Formula:KLS:14.07:20

From DRMF
Jump to navigation Jump to search


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 = q - x + q x + γ + δ + 1 = q - x + γ δ q x + 1 𝜇 𝑥 superscript 𝑞 𝑥 𝛾 𝛿 superscript 𝑞 𝑥 1 superscript 𝑞 𝑥 superscript 𝑞 𝑥 𝛾 𝛿 1 superscript 𝑞 𝑥 𝛾 𝛿 superscript 𝑞 𝑥 1 {\displaystyle{\displaystyle{\displaystyle\mu(x)=q^{-x}+\gamma\delta q^{x+1}=q% ^{-x}+q^{x+\gamma+\delta+1}=q^{-x}+\gamma\delta q^{x+1}}}} &

μ ( n ) = q - n + α β q n + 1 𝜇 𝑛 superscript 𝑞 𝑛 𝛼 𝛽 superscript 𝑞 𝑛 1 {\displaystyle{\displaystyle{\displaystyle\mu(n)=q^{-n}+\alpha\beta q^{n+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 = q - x + q x + γ + δ + 1 = q - x + γ δ q x + 1 𝜇 𝑥 superscript 𝑞 𝑥 𝛾 𝛿 superscript 𝑞 𝑥 1 superscript 𝑞 𝑥 superscript 𝑞 𝑥 𝛾 𝛿 1 superscript 𝑞 𝑥 𝛾 𝛿 superscript 𝑞 𝑥 1 {\displaystyle{\displaystyle{\displaystyle\mu(x)=q^{-x}+\gamma\delta q^{x+1}=q% ^{-x}+q^{x+\gamma+\delta+1}=q^{-x}+\gamma\delta q^{x+1}}}} &

μ ( n ) = q - n + α β q n + 1 𝜇 𝑛 superscript 𝑞 𝑛 𝛼 𝛽 superscript 𝑞 𝑛 1 {\displaystyle{\displaystyle{\displaystyle\mu(n)=q^{-n}+\alpha\beta q^{n+1}}}}


Proof

We ask users to provide proof(s), reference(s) to proof(s), or further clarification on the proof(s) in this space.

Symbols List

& : logical and

Bibliography

Equation in Section 14.7 of KLS.

URL links

We ask users to provide relevant URL links in this space.