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Taylor Series Error Integration

  1. Nov 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Using Taylor series, Find a polynomial p(x) of minimal degree that will approximate F(x) throughout the given interval with an error of magnitude less than 10-4

    F(x) = ∫0x sin(t^2)dt

    2. Relevant equations

    Rn = f(n+1)(z)|x-a|(n+1)/(n+1)!


    3. The attempt at a solution

    I am confused about the integral and have never done a problem like this
    Do I use the fundamental theorem of calculus to solve the nth derivative of the function and see at what point the error is below the required value?
    Do I somehow integrate the taylor series for sin(x) and somehow use that?
    I just do not know how to go about this problem
     
  2. jcsd
  3. Nov 15, 2016 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You did not actually specify an interval over which you are approximating F(x).

    Anyway, instead of agonizing about how to do it, just start doing it. If one way does not work, try another.
     
  4. Nov 15, 2016 #3

    Mark44

    Staff: Mentor

    Yes. Write a few terms of the Maclaurin series for ##\sin(t^2)##, and then integrate that series. Note that the series you get is an alternating series. There should be a theorem in your book on the error that results from truncating an alternating series.
     
  5. Nov 15, 2016 #4
    I worked out some stuff let me know what to fix so far

    F(x) = INT(0 to x) sin(t^2)dt | on the interval [0, 1]

    Polynomial degree with error of magnitude less than 10^-4

    sin(x) = x - x^3/3! + x^5/5! >>>

    sin(x^2) = x^2 - x^6/3! + x^10/5! >>>

    sin(x) = (-1)^n * x^(2n+1)/(2n+1)!

    sin(x^2) = (-1)^n * x^(4n+2)/(2n+1)!

    INT sin(x^2) = (-1)^n * x^(4n+3)/(4n+3)(2n+1)!

    |x^(4n+3)/(4n+3)(2n+1)!| < 10^-4

    x max of 1

    10000 < |(4n+3+1)(2n+1+1)!|

    (4n+7)(2n+3)! = h(n)

    h(0)=7*3!
    h(1)=11*5!
    h(2)=15*7! > 10000

    n = 2

    The degree of the polynomial should be 2 right?
     
  6. Nov 15, 2016 #5
    x^3/3 - x^7/(7*3!) + x^11/(11*5!)
    was accepted as the correct answer on my homework
    been on that problem all day good to finally solve it
    so I guess you integrate the summation of the series
    find when the term of n+1 is bounded above by the error
    then use the McLaurin Series up to the nth term?
     
  7. Nov 15, 2016 #6

    Mark44

    Staff: Mentor

    The two lines above are incorrect -- you have omitted the summation sign.
    This is also incorrect. On the right side there should be a summation.
    No. n = 2 indicates the index of the term in the polynomial. n = 0 refers to the first term, which is of degree 3. n = 1 refers to the second term, which is of degree 7, and so on.

    Rather than working with the general terms in the Maclaurin expansion, it's easier to write the polynomial as you did above, like so: ##\sin(t^2) = t^2 - \frac{t^6}{3!} + \frac{t^{10}}{5!} -+ \dots##.
     
  8. Nov 15, 2016 #7
    Thanks for the feed back
    Yeah I totally forgot the summation sign that was lazy of me
    The index instead of degree is good to know as I was mistaken about the terminology
     
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