Mathematical Analysis: Supplementary notes I


 Norman Mosley
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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 F1F7 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 Mtest) 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 ntuples 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 zxis 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 onedimensionl 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
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