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at the simple continued fraction of Jd has period length two if and only if d : a2 + b where a and b are integers irrationals, then ) ca1-a2)' : d'r - d2 (c''c.z)' : ot't'or2 . 1 1 . Which of the following quadratic irrationals have purely periodic continued fractions l + . 6 a ) 2 + ,/-B b) c ) 4 + ' , m c) d) e) (tt ... |
this statement many people have searched for a proof of this assertion without success. Even trrouitr no ,or...t proof has yet been discovered, the foilowing conjecture is knowi as Fermat,s rast theorem. Fermat's Last Theorem. The diophantine equation x ' + l n : z n has no solutions in nonzero integers x, r, z when n... |
n is odd, the positive ' : even, the positive solutions of * : Pyoof. Theorem 1r.3 tells us that if xo,ro is a positive solution of simple continued fraction of ,/7 . On the other hand, from Theorem I 1.4 we know that t l , 1 1 .3 Pell's Equation 405 . Because the period cf the continued expansion oL"/j Qjn : Qo:I for... |
7 768 769 770 771 772 773 774 775 776 777 7 7 8 779 Ap pendix Table 1. (Continued)' 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 800 8 01 802 803 804 805 806 807 808 809 8 1 0 8 1 1 812 8 1 3 8 1 4 8 1 5 700 701 '702 703 704 705 706 707 708 709 7r0 7 t l 7 t 2 713 7 t 4 7 1 5 7 t 6 71... |
on Section 3.) b) no soluti on c) (x ,y) = (0,2) , (1,3) , (2,4 ), (:, 0) or (2,5), (3,2), (4,6), (5,3) , (6,0) (mod 7) b) no sol uti on . a) x = 28 (mod 30) b) no solution ) x : 44 (mod g40) e) no solution Section 3. rl ls rl fr rJ ls l ) l5 ) An swers to Selected Problems 4 3 1 Section 4.. a) by 3, not by 9 b) by 3, ... |
. Hagis, Jr., "Sketch of a proof that an odd perfect number relatively factors," Mathematics of least eleven prime prime Computations, Volume 46 0983), 399-404. to 3 has at J. C. Lagarias and A. M. Odlyzko, "New algorithms for computing n(ff)," Bell Laboratories Technical Memorandum TM-82-1 I 218-57. H. P. Lawther, Jr.... |
r, 198 l2l Identity matrix modulo z, Inclusion-exclusion, principle of, 17,51 Incongruent, 9l Index of an integer, 252,421 Index of summation, 5 Index system, 262 Induction, mathematical, 4 Infinite simple continued fraction, Infinitude of primes, 45,82 Integer, 362 deficient, 185 palindromic, 133 powerful, 16 t73 Inve... |
ons, the likelihood ratio test is still a good idea. In fact, you will show in the example sheets that for a discrete distribution, as long as a likelihood ratio test of exactly size α exists, the same result holds. Example. Suppose X1, · · · , Xn are iid N (µ, σ2 0 is known. We want to find the best size α test of H0 ... |
− |Θ0|. 2.3.2 Pearson’s chi-squared test Notice that the two log likelihoods are of the same form. In general, let oi = ni (observed number) and let ei = n ˜pi or npi(ˆθ) (expected number). Let δi = oi − ei. Then 2 log Λ = 2 oi log oi ei = 2 (ei + δi) log = 2 (ei + δi) δi ei = 2 δi + δ2 i ei − − δ2 i 2ei 26 + O(δ3 i )... |
}. So a 95% √ √ √ confidence set for µ is ( ¯X − 1.96/ n, ¯X + 1.96/ n). 2.5 Multivariate normal theory 2.5.1 Multivariate normal distribution So far, we have only worked with scalar random variables or a vector of iid random variables. In general, we can have a random (column) vector X = (X1, · · · , Xn)T , where the ... |
∥Xt∥ = 0, then we would have produced a linear combination of the columns of X that gives 0). So X T X is positive definite, and hence has an inverse. So ˆβ = (X T X)−1X T Y, (4) which is linear in Y. We have E(ˆβ) = (X T X −1)X T E[Y] = (X T X)−1X T Xβ = β. So ˆβ is an unbiased estimator for β. Also cov(ˆβ) = (X T X)−... |
2 m and V ∼ χ2 V /n is said to have an F -distribution on m and n degrees of freedom. We write X ∼ Fm,n n. The X = U/m Since U and V have mean m and n respectively, U/m and V /n are approxi- mately 1. So F is often approximately 1. It should be very clear from definition that Proposition. If X ∼ Fm,n, then 1/X ∼ Fn,m. ... |
