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Math Help - maclaurin series problem

  1. #1
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    maclaurin series problem

    Given that sinh(x) = (e^x - e^-x) / 2, find a series representation for arcsinh(x).

    So my book did a Maclaurin series expansion of sinh(x) = x + x^3 / 3! + x^5 / 5! + ...

    Then it said: the inverse will have some series expansion which we will write as arcsinh(x) = b0 + b1 x + b2 x^2 + b3 x^3 + ... where the "b"s are coefficients.

    we label the coeficients in the series expansion of sinh(x) as a_j. we find that,
    b0 = a0, b1 = 1 / a1 = 1, b2 = -a2/(a1)^3 = 0, b3 = [1 / (a5)^6] (2(a2)^2 - a1a3) = -1/6

    therefore it follows that arcsinh(x) = x - (1/6)x^3 + ...

    how did they get the coeffients? i have no idea where they came from.
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  2. #2
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    Quote Originally Posted by oblixps View Post
    Given that sinh(x) = (e^x - e^-x) / 2, find a series representation for arcsinh(x).

    So my book did a Maclaurin series expansion of sinh(x) = x + x^3 / 3! + x^5 / 5! + ...

    Then it said: the inverse will have some series expansion which we will write as arcsinh(x) = b0 + b1 x + b2 x^2 + b3 x^3 + ... where the "b"s are coefficients.

    we label the coeficients in the series expansion of sinh(x) as a_j. we find that,
    b0 = a0, b1 = 1 / a1 = 1, b2 = -a2/(a1)^3 = 0, b3 = [1 / (a5)^6] (2(a2)^2 - a1a3) = -1/6

    therefore it follows that arcsinh(x) = x - (1/6)x^3 + ...

    how did they get the coeffients? i have no idea where they came from.
    I would advise you to convert \sinh^{-1}{x} into its logarithmic equivalent and use a Taylor series expansion of the logarithm.


    If x = \sinh{y}

    Then y = \sinh^{-1}{x}.


    But \sinh{y} = \frac{e^y - e^{-y}}{2}

    So x = \frac{e^y - e^{-y}}{2}

    2x = e^{y} - e^{-y}

    2x = e^y - \frac{1}{e^y}

    0 = e^y - \frac{1}{e^y} - 2x

    0 = e^{2y} - 1 - 2x\,e^y

    0 = e^{2y} - 2x\,e^y - 1

    0 = e^{2y} - 2x\,e^y + (-x)^2 - (-x)^2 - 1

    0 = \left(e^{y} - x\right)^2 - x^2 - 1

    x^2 + 1 = \left(e^y - x\right)^2

    \sqrt{x^2 + 1} = e^y - x

    x + \sqrt{x^2 + 1} = e^y

    y = \ln{\left(x + \sqrt{x^2 + 1}\right)}.


    Therefore \sinh^{-1}{x} = \ln{\left(x + \sqrt{x^2 + 1}\right)}.

    You should now be able to use the Taylor Series expansion of a Logarithm to find an expansion for \sinh^{-1}{x}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    I would advise you to convert \sinh^{-1}{x} into its logarithmic equivalent and use a Taylor series expansion of the logarithm.


    If x = \sinh{y}

    Then y = \sinh^{-1}{x}.


    But \sinh{y} = \frac{e^y - e^{-y}}{2}

    So x = \frac{e^y - e^{-y}}{2}

    2x = e^{y} - e^{-y}

    2x = e^y - \frac{1}{e^y}

    0 = e^y - \frac{1}{e^y} - 2x

    0 = e^{2y} - 1 - 2x\,e^y

    0 = e^{2y} - 2x\,e^y - 1

    0 = e^{2y} - 2x\,e^y + (-x)^2 - (-x)^2 - 1

    0 = \left(e^{y} - x\right)^2 - x^2 - 1

    x^2 + 1 = \left(e^y - x\right)^2

    \sqrt{x^2 + 1} = e^y - x

    x + \sqrt{x^2 + 1} = e^y

    y = \ln{\left(x + \sqrt{x^2 + 1}\right)}.


    Therefore \sinh^{-1}{x} = \ln{\left(x + \sqrt{x^2 + 1}\right)}.

    You should now be able to use the Taylor Series expansion of a Logarithm to find an expansion for \sinh^{-1}{x}.
    the maclaurin series for ln(1+x) is (-1)^(n+1) x^n / n so i did ln(1 + (x+ sqrt(x^2+1) -1) so i got (-1)^(n+1) (x+sqrt(x^2+1)-1)^n / n. is that correct? by the way do you know what my book is doing?
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  4. #4
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    is this right? i looked up the series representation of arcsinh(x) and it looks nothing like what i got in the post above.
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