), clled the upper nd lower Riemnn sum (with respect to P ), respectively Since f is bounded, sy M f(x) M for every x [, b], we see tht (61) (b )M L(f, P ) U(f, P ) (b )M for every prtition P of [, b] It seems resonble to expect tht s the size of the intervls in P gets smller, U(f, P ) decreses while L(f, P ) increses DEFINITION If P nd P re prtitions of [, b] with P P, then P is clled refinement of P Lemm 1 If P is refinement of P, then L(f, P ) L(f, P ) nd U(f, P ) U(f, P ) According to the inequlity (61) Riemnn sums re bounded: therefore we cn introduce the following nottion: the upper Riemnn integrl, nd f(x) dx = inf{u(f, P ); P is ny prtition of [, b]}, f(x) dx = sup{l(f, P ); P is ny prtition of [, b]}, the lower Riemnn integrl Lemm 2 Let P 1 nd P 2 be ny prtitions of [, b] Then L(f, P 1 ) U(f, P 2 ) Corollry 3 f(x) dx f(x) dx 11

12 DEFINITION We sy tht f : [, b] R is Riemnn integrble or tht the Riemnn integrl exists, if f(x) dx = The common vlue is denoted by f(x) dx f(x) dx Theorem 4 A function f : [, b] R is integrble on [, b] if given ny ε > 0 there is prtition P such tht U(f, P ) L(f, P ) < ε Theorem 5 (i) If f : [, b] R is bounded nd continuous t ll but finitely mny points of [, b], then f is integrble on [, b] (ii) Any incresing (decresing) function on [, b] is integrble on [, b] PROPERTIES OF INTEGRALS Theorem 6 (i) If f is bounded nd integrble on [, b] nd k R, then k f is integrble on [, b] nd k f(x) dx = k f(x) dx (ii) If f nd g re bounded nd integrble on [, b], then f + g is integrble on [, b] nd (f + g) dx = f dx + g dx (iii) If f nd g re bounded nd integrble on [, b] nd f(x) g(x) for every x [, b], then f(x) dx g(x) dx (iv) If f is bounded nd integrble on [, b] nd [b, c], then f is integrble on [, c] nd c f dx = f dx + c b f dx Theorem 7 (Men vlue theorem) If f is continuous on [, b], then f(x) dx = f(c)(b ) 12

13 for some c [, b] Let f : [, b] R be continuous function DEFINITION An ntiderivtive of f is continuous function F : [, b] R such tht F is differentible on (, b) nd F (x) = f(x) for < x < b Theorem 8 Let f be bounded nd integrble on [, b] Then f is integrble on [, b], nd f dx f dx Let f : [, b] R be contin- Theorem 9 (The fundmentl theorem of Clculus) uous function Then f hs n ntiderivtive F nd f(x) dx = F (b) F () If G is ny other ntiderivtive of f, we lso hve f(x) dx = G(b) G() Theorem 10 (Integrtion by prts) If du dx nd dv dx re continuous on [, b], then u dv du dx = u(b)v(b) u()v() dx dx v dx We often need to integrte unbounded functions or to integrte over unbounded regions The resulting improper integrls led to convergence problems nlogous to those for n infinite series DEFINITION (improper integrls - first kind) Let f : (, b] R nd suppose tht f is not necessrily bounded t (ner ) but f is integrble on [ + ε, b] for every ε > 0 sufficiently smll We sy tht f is improperly integrble (or f(x) dx exists) if f(x) dx exists ε 0 + +ε If this it exists it is denoted by b f(x) dx (tht is, f(x) dx = f(x) dx) ε 0 + +ε In similr mnner we define improper integrls f : [, b) R (if f is unbounded ner b) DEFINITION (improper integrls - second kind) Let f : [, ) R nd suppose tht f(x) dx exists for every b > We sy tht f is improperly integrble if b f(x) dx exists 13

14 If this it exists, it is denoted by f(x) dx Similrly we define improper integrls for f : (, ] R, If f : (, ) R, we set f(x) dx = f(x) dx = b 0 b f(x) dx f(x) dx + b 0 f(x) dx The integrl f(x) dx diverges if one of these its does not exist Theorem 11 (Comprison test for improper integrls) Suppose f(x) 0 nd g(x) 0 for x (i) If g(x) f(x), then the convergence of (ii) If g(x) f(x), then the divergence of f(x) dx implies the convergence of g(x) dx f(x) dx implies the divergence of g(x) dx The nlogy between positive - term series nd improper integrls of positive functions is the key to the integrl test Theorem 12 If f is continuous, nonnegtive nd nonincresing on [1, ), then n=1 f(n) nd f(x) dx converge or diverge together 1 14

15 7 TAYLOR SERIES TAYLOR S FORMULA Theorem 1 (Tylor s formul) Suppose tht the first (n + 1) derivtives of the function f exist on n intervl contining points nd b Then (71) f(b) = f() + f ()(b ) + f () 2! + + f (n) () n! (b ) 2 + f (3) () (b ) 3 3! (b ) n + f (n+1) (ξ) (b )n+1 (n + 1)! for some number ξ between nd b REMARK Tylor s formul with the Cuchy form of the reminder: f(b) = f() + f ()(b ) + f () (b ) f (n) () (b ) n 2! n! + 1 n! for some number t between nd b TAYLOR SERIES (b t) n f (n+1) (t) dt Suppose tht f is function with continuous derivtives of ll orders in n intervl (c, d) Let (c, d) nd let n be n rbitrry positive integer We know by Tylor s formul tht where f(x) = P n (x) + R n (x), P n (x) = f() + f ()(x ) + f (x )2 () 2! + + f (n) (x )n () n! nd R n is either the Lgrnge or Cuchy reminder Now suppose tht, for some prticulr fixed vlue of x, we cn show tht Then it follows from (71) tht R n(x) = 0 n f(x) = n P n (x) = n = k=0 f (k) () (x ) k k! ( n k=0 ) f (k) () (x ) k k! The infinite series in this eqution is clled the Tylor series (of f t ) 15

16 EXAMPLES We hve the following Tylor formule for the exponentil nd trigonometric functions: e x = 1 + x + x2 2! + x3 3! + + xn n! + e ξ (n + 1)! xn+1, cos x = 1 x2 2! + x4 x2n + ( 1)n 4! (2n)! + cos ξ ( 1)n+1 (2n + 2)! x2n+2, sin x = x x3 3! + x5 5! + ( 1)n x 2n+1 (2n + 1)! + ( 1)n+1 cos ξ (2n + 3)! x2n+3, In ech cse ξ is some number between 0 nd x Since ξ is between 0 nd x, it follows tht 0 < e ξ e x in Tylor s formul for e x In the formuls for the sine nd cosine functions, 0 cos ξ 1 Therefore the fct tht x n n n! = 0 for ll x implies tht n R n (x) = 0 in ll three cses bove This gives the following Tylor series: e x x n = n! = 1 + x + x2 2! + x3 3! + x4 4! +, cos x = sin x = n=0 n=0 ( 1) n x2n (2n)! = 1 x2 2! + x4 4! x6 6! +, n=0 ( 1) n x 2n+1 (2n + 1)! = x x3 3! + x5 5! x7 7! + 8 VECTOR FUNCTIONS - FUNCTIONS OF SEV- ERAL VARIABLES Addition nd sclr multipliction of n-tuples re defined by (x 1,, x n ) + (y 1,, y n ) = (x 1 + y 1,, x n + y n ) nd (x 1,, x n ) = (x 1,, x n ) for R The length or norm of vector x in R n is defined by x = x = { n i=1 x 2 i }

17 The distnce between two vectors x nd y is defined by { n } 1 2 x y = x y = (x i y i ) 2 The inner (sclr) product of x nd y is defined by i=1 (x, y) = n x i y i i=1 We hve: 17

18 Theorem 1 For x, y, z R n there holds: (i) (x, x) = x 2, (ii) (x, y) x y (Cuchy - Schwrz inequlity), (iii) x + y x + y (iv) x y x z + z y ((iii) nd (iv) re clled tringle inequlities) DEFINITION Let S nd T be given sets A function f : S T consists of two sets S nd T together with rule tht ssigns to ech x S specific element of T, denoted by f(x) One often writes x f(x) to denote tht x is mpped to the element f(x) For function f : S T, the set S is clled the domin of F The rnge, or imge, of f is the subset of T defined by f(s) = {f(x) T ; x S} If f : Ω R n R m, we write f(x 1,, x n ) = (f 1 (x 1,, x n ),, f m (x 1,, x n )) The f i re clled coordinte functions, or components of f Composite functions If two functions f nd g re so relted tht the rnge spce of f is the sme s the domin spce of g, we my form the composite function g f by first pplying f nd then g Thus g f(x) = g(f(x)) for every vector x in the domin of f The opertion of ddition nd multipliction of vector functions Let f nd g be functions with the sme domin nd hving the sme rnge spce Then the function f + g is the sum of f nd g defined by for ll x in the domin of both f nd g (f + g)(x) = f(x) + g(x) Similrly, if r R, then rf is the numericl multiple of f by r nd is defined by rf(x) = r f(x) LIMITS AND CONTINUITY OF VECTOR FUNCTIONS Let f : Ω R n R m We use x y = n i=1 (x i y i ) 2, x = (x 1,, x n ), y = (y 1,, y n ) DEFINITION Let Ω R n Then is it point of Ω if, for every ε > 0, there exists point y Ω such tht 0 < y < ε In other words, the definition sys tht is it point (or ccumultion point) of Ω if there re points in Ω other thn tht re contined in bll of rbitrrily smll rdius with centre t 18

19 We come now to the definition of it for function f : Ω R m, Ω R n DEFINITION Let y 0 R m, nd let x 0 R n be it point of Ω Then y 0 is the it of f t x 0 if, for every ε > 0, there is δ > 0 such tht f(x) y 0 < ε whenever x stisfies 0 < x x 0 < δ nd x Ω (We write f(x) = y 0 ) x x0 Less formlly, the definition sys tht for x x 0, f(x) cn be mde rbitrrily close to y 0 by choosing x sufficiently close to x 0 Geometriclly, the ide is this: given ny bll B(y 0, ε) in R m, there exists bll B(x 0, δ) R n whose intersection with Ω (the domin of f), except possibly for x 0 itself, is sent by f into B(y 0, ε) Theorem 2 Let f : Ω R m, Ω R n nd let x 0 be it point of Ω Then f(x) = y0 x x0 if nd only if f i(x) = y 0 x x 0 i, i = 1,, m DEFINITION A function f is continuous t x 0 if x 0 is in the domin of f nd x x0 f(x) = f(x 0 ) At nonit or isolted point of the domin, we cnnot sk for it; insted we simply define f to be utomticlly continuous t such point Theorem 3 A vector function is continuous t point x 0 if nd only if its coordinte functions re continuous there Theorem 4 Every liner function L : R n R m is continuous on R n, nd for such n L there is number k such tht L(x) k x for every x R n The continuity of mny functions cn be deduced from repeted pplictions of the following theorem: Theorem 5 (1) The functions P k : R n R, where P k (x) = x k, (ie P k : (x 1,, x n ) x k ) re continuous for k = 1,, n (2) The functions S : R 2 R nd M : R 2 R, defined by S(x, y) = x + y nd M(x, y) = xy re continuous (3) If f : R n R m nd g : R m R p re continuous, then the composition g f given by g f(x) = g(f(x)) is continuous wherever it is defined 19

