Anlysis I Prof. T. W. Körner Lent 003 Contents Why do we bother? The xiom of Archimedes 3 3 Series nd sums 6 4 Lest upper bounds 0 5 Continuity 4 6 Differentition 8 7 The men vlue theorem 8 Complex vrible 7 9 Power series 9 0 The stndrd functions 3 Onwrds to the complex plne 40 The Riemnn integrl 45 3 Some properties of the integrl 5 4 Infinite integrls 59 5 Further reding 63
Lecturer s premble This course centres round ide, the re-founding of clculus on rigorous bsis. This is essentilly bsed on the definition of limit: ɛ > 0 n 0 (ɛ) such tht n < ɛ n n 0 (ɛ) From this we get the FUNDAMENTAL THEOREM OF ANALYSIS Every strictly incresing sequence bounded bove tends to limit This does not work in Q, e.g. look t the deciml expnsion of, s you dd more nd more digits it is clerly incresing, it is bounded bove by, but its limit ( ) does not exist in Q. However, it lwys works on R Using this xiom nd the lws of lgebr (e.g. + b = b + ), we cn now refound the clculus. NOTE: We cn now split proofs nd results into two groups, those which re mere lgebr (i.e. proof lso works on Q) nd those using nlysis. Why do we bother? It is surprising how mny people think tht nlysis consists in the difficult proofs of obvious theorems. All we need know, they sy, is wht limit is, the definition of continuity nd the definition of the derivtive. All the rest is intuitively cler. If pressed they will gree tht these definitions pply s much to the rtionls Q s to the rel numbers R. They then hve to explin the following interesting exmple. Exmple.. If f : Q Q is given by f(x) = if x <, f(x) = otherwise, then (i) f is continuous function with f(0) =, f() = yet there does not exist c with f(c) = 0, (ii) f is differentible function with f (x) = 0 for ll x yet f is not constnt. Wht is the difference between R nd Q which mkes clculus work on one even though it fils on the other? Both re ordered fields, tht is, both support opertions of ddition nd multipliction together with reltion greter thn ( order ) with the properties tht we expect. If the reder is interested she will find complete list of the pproprite xioms in texts like the ltogether excellent book of Spivk [5] nd its mny rther less excellent competitors, but, interesting s such things my be, they re irrelevnt to our purpose which is not to consider the shred properties of R nd Q but to identify difference
between the two systems which will enble us to exclude the possibility of function like tht of Exmple. for functions from R to R. To stte the difference we need only recll definition from course C3. Definition.. If n R for ech n nd R then we sy tht n if given ɛ > 0 we cn find n n 0 (ɛ) such tht n < ɛ for ll n n 0 (ɛ) The key property of the rels, the fundmentl xiom which mkes everything work ws lso stted in the course C3. Axiom.3 (The fundmentl xiom of nlysis). If n R for ech n, A R nd 3... nd n < A for ech n then there exists n R such tht n s n. Less ponderously, nd just s rigorously, the fundmentl xiom for the rel numbers sys every incresing sequence bounded bove tends to limit. Everything which depends on the fundmentl xiom is nlysis, everything else is mere lgebr. The xiom of Archimedes We strt by proving the following results on limits, some of which you sw proved in course C3. Lemm.. (i) The limit is unique. Tht is, if n nd n b s n then = b. (ii) If n s n nd n() < n() < n(3)... then n(j) s j. (iii) If n = c for ll n then n c s n. (iv) If n nd b n b s n then n + b n + b. (v) If n nd b n b s n then n b n b. (vi) If n s n, n 0 for ech n nd 0, then n. (vii) If n A for ech n nd n s n then A. Proof. i) If b then b > 0 so set ɛ = b 4 There exists n n 0 such tht n < ɛ n > n 0 There exists n n such tht n b < ɛ n > n Set n = mx{n 0, n } We hve n < ɛ nd n b < ɛ so b < ɛ = b < b # ii) Suppose n Then given ɛ > 0 n 0 such tht n < ɛ n n 0 If j n 0 then n(j) n 0, genertes subsequence, so n(j) < ɛ, thus n(j) 3
iii) If n = c then given ɛ > 0 set n 0 (ɛ) =. Then n c = c c = 0 < ɛ n n 0 (ɛ) s required iv) We know tht Given ɛ > 0 n 0 (ɛ) such tht n < ɛ n n 0 (ɛ) Given ɛ > 0 n (ɛ) such tht b n b < ɛ n n (ɛ) Thus if n n (ɛ) = mx{n 0 ( ɛ ), n ( ɛ )} for n n (ɛ) ( n + b n ) ( b) = ( n ) + (b n b) n + b n b < ɛ + ɛ v) As before, we hve tht There exists n n 0 such tht n < ɛ n > n 0 There exists n n such tht n b < ɛ n > n Thus if n n ( ɛ 3 ) = mx{n (), n 0 ( ( b +) ), n ( ɛ ɛ ( +) )} Then n b n b n b n b n + b n b = ( n )b n + (b n b) ( n )b n + (b n b) = n b n + b n b n ( b + ) + b n b < ɛ ( b + ) + ɛ ( + ) ɛ + ɛ = ɛ Note: If n n (), b n b < so b n < b + ɛ We use insted of in cse = 0 ɛ ( +) vi) Agin, given ɛ > 0 n 0 (ɛ) such tht n < ɛ Thus if n n (ɛ) = mx{n 0 ( ), n 0( ɛ )} 4
= n n = n n n < n < ɛ.( ) = ɛ Note: If n n (ɛ) then n < so n > vii) If > A then set ɛ = A. Then n A > ɛ n, so n # [Remrk: n < A, n < A, observe n < 0 but n 0] We need the following vrition on the fundmentl xiom. Exercise.. A decresing sequence of rel numbers bounded below tends to limit. [Hint. If b then b.] Proof. If b b b 3... nd b n B then b b b 3... nd b n B so by the fundmentl xiom b n β, sy, so b n = ( ).( b n ) ( ).β = β Useful s the results of Lemm. re, they re lso true of sequences in Q. They re therefore mere, if importnt, lgebr. Our first truly nlysis result my strike the reder s rther odd. Theorem.3 (Axiom of Archimedes). n 0 s n Proof. The sequence,, 3,... is decresing, nd bounded below by 0, so n l for some l. Observe tht n is subsequence of n so n l. But n = n l = l l = l l = 0 Theorem.3 shows tht there is no exotic rel number ג sy (to choose n exotic symbol) with the property tht /n > ג for ll integers n yet ג > 0 (tht is, ג is strictly positive nd yet smller thn ll strictly positive rtionls). There exist number systems with such exotic numbers (the fmous non-stndrd nlysis of Abrhm Robinson nd the surrel numbers of Conwy constitute two such systems) but, just s the rtionls re, in some sense, too smll system for the stndrd theorems of nlysis to hold so these non-archimeden systems re, in some sense, too big. Archimedes nd Eudoxus relised the need for n xiom to show tht there is no exotic number ℸ bigger thn ny integer Footnote for pssing historins, this is course in mthemtics. 5
