1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Numerical Analysis: Taylor Polynomials, Error, Bounds

  1. Mar 23, 2006 #1
    (a) I found the answer to be:
    1/(1-x) = 1 + x + x^2 + x^3 + ... + [x^(n+1)]/(1-x) for x != 1

    *Note: "^" precedes a superscript, "!=" means "does not equal"

    (b) Use part (a) to find a Taylor polynomial of a general (3n)th degree for:
    f(x) = (1/x)*Integral[(1/(1 + t^3), t, 0, x]

    *Note: Integral[f(t), variable, lower bound, upper bound]

    (c) What is the error for the approximation you found in part (b)?

    (d) Find a bound for the error when a 6th order Taylor polynomial is used to approximate f(0.1).

    ====================================

    For (b), I used (a)'s equation to find that 1/[1 - (-t^3)] = 1 - t^3 + t^6 - t^9 + t^12 - ... + (-t)^(3n) + (-t)^[3(n + 1)]
    Then I integrated with respect to t with lower and upper bounds of 0 and x, respectively. I ended up with:
    f(x) = 1 - (x^3)/4 + (x^6)/7 - (x^9)/10 + ... + Integral[(-t)^(3n), t, 0, x] + Integral[(-t)^[(3(n+1)]/(1+t^3, t, 0, x]

    I'm confused about finding the (3n)th polynomial...

    For (c), I said that the error is R_(3n)(x) = f(x) - p_(3n)(x)... But I need to find the (3n)th polynomial to finish this one...

    For (d), I let n = 6, because it is the 6th order Taylor polynomial (??) to be p_6(x) = 1 - (x^3)/4 + ... The professor said to use n = 2, but I am not sure why...
     
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted