NWI: Mathematics. Various books in the library with the title Linear Algebra I, or Analysis I. (And also Linear Algebra II, or Analysis II.

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1 NWI: Mthemtics Literture These lecture notes! Vrious books in the librry with the title Liner Algebr I, or Anlysis I (And lso Liner Algebr II, or Anlysis II) The lecture notes of some of the people who hve given these lectures in the pst In prticulr: mspiess/lehrelthtml froyshov/nwi2/indexhtml froyshov/nwi/indexhtml Some stndrd logicl symbols commonly used in mthemtics X mens, X is set, nd is n element of X is the empty set, which contins no elements X Y is the union of the sets X nd Y It is the set which contins the elements of X nd lso the elements of Y X Y is the intersection It is the set consisting of the elements which re in both X nd Y X \ Y is the set difference It is the set contining the elements of X which re not in Y X Y mens tht X is subset of Y All the elements of X re lso elements of Y Note tht mny people use the nottion X Y to expressly sy tht equlity X = Y is lso possible But I will ssume tht when writing X Y, the cse X = Y is lso possible mens for ll, s for exmple: x, x 0 Tht mens: for ll x, we hve the condition x 0 mens there exists P Q mens tht P nd Q re logicl sttements, nd if P is true, then Q must lso be true (If P is flse, then the combined sttement P Q is true, regrdless of whether or not Q is true) P Q mens tht both P Q nd lso Q P re true Tht is, P nd Q re logiclly equivlent; they re simply different wys of sying the sme thing (Although often it is not immeditely cler tht this is the cse Thus we need to think bout why it is true, constructing proof) i

2 Contents 1 Numbers, Arithmetic, Bsic Concepts of Mthemtics 1 11 How computers del with numbers 1 12 The system Z/nZ for n = Equivlence reltions, equivlence clsses 4 14 The system Z/nZ revisited 5 15 The gretest common divisor function 5 16 Mthemticl induction 7 17 The binomil theorem: using mthemticl induction 8 18 The bsic structures of lgebr: groups, fields Anlysis nd Liner Algebr 14 2 Anlysis 1 21 Injections, Surjections, Bijections 1 22 Constructing the set of rel numbers R Dedekind cuts Deciml expnsions Convergent sequences 4 23 Convergent sequences Bounded sets Subsequences Cuchy sequences Sums, products, nd quotients of convergent sequences 8 24 Convergent series 9 25 The stndrd tests for convergence of series The Leibniz test The comprison test Absolute convergence The quotient test Continuous functions Sums, products, nd quotients of continuous functions re continuous The exponentil function Some generl theorems concerning continuous functions Differentibility Tking nother look t the exponentil function The logrithm function The men vlue theorem Complex numbers The trigonometric functions: sin nd cos 33 ii

3 215 The number π The geometry of the complex numbers The Riemnn integrl Step functions Integrls defined using step functions Simple consequences of the definition Integrls of continuous functions Axiomtic chrcteriztion of the Riemnn integrl The fundmentl theorem of clculus Anti-derivtives, or Stmmfunktionen Another look t the fundmentl theorem Prtil integrtion The substitution rule Tylor series; Tylor formul The Tylor formul The Tylor series Power series, Fourier series, etc More generl integrls Mesure theory, generl integrls: brief sketch Integrls in R n ; Fubini s theorem Fubini s theorem Axiomtic chrcteriztion of integrls in R n Regions in R n ; open sets, closed sets The topology of metric spces Prtil derivtives Prtil derivtives commute if they re continuous Totl derivtives Further results involving prtil derivtives The chin rule in higher dimensions The directionl derivtive The trnsformtion formul for higher dimensionl integrls Uniformly convergent sequences of functions Ordinry differentil equtions Seprtion of vribles An exmple: y = x y Another exmple: homogeneous liner differentil equtions Vrition of constnts The theorem of Picrd nd Lindelöf Systems of first order differentil equtions The Lipschitz condition Uniqueness of solutions The Bnch fixed point theorem Existence of solutions The eqution y = f ( y x) Ordinry differentil equtions of higher order Numericl methods for solving ordinry differentil equtions Euler s method The Rung-Kutt method 73 iii

4 231 The vritionl clculus: quick sketch 74 3 Liner Algebr 1 31 Bsic definitions 2 32 Subspces 3 33 Liner independence nd dimension 4 34 Liner mppings 8 35 Liner mppings nd mtrices Mtrix trnsformtions Systems of liner equtions Invertible mtrices Similr mtrices; chnging bses Eigenvlues, eigenspces, mtrices which cn be digonlized The elementry mtrices The determinnt Leibniz formul Specil rules for 2 2 nd 3 3 mtrices A proof of Leibniz formul The chrcteristic polynomil Sclr products, norms, etc Orthonorml bses Orthogonl, unitry nd self-djoint liner mppings Chrcterizing orthogonl, unitry, nd Hermitin mtrices Orthogonl mtrices Unitry mtrices Hermitin nd symmetric mtrices Which mtrices cn be digonlized? 43 iv

5 Chpter 1 Numbers, Arithmetic, Bsic Concepts of Mthemtics To begin with, we hve the usul nd simple systems of numbers: The nturl numbers N = {1, 2, 3, 4, } The whole numbers, or integers Z = {, 3, 2, 1, 0, 1, 2, 3, } The rtionl numbers Q = { b : Z, b N} But then we hve somewht more unusul systems: The rel numbers R The complex numbers C The residue clsses modulo n, for n N, nmely Z/nZ 11 How computers del with numbers A computer dels with informtion by mnipulting mny tiny trnsistors, ech of which either does, or does not, hve prticulr electricl potentil pplied to it We cn think of this s the trnsistor representing either the digit 0, or else the digit 1 In this wy, nturl numbers cn be represented ccording to the following scheme 1 : 1 2 : 10 3 : 11 4 : : : : : 1000 etc This is rithmetic to the bse 2 The binry system of rithmetic Erlier computers : mechnicl dding mchines, often hd toothed wheels for representing numbers Thus some of the wheels 1