(i.e. ℸ > n for ll integers n ; to see the connection with our form of the xiom consider ג = /ℸ). However, in spite of its nme, wht ws n xiom for Archimedes is theorem for us. Theorem.4. Given ny rel number K we cn find n integer n with n > K. Proof. Assume we cn find K tht is greter thn ll numbers. Clerly it is greter thn. Thus 0 < K n, nd thus n 0# By reductio d bsurdum, K such tht K n n Z 3 Series nd sums There is no need to restrict the notion of limit to rel numbers. Definition 3.. If n C for ech n nd C then we sy tht n if given ɛ > 0 we cn find n n 0 (ɛ) such tht n < ɛ for ll n n 0 (ɛ) Exercise 3.. We work in C. (i) The limit is unique. Tht is, if n nd n b s n then = b. (ii) If n s n nd n() < n() < n(3)... then n(j) s j. (iii) If n = c for ll n then n c s n. (iv) If n nd b n b s n then n + b n + b. (v) If n nd b n b s n then n b n b. (vi) If n s n, n 0 for ech n nd 0, then n. Exercise 3.3. Explin why there is no result in Exercise 3. corresponding to prt (vii) of Lemm.. We illustrte some of the ides introduced by studying infinite sums. Definition 3.4. We work in F where F = R or F = C. If j F we sy tht j= j converges to s if N j s j= s N. We write j= j = s. If N j= j does not tend to limit s N, we sy tht the sum diverges. j= j Note: Work s much s possible with N n nd s little s possible with n until you know tht n exists Exmple: Let u n = ( ) n+, S = u n i) S = u + n= u n = u + n= ( )u n = u n= u n = S so S = ii) S = r= (u r + u r ) = r= ( ) = r= 0 = 0 6
The bove logic is fulty{ becuse S does not exist N 0 if N even u n = + = does not converge if N odd Lemm 3.5. We work in F where F = R or F = C. (i) Suppose j, b j F nd λ, µ F. If j= j nd j= b j converge then so does j= λ j + µb j nd λ j + µb j = λ j + µ j= (ii) Suppose j, b j F nd there exists n N such tht j = b j for ll j N. Then, either j= j nd j= b j both converge or they both diverge. (In other words, initil terms do not mtter.) Proof. i) N j= (λ j + µb j ) = λ N j= j + µ N j= b j λ j= j + µ j= b j ii) Suppose j = b j for j M, then j = b j + c j where c j = 0 for j M, so N c j = M c j for N M M c j So c j exists, so if b j exists, then (b j+c j ) exists, j exists Exercise 3.6. Any problem on sums j= j cn be converted into one on sequences by considering the sequence s n = n j= j. Show conversely tht sequence s n converges if nd only if, when we set = s nd n = s n s n [n ] we hve j= n j convergent. Wht cn you sy bout lim n j= j nd lim n s n if both exist? j= j= The following results re fundmentl to the study of sums. Theorem 3.7 (The comprison test). We work in R. Suppose tht 0 b j j for ll j. Then, if j= j converges, so does j= b j. Proof. Observe tht N j= j is incresing, nd recll tht if c N k N nd c N c then c k, It follows tht N j= j j= j 0 N j= b j N j= j j= N j. j= b j is n incresing sequence, since b j 0 so since it is bounded bove the fundmentl theorem tells us tht N j= b j tends to limit s n i.e. j= b j exists Note tht ws cn use this n nother proof of summing geometric progression. Let x < b j N n=0 x n = x x N n=0 x n = x + x x + x... x N+ x = xn+ x x s N 7
Now look t N Since 0 n= x n n +3+sin x for 0 x < x n n +3+sin x xn, n= xn exists, n= x n n +3+sin x exists. Theorem 3.8. We work in F where F = R or F = C. If j= j converges, then so does j= j. This is sying tht, if we tke sequence, nd plot the points of the sequence nd join them up, j= j converges mens the length of the line produced is bounded, nd thus it converges to point. Proof. First { consider it over R Set + j = j if j 0 0 if j < 0 { Set 0 if j = j 0 j if j < 0 j = + j j nd 0 + j j, 0 j j + j, j converge becuse j does, using the comprison test. Thus j= j = j= (+ j j ) convergses to + j j If F = C with j = x j + iy j, x j, y j R x j j so by comprison x j converges, nd so by the first prt x j lso converges. Similrly y j converges So j= j = j= (x j = iy j ) converges to j= x j + i j= y j Theorem 3.8 is often stted using the following definition. Definition 3.9. We work in F where F = R or F = C. If j= j converges we sy tht the sum j= j is bsolutely convergent. Theorem 3.8 then becomes the sttement tht bsolute convergence implies convergence. Here is trivil but useful consequence of Theorems 3.7 nd 3.8. Lemm 3.0 (Rtio test). Suppose j C nd j+ / j l s j. If l <, then j= j converges. If l >, then j= j diverges. Of course Lemm 3.0 tells us nothing if l = or l does not exist. Proof. If j+ j l < then choose ɛ = l. j(0) such tht j+ j l < ɛ for j j(0) so j+ j < l + ɛ = +l = K < for j j(0), so j j(0) K j j(0) for j j(0) We know tht j(0)k j j(0) converges (s it is geometric progression with rtio 0 K < ). We lso know tht the first j(0) terms re irrelevnt to convergence, nd we know tht so by comprison j converges j j(0) K j j(0) j j(0) 8
To prove the second prt (if n+ n l, l > then n ) set ɛ = l. N such tht n+ n l < ɛ = l for n N, so n+ n > l+ for n N n > K( l+ )n for n N, so n unbounded, so n cnnot converge Sums which re not bsolutely convergent re much hrder to del with in generl. It is worth keeping in mind the following trivil observtion. Lemm 3.. We work in F where F = R or F = C. If j= j converges, then j 0 s j. Proof. Set S n = n j= j. n+ = s n+ s n j j = 0 At deeper level the following result is sometimes useful. Lemm 3. (Alternting series test). We work in R. If we hve decresing sequence of positive numbers n with n 0 s n, then j= ( )j+ j converges. Further n ( ) j+ j ( ) j+ j N+ for ll N. Proof. Set j= j= S = S = S 3 = + 3. Obeserve tht S N+ = S N + ( N + N+ ) S N so the sequence S N+ is decresing. But lso S N+ = + 3 4 + = ( ) + ( 3 4 ) +... + ( N N ) + N+ So S N+ is decresing sequence bounded below, so S N+ l. Similrly S N is n incresing sequence bounded bove so S N l. Now N+ = S N+ S N l l nd N+ 0 so l l = 0 so l = l nd S n l s n To finish observe tht S N+ is decresing, S N is incresing, so S N+ l S N so l S N S N S N+ = N+. Similrly l S N+ N+ 9
The lst lemm is sometimes expressed by sying the error cused by replcing convergent infinite lternting sum of decresing terms by the sum of its first few terms is no greter thn the bsolute vlue of the first term neglected. Lter we will give nother test for convergence clled the integrl test (Lemm 4.4) from which we deduce the result known to mny of you tht n= n diverges. I will give nother proof in the next exmple. Exmple 3.3. (i) n diverges. n= ( ) n (ii) is convergent but not bsolutely convergent. n n= (iii) If v n = /n, v n = /(n) then n= v n is not convergent. Proof. i) This shows tht n 0 n converges + + 3 + 4 + = + + ( 3 + 4 ) + ( 5 + 6 + 7 + 8 ) + + ( N + + N + + + N ) + + + ( 4 + 4 ) + ( 8 + 8 + 8 + 8 ) + + ( n + n + n + + n ) + + + + + So N r is not bounded nd diverges Note: + + 3 + = + ( + 3 ) + ( 4 + 5 + 6 + 7 ) + + ( + ) + ( 4 + 4 + 4 + 4 ) + + N +. Note tht 0 0 6, so 0 6 N so it is very slow to diverge ii) n is decresing, n 0 nd n 0 so we cn pply the lternting series test. Thus ( ) n n is convergent, nd i) tells us tht is it not bsolutely convergent. iii) This is similr to i) v + v +... + v n = (v + v ) + (v 3 + v 4 ) + So v j does not converge 4 Lest upper bounds = ( ) + ( 4 ) + + ( n n ) = ( + + 3 + ) A non-empty bounded set in R need not hve mximum. 0
