1:37 AM
Just looking to understand some basics about a function (really a limit of a sequence of functions tending to infinity) $f: \Bbb Q \to \Bbb Q$ where $f$ is a 'normalized $\pi(x)$' (prime counting function)
that's an example
Any good ways to think about $f$ as a function from the rationals to the rationals?
import numpy as np
import matplotlib.pyplot as plt
from sympy import primerange, primepi

# Define new limits for generating prime counts
limits = [10, 100, 1000, 10000]
colors = ['blue', 'green', 'purple', 'orange']

plt.figure(figsize=(12, 8))

for i, limit in enumerate(limits):
primes = list(primerange(1, limit + 1))
prime_counts = [primepi(n) for n in range(1, limit + 1)]
max_prime_count = prime_counts[-1]
normalized_prime_counts = [float(count) / max_prime_count for count in prime_counts]
x_values = np.linspace(0, 1, len(normalized_prime_counts))
code, in case it help

4 hours later…
5:47 AM
@SineoftheTime But to find the speed with which the stone hits the ground, I can do $v = \sqrt{v_x^2 + v_y^2}$ ?
$v_{0x} = v_x , v_y = v_{0y} - gt$
how do you find $t$?
@SineoftheTime $x(t) = v_0 \cdot \cos(\theta) \cdot t \to t = \frac{x}{v_0 \cdot \cos(\theta)}$
Wait, in the exercise I already know the initial angle
so you have to find $x$
use the conservation of energy to find the final speed
Yes x , I find it from $y(t) = h_0 + v0 \cdot \cos(\theta) \cdot t - \frac{1}{2}gt^2$
$y(t) = h_0 + v0 \cdot \sin(\theta) \cdot \frac{x}{v_0 \cdot \cos(\theta)} - \frac{1}{2}g \frac{x^2}{v_0^2 \cdot \cos(\theta)^2}$
why don't you use conservation? are you not allowed?
6:02 AM
I have replace t in the formula of y
@SineoftheTime When I did this topic, I had to use the parabolic motion formulas
then it's better to find $t$ and substitute instead of finding $x$
But I can see if using energy conservation I can get the same results
good idea
@SineoftheTime I take t, then replace in the y(t) equation, otherwise how can I do it?
By using the energy I will have : $\frac{1}{2}mv_0^2 + mgh = \frac{1}{2}mv_f^2$
$v_0^2 + 2gh = v_f^2 \to v_f = \sqrt{v_0^2 + 2gh}$
I always get 36, with both formulas
6:30 AM
So to find the angle of impact: $v_x = v_f \cos(\theta) \to \theta = \arccos \left(\frac{v_x}{v_f}\right)$
6:47 AM
18 hours ago, by Xander Henderson
@ThomasFinley In general, "linearity" has to do with the way that linear combinations behave. Indeed, the usual exception to this is in precalc/calculus classes, in which functions whose graphs are lines are often called "linear". For what it is worth, I think that this is not really correct, and such functions should be called "affine".
@XanderHenderson Ok, so reading your comment I feel like that we say a PDE say, $F(u)=0$ (, where $u=u(x_1,...,x_n)$ is a function of n independent variables and $F$ is an operator) is linear iff $F(p+q)=F(p)+F(q)$ and $F(cp)=cF(p)$ for any two functions $p,q$ of n- independent variables and a real number $c.$
Is this what you meant?
Also, then what does a quasi-linear PDE means?
7:44 AM
this proof here math.stackexchange.com/a/858656/1118236 seems hard to intuit doesnt it? If you were asked to prove this how would you approach? I was going down dead end roads throughout my attempts
8:01 AM
Sorry perhaps I misremembered it for statement about finite unions of subspaces
this is perhaps one of the proofs where I just read the argument and try to work it out again myself
@Jakobian no worries