6 might hve hd 10 teeth, representing the numbers in our deciml system Also persons counting numbers on the fingers of their two hnds represent numbers in the deciml system Mechnicl clocks hve toothed wheels with 60 teeth, representing the numbers to the bse 60 This system continues the trdition of rithmetic used by the Bbylonins in the ncient world All of these mchines clocks, mechnicl clcultors, fingers, computers re finite systems Therefore, since the set of nturl numbers N is infinite, it follows tht there is limit to how much cn be represented in ech mchine For exmple, mechnicl clock generlly goes through cycle of 12 hours, then t midnight, or noon, it strts over gin from the beginning Similrly, the stndrd unit of computer, byte, consists of 8 trnsistors, so it cn only represent 2 8 = 256 different numbers The usul convention is to think of these s being the integers from 0 to 255 How do we do rithmetic in the system of numbers from 0 to 255? Well, for exmple 2+2=4 is OK, since both the numbers 2 nd 4 re represented in one byte Nmely, we hve = in binry rithmetic But how bout ? Unfortuntely, 400 is not represented in the byte, since the number 400 is greter thn 255 In fct, when the computer reches 256, it simply cycles bck to zero Specificlly, we hve tht 200 is in the binry system Therefore is = 1 } {{} 8 plces Unfortuntely, there re 9 plces (or bits) in This is too much to fit into byte Therefore the lst bit is simply removed, discrded So we hve = Or, expressed in the more usul deciml nottion, we hve = 144 Wht we re doing here is modulr rithmetic, modulo The system Z/nZ for n = 256 We hve seen tht in the system of modulr rithmetic modulo 256, we hve the eqution = 144 Another wy to think bout this is to sy tht in the usul integer rithmetic of Z we hve = 400 = More generlly, given ny integer x, nd ny nturl number n, we hve two unique integers nd b, such tht x = n + b, where 0 b < n The number is the result of the whole number division of x by n, nd b is the reminder which results from this whole number division For exmple, in the C lnguge, the whole number division of x by n is given by the instruction 2

7 = x / n; The reminder term is given by the instruction b = x % n; In mthemtics, we use the nottion In prticulr then, we hve the eqution 1 b = x mod n 144 = 400 mod 256 Arithmetic generlly hs four opertions: ddition, subtrction, multipliction, nd division So let us sy we hve two numbers, x nd y in our system Z/nZ Tht is, we cn ssume tht 0 x, y < n Then we simply define the sum of x nd y to be Similrly, the difference is nd the product is (x + y) mod n (x y) mod n, (x y) mod n All of this is esy, since x ± y nd x y re lwys integers However, wht bout division? The number x y is only occsionlly n integer And wht do we do when y = 0? The solution to this problem is to think of division s being the problem of solving simple eqution Thus the number x y is relly the solution z of the eqution z y = x For exmple, wht is 1 3 in our modulr rithmetic modulo 256? Tht is, the problem is to find some number z with 0 z < 256, such tht 1 = (z 3) mod 256 The nswer? It is z = 171, since = 513, nd 1 = 513 mod 256 On the other hnd, wht is 1 2 modulo 256? Tht is, let z be such tht 1 = (z 2) mod 256 Wht is z? The nswer is tht there is no nswer! Tht is to sy, the number 1 2 does not exist in the modulr rithmetic modulo 256 The reson for this is tht for ll z we lwys hve z 2 being n even number, yet since 256 is lso n even number, it must be tht the eqution 1 = y mod 256 cn only hve solution when y is n odd number So we see tht rithmetic modulo 256 gives us problems when it comes to division In fct, ll of this discussion just shows tht rithmetic in computers is much more complicted thn we might t first hve thought 1 Actully, for historicl resons, it is more correct to write b x mod n here Tht is, we hve n equivlence reltion I will come bck to this lter But for the moment, we should just think of the number x mod n s being the result of the rithmeticl opertion of finding the reminder fter dividing x by n This is lwys number between 0 nd n 1 3

8 13 Equivlence reltions, equivlence clsses Definition Let M be set The set of ll pirs of elements of M is denoted by M M Thus M M = {(, b) :, b M} This is clled the Crtesin product of M with itself 2 An equivlence reltion on M is subset of M M Given two elements, b M, we write b to denote tht the pir (, b) is in the subset For n equivlence reltion, we must hve: 1, for ll M (reflectivity) 2 if b, then we lso hve b (symmetry) 3 if b nd b c the we lso hve c (trnsitivity) If b, then we sy tht is equivlent to b Exmples 1 Given ny set M, the most trivil possible equivlence reltion is simply equlity Nmely b only when = b 2 In Z, the set of integers, let us sy tht for two integers nd b, we hve b if nd only if b is n even number Then this is n equivlence reltion on Z 3 Agin in Z, this time tke some nturl number n N Now we define to be equivlent to b if nd only if there exists some further number x Z with b = xn Tht is, the difference b is divisible by n And gin, this is n equivlence reltion on Z (Obviously, the exmple 2 is just specil cse of exmple 3 In fct, it is the equivlence reltion which results when we tke n = 2) Definition Given set M with n equivlence reltion, then we hve M being split up into equivlence clsses For ech M, the equivlence clss contining is the set of ll elements of M which re equivlent to The equivlence clss contining is usully denoted by [] Therefore [] = {x M : x } Note tht if we hve two equivlence clsses [] nd [b] such tht their intersection is not empty [] [b], then we must hve [] = [b] To see this, ssume tht x [] [b] Then x nd x b But x mens tht x, since the equivlence reltion is symmetric Then b since it is trnsitive If then y [b], then we hve y b But lso b, nd so using the trnsitivity of the equivlence reltion gin, we hve y Thus y [] So this shows tht [b] is contined in [] ie [b] [] A similr rgument shows tht lso [] [b] Therefore we hve shown tht: Theorem 11 Given n equivlence reltion on set M, then the equivlence reltion splits M into set of disjoint equivlence clsses 2 More generlly, if X nd Y re two different sets, then the Crtesin product X Y is the set of ll pirs (x, y), with x X nd y Y 4