Exmple 4.. The set E = { /n : n } is non-empty nd ny e E stisfies the inequlities e 0 but E hs no lrgest member. Proof. If x 0 then n such tht x > n > 0 so n mximum If x 0 then x / E Thus E contins no lrgest member > x nd x is not However, s we shll see every non-empty bounded set in R hs lest upper bound (or supremum). Definition 4.. Let E be non-empty set in R. We sy tht α is lest upper bound for E if (i) α e for ll e E [tht is, α is n upper bound for E] nd (ii) If β e for ll e E then β α [tht is, α is the lest such upper bound] If E hs supremum α we write sup e E e = sup E = α. Lemm 4.3. If the lest upper bound exists it is unique. Proof. Suppose E is non-empty subset of R nd α, α re lest upper bounds for E. Then since α is n upper bound, nd α is lest upper bound α α. Similrly α α nd so α = α The following remrk is trivil but sometimes helpful. Lemm 4.4. Let E be non-empty set in R. Then α is lest upper bound for E if nd only if we cn find e n E with e n α nd b n such tht b n e for ll e E nd b n α s n. Proof. First show tht α is n upper bound. If e E then b n e. But b n α so α e. Then show α is the lest upper bound i.e. if β is n upper bound β α If β is n upper bound then, in prticulr, β e n, but e n α so β α Here is the promised result. Theorem 4.5. Any non-empty set in R with n upper bound hs lest upper bound. Proof. Let E be such set. Since E is non-empty, we cn find 0 E. Observe tht e E such tht e 0 (e.g. 0 itself). Since E is bounded, we cn find b 0 such tht b 0 e e E. Set c 0 = 0+b0. Either e E such tht e c 0 nd we put = c 0, b = b 0 Or e E c 0 > e nd we put = 0, b = c 0 Observe tht
0 b b 0 b = b0 0 nd ) e E such tht e ) b e e E Repeting we obtin 0... n b n... b b 0 b n n = b0 0 n nd ) e n E such tht e n ) b n e e E The n s form n incresing sequence bounded bove, so n α by the fundmentl theorem b n = n + (b n + n ) = n + n (b 0 0 ) α + 0 = α so n e n b n with ll the e n E, so e n α nd b n e e E so α is the lest upper bound of E We observe tht this result is ctully equivlent to the fundmentl xiom. Theorem 4.6. Theorem 4.5 implies the fundmentl xiom. Proof. On the ssumption tht every bounded non-empty set hs lest upper bound, suppose... n A 0 n Set A = {,,...} is non-empty nd bounded, so it hs supremum which we cll α. Given ɛ > 0 we know α ɛ is not n upper bound, so N such tht N > α ɛ. Now if n N, α n N α ɛ, so n α < ɛ nd n α Of course we hve the notion of gretest lower bound or infimum. Exercise 4.7. Define the gretest lower bound in the mnner of Definition 4., prove its uniqueness in the mnner of Lemm 4.3 nd stte nd prove result corresponding to Lemm 4.4. If E hs n infimum β we write inf e E e = inf E = β. One wy of deling with the infimum is to use the following observtion. Lemm 4.8. Let E be non-empty set in R nd write E = { e : e E}. Then E hs n infimum if nd only if E hs supremum. If E hs n infimum inf E = sup( E). Proof. If E hs supremum α i) α e e E ii) If β e e E then β α So i) e e E ii) If β e e E then β α So α is the infimum of E
Exercise 4.9. Use Lemm 4.8 nd Theorem 4.5 to show tht ny non-empty set in R with lower bound hs gretest lower bound. The notion of supremum will ply n importnt rôle in our proofs of Theorem 5. nd Theorem 9.. The following result is lso equivlent to the fundmentl xiom (tht is, we cn deduce it from the fundmentl xiom nd conversely, if we tke it s n xiom, rther thn theorem, then we cn deduce the fundmentl xiom s theorem). Theorem 4.0 (Bolzno-Weierstrss). If x n R nd there exists K such tht x n K for ll n, then we cn find n() < n() <... nd x R such tht x n(j) x s j. Proof. Set 0 = K, b 0 = K. {m : x m [ 0, b 0 ]} is infinite. Set c 0 = 0+b0 If {m : x m [ 0, c 0 ]} is infinite we set = 0, b = c 0, otherwise set = c 0, b = b 0 Observe tht 0 b b 0 b = (b 0 0 ) {m : x m [, b ]} is infinite Repet. n is incresing nd bounded bove so tends to limit, cll it α b n = n + (b n n ) = n + (b 0 0 ) n α Since [ n, b n ] contins n x m for infinitely mny m I cn choose m() < m() < m(3) <... such tht x m(j) [ j, b j ]. Thus j x m(j) b j nd s j, b j α, x m(j) α The Bolzno-Weierstrss theorem sys tht every bounded sequence of rels hs convergent subsequence. Notice tht we sy nothing bout uniqueness; if x n = ( ) n then x n but x n+ s n. We proved the theorem of Bolzno-Weierstrss by lion hunting but your supervisor my well show you nother method. We shll use the Bolzno-Weierstrss theorem to prove tht every continuous function on closed bounded intervl is bounded nd ttins its bounds (Theorem 5.). The Bolzno-Weierstrss theorem will be much used in the next nlysis course becuse it generlises to mny dimensions. Note lso tht the Bolzno-Weierstrss theorem implies the fundmentl xiom Proof. Suppose... nd n A. Then { j } is bounded sequence so by Bolzno-Weierstrss it hs convergent subset i.e. n() < n() <... such tht n(k) α Since n(k) j j n(k) we hve α n (k) k so α j j Now given ɛ > 0 n(k) such tht α n(k) < ɛ so α j n(k) j n(k), so α j < ɛ j n(k) nd thus j α 3
5 Continuity Figure : Equivlence of fundmentl xioms We mke the following definition. Definition 5.. A function f : R R is continuous t x if given ɛ > 0 we cn find δ(ɛ, x) > 0 [red delt depending on epsilon nd x ] such tht f(x) f(y) < ɛ for ll y with x y < δ(ɛ, x). If f is continuous t ech point x R we sy tht f is continuous function on R. I shll do my best to mke this seem resonble definition but it is importnt to relise tht I m relly stting rule of the gme (like knights move in chess or the definition of offside in footbll). If you wish to ply the gme you must ccept the rules. Results bout continuous functions must be derived from the definition nd not stted s obvious from the notion of continuity. In prctice we use slightly more generl definition. Definition 5.. Let E be subset of R. A function f : E R is continuous t x E if given ɛ > 0 we cn find δ(ɛ, x) > 0 [red delt depending on epsilon nd x ] such tht f(x) f(y) < ɛ for ll y E with x y < δ(ɛ, x). If f is continuous t ech point x E, we sy tht f is continuous function on E. However, it will do no hrm nd my be positively helpful if, whilst you re getting used to the ide of continuity, you concentrte on the cse E = R. 4