A1 = In − P , A2 = P − P0, and apply our previous lemma, and the fact that ZT AiZ ∼ σ2χ2 r, then RSS = YT (In − P )Y ∼ χ2 RSS0 − RSS = YT (P − P0)Y ∼ χ2 n−p p−p0 and these random variables are independent. So under H0, F = YT (P − P0)Y/(p − p0) YT (In − P )Y/(n − p) = (RSS0 − RSS)/(p − p0) RSS/(n − p) ∼ Fp−p0,n−p, Hen... |
s+1 1/2 N, N t = M, M t s 1/2 , s (Ni2−n − N(i−1)2−n )2 1/2 (Cauchy–Schwarz) where all equalities are u.c.p. Claim. For all 0 ≤ s < t, we have t s |dM, N u| ≤ M, M t s N, N t s. Indeed, for any subdivision s = t0 < t1 < · · · tn = t, we have n i=1 |M, N ti ti−1| ≤ n M, M ti ti−1 N, N ti ti−1 i=1 n i=1 ≤ M, M ti ti−1 1/... |
Extension to semi-martingales Definition (Locally boounded previsible process). A previsible process H is locally bounded if for all t ≥ 0, we have |Hs| < ∞ a.s. sup s≤t Fact. (i) Any adapted continuous process is locally bounded. (ii) If H is locally bounded and A is a finite variation process, then for all t ≥ 0, we ha... |
e quadratic variation of a process doesn’t have to be linear in t. It turns out if the quadratic variation increases to infinity, then the martingale is still a Brownian motion up to reparametrization. Theorem (Dubins–Schwarz). Let M be a continuous local martingale with M0 = 0 and M ∞ = ∞. Let Ts = inf{t ≥ 0 : M t > s}... |
o show its quadratic variation is t. We simply compute B, Bt = t 0 dXs, Xs = t. So there is weak existence. The same argument shows that any solution is a Brownian motion, so we have uniqueness in law. Finally, observe that if x = 0 and X is a solution, then −X is also a solution with the same Brownian motion. Indeed, ... |
ion to Ex(σ, b), and then simply check that it works. We can indeed do that, but it takes a bit more time than we have. Instead, we shall prove a slightly weaker result that if we happen to have a solution, it must be given by our formula involving the SDE. So we first note the following theorem without proof: Theorem. ... |
dzk = i 2 j,k hjk dzk ∧ ¯zj. So we have hkj = hjk. The non-degeneracy condition ω∧n = 0 is equivalent to det hjk = 0, since ωn = n i 2 n! det(hjk) dz1 ∧ d¯z1 ∧ · · · ∧ dzn ∧ d¯zn. So hjk is a non-singular Hermitian matrix. Finally, we take into account the compatibility condition ω(v, Jv) > 0. If we write then we have ... |
eled their systems by a symplectic manifold. The Hamiltonian function H ∈ C∞(M ), which returns the energy of a particular configuration, generates a vector field on M which is the equation of motion of the system. In this chapter, we will discuss how this process works, and further study what happens when we have famili... |
iltonian function can be defined exactly as before: H(x, y) = k 1 2mk |yk|2 + V (x), where x(t) = (x1, . . . , xn) and each xi is a 3-vector; and similarly for y with dxt yk = mk dt . Then in Hamiltonian mechanics, we say (x, y) evolves under Hamiltonian flow. Now fix a, b ∈ R and p, q ∈ R3N . Write P for the space of all... |
preserved by G, then we can instead consider µ−1(ξ)/Gξ, or equivalently take µ−1(Oξ)/G. We check that µ−1(ξ)/Gξ ∼= µ−1(Oξ)/G ∼= ϕ−1(0)/G, where ϕ : M × Oξ → g∗ (ρ, η) → µ(p) − η is a moment map for the product action of G on (M × Oξ, ω × ωξ). So in fact there is no loss in generality for considering just µ−1(0). 40 3 ... |
a line of the form π−1(c) with π integral. Then the line between p 0 and p 1 is contained in µ(M ) by the above argument, and we are done since µ(M ) is compact, hence closed. 0, p It thus remains to prove (i), where we have to put in some genuine work. This requires Morse–Bott theory. Let M be a manifold, dim M = m, a... |
ytope as well. Thus, we want to have as large of a torus action as we can. The following proposition puts a bound on “how much” torus action we can have: Proposition. Let (M, ω) be a compact, connected symplectic manifold with moment map µ : M → Rn for a Hamiltonian T n action. If the T n action is effective, then (i) T... |