9 14 The system Z/nZ revisited In fct, rther thn thinking bout Z/nZ s the set of numbers {0,, n 1}, it is more usul to sy tht Z/nZ is the set of equivlence clsses with respect to the equivlence reltion given by x y if nd only if x y is divisible by n Thus Z/nZ = {[0],, [n 1]} But rther thn writing x y, it is more usul to write x y mod n when describing this equivlence reltion One sys tht x is congruent to y modulo n Addition nd multipliction in Z/nZ re given by the simple rules nd [x] + [y] = [x + y] [x] [y] = [x y], for ny two numbers x, y Z However, we re still left with the problem of division in Z/nZ Tht is, given, b Z, does there exist n x Z such tht x b mod n? 15 The gretest common divisor function To solve this eqution, we first need to think bout gretest common divisors Definition Let x, y Z Then we sy tht x is divisor of y if there exists z Z with y = xz Given two numbers, b Z, the number d is common divisor of nd b if d is divisor of both nd b The gretest common divisor of nd b, is denoted by gcd(, b) Obviously, every integer is divisor of the number zero Furthermore, if x divides y, then obviously x lso divides y Thus we cn restrict our thinking to the integers which re either zero, or else positive Given two integers nd b, not both zero, then obviously the number 1 is common divisor Therefore we lwys hve gcd(, b) 1 Theorem 12 Given ny two integers nd b, not both zero, then there exist two further integers x nd y, such tht x + yb = gcd(, b) Proof If one of the integers is zero, sy = 0, then obviously gcd(, b) = b (we ssume here tht b is positive) So we hve 3 gcd(, b) = b = b, nd the theorem is true in this cse Let us therefore ssume tht nd b re both positive integers If the theorem were to be flse, then it must be flse for some pir of integers, b N Assume tht b, nd tht this pir is the smllest possible counterexmple to the theorem, in the sense tht the theorem is true for ll pirs of integers b, with b < b 3 From now on I will use the more usul nottion b, or even just b, for multipliction, rther thn the nottion b, which I hve been using up till now 5

10 But we cn immeditely rule out the possibility tht = b, since in tht cse we would hve gcd(, b) = b, nd gin we would hve the solution gcd(, b) = b = b Thus the pir, b would not be counterexmple to the theorem Therefore we must hve < b So let c = b Then c N nd the theorem must be true for the smller pir c, Thus there exist x, y Z with gcd(, c) = x + y c = x + y (b ) = (x y ) + y b But wht is gcd(, c) = gcd(, b )? Obviously, ny common divisor of nd b is lso common divisor of nd b Also ny common divisor of nd b must be common divisor of both nd b Therefore gcd(, c) = gcd(, b), nd so we hve gcd(, b) = (x y ) + y b, which contrdicts the ssumption tht the pir, b is counterexmple to the theorem It follows tht there cn be no counterexmple, nd the theorem must lwys be true Solving the eqution x b mod n So let, b Z be given, together with nturl number n N The question is, does there exist some x Z with x b mod n? Tht is to sy, does n divide the number x b? Or put nother wy, does there exist some y Z with Tht is the sme s x b = yn? b = x + ( y)n Therefore, we see tht the eqution x b mod n cn only hve solution if every common divisor of nd n is lso divisor of b Tht is, we must hve gcd(, n) being divisor of b On the other hnd, ssume tht gcd(, n) does, in fct, divide b Sy b = z gcd(, n) Then, ccording to the previous theorem, there must exist u, v Z with Therefore, we hve gcd(, n) = u + vn b = z gcd(, n) = z(u + vn) = (zu) + (zv)n = x + ( y)n, when we tke x = zu nd y = zv To summrize: Theorem 13 The eqution x b mod n hs solution if nd only if gcd(, n) is divisor of b If b = z gcd(, n) then solution is x = zu, where gcd(, n) = u + vn 6

11 The system Z/pZ, when p is prime number The prime numbers re 2, 3, 5, 7, 11, 13, 17, 19, 23, A prime number p N is such tht it hs no divisors in N other thn itself nd 1 Or put nother wy, for ll 1 < p we hve gcd(, p) = 1 Therefore, ccording to the previous theorem, for ll [] Z/pZ with [] [0] there must exist some [b] Z/pZ with [][b] = [1] Tht is to sy, b 1 mod p so tht in the modulr rithmetic modulo p, we hve tht 1 is b Therefore it is lwys possible to divide numbers by In fct, dividing by is simply the sme s multiplying by b On the other hnd, if n is not prime number, then there exists some with 1 < < n nd gcd(, n) > 1 In this cse, ccording to the theorem, there cn be no solution to the eqution x 1 mod n Therefore it is impossible to divide numbers by in modulr rithmetic modulo n when n is not prime number nd gcd(, n) > 1 16 Mthemticl induction An exmple The formul n k=1 1 k(k + 1) = is true for ll n N How do I know tht this is true?? n n + 1 Well, first of ll, I know tht it is true in the simple cse n = 1 For here we just hve 1 1 k(k + 1) = 1 1(1 + 1) = k=1 k=1 But then I know it s true for n = 2 s well, since 2 1 = k(k + 1) 1 1 2(2 + 1) + k=1 1 k(k + 1) 1 = 2(2 + 1) = Note tht the second eqution follows, since I lredy know tht the formul is true for the cse n = 1 More generlly, ssume tht I know tht the formul is true for the cse n, for some prticulr n N Then, exctly s before, I cn write n+1 k=1 1 k(k + 1) = = = 1 n (n + 1)((n + 1) + 1) + 1 k(k + 1) k=1 1 (n + 1)((n + 1) + 1) + n n + 1 (n + 1) (n + 1) + 1 7

12 Therefore, the proof tht the formul is true progresses stepwise through the numbers 1, 2, 3,, nd so we cn conclude tht the formul is true for ll n N This is the principle of mthemticl induction (or vollständige Induktion in Germn) Let P (n) be some sttement which depends on the number n, for rbitrry n N Then P (n) is true for ll n N if: First of ll, the specil cse P (1) cn be proved, nd then it cn be proved tht if P (n) is true for some rbitrrily given n N, then lso P (n+1) must be true We will be using mthemticl induction very often here in these lectures! It is one of the most bsic principles of mthemtics 17 The binomil theorem: using mthemticl induction The binomil theorem is concerned with wht hppens when the expression ( + b) n is multiplied out For exmple, we hve 1 : + b 2 : b 2 3 : b + 3b 2 + b 3 4 : b b 2 + 4b 3 + b 4 5 : b b b 3 + 5b 4 + b 5 etc Grdully we see pttern emerging, nmely Pscl s tringle: nd so on Writing out the expression ( + b) n s sum, one uses the binomil coefficients, ( n k) Thus one writes n ( ) n ( + b) n = n k b k k k=0 So looking t Pscl s tringle, we see tht ( ( 2 0) = 1, 7 4) = 35, nd so forth The binomil theorem is the formul tht sys tht for ll n N nd 0 k n, we hve ( n k ) = n! k!(n k)! But for the moment, let us simply define the number ( ) n k to be n! truly the binomil coefficients 8 k!(n k)!, nd then see if these re

