Originally Posted by
oblixps 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.