Lemm 5.3. Suppose tht E is subset of R, tht x E, nd tht f nd g re functions from E to R. (i) If f(x) = c for ll x E, then f is continuous on E. (ii) If f nd g re continuous t x, then so is f + g. (iii) Let us define f g : E R by f g(t) = f(t)g(t) for ll t E. Then if f nd g re continuous t x, so is f g. (iv) Suppose tht f(t) 0 for ll t E. If f is continuous t x so is /f. Proof. i) Trivil ii) As f, g re continous t x Given ɛ > 0 δ (ɛ) > 0 such tht f(x) f(y) < ɛ for x y < δ (ɛ) Given ɛ > 0 δ (ɛ) > 0 such tht g(x) g(y) < ɛ for x y < δ (ɛ) Now set δ 3 (ɛ) = min{δ ( ɛ ), δ ( ɛ )} Then if x y < δ 3 (ɛ) (f + g)(x) (f + g)(y) = f(x) + g(x) f(y) + g(y) = (f(x) f(y)) + (g(x) g(y)) f(x) f(y) + g(x) g(y) ɛ + ɛ = ɛ iii) Is done similrly iv) f is continuous t x Thus given ɛ > 0 δ(ɛ) > 0 such tht f(x) f(y) < ɛ for x y < δ(ɛ) Set δ (ɛ) = min{δ( f(x) ), δ( ɛ f(x) )} Then if x y < δ (ɛ) f(x) f(y) f(x) f(y) f(x) f(y) f(x) f(y) f(x) < ɛ Using f(y) f(x) < f(x) so f(y) > f(x) Lemm 5.4. Let U nd V be subsets of R. Suppose f : U R is such tht f(t) V for ll t U. If f is continuous t x U nd g : V R is continuous t f(x), then the composition g f is continuous t x. Proof. Given ɛ > 0 δ (ɛ) > 0 such tht f(x) f(y) < ɛ for x y < δ (ɛ) Given ɛ > 0 δ (ɛ) > 0 such tht g(f(x)) g(u) < ɛ for u f(x) < δ (ɛ) Set δ 3 (ɛ) = δ (δ (ɛ)) Then if x y < δ 3 (ɛ) we hve f(x) f(y) < δ (ɛ) nd so g(f(x)) g(f(y)) < ɛ By repeted use of prts (ii) nd (iii) of Lemm 5.3 it is esy to show tht polynomils P (t) = n r=0 rt r re continuous. The detils re spelled out in the next exercise. 5
Exercise 5.5. Prove the following results. (i) Suppose tht E is subset of R nd tht f : E R is continuous t x E. If x E E then the restriction f E of f to E is lso continuous t x. (ii) If J : R R is defined by J(x) = x for ll x R, then J is continuous on R. (iii) Every polynomil P is continuous on R. (iv) Suppose tht P nd Q re polynomils nd tht Q is never zero on some subset E of R. Then the rtionl function P/Q is continuous on E (or, more precisely, the restriction of P/Q to E is continuous.) The following result is little more thn n observtion but will be very useful. Lemm 5.6. Suppose tht E is subset of R, tht x E, nd tht f is continuous t x. If x n E for ll n nd x n x s n, then f(x n ) f(x) s n. Proof. Given ɛ > 0 δ(ɛ) > 0 such tht f(x) f(y) < ɛ for x y < δ(ɛ) Given ɛ > 0 N(ɛ) such tht x n x < ɛ for n N(ɛ) If we set M(ɛ) = N(δ(ɛ)) then n M(ɛ) x n x < δ(ɛ) f(x n ) f(x) < ɛ So fr in this section we hve only done lgebr but the next result depends on the fundmentl xiom. It is one of the key results of nlysis nd lthough my recommendtion runs contrry to century of enlightened pedgogy I cn see no objections to students lerning the proof s model. Notice tht the theorem resolves the problem posed by Exmple. (i). Theorem 5.7 (The intermedite vlue theorem). If f : [, b] R is continuous nd f() 0 f(b) then there exists c [, b] such tht f(c) = 0 Proof. This relies on the fct tht we re working in R Set 0 =, b 0 = b, c 0 = 0+b0 If f(c 0 ) 0 set = c 0, b = b 0, otherwise set = 0, b = c 0. Observe tht 0 b b 0 b = (b 0 0 ) f( ) 0 f(b ) Repet, obtining 0... n b n... b b 0 b n n = n (b 0 0 ) f( n ) 0 f(b n ) n is n incresing sequence bounded bove, so n α with α b n n s n b n b n = n + (b n n ) = n + n (b 0 0 ) α + 0 = α, nd s b n n n, α Now n α so f( n ) f(α) nd similrly f(b n ) f(α). But f( n ) 0 so f(α) 0 nd f(b n ) 0 so f(α) 0. Thus f(α) = 0 6
The next three exercises re pplictions of the intermedite vlue theorem. Exercise 5.8. Show tht ny rel polynomil of odd degree hs t lest one root. Is the result true for polynomils of even degree? Give proof or counterexmple. Proof. Let p(x) = x 3 +x +bx+c = x 3 ( x + b x + c x ) x 3 ( 3 x b x c x ) > x5 3 4 If we set x 4( + + b + c ), p(x) 0 nd for x sufficiently lrge nd negtive p(x) is negtive, so by continuity hs root. Exercise 5.9. Suppose tht g : [0, ] [0, ] is continuous function. By considering f(x) = g(x) x, or otherwise, show tht there exists c [0, ] with g(c) = c. (Thus every continuous mp of [0, ] into itself hs fixed point.) Give n exmple of bijective (but, necessrily, non-continuous) function h : [0, ] [0, ] such tht h(x) neqx for ll x [0, ]. [Hint: First find function H : [0, ] \ {0,, /} [0, ] \ {0,, /} such tht H(x) x.] Proof. Let f(x) = g(x) x f(0) 0 (if f(0) = 0, 0 = c nd done) f() 0 (if f() = 0, 0 = c nd done) Therefore using the Intermedite Vlue Theorem c such tht f(c) = 0 nd thus g(c) = c Exercise 5.0. Every mid-summer dy t six o clock in the morning, the youngest monk from the monstery of Dmt strts to climb the nrrow pth up Mount Dipmes. At six in the evening he reches the smll temple t the pek where he spends the night in medittion. At six o clock in the morning on the following dy he strts downwrds, rriving bck t the monstery t six in the evening. Of course, he does not lwys wlk t the sme speed. Show tht, none the less, there will be some time of dy when he will be t the sme plce on the pth on both his upwrd nd downwrd journeys. We proved the intermedite vlue theorem (Theorem 5.7) by lion hunting. We prove the next two theorems by using the Bolzno-Weierstrss Theorem (Theorem 4.0). Agin the results re very importnt nd I cn see no objection to lerning the proofs s model. Theorem 5.. If f : [, b] R is continuous then we cn find n M such tht f(x) M for ll x [, b]. Proof. Note tht this needs to be on closed intervl, on (0, ) f : x x is unbounded Suppose f is unbounded Then we cn find x n [, b] such tht f(x n ) n The x n lie in bounded intervl so they hve convergent subsequence x n(j) α sy. Since x n(j) b, α b nd since x n(j), α so α [, b] Now f is continuous t α so δ > 0 such tht f(α) f(x) < x α < δ x [, b] so f(x) f(α) + x α < δ, x [, b] 7
But x n(j) α. So J such tht if j J x n(j) α < δ so n(j) f(x n(j) f(α) + j J # Thus f must be bounded In other words continuous function on closed bounded intervl is bounded. We improve this result in the next theorem. Theorem 5.. If f : [, b] R is continuous then we cn find x, x [, b] such tht f(x ) f(x) f(x ) for ll x [, b]. Proof. Strt by observing tht {f(x) : x [, b]} is non-empty bounded set, so it hs supremum M. Wht we need to show is tht α [, b] such tht f(α) = M Since M is the supremum of {f(x) : x [, b]} we cn certinly find x n [, b] such tht f(x n ) M n (if not, then (M n ) would be n upper bound, but M is the lest upper bound). Automticlly M f(x n ) By Bolzno-Weierstrss there is convergenet subsequence x n(j) α. As before α [, b] nd f(x n(j) ) f(α). But M f(x n(j) ) M n(j), so f(x n(j)) M so f(α) = M In other words continuous function on closed bounded intervl is bounded nd ttins its bounds. 6 Differentition In this section it will be useful to hve nother type of limit. Definition 6.. Let E be subset of R, f be some function from E to R, nd x some point of E. If l R we sy tht f(y) l s y x [or, if we wish to emphsise the restriction to E tht f(y) l s y x through vlues y E] if, given ɛ > 0, we cn find δ(ɛ) > 0 [red delt depending on epsilon ] such tht f(y) l < ɛ for ll y E with 0 < x y < δ(ɛ). As before there is no rel loss if the reder initilly tkes E = R. The following two exercises re esy but useful. Exercise 6.. Let E be subset of R. Show tht function f : E R is continuous t x E if nd only if f(y) f(x) s y x. Exercise 6.3. Let E be subset of R, f, g be some functions from E to R, nd x some point of E. (i) The limit is unique. Tht is, if f(y) l nd f(y) k s y x then l = k. 8
(ii) If x E E nd f(y) l s y x through vlues y E, then f(y) l s y x through vlues y E. (iii) If f(t) = c for ll t E then f(y) c s y x. (iv) If f(y) l nd g(y) k s y x then f(y) + g(y) l + k. (v) If f(y) l nd g(y) k s y x then f(y)g(y) lk. (vi) If f(y) l s y x, f(t) 0 for ech t E nd l 0 then f(t) l. (vii) If f(t) L for ech t E nd f(y) l s y x then l L. Proof. iii) f(t) = c t. Given ɛ > 0 set δ =. f(y) c = c c = 0 = 0 < ɛ iv) Since f(y) l, g(y) k s y x Given ɛ > 0 δ (ɛ) > 0 such tht f(y) l < ɛ for y x < δ (ɛ) Given ɛ > 0 δ (ɛ) > 0 such tht g(y) k < ɛ for y x < δ (ɛ) ɛ Thus given ɛ > 0 if we set δ 3 (ɛ) = min{(δ (), δ ( (+ k ) ), δ ɛ ( (+ l ) )} then f(y)g(y) lk = f(y)g(y) f(y)k + f(y)k lk f(y)g(y) f(y)k + f(y)k lk = f(y) g(y) k + f(y) l k ( l + ) g(y) k + f(y) l k We cn now define the derivtive. Definition 6.4. Let E be subset of R. A function f : E R is differentible t x E with derivtive f (x) if f(y) f(x) f (x) y x s y x. If f is differentible t ech point x E, we sy tht f is differentible function on E. As usul, no hrm will be done if you replce E by R. You cn think of differentition s f(y) = f(x) + f (x)(y x) +error term }{{} liner to be useful the error hs to 0 fster thn the liner term. Note: We cn lso use this in 3-D with grd δf = f.δx δf = δf δx + δf δx + δf δx 3 δx δx δx 3 f(y) = f(x) + f 0.(y x) + error term nd s before error 0 fster thn the liner term Here re some esy consequences of the definition. 9
Exercise 6.5. Let E be subset of R, f some function from E to R, nd x some point of E. Show tht if f is differentible t x then f is continuous t x. Proof. Given ɛ > 0 δ(ɛ) > 0 such tht f(x) f(y) y x f (x) < ɛ for y x < δ(ɛ), so f(y) f(x) f (x)(y x) < ɛ y x thus f(y) f(x) < ɛ y x + f (x)(y x) = ɛ y x + f (x) y x All of the terms in the lst expression re smll except f (x) which is fixed. So if we set δ (ɛ) = min{δ(), ɛ, ɛ ( f (x)+ ) } for ɛ <. Then if 0 < x y < δ (ɛ) f(x) f(y) <. y x + f (x) y x < ɛ + ɛ = ɛ Exercise 6.6. Let E be subset of R, f, g be some functions from E to R, nd x some point of E. Prove the following results. (i) If f(t) = c for ll t E then f is differentible t x with f (x) = 0. (ii) If f nd g re differentible t x then so is their sum f + g nd (f + g) (x) = f (x) + g (x) (iii) If f nd g re differentible t x then so is their product f g nd (f g) (x) = f (x)g(x) + f(x)g (x) (iv) If f is differentible t x nd f(t) 0 for ll t E then /f is differentible t x nd (/f) (x) = f (x)/f(x) (v) If f(t) = n r=0 rt r on E then f is differentible t x nd Proof. iii) f(y)g(y) f(x)g(x) y x f (x) = = n r r x r r= f(y)(g(y) g(x)) y x + = f(x)g (x) + g(x)f (x) g(x)(f(y) f(x)) y x 0
(Note: We hve used tht, s we pss to the limit, f(y) f(x)) iv) f(y) f(x) y x = f(y)f(x) = f(y)f(x) f (x) (f(x)).f(x) f(y) y x.f(y) f(x) y x The next result is slightly hrder to prove thn it looks. (we split the proof into two hlves depending on whether f (x) 0 or f (x) = 0). Lemm 6.7. Let U nd V be subsets of R. Suppose f : U R is such tht f(t) V for ll t U. If f is differentible t x U nd g : V R is differentible t f(x), then the composition g f is differentible t x with (g f) (x) = f (x)g (f(x)) Proof. If f (x) 0 then f(x + h) f(x) 0 for h smll nd thus g(f(x + h)) g(f(x)) h = g(f(x + h)) g(f(x)) f(x + h) f(x). f(x + h) f(x) h = g (f(x)).f (x) (since f is continuous t x 0, f(x + h) f(x)) We hve lots of different wys of deling with f (x) = 0. Here is one Since g is differentible t f(x) g(f(x) + h) g(f(x)) k g (f(x)) so if k is sufficiently smll, sy for 0 < k < µ g(f(x) + k) g(f(x)) g (f(x)) < k So g(f(x) + k) g(f(x)) k ( + g (f(x)) ) Now f is continuous t x so δ(µ) > 0 such tht f(x + h) f(x) < µ for h < δ(µ) Tke k = f(x + h) f(x). We hve g(f(x + h)) g(f(x)) f(x + h) f(x) ( + g (f(x)) ) so g(f(x + h)) g(f(x)) f(x + h) f(x) ( + g (f(x)) ) h }{{ h } 0 Thus g f is differentible with derivtive 0
7 The men vlue theorem We hve lmost finished our project of showing tht the horrid sitution reveled by Exmple. cn not occur for the rels. Our first step is to prove Rolle s theorem. Theorem 7. (Rolle s theorem). If g : [, b] R is continuous function with g differentible on (, b) nd g() = g(b), then we cn find c (, b) such tht g (c) = 0. Proof. A continuous function on closed bounded intervl is bounded nd ttins its bounds (by theorem 5.) In other words, we cn find c, c [, b] such tht g(c ) g(x) g(c ) x [, b] If both re end points (i.e. c, c {, b}) then g(c ) = g(c ) so g(x) = g(c ) x [, b] so g is constnt nd we could tke c = +b. Hence we my ssume tht t lest one of c, c is not n end point. WLOG ssume tht c is not n end point (otherwise consider g Let c = c, we hve (c δ, c + δ) [, b] for some δ > 0 Observe tht f(c + h) g(c) 0 (s c is mximum point) g(c+h) g(c) So, if h > 0 h 0 so llowing h 0 through positive vlues g (c) 0 g(c+h) g(c) Similrly if h < 0 h 0 so llowing h 0 through negtive vlues g (c) 0. Thus g (c) = 0 A simple tilt gives the fmous men vlue theorem. Theorem 7. (The men vlue theorem). If f : [, b] R is continuous function with f differentible on (, b), then we cn find c (, b) such tht f(b) f() = (b )f (c) Proof. Set g(t) = (f(t) f()) t t b (f(b) f()) g(t) = Af(t)+B where A, B re such tht g() = g(b) = 0. As g is differentible on (, b) nd continuous on [, b], by Rolle s Theorem c (, b) such tht g (c) = 0 i.e. 0 = f (c) f(b) f() b We now hve the results so long desired. Lemm 7.3. If f : [, b] R is continuous function with f differentible on (, b), then the following results hold. (i) If f (t) > 0 for ll t (, b) then f is strictly incresing on [, b]. (Tht is, f(y) > f(x) whenever b y > x.) (ii) If f (t) 0 for ll t (, b) then f is incresing on [, b]. (Tht is, f(y) f(x) whenever b y > x.) (iii) [The constnt vlue theorem] If f (t) = 0 for ll t (, b) then f is constnt on [, b]. (Tht is, f(y) = f(x) whenever b y > x.)