13 The expression n! is clled n-fctoril For n N it is defined to be n! = n (n 1) (n 2) Tht is, just the product of ll the numbers from 1 up to n In the specil cse tht n = 0, we define 0! = 1 So let s see how this works out in the cse ( 7 4) We hve ( ) 7 7! = 4 4!(7 4)! = ( ) (3 2 1) = 35, in greement with Pscl s tringle But how do we prove it in generl? Theorem 14 As in Pscl s tringle, we hve ( ) ( ) n + 1 n = + k k 1 tht is for ll n N nd 1 k n Proof ( ) n, k (n + 1)! k!((n + 1) k)! = n! (k 1)!(n (k 1))! + n! k!(n k)!, n! (k 1)!(n (k 1))! + n! k!(n k)! = = = = k n! (n k + 1) n! + k!(n k + 1))! k!(n k + 1)! k n! (n + 1) n! k n! + k!(n k + 1))! k!(n k + 1)! (n + 1) n! k!(n k + 1))! (n + 1)! k!((n + 1) k)! Theorem 15 For ll n N nd 0 k < n, we hve ( + b) n = n k=0 ( ) n n k b k, k with ( ) n = k n! k!(n k)! 9

14 Proof Induction on n For the cse n = 1, the theorem is trivilly true Therefore we ssume tht the theorem is true in the cse n, nd so our tsk is to prove tht under this ssumption, the theorem must lso be true in the cse n + 1 We hve: Here we hve: ( + b) n+1 = ( + b) ( + b) n ( n ( ) n = ( + b) ) n k b k k k=0 ( n ( ) n = ) n k b k + b k = = = = n k=0 n k=0 k=0 ( ) n n k+1 b k + k ( ) n n k+1 b k + k ( ) n n n k=1 (( n k n k=0 n+1 k=1 ) + n+1 ( ) n + 1 (n+1) k b k k k=0 the first eqution is trivil, the second eqution is the inductive hypothesis, the third nd fourth equtions re trivil, ( n k=0 ( ) n n k b k+1 k ( ) n ) n k b k k ( ) n n (k 1) b k k 1 ( )) n (n+1) k b k + k 1 the fifth eqution involves substituting k 1 for k in the second term, the sixth eqution is trivil, nd ( ) n b n+1 n the ( seventh eqution uses the theorem which we hve just proved nd, lso the fct tht n ) ( 0 = n ) n = 1, for ll n N 18 The bsic structures of lgebr: groups, fields Now tht we hve gotten the binomil theorem out of the wy, let us return to thinking bout numbers We hve N Z Q The set of nturl numbers N hs ddition nd multipliction, but not subtrction nd division 4 The set of integers Z hs ddition, subtrction nd multipliction, but division fils However, in the set of rtionl numbers Q, ll of these four bsic opertions cn be crried out (Of course, we exclude the specil number zero when thinking bout division) 4 Subtrction fils in N: for exmple 1 2 = 1, but 1 is not n element of N Also division obviously fils: for exmple 1/2 is lso not n element of N 10

15 Furthermore, in the rithmeticl system Z/nZ we hve ddition, subtrction nd multipliction If (nd only if) n is prime number, then we lso hve division Arithmeticl systems in which these four opertions cn be sensibly crried out re clled fields (In Germn, Körper) In order to define the concept of field, it is best to strt by defining wht we men in mthemtics when we spek of group But in order to do tht, we should first sy wht is ment when we spek of function, or mpping Definition Let X nd Y be non-empty sets A function f : X Y is rule which ssigns to ech element x X unique element f(x) Y Exmples For exmple, f : N N with f(n) = n 2 is function But f(n) = n is not function from N to N, since n N, for ll n N On the other hnd, f(n) = n is function from N to Z Tht is, f : N Z Definition A group is set G, together with mpping f : G G G stisfying the following three conditions: f((f(, b), c)) = f((, f(b, c))), for ll, b nd c in G There exists n element e G with f((e, g)) = f((g, e)) = g, for ll g G For ll g G there exists element, usully denoted by g 1 G, such tht f((g 1, g)) = f((g, g 1 )) = e Actully, this mpping f : G G G is usully thought of s being n bstrct kind of multipliction Therefore, we usully write b or b, rther thn this cumbersome f((, b)) With this nottion, the group xioms become (b)c = (bc), for ll, b nd c in G (The Associtive Lw) There exists specil element (the unit element ) e G, with eg = ge = g, for ll g G (The existence of the unit, or neutrl element) For ll g G there exists n inverse g 1 G with g 1 g = gg 1 = e (The existence of inverses) If, in ddition to this, the Commuttive Lw holds: b = b, for ll nd b in G, then the group G is clled n Abelin group Remrk When thinking bout numbers, you might think tht it is entirely nturl tht ll groups re Abelin groups However this is certinly not true! Mny of the groups we will del with in these lectures re definitely not Abelin For exmple the mtrix groups which re used continuously when computer clcultes 3-dimensionl grphics re non-abelin groups But now we cn define the ide of field 11

16 Definition A field is set F, together with two opertions, which re clled ddition nd multipliction They re mppings + : F F F : F F F stisfying the following conditions (or xioms ) F is n Abelin group with respect to ddition The neutrl element of F under ddition is clled zero, denoted by the symbol 0 For ech element F, its inverse under ddition is denoted by Thus, for ech, we hve + ( ) = 0 Let F \ {0} denote the set of elements of F which re not the zero element Tht is, we remove 0 from F Then F \ {0} is n Abelin group with respect to multipliction The neutrl element of multipliction is clled one, denoted by the symbol 1 For ech F with 0, the inverse is denoted by 1 Thus 1 = 1 The Distributive Lw holds: For ll, b nd c in F we hve both (b + c) = b + c, nd ( + b)c = c + bc Some simple consequences of this definition re the following Theorem 16 Let F be field Then the following sttements re true for ll nd b in F 1 Both nd 1 (for 0) re unique 2 0 = 0 = 0, 3 ( b) = ( b) = ( ) b, 4 ( ) =, 5 ( 1 ) 1 =, if 0, 6 ( 1) =, 7 ( )( b) = b, 8 b = 0 = 0 or b = 0 Proof This involves few simple exercises in fiddling with the definition 1 If + = 0 nd + = 0 then + ( + ) = + 0 Therefore = 0 + = ( + ) + = + ( + ) = + 0 = The fct tht 1 is unique is proved similrly 12

