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Thread: maclaurin series question

  1. #1
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    maclaurin series question

    hello, if possible hep with the following:

    Q: By expanding (1-4x)^1/2 and integrating term by term, or otherwise, find the series expansion for arcsin(2x), when |x|<1/2, as far as the term in x^7.

    A: I have done it by consequatively differentiating until the seventh order and then evaluating at x = 0 to produce the desired series.

    However I am at a loss as to how to do it by "expanding (1-4x)^1/2 and integrating term by term"..

    ...I know its basic stuff but I would much appreciate any help!
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  2. #2
    Super Member Random Variable's Avatar
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    Do you mean $\displaystyle \frac{1}{\sqrt{1-4x^{2}}} $ ?
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  3. #3
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    I do! I am ever so sorry... Thanks for spotting it
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by pb6883 View Post
    hello, if possible hep with the following:

    Q: By expanding (1-4x)^1/2 and integrating term by term, or otherwise, find the series expansion for arcsin(2x), when |x|<1/2, as far as the term in x^7.

    A: I have done it by consequatively differentiating until the seventh order and then evaluating at x = 0 to produce the desired series.

    However I am at a loss as to how to do it by "expanding (1-4x)^1/2 and integrating term by term"..

    ...I know its basic stuff but I would much appreciate any help!
    Are you sure that you were asked to expand $\displaystyle (1-4x)^{1/2}$ and integrate the resulting series to get the expansion for $\displaystyle \mathrm{arcsin}(2x)$? - Because it is not at all clear what $\displaystyle \int (1-4x)^{1/2}$ has got to do with $\displaystyle \mathrm{arcsin}(2x)$.
    Maybe you were asked to expand $\displaystyle (1-4x^2)^{-1/2}$ instead? Because in this case, it is quite obvious how the integral of this term is related to $\displaystyle \mathrm{arcsin}(2x)$.
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by pb6883 View Post
    I do! I am ever so sorry... Thanks for spotting it
    Ok, in that case you have that $\displaystyle \int (1-4x^2)^{-1/2}\, dx=\frac{\mathrm{arcsin}(x)}{2}+C$. So multiply the integrated series by 2 and check for a possibly necessary adjustment of the C. (Actually, C=0 will do - I think.)
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  6. #6
    Super Member Random Variable's Avatar
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    You could use the generalized binomial theorem to expand $\displaystyle \frac{1}{\sqrt{1-4x^{2}}} = (1-4x^2)^{-1/2} $

    $\displaystyle (1+x)^{n} = 1 +nx +\frac{n(n-1)x^{2}}{2!} + \frac{n(n-1)(n-2)x^{3}}{3!} + ... $
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  7. #7
    MHF Contributor Bruno J.'s Avatar
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    You can show easily by induction that

    $\displaystyle \frac{1}{\sqrt{1-t}} = 1+\sum_{j=1}^\infty\frac{1\cdot 3 \cdot ... \cdot (2j-1)}{2^jj!}t^j$

    Now if you use

    $\displaystyle \int_0^x\frac{1}{\sqrt{1-t^2}}\: dt = \arcsin x$

    you can get the series for arcsin by replacing $\displaystyle t$ by $\displaystyle t^2$ in the first identity and integrating the resulting series term by term.
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  8. #8
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    thanks all, your comments have really helped (p.s. i told you it was basic )
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