20 9 DIFFERENTIABILITY OF VECTOR FUNCTIONS DEFINITION Let Ω R n A point x 0 Ω is n interior point of set S if there exists positive number δ such tht {x : x x 0 < δ} Ω (equivlently B(x 0, δ) Ω) A subset of R n, ll of whose points re interior, is clled open Mny of the techniques of clculus hve s their foundtion the ide of pproximting vector function by liner function or by n ffine function Recll tht function A : R n R m is ffine if there exists liner function L : R n R m nd vector y 0 R m such tht A(x) = L(x) + y 0 for every x R n ( ) 1 EXAMPLE Consider point y 0 = nd liner function L : R 3 3 R 2 given by ( ) x y = L(x) = x , x 3 equivlently, y 1 = x 1 + 2x 2 + x 3 y 2 = x 1 + 4x 2 + 5x 3 The ffine function A(x) = L(x) + y 0 is defined by the equtions y 1 = x 1 + 2x 2 + x y 2 = 3x 1 + 4x 2 + 5x We shll now study the possibility of pproximting n rbitrry vector function f : Ω R m, with Ω R n, ner point x 0 of Ω by n ffine function A We begin by requiring tht f(x 0 ) = A(x 0 ) Since A(x) = L(x) + y 0, where L is liner, we obtin f(x 0 ) = L(x 0 ) + y 0 nd so (71) A(x) = L(x x 0 ) + f(x 0 ) A nturl requirement is tht (72) x x0 ( f(x) A(x) ) = 0 We observe tht from (71) we hve Since L is continuous, (72) sys tht f(x) A(x) = f(x) f(x 0 ) L(x x 0 ) 0 = x x0 ( f(x) A(x) ) = x x0 ( f(x) f(x0 ) ), 20

21 which is precisely the sttement tht f is continuous t x 0 This is significnt, but it sys nothing bout L Thus, in order for our notion of pproximtion to distinguish one ffine function from nother or to mesure how well A pproximtes f, some dditionl requirement is necessry We require tht f(x) A(x) pproches 0 fster thn x pproches x 0 Tht is, we demnd tht f(x) f(x 0 ) L(x x 0 ) = 0 x x 0 x x 0 Equivlently, we cn sk tht f be representble in the form f(x) = f(x 0 ) + L(x x 0 ) + x x 0 z(x x 0 ), where z(y) is some function tht tends to 0 s y tends to 0 DEFINITION A function f : D R n R m is sid to be differentible t x 0, if (1) x 0 is n interior point of the domin D of f, (2) there is n ffine function tht pproximtes f ner x 0 Tht is, there exists liner function L : R n R m such tht f(x) f(x 0 ) L(x x 0 ) x x 0 x x 0 The liner function L is clled the differentil of f t x 0 The function f is sid to be differentible if it is differentible t every point of its domin If n = m = 1, n ffine function hs the form x+b Hence f : R R tht is differentible t x 0 cn be pproximted ner x 0 by function A(x) = x+b Since f(x 0 ) = A(x 0 ) = x 0 +b, we obtin b = f(x 0 ) x 0 nd A(x) = x + b = (x x 0 ) + f(x 0 ) The liner prt of A, denoted erlier by L, is L(x) = x The condition (2) of the definition becomes f(x) f(x 0 ) (x x 0 ) = 0 x x 0 x x 0 This is equivlent to f(x) f(x 0 ) = x x 0 x x 0 The number is commonly denoted by f (x 0 ) nd it is the derivtive of f t x 0 The ffine function A is therefore given by A(x) = f(x 0 ) + f (x 0 )(x x 0 ) Its grph is the tngent line to the grph of f t x 0 = 0 21

22 Generl cse: n, m 1 A liner function L : R n R m must be representble by n m - by - n mtrix, tht is, L(x) = Ax We shll show tht the mtrix A of L stisfying (1) nd (2) of the definition cn be computed in terms of prtil derivtives of f To find the mtrix A, we consider the stndrd bsis (e 1,, e n ) of R n If x 0 is n interior point of the domin of f, the vectors x j = x 0 + te j, j = 1,, m, re ll in the domin of f for sufficiently smll t By condition (2) of the definition we hve Since L is liner, we deduce from this f(x j ) f(x 0 ) L(te j ) t 0 t f(x j ) f(x 0 ) t 0 t = L(e j ) L(e j ) is the jth column of the mtrix A of L On the other hnd the vector x j differs from x 0 only in the jth coordinte, tht is, f(x j ) f(x 0 ) t 0 t nd the entire mtrix of L hs the form f 1 (x 0 ) f, 1 (x 0 ),, x 1 x 2 f 2 (x 0 ) f, 2 (x 0 ),, x 1 x 2 f m(x 0 ) f, m(x 0 ),, x 1 x 2 = f 1 (x 0 ) x n f 2 (x 0 ) x n f m(x 0 ) x n = 0 f 1 (x 0 ) x j f m(x 0 ) x j, ij = f i(x 0 ) x j This mtrix is clled the Jcobin mtrix or the derivtive of f t x 0, nd it is denoted by f (x 0 ) It follows tht L is uniquely determined by the prtil derivtives f i(x 0 ) x j The differentil of L t x 0 is lso denoted by d x0 f or D x0 f We summrize wht we hve just proved s follows Theorem 1 If the function f : R n R m is differentible t x 0, then the differentil d x0 f is uniquely determined, nd its mtrix is the Jcobin mtrix of f Tht is, for every vector y in R n we hve D x0 f(y) = f (x 0 )y We interpret y = f(x 0 ) + f (x 0 )(x x 0 ) s the eqution of the tngent plne to the grph of f t (x 0, f(x 0 )) 22

