DLMF:19.14.E3 (Q6308): Difference between revisions

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cos ϕ = 1 - x 2 1 + x 2 italic-ϕ 1 superscript 𝑥 2 1 superscript 𝑥 2 {\displaystyle{\displaystyle\cos\phi=\dfrac{1-x^{2}}{1+x^{2}}}}

\cos@@{\phi}=\dfrac{1-x^{2}}{1+x^{2}}
Property / constraint: cos ϕ = 1 - x 2 1 + x 2 italic-ϕ 1 superscript 𝑥 2 1 superscript 𝑥 2 {\displaystyle{\displaystyle\cos\phi=\dfrac{1-x^{2}}{1+x^{2}}}} / rank
 
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Revision as of 18:06, 30 December 2019

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DLMF:19.14.E3
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    0 x d t 1 + t 4 = sign ( x ) 2 F ( ϕ , k ) , superscript subscript 0 𝑥 𝑡 1 superscript 𝑡 4 sign 𝑥 2 elliptic-integral-first-kind-F italic-ϕ 𝑘 {\displaystyle{\displaystyle\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{1+t^{4}}}=% \frac{\operatorname{sign}\left(x\right)}{2}F\left(\phi,k\right),}}
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    DLMF:19.14.E3
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    cos ϕ = 1 - x 2 1 + x 2 italic-ϕ 1 superscript 𝑥 2 1 superscript 𝑥 2 {\displaystyle{\displaystyle\cos\phi=\dfrac{1-x^{2}}{1+x^{2}}}}
    0 references