Difference between revisions of "Jacobi cn"

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(Created page with "Let $u=\displaystyle\int_0^x \dfrac{1}{\sqrt{(1-t^2)(1-mt^2)}}dt = \displaystyle\int_0^{\phi} \dfrac{1}{\sqrt{1-m\sin^2 \theta}} d\theta.$ Then we define $$\mathrm{cn \hspace{...")
 
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Let $u=\displaystyle\int_0^x \dfrac{1}{\sqrt{(1-t^2)(1-mt^2)}}dt = \displaystyle\int_0^{\phi} \dfrac{1}{\sqrt{1-m\sin^2 \theta}} d\theta.$ Then we define
 
Let $u=\displaystyle\int_0^x \dfrac{1}{\sqrt{(1-t^2)(1-mt^2)}}dt = \displaystyle\int_0^{\phi} \dfrac{1}{\sqrt{1-m\sin^2 \theta}} d\theta.$ Then we define
 
$$\mathrm{cn \hspace{2pt}} u = \cos \phi = \sqrt{1-x^2}.$$
 
$$\mathrm{cn \hspace{2pt}} u = \cos \phi = \sqrt{1-x^2}.$$
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=Properties=
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#$\mathrm{sn \hspace{2pt}}^2u+\mathrm{cn \hspace{2pt}}^2u=1$

Revision as of 07:26, 10 March 2015

Let $u=\displaystyle\int_0^x \dfrac{1}{\sqrt{(1-t^2)(1-mt^2)}}dt = \displaystyle\int_0^{\phi} \dfrac{1}{\sqrt{1-m\sin^2 \theta}} d\theta.$ Then we define $$\mathrm{cn \hspace{2pt}} u = \cos \phi = \sqrt{1-x^2}.$$

Properties

  1. $\mathrm{sn \hspace{2pt}}^2u+\mathrm{cn \hspace{2pt}}^2u=1$