with the maclaurin series is the expansion or answer wateva to arsinh(4/3+x) = f(0)+(f'(0))/2! x (x)^2 + (f''(0))/3! x (x)^3.... or is it f(0)+(f'(0))/2! x (4/3+x)^2 + (f''(0))/3! x (4/3+x)^3.....???
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The latter. The Maclaurin series is given as: $\displaystyle f(x) = 1 + f'(0)x + \frac{f''(0)}{2!}x^{2} + \frac{f^{(3)}(0)}{3!}x^{3} + ...$ Here, you are dealing with $\displaystyle f\left(\frac{4}{3} + x\right)$ with $\displaystyle f(x) = sinh^{-1} x$
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