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location Singapore
age 37
visits member for 2 years, 8 months
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C'est par la logique qu'on démontre, c'est par l'intuition qu'on invente.


Nov
12
revised Martin Gardener's best books?
added 12 characters in body
Nov
12
asked Martin Gardener's best books?
Oct
31
comment An introduction for integral tricks.
thanks @David , it does look interesting!
Oct
27
answered Rate of convergence for 'Law of large numbers'
Oct
25
comment Posterior probability or Prior probability
basically, our difference is, what Jane met (10 or 50 white balls in a row), should it affect the estimation of the bowl B black-white distribution? your answer is no, but I doubt -- if John is a dealer, Jane has reason to believe he's cheating!
Oct
25
comment Posterior probability or Prior probability
pls see my update of the question. hopefully I explained myself better this round.
Oct
25
comment Posterior probability or Prior probability
where is the 10 from?
Oct
25
revised Posterior probability or Prior probability
added 1542 characters in body
Oct
25
asked Posterior probability or Prior probability
Oct
25
comment why $f$ is holomorphic if $f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)\, d \zeta}{\zeta - z}$?
i've updated the full content of the theorem and proof in the question.
Oct
25
revised why $f$ is holomorphic if $f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)\, d \zeta}{\zeta - z}$?
added 1565 characters in body
Oct
23
revised why $f$ is holomorphic if $f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)\, d \zeta}{\zeta - z}$?
deleted 1 character in body
Oct
23
comment holomorphic function power series: $f(z)=\sum_{j=0}^\infty a_j(z-z_0)^j$, can it be extended to $U$?
thank you Martin
Oct
23
asked why $f$ is holomorphic if $f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)\, d \zeta}{\zeta - z}$?
Oct
22
accepted why $G(z)=\frac{f(z)}z, z\ne 0; f'(0), z=0$ is holomorphic?
Oct
22
comment holomorphic function power series: $f(z)=\sum_{j=0}^\infty a_j(z-z_0)^j$, can it be extended to $U$?
is there a theorem for this?
Oct
22
accepted holomorphic function power series: $f(z)=\sum_{j=0}^\infty a_j(z-z_0)^j$, can it be extended to $U$?
Oct
22
revised why $G(z)=\frac{f(z)}z, z\ne 0; f'(0), z=0$ is holomorphic?
deleted 134 characters in body; edited title
Oct
22
asked why $G(z)=\frac{f(z)}z, z\ne 0; f'(0), z=0$ is holomorphic?
Oct
22
asked holomorphic function power series: $f(z)=\sum_{j=0}^\infty a_j(z-z_0)^j$, can it be extended to $U$?