23 EXAMPLE The function ( x 2 + e y ) x + y sin z hs coordinte functions f 1 (x, y, z) = x 2 + e y nd f 2 (x, y, z) = x + y sin z The Jcobin mtrix of f t (x, y, z) is [ f1 f x (x, y, z) =, f 1, ] f 1 [ ] y z 2x, e f 2, f 2, f 2 = y, 0 1, sin z, y cos z x y z The differentil of f t (1, 1, π) is d (1,1,π) f(x, y, z) = [ 2 e ] x y z = The ffine mpping tht pproximtes f ner (1, 1, π) is [ ] x 1 2 e 0 A(x, y, z) = y 1 + f(1, 1, π) z π Remrk: = = [ ] [ ] x e 2 e 0 + y z π [ ] 1 + e + 2(x 1) + e(y 1) = 1 + (x 1) (z π) [ 2x + ey x z ] [ 2x + ey 1 x z + π If f : R n R, then the Jcobin mtrix is reduced to grdient f (x) = ( f x 1, f x 2,, f x n ) The nottion f(x) is lso used In this cse the differentil of f t x 0 is given by L(x) = d x0 f(x) = x 1 f(x 0 ) x 1 + x 2 f(x 0 ) x x n f(x 0 ) x n EXAMPLE Let f(x) = n i=1 x2 i = x 2 Show tht f is differentible t every point x 0 R n If f is differentible t x 0, then the differentil of f t x 0 must be given by n L(x) = d x0 f(x) = 2x 0 i x i = 2x 0 x Then f(x) f(x 0 ) L(x x 0 ) x x 0 = x 2 x 0 2 2x 0 x + 2 x 0 2 x x 0 = x x 0 0 s x x 0 i=1 = x 2 x 0 2 2x 0 (x x 0 ) x x 0 23 = x 2 + x 0 2 2x 0 x x x 0 ] = x x 0 2 x x 0

24 Therefore f is differentible t ech point x 0 R n How one cn tell whether or not vector function is differentible? We only know tht if f is differentible then the differentil is represented by the Jcobin mtrix Theorem 2 If the domin of f : R n R m is n open set D R n on which ll prtil derivtives f i x j re continuous, then f is differentible t every point of D For exmple f(x 1, x 2, x 3 ) = ( x x 1 x 2 x 3, e x 1+x 2 +x 3, sin(x 1 + x 2 + x 3 ) ) is differentible t every point of R 3 It follows from the definition of differentible vector function tht: Theorem 3 If f is differentible t x 0, then f is continuous t x 0 The converse is not true EXAMPLE Consider the function { xy for (x, y) (0, 0) f(x, y) = x 2 +y 2 0 for (x, y) = (0, 0) Since (x,y) (0,0) (0, 0), then Since nd xy = 0 = f(0, 0), f is continuous t (0, 0) If f were differentible t x 2 +y2 d (0,0) f(x, y) = (f x (0, 0), f y (0, 0)) f x (0, 0) = h 0 f(h, 0) f(0, 0) h f y (0, 0) = h 0 f(0, h) f(0, 0) h d (0,0) f(x, y) = 0 for ll (x, y) R 2 On the other hnd = 0 = 0, f(x, y) f(0, 0) d (0,0) f(x, y) xy = (x,y) (0,0) x2 + y 2 (x,y) (0,0) x 2 + y = 0, 2 which is impossible s the function xy x 2 +y 2 does not hve it t the origin Notice tht f hs prtil derivtives t (0, 0): f(0,0) = 0 nd f(0,0) = 0 This mens tht the x y grdient (the Jcobin mtrix) exists However, this does not gurntee the differentibility of f t (0, 0) Let f : R n R, x 0 R n nd u R n be unit vector This mens u = 1, where u = (u 1,, u n ), u = ( n i=1 u 2 i )

25 DEFINITION The directionl derivtive of f t x 0 in the direction u, denoted by f(x 0) u or D u f(x 0 ), is defined by f(x 0 ) u f(x 0 + tu) f(x 0 ) = t 0 t From this definition we see tht the directionl derivtive is the rte of chnge of f in the direction u Theorem 4 If f is differentible t x, then for every unit vector in R n f(x) u = f (x)u EXAMPLE Let f(x 1, x 2, x 3 ) = x x e x 2, u = ( 1 2, 1 2, 1 2 ) Then f(1, 1, 1) u = f(1, 1, 1) f(1, 1, 1) f(1, 1, 1) u 1 + u 2 + u 3 x 1 x 2 x 3 = e = 1 + e EXAMPLE Show tht the existence of ll directionl derivtives t point x does not imply the differentibility t this point Let { x y for (x, y) (0, 0), f(x, y) = x 2 +y 2 0 for (x, y) = (0, 0) We know tht f is not differentible t (0, 0) u = (u 1, u 2 ) Indeed, However, f(0,0) u exists t ech direction f(0, 0) u = t 0 tu 1 tu 2 t t 2 u t 2 u 2 2 = t 0 t t u 1 u 2 t t = u 1 u 2 We hve tht f(x 0) is the slope of the tngent line t (x u 0, f(x 0 )) to the curve formed by the intersection of the grph of f with the plne tht contins (x 0 + tu) nd x 0, nd is prllel to the z-xis We lwys hve For u = f(x 0) f(x 0, we hve ) f(x 0 ) u = f(x 0 )u f(x 0 ) u = f(x 0 ) f(x 0 ) u = f(x 0) f(x 0 ) f(x 0 ) = f(x 0) 2 f(x 0 ) = f(x 0 ) 25

