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For each of the following (a) (2 pts) Find S1o(b) (5 pts) then find a bound on theremainder and then (c) 5 pts) Find a N so that Sw is within 0.001 of the limit. Sh...

Question

For each of the following (a) (2 pts) Find S1o(b) (5 pts) then find a bound on theremainder and then (c) 5 pts) Find a N so that Sw is within 0.001 of the limit. Show all n=ll work s0 / can see how you got your bounds and N.4(3" ) n_1 5" +2

For each of the following (a) (2 pts) Find S1o (b) (5 pts) then find a bound on the remainder and then (c) 5 pts) Find a N so that Sw is within 0.001 of the limit. Show all n=ll work s0 / can see how you got your bounds and N. 4(3" ) n_1 5" +2



Answers

Consider the following sequences defined by a recurrence relation. Use a calculator, analyti- cal methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=\frac{1}{2} a_{n}+2 ; a_{0}=5, n=0,1,2, \ldots$$

So for part A, let's use a graph to find this limit and to the fifth divided by in factorial. So let's go to our graph here. And you can see once you get to about two ten term in the sequence, it looks like they started conversion to zero quickly. So my guess here is that the limit is zero. Yeah, So for part B, we want the smallest value of end such that the following is true in this case. In general, you should write and minus the limit, but from party A we know that this equal zero So we're looking for the smallest value of capital and such that this implies a M wishes already positive is less than epsilon. So for the first case, we'LL do absalon equals zero point one and let's go ahead and find the value. And in the calculator so zero point one, you can see that the first time that happens the first time you're below point one is when n equals ten. However, the definition requires little and bigger than end. So we should really do and equal to nine. And if you take any little and bigger than this And nine, That means you're starting at ten. And that's what we want. Now we'LL do epsilon equals zero point zero zero one So we'LL go ahead and have to zoom in a little more here until we get a point zero zero one showing up. No, zoom in a little more And there we go. Point zero zero one we see it. Now let's go back a little bit to the graph and here we are. We see an eleven, we're still above point zero zero one. But if we go over to twelve, then we're below point zero zero one. So the first time we get below point zero zero one is that twelve? So we should go ahead and make this eleven and then any little and bigger than eleven. This is a problem to say and is bigger than her equals a twelve. This is why we're subtracting what? From the end. Wait. And that results part B

This question sequence under form, I only go, Joe for about and plus five and cantorial, you're running by and fraternal just to end. And where did you find a limit? Uh, I am here and we come from a tea room 10.6. It would have the limit ever him that I in is the girl room that I and would be most modern and api and and then we have the limit off the i n m a p in we go ju zero. But if the gravel rate I'm the, uh do chickens under seven right here. And then they found a limit were I am a B and we go to one as in just infinity. And here look in the function here we are given, uh, exponential and a factorial here for us. We see that the exponential form about and from the list of the function in a serum 10.6 the exponential would be must on the graboid when we must Smarter than the factory. Oh, and therefore in discussion. Here is Angus to infinity Doesn't will be the same. Is the limit off the this phone about Ah, And here we be dominated by the inventory of there. So we have 25 inch Victorio and now something the to about and is much smaller than, in fact Oreo. Therefore, we have only in the in Vitoria. Yeah, it's in Vegas to infinity and then we'll see. Then we can against in the end for during with in Victoria Andi and we have only you go to five year, a communal limit of a sequence.

Conduct the Inter Go from a to B How the function f x t x can be turning to the limit form off the norm of the delta goes to zero So much angles from one of your n f on the c I times Delta x i in this question we even limit after submission I goes from one of Japan. The number Datta goes to zero off the sixth See, I terms the form on the C I square times that the X I yeah, we need to turn it on in June the integral form. So we're given that the interval and is between the show and far therefore which you get now this will be the lower limit and this will be the upper limit Yeah, the function in suburb it on this form So whenever we see the sea, we need to replace it as an ex to be negative The six x times four minus x square T x and this will be the answer

Okay, So for this problem, we're finding the limit as an approaches. Infinity, um, for a right remind some with ends of intervals. So the first step is to realize what, how to represent the right response. Amber and seven jiggles. So are are, they said is equal to Dr X Times the sum of the function values that right endpoints of the 70 bolt. So that could be written As this sum room J is equal to one to end of, uh, a plus J. Delta X. And if you need to review this formula or help understand this, look through the chapter, they give a very thorough explanation of what this means. So now that we have this of, let's write what our formula is what our questions in the key for this particular problem only one thing to mention Delta X is equal to be minus a over and and B minus. The Vienna are the end points of the interval that were working with. So in this case, f of X is equal to is equal to three x plus six, and our interval is from 1 to 4. Okay, so first, let's frying del tax Delta X is going to be four minus one over end, Sam, is that it's equal to three over. So knowing that we can rewrite this using these values so in this case, are based on Is there the equal to Rio over and Giselle to x times This sum, Brian M. J. Is equal to one. Um, uh, yeah, of one waas, uh, three times J. So now we know what f back is, so we can actually just plug this into after backs. That's gonna give us three and times this sum, uh, that you were gonna have three times. One less three j over end plus six. So now let's continue to simplify this further. We can distribute this through here and at the six. Um, and that's gonna give us three over end. I'm cece some, and we're gonna have nine plus nine j over at. Okay, so now we can do is we can, um, take the sum of each of these turns so we can split it up into two different sums. We do that, we get three over end times so we can rewrite this, but we have the sum of nine. Um So that's just gonna give us nine n plus. And we can factor this nine over and out. So you have nine over ed times. Um, the son of J equals one Thio a J. So now that we have this, we can apply our power some formulas to figure with this, figure out what this is gonna be. So we still have three over end times nine and plus nine over end times. And this is coming from the powers of formula we get and times and plus one over, too. Let me go to the next page. Um, so now we have When we worked that out, we get 27 times nine and plus Sorry, we get 27 times one plus and over to, uh, plus one over to it. Um, and just through the reminder, this is what you're representing as our basement or the right three months on end of intervals. So now that we have this, we're ready to take the limit. So the live it as an approaches, infinity of this is going to be equal to 27 times re over too, plus zero. Um, and that equals 81 over, too. And so this is our final


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