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Use the Taylor polynomial Ta(x) to estimate the following expression correct to five decimal places_ cos(80o _...

Question

Use the Taylor polynomial Ta(x) to estimate the following expression correct to five decimal places_ cos(80o _

Use the Taylor polynomial Ta(x) to estimate the following expression correct to five decimal places_ cos(80o _



Answers

(a) use a Taylor polynomial of degree 4 to approximate the given number, (b) estimate the error in the approximation and (c) estimate the number of terms needed in a Taylor polynomial to guarantee an accuracy of $10^{-10}$ $$\ln (0.9)$$

It's freezing. Taylor polynomial is to estimate what this would approximate to Within .01 here. So first let's get this into radiant format. We have sign of so 10 degrees, multiplied by pi over 1 80 is equal to pi over 18. And the taylor polynomial that we're gonna be using, it's going to be from sine of X, which is equivalent to x minus X. To the third over three factorial Plus X to the 5th over five factorial -X to the 7th over seven factorial. So we have this here and so we're going to be using is plugging in pi over 18 for X. And so we put that in Get Pie over 18. It's a pie over 18. So the third power To buy three Victoria which is defined by six. So plug that in here. This result here gives us points through zero 0008 So on which is less than one over 100. So then if we take just this first part into account here Is Equal 2.174 or five. The approximation here incident sign of pie or 18. Approximately 1745 here. So if we take a look at these two, first terms, if we plug it into the calculator, get that it's exactly equal to .173648. So on we see that the first two terms are indeed 1st 2 decimal places are indeed the same. And so that's our approximation there

