# Inverse Jacobi elliptic arcdn(x,k) Calculator

## Calculates the Inverse Jacobi elliptic arcdn(x,k).

 x √(1-k2)≦x≦1 k 0≦k≦1 6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit
 $\normal Inverse\ Jacobi\ elliptic\ function\ arcdn(x,k)\\[10](1)\ x=dn(u,k)\hspace{5}\rightarrow\hspace{5} u=arcdn(x,k)\equiv dn^{\tiny -1}(x,k)\\[10](2)\ arcdn(x,k)=F(\sqrt{1-x^2}/k,k)\ ,\hspace{20} \sqrt{1-k^2}\le x\le1\\\hspace{20}F(x,k)={\large\int_{\small 0}^{\hspace{25}\small x}\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}}\ ,\hspace{20} -1\le x\le1\\(3)\ k^2sn^2(u,k)+dn^2(u,k)=1\\$

Inverse Jacobi elliptic arcdn(x,k)
 [1-1] /1 Disp-Num5103050100200
[1]  2017/05/27 20:17   Male / 20 years old level / High-school/ University/ Grad student / Very /
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This online calculator is very useful. I would like to ask about the unit of this operation. Is the unit of arcdn(x,k) in this calculator in radians or unitless?
from Keisan
Thank you for using keisan calculator.
The unit of arcdn(x,k) is unitless.

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