Difference between revisions of "Arctan"
From specialfunctionswiki
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The $\mathrm{arctan}$ function is the inverse function of the [[tangent]] function.<br /> | The $\mathrm{arctan}$ function is the inverse function of the [[tangent]] function.<br /> | ||
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| + | File:Arctan.png|Graph of $\mathrm{arctan}$ on $[-1,1]$. | ||
| + | File:Complex arctan.jpg|[[Domain coloring]] of the [[analytic continuation]] of $\mathrm{arctan}$. | ||
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| + | </div> | ||
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=Properties= | =Properties= | ||
Revision as of 05:11, 31 October 2014
The $\mathrm{arctan}$ function is the inverse function of the tangent function.
- Arctan.png
Graph of $\mathrm{arctan}$ on $[-1,1]$.
- Complex arctan.jpg
Domain coloring of the analytic continuation of $\mathrm{arctan}$.
Properties
Proposition: $$\dfrac{d}{dz} \mathrm{arctan}(z) = \dfrac{1}{z^2+1}$$
Proof: If $y=\mathrm{arctan}(z)$ then $\tan y = z$. Now use implicit differentiation with respect to $z$ yields $$\sec^2(y)y'=1.$$ Substituting back in $y=\mathrm{arccos(z)}$ yields the formula $$\dfrac{d}{dz} \mathrm{arccos(z)} = \dfrac{1}{\sec^2(\mathrm{arctan(z)})} = \dfrac{1}{z^2+1}. █$$
Proposition: $$\int \mathrm{arctan}(z) = z\mathrm{arctan}(z) - \dfrac{1}{2}\log(1+z^2)+C$$
Proof: █
Proposition: $$\mathrm{arctan}(z) = \mathrm{arccot}\left( \dfrac{1}{z} \right)$$
Proof: █