26 This shows tht the rte of chnge of f(x 0) u in the direction of the grdient is never greter thn f(x 0 ) nd is equl to it CHAIN RULE As in the one-dimensionl cse, the chin rule is rule for differentiting composite functions Theorem 5 Let g be continuously differentible function on n open set Ω R n nd let f be defined nd differentible for < t < b, tking its vlues in Ω Then the composite function F (t) = g(f(t)) is differentible for < t < b nd F (t) = g(f(t)) f (t) EXAMPLE Let g(x, y) = x 2 y + e x+y for (x, y) R 2 nd let f(t) = (t, t 2 ) Then nd f (t) = (1, 2t), g(x, y) = (2xy + e x+y, x 2 + e x+y ) F (t) = (2t 3 + e t+t2, t 2 + e t+t2 ) (1, 2t) = 2t 3 + e t+t2 + 2t 3 + 2te t+t2 = 4t 3 + e t+t2 + 2te t+t2 The following theorem gives the extension to ny dimension for the domin nd rnge of g nd f Theorem 6 (the Chin Rule) Let f : R n R m be continuously differentible t x nd let g : R m R p be continuously differentible t f(x) If g f is defined on n open set contining x, then g f is continuously differentible t x, nd ( g f ) (x) = g (f(x))f (x) Proof The mtrices here hve the form g 1 (f(x)) y 1,, g p(f(x)) y 1,, g 1 (f(x)) y m g p(f(x)) y m nd f 1 (x) x 1,, f m(x) x 1,, The product of the mtrices hs s its ij th entry the sum of products m g i (f(x)) f k (x) y k x j k=1 f 1 (x) x n f m(x) x n This expression is the sclr product of two vectors g i (f(x)) nd f(x) x j It follows from Theorem 5 tht g i (f(x)) f(x) x j becuse we differentite with respect to the single vrible x j EXAMPLES 26 = (g i f)(x) x j,

27 (1) Let f(x, y) = (x 2 + y 2, x 2 y 2 ) nd g(u, v) = (uv, u + v) Find (g f) (2, 1) First we compute g nd f [ ] [ ] v, u 2x, 2y g (u, v) =, f (x, y) = 1, 1 2x, 2y To find (g f) (2, 1), we note tht f(2, 1) = (5, 3), [ ] 3, 5 g (5, 3) = f (2, 1) = 1, 1 [ 4, 2 4, 2 Then the product of the mtrices g (5, 3) nd f (2, 1) gives [ ] [ ] [ ] 3, 5 4, , 6 10 (g f)(2, 1) = = = 1, 1 4, , 2 2 (2) Let w = g(x, y, z) : R 3 R nd (x, y, z) = f(s, t) = ( f 1 (s, t), f 2 (s, t), f 3 (s, t) ) : R 2 R 3 Compute the prtil derivtives w s By the chin rule we hve Mtrix multipliction yields w s w t w nd t ] of the composite function w = g ( f 1 (s, t), f 2 (s, t), f 3 (s, t) ) ( w s, w ) ( g = t x, g y, g ) z x, s y, s z, s x t y t z t = g x x s + g y y s + g z z s = g x x t + g y y t + g z z t (3) Let g : R 3 R 3, f : R 2 R 3 We introduce nottion w = (w 1, w 2, w 3 ) = ( g 1 (x, y, z), g 2 (x, y, z), g 3 (x, y, z) ), (x, y, z) = ( f 1 (s, t), f 2 (s, t), f 3 (s, t) ) Compute w 1, w 1, w 2, w 2, w 3 nd w 3 We hve s t s t s t w 1, w 1 s t w 2, w 2 s t w 3, w 3 s t = Mtrix multipliction yields, for j = 1, 2, 3: g 1, g 1, g 1 x y z g 2, g 2, g 2 x y z g 3, g 3, g 3 x y z f 1, f 1 s t f 2, f 2 s t f 3, f 3 s t [ 32, 4 8, 0 ] w j s w j t = g j f 1 x = g j x s + g j f 2 y s + g j f 3 z s, f 1 t + g j f 2 y t + g j f 3 z t 27

28 10 THE IMPLICIT AND INVERSE FUNCTION THE- OREMS THE IMPLICIT FUNCTION THEOREM An eqution in two vribles x nd y my hve one or more solutions for y in terms of x or for x in terms of y We sy tht these solutions re functions implicitly defined by the eqution For exmple the eqution of the unit circle, x 2 + y 2 = 1, implicitly defines four function (mong others): y = 1 x 2 for x [ 1, 1], y = 1 x 2 for x [ 1, 1], x = 1 y 2 for y [ 1, 1], x = 1 y 2 for y [ 1, 1] In generl cse, we consider function F : R n R m R m, nd study the reltion F (x, y) = 0, or, written out, F 1 (x 1,, x n, y 1,, y m ) = 0 F m (x 1,, x n, y 1,, y m ) = 0 The gol is to solve for the m unknowns y 1,, y m from the m equtions in terms of x 1,, x n Theorem 1 (The Implicit Function Theorem) Let Ω R n R m be n open set, nd let F : Ω R m be function of clss C 1 Suppose (x 0, y 0 ) Ω nd F (x 0, y 0 ) = 0 Assume tht F 1 F y 1,, 1 y m det 0 evluted t (x 0, y 0 ), F m F y 1,, m y m where F = (F 1,, F m ) Then there re open sets U R n nd V R m, with x 0 U nd y 0 V, nd unique function f : U V such tht for ll x U Furthermore, f is of clss C 1 F (x, f(x)) = 0 THE INVERSE FUNCTION THEOREM If function f is thought of s sending vectors x into vectors y in the rnge of f, then it is nturl to strt with y nd sk wht vector or vectors x re sent by f into y 28

29 More prticulrly, we my sk if there is function tht reverses the ction of f If there is function f 1 with the property f 1 (y) = x if nd only if f(x) = y, then f 1 is clled the inverse function of f It follows tht the domin of f 1 is the rnge of f nd tht the rnge of f 1 is the domin of f Given function f : Ω R n R n, where Ω R n, one my sk: (1) Does it hve n inverse? (2) If it does, wht re its properties? In generl it is not esy to nswer these questions just by looking t the function However, in certin circumstnces we get useful result Theorem 2 (The inverse function theorem) Let f : R n R n be continuously differentible function such tht f (x 0 ) hs n inverse Then there is n open set Ω contining x 0 such tht f, when restricted to Ω, hs continuously differentible inverse The imge set f(ω) is open In ddition, [ f 1 (y 0 ) ] = [f (x 0 )] 1, where y 0 = f(x 0 ) Tht is, the differentil of the inverse function t y 0 is the inverse of the differentil of f t x 0 29

