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MacLaurin Polynomial for sin(x)

  1. Mar 19, 2009 #1
    1. The problem statement, all variables and given/known data
    The formula for the Maclaurin Polynomial of sin(x) can be found on this page: http://www.tvalx.com/MathArticles/ExploringTaylorPolynomials/ExploringTaylorPolynomials.htm

    (close to the top).

    Find the Mauclaurin Polynomial of degree 4. Use it to estimate sin(0.5).


    2. Relevant equations



    3. The attempt at a solution
    I tried evaluating the formula for n=0, 1, 2, 3, and 4.

    0 --> x
    1 --> (-1/6)x^3
    2 ---> (1/120)x^5
    3 ---> (-1.98E-4)x^7
    4 ---> (2.76E-6)x^9

    Now I'll plug in x = 0.5 for each one, and add them all up. When I do that, I get 0.479. Which seems to be right, because when I actually evaluate sin(0.5) I get the same thing.

    Am I doing something wrong? The answer key gives a different answer =/
     
    Last edited: Mar 19, 2009
  2. jcsd
  3. Mar 19, 2009 #2

    Mark44

    Staff: Mentor

    How much different?

    I don't seen anything obviously wrong, but you have rounded off some of your fractions to only 3 significant digits, and that will affect your approximation. Keep all of your fractions intact (don't round them), and keep all of the precision your calculator will give you.
     
  4. Mar 19, 2009 #3
    I did keep all the precision my calculator gave me, but still got the wrong answer.

    The answer on the key is 0.432.
     
  5. Mar 20, 2009 #4

    Mark44

    Staff: Mentor

    You have too many terms (so your answer is closer than what is asked for). The problem asks for the Maclaurin polynomial of degree 4, and yours goes up to degree 9. Take your first two terms and call it good.

    The Maclaurin series approximation of degree 4 is sin(x) [itex]\approx[/itex] 0 + x + 0x^2 - 1/6 x^3 + 0 x^4.
     
  6. Mar 21, 2009 #5
    Hmm, I'm a bit confused now. Is the following right:

    What if the question asks for the third taylor polynomial of some function? In that case you would take the first, second, and third derivatives, right?

    But if it asks for a polynomial of degree y, then you go up to x^y?
     
  7. Mar 21, 2009 #6

    Mark44

    Staff: Mentor

    Looking at Taylor polynomials as a sequence of polynomials of increasing degree, the third Taylor polynomial would be the one whose highest degree term is x^2, so you'd only need to go up to the second derivative.
    f0(x) = f(a)
    f1(x) = f(a) + f'(a)(x - a)
    f2(x) = f(a) + f'(a)(x - a) + 1/2! * f''(a)*(x - a)^2

    It's only tricky because there is a mismatch between the indexes, which start from zero, and their order, which we usual start with 1 (for first).

    For your second question, yes.
     
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