action, 40 Fubini–Study form, 24 gauge group, 45 generalized capacity, 56 generating function, 8, 10 Hamilton equations, 32 Hamiltonian, 35 Hamiltonian action, 37, 38 Hamiltonian function, 30, 32 Hamiltonian system, 32 Hamiltonian vector field, 30 harmonic form, 27 Hessian matrix, 47 Index III Symplectic Geometry Hodge ... |
art of our usual picture of the reals is the sense that some numbers are “bigger” than others or more to the “right” than others. We express this by using inequalities x < y or x y. The order structure is closely related to the field structure. For example, when we use inequalities in elementary courses we frequently us... |
er bound (i.e., inf E exists and is a real number). 1.6.20 A function is said to be bounded if its range is a bounded set. Give examples of functions f : R R that are bounded and examples of such functions that are unbounded. Give an example of one that has the property that → is finite but max { f (x) : x R } ∈ sup { d... |
just less than nx + 1; more precisely (using Exercise 1.7.4), find m so that m Now some arithmetic on these inequalities shows that nx + 1 < m + 1. ≤ and then m 1 − ≤ nx < ny thus exhibiting a rational number m/n in the interval (x, y). Exercises 1.9.1 Show that the definition of “dense” could be given as A set E of rea... |
ppose a > 0 and b > 0 and a = b. Establish that √a = √b. Establish that (√a − √b)2 > 0. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 28 NOTES Carry on. What have you proved? Now what if a = b? 4Exercise 1.6.4. You can use induction on the size of E, that is, prove fo... |
iant on the arithmetic progression is obtained by replacing the addition of a fixed amount by the multiplication by a fixed amount. These sequences are called geometric progressions. The sequence c, cr, cr2, cr3, cr4, . . . , crn 1, . . . − is the most general geometric progression. The number r is called the common rati... |
he front of the list.” What is your rejoinder? 2.3.3 A set (any set of objects) is said to be countable if it is either finite or there is an enumeration (list) of the set. Show that the following properties hold for arbitrary countable sets: (a) All subsets of countable sets are countable. (b) Any union of a pair of co... |
equence (sometimes called the “tail” of the sequence) by writing where M is some integer (perhaps large). Show that tn = sM +n for n = 1, 2, 3, . . . sn} { converges if and only if tn} { converges. sn} { converge? sn} { ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Secti... |
ary Real Analysis, 2nd Edition (2008)6 6 55 Section 2.7. Algebra of Limits if n ≥ N . Note that (sn + tn) (S + T ) | − sn − S + | tn − | T | | ≤ | by the triangle inequality. Thus we can make this expression smaller than ε by making each of the two expressions on the right smaller than ε/2. This provides the method. Su... |
this might seem hopeless at first sight since the values of sin nθ are quite unpredictable, we recall that none of these values lies outside the interval [ 1, 1]. Hence sin nθ n . lim n →∞ 1 n ≤ − − sin nθ 1 n . n ≤ The two outer sequences converge to the same value 0 and so the inside sequence (the “squeezed” one) must... |
.1 Define a sequence sn} { recursively by setting s1 = α and sn = (sn−1)2 + β 2sn−1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.9. Monotone Convergence Criterion 71 where α, β > 0. (a) Show that for n = 1, 2, 3, . . . √β)2 (sn − 2sn = sn+1 − β. (b) Show that s... |
es Our first question is easy to answer for any specific sequence, but harder to settle in general. Given a sequence can we always select a subsequence that is monotonic, either monotonic nondecreasing or monotonic nonincreasing? Theorem 2.39: Every sequence contains a monotonic subsequence. Proof. We construct first a no... |
s not necessarily equivalent with being convergent. Our proof is a bit lengthy and will require an application of the Bolzano-Weierstrass theorem. The proof in one direction, however, is easy. Suppose that is convergent to a number L. Let ε > 0. Then there must be an integer N so that ε 2 N . Thus if both m and n are l... |
