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Calculus: Early Transcendentals

James Stewart 8th Edition

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06/20/2023

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7. Let {an} be a sequence and let set S = {an | n ∈ N}. If s0 is an accumulation point of S, prove that there exists a subsequence of {an} that converges to s0. (Compare to Exercise 4 from Section 2.5.) 8. Suppose that the sequence {an} converges to A, and suppose that B is an accumulation point of the set S = {an | n ∈ N}. Prove that A = B. 9. Use subsequences to find the limit of the sequence {an} with an as given. (a) an = n∑c, with any constant c ∈ (0, 1). (See Exercise 14 from Section 2.1.) (b) an = r^n with any constant r ∈ (0, 1). (See Exercise 2(c) above and Theorem 2.1.13.) (c) a1 = 1 and an+1 = ∑2an for all n ∈ N. (See Exercise 11(c) of Section 2.4.) Complete the proof in Example 2.6.6. Complete the proof in Theorem 2.6.4. THEOREM 2.1.13. If a real number, r, satisfies |r| < 1, then lim n→∞ r^n = 0. Proof. If r = 0, then r^n = 0 for any n ∈ N, and so lim n→∞ r^n = lim n→∞ 0 = 0, which is what we wanted to prove. If r ≠ 0, then we write |r| = 1 / (1 + b) for some b > 0, since |r| < 1. Why? Using the binomial theorem (Theorem 1.3.7) and the fact that b > 0, we obtain the expression (1 + b)^n = 1 + nb + n(n - 1) / 2 b^2 + ... + b^n ≥ 1 + nb (Bernoulli's inequality; also see Exercise 5 from Section 1.3 and ...

Adi S.

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Exercise 1.5.6. (a) Give an example of a countable collection of disjoint open intervals. (b) Give an example of an uncountable collection of disjoint open intervals, or argue that no such collection exists. Exercise 1.5.7. Consider the open interval (0,1), and let S be the set of points in the open unit square; that is, S = {(x, y) : 0 < x, y < 1}. (a) Find a 1-1 function that maps (0, 1) into, but not necessarily onto, S. (This is easy.) (b) Use the fact that every real number has a decimal expansion to produce a 1-1 function that maps S into (0, 1). Discuss whether the formulated function is onto. (Keep in mind that any terminating decimal expansion such as .235 represents the same real number as .234999...) The Schroder-Bernstein Theorem discussed in Exercise 1.5.11 can now be applied to conclude that (0, 1) ~ S. Exercise 1.5.8. Let B be a set of positive real numbers with the property that adding together any finite subset of elements from B always gives a sum of 2 or less. Show B must be finite or countable. Exercise 1.5.9. A real number x ∈ R is called algebraic if there exist integers a0, a1, a2, ..., an ∈ Z, not all zero, such that anx^n + an-1x^n-1 + ... + a1x + a0 = 0. Said another way, a real number is algebraic if it is the root of a polynomial with integer coefficients. Real numbers that are not algebraic are called transcendental numbers. Reread the last paragraph of Section 1.1. The final question posed here is closely related to the question of whether or not transcendental numbers exist. (a) Show that √2, 3√2, and √3 + √2 are algebraic. (b) Fix n ∈ N, and let An be the algebraic numbers obtained as roots of polynomials with integer coefficients that have degree n. Using the fact that every polynomial has a finite number of roots, show that An is countable. (c) Now, argue that the set of all algebraic numbers is countable.

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In section 1.3 we fill the gaps in ℑ by using numbers that satisfy the Completeness Axiom. (They turn out to be real numbers, but it's not obvious.) We could alternatively try to fill the gaps directly. Let's say that a number system containing ℑ has the Gapless Property if the following is true: whenever we have two sets A and B with a < b for all a ∈ A and b ∈ B, then there exists a number c such that x ≤ c ≤ y for all x ∈ A, y ∈ B. (This is very, very similar to the Cut Property in exercise 1.3.10.) Intuitively, we have the following picture. A hole in a geometric line divides the line into two parts: the part to the left of the hole (described by small numbers) and the part to the right of the hole (described by big numbers). The gapless property says that there is actually a number in between these two parts, thus plugging the hole. . . (Well, as long as the hole is not too big.) Now prove that a number system with the gapless property automatically satisfies the Completeness Axiom. Hint: Let A be a nonempty set that has an upper bound. You need to prove that A has a least upper bound. Define B to be the set of all upper bounds of A. There are two cases: A ∩ B = ∅ and A ∩ B ≠ ∅. Conversely, prove that a number system satisfying the Completeness Axiom (such as ℑ) automatically has the gapless property.

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Transcript

00:01 In this question, we are given a function f of x is equals to 3x squared separated by 4x and we are given the interval of x as 0 to 3 and for n is equals to 6 rectangles, we are asked to calculate the riemann sum at right end points.

00:25 So for that, first we have to calculate what is del x that is change in the interval.

00:32 So it's going to be 3 subtracted by 0 over the number of rectangles that is 6.

00:40 So that is going to be 1 over 2.

00:43 So the interval will be like 1 over 2 and adding 1 over 2 again, it's 1 and again adding 1 over 2, it's going to be 3 over 2 and next we'll be having 2 and 5 over 2 and 3.

01:04 So we are going to stop this with this as n is equals to 6 rectangles is the given one.

01:10 So to find the riemann sum at the right end points, we'll be having rectangles as 0 comma 1 over 2.

01:30 Next we'll be having 1 over 2 comma 1, 1 comma 3 over 2, 3 over 2 comma 2, 2 comma 5 over 2, 5 over 2 comma 3.

01:47 So these are the given rectangles.

01:51 So now let's moving on, let's calculate what is the riemann sum for the rectangle 6.

01:59 So this is going to be f times of 1 over 2, the only interval ranges plus f times of 1 plus f times of 3 over 2 plus f times of 2 plus f times of 5 over 2 plus f times of 3 and this is multiplied by the del x that is the difference...

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