Difference between revisions of "Elliptic K"
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(Created page with "If $m=k^2$ we define the complete elliptic integral of the first kind, $K$ to be $$K(k)=K(m)=\displaystyle\int_0^{\frac{\pi}{2}} \dfrac{1}{\sqrt{1-k^2\sin^2 \theta}} d\theta.$...") |
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| − | If $m=k^2$ we define the complete elliptic integral of the first kind, $K$ to be | + | The Elliptic $K$ function is also known as the complete Elliptic integral of the first kind. If $m=k^2$ we define the complete elliptic integral of the first kind, $K$ to be |
$$K(k)=K(m)=\displaystyle\int_0^{\frac{\pi}{2}} \dfrac{1}{\sqrt{1-k^2\sin^2 \theta}} d\theta.$$ | $$K(k)=K(m)=\displaystyle\int_0^{\frac{\pi}{2}} \dfrac{1}{\sqrt{1-k^2\sin^2 \theta}} d\theta.$$ | ||
The incomplete elliptic integral of the first kind is | The incomplete elliptic integral of the first kind is | ||
Revision as of 18:07, 25 July 2015
The Elliptic $K$ function is also known as the complete Elliptic integral of the first kind. If $m=k^2$ we define the complete elliptic integral of the first kind, $K$ to be $$K(k)=K(m)=\displaystyle\int_0^{\frac{\pi}{2}} \dfrac{1}{\sqrt{1-k^2\sin^2 \theta}} d\theta.$$ The incomplete elliptic integral of the first kind is $$K(\phi |k) = K(\phi |m) = \displaystyle\int_0^{\phi} \dfrac{1}{\sqrt{1-k^2\sin^2 \theta}} d\theta.$$
