Hello one and all! Is anyone here familiar with planar prolate spheroidal coordinates? I am reading a book on dynamics and the author states If we introduce planar prolate spheroidal coordinates $(R, \sigma)$ based on the distance parameter $b$, then, in terms of the Cartesian coordinates $(x, z)$ and also of the plane polars $(r , \theta)$, we have the defining relations $$r\sin \theta=x=\pm R^2−b^2 \sin\sigma, r\cos\theta=z=R\cos\sigma$$ I am having a tough time visualising what this is?
Consider the function $f(z) = Sin\left(\frac{1}{cos(1/z)}\right)$, the point $z = 0$
a removale singularity
a pole
an essesntial singularity
a non isolated singularity
Since $Cos(\frac{1}{z})$ = $1- \frac{1}{2z^2}+\frac{1}{4!z^4} - ..........$
$$ = (1-y), where\ \ y=\frac{1}{2z^2}+\frac{1}{4!...
This is very dumb but I don't see how the translation operator, translating a vector by a constant vector c is linear. T(x + y) = x + y + c which is not the same as T(x) + T(y) = x + y + 2c
i.e., $T_x(\psi + \phi) = T_x(\psi) + T_x(\phi)$ as wavefunctions. That's the same thing as saying $T_x$ is a linear transformation on the space of wavefunctions
I am having trouble understanding non-isolated singularity points. An isolated singularity point I do kind of understand, it is when: a point $z_0$ is said to be isolated if $z_0$ is a singular point and has a neighborhood throughout which $f$ is analytic except at $z_0$. For example, why would $...
@Simone We don't have automatic MathJax in chat, but there are scripts that can be used. The bookmarklets kind of work on mobile. See meta.stackexchange.com/a/220976/334566
No worries. There's currently some kind of technical problem affecting the Stack Exchange chat network. It's been pretty flaky for several hours. Hopefully, it will be back to normal in the next hour or two, when business hours commence on the east coast of the USA...
@Simone Mathoverflow is primarily for research level mathematics, while Mathstack is for mathematics of any level (and if it is deemed worthy it will be migrated to Mathoverflow)
The absolute value of a complex number $z=x+iy$ is defined as $\sqrt{x^2+y^2}$. Hence, when evaluating the absolute value of $x+i$ I get the number $\sqrt{x^2 +1}$; but the answer to the problem says it's actually just $x^2 +1$. Why?
mmh, I probably should ask this on the forum. The full problem asks me to show that we can choose $log(x+i)$ to be $$log(x+i)=log(1+x^2)+i(\frac{pi}{2} - arctanx)$$ So I'm trying to find the polar coordinates (absolute value and an argument $\theta$) of $x+i$ to then apply the $log$ function on it
@Pie pinging everyone in the chat is bad practice, if you want to ask a question just post it and somebody will look at it if they decide to, if not just post it on main
Consider the function $f(z) = Sin\left(\frac{1}{cos(1/z)}\right)$, the point $z = 0$
a removale singularity
a pole
an essesntial singularity
a non isolated singularity
Since $Cos(\frac{1}{z})$ = $1- \frac{1}{2z^2}+\frac{1}{4!z^4} - ..........$
$$ = (1-y), where\ \ y=\frac{1}{2z^2}+\frac{1}{4!...