### Lecture 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

### Math 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

### The 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

### Math 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:

### 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

### 63. 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 =

### Chapter 6. Riemann Integral

Introduction to Riemnn integrl Chpter 6. Riemnn Integrl Won-Kwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl

### Partial Derivatives. Limits. For a single variable function f (x), the limit lim

Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the right-hnd side limit equls to the left-hnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles

### Lecture 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

### MA 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,

### For 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

### Math 100 Review Sheet

Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s

### 1 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

### MAT612-REAL ANALYSIS RIEMANN STIELTJES INTEGRAL

MAT612-REAL ANALYSIS RIEMANN STIELTJES INTEGRAL DR. RITU AGARWAL MALVIYA NATIONAL INSTITUTE OF TECHNOLOGY, JAIPUR, INDIA-302017 Tble of Contents Contents Tble of Contents 1 1. Introduction 1 2. Prtition

Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be rel-vlues nd smooth The pproximtion of n integrl by numericl

### MATH 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

### Best 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

### Integrals along Curves.

Integrls long Curves. 1. Pth integrls. Let : [, b] R n be continuous function nd let be the imge ([, b]) of. We refer to both nd s curve. If we need to distinguish between the two we cll the function the

### Week 10: Riemann integral and its properties

Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the

### Czechoslovak 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 HENSTOCK-KURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When rel-vlued

### Math 8 Winter 2015 Applications of Integration

Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl

### FINALTERM EXAMINATION 2011 Calculus &. Analytical Geometry-I

FINALTERM EXAMINATION 011 Clculus &. Anlyticl Geometry-I Question No: 1 { Mrks: 1 ) - Plese choose one If f is twice differentible function t sttionry point x 0 x 0 nd f ''(x 0 ) > 0 then f hs reltive...

### 38 Riemann sums and existence of the definite integral.

38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These

### NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.

NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with

### Math 4200: Homework Problems

Mth 4200: Homework Problems Gregor Kovčič 1. Prove the following properties of the binomil coefficients ( n ) ( n ) (i) 1 + + + + 1 2 ( n ) (ii) 1 ( n ) ( n ) + 2 + 3 + + n 2 3 ( ) n ( n + = 2 n 1 n) n,

### Continuous 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 rel-vlued function defined by : { 1 if x S 1 S (x) if x S Suppose tht

### MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

### Big idea in Calculus: approximation

Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:

### 8 Laplace s Method and Local Limit Theorems

8 Lplce s Method nd Locl Limit Theorems 8. Fourier Anlysis in Higher DImensions Most of the theorems of Fourier nlysis tht we hve proved hve nturl generliztions to higher dimensions, nd these cn be proved

### The problems that follow illustrate the methods covered in class. They are typical of the types of problems that will be on the tests.

ADVANCED CALCULUS PRACTICE PROBLEMS JAMES KEESLING The problems tht follow illustrte the methods covered in clss. They re typicl of the types of problems tht will be on the tests. 1. Riemnn Integrtion

### Riemann Stieltjes Integration - Definition and Existence of Integral

- Definition nd Existence of Integrl Dr. Adity Kushik Directorte of Distnce Eduction Kurukshetr University, Kurukshetr Hryn 136119 Indi. Prtition Riemnn Stieltjes Sums Refinement Definition Given closed

### Section 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

### Week 7 Riemann Stieltjes Integration: Lectures 19-21

Week 7 Riemnn Stieltjes Integrtion: Lectures 19-21 Lecture 19 Throughout this section α will denote monotoniclly incresing function on n intervl [, b]. Let f be bounded function on [, b]. Let P = { = 0

### Section 14.3 Arc Length and Curvature

Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in

### Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230

Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given

### SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 014 Mrk Scheme: Ech prt of Question 1 is worth four mrks which re wrded solely for the correct nswer.

### Convex 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

### Summer MTH142 College Calculus 2. Section J. Lecture Notes. Yin Su University at Buffalo

Summer 6 MTH4 College Clculus Section J Lecture Notes Yin Su University t Bufflo yinsu@bufflo.edu Contents Bsic techniques of integrtion 3. Antiderivtive nd indefinite integrls..............................................

### Sections 5.2: The Definite Integral

Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)

### Riemann Integrals and the Fundamental Theorem of Calculus

Riemnn Integrls nd the Fundmentl Theorem of Clculus Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University September 16, 2013 Outline Grphing Riemnn Sums

### Mapping the delta function and other Radon measures

Mpping the delt function nd other Rdon mesures Notes for Mth583A, Fll 2008 November 25, 2008 Rdon mesures Consider continuous function f on the rel line with sclr vlues. It is sid to hve bounded support

### Calculus and linear algebra for biomedical engineering Week 11: The Riemann integral and its properties

Clculus nd liner lgebr for biomedicl engineering Week 11: The Riemnn integrl nd its properties Hrtmut Führ fuehr@mth.rwth-chen.de Lehrstuhl A für Mthemtik, RWTH Achen Jnury 9, 2009 Overview 1 Motivtion:

### 13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS

33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in

### If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du

Integrtion by Substitution: The Fundmentl Theorem of Clculus demonstrted the importnce of being ble to find nti-derivtives. We now introduce some methods for finding ntiderivtives: If u = g(x) is differentible

### u(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C.

Lecture 4 Complex Integrtion MATH-GA 2451.001 Complex Vriles 1 Construction 1.1 Integrting complex function over curve in C A nturl wy to construct the integrl of complex function over curve in the complex