Proof. i) f(y) f(x) = (y x)f (c) with c between x nd y by the MVT, nd thus s f (c) is positive nd y x is f(y) > f(x) ii) Sme s bove, except tht f (c) might be 0, so cn get insted of > iii) f(y) f(x) = (y x)f (c) = 0 The prticl converses re esy: If f is differentible, nd incresing, then f(x+h) f(x) h 0 h 0 so f f(x + h) f(x) (x) = lim 0 h 0 h Notice tht since we deduce Lemm 7.3 from the men vlue theorem we cn not use it in the proof of Rolle s theorem. The men vlue theorem hs mny importnt consequences, some of which we look t in the reminder of the section. We strt by looking t inverse functions. Lemm 7.4. (i) Suppose f : [, b] R is continuous. Then f is injective if nd only if it is strictly incresing (tht is f(t) > f(s) whenever s < t b) or strictly decresing. (ii) Suppose f : [, b] R is continuous nd strictly incresing. Let f() = c nd f(b) = d. Then the mp f : [, b] [c, d] is bijective nd f is continuous on [c, d]. Proof. If f is strictly incresing, y x then either y > x or x > y. If y > x, f(y) > f(x) so f(y) f(x), sme for x > y Conversely, suppose f is not strictly incresing. Then either A) We cn find x < x < x 3 such tht f(x ) f(x ) nd f(x 3 ) f(x ) B) we cn find x < x < x 3 such tht f(x ) f(x ) nd f(x 3 ) f(x ) WLOG ssume cse A (otherwise look t f) If f(x ) = f(x ) or f(x ) = f(x 3 ) then we hve contrdiction (f would not be injective). So suppose ll re strict inequlities Choose c with f(x ) > c > mx{f(x ), f(x )}. By the Intermedite Vlue Theorem we cn find α, α such tht x < α < x 3, x < α < x 3 nd f(α ) = x = f(α ). Thus f is not injective ii) The injectivity follows from i) Surjectivity follows from the Intermedite Vlue Theorem: If x γ d then since f() = c, f(b) = d t with t b such tht f(t) = γ Thus f is bijective nd f is defined s function, f : [c, d] [, b] Clim tht f is continuous Let y [c, d]. To simplify mtters, suppose y c, y d If δ > 0 consider the intervl [θ, θ ] = [, b] [f (y δ ), f (y + δ )] θ < f (y) < θ 3
f(θ ) < y < f(θ ) so choose ɛ > 0 such tht f(θ ) < y ɛ < y < y + ɛ < f(θ ) If y z < ɛ, f(θ ) < z < f(θ ) so θ < f (z) < θ nd so f (z) f (y) < δ < δ The proof if y = c or y = d is similr, but one-sided Lemm 7.5 (Inverse rule). Suppose f : [, b] R is differentible on [, b] nd f (x) > 0 for ll x [, b]. Let f() = c nd f(b) = d. Then the mp f : [, b] [c, d] is bijective nd f is differentible on [c, d] with Proof. f (x) = f (y + h) f (y) h = f (f (x)) f (y + h) f (y) f(f (y + h)) f(f (y)) But f is continuous, so f (y + h) f (y) 0 nd thus f (y + h) f (y) h f (f (y)) Note: Suppose we know tht f nd f re differentible, then t = f(f (t)) = f f (t). By the chin rule = f (f (t)).(f ) (t) so (f ) (t) = f (f (t)) This does not constitute proof! In the opinion of the uthor the true mening of the inverse rule nd the chin rule only becomes cler when we consider higher dimensions in the next nlysis course. We now prove form of Tylor s theorem. Theorem 7.6 (nth men vlue theorem). Suppose tht b > nd f : [, b] R is n + times differentible. Then f(b) n j=0 for some c with < c < b. f (j) () j! (b ) j = f (n+) (c ) (b ) n+ (n + )! Proof. This is complicted, so we try to simplify little Step : By trnsltion x x we my suppose = 0, nd our sttement becomes f(b) n n=0 f (j) (0) j! b j = f (n+) (c ) b n+ (n + )! with 0 < c < b Step : Set g(t) = f(t) n f (r) (0) r=0 r! t r Then g(0) = g (0) = = g (n) (0) = 0 nd our theorem becomes 4
If g is (n + ) times differentible, nd g(0) = g (0) = = g (n) (0) = 0 then g(b) = g(n+) (c) (n + )! bn+ (g theorem) We now need to prove the g theorem Let G(x) = g(x) xn+ b g(b). Then G(0) = G (0) = = G (n) (0) = 0 nd n+ G(b) = 0, so by Rolle, 0 < b < b such tht G (b ) = 0 And becuse G (0) = 0 0 < b < b such tht G (b ) = 0 And becuse G (0) = 0 0 < b 3 < b such tht G (b 3 ) = 0 nd so on Until we get to 0 < b n < b n with G (n) (b n ) = 0. But G (n) (0) = 0 so 0 < c < b n such tht G (n+) (c) = 0 G (n+) (x) = g (n+) (c) g(b)(n + )! b n g(b) = g(n+) (c)b n+ (n + )! This gives us globl nd locl Tylor s theorem. Theorem 7.7 (Globl Tylor theorem). Suppose tht b > nd f : [, b] R is n + times differentible. If x, t [, b] f(t) = n j=0 for some θ (0, ). f (j) (x) j! (t x) j + f (n+) (x + θ(t x)) (t x) n+ (n + )! Theorem 7.8 (Locl Tylor theorem). Suppose tht δ > 0 nd f : (x δ, x + δ) R is n times differentible on (x δ, x + δ) nd f (n) is continuous t x. Then n f (j) (x) f(t) = (t x) j + ɛ(t)(t x) n j! where ɛ(t) 0 s t x. j=0 Proof. Move to the origin, f(t) = f(0) + f (0)t + + f (n ) (0) (n )! t n + f (n) (c)t n n! If f (n) is continuous t O f (n) (c) f (n) (0) s t 0 so c 0 Write ɛ(t) = (f (n) (c) f (n) (0)) n! f(t) = f(0) + f (0)t + + f (n ) (0)t n (n )! + f (n) (0)t n ɛ(t)t n n! 5
Notice tht the locl Tylor theorem lwys gives us some informtion but tht the globl one is useless unless we cn find useful bound on the n + th derivtive. To reinforce this wrning we consider fmous exmple of Cuchy. Exercise 7.9. Consider the function F : R R defined by F (0) = 0 F (x) = exp( /x ) otherwise. (i) Prove by induction, using the stndrd rules of differentition, tht F is infinitely differentible t ll points x 0 nd tht, t these points, F (n) (x) = P n (/x) exp( /x ) where P n is polynomil which need not be found explicitly. (ii) Explin why x P n (/x) exp( /x ) 0 s x 0. (iii) Show by induction, using the definition of differentition, tht F is infinitely differentible t 0 with F (n) (0) = 0 for ll n. [Be creful to get this prt of the rgument right.] (iv) Show tht F (j) (0) F (x) = x j j! j=0 if nd only if x = 0. (The reder my prefer to sy tht The Tylor expnsion of F is only vlid t 0.) (v) Why does prt (iv) not contrdict the locl Tylor theorem (Theorem 7.8)? Since exminers re fonder of the globl Tylor theorem thn it deserves I shll go through the following exmple. Exmple 7.0. Assuming the stndrd properties of the exponentil function show tht x j exp x = j! for ll x. j=0 Proof. If f(x) = e x then f (r) x = e x nd using the formul f(x) = f(0) + f (0)! which becomes x + + f (n) (0) x n + f (n+) (c) n! (n + )! xn+ c < x e x = + x! + x! + + xn n! + ec x n+ (n + )! Look t R n = e c xn+ (n+)!, the error term R n e x x n+ x (n+)! 0 (becuse if n x +, n So e x = + x + x! + 6
