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Homework Help: Power series applications

  1. Sep 26, 2008 #1


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    I'm having trouble with a few homework problems, so here are the problems and my thoughts.

    1. The problem statement, all variables and given/known data
    Use the power series to evaluate the function
    [tex] f(x)= \frac{1}{\sqrt{1+x^4}}-cos(x^2)[/tex]
    at x=0.01. Use the first two terms in the series to approximate the function, but estimate the error introduced by truncating the series.

    The attempt at a solution
    My main problem with this question is that it appears the function is equal to zero at, and in the neighborhood of, x=0. Also, the values of all the derivatives of the function are zero at x=0, so my power series expansion looks like this: f(x)=0. Am I missing something here, or is this really a "trick" question?

    2. The problem statement, all variables and given/known data
    Find a two term approximation and an error bound for the integral
    [tex]\int_0^t e^{-x^2}dx[/tex]
    in the interval 0<t<0.1

    The attempt at a solution
    I'm not sure how to start this one...should I treat the integrand as the function or is the integral included? If the function is just the integrand, I don't see any problems. However, if the integral is included in the function then how would I proceed?

    Any thoughts or hints you all could provide would be most appreciated. This HW is due tomorrow so quick replies are welcome!

  2. jcsd
  3. Sep 26, 2008 #2
    The first function is in first two aproximations always zero.
    For the second i would suggest
    [tex]\int_0^t e^{-x^2}dx = 0.5\int e^{-s}s^{-1/2}ds[/tex]
    then approximate [tex]e^{-s}[/tex] which would lead to
    [tex]0.5\int \sum_0^\infty((-1)^n (1/n!)s^{n})s^{-1/2}ds[/tex]
    which can be easily integrated
  4. Sep 26, 2008 #3


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    What formula did you use to convert the integral? Also, are the limits of integration the same? Where does the two term approximation come in? Is estimating the error in the expansion of the e-fcn enough?
  5. Sep 27, 2008 #4
    I used McLaurin formula for [tex]e^x[/tex]. The limits aren't the same but you can get back to x after integrating. The two term approximations means you use only two first terms in the sum. The sum is alternating so the error wont be greater than the absolute value of the third term in this case.
  6. Sep 27, 2008 #5


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    how would i go about switching back to x after the following:

    [tex]0.5\int (1-s)s^{-1/2}ds=-\frac{\sqrt{s}}{3}(s-3)[/tex]
  7. Sep 29, 2008 #6
    The substitution used was [tex]s=x^{2}[/tex]
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