My goal is to find a Mobius transformation $g$ that sends $K : |z| < R$ bijectively to itself, and also sends $0$ to $a \in R$. To my knowledge, there is a theorem that says the following:
For a disc $K: |z| \leq 1$ and with $f:K \rightarrow K$ being analytic on $|z| <1$, $f$ is of the following...
$g_a$ is the desired function $D_R\to D_R$ which sends $0$ to $a$, and $f_{a/R}$ is the specified function with $a/R$ instead of $a$ (it is $D_1\to D_1$ and sends $0$ to $a/R$
(where $D_r$ is the disk of radius $r$ centered at $0$)
I think you want $\displaystyle f_a(z)=\frac{a-z}{1-\bar{a}z}$ to be the mobius transformation of the unit disk that sends $0$ to $a$ (where $|a|\le 1$)
then $g_a=Rf_{a/R}(z/R)$ is the mobius transformation of the disk of radius $R$ which sends $0$ to $a$ (where $|a|\le R$)
Is there a book which covers all of the core graduate school mathematics in a dense and compact manner? This includes all core material: algebra, analysis, combinatorics and graph theory. I have found a book by Hua but it is geared towards undergraduate mathematics.
@Forever: It was after I already had a job offer after grad school that I discovered a major gap in the proof of my main theorem for my Ph.D. You can imagine a week of sleepless nights.
I had a group action on a space and a function (well, actually differential forms, but that doesn't matter) invariant under the action. I claimed the function had to be constant.
Grad school takes a lot of time, @Shahab. I don't get your point.
@Forever: Honestly, the chances are probably infinitesimal. But, even if it happens, if you know nothing about the other person (and vice versa), it is still degree-worthy (if it was in the first place).
No, @Simeon.
Unless you know the set of roots is a discrete set.
Real quick... Is it possible to write $$\int_{c}^{d}\int_{a}^{b}(x^2y+2x^3y+3x^2y^2)\,\text{d}x\text{d}y$$ as a sum of three double integrals and then "split" each by $x$ and $y$?
I still don't quite understand from earlier @ForeverMozart
I believe the line integral can be thought of as drawing a "curtain" under the scalar function, giving the area. Much like an ordinary univariate def. integral
Right... Probably overkill anyhow. I suppose a surface integral simply sums all the scalar values along the surface like any integral of this type would.
the one i've got right now though reminds me of Mr. Mackey from South Park in the "drugs are bad, mmmkay?" episode when his head turns into a balloon and flies away
Just out of interest I tried to solve a problem from "Recent" category of Project Euler ( the link Digit Sum sequence ). But I am unable to think of a way to solve the problem efficiently. The problem is as follows ( in the original question sequence has two ones in beginning , but it does not ch...
@AlexClark Hey, I see that Peter McNamara answered an question on MO that was so old that Ben Webster (who had asked it) had forgotten why he was interested in it in the first place :)
the only eccentric prof i had in primary school, he wore a pair of big oval glasses that made his eyes unproportionally big relatively to his face
and he talks in cryptic way, he uses mathematical symbols like overall, it exists, equivalence , induction, inclusion, deduction, all these logical notations that make his interlocutor waiting the opportun moment to give up the convo
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Book - An Introduction To Algebraic Topology by Joseph J. Rotman
A fellow student was talking to me today about infinitesimals today, and I guess I never really knew or considered the true definition of an infinitesimal... Is one allowed to use them practically in mathematics? I'd like to learn more about this
@datalava Indeed, the actual definition is easy enough, it is just a matter of constructing a place where it lives ($x$ is an infinitesimal if $0< x < a$ for all positive reals $a$)
Could some give me a hint in proving that if $f$ is not continuous at $c$ then it is not necessary that $\lim_{n \rightarrow \infty}f(x_n) = f(c) $ even if $\lim_{n \rightarrow \infty}x_n =c$. I think I need to construct a counterexample and then use the definition of discontinuous but I am not too sure exactly how.
So if we pick $\delta = \frac1k$ and then we construct a sequence $x_k$ such that $|x_k -c | < \frac1k$ then we would have $|f(x_n)-f(c)| \geq \epsilon $ given $|x-c|<\delta$ ?
How can I go about finding two continuous, non-differentiable (at $x=0$) functions such that $f(0)=g(0)=0$ and $f(g(0))=0$ but $f(g(x))$ is differentiable at $x=0$. I feel like you'll need $f=g^{-1}$ but I haven't made any progress in finding one.