Is there a trick to finding the Maclaurin series for $f(x)=\ln(x+\sqrt{1+x^2})$ fast? Vaguely, I recall this being some sort of inverse hyperbolic function, but I'm not sure about which one, and what its derivatives are. This is a past exam question and I would like to know, should something similar appear again, if there is a quick method of finding this series without using inverse hyperbolic functions. The derivatives of this look absolutely painful to calculate, although easy at $x=0$, but I'm not sure if I could easily see a pattern for the $n$-th derivative at $0$.
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asked Jun 22, 2013 at 20:48
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$\begingroup$ It's the inverse of $\sinh$. $\endgroup$– Git GudCommented Jun 22, 2013 at 20:56
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1 Answer
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The easiest method: note that $$ \frac{d}{dx}\left[\ln(x+\sqrt{1+x^2})\right]=\frac{1}{\sqrt{1+x^2}}=(1+x^2)^{-1/2}. $$ This last can be expressed through a binomial series; you can then integrate term by term, and solve for the constant of integration, to get the series you want.
answered Jun 22, 2013 at 20:52
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$\begingroup$ Haha, of course! I calculated the first derivative but failed to see the simplification so I thought it's much uglier! Thanks. $\endgroup$ Commented Jun 22, 2013 at 20:57
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$\begingroup$ Sure, couldn't accept straight away because I accepted one recently. $\endgroup$ Commented Jun 22, 2013 at 21:04
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$\begingroup$ Note that the derivative is an even function, so alternating terms of the Maclaurin Series will be 0. $\endgroup$– DanCommented Jun 22, 2013 at 21:33