They want us to estimate natural log of 1.5 using 1/4 degree Taylor polynomial. So what we're going to do? So this is going to be approximately equal to well, in number 23 we found earliest. They just tell us that natural log of X is equal to this year. So I say it would be valid for us. Just go ahead and use that since it was from a previous problem. That way we just save a little bit of time from actually deriving this ourselves. So if it's the fourth degree, remember, we're just gonna change that infinity of there to afford and then plug in that started before we do that. Let's just figure out what is the fourth degree polynomial so p sub four of X is going to be equal to. So we're going to start plugging everything into here, and it looks like it would be so first positive. Eso just be X minus one minus X minus one squared over two plus X minus one Cute over three and then my sex My swan to the fourth over four. All right, so now we can go ahead and plug this end to here that the knows all the X minus ones are just going to be 0.5 article Unplugged added. And so I remember the reason why we want to do this or why we can use this Taylor polynomial is because it's a lot easier to plug in point or 1.5 into this Taylor expansion than it is to directly plug it in four. Just a natural log of it, because I honestly have no idea how to plug in 1.5 international log of X by hand. So using this Taylor polynomial to estimate it gives us a good way to do something by hand to figure it out. And I plugged this in earlier and least when I plugged into desk most, it told me it was about 0.48 Ah, 79 and it will just keep on going. But this is just where symbol of cut it off when I plugged it it. So this is going to be our estimation or natural. That 1.5 now, the next thing they want us to do is to estimate, so I'll just put a appearance so the next part for B. They wanted to estimate the error. So that's why I have this remainder up here. And remember, the think we're going to do is go ahead and first take the absolute value of this. And then this will give us an estimation of our error. All right, so for us, we are interested in the fourth degree. So let's just go ahead and plug that in. So in is going to be four here. And so that means we're going to get the fifth derivative. We'll have five factorial or very good. Instead of plugging that all there, I'll just do this. So I'll say and is equal to four. And then we also know that C was equal toe one and X was equal to 1.5. So I'm just gonna plug all these India just so we kind of have that still so b r in of are four actually are four of 1.5 is equal to so the this derivative at Z, whatever that value might be. But I never got put out some values around that, and then it would be over five factorial times 0.5 to the fifth. Okay? Now what we need to do is figure out what is this f to the fifth of Z. So, actually, something else we also found in number 23 Waas that or natural log of X, the inst Areva tive. And actually, I'll just go ahead and say in E no. Yeah, we'll do that. Derivative of this is going to be X to the minus one in plus one than and minus one factorial all over um X to the so, actually, for us, we cared about ZIS. Let me just plug Z and wall about it. Yeah, So if you want to see how this was actually derived other than we just kind of throwing it up on the board, Yeah, so go check out number 23 because I kind of go through how we can find this pattern. All right, But for us, we care about f of five. So let's go ahead and plug that in over here. So it's going to be so if I plug in five would be negative one to the six. So that just goes away. Then we have five factorial, So I'm sorry for factorial because it's five minus 14 factorial and then just over Z to the fifth and then the green parts. So all I did was write this. Now the rest of the green parts are just going to be 0.5 raised to the fifth over five factorial. All right, so now notice the 54 factory will just simplify down toe 1/5. And then we got to do something with this Z here. So I'm gonna across those out and then just go five right there, so, yeah, we need to do something with this Z now so we can get a good estimation of it. Well, if we go back up and look, we see the Z. Since it's in between, X and C has to be some number in between one and two since sexism between one and two. So what I'm going to do down here, So I'm going to write one less than Z less this, uh, see, but two. Okay, well, a trick to do when you have something like this is to reciprocate it since we have one over Z essentially over here and then So the two is going to be less than now. So be one half and then on the other side to be 1/1, which is just one. So, for our intents and purposes, one of Rosie is going to be strictly less than one. So we can replace that Z here with a one. So it be less than so. 0.5 raised to the fifth over, one to the fifth. And so that's just going to be 0.5 raised to the fifth. And then let's see what we get for this. So five raised to the fifth power that looks like we get something. Actually, maybe this is actually equal to 3.125 times, 10 to the center. And I don't think that's rounded it all. Let me double check. Yeah, so that should actually be exact. So our error here, we expect to be less than this. 3.1 25 times 10 walked up to the seventh, but to the negative. Especially a negative right there. Yeah, right. Yeah. And if we look at that, that that's a really small number, and it kind of makes sense. Why? If we looked up here, why the 1st, 2nd, 3rd, 4th, 5th decimal place of this was exactly like what our calculator would give. Right. So this would be an error for this now for the last part for C, they want us to find when error was less than 10 to the 10th. What value for end We would need for that. Well, actually, one thing I just realized I didn't do over here, um, forgot about this five. So there should be times five over here, divided by five. So actually, let me fix that really quickly. So divided by five, they onto you five. So it looks like this is actually let me just erase this. I'm glad I caught that. So 6.25 times 10 to the negatively. And then the error again is going to be 6.25 times 10 to the negative, eight less than that right now that looks more like what I got earlier. I worked this out right now for the error here to be less than to the 10. We're going to pretty much do what we did up here. But just use this formula now without having to do everything else So what? I'm gonna do it. So I was kind of kind of pick this up and bring it down. Uh, just cause I'm too lazy to write it out again and what we're going to do is set this less than or equal to 10 to the 10th. So let's figure this out in general and then we can go from there. Actually, one thing I'm going to do again since I'm lazy is picked this up and bring it down with me just because it will save us a little bit of time for writing it down again. All right, so we're going to go ahead and just plug this in to here and doing that will give us. So be negative one to the actually. So that is for the insta riveted. We want the plus one derivative. So that's actually going to be so We just add one toe. All those two B minus one to the and plus two, which would be the same thing, is just to the end in factorial over Z to the end, plus one more than times. So again we would have X minus one. Because our center is that one, then in plus one all over and plus one factorial. Okay, um so knows we could go ahead and simplify this down a little bit, just like before. The factorial is will simplify down. So we get negative one to the and actually let's go in and take the absolute value of this while we're at it. So take absolute value of all of us. So again that negative one goes away just like it did before. The factorial will cancel down. And then we're going to get next by this one on since Z and exploits would have the same power I was going to write it. Is this here? And then we have one over and plus one. And then we still have our absolute values right here. Right now. Since we have this, we can go ahead and plug in, but our exes. So if X is equal to 1.5, so again we get that 0.5 So this is going to be equal to 0.5 over Z raised to the M plus 1/1 plus one. And just like we did before, where we said that one over Z was going to be strictly less than one will apply that again. All right, so we want this to be less than 0.5. So the Z just becomes one Essentially. So just be Let me scoop that down world 0.5, raise to the end, plus one, and then they'll just be over in plus one, because again, that's just going to be one. Once we apply this inequality right here, right? And now we want this expression to be strictly less than 10 to the negative tents. But to be honest, all right, I don't know how to solve this exactly without, just, like, plugging in a bunch of numbers. So what we're going to do is the following, So I'm going to go ahead, scoop this down one. And we know that if we replace in plus one with just one, since it is going to be some large number, this expression should be larger than this. Here and now. From here, we can actually go about solving this algebraic lee. So, um, let's actually just figure out where this is equal to it, since everything else is strictly less than at some point right. So first that we would do for this is just take, uh, our favorite log on either side. So I'll just do natural log. So that's going to give us in, plus one natural law of 0.5 and then this is equal to over here. Never yet. Listen to log Base 10. And the reason why I want to log based in is because over here, the 10 and the negative 10 counts out. We'll just get minus 10 and then to solve for N. We would divide that over so in plus one is equal to negative 10 over log of 0.5 and then we would just subtract one over and that's going to give us in is equal to negative 10 over log of 0.5 minus one. And if we were to plug this in, let's see what we get. So negative 10 divided by log of point. Oh, all right, so that's about seven. We subtract one off that, So this is going to give us something around 6.68 So what? This implies. So this implies that if n is equal to seven, then the absolute value of our of in 1.5 is going to be strictly less than, um, 10 to the negative 10. So, yeah, What we wanted to find here was this value right here and so again, Like I was saying, if you really don't want to, like, do this algebraic Lee at this point right here, you could just keep plugging in values of in to see which one works. So if you were to plug in six, it will still be larger than 10th of negative 10th. But if you plug in seven, then that's when it would actually roll over to that next one. But sometimes, if it's really large, won't be able to do that. So doing a little bit of algebra tricks here to actually solve it Algebraic lee or get a algebraic estimation for it will really help us, but yeah, so that's how we could go about finding what degree polynomial we will need