equality follows, and the theorem is proved. The connection between limits and extreme limits is close. If a limit exists then the upper and lower limits must be the same. Theorem 2.48: Let lim supn xn} { xn = lim inf n be a sequence of real numbers. Then xn and these are finite. In this case →∞ →∞ xn} { is convergent ... |
s one ends in a string of zeros and the other in a string of nines; and (iii) every string of 5’s and 6’s a defines a real number with that decimal expansion. · · · 15Exercise 2.3.10. Try to find a way of ranking the algebraic numbers in the same way that the rational numbers were ranked. 16Exercise 2.4.6. You will need ... |
Sums. s1, s2, etc.) gives the following closed form: s1) + (s3 − s0) + (s2 − s2) + (s4 − It is convenient to call such a sum “telescoping” as an indication of the method that can be used to compute it. 1) = sn − + (sn − (s1 − s3) + s0. · · · sn − Example 3.1: For a specific example of a sum that can be handled by consi... |
K Xi ∈ I0 and consider taking a sum over K = I0 ∪ \ J. for every finite set I0 ⊂ Then K ⊂ and By subtracting these two inequalities and remembering that C ai − I0 Xi ∪ ∈ J C ai − I0 Xi ∈ < ε/2 < ε/2. ai + ai = ai I0 Xi ∪ ∈ J J Xi ∈ I0 Xi ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd ... |
see immediately that the study of such a series reduces to the computation of the limit 1 + r + r2 + · · · + rn − 1 + rn = 1 − 1 rn+1 r − (r = 1) 1 1 < r < 1 (which is usually expressed as − 1 rn+1 r − 1 = lim n →∞ 1 r | − r | < 1) and invalid for all other values of r. ∞ rk − 1 = 1 + r + r2 + 1 = · · · 1 r Xk=1 − 1 t... |
: Divide the interval [0, 1) into p intervals of equal length [0, 1/p), [1/p, 2/p), . . . , [(p 1)/p, 1) − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 132 Infinite Sums Chapter 3 1. Then k1 is chosen so that x belongs to the k1th interval. and label them from left to ri... |
the Cauchy-Schwarz inequality: For any finite sequences the inequality must hold. a1, a2, . . . , an} { and b1, b2, . . . , bn} { n (ak)2 ≤ Xk=1 1 2 n ! (bk)2 Xk=1 1 2 ! n Xk=1 akbk ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 139 3.5.1... |
r, that sequence is bounded Thus an application of the direct comparison test shows that the series ak ≤ Cbk. 3.6.5 d’Alembert’s Ratio Test ∞k=1 ak converges. P ∞k=1 rk for The ratio comparison test requires selecting a series for comparison. Often a geometric series some 0 < r < 1 may be used. How do we compute a numb... |
the ratios ak+1 ak but often fails. We know that if this tends to 1, then the ratio test says nothing about the convergence or divergence of the series ∞k=1 ak. Kummer’s tests provide a collection of ratio tests that can be designed by taking different choices of . The choices Dk = 1, Dk = k and Dk = k ln k are used in... |
ergence 163 3.43 (Alternating Series Test) The series ∞ ( 1)k − 1ak, − Xk=1 whose terms alternate in sign, converges if the sequence decreases monotonically to zero. Moreover, the value of the sum of such a series lies between the values of the partial sums at any two consecutive stages. ak} { The proof is just exactly... |
on 3.6. Tests for Convergence 171 3.6.24 Let F be a positive function on [1, ) with a positive, decreasing and continuous derivative F ′. ∞ ∞ k=1 F ′(k) converges if and only if (a) Show that P converges. F ′(k) F (k) ∞ Xk=1 (b) Suppose that ∞ k=1 F ′(k) diverges. Show that P converges if and only if p > 1. See Note 59... |
onnection with the ordered sum ai, IN Xi ∈ ∞ ai. Xi=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 178 Infinite Sums Chapter 3 We should expect the two to be the same when both converge, but is it possible that one converges and not the other? The answer is that the conv... |
rmonic series with p = 1 2 . ∞ ( 1)k − 1 √k + 1 Xk=0 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.8. Products of Series Let ∞ ck be the formal product of this series with itself. By definition the term ck is given by Xk=0 1)k ( − " 1 (k + 1) 1 · + 1 · 2 + (k) 1... |
