Help with Sequences and Subsequences?
The first three terms of {A(n + 4)} are A(1 + 4) = A5 = 5² = 25, A(2 + 4) = A6 = 6² = 36, and A(3 + 4) = A7 = 7² = 49. The first three terms of {A(3n - 1)} are A(3⋅1 - 1) = A2 = 2² = 4, A(3⋅2 - 1) = A5 = 25, and A(3⋅3 - 1) = A8 = 8² = 64. The first three terms of {A(2n)} are A(2⋅1) = A2 = 4, A(2⋅2) = A4 = 4² = 16, and A(2⋅3) = A6 = 36. I hope that helps!
Sequences and Subsequences?
Since S_0 is an accumulation point, for every positive integer k we can find a a_(n_k) in S such that | a_(n_k) - S_0 | < 1/k. Now we claim {a_(n_k) | k in N} converges to S_0. Let e > 0 (instead of epsilon) be given. By the Archimedian Property there is a positive integer k such that k > 1/e -> 1/k < e. Then | a_(n_k) - S_0 | < 1/k < e. This holds for any e, thus the subsequence converges to S_0.
Sequences and Subsequences?
define f(x) = 0,one million/(0,one million+x) for x > 0,one million. the form of f is the open era (d192e0c4ad64a9c35fe32972477e4cd8d192e0c... f'(x) = -- 0,one million / (0,one million+x)^2 so f is strictly reducing . for the reason that f(0,one million) = 0,one million and f(0,one million) = 0,one million/2d192e0c4ad64a9c35fe32972477e4cd... f maps (d192e0c4ad64a9c35fe32972477e4cd8d192e0c... = J_d192e0c4ad64a9c35fe32972477e4cd8 0,one million-0,one million onto J_2 = (0,one million/2d192e0c4ad64a9c35fe32972477e4cd... 0,one million)0,one million f maps J_2 to J_3 = (0,one million/2d192e0c4ad64a9c35fe32972477e4cd... 2/3)0,one million and so on. observe that |f'(x)| < 4/9 < 0,one million/2 if x > 0,one million/2. consequently if we define durations J_(n+0,one million) = f(J_n)0,one million J_(n+0,one million) is a subinterval whose length is under 0,one million/2 the dimensions of J_n. If m & n are > N then x_m & x_n are in J_Nd192e0c4ad64a9c35fe32972477e4cd8 so |x_m -- x_n| < 0,one million/2^(N-0,one million) consequently {x_n: n > 0,one million} is a Cauchy sequenced192e0c4ad64a9c35fe32972477e4cd8 so converges to a decrease L in [d192e0c4ad64a9c35fe32972477e4cd8d192e0c... of course L = f(L) = 0,one million/(0,one million + L)0,one million so L^2 +L -- 0,one million = 0,one million. The useful root of this quadratic is ( -- 0,one million + sqrt(5))/2 . ------------
Sequences and Subsequences?
1. If A is an infinite and boudned subset of R, A has a limit point. 2. If {a_n} is a bounded sequence, {a_n} has a converging subsequence. The two statements, based on the Bolzano-Weierstrass theorem, are actually equivalent. To prove the (→) direction, let {a_n} be bounded. If {a_n} is finite .... (skipping since this is not relevant to my question) If {a_n} is infinite, let A = {a_n}, which has a limit point by statement (1). For all ε, N(p, ε)\{p}∩A is not empty. Thus, for all ε, select k such that |a_nk - p| < 1/k. Then lim k→∞ a_nk = p. Elsewhere in the book, the author used the same contruction to prove that if a set has a limit point, q, there is a sequence {q_n} converving to q. for all ε, select n such that |q_n - q| < 1/n. Then lim n→∞ q_n = q. In the first case, the author proved the existence of a converging subsequence. Then, in the second case, the author proved the existence of a converging seqence using the same construction. I recognize how, in a bounded real set, you can pick out points to build a sequence to converge to any point in the set, and sometimes to a point outside the set. But I am not comfortable with flipping between sequences and subsequences. When should you be building sequence, or subsequence?
HELP ASAP Do subsequences of infinite sequences have to be infinite? Or can they be finite?
And then... I need an example of the following or why it is not possible: a) A sequence that does not contain 0 or 1 as a tern but contains subsequences converging to each of these values. Does this work? {1/2, 2, 1/4, 2^(1/2), 1/8, 2^(1/3), ...} bc { 1/2, 1/4, 1/8,...} converges to 0 and {2, 2^(1/2), 2^(1/3)} , converges to 1. b) A monotone sequence that diverges but has a convergent subsequence. Would something as simple as {1, 2, 3, 4, ...} and then {1,2,3} is convergent work?? c) A sequence that contains subsequences converging to every point in the infinite set { 1, 1/2, 1/3, 1/4, ...} No clue. d) An unbounded sequence with a convergent subsequence. I feel like this doesn't work unless the subsequence can be finite.... e) A sequence that has a subsequence that is bounded but contains no subsequence that converges. It would have to not be monotone if at all possible??
