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2d
comment What general mobius transformation maps $|z-1|=1$ to itself and $|z+1|=1$ to $|w-3|=3$.
Thank you, that is helpful. Another question: how do you know to choose $D$ and $G$ as you have chosen? It seems a little arbitrary to me. For example, how do you know not to to choose the interior of the two circles in the domain? And why is $G$ chosen to be the space between the two circles in the range?
2d
accepted What general mobius transformation maps $|z-1|=1$ to itself and $|z+1|=1$ to $|w-3|=3$.
2d
comment What general mobius transformation maps $|z-1|=1$ to itself and $|z+1|=1$ to $|w-3|=3$.
Thanks for this. The mathematics makes sense to me but how do you motivate such an approach? For example, how did you know we needed to consider a transformation $h$ from $E$ to $F$?
Feb
8
comment If $T$ is an orthogonally diagonalizable linear operator in an inner product space, show that $T^*$ is also orthogonally diagonalizable.
@Omnomnomnom But aren't the $PDP^*$ meant to be understood as matrices in such a representation? Or as the composition of linear operators? In which case, it could also be written $P \circ D \circ P^*$?
Feb
8
asked If $T$ is an orthogonally diagonalizable linear operator in an inner product space, show that $T^*$ is also orthogonally diagonalizable.
Feb
7
comment Let $\{e_1,\ldots,e_n\}$ be an arbitrary basis in a finite dimensional inner product space. Prove $\exists \{f_1,\ldots,f_n\}: (e_i,f_j)=\delta_{ij}$
Does this really show they exist or just that $(A^T)^{-1}$ would have the $f_i$ as columns if they did exist?
Feb
6
asked Let $\{e_1,\ldots,e_n\}$ be an arbitrary basis in a finite dimensional inner product space. Prove $\exists \{f_1,\ldots,f_n\}: (e_i,f_j)=\delta_{ij}$
Feb
4
asked What general mobius transformation maps $|z-1|=1$ to itself and $|z+1|=1$ to $|w-3|=3$.
Feb
3
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 $z:|z|<1$, $|f(z)| \leq M |\frac{z-a}{\bar a z - 1}|$
deleted 38 characters in body
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?
added 87 characters in body
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?