using the fact that we obtain F (x) x 1 − ∞ = (a0 + a1 + a2 + Xk=0 + ak)xk. · · · sk = (a0 + a1 + a2 + + ak) · · · sk → A = ∞ ak, Xk=0 F (x) = (1 x) − Let ε > 0 and choose N so large that ∞ Xk=0 skxk = A (1 − − x) A)xk. (sk − ∞ Xk=0 sk − | A | < ε/2 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Anal... |
served us through most of this chapter: A sequence that is monotonic is convergent if and only if it is bounded. P Note that a1 + a2 + a3 + + an ≤ · · · (1 + a1)(1 + a2)(1 + a3) (1 + an) × · · · × so that the convergence of the product gives an upper bound for the partial sums of the series. It follows that if the prod... |
es must converge. 50Exercise 3.4.26. As a first step show that 2kπ+3π/4 2kπ+π/4 Z | sin x x | dx ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 214 NOTES (Remember that in calculus an integral 51Exercise 3.4.28. Establish that ∞ 0 R 1 √2 ≥ 2kπ+3π/4 2kπ+π/4 Z 1 x dx. is int... |
set IN of natural numbers also has infinitely many points, but as we look closely at any one of these points we see that each point is all alone, at a certain distance away from every other point in the set. We call these points isolated points of the set. Definition 4.3: (Isolated Point) Let E be a set of real numbers.... |
lementary Real Analysis, 2nd Edition (2008)6 6 6 Section 4.3. Sets 227 many areas of mathematics. On the real line we can master open and closed sets and describe precisely what they are. 4.3.1 Closed Sets In many parts of mathematics the word “closed” is used to indicate that some operation stays within a system. For ... |
r one chosen was required to have a point rmk that did not belong to any of the preceding choices. But that means then that the new component chosen is disjoint from all the previous ones. G. Thus every point in one of the intervals belongs to G. This completes the checking of the details and so the proof is done. Exer... |
closed sets is closed. 4.4.4 Give an example of a sequence of open sets G1, G2, G3, . . . whose intersection is neither open nor closed. Why does this not contradict Theorem 4.17? 4.4.5 Give an example of a sequence of closed sets F1, F2, F3, . . . whose union is neither open nor closed. Why does this not contradict Th... |
s in the sequence are closed (but not bounded). E1 ⊃ E2 ⊃ . . . , { En} so { Fn = [n, the sequence ∞ Fn} { ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 244 Sets of Real Numbers Chapter 4 In a paper in 1879 Cantor described the following theorem and the role it plays in ... |
bserve in these examples that IN is closed (but not bounded) while A is bounded (but not closed). We shall prove, in Theorem 4.33, that a set A has the Heine-Borel property if and only if that set is both closed and bounded. Theorem 4.33 (Heine-Borel) A set A and bounded. ⊂ R has the Heine-Borel property if and only if... |
of E also compact? 4.5.18 " Prove Lindel¨off’s covering theorem: be a collection of open intervals such that every point of a set E belongs to at least one of the Let intervals. Then there is a sequence of intervals I1, I2, I3, . . . chosen from C that also covers E. C See Note 82 ClassicalRealAnalysis.comThomson*Bruck... |
irrational and f (x) = q if x = p/q where p/q is a rational with p, q integers and with no common factors. 79Exercise 4.5.5. Take compact to mean closed and bounded. Show that a finite union or arbitrary intersection of compact sets is again compact. Check that an arbitrary union of compact sets need not be compact. Sho... |
suring that of a limit statement. To keep it simple let us show that limx the expression → is smaller than ε. Some arithmetic converts this to = x2 9 If we insist that x2 /M, where M is bigger than any value of how big might x + 3 values of x), then this is not too big. For example, if x stays inside (2, 4), then enoug... |
all three of our definitions are equivalent we can use either a sequential argument, a mapping argument, or an ε-δ argument in our discussions of limits. Exercises x 5.1.23 Show that limx→0 | 5.1.24 Prove directly that the sequential definition of limit is equivalent to the mapping definition. /x does not exist using the ... |
