5

17. Fix & > 1. Take x > 4, and define Jkvini; 66 } '72 & +xn a-x Xn+[ = =Xn + T+xn T+x (a) Prove that x1 >xa>xs> (6) Prove that Xz <...

Question

17. Fix & > 1. Take x > 4, and define Jkvini; 66 } '72 & +xn a-x Xn+[ = =Xn + T+xn T+x (a) Prove that x1 >xa>xs> (6) Prove that Xz <x <xs < (c) Prove that lim x = Va_ (d) Compare the rapidity of convergence of this process with the one described in Exercise 16. 18. Replace the recursion formula of Exercise 16 byXn+1P_ Xn + X;p+1 pwhere p is a fixed positive integer, and describe the behavior of the resulting sequences {xa}:

17. Fix & > 1. Take x > 4, and define Jkvini; 66 } '72 & +xn a-x Xn+[ = =Xn + T+xn T+x (a) Prove that x1 >xa>xs> (6) Prove that Xz <x <xs < (c) Prove that lim x = Va_ (d) Compare the rapidity of convergence of this process with the one described in Exercise 16. 18. Replace the recursion formula of Exercise 16 by Xn+1 P_ Xn + X;p+1 p where p is a fixed positive integer, and describe the behavior of the resulting sequences {xa}:



Answers

In Problems $61-64,$ use the Squeeze Theorem to establish convergence of the given sequence. $\left\{\frac{\sin ^{2} n}{4^{n}}\right\}$

All right. In this problem, we want to show that each sequence converges finding its limit. We are related function or via the squeeze zero. We have the SN has term and squared minus four over n squared plus and minus to this Russian challenges our understanding of sequences and convergence. So remember that you can use the following properties of conversion sequences to 89. Evaluation of the sequence convergence will utilize these properties in conjunction with As a question suggests a related function and squeeze here and that's probably going to use the related functions. So we take a little bit as X approaches infinity of related functions, effects with limited X approaches infinity fx equals L is the same as limited. That approaches infinity SN If the limit converges, we know the sequence converges to that value. Thus we have limited and approach infinity of our sequence equals limited X approach infinity x minus four over X. Where x minus two, dividing in return by x squared gives the following limit, which evaluates to one minus four. Infinity squared over once was one of infinity minus two infinity squared, divided constantly infinity, go to zero. That's our sequence converges as one minus 0/1 plus zero minus zero equals one a convergent sequence.

In this problem, we wish to show that each sequence converges, finding this limit via related function or the squeeze through. The sequence in question is S. N. Um I. M. Term six under the fourth minus 5/7 under the fourth plus three. So this question is child, your understanding of convergent sequences in order to solve what we want to do is utilize properties of convergent sequences. These are properties one through 60%. Here we utilize this juncture with for this problem, the related function approach to limit convergence to do so, we can be limited. X approaches infinity fx related function for S. M. It's a limited exposure to the F X. L. That's the same convergent value for S. M. That is the limit converges for the function, the sequence converges to the same value so limited and approaches infinity of sm is limited expertise in 36 to the fourth minus 5/7 X to the fourth plus three, dividing every turn by extra the fourth, transformed our limit to six minus five women fit into the fourth over seven plus three or infinity. Fourth, dividing a constant by infinity, goes to zero. So we're left with convergent value. Six minutes, 0/7 plus zero equals 6/7.

Something from Anna and about you and your money, My amber big a boy and a guitarist Regimental, infinity, infinity and be here will be which is owned one. And now we wouldn't use their limit comparison. That's where I could bear when there's a mission on one of them. And er on, then Yeah, listen, goes from two to infinity on my aunt will be, ah, being good and be and smaller. And And the one therefore, I wouldn't limit here and goes to infinity off the, uh and I'll kill anybody. But in about big terms of the reciprocal, this one be any one on one so that you get a limit and goes to infinity. And here we're having a and about you And who had end off being Minister, huh? Okay on. Then I from hand again, huh? Because the gladness ah, and will be a smart smarter than zero. Now, founders, one *** too Infinity on dhe means stop. Doesn't implies that the seriously Yeah, by Lim comparison test part I see me again deserves Naomi divergent

Hello and welcome. We are looking at Chapter two, Section one Problem 55. Now we're given a new type of equation. Well, new to me, maybe not new to you. The Ricker equation. It looks like this. The general form at least. All right, so we're just introducing an exponential. Really? This is be here and were given some parameters values to plug in here, this specific case. So this is what specific to this question here we are asked to plot some terms to see how it behaves. And then kind of in further convergence, Convergence s O. I did use a spreadsheet to graft this. And so, uh, it looks like this. Ah, sequence is converging. Thio this decimal value here Something to do. 0.6931 The last six terms here are all 6.6931 So it's getting closer and closer to some value. This 0.6931 fish. Um, so you can say this appears to converge? We're not sure exactly. Toe what? It's just near 0.6931 All right. Uh, so moving forward from there. It says if so, if it appears to be convergent, estimate the limit and then assuming the limit exists, that is its exact value. So we're gonna estimate 0.6931 and we're gonna assume the limit exists. So just like we've done with a few different problems here, we're gonna make an assumption if we assume the limit exists, we're assuming that the limit as C goes to infinity of except e converges to some value. We're assuming the limit exists. So it goes to l just call it hell. And if we make that assumption that it follows logically that this also except E plus one the next term, uh, the limit as to goes to infinity of that same sequence, basically just one term down the line, it's gonna obviously converge to the same value. All right, So what we do then, is we take the limit of both sides of our equation so I could plug in my see value of two, plug everything in. I'm taking the limit as he goes to infinity of both sides. So the left hand side is definitely getting go to l. That was part of my assumption. My right hand side, it's going to, um Well, I cannot hold the two out of the limit on the right hand side is a little bit more complicated, so let's focus on that first. Yeah. So let's continue this on. Ah, the next page here. So I can sure the work more clearly. All right. So, uh, limits using properties of limits, they're pretty flexible things. Um, So I can write this as the limit s t goes to infinity of except E. And then the limit of a product, you can take the limit of each of the factors. That's one of our linen properties. So by our assumption, we have l equals two. And then this here goes to el by our assumption. And then what weaken do, uh, with this limit here, we can actually move. The limit is just and a constant value 2.7 something so I can actually move the limit into the exponents. That's one of our limit properties as well. So little squished here, but, um, the the limit is in the exponents with the negative x two x sub d. So, after we done resulting all of the limits, it just looks like this. L equals two times at all times E to the negative, Al. So remembering what we know about negative exponents this we could also write this. I would suggest it is running this times one over easy bl or that's the same is over e to be out. So if I want to get, um, l by itself, you could cross, multiply, could do different things. We just want, um, everything in the new greater so I'll go ahead and do it and more steps than I need to. Just to be clear, I just multiply e the lt other side. Now I'm gonna divide both sides by l get you to the l equals two. And then to get rid of that e the exponential I need to use natural logs or natural log cancels with E. If I do it to one side, I have to do it to both sides. And so then I end up with l equals natural law of two s o weaken. Check to see if this works with our our graph by typing in natural log of two into our calculator. If you type in natural, log on to that's what it's approximately 0.6931 So, this, um, this is consistent with our graph. I'll just show you that one more time. This is about 0.6931 from our grasp. All right, so, uh, double checking the question, make sure we answer the whole thing. Um, if it seems to be convergent, estimate the limit. Assuming Dylan just calculate its exact value. And that's exactly what we did. We are.


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