17 2 Since = 0, we hve (0 + 0) = = 0 Then 0 = 0 + ( ( 0)) = ( 0 + 0) + ( ( 0)) = 0 + ( 0 + ( ( 0))) = = 0 The fct tht 0 = 0 is proved similrly 3 0 = 0 = (b + ( b)) = b + ( b) Therefore we must hve b = ( b) The other cses re similr 4 + ( ( )) = 0 But lso + = 0, nd from (1) we know tht dditive inverses re unique Therefore = ( ) 5 ( 1 ) 1 = is similr 6 We hve 0 = 0 = (1 + ( 1)) = 1 + ( 1) = + ( 1) Therefore ( 1) = 7 0 = 0 ( 1) = (1 + ( 1))( 1) = 1 + ( 1)( 1) Therefore Then 8 If 0 then = 1 = 1 + ( 1) + ( 1)( 1) = ( 1)( 1) ( )( b) = (( 1))(( 1)b) = (( 1)( 1))b = 1 b = b b = 1 b = ( 1 )b = 1 (b) = 1 0 = 0 Which groups nd fields re importnt for these lectures? The groups we will use: Of course fields re themselves groups under ddition So ll fields tht is to sy, ll the number systems we will consider re themselves groups Liner lgebr is concerned with vectors A system of vectors is clled vector spce Ech vector spce is group, with respect to vector ddition The set of liner trnsformtions (rottions, inversions, chnges of perspective) of vector spce re described using invertible, squre mtrices These mtrices form non-abelin group under mtrix multipliction When deling with determinnts of mtrices, we will consider the group of permuttions of n objects This is lso non-abelin group 13

18 The fields we will use: We hve lredy seen the two fields which give us the most bsic number systems in mthemtics: nmely the rtionl numbers Q nd the modulr system Z/pZ, for prime numbers p The rel numbers R re constructed by filling in the gps in Q This is the bsis of rel nlysis, which will constitute hlf of these lectures But fter constructing R, we see tht something is still missing Mny polynomils, for exmple the polynomil x 2 + 1, hve no roots in the system of numbers R To solve this problem, the system of complex numbers C will be constructed 19 Anlysis nd Liner Algebr Mthemtics, s it is tught tody in universities, lwys begins with two seprte series of lectures Nmely Anlysis nd Liner Algebr Only lter, prticulrly when it comes to the subject of functionl nlysis, do we see tht nlysis nd liner lgebr re, in mny wys, just two different spects of the sme thing But, unfortuntely (or fortuntely??) you, s students of informtion technology, will probbly leve pure mthemtics before reching tht stge In ny cse, I will continue these lectures by tlking bout nlysis nd liner lgebr s if they were two entirely different subjects In these lecture notes, they will be delt with in two different chpters, which will be developed simultneously Anlysis Anlysis cn be thought of s being the study of the rel nd the complex numbers The ide of functions plys the importnt role Which functions re continuous, or differentible? How does integrtion work? How do we solve simple systems of differentil equtions? How should we define the bsic functions, such s the exponentil function, the logrithm function, the trigonometric functions, etc? For this, we need to think bout whether or not given infinite series of numbers converges, or not Liner Algebr This is concerned with geometry How cn computer work out movements, perspective in 3- dimensionl spce, nd then represent these on 2-dimensionl screen? Wht is bsis for coordinte system? When is set of vectors linerly independent? How re liner mppings of vector spces represented by mtrices? And then, tking step wy from geometry, how do we solve systems of liner equtions using the methods of liner lgebr? This lst question is very importnt when it comes to the use of computers in economics 14

19 Chpter 2 Anlysis 21 Injections, Surjections, Bijections The subject of mthemticl nlysis is minly concerned with functions, or mppings 1 We hve lredy seen tht function is rule f, which ssigns to ech element x X of set X, unique element f(x) Y of set Y One writes f : X Y Given such function f from X to Y, one sys tht X is the domin of f Furthermore, the set {f(x) : x X} Y is the rnge of f One writes f(x) for the rnge of X Thus, f(x) = {f(x) : x X} Given ny element y Y, one writes f 1 (y) to denote the subset of X consisting of ll the elements which re mpped onto y Tht is, f 1 (y) = {x X : f(x) = y} Of course, if f is not surjection, then f 1 (y) must be the empty set, for some of the elements of Y Definition Let X nd Y be sets, nd let f : X Y be function Then we sy tht: f is n injection if, given ny two different elements x 1, x 2 X with x 1 x 2, we must hve f(x 1 ) f(x 2 ) Or put nother wy, the only wy we cn hve f(x 1 ) = f(x 2 ) is when x 1 = x 2 f is surjection if, for ll y Y, there exists some x X with f(x) = y Tht is, if f : X Y is surjection, then we must hve f(x) = Y f is bijection if it is both n injection, nd lso surjection 1 Tht is, Funktionen nd Abbildungen in Germn The words function nd mpping both men the sme thing in mthemtics Perhps some people would sy tht mpping f : X Y is function if the set Y is some sort of system of numbers, otherwise it is mpping But we certinly needn t mke this distinction 1

20 Exmples Consider the following functions f : Z Z: f() = 2, for ll Z This is n injection, but it is not surjection since only even numbers re of the form 2, for Z For exmple, the number 3 is in Z, yet there exists no integer with 2 = 3 { /2, if is even, f() = ( + 1)/2, if is odd, is surjection, but it is not n injection For exmple, f(0) = 0 = f( 1) f() =, for ll Z, is bijection Theorem 21 Let f : X Y be n injection Then there exists surjection g : Y X Conversely, if there exists surjection f : X Y, then there exists n injection g : Y X Proof Assume tht there exists n injection f : X Y A surjection g : Y X cn be constructed in the following wy First choose some prticulr element x 0 X Then surjection g : Y X is given by the rule { x, where f(x) = y if y f(x), g(y) = x 0, if y f(x), for ll y Y Going the other wy, ssume tht there exists surjection f : X Y Then n injection g : Y X cn be constructed in the following wy Since f is surjection, we know tht the set f 1 (y) X is not empty, for ech y Y Therefore, for ech y Y, choose some prticulr element x y f 1 (y) Then the injection g : Y X is given by the rule g(y) = x y, for ll y Y Remrk: This procedure of choosing elements from collection of sets is only vlid if we use the xiom of choice in the theory of sets This is certinly the usul kind of mthemtics which lmost ll mthemticins pursue However it is perfectly possible to develop n lterntive theory of mthemtics in which the xiom of choice is not true In this lterntive mthemtics, this proof would not be vlid Furthermore, we hve the following theorem bout bijections Theorem 22 (Schröder-Bernstein) Let X nd Y be sets Assume tht there exists n injection f : X Y, nd lso there exists surjection g : X Y Then there exists bijection h : X Y Proof An exercise 22 Constructing the set of rel numbers R 221 Dedekind cuts The simplest method for defining rel numbers is to use the technique of Dedekind cuts 2

21 Definition A Dedekind cut of the rtionl numbers Q is pir of nonempty subsets A, B Q, such tht if A nd x <, then x A s well Furthermore, if b B nd y > b, then y B s well Also A B = Q nd A B = Finlly, we require tht the subset A hs no gretest element Then the set of rel numbers R cn be defined to be the set of Dedekind cuts of the rtionl numbers One my think of ech rel number s the point between the upper set B nd the lower set A If the given rel number hppens to be rtionl number, then it is the smllest number in the set B For exmple, it is well known tht the number 2 is irrtionl Theorem 23 There exists no rtionl number b with ( b ) 2 = 2 Proof Assume to the contrry tht there does indeed exist such rtionl number b Perhps there exist mny such rtionl squre roots of 2 If so, choose the smllest one, b, in the sense tht if b is lso squre root of 2, then we must hve b b Now, since b is squre root of 2, we must hve ( ) 2 = 2 b Therefore, 2 = 2b 2 But this cn only be true if is n even number So let us write = 2c, with c Z Then we hve Or 2 = 4c 2 = 2b 2 b 2 = 2c 2 Therefore b is lso n even number, sy b = 2d But in this cse we must hve c d = b, so c d is lso squre root of 2 But this is impossible, since d < b nd we hve ssumed tht b ws smllest possible squre root of 2 Given ny rtionl number q Q, we hve q 2 being lso rtionl number So we cn mke Dedekind cut by tking the pir (A, B), with B being ll the positive rtionl numbers b with b 2 > 2 Then A is the rest of the rtionl numbers Tht is, A is the set of rtionl numbers less thn 2, nd B is the set of rtionl numbers greter thn 2 So this Dedekind cut defines the rel number 2 Of course the rtionl numbers themselves cn lso be represented in terms of Dedekind cuts For exmple the number 2 is simply the Dedekind cut (A, B), with A = {q Q : q < 2} nd B = {q Q : q 2} So here, the number 2 is the smllest number in the set B The reson Dedekind brought in this definition in the 19th century is tht with it, it is possible to define the rel numbers without, hving to use the xiom of choice 222 Deciml expnsions For exmple, we hve 1 3 = Also 2 =

22 Another well-known irrtionl number is π = As we know, rtionl number hs repeting deciml expnsion On the other hnd, irrtionl numbers do not repet when written out s deciml expnsions One might sy tht, for exmple, the number is the sme s the number , which, of course, is relly just the number one But if we exclude deciml expnsions which end in never-ending sequence of 9s, then the deciml expnsion for ech rel number is unique Therefore, n lterntive wy to define the rel numbers is to sy tht they re nothing more thn the set of ll possible deciml expnsions which do not end with n infinite sequence of 9s 223 Convergent sequences But the most usul method of defining the rel numbers is s equivlence clsses of convergent sequences We need the ide of convergent sequences in ny cse, so let us tke the set of rel numbers R s given (using either of the previous definitions), nd consider the theory of sequences, either in Q or in R itself 2 23 Convergent sequences A sequence is simply n infinite list of numbers For exmple, the sequence 1, 2, 3, 4, 5, 6, 7, is certinly esy to think bout, but obviously it doesn t converge The numbers in the sequence get lrger nd lrger, incresing beyond ll possible finite bounds Another exmple is the sequence 1, 1, 1, 1, 1, 1, 1, 1, This sequence remins bounded, just jumping bck nd forth between the two numbers 1 nd 1 But it never converges to nything; it lwys keeps jumping bck nd forth An exmple of convergent sequence is 1, 1 2, 1 3, 1 4, 1 5, 1 6, 1 7, This sequence obviously converges down to zero In generl, when thinking bout bstrct sequences of numbers, we write 1, 2, 3, So 1 is the first number in the sequence 2 is the second number, nd so forth A shorter nottion, for representing the whole sequence is ( n ) n N But when thinking bout the concept of convergence, it is cler tht we lso need n ide of the distnce between two numbers 2 Agin nd this is the lst time I will mention this fct the theory of convergent sequences requires the xiom of choice 4

23 Definition Given rel (or rtionl) number x, the bsolute vlue of x is given by { x, if x 0, x = x, if x < 0 So one cn think of x s being either zero, if x is zero, otherwise x is the distnce of x from zero More generlly, given two numbers nd b, the distnce between them is b It is simple mtter to verify tht the tringle inequlity lwys holds Tht is, for ll x, y R, we lwys hve x + y x + y Definition The sequence ( n ) n N converges to the number if, for ll positive numbers ɛ > 0, there exists some sufficiently lrge nturl number N ɛ N, such tht n < ɛ, for ll n N ɛ In this cse, we write lim n n = If the sequence does not converge, then one sys tht it diverges This definition is rther bstrct But, for exmple, it doesn t relly tell us wht is hppening with the simple sequence 1, 1, 1, 1, 1, 1, Although this sequence does not converge ccording to our definition still, in wy it relly converges to the two different points 1 nd Bounded sets Given the set of ll rel numbers R, let us consider some rbitrrily given subset A R Definition We will sy tht A R is bounded bove, if there exists some K R, such tht K, for ll A The number K is clled n upper bound for A Similrly, A is bounded below if there exists some L R with L, for ll A Then L is lower bound for A If A is bounded both bove nd below, then we sy tht A is bounded In this cse, clerly there exists some M 0 with M, for ll A If A, nd if A is bounded bove, then the smllest upper bound is clled the lest upper bound, written lub(a) Similrly, glb(a) is the gretest lower bound The lest upper bound is lso clled the Supremum, tht is, sup(a) The gretest lower bound is clled the Infimum, written inf(a) Exmples Let [0, 1] = {x R : 0 x 1} Then [0, 1] is bounded, nd the lest upper bound is 1; the gretest lower bound is 0 This time, tke [0, 1) = {x R : 0 x < 1} This is of course lso bounded, nd the lest upper bound is gin 1, even though 1 is not contined in the subset [0, 1) N R is bounded below (with gretest lower bound 1), but it is not bounded bove Z R is not bounded below, nd lso not bounded bove 5