ement can be made about boundedness away from zero. This shows that if a function has a nonzero limit, then close by to the point the function stays away from zero. The proof uses similar ideas and is left for the exercises. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) ... |
im x0 x → lim x → x0 | f (x) = L . | | | Since maxima and minima can be expressed in terms of absolute values, there is a corollary that is sometimes useful. Corollary 5.24 (Max/Min of Limits) Suppose that the limits exist and that x0 is a point of accumulation of dom(f ) and lim x0 x → max { f (x) = L and lim x0 x → g... |
ition (2008) 298 Continuous Functions Chapter 5 will hold at every irrational point x0 but must fail at every rational point. In the language of continuity we have proved that this function is continuous at every irrational point but discontinuous at every rational point. A curious function: It appears to be continuous... |
nd to focus attention on the concept that has now assumed such an important role in analysis. 5.4.1 How to Define Continuity " Enrichment section. May be omitted. Before we proceed to the present day definition, let us consider another notion. Even as late as the middle of the nineteenth century, some mathematicians beli... |
ow again < δ and x > 0, then ∞ ∞ → ) Let x0 ∈ 1/x0| − 1/x | (0, < ε. Rewriting this last inequality as ∞ | − x0| x0| suggests we try δ = εxx0. But δ should depend only on ε and x0, not on x. There is no δ > 0 for which the inequality < δ implies the inequality < εxx0 − x x | (3) x0| | − x | − x0| < εxx0 ClassicalRealAn... |
e leave this as Exercise 5.4.23. As a corollary let us point out that we can replace open sets by open intervals; thus to check continuity of a function f it is enough to show that f − 1((α, β)) is open for every interval (α, β). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2... |
between continuity and uniform continuity. Theorem 5.48: Let f be continuous on [a, b]. Then f is uniformly continuous. Proof. Our proof invokes a compactness argument. We recall from our investigations of compactness in Section 4.5 that there are several equivalent formulations possible. We shall use the Bolzano-Weier... |
then f (t) > c < xn = b < ε whenever − − /2, x + δ/2]. By Cousin’s lemma there exists a partition of [a, b], a = x0 < x1 < Suppose now that f (a) < c. The argument is similar if f (a) > c. Since [a, x1] = [x0, x1] for all x this way, we see that f (x) < c for all x [x0, x1]. Analogously, since [x1, x2] [a, b]. ∈ , and ... |
tinuity whenever k ≥ N . For all x satisfying xN ≤ L − It follows that x ≤ f (x) x0 we thus have L − ≤ f (xN ) < ε. lim x0 − → so f has a left-sided limit at x0. A similar argument shows that f also has a right-sided limit at x0. f (x) = L, x 335 Monotonic Functions Have Jump Discontinuities Recall that a function f is... |
ne with the property that f (x + y) = f (x) + f (y) for all x, y. Suppose that f is continuous at 0. Show that f (x) = Cx for all x and some number C. See Note 143 5.10.3 Suppose that f is a function defined on the real line with the property that f (x + y) = f (x)f (y) for all x, y. Suppose that f is continuous at 0. S... |
rucknerElementary Real Analysis, 2nd Edition (2008) NOTES 347 If x < y and x | y | − < δ, then either there exists i for which in which case or there exists i such that in which case xi−1 ≤ x, y ≤ xi, f (x) | − f (y) < ε/2, xi−1 ≤ x ≤ xi ≤ y ≤ xi+1, f (y) | + − f (xi) | f (x) | ≤ | f (x) | − f (y) ε 2 < − + f (xi) | ε ... |
nd A is dense in B, then A = B. Is the statement correct without the assumption that ⊂ B? A ⊂ 6.2.5 Is R Q dense in Q? \ 6.2.6 The following are several pairs (A, B) of sets. In each case determine whether A is dense in B. (a) A = IN, B = IN (b) A = IN, B = Z (c) A = INd) A = 2n , m Z, n ∈ ∈ IN , B = Q 6.2.7 Let A and ... |
