Difference between revisions of "Jacobi dn"
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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 $$\mathrm{dn \hspace{...") |
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Let $u=\displaystyle\int_0^x \dfrac{1}{\sqrt{(1-t^2)(1-mt^2)}}dt = \displaystyle\int_0^{\phi} \dfrac{1}{\sqrt{1-m\sin^2 \theta}} d\theta.$ Then we define | Let $u=\displaystyle\int_0^x \dfrac{1}{\sqrt{(1-t^2)(1-mt^2)}}dt = \displaystyle\int_0^{\phi} \dfrac{1}{\sqrt{1-m\sin^2 \theta}} d\theta.$ Then we define | ||
$$\mathrm{dn \hspace{2pt}} u = \sqrt{1-m\sin^2 \phi} = \sqrt{1-mx^2}.$$ | $$\mathrm{dn \hspace{2pt}} u = \sqrt{1-m\sin^2 \phi} = \sqrt{1-mx^2}.$$ | ||
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| + | =Properties= | ||
| + | #$m \mathrm{sn \hspace{2pt}}^2 u + \mathrm{dn \hspace{2pt}}^2u=1$ | ||
Revision as of 07:28, 10 March 2015
Let $u=\displaystyle\int_0^x \dfrac{1}{\sqrt{(1-t^2)(1-mt^2)}}dt = \displaystyle\int_0^{\phi} \dfrac{1}{\sqrt{1-m\sin^2 \theta}} d\theta.$ Then we define $$\mathrm{dn \hspace{2pt}} u = \sqrt{1-m\sin^2 \phi} = \sqrt{1-mx^2}.$$
Properties
- $m \mathrm{sn \hspace{2pt}}^2 u + \mathrm{dn \hspace{2pt}}^2u=1$