24 232 Subsequences Definition Let i : N N be mpping such tht for ll n, m N with m < n, we hve i(m) < i(n) Then given sequence ( n ) n N, subsequence, with respect to the mpping i, is the sequence ( i(n) ) n N For exmple, let s look gin t the sequence (( 1) n ) n N Then tke the mpping i : N N with i(n) = 2n In this cse, we hve the subsequence (( 1) i(n) ) n N = (( 1) 2n ) n N = (( ( 1) 2) n)n N = (1n ) n N = (1) n N But this is just the trivilly convergent constnt sequence of 1s, which obviously converges to 1 So we see tht in this exmple, the sequence relly consists of two convergent subsequences, one of them converges to the number 1, nd the other converges to the number 1 On the other hnd, the sequence (n) n N hs no convergent subsequences All subsequences simply diverge to infinity The problem is tht it just keeps getting bigger, incresing beyond ll bounds To void this, we hve the following definition Definition The sequence ( n ) n N is clled bounded if the set { n : n A} is bounded in R (Similrly, we sy the sequence is bounded bove, or below, if those conditions pply to this set) Theorem 24 Let ( n ) n N be bounded sequence in R Then there exists convergent subsequence, converging to number in R Proof Since the sequence is bounded, there must exist two rel numbers x < y, such tht x n y, for ll n N Let z = (x + y)/2 Tht is, z is the point hlf wy between x nd y So now the originl intervl from x to y hs been split into two equl subintervls, nmely the lower one from x to z, nd the upper one from z to y Since our sequence contins infinitely mny elements, it must be tht there re infinitely mny in one of these two subintervls For exmple, let s sy there re infinitely mny elements of the sequence in the lower subintervl In this cse, we set x 1 = x nd y 1 = z If only finitely mny elements of the sequence re in the lower subintervl, then there must be infinitely mny in the upper subintervl In this cse, we set x 1 = z nd y 1 = y Then the intervl from x 1 to y 1 is divided in hlf s before, nd subintervl x 2 to y 2 is chosen which contins infinitely mny elements of the sequence And so on By this method, we construct two new sequences, (x n ) n N nd (y n ) n N, nd we hve x x 1 x 2 x 3 x 4 y 4 y 3 y 2 y 1 y We hve y n x n = y x 2 n Therefore the two sequences pproch ech other more nd more nerly s n gets lrger Now tke (A, B) to be the following Dedekind cut of the rtionl numbers Q B = {q Q : q x n, n} Then set A = Q \ B Let us sy tht R is the rel number which is given by the Dedekind cut (A, B) Then clerly there is subsequence ( i(n) ) n N with lim i(n) = n 6

25 Definition The sequence ( n ) n N is clled monotoniclly incresing if n n+1, for ll n; it is monotoniclly decresing if n n+1, for ll n; finlly, one simply sys tht it is monotonic if it is either monotoniclly incresing, or monotoniclly decresing It is simple exercise to show tht theorem 24 implies tht the following theorem is lso true Theorem 25 Every bounded, monotonic sequence in R converges Conversely, we hve tht Theorem 26 Every convergent sequence is bounded Proof This is relly rther obvious Let the sequence ( n ) n N converge to the point R Choose ɛ = 1 Then there exists some N(1) N with n < 1, for ll n N(1) We hve the numbers 1, 2,, N(1) Let M be either the lrgest of these numbers, or else + 1, whichever is lrger Then we must hve n M, for ll n N Thus the sequence is bounded below by M, nd bove by M 233 Cuchy sequences Definition A sequence ( n ) n N is clled Cuchy sequence if for ll ɛ > 0, there exists number N(ɛ) N such tht n m < ɛ, for ll m, n N(ɛ) It is gin n exercise to show tht: Theorem 27 Every convergent sequence is Cuchy sequence The lterntive, nd more usul wy to define the rel numbers is s equivlence clsses of Cuchy sequences of rtionl numbers The equivlence reltion is the following Let ( n ) n N nd (b n ) n N be two Cuchy sequences, with n nd b n Q, for ll n Then we will sy tht the they re equivlent to one nother if nd only if for ll ɛ > 0, there exists some N(ɛ) N, with n b n < ɛ, for ll n N(ɛ) The fct tht this is, in fct, n equivlence reltion is lso left s n exercise Then R is defined to be the set of equivlence clsses in the set of Cuchy sequences in Q But not ll Cuchy sequences converge!! If we lwys think bout the set of rel numbers R, then of course every Cuchy sequence converges As we hve seen, this is simply wy of defining the set of rel numbers! But if we think bout other sets which re not simply ll of R, then it is definitely not true tht ll Cuchy sequences converge For exmple, let us consider the set (0, 1] = {x R : 0 < x 1} Within this set, the sequence (1/n) n N is Cuchy sequence Considered in R, it converges to the number 0 But considered within (0, 1] lone, it doesn t converge, since 0 is not n element of (0, 1] Similrly, if we consider the set of rtionl numbers Q, then there re mny Cuchy sequences which converge to irrtionl numbers, when considered in R Yet those irrtionl numbers do not belong to Q Therefore they do not converge in Q 7