Okay. So the first thing we're gonna want to do and we're finding a 4th taylor polynomial is we're gonna need to find the first four derivatives of our function and we're told that we're doing this at X is equal to zero. So we need to find what those values of those derivatives are at X is equal to zero. Good thing for us is our derivative of fx. Since our function is just fx just gonna B E to the X. This is gonna be the same for all of our derivatives. So that means all of our derivative values are going to be the same at X is equal to zero and they're all going to be equal to eat a zero or one. So we can say f of zero, it's equal to f prime of zero, sequel to double prime of zero and so on. Which is just equal to one. If we go ahead and look at the formula to find a taylor polynomial, You go from N is equal to zero to in this case force and you want the 4th Taylor polynomial And it's the interim motive of f at in this case zero. Right about in factorial, multiplied by X to the and power. So this is going to be equal to one plus X plus x squared divided by two plus Next the 3rd divided by six and then plus X to the 4th. But by four factorial, which is 24. So now that we have this fourth taylor polynomial of our function F of X. We want to use it to try and estimate. But E to the .01 power is so we can just plug in .01 into our taylor polynomial. See what we get as an estimate. So Can say P four, Which is the Taylor. Polynomial. of 0, 1 is equal to one plus 0.1 Plus point of one Square divided by two Plus Point No 1 to the 3rd divided by six. And then post point no one to the fourth divided by 24. And if you plug this into a calculator you should get a value that's equal to about one point oh one oh five. So this is a pretty good estimation of our function E to the X at X 02 point oh one. I'm actually just gonna plug that into a calculator to see how close we really were. Um so we have e to the .1 power and that is very close. It's actually The the .1 power is 1.0105 like we put down but then it has some more stuff which is where our error would come in. So we're very close to each point no one but still off by a very small decimal

Taylor polynomial to figure out exactly what this evaluates to And get it close to .011 of an approximation there. Incident let's go ahead and use cosign effects first. The taylor polynomial for that which is equivalent to on minus X squared over two Plus. Next to the 4th over four factorial I guess XS six or 6 Victoria and so on. And then we're going to plug in. So now six in terms of degrees multiplied by that point pi over 1 80 Converting it to radiance. We have pie over 30 and so we plug in Pie for 30 for X. So we have one minus five or 30 squared over to Look supply over 30 to the fourth of her 24. Excellent. It's about me see here talking it in there. So we get that this term right here. Actually if we look at the This term pi over 30 sixth over 7 20. So that's kept. This term is actually less than one over 100 here. Which case we add up these three terms Get the approximation to be .9947. Yeah, If we plug in, co sign off by over 30 into the calculator gets equal 2.9945. So on you see that the first two terms Are indeed accurate in terms of the approximation two points or one. So this is that there


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