conformal map f(z) mapping two curves through
z0 to two curves through w0 = f(z0). The tangent vectors to each of the original curves are
both rotated and scaled by the same amount. How to plot this?
Tag: functions
Rouches Theorem in Complex Analysis on the relation of the number of zeros and poles of meromorphic functions in a region
This question is from my son referenced in my earlier question, Need advice or assistance for son who is in prison. His interest is scattering theory . He asked me to post this question:
Hello and thanks to everyone for help finding papers thus far. I am currently looking for some further information and applications of Rouches Theorem in Complex Analysis on the relation of the number of zeros and poles of meromorphic functions in a region. I have the basic statement, but am looking for some more advanced or peripheral results, reformolulations, extensions, etc. Any other theorems with conditions for the relation of the poles and zeros of two functions in a region would also be helpful.
To be very specific, if f=g+h, with all functions meromorphic in the plane, I’m looking for conditions on f, g, h, so that f and g have the same number of poles and zeros in a region. The form of the particular functions I’m dealing with are generally highly oscillatory, nonlinear Fourier transforms of smooth, compactly supported functions where the nonlinearity can cause poles, but sometimes their real and imaginary parts can be controlled well, so conditions relating their arguments, or real/im parts might be useful. Thanks.
-Travis.
real analysis – Looking for non-polynomial functions: with the growth condition: $phibig(theta frac{s}{t}big) leq frac{phi(s)}{phi(t)}$
I am for example(s) of an invertible increasing function $phi: (0,infty)to (0, infty)$ such that $phi(0)=0$ and there exists $theta>0$ and for all $sleq t$ we have
begin{align}label{EqI}tag{I}
phibig(theta frac{s}{t}big) leq frac{phi(s)}{phi(t)} qquadtext{or equaly} qquad theta leq phi^{-1}big(frac{s}{t}big)frac{phi^{-1}(t)}{phi^{-1}(s)}
end{align}
The most simple class consists of polynomial functions of the form $phi(t)= ct^p$ with $c>0$ and $p>0$.
Question: Are there other possible non-polynomial examples satisfying $eqref{EqI}$?
As an attempt with $phi(t)= e^{t^alpha}-1$, I wonder if there is a constant $c>0$ such that
$$ ln(t+1)lnleft(frac{s}{t}+1right)geq cln(s+1),qquad text{for all $0leq sleq t$}.$$
asynchronous – Syntax for TypeScript Arrow Functions With Generics
I don’t know if this is done properly or not, but I posted an answer in StackOverflow back in Sept. 2020 about implementing generics in TypeScript. I got downvoted 3 times, and I was wondering if the following piece of code was problematic:
This is how I defined an async generic function expression in TypeScript
const request = async <T>(param1: string, param2: number) => {
const res = await func();
return res.response() as T;
}
And a more complex pattern, in case you’d like to wrap your function inside a generic counterpart, such as memoization (Example uses fast-memoize):
const request = memoize(
async <T>(
url: string,
token?: string
) => {
// Perform your code here
}
);
See how you define the generic after the memoizing function.
This function has worked well for me, so I don’t see any issues.
This is the original SO question.
complexity theory – Classes of Functions Closed Under Polynomial Composition – Papadimitriou Exercise 7.4.4
I am studying Computation complexity using Papadimitrious’s book: “Computational Complexity”.
I am trying to do Problem 7.4.4:
“Let $C$ be a class of functions from nonnegative integers to nonnegative integers. We say that $C$ is closed under left polynomial composition if $f(n) in C$ implies $p(f(n))=O(g(n))$ for some $g(n) in C$, for all polynomials $p(n)$. We say that $C$ is closed under right polynomial composition if $f(n) in C$ implies $f(p(n))=O(g(n))$ for some $g(n) in C$, for all polynomials $p(n)$.
Intuitively, the first closure property implies that the corresponding complexity class is “computational model-independent”, that is, it is robust under reasonable changes in the underlying model of computation (from RAM’s to Turing machines, to multistring Turing machines, etc.) while closure under right polynomial composition suggests closure under reductions (see the next chapter).”
Which of the following classes of functions are closed under left polynomial composition, and which under right polynomial composition?
(a) – ${n^k: k > 0 }$
(b) – ${k cdot n: k > 0 }$
(c) – ${k^n : k > 0 }$
(d) – ${2^{n^k} : k > 0 }$
(e) – ${log^k n: k > 0 }$
(f) – ${log n}$
After understanding the definition of closed under left/right polynomial composition, I think I was able to solve items (a), (b), (c) and (f). However, I was not able to solve items (d) and (e).
My solution for item (a):
Closed Under Left Polynomial Composition: consider an arbitrary $f(n) in C$ and an arbitrary polynomial $p(n)$. Then, $f(n)$ is of the form $n^{k’}$, for some $k’ > 0$ and therefore $p(f(n))$ is a polynomial. Let $k”$ be the degree of the polynomial $p(f(n))$. Take $g(n) = n^{k”} in C$ and we have $p(f(n)) = O(g(n))$.
Closed Under Right Polynomial Composition: same reasoning.
My solution for item (b):
Not Closed Under Left Polynomial Composition: consider as a counterexample $f(n) = n in C$ and $p(n) = n^2$. Then, $p(f(n)) = n^2$. For every $g(n) = k n in C$ we have $O(g(n)) = O(n)$. Since $n^2 neq O(n)$ we conclude.
Not Closed Under Right Polynomial Composition: the previous counterexample applies.
My solution for item (c):
Closed Under Left Polynomial Composition: Consider an arbitrary $f(n) = k_1^n$ and a polynomial $p(n)$. Notice that $p(f(n))$ is a polynomial in $k_1^n$. For sufficiently large $n$, there exists some $k_2$ such that $p(n) leq n^{k_2}$ and therefore $p(f(n)) leq (f(n))^{k_2} = (k_1^{n})^{k_2} = (k_1^{k_2})^n$. Therefore, taking $g(n) = (k_1^{k_2})^n in C$ we obtain that $p(f(n)) = O(g(n))$.
Not Closed Under Right Polynomial Composition: Consider as a counterexample $f(n) = 2^n$ and $p(n) = n^2$. Then, $f(p(n)) = 2^{n^2}$, which is greater than $g(n) = k^n$, for every fixed value of $k$, if $n$ is sufficiently large. Therefore, $f(p(n)) not in O(g(n))$.
My solution for (f):
Not Closed Under Left Polynomial Composition: Consider as a counterexample $f(n) = log n$ and $p(n) = n^2$. Then, $p(f(n)) = log^2 n$. Also, $g(n) in C$ implies that $g(n) = O(log n)$. We have $log^2 n not in O(log n)$.
Closed Under Right Polynomial Composition: If $f(n) in C$ then $f(n) = log n$. Given an arbitrary polynomial $p(n)$, we have that there exists some $k’$ such that, for sufficiently large $n$, $p(n) < n^{k’}$. Then, for sufficiently large $n$:
$ f(p(n)) leq f(n^{k’}) = log n^{k’} = k’ log n = O(log n) = O(g(n)).$
Can anyone help me with items (d) and (e)?
Thanks in advance. Of course, corrections/comments on the other items are also welcomed.
rt.representation theory – The product of $Z(mathfrak{g})$-finite functions is also $Z(mathfrak{g})$-finite?
Let $G$ be a classical group defined over $mathbb{Q}$.
Let $mathfrak{g}$ be the Lie algebra of $G(mathbb{R})$ and $U(mathfrak{g}_{mathbb{C}})$ its universal enveloping algebra of $mathfrak{g}_{mathbb{C}}$.
Let $Z(mathfrak{g})$ be the center of $U(mathfrak{g}_{mathbb{C}})$. We regard the elements of $U(mathfrak{g}_{mathbb{C}})$ as differential operators on $C^{infty}(G)$, the space of smooth functions on $G(mathbb{R})$, acting by right infinitesimal translation.
Let $f,g in C^{infty}(G)$ be $Z(mathfrak{g})$-finite. (I.e. $<zcdot f | zin Z(mathfrak{g})>, <zcdot g | zin Z(mathfrak{g})>$ are finite dimensional vector space.)
Then I am wondering whether $f cdot g$ is also $Z(mathfrak{g})$-finite.
Any comments are appreciated!
built in symbols – What could go wrong in redefining Power to act on functions?
Inspired by a tweet (from a locked account, unfortunately, so I can’t link it), we can get notation like $cos^2(x)$ to work in Mathematica in general:
Unprotect(Power);
Power(f_, n_)(x___) := Power(f(x), n);
Protect(Power)
Weirdly, so far, I can’t think of anything this would break, since—as far as I can tell—Power(f_, n_) is never otherwise expected to appear as the head of some other expression. It seems like this might actually be a convenient choice.
Is there any chaos this could cause that I’m not thinking of?
How do you use ad hoc polymorphism/function overloading with functions in Python?
So, let’s say you’ve got a function foobar() which can function with a variable number of parameters inputted into it, and has different behavior for each of them. How do you get this to function properly in Python? Are you forced to name every possible combination of parameters something different?
For instance, let’s take the following code:
def foobar(foo):
print(foo)
def foobar(foo,bar):
print(foo+bar)
foobar("Hello")
foobar("Hello","World")
This gives the following error: “TypeError: foobar() missing 1 required positional argument: ‘bar'”. If you reverse the order of foobar(foo) and foobar(foo,bar), you instead get the error “TypeError: foobar() takes 1 positional argument but 2 were given”. From this, it seems like Python only stores one copy of foobar(), related to whichever version was the latest to be defined.
How do you get code like this to function properly in Python? Is there a “Pythonic” way of accomplishing this goal? I can see a bunch of other questions asking about polymorphism in Python, but none of them seem to address this issue; they all seem to be talking about different classes with methods that use the same name, or passing different sorts of variables into a function as a parameters.
magento2 – Magento 2 Uncaught TypeError: Constr is not a constructor when returning functions on Knockout Js
i’m trying to get functions from a JS file in a Knockou file, like this:
quote.js
define(('jquery', 'uiComponent', 'ko','Vendor_GetQuote/js/ajax-request'), function ($, Component, ko, ajaxrequest) {
'use strict';
And I try to access like this
ajaxrequest.functionA();
ajaxrequest.functionB();
With just one function it works, but with 2 or more it does not.
This is my ajax-request.js file
define(
(
'jquery'
),
function(
$
) {
'use strict';
var methods = {
functionA: function() {
console.log('This is from module A, functionA');
},
functionB: function() {
console.log('This is from module A, function B');
}
};
return methods;
}
);
This is the error I’m getting:
Uncaught TypeError: Constr is not a constructor
Can someone orient me?
Thanks!
inverse trigonometric functions in traditional form
Is there a way to have ‘arcsin’ rather than ‘sin^-1’ as arcsin in TraditionalForm? I couldn’t find any way to set options for customizing the way in which expressions are turned into traditional form, and feel that ‘arcsin’ is widely more used among mathematicians than sin^-1.
Thanks in advance to everybody.
