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Taylor polynomials

  1. Dec 3, 2015 #1
    1. The problem statement, all variables and given/known data
    In the attached file. (a,b)

    2. Relevant equations
    [tex]\cos(x)=\sum_{k=0}^{n}\frac{(-1)^kx^{2k}}{(2k)!}[/tex]

    Pn- Taylor expansion of order n

    3. The attempt at a solution
    I know that in this case, in order to get an error less than 1/100, I need 18 terms/order 18(according to Wolfram Alpha). But does it mean that if I take the integral of cos(x) so I can take the integral of Pn and solve part (a)?

    I am not really sure regarding part (b). Am I supposed to compute the integral of Pn that I (hopefully) found in part (a) and express it in sigma notation?

    Thank you,
    Thomas
     

    Attached Files:

  2. jcsd
  3. Dec 3, 2015 #2

    RUber

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    For part a, it is only asking for the polynomial. Part b is asking you to integrate the polynomial (leaving it in sum form).
    You can take the integral for the polynomial and evaluate it at your endpoints to get your integral value.
    I get error under 1/100 with k = 9, i.e. order 18 polynomial as you pointed out.
     
  4. Dec 3, 2015 #3

    Ray Vickson

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    Do NOT write ##\cos(x) = \sum_{k=0}^n (-1)^k x^{2k}/(2k)!,## because that is not true. You need to replace the "##=##' sign by an approximation sign such as "##\approx##" or "##\doteq##".

    When I do it I need fewer than 10 terms---nowhere near the 18 that you claim.

    As to your question: what do YOU think?
     
  5. Dec 3, 2015 #4
    Well, it is 18 terms if you count the zeros. but it is 10 terms if you don't count them, but still, order 18.

    Regarding the sign, I know that in order to have an equal sign, I need to add the error I make. But for some reason decided not to change the equal sign... I don't know why.

    So if I understand correctly, the required polynomial is just the polynomial that approximate cos(x) up to the error?

    Thank you,
    Thomas
     
  6. Dec 3, 2015 #5

    RUber

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    Yes. Part a says find P(x). Part b says find ##\int_0^6 P(x) dx## leaving it in sum form and check your error.
     
  7. Dec 3, 2015 #6

    Ray Vickson

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    The series for ##\cos(x)## has infinitely many terms; when you integrate it, you obtain a new infinite series. When you truncate the ##\cos## series at terms of order ##x^{2n}## you have some error ##e_n(x)##. The error in ##\int \cos(x) \, dx## is the integral of the error in ##\cos(x)##. You want to choose ##n## so that the error in the integrated series is provably < 0.01. This is difficult to determine exactly, so you need to be satisfied with having an upper bound on the error < 0.01 for sure. Finding that is a lot easier.
     
  8. Dec 3, 2015 #7
    Due to the regularity of ##\cos (x) ## on the interval ##[0,6]##, you have :

    ##\forall x\in [0,6],\ \forall n\in\mathbb{N}: \ \cos(x) = \sum_{k = 0}^n \frac{ \cos^{(k)}(0) }{k!} x^k + \int_0^x \frac{(x-t)^n}{n!} \cos^{(n+1)}(t) \ dt ##

    So your job is to determine the smallest ##n## such that:

    ##|\int_0^6 \int_0^x \frac{(x-t)^n}{n!} \cos^{(n+1)}(t) \ dt dx | \le 0.01 ##
     
  9. Dec 3, 2015 #8

    Ray Vickson

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    In a perfrect world, yes, that would be his problem. But in the real world there is a much simpler way, but which might give him a value of ##n## just a little bit too large.
     
  10. Dec 3, 2015 #9
    You could say the same thing using a sentence like ' I agree, but I think a computer program is more appropriate for this problem, as there is no specific restriction on the precision, etc ... ' :smile:
     
  11. Dec 3, 2015 #10

    Ray Vickson

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    But I had already said essentially the same thing myself in post #6 (without giving a formula for the error) and then dismissed the method as impractical. So, no, I could not have used a sentence similar to what you suggest, because I most definitely do not agree.
     
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