### Calculus MATH 172-Fall 2017 Lecture Notes

Clculus MATH 172-Fll 2017 Lecture Notes These notes re concise summry of wht hs been covered so fr during the lectures. All the definitions must be memorized nd understood. Sttements of importnt theorems

### Necessary and Sufficient Conditions for Differentiating Under the Integral Sign

Necessry nd Sufficient Conditions for Differentiting Under the Integrl Sign Erik Tlvil 1. INTRODUCTION. When we hve n integrl tht depends on prmeter, sy F(x f (x, y dy, it is often importnt to know when

### Chapter 4 Contravariance, Covariance, and Spacetime Diagrams

Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz

### New Expansion and Infinite Series

Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06-073 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University

### Pre-Session Review. Part 1: Basic Algebra; Linear Functions and Graphs

Pre-Session Review Prt 1: Bsic Algebr; Liner Functions nd Grphs A. Generl Review nd Introduction to Algebr Hierrchy of Arithmetic Opertions Opertions in ny expression re performed in the following order:

### (a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P.

Chpter 7: The Riemnn Integrl When the derivtive is introdued, it is not hrd to see tht the it of the differene quotient should be equl to the slope of the tngent line, or when the horizontl xis is time

### 1 Sequences. 2 Series. 2 SERIES Analysis Study Guide

2 SERIES Anlysis Study Guide 1 Sequences Def: An ordered field is field F nd totl order < (for ll x, y, z F ): (i) x < y, y < x or x = y, (ii) x < y, y < z x < z (iii) x < y x + z < y + z (iv) 0 < y, x

### Summary of Elementary Calculus

Summry of Elementry Clculus Notes by Wlter Noll (1971) 1 The rel numbers The set of rel numbers is denoted by R. The set R is often visulized geometriclly s number-line nd its elements re often referred

### 20 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

### Chapter 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.

### The 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

### MATH 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 rel-vlued function on I), nd let L 1 (I) denote the completion

### E1: CALCULUS - lecture notes

E1: CALCULUS - lecture notes Ştefn Blint Ev Kslik, Simon Epure, Simin Mriş, Aureli Tomoiogă Contents I Introduction 9 1 The notions set, element of set, membership of n element in set re bsic notions of

### 440-2 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam

440-2 Geometry/Topology: Differentible Mnifolds Northwestern University Solutions of Prctice Problems for Finl Exm 1) Using the cnonicl covering of RP n by {U α } 0 α n, where U α = {[x 0 : : x n ] RP

### MATH 409 Advanced Calculus I Lecture 18: Darboux sums. The Riemann integral.

MATH 409 Advnced Clculus I Lecture 18: Drboux sums. The Riemnn integrl. Prtitions of n intervl Definition. A prtition of closed bounded intervl [, b] is finite subset P [,b] tht includes the endpoints

### Math 324 Course Notes: Brief description

Brief description These re notes for Mth 324, n introductory course in Mesure nd Integrtion. Students re dvised to go through ll sections in detil nd ttempt ll problems. These notes will be modified nd

### More Properties of the Riemann Integral

More Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologil Sienes nd Deprtment of Mthemtil Sienes Clemson University Februry 15, 2018 Outline More Riemnn Integrl Properties The Fundmentl

### 8. Complex Numbers. We can combine the real numbers with this new imaginary number to form the complex numbers.

8. Complex Numers The rel numer system is dequte for solving mny mthemticl prolems. But it is necessry to extend the rel numer system to solve numer of importnt prolems. Complex numers do not chnge the

### different methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).

Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different

### A Convergence Theorem for the Improper Riemann Integral of Banach Space-valued Functions

Interntionl Journl of Mthemticl Anlysis Vol. 8, 2014, no. 50, 2451-2460 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/10.12988/ijm.2014.49294 A Convergence Theorem for the Improper Riemnn Integrl of Bnch

### Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University

U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions

### MATH 409 Advanced Calculus I Lecture 22: Improper Riemann integrals.

MATH 409 Advned Clulus I Leture 22: Improper Riemnn integrls. Improper Riemnn integrl If funtion f : [,b] R is integrble on [,b], then the funtion F(x) = x f(t)dt is well defined nd ontinuous on [,b].

### A. Limits - L Hopital s Rule ( ) How to find it: Try and find limits by traditional methods (plugging in). If you get 0 0 or!!, apply C.! 1 6 C.

A. Limits - L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( x) lim where lim f x x! c g x ( ) = or lim f ( x) = limg( x) = ". ( ) x! c limg( x) = 0 x! c x! c

### KOÇ UNIVERSITY MATH 106 FINAL EXAM JANUARY 6, 2013

KOÇ UNIVERSITY MATH 6 FINAL EXAM JANUARY 6, 23 Durtion of Exm: 2 minutes INSTRUCTIONS: No clcultors nd no cell phones my be used on the test. No questions, nd tlking llowed. You must lwys explin your nswers

### 2 Definitions and Basic Properties of Extended Riemann Stieltjes Integrals

2 Definitions nd Bsic Properties of Extended Riemnn Stieltjes Integrls 2.1 Regulted nd Intervl Functions Regulted functions Let X be Bnch spce, nd let J be nonempty intervl in R, which my be bounded or

### Numerical Integration

Chpter 1 Numericl Integrtion Numericl differentition methods compute pproximtions to the derivtive of function from known vlues of the function. Numericl integrtion uses the sme informtion to compute numericl

### Line and Surface Integrals: An Intuitive Understanding

Line nd Surfce Integrls: An Intuitive Understnding Joseph Breen Introduction Multivrible clculus is ll bout bstrcting the ides of differentition nd integrtion from the fmilir single vrible cse to tht of

### Power Series, Taylor Series

CHAPTER 5 Power Series, Tylor Series In Chpter 4, we evluted complex integrls directly by using Cuchy s integrl formul, which ws derived from the fmous Cuchy integrl theorem. We now shift from the pproch