Plese note tht in pure mthemtics question mny (or even most) of the mrks in question of this type will depend on estimting the reminder term. [In methods questions you my simply be sked to find the Tylor s series without being sked to prove convergence.] 8 Complex vrible The field C of complex numbers resembles the field R of rel numbers in mny wys but not in ll. Lemm 8.. We cn not define n order on C which will behve in the sme wy s > for R. However there is sufficient similrity for us to define limits, continuity nd differentibility. (We hve lredy seen some of this in Definition 3. nd Exercise 3..) Definition 8.. Let E be subset of C, f be some function from E to C, nd z some point of E. If l C we sy tht f(w) l s w z [or, if we wish to emphsise the restriction to E tht f(w) l s w z through vlues w E] if, given ɛ > 0, we cn find δ(ɛ) > 0 [red delt depending on epsilon ] such tht f(w) l < ɛ for ll w E with 0 < w z < δ(ɛ). As usul there is no rel loss if the reder initilly tkes E = C. Definition 8.3. Let E be subset of C. We sy tht function f : E C is continuous t z E if nd only if f(w) f(z) s w z. Exercise 8.4. Let E be subset of C, f, g be some functions from E to C, nd z some point of E. (i) The limit is unique. Tht is, if f(w) l nd f(w) k s w z then l = k. (ii) If z E E nd f(w) l s w z through vlues w E, then f(w) l s w x through vlues w E. (iii) If f(u) = c for ll u E then f(w) c s w z. (iv) If f(w) l nd g(w) k s w z then f(w) + g(w) l + k. (v) If f(w) l nd g(w) k s w z then f(w)g(w) lk. (vi) If f(w) l s w z, f(u) 0 for ech u E nd l 0 then f(w) l. Exercise 8.5. Suppose tht E is subset of C, tht z E, nd tht f nd g re functions from E to C. (i) If f(u) = c for ll u E, then f is continuous on E. (ii) If f nd g re continuous t z, then so is f + g. (iii) Let us define f g : E C by f g(u) = f(u)g(u) for ll u E. Then if f nd g re continuous t z, so is f g. 7
(iv) Suppose tht f(u) 0 for ll u E. If f is continuous t z so is /f. (v) If z E E nd f is continuous t z then the restriction f E of f to E is lso continuous t z. (vi) If J : C C is defined by J(z) = z for ll z C, then J is continuous on C. (vii) Every polynomil P is continuous on C. (viii) Suppose tht P nd Q re polynomils nd tht Q is never zero on some subset E of C. Then the rtionl function P/Q is continuous on E (or, more precisely, the restriction of P/Q to E is continuous.) Exercise 8.6. Let U nd V be subsets of C. Suppose f : U C is such tht f(z) V for ll z U. If f is continuous t w U nd g : V C is continuous t f(w), then the composition g f is continuous t w. Definition 8.7. Let E be subset of C. A function f : E C is differentible t z E with derivtive f (z) if f(w) f(z) f (z) w z s w z. If f is differentible t ech point z E we sy tht f is differentible function on E. Exercise 8.8. Let E be subset of C, f some function from E to C, nd z some point of E. Show tht if f is differentible t z then f is continuous t z. Exercise 8.9. Let E be subset of C, f, g be some functions from E to C, nd z some point of E. Prove the following results. (i) If f(u) = c for ll u E then f is differentible t z with f (z) = 0. (ii) If f nd g re differentible t z then so is their sum f + g nd (f + g) (z) = f (z) + g (z) (iii) If f nd g re differentible t z then so is their product f g nd (f g) (z) = f (z)g(z) + f(z)g (z) (iv) If f is differentible t z nd f(u) 0 for ll u E then /f is differentible t z nd (/f) (z) = f (z)/f(z) (v) If f(u) = n r=0 ru r on E then f is differentible t z nd f (z) = n r r z r r= Proof. v) If f n (z) = z n then f is differentible with f n = nf n So prove by induction, lredy proved for 0, Suppose true for n. f n+ = f n.f so by the multiplictive rule f n+ = f n.f + f n.f = nf n.f + f n. = (n + )f n 8
Exercise 8.0 (Chin rule). Let U nd V be subsets of C. Suppose f : U C is such tht f(z) V for ll t U. If f is differentible t w U nd g : V R is differentible t f(w), then the composition g f is differentible t w with (g f) (w) = f (w)g (f(w)) In spite of these similrities the subject of complex differentible functions is very different from tht of rel differentible functions. It turns out tht well behved complex functions need not be differentible. Exmple 8.. Consider the mp Γ : C C given by Γ(z) = z. The function Γ is nowhere differentible. Proof. Γ(z + h) Γ(z) h = (z + h) z h Clim tht h h does not tend to limit s h 0 h First, let h R, h 0. h = h h = h Now let h = ik, j R, k 0. h = ik ik = # We cn view this nother wy (this is not proof) = h h (z + h) z h = (z + reiθ ) z re iθ = re iθ re iθ = e iθ so the derivtive depends on the direction tken f(z + iw) = f(z) + Re iθ + error Also, given tht z chnges sense it is not surprising tht it is not differentible Becuse complex differentibility is so much more restrictive thn rel differentibility we cn prove stronger theorems bout complex differentible functions. For exmple it cn be shown tht such functions cn be written loclly s power series (contrst the sitution in the rel cse reveled by Exmple 7.9). To lern more go to the course P3 on complex methods. 9 Power series In this section we work in C unless otherwise stted. We strt with very useful observtion. Lemm 9.. If n=0 nz n 0 converges nd z < z 0 then n=0 nz n converges. The syllbus sys tht this fct is prt of this course. In this one instnce I dvise you to ignore the syllbus. 9