5 hours later…
12:54 PM
@ThomasFinley Yes. That is the idea of linearity.
@ThomasFinley You can prove, as a theorem, that if $F$ is a linear differential operator, then $F(u) = a_1 u + a_2 u' + a_3 u'' + \dotsc + a_n u^{(n)}$ for some functions $a_j$ which do not depend on $u$.
Which is what you were originally given as a definiion.
So you can phrase linearity in terms of the form of the DE.
Quasilinearity is a breaking of that form, but only in the lower order terms.
Because I am lazy, that is the definition from Evans' book.
1:29 PM
> Theorem 7.32 Let $\mathcal{A}$ be an algebra of real continuous functions on a compact set $K$. If $\mathcal{A}$ separates points on $K$ and if $\mathcal{A}$ vanishes at no point of $K$, then the uniform closure $\mathcal{B}$ of $\mathcal{A}$ consists of all real continuous functions on $K$.
The proof is proved in four claims, where the fourth claim reads as follows:
> Claim 4 Given a real function $f$, continuous on $K$, and $\epsilon>0$, there exists a function $h\in\mathcal B$ such that $$|h(x)-f(x)|<\epsilon,\quad (x\in K).$$
It is claimed claim 4 is equivalent to the conclusion of the theorem. I don't see how. Is this clear to someone?
For me, claim 4 looks more like saying that $\mathcal B$ is dense in $C(K)$, but I may be wrong.
2:04 PM
Staring at this for some time, I think I've made some progress in my understanding. Claim 4 states that an arbitrary function $f\in C(K)$ is an adherent point of $\mathcal B$, i.e. $$B_\epsilon(f)\cap \mathcal B\neq\varnothing.$$ The closure of $\mathcal B$ (which is $\mathcal B$, as it is closed) are all the adherent points to $\mathcal B$, i.e. all $f\in C(K)$ and thus $\mathcal B=C(K)$.
Although I'd still be curious to hear if Claim 4 actually has something to do with density. This would help my understanding further.
@psie the statement I usually see is that $\mathcal{A}$ contains all constants, instead of the condition of vanishing at points
not sure if this is equivalent
@psie $\mathcal{B}$ is closed, so being dense would mean the two are equal
2:20 PM
ah yeah, right, now I remember
@psie it is density. I don't really like your talk with "adherent points", I don't feel like that brings anything to the discussion
ok, I'll have to think this through again
For a set to be dense it means that its closure is whole space, or equivalently, given some base, that it intersects all elements of it
open balls are an example of such base
@AlessandroCodenotti @Thorgott You can prove that bounded uniformly continuous functions extend to uniformly continuous functions by first restricting to the case of one pseudometric (uniform continuity of $f$ depends on countable amount of pseudometrics on $X$, and then you can construct just one admissible one for which it does, similarly to how you prove a countable product of metrizable spaces is metrizable). Then you can define a new pseudometric on which $f$ is Lipschitz, and extend it
This is by a known argument that Lipschitz functions on subspace extend to Lipschitz functions with the same constant. You construct it by using the modulus of continuity $\omega_0(t) = \inf\{|f(x)-f(y)|: d(x, y)\leq t\}$ by modifying it to a function $\omega$ so that its subadditive, continuous, $\omega(0) = 0$ and increasing.
2:47 PM
Actually you don't need that $\omega$ is continuous, just that its continuous at $0$. Although it is continuous
Hello, everyone. How to solve the differential equation $\frac{dy}{dx} = \sqrt{2}y$? I was going to do it via separation of variables, but then an answer on Math.SE changed my mind and now I'm in difficulty...
@Bml Well, it is a separable equation, so that seems like the right approach. But you haven't given your attempt, nor a link to the confusing Math SE answer, so it is hard to know where your problem lies...
18

Most explanations of the method of separating variables do not make clear that it only works on a region where the arithmetic operations are all valid, including the division by $y$. Here is an example where the method fails to find the correct answers if you anyhow perform invalid operations. (W...