### Section 6.1 Definite Integral

Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined

### Intuitionistic Fuzzy Lattices and Intuitionistic Fuzzy Boolean Algebras

Intuitionistic Fuzzy Lttices nd Intuitionistic Fuzzy oolen Algebrs.K. Tripthy #1, M.K. Stpthy *2 nd P.K.Choudhury ##3 # School of Computing Science nd Engineering VIT University Vellore-632014, TN, Indi

### 2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).

AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following

### Properties of the Riemann Stieltjes Integral

Properties of the Riemnn Stieltjes Integrl Theorem (Linerity Properties) Let < c < d < b nd A,B IR nd f,g,α,β : [,b] IR. () If f,g R(α) on [,b], then Af +Bg R(α) on [,b] nd [ ] b Af +Bg dα A +B (b) If

### Orthogonal Polynomials and Least-Squares Approximations to Functions

Chpter Orthogonl Polynomils nd Lest-Squres Approximtions to Functions **4/5/3 ET. Discrete Lest-Squres Approximtions Given set of dt points (x,y ), (x,y ),..., (x m,y m ), norml nd useful prctice in mny

### Math 113 Exam 2 Practice

Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.-7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number

### 7.2 The Definition of the Riemann Integral. Outline

7.2 The Definition of the Riemnn Integrl Tom Lewis Fll Semester 2014 Upper nd lower sums Some importnt theorems Upper nd lower integrls The integrl Two importnt theorems on integrbility Outline Upper nd

### BIFURCATIONS IN ONE-DIMENSIONAL DISCRETE SYSTEMS

BIFRCATIONS IN ONE-DIMENSIONAL DISCRETE SYSTEMS FRANCESCA AICARDI In this lesson we will study the simplest dynmicl systems. We will see, however, tht even in this cse the scenrio of different possible

### Henstock Kurzweil delta and nabla integrals

Henstock Kurzweil delt nd nbl integrls Alln Peterson nd Bevn Thompson Deprtment of Mthemtics nd Sttistics, University of Nebrsk-Lincoln Lincoln, NE 68588-0323 peterso@mth.unl.edu Mthemtics, SPS, The University

### dx dt dy = G(t, x, y), dt where the functions are defined on I Ω, and are locally Lipschitz w.r.t. variable (x, y) Ω.

Chpter 8 Stility theory We discuss properties of solutions of first order two dimensionl system, nd stility theory for specil clss of liner systems. We denote the independent vrile y t in plce of x, nd

### UniversitaireWiskundeCompetitie. Problem 2005/4-A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that

Problemen/UWC NAW 5/7 nr juni 006 47 Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de

### a n+2 a n+1 M n a 2 a 1. (2)

Rel Anlysis Fll 004 Tke Home Finl Key 1. Suppose tht f is uniformly continuous on set S R nd {x n } is Cuchy sequence in S. Prove tht {f(x n )} is Cuchy sequence. (f is not ssumed to be continuous outside

### 5: The Definite Integral

5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce

### STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.

STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t

### CS667 Lecture 6: Monte Carlo Integration 02/10/05

CS667 Lecture 6: Monte Crlo Integrtion 02/10/05 Venkt Krishnrj Lecturer: Steve Mrschner 1 Ide The min ide of Monte Crlo Integrtion is tht we cn estimte the vlue of n integrl by looking t lrge number of

### Numerical integration

2 Numericl integrtion This is pge i Printer: Opque this 2. Introduction Numericl integrtion is problem tht is prt of mny problems in the economics nd econometrics literture. The orgniztion of this chpter

### approaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below

. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.

### Anonymous Math 361: Homework 5. x i = 1 (1 u i )

Anonymous Mth 36: Homewor 5 Rudin. Let I be the set of ll u (u,..., u ) R with u i for ll i; let Q be the set of ll x (x,..., x ) R with x i, x i. (I is the unit cube; Q is the stndrd simplex in R ). Define

### The Dirac distribution

A DIRAC DISTRIBUTION A The Dirc distribution A Definition of the Dirc distribution The Dirc distribution δx cn be introduced by three equivlent wys Dirc [] defined it by reltions δx dx, δx if x The distribution

### NOTES AND PROBLEMS: INTEGRATION THEORY

NOTES AND PROBLEMS: INTEGRATION THEORY SAMEER CHAVAN Abstrct. These re the lecture notes prepred for prticipnts of AFS-I to be conducted t Kumun University, Almor from 1st to 27th December, 2014. Contents

### Keywords : Generalized Ostrowski s inequality, generalized midpoint inequality, Taylor s formula.

Generliztions of the Ostrowski s inequlity K. S. Anstsiou Aristides I. Kechriniotis B. A. Kotsos Technologicl Eductionl Institute T.E.I.) of Lmi 3rd Km. O.N.R. Lmi-Athens Lmi 3500 Greece Abstrct Using

### Discrete Least-squares Approximations

Discrete Lest-squres 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

### Sum: lim( f + g) = lim f + limg. Difference: lim( f g) = lim f limg. Product: lim( f g) = (lim f )(limg)

MATH 5: Clculus, SET8 SUMMARIES [Belmonte, 207] 2 Limits nd Derivtives 2.2 The Limit of Function Limit of sclr function: We write lim x f (x) = L nd sy the limit of f (x) s x pproches equls L if nd only

### ECO 317 Economics of Uncertainty Fall Term 2007 Notes for lectures 4. Stochastic Dominance

Generl structure ECO 37 Economics of Uncertinty Fll Term 007 Notes for lectures 4. Stochstic Dominnce Here we suppose tht the consequences re welth mounts denoted by W, which cn tke on ny vlue between

### 5.5 The Substitution Rule

5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n nti-derivtive is not esily recognizble, then we re in