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Estimates using Power Series

  1. Apr 13, 2004 #1

    This was a question on a past exam, and I'm very confused about how to do this. We're not allowed calculators, so I'm sure there must be some simple solution that I'm missing.

    It asks: Use power series to estimate [tex]\int_{0}^{1/2}\frac{1}{\sqrt{1+x^3}}dx[/tex] within 10^-3.

    I started off with the binomial series:

    [tex](1+x^3)^{-1/2} = \sum_{k=0}^{\infty}\left(\begin{array}{ccc}
    -1/2 \\
    k \end{array}\right)x^{3k}[/tex]

    I then integrated from 0 to 1/2 to get:

    [tex]\int_{0}^{1/2}(1+x^3)^{-1/2}dx = \sum_{k=0}^{\infty}\left(\begin{array}{ccc}
    -1/2 \\
    k \end{array}\right)\frac{(1/2)^{3k+1}}{3k+1}[/tex]

    Now I'm stuck because I don't know how to make it accurate within 0.01. The remainder has a formula:

    [tex]R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}[/tex]

    But I don't know how to find M, such that [tex]f^{(n+1)}(c)<=M[/tex].

    Did I do this one completely wrong? Is there a power series that's easier to work with than the one I came up with? To me, it seems almost impossible to do this question without a lot of tedious work and using a calculator, but our exam is strictly no calculators allowed.

    Any help would be appreciated!
  2. jcsd
  3. Apr 13, 2004 #2

    matt grime

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    The series you have there is for the function inside the integral. You still need to do some integration. You can estimate the error because, say, the integral of g over the interval [0,1/2] is no more than 1/2 sup g(x), where g is some positive function.

    So you can do the integration of the powers of x, and get an estimate of the integral of the remainder.
  4. Apr 13, 2004 #3
    But I thought I did the integration already?

    [tex]\int_{0}^{1/2}\sum_{k=0}^{\infty}\left(\begin{array}{ccc}-1/2 \\k \end{array}\right)x^{3k} = \sum_{k=0}^{\infty}\left(\begin{array}{ccc}-1/2 \\k \end{array}\right) \int_{0}^{1/2}x^{3k}dx[/tex]

    And that is [tex] = \sum_{k=0}^{\infty}\left(\begin{array}{ccc}-1/2 \\k \end{array}\right)\frac{(1/2)^{3k+1}}{3k+1}[/tex]
  5. Apr 13, 2004 #4

    matt grime

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    you did, i misread, but the thing about being able to estimate the error is still valid:

    proper integral - integral; of first n terms = integral of error


    integral of error can be bound by 1/2 sup possible error.

    ie take the worst cas scenario
  6. Apr 13, 2004 #5
    Sorry to be a bother, but I'm really unclear on this error bounds stuff. Can you explain what you mean by the phrase "integral of error can be bound by 1/2 sup possible error"?

    Do you mean that I should do something like:

    [tex]\sum_{k=0}^{\infty}\left(\begin{array}{ccc}-1/2 \\k \end{array}\right)\frac{(1/2)^{3k+1}}{3k+1} - \sum_{k=0}^{n}\left(\begin{array}{ccc}-1/2 \\k \end{array}\right)\frac{(1/2)^{3k+1}}{3k+1} = 0.001[/tex]

    ? If so, how do I go about figuring out what n is supposed to be from that?
  7. Apr 13, 2004 #6

    matt grime

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    Let me do it this way, the detail is annoying to type but I think I can't avoid it.
    Let f be the function, f_n the taylor expansion up to the n'th term

    [tex]f(x) - f_n(x) = R_n(x)[/tex]


    [tex]R_n= f^{(n+1)}(c)x^{n+1}/n![/tex]

    what are the derivatives of f? what is the maximal value of the n'th derivative? It should be quite small, and you can get any bound that works. Call this number M.

    then the error in the integral is

    M/(n+1)! times the integral of x^{n+1} between 0 and 1/2 which is intself quite small, so just find an n to make it as small as you want.

    Alternatively you could have just made the substitution x^3 = u in the original integral and it would make it a lot easier - the derivatives would be so much easier to work out.
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