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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
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