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This is actually a technical question about why I'm getting different results when trying to do the same thing using Maxima and WolframAlpha.

When I enter

expand arctan(x)

in WolframAlpha I get


This is what I enter into maxima to get the 5th degree expansion for example


The result is littered with terms containing derivative symbols, unlike the nice-looking polynomial WolframAlpha returns.

My question is how to get Maxima to evaluate the expansion at the point 0 and return the first polynomial?

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Repeated differentiation of $\dfrac{1}{1+x^2}$ and evaluation at $x=0$ won't give a $1$. Try it. The second derivative of $\arctan x$ is $0$ at $0$. The third derivative is $-2$ at $x=0$. –  André Nicolas Feb 27 '12 at 14:20
I don't have maxima, and you haven't shown what its output is. But maybe "arctan" is not known to it... perhaps "atan" or something is. –  GEdgar Feb 27 '12 at 15:11
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1 Answer

up vote 6 down vote accepted

Maxima uses that name "atan", not "arctan". Thus "taylor(atan(x),x,0,9)" will give you the same result as Alpha. When you use arctan, Maxima simply returns the generic Taylor expansion of an unknown function, hence all the derivatives.

You can execute "alias(arctan,atan)" if you wish to use the arctan as an alias to the built-in atan.

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