Formula:KLS:14.02:26: Difference between revisions

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Latest revision as of 08:36, 22 December 2019


μ := μ ( x ) assign subscript 𝜇 𝜇 𝑥 {\displaystyle{\displaystyle{\displaystyle\nabla_{\mu}:=\frac{\nabla}{\nabla% \mu(x)}}}}

Substitution(s)

μ ( x ) = q - x + γ δ q x + 1 = λ ( x ) = q - x + c q x - N = q - x + q x + γ + δ + 1 = 2 a cos θ 𝜇 𝑥 superscript 𝑞 𝑥 𝛾 𝛿 superscript 𝑞 𝑥 1 𝜆 𝑥 superscript 𝑞 𝑥 𝑐 superscript 𝑞 𝑥 𝑁 superscript 𝑞 𝑥 superscript 𝑞 𝑥 𝛾 𝛿 1 2 𝑎 𝜃 {\displaystyle{\displaystyle{\displaystyle\mu(x)=q^{-x}+\gamma\delta q^{x+1}=% \lambda(x)=q^{-x}+cq^{x-N}=q^{-x}+q^{x+\gamma+\delta+1}=2a\cos\theta}}} &

λ ( x ) = x ( x + γ + δ + 1 ) 𝜆 𝑥 𝑥 𝑥 𝛾 𝛿 1 {\displaystyle{\displaystyle{\displaystyle\lambda(x)=x(x+\gamma+\delta+1)}}} &
μ ( x ) := q - x + γ δ q x + 1 assign 𝜇 𝑥 superscript 𝑞 𝑥 𝛾 𝛿 superscript 𝑞 𝑥 1 {\displaystyle{\displaystyle{\displaystyle\mu(x):=q^{-x}+\gamma\delta q^{x+1}}}} &
μ ( x ) = q - x + γ δ q x + 1 = λ ( x ) = q - x + c q x - N = q - x + q x + γ + δ + 1 = 2 a cos θ 𝜇 𝑥 superscript 𝑞 𝑥 𝛾 𝛿 superscript 𝑞 𝑥 1 𝜆 𝑥 superscript 𝑞 𝑥 𝑐 superscript 𝑞 𝑥 𝑁 superscript 𝑞 𝑥 superscript 𝑞 𝑥 𝛾 𝛿 1 2 𝑎 𝜃 {\displaystyle{\displaystyle{\displaystyle\mu(x)=q^{-x}+\gamma\delta q^{x+1}=% \lambda(x)=q^{-x}+cq^{x-N}=q^{-x}+q^{x+\gamma+\delta+1}=2a\cos\theta}}} &

λ ( x ) = x ( x + γ + δ + 1 ) 𝜆 𝑥 𝑥 𝑥 𝛾 𝛿 1 {\displaystyle{\displaystyle{\displaystyle\lambda(x)=x(x+\gamma+\delta+1)}}}


Proof

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Symbols List

& : logical and
cos cos {\displaystyle{\displaystyle{\displaystyle\mathrm{cos}}}}  : cosine function : http://dlmf.nist.gov/4.14#E2

Bibliography

Equation in Section 14.2 of KLS.

URL links

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