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Homework Help: Taylor Polynomial

  1. Feb 18, 2010 #1
    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.
     
  2. jcsd
  3. Feb 18, 2010 #2

    Dick

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    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?
     
  4. Feb 18, 2010 #3
    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.
     
  5. Feb 18, 2010 #4

    Dick

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    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?
     
  6. Feb 18, 2010 #5
    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.
     
  7. Feb 18, 2010 #6

    Dick

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    Sure. What is M? That's the thing you really need to estimate. How is it defined?
     
  8. Feb 18, 2010 #7
    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?
     
  9. Feb 18, 2010 #8

    Dick

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    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?
     
  10. Feb 18, 2010 #9
    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] ?
     
  11. Feb 18, 2010 #10
    Wait, so does that mean the fx+1(x) must be less than M.

    So the nth derivative being ex:

    ex<M
    So, e0.5 < M
    1.6487 < M

    I may be completely wrong, just throwing something out there that i saw.
     
  12. Feb 18, 2010 #11

    Dick

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    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.
     
  13. Feb 18, 2010 #12
    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.
     
  14. Feb 18, 2010 #13

    Dick

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    "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?
     
  15. Feb 18, 2010 #14
    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!
     
  16. Feb 18, 2010 #15

    Dick

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    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.
     
  17. Feb 18, 2010 #16
    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?
     
  18. Feb 18, 2010 #17

    Dick

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    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.
     
  19. Feb 18, 2010 #18
    Ahh, ok. I have a pretty incompetent teacher and trying to make sense of the proofs represented in the book is often hard. Thanks for your help and care :)
     
  20. Feb 18, 2010 #19
    What if f(x) is equal cos(0.5)?
     
  21. Feb 18, 2010 #20

    Dick

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    Then f(x) is a constant. Don't be silly. Do you mean f(x)=cos(0.5*x)?
     
  22. Feb 18, 2010 #21
    No I am just asking because would that mean that M is >= 0? Because the n+1'th derivative of cos(0.5) is just 0. Or, is M just the function itself?...

    M >= cos(0.5)
     
  23. Feb 18, 2010 #22

    Dick

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    Oh, I see, you want cos(0.5) using the Taylor series for cos(x) around x=0. Not f(x)=cos(0.5). Do the same thing IntegrateMe did. M is related to a derivative of cos(x). Isn't it bounded by 1?
     
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