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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It's the inverse of $\sinh$. – Git Gud Jun 22 '13 at 20:56
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2This question is similar to: Power series of $\ln(x+\sqrt{1+x^2})$ without Taylor. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Integreek Nov 09 '24 at 11:12
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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.
Nick Peterson
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Haha, of course! I calculated the first derivative but failed to see the simplification so I thought it's much uglier! Thanks. – Spine Feast Jun 22 '13 at 20:57
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Sure, couldn't accept straight away because I accepted one recently. – Spine Feast Jun 22 '13 at 21:04
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Note that the derivative is an even function, so alternating terms of the Maclaurin Series will be 0. – Dan Jun 22 '13 at 21:33