Proof. Since n z n 0 converges, n z n 0. Since the series n z n 0 0 it is bounded i.e. K such tht n z n 0 K n (this follows: n z n 0 0 N : n z n 0 n N, tke K = mx{, 0, z, z 0,..., N z N 0 } Thus n z n = n z n 0 z z 0 n K z z 0 n Since z z 0, z z 0 n converges over the rels (s it is geometric progression) (note, we sy nothing bout C), so by the comprison test n z n converges, nd since n bsolutely convergent series converges, n z n converges This gives us the following bsic theorem on power series. Theorem 9.. Suppose tht n C. Then either n=0 nz n converges for ll z C, or there exists rel number R with R 0 such tht (i) n=0 nz n converges if z < R, (ii) n=0 nz n diverges if z > R. Proof. If n z n converges z, we set R =, nothing to prove. If not, consider E = { z : n z n diverges} E is subset of R, E φ, E bounded below by 0. R = E is thus defined. We must now prove tht it hs the required properties. If z < R then n z n converges by definition If z > R then by definition we cn find w C such tht z > w nd n w n diverges. Now, by the previous lemm, if n z n converged then n w n would converge, but it does not, so n z n diverges We cll R the rdius of convergence of n=0 nz n. If n=0 nz n converges for ll z we write R =. Wrning: Although the rtio test is useful first test for trying to find R it is not lwys successful nd it cnnot be used in the definition. Look t + z + z + z 3 + z 4 + z 5 + is lterntely z nd z n+ n However, +z +z 4 + converges, nd z+z 3 +z 5 + = z(+z +z 4 + ) converges too (if z <, so + z + z + z 3 + z 4 + z 5 + converges for z < z n if z >, so + z + z + z 3 + z 4 + must diverge if z >, so the rdius of convergence if We never sy nything bout behviour on the rdius of convergence. The following useful strengthening is left to the reder s n exercise. Exercise 9.3. Suppose tht n=0 nz n hs rdius of convergence R. Then the sequence n z n is unbounded if z > R nd n=0 nz n converges bsolutely if z < R. Note tht we sy nothing bout wht hppens on the circle of convergence. Exmple 9.4. (i) n= n z n hs rdius of convergence nd converges for ll z with z =. (ii) n= zn hs rdius of convergence nd diverges for ll z with z =. 30
Proof. i) If z =, zn n = n Using the rtio test: zn+ (n+) / zn so n z n is bsolutely convergent z z so the rdius of convergence is (+ n ) n =, nd we hve convergence on the circle of convergence ii) N 0 zn = zn+ z for z which tends to limit iff z < nd N 0 n = N so z n hs rdius of convergence nd diverges t every point on the circle of convergence A more complicted exmple is given in question 6 on the 3rd exmple sheet. It is remrkble fct tht we cn operte with power series in the sme wy s polynomils (within the rdius of convergence). In prticulr we shll show tht we cn differentite term by term. Theorem 9.5. If n=0 nz n hs rdius of convergence R nd we write f(z) = n=0 nz n then f is differentible t ll points z with z < R nd f (z) = n n z n The proof is strred in the syllbus. We use three simple observtions. n= Lemm 9.6. If n=0 nz n hs rdius of convergence R then given ny ɛ > 0 we cn find K(ɛ) such tht n=0 nz n < K(ɛ) for ll z R ɛ. Lemm 9.7. If n=0 nz n hs rdius of convergence R then so do n= n nz n nd n= n(n ) nz n. Proof. Suppose n z n hs rdius of convergence R. If z < R we cn find w 0 with z < w 0 < R Since w 0 < R, n w0 n converges, so M such tht n w0 n M n Now n n z n = n w0 n z w 0 n n w Mn 0 w 0 z w 0 n = u n Clim tht u n converges un+ u n = z w 0 ( + n ) z w 0 < so rtio test works. Since u n converges the comprison test tells us tht n n z n converges, so n n z n converges. Thus n n z n hs rdius of convergence t lest R, since n n z n n n z n = z n nz n n(n )n z n hs rdius of convergence R by differentiting gin. ( ) ( ) n n Lemm 9.8. (i) n(n ) for ll r n. r r (ii) (z + h) n z n nhz n n(n )( z + h ) n h for ll z, h C. Proof. i) ( ) n n(n ) = r n(n )(n ) (r )!(n r)! = n! (r )!(n r)! n! r!(n r)! = ( ) n r 3
ii) n ( ) n (z + h) n nz n h z n = z n r h r r= n ( n n(n ) r r= n ( n = n(n )h r r=0 ) z n r h r = n(n ) h ( z + h ) n ) z n r h r Thus if z + h < R n (z + h) n n z n n n z n = h h ( n ((z + h) n z n nz n h)) n (z + h) n z n nz n h h n h ( z + h ) n n(n ) h = h n(n ) n ( z + h ) n nd given tht we re within the rdius of convergence this sum converges to bounded vlue, so n(n ) n w n K for w < R ɛ In this course we shll minly work on the rel line. Restricting to the rel line we obtin the following result. Theorem 9.9. (i) If j R there exists unique R (the rdius of convergence) with 0 R such tht n=0 nx n converges for ll rel x with x < R nd diverges for ll rel x with x > R. (ii) If n= nx n hs rdius of convergence R nd we write f(x) = n=0 nx n then f is differentible t ll points x with x < R nd f (x) = n n x n n= 0 The stndrd functions In school you lerned ll bout the functions exp, log, sin nd cos nd bout the behviour of x α. Nothing tht you lerned ws wrong (we hope) but you might be hrd pressed to prove ll the fcts you know in coherent mnner, To get round this problem, we strt from scrtch mking new definitions nd ssuming nothing bout these vrious functions. One of your tsks is to mke sure tht the lecturer does not slip in some unproved fct. On the other hnd 3
you must llow your lecturer to choose definitions which llow n esy development of the subject rther thn those tht follow some historic, intuitive or pedgogiclly pproprite pth 3. Let us strt with the exponentil function. Throughout this section we shll restrict ourselves to the rel line. Lemm 0.. The sum n=0 xn n! hs infinite rdius of convergence. Proof. x n+ (n + )! / xn n! = x n + 0 so by the rtio test n=0 xn n! converges x so the rdius of convergence is We cn thus define function e : R R by e(x) = (We use e(x) rther thn exp(x) to help us void mking unjustified ssumptions.) Theorem 0.. (i) The function e : R R is everywhere differentible with e (x) = e(x). (ii) e(x + y) = e(x)e(y) for ll x, y R. (iii) e(x) > 0 for ll x R. (iv) e is strictly incresing function. (v) e(x) s x, e(x) 0 s x. (vi) e : R (0, ) is bijection. Proof. i) Since power series cn be differentited term by term within its rdius of convergence e(x) is differentible n=0 x n n! e (x) = n. xn n! = x n (n )! = 0 x n n! = e(x) ii) Let f(x) = e( x)e(x) with constnt f (x) = e ( x)e(x) + e( x)e (x) = e( x)e(x) + e( x)e(x) = 0 so by the men vlue theorem, f is constnt, f(x) = f(0) i.e. e( x)e(x) = e()e(0) Looking t power series e(0) = + 0! + 0! + = so e( x)e(x) = e(), x Set = x + y to get e(y)e(x) = e(x + y) 3 If you wnt to see tretment long these lines see the excellent text of Burn[3]. 33