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Change A
Δ{\displaystyle \Delta }
δ {\displaystyle \delta }
[zn](1−z)α(1zlog11−z)β(α,β∉0,1,2,…{\displaystyle [z^{n}](1-z)^{\alpha }\left({\frac {1}{z}}\log {\frac {1}{1-z}}\right)^{\beta }\quad (\alpha ,\beta \notin {0,1,2,\ldots }}
f(z)=∑n=0∞an(z−z0)n{\displaystyle f(z)=\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}}
[xn](1−x)α(1xlog11−x)β(α,β∉0,1,2,…{\displaystyle [x^{n}](1-x)^{\alpha }\left({\frac {1}{x}}\log {\frac {1}{1-x}}\right)^{\beta }\quad (\alpha ,\beta \notin {0,1,2,\ldots }}
f(x)=∑n=0∞an(x−x0)n{\displaystyle f(x)=\sum _{n=0}^{\infty }a_{n}(x-x_{0})^{n}}
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