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Doing a normal Taylor expansion of $\arccos(1-x)$ around $x=0$ to two terms by taking derivatives doesn't work because of division by zero.

I've put this into wolfram alpha: http://www.wolframalpha.com/input/?i=taylor+series+arccos%281-x%29. It's nice but I need to show that I can do it myself. This isn't an analysis class so I have never seen square roots in a series expansion before.

I found this in my search: Some approximations for $\arccos(1/(1+x))$ but I need the expansion to two terms.

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    What value are you expanding "about"?2012-12-11
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    "Taylor series about $x=0$" means a series in nonnegative integer powers of $x$. This isn't. It is still a series expansion, just not a Taylor series. And that explains why the derivative formulas for Taylor series don't work here.2012-12-11

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