26 On the other hnd, ll Cuchy sequences do converge in R For let ( n ) n N be Cuchy sequence in R Let A = {q Q : N N, n N, with q < n } Then let B = Q \ A If it hppens to be the cse tht A hs lrgest element, sy x 0, then tht should be trnsferred over to B Tht is, we chnge A to A \ {x 0 } nd B to B {x 0 } This gives Dedekind cut of Q, representing the rel number R, sy, nd then we must hve the Cuchy sequence ( n ) n N converging to To see this, let ɛ > 0 be chosen The problem is to show tht there exists some N N, such tht n < ɛ, for ll n N Let us strt by choosing some rtionl number q A with q < ɛ/6 Then there must exist some other rtionl number p B, with p q < ɛ/3 Looking t the definition of the set A, we see tht the number p must be such tht for sufficiently lrge n, ll the numbers n re less thn p On the other hnd, given such n n which is greter thn q, then it must be between q nd p Tht mens tht the distnce between q nd n must be less thn ɛ/3 Since the sequence ( n ) n N is Cuchy sequence, there exists number N(ɛ/3) N such tht for ll n, m N(ɛ/3), we hve n m < ɛ/3 Then setting N = N(ɛ/3), nd tking m N with q < m, we hve for ll n N Therefore we hve the theorem: n = ( q) + (q m ) + ( m n ) q + q m + m n < ɛ 3 + ɛ 3 + ɛ 3 = ɛ Theorem 28 All Cuchy sequences converge in R 234 Sums, products, nd quotients of convergent sequences Let ( n ) n N nd (b n ) n N be two convergent sequences in R with lim n = nd lim b n = b n n Then the sequence ( n + b n ) n N lso converges, nd lim ( n + b n ) = + b n To see this, let ɛ > 0 be given, nd let N (ɛ), N b (ɛ) N with n < ɛ/2 nd b b m < ɛ/2, for ll n N (ɛ) nd m N b (ɛ) Then tke N(ɛ) = mx{n (ɛ), N b (ɛ)}, tht is, the lrger of the two numbers For ny k N(ɛ) we then hve ( + b) ( k b k ) = ( k ) + (b b k ) ( k ) + (b b k ) < ɛ 2 + ɛ 2 = ɛ Here, we hve used the tringle inequlity for the bsolute vlue function Obviously, the difference of two sequences lso converges to the difference of their limit points As for multipliction, gin tke the convergent sequences ( n ) n N nd (b n ) n N s before We hve lim n n = nd lim n b n = b Now let M > 0 be such tht nd n M, 8

27 for ll n N Also let M b > 0 be such tht b nd b m M b, for ll m N (These numbers must exist, since convergent sequences re bounded) Then, given ɛ > 0, choose N (ɛ) such tht for ll n N (ɛ) we hve n < ɛ 2M b Similrly, N b (ɛ) is chosen such tht for ll m N b (ɛ) we hve b b n < ɛ 2M Then tke N(ɛ) = mx{n (ɛ), N b (ɛ)} So gin, For ny k N(ɛ) we hve b k b k = b b k + b k k b k b b k + b k k b k = b b k + b k k ɛ ɛ < + b k 2M 2M b ɛ 2 + ɛ 2 = ɛ Finlly, ssume tht ( n ) n N is convergent sequence such tht the limit is not zero Then the sequence (1/ n ) n N (t most finitely mny elements of the sequence cn be zero, nd so we disregrd these zero elements) converges to 1/ In order to see this, let M > 0 be lower bound of the sequence of bsolute vlues ( n ) n N, together with Given ɛ > 0, this time choose N(ɛ) N to be so lrge tht for ll n N(ɛ), we hve n < ɛm 2 Then 1 1 n = n n 1 = n n < ɛm 2 n ɛ Then, in order to divide convergent sequence by convergent sequence which does not converge to zero, we first tke the convergent sequence of the inverses, then multiply with tht In summry, we hve Theorem 29 Convergent series cn be dded, subtrcted, multiplied nd divided (s long s they do not converge to zero), to obtin new convergent sequences which converge to the sum, difference, product, nd quotient of the limits of the given sequences 24 Convergent series Given sequence ( n ) n N, we cn imgine trying to find the sum of ll the numbers in the sequence Thus writing n, n=1 9

28 we hve the series given by the sequence ( n ) n N Obviously, mny series do not converge For exmple the series n = n=1 does not converge Also the series ( 1) n = n=1 does not converge Why is this? Definition Given the series n=1 n, the n-th prtil sum (for ech n N) is the finite sum S n = n n k=1 The series n=1 n converges, if the sequence of its prtil sums (S n ) n N converges If the series does not converge, then one sys tht it diverges So wht re the prtil sums for the series n=1 ( 1)n? Clerly, we hve { 1, if n is odd, S n = 0, if n is even Therefore, the prtil sums jump bck nd forth between -1 nd 0, never converging A delicte cse: the series n=1 1/n But wht bout the series n=1 1 n = Obviously the prtil sums get lrger nd lrger: S n+1 > S n, for ll n N But the growth of the sequence of prtil sums keeps slowing down So one might think tht this series could converge But does it? In fct, it ctully diverges We cn see this by looking t the sum split into blocks of ever incresing length 1 n = n=1 }{{} }{{ 8} >1/2 >1/2 Tht is to sy, for ech n N, we hve 2 n k=2 n k > 2n k=2 n n = 1 2, so we hve n infinite series of blocks, ech greter thn 1/2 Therefore it must diverge 10

29 The geometric series This is the series n, n=0 for vrious possible numbers R (Note tht it is sometimes convenient to tke the sum from 0 to infinity, rther thn from 1 to infinity Also note tht by convention, we lwys define 0 = 1, even in the cse tht = 0) Theorem 210 For ll rel numbers with < 1, the sequence ( n ) n N converges to zero For 1, the sequence diverges Proof Without loss of generlity, we my ssume tht > 0 If < 1 then the sequence ( n ) n N is strictly decresing sequence Tht is, n+1 < n, for ll n N This follows, since n+1 = n, nd 0 < < 1 So the sequence ( n ) n N gets smller nd smller, s n gets bigger And of course, it strts with, so it is confined to the intervl between 0 nd We cn define Dedekind cut (A, B) of Q s follows A = {x Q : x < n, n N}, nd B = Q \ A (the set difference) Finlly, if A hs gretest element, sy x 0, then tke A = A \ {x 0 } nd B = B {x 0 } Otherwise simply tke A = A nd B = B The pir (A, B) is then Dedekind cut So let ξ be the rel number represented by this Dedekind cut Then we must hve 0 ξ < 1 If ξ = 0 then the sequence converges to zero, nd we re finished Otherwise, we must hve ξ > 0 Now since 0 < < 1, it must be tht the number 1/ is greter thn 1 Thus ξ < ξ 1 But from the definition of ξ, there must be some m N with ξ < m < ξ 1 However, then we hve m+1 = m < ξ 1 = ξ, nd this contrdicts the definition of ξ Therefore the ide tht we might hve ξ > 0 simply leds to contrdiction The only conclusion is tht ξ = 0, nd so the sequence converges If > 1, then, using wht we hve just proved, we see tht the sequence ( 1 )n N converges n to zero Clerly, this implies tht ( n ) n N diverges (or, in this cse, converges to infinity ) Theorem 211 The geometric series converges for < 1, nd it diverges for 1 Proof We hve Therefore, if 1, we hve ( n ) ( 1) k = n+1 1 n k=0 k=0 k = n+1 1 1, 11