on (2008) Section 6.4. The Baire Category Theorem 359 6.4.2 The Baire Category Theorem " Advanced section. May be omitted. We can formulate our result from our discussion of the game in several ways: 1. R cannot be expressed as a countable union of nowhere dense sets. 2. The complement of a countable union of nowhere d... |
that x0 is a limit point of K. Thus K has no isolated points. K such that 0 < L1. Then 0 < x0 − x1| ∩ | | The set K is called the Cantor set. Because of its construction, it is often called the Cantor middle third set. In a moment we shall present a purely arithmetic description of the Cantor set that suggests another... |
function. Exercises 6.5.11 In the construction of the Cantor function complete the verification of details. (a) Show that f (G) is dense in [0, 1]. (b) Show that f is nondecreasing on [0, 1]. (c) Infer from (a) and (b) that f is continuous on [0, 1]. (d) Show that f (K) = [0, 1] and thus (again) conclude that K is uncou... |
of points of \ We answer this question in this section. The principal tool is that of oscillation of a function at a point. 6.7.1 Oscillation of a Function " Advanced section. May be omitted. In order to describe a point of discontinuity we need a way of measuring that discontinuity. For monotonic functions the jump wa... |
− − − 1/3 + 2(1/9) + 4(1/27) + = · · · 1/3 (1 + 2/3 + (2/3)2 + (2/3)3 + . . . ) = 1. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 386 More on Continuous Functions and Sets Chapter 6 From the interval [0, 1] we appear to have removed all of the length. What is left over,... |
r every ε > 0 there is a finite collection of open intervals (a1, b1), (a2, b2), (a3, b3), (a4, b4), . . . , (aN , bN ) that covers the set E and so that (These sets are said to have zero content.) N (bk − Xk=1 ak) < ε. 6.8.15 Show that a set E has measure zero if and only if there is a sequence of intervals (a1, b1), (... |
The function has different right-hand and left-hand derivatives at x0 = 0 so is not differentiable at x0 = 0. ◭ f ′ − = − 1. → − x | x | x ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 400 Differentiation Chapter 7 Figure 7.2. A function trapped between x2 and x2. − Example... |
gnification / | J | | | In Exercise 7.2.26 we ask you to prove a precise statement covering the preceding discussion. Exercises 7.2.24 What is the ratio f (J) | | J | | for the function f (x) = x2 if J = [2, 2.001], J = [2, 2.0001], J = [2, 2.00001]? 7.2.25 In this section we have interpreted f ′(x0) as a magnification f... |
ussion we could begin by writing If we wished to formulate a proof of the chain rule based on the preceding g(f (x)) x − − g(f (x0)) x0 = g(f (x)) f (x) − − g(f (x0)) f (x0) f (x) x − − f (x0) x0 (5) which compares to our formula | | = | h(J) J | | g(f (J)) | f (J) | f (J) J | . | | | | ClassicalRealAnalysis.comThomson... |
a and the rule for sums. Note that the derivative of a polynomial ◭ is again a polynomial. Example 7.16: A rational function is a function R(x) that can be expressed as the quotient of two polynomials, R(x) = p(x) q(x) . This would be defined at every point at which the denominator q(x) is not equal to zero. Every ratio... |
of the function. The other direction is more subtle. How does information about the derivative provide us with information about the function? One of the keys to providing that information is the mean value theorem. Suppose f is continuous on an interval [a, b] and is differentiable on (a, b). Consider a point x in (a, ... |
e exists c (a, b) such that ∈ [f (b) − f (a)]g′(c) = [g(b) g(a)]f ′(c). − (12) Proof. Let φ(x) = [f (b) Then φ is continuous on [a, b] and differentiable on (a, b). Furthermore, f (a)]g(x) [g(b) g(a)]f (x). − − − By Rolle’s theorem, there exists c ∈ (a, b) for which φ′(c) = 0. It is clear that this point c satisfies (12)... |
of that theorem. Theorem 7.30: Let f be continuous on [a, b]. If D+f (x) > 0 at each point x on [a, b]. ∈ [a, b), then f is increasing Proof. Let us first show that f is nondecreasing on [a, b]. We prove this by contradiction. If f fails to be nondecreasing on [a, b], there exist points c and d such that a b and f (c) >... |
is differentiable and monotonic, then f ′ must be continuous on R. f ′(ξ1)f ′(ξ2) = 1. 7.10 Convexity In elementary calculus one studies functions that are concave-up or concave-down on an interval. A knowledge of the intervals on which a function is concave-up or concave-down is useful for such purposes as sketching t... |
x → limx limx → → x0 f (x) x0 g(x) = A B , unless B = 0. But what happens if B = 0, which is often the case? A number of possibilities exist: If B = 0 and = 0, then the limit does not exist. The most interesting case remains: If both A and B are zero, then A the limiting behavior depends on the rates at which f (x) an... |
there exists δ1 such that f ′(x)/g′(x) < r whenever (a, a + δ1). If a < x < y < a + δ1, then we infer from Theorem 7.21, Cauchy’s form of the mean value ∈ x theorem, and our assumption (ii) that there exists c ∈ (x, y) such that ∈ f (x) g(x) − − f (y) g(y) = f ′(c) g′(c) < r. (22) Fix y in (22). Since limx a+ g(x) = ,... |
e functions f (x) = x5 and g(x) = ln x. 7.12.4 Let f (x) = 1 1, and n = 2. Show that x+2 , c = − where, for some z between x and 1 x + 2 1, − = 1 − (x + 1) + (x + 1)2 + R3 R3 = (x + 1)3 (2 + z)4 . − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.13. Challenging ... |
nt line? 177Exercise 7.3.21. Use the idea in the example. If f (x) = x1/m, then [f (x)]m = x and use the chain rule. If then and use the chain rule. 178Exercise 7.3.22. Once you know that you can determine that F (x) = xn/m, [F (x)]m = xn d dx ex = ex d dx ln x = 1/x using inverse functions. Then consider xp = ep(ln x)... |
l Analysis, 2nd Edition (2008) 484 NOTES 221Exercise 7.13.13. This is from the 1946 Putnam Mathematical Competition. 222Exercise 7.13.14. This is from the 1958 Putnam Mathematical Competition. 223Exercise 7.13.15. This is from the 1962 Putnam Mathematical Competition. 224Exercise 7.13.16. This is from the 1992 Putnam M... |
two partitions π and πN to form a partition containing all points in either partition. The Riemann sum over such a partition belongs to the interval [m(π), M (π)] and also to the interval [m(πN ), M (πN )]. This completes the proof. (That I is unique is left as Exercise 8.2.2.) A special case of this definition and this... |
f f (x) b a Z f (x) dx ≤ a Z b g(x) dx. g(x) for all a x ≤ ≤ b, ≤ Proof. Consider a sequence of partitions πn chosen so that the points in the partition are closer together than 1/n. If S(πn, f ) denotes a Riemann sum taken over this partition for the function f , then In the same way for g we would have S(πn, f ) = li... |
be continuous on (a, b] and such that f (x) | absolutely convergent, show that so also is the integral g(x) | ≤ | | b a f (x) dx. for all a < x b. If the integral b a g(x) dx is ≤ R 8.4.5 For what continuous functions f must the integral 1 R f (x) converge? See Note 229 −1 Z √1 x2 − dx 8.4.6 Let f be a bounded functio... |
RealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 513 (3) ◭ Section 8.6. The Riemann Integral while if we choose associated points ηk ∈ [xk n − 1, xk] so that ηk is irrational Because of (2) and (3), the integral f (ηk)(xk − xk − 1) = 0. Xk=1 1 0 f (x) dx cannot exist. R Example 8.14... |
n add a few sufficiently short intervals to our collection. Thus we have proved that for any ε > 0 the set N (e) can be covered by a collection of open intervals of total length less than ε. It follows that N (e) has measure zero. But the set of points of discontinuity of f is the union of the sets N (1), N (1/2), N (1/4... |
. | | b a Z f (x) dx b ≤ a | Z f (x) dx. | Fundamental Theorem of Calculus The next two properties, 8.26 and 8.27, are important. They show how the processes of integration and differentiation are inverses of each other. Together they are known as the fundamental theorem of calculus for the Riemann integral. You should... |
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