Subsequence question?
define f(x) = 0,a million/(0,a million+x) for x > 0,a million. the form of f is the open era (d192e0c4ad64a9c35fe32972477e4cd8d192e0c... f'(x) = -- 0,a million / (0,a million+x)^2 so f is precisely lowering . because f(0,a million) = 0,a million and f(0,a million) = 0,a million/2d192e0c4ad64a9c35fe32972477e4cd... f maps (d192e0c4ad64a9c35fe32972477e4cd8d192e0c... = J_d192e0c4ad64a9c35fe32972477e4cd8 0,a million-0,a million onto J_2 = (0,a million/2d192e0c4ad64a9c35fe32972477e4cd... 0,a million)0,a million f maps J_2 to J_3 = (0,a million/2d192e0c4ad64a9c35fe32972477e4cd... 2/3)0,a million and so on. observe that |f'(x)| < 4/9 < 0,a million/2 if x > 0,a million/2. hence if we define durations J_(n+0,a million) = f(J_n)0,a million J_(n+0,a million) is a subinterval whose length is under 0,a million/2 the dimensions of J_n. If m & n are > N then x_m & x_n are in J_Nd192e0c4ad64a9c35fe32972477e4cd8 so |x_m -- x_n| < 0,a million/2^(N-0,a million) hence {x_n: n > 0,a million} is a Cauchy sequenced192e0c4ad64a9c35fe32972477e4cd8 so converges to a cut back L in [d192e0c4ad64a9c35fe32972477e4cd8d192e0c... for sure L = f(L) = 0,a million/(0,a million + L)0,a million so L^2 +L -- 0,a million = 0,a million. The beneficial root of this quadratic is ( -- 0,a million + sqrt(5))/2 . ------------
Convergence and Subsequences Help!!?
define f(x) = 0,a million/(0,a million+x) for x > 0,a million. the variety of f is the open era (d192e0c4ad64a9c35fe32972477e4cd8d192e0c... f'(x) = -- 0,a million / (0,a million+x)^2 so f is strictly reducing . considering the fact that f(0,a million) = 0,a million and f(0,a million) = 0,a million/2d192e0c4ad64a9c35fe32972477e4cd... f maps (d192e0c4ad64a9c35fe32972477e4cd8d192e0c... = J_d192e0c4ad64a9c35fe32972477e4cd8 0,a million-0,a million onto J_2 = (0,a million/2d192e0c4ad64a9c35fe32972477e4cd... 0,a million)0,a million f maps J_2 to J_3 = (0,a million/2d192e0c4ad64a9c35fe32972477e4cd... 2/3)0,a million etc. word that |f'(x)| < 4/9 < 0,a million/2 if x > 0,a million/2. subsequently if we define durations J_(n+0,a million) = f(J_n)0,a million J_(n+0,a million) is a subinterval whose length is under 0,a million/2 the size of J_n. If m & n are > N then x_m & x_n are in J_Nd192e0c4ad64a9c35fe32972477e4cd8 so |x_m -- x_n| < 0,a million/2^(N-0,a million) subsequently {x_n: n > 0,a million} is a Cauchy sequenced192e0c4ad64a9c35fe32972477e4cd8 so converges to a decrease L in [d192e0c4ad64a9c35fe32972477e4cd8d192e0c... needless to say L = f(L) = 0,a million/(0,a million + L)0,a million so L^2 +L -- 0,a million = 0,a million. The constructive root of this quadratic is ( -- 0,a million + sqrt(5))/2 . ------------
Real Analysis Help (Sequences and Subsequences) Prove or provide a counterexample of the statements?
Hello Everyone! I am currently taking a Advanced Calculus a.k.a Real Analysis course and am studying for an upcoming exam. Fortunately, I've been able to understand most of the concepts I've reviewed, but as I was going through the questions at the of of the section of subsequences, I noticed there were a question I could not answer. Here it is: (It has several sections to it). Prove or give a counter example: a. Every oscillating sequence has a congruent subsequence. b. Every oscillating sequence diverges. c. Every divergent sequence oscillates. d. Every bounded sequence has a Cauchy subsequence. e. Every monotone sequence has a bounded subsequence. f. Every convergent sequence can be represented as the sum of two oscillating subsequences. Can some one please help me out with these questions? I want to do really good on this test. :) Thank you! :)