## Taylor Polynomial

I just need help on how to start the problem, I'm not asking anyone to do it for me, I'm just slightly confused.

What is the degree of the Taylor polynomial needed to approximate sqrt(e) with error < 0.001. Use ex as your function, with x = 0.5.

I'm just honestly confused on where to even start, any help is greatly appreciated. Thanks.

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 Recognitions: Homework Help Science Advisor Look up some form of the remainder term for a Taylor series and try to bound it by 0.001. I'm guessing you know the Taylor series for e^x, right?
 Yes, by "Taylor series" I'm assuming you mean the Maclaurin series centered at 0, so: xn/n! By "remainder form" do you mean Taylor's inequality: R(x) = [M/(n+1)!][x-a]n+1 ? Thanks for the help.

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## Taylor Polynomial

You asked "how to even start". I told you how to start. If you had presented that info to begin with we could have skipped the preliminaries. Ok, so now try to bound your R by 0.001. What's the largest M can be if x=1/2?

 I'm sorry. Don't get mad at me, I'm really trying. How would i find out the M value without knowing what a or n equal? Taylor's inequality is what i really struggle with, I'm pretty good with figuring out series, but when it comes to "error" problems I'm a little shakey.

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 Quote by IntegrateMe I'm sorry. Don't get mad at me, I'm really trying. How would i find out the M value without knowing what a or n equal? Taylor's inequality is what i really struggle with, I'm pretty good with figuring out series, but when it comes to "error" problems I'm a little shakey.
Sure. What is M? That's the thing you really need to estimate. How is it defined?

 We want to show that as n goes to infinity, the remainder will go to 0, correct? Because doesn't that show that we have a series that adequately represents our function? But in this case it's asking for the "degree" and isn't "n" the degree of the polynomial? I'm not sure what "M" represents, is it the upperbound?
 Recognitions: Homework Help Science Advisor Ok, sure M will go to zero if it's a convergent Taylor series. But you want to know how fast it goes to zero so you can decide how many terms to keep. Doesn't M have something to do with the (n+1)th derivative of e^x on the interval [0,1/2]? Surely you can estimate that?
 The derivative of e^x will always be e^x, so no matter how many terms we add ( (n+1)th ), we will always have the same function. So, am i trying to calculate the values of x that will make e^x bound within [0,1/2] ?
 Wait, so does that mean the fx+1(x) must be less than M. So the nth derivative being ex: ex
 Recognitions: Homework Help Science Advisor Yes. That will bound M. Now write the whole remainder term in terms of n and try to find an n so it's less than 0.001.
 You lost me there, I'm sorry :/ I'm trying to find "n." I have all of my values except "n" and "a" so: 0.001 = [1.648/(n+1)!](0.5 - a)^(n+1) I'm up to this point...if I'm correct.
 Recognitions: Homework Help Science Advisor "a" is 0, isn't it? You were the one who said "Maclaurin series". So ok, 1.648*(0.5)^(n+1)/(n+1)!. Find an n so that's less than 0.001. This isn't all that hard, right?
 Haha, i get frightened to assume things that i don't know are 100% fact. Thanks for reassuring :) Oh, ok. So i just set 1.648*(0.5)^(n+1)/(n+1)! < 0.001 and solve for n? I get n = 4 by just plugging in different values for "n." Is there a more efficient way of doing this or is testing different values optimal? btw, thanks a ton for the help...life saver!
 Recognitions: Homework Help Science Advisor Testing different values of n is the perfect way to do it. Just looking at that expression you know n can't be too big. You're welcome.
 Thanks, i just have one more question. When we did this problem, our function e^x always had the save derivative. What happens if we get a different function where the f^n+1 isn't always the same?
 Recognitions: Homework Help Science Advisor Then you have to find some other way to estimate the max of f^(n+1) by some other method. Like if f(x)=sin(x), then it's pretty safe to say |f^(n+1)(x)|<=1. Stuff like that.