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2d
asked What general mobius transformation maps $|z-1|=1$ to itself and $|z+1|=1$ to $|w-3|=3$.
2d
accepted If $f$ is analytic in $D$ and $|f(z)|<M$ everywhere on $|z|=1$, show for all $z:|z|<1$, $|f(z)| \leq M |\frac{z-a}{\bar a z - 1}|$
Feb
3
revised If $f$ is analytic in $D$ and $|f(z)|<M$ everywhere on $|z|=1$, show for all $z:|z|<1$, $|f(z)| \leq M |\frac{z-a}{\bar a z - 1}|$
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Feb
2
asked If $f$ is analytic in $D$ and $|f(z)|<M$ everywhere on $|z|=1$, show for all $z:|z|<1$, $|f(z)| \leq M |\frac{z-a}{\bar a z - 1}|$
Feb
1
comment Let $V$ be the space of complex polynomials on $[0,1]$. Is the differentiation operator self-adjoint?
@Qidi, I have revised my question to clearly define $D$. I don't think this aligns with your suggestion, does it? Where can I find that definition of the differential operator, which you say is the usual convention?
Feb
1
revised Let $V$ be the space of complex polynomials on $[0,1]$. Is the differentiation operator self-adjoint?
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Feb
1
comment Let $V$ be the space of complex polynomials on $[0,1]$. Is the differentiation operator self-adjoint?
Integration by parts gives me $\langle Df, g \rangle = \sum\sum a_i \overline b_j - a_0 \overline b_0 - \int_0^1 f \frac{\partial \overline g}{\partial t} dt$. Does that give us something useful?
Feb
1
asked Let $V$ be the space of complex polynomials on $[0,1]$. Is the differentiation operator self-adjoint?
Feb
1
accepted Suppose $T$ is diagonalizable in $\mathbb{C}$. Show $e^T = \sum_{\lambda \in sp(T)} e^\lambda P_\lambda$ is the matrix exponential series.
Jan
31
comment Suppose $T$ is diagonalizable in $\mathbb{C}$. Show $e^T = \sum_{\lambda \in sp(T)} e^\lambda P_\lambda$ is the matrix exponential series.
Thanks for the hint. Is there a reason to do this over the other suggested answers which use the series representation of the exponential function?
Jan
30
asked Suppose $T$ is diagonalizable in $\mathbb{C}$. Show $e^T = \sum_{\lambda \in sp(T)} e^\lambda P_\lambda$ is the matrix exponential series.
Jan
18
accepted (Ahlfors, p198) Why is it clear we can write $G(z-1)=ze^{\gamma(z)}G(z)$ when deriving the Gamma function?
Jan
17
asked (Ahlfors, p198) Why is it clear we can write $G(z-1)=ze^{\gamma(z)}G(z)$ when deriving the Gamma function?
Jan
17
accepted Laurent expansion of $1/(1+z^n)$ for $n \in \mathbb{N}$.
Jan
16
asked Laurent expansion of $1/(1+z^n)$ for $n \in \mathbb{N}$.
Jan
13
accepted Let $f(x) = (x^n-1)/(x-1)$. Why does $f(1)=n$?
Jan
13
comment Let $f(x) = (x^n-1)/(x-1)$. Why does $f(1)=n$?
See my edit to the original post. $f$ is defined as a polynomial so I think you are right in that in order for $f$ to be a polynomial, it must be defined that $f(1)=n$.
Jan
13
revised Let $f(x) = (x^n-1)/(x-1)$. Why does $f(1)=n$?
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Jan
13
asked Let $f(x) = (x^n-1)/(x-1)$. Why does $f(1)=n$?
Jan
3
comment Let $f(z) = \frac{z^{-2}}{\sin( \pi z )}$. What is the residue for $z \neq 0$?
Thank you for the clarification. Understood