@Bml Answers are not always presented in the same order, so "the second answer" is not meaningful. But you have linked to an answer---is that the one you mean?
Can you be more specific?
@XanderHenderson Yes
@XanderHenderson My attempted resolution simply gives $f(x) = c_1 e^{\sqrt{2}x}$, in accordance with all differential equations of the type $y' = f(x) y$. I do not understand what is wrong with this, because the answer seems to suggest that this is not right.
2:58 PM
I don't see how the answer suggests that you are wrong. The answer only suggests that you need to be careful about the domain of your solution.
@XanderHenderson The answer says WolframAlpha is also wrong about the solution of the ODE. I put this differential equation on WolframAlpha and it gives exactly the result I said before...
@XanderHenderson After that, the answer says "The answer is not $y = (x+a)^2$, which you would get by the method of separating variables.". How do we obtain that result - even wrong - by method of separating variables?
It is about the domains of solutions, not the solutions themselves.
I have no idea how the user got $(x+a)^2$, but that isn't really the point they are trying to make.
@XanderHenderson I read it, but I still don't see where that $(x+a)^2$ could come from, even if it should be wrong as the solution says.
@XanderHenderson Exactly...
Oh, no. I do see where it comes from. It just isn't the same question which is being asked at the top of that thread.
I still haven't seen your solution.
In any event, your equation seems to be $y' = \sqrt{2} y$, while the one you cite is $y'= \sqrt{y}$.
Not the same...
3:45 PM
@XanderHenderson Yes, I said $y' = \sqrt{2} y$, while it was $y'= 2 \sqrt{y}$. However, I still don't see how he obtains $y= (x+a)^2$...
@Bml Well, again, you haven't shown your work, so I have no idea where your confusion is.
@XanderHenderson Mathematica said $y= 1/4 (2x+a)^2$
Mathematica said that was a solution to $y' = \sqrt{y}$?
@XanderHenderson Mathematica/WolframAlpha says that a solution of $y'= 2 \sqrt{y}$ is $y= 1/4 (2x+a)^2$.
@Bml That isn't the DE I've been talking about.
3:52 PM
37 mins ago, by Xander Henderson
In any event, your equation seems to be $y' = \sqrt{2} y$, while the one you cite is $y'= \sqrt{y}$.
But you still haven't shown your work.
@XanderHenderson All right, I will now send my work.
What have you done? I have done nothing to solve this problem, because it isn't my problem. I don't actually care what the solution is, and I haven't worked it out for myself.
@XanderHenderson I'm sending it, only two minutes :-)
Also $\frac{1}{4} (2x + a )^2 = (x+a/2)^2$, no?
And $a$ is an arbitrary constant.
So $\frac{1}{4} (2x+a)^2 = (x+a)^2$.
They are the same solution.
3:59 PM
@XanderHenderson OK, so problem solved. If it weren't for the fact that this isn't the correct result either, right?
YOU STILL HAVEN'T SHOWN ME YOUR WORK!
Is it reasonable?
That looks like a reasonable way of solving the problem as that kind of solution technique is usually taught in elementary calculus classes. But you fail to indicate the domain of the solution.
@XanderHenderson I was sending but my phone didn't receive the command, sorry...
@XanderHenderson How can I achieve this?
How can you "achieve" what?
4:12 PM
@XanderHenderson You said that I fail to indicate the domain of the solution. How can I achieve the correct result by indicate the domain of the solution?
Well, you would indicate the domain on which the solution is valid.
@XanderHenderson The answer says "First prove that for any point where y≠0, there is an open interval around x for which y≠0." How to prove that?

1 hour later…
5:26 PM
@XanderHenderson Is the proof straightforward?

4 hours later…
9:19 PM
I'm studying a theorem that states that the Laplace transform of $X$ determines the law $\mathbb P_X$ of $X$. In the proof of the theorem, we are considering the vector space $H$ of functions $\psi_{\lambda}=e^{-\lambda x}$ for $\lambda\geq 0$. By Stone-Weierstrass, $H$ is dense in $C([0,\infty],\mathbb R)$ (we include $\infty$ by extending $\psi_\lambda$ by continuity).
If we denote the Laplace transform by $L_X(\lambda)=\mathbb E[e^{-\lambda X}]$, then $L_X=L_{X'}$ implies $\mathbb E[\phi(X)]=\mathbb E[\phi(X')]$ for every $\phi\in H$. Then it is claimed that the mapping $$\phi\mapsto\mathbb E[\phi(X)]=\int \phi\,\mathrm{d}\mathbb P_X,$$ is clearly continuous on $C([0,\infty],\mathbb R)$. Why is it "clearly continuous" on $C([0,\infty],\mathbb R)$. I don't see this.
@psie by vector space $H$ of functions $\psi_\lambda$, do you mean linear combinations of them
@Jakobian yeah, "let $H$ be the vector space spanned by the functions $\psi_\lambda,\lambda\geq 0$."
that's what the book says
sure, $H$ contains all constants, is a subring of $C([0, \infty], \mathbb{R})$ and separates points
indeed, so it is dense in $C([0, \infty], \mathbb{R})$ by SW.
$\left |\int \phi_1 - \int\phi_2 \right | \leq \int |\phi_1-\phi_2| \leq \|\phi_1-\phi_2\|_\infty$
Its Lipschitz with constant $1$
9:34 PM
alright, cool :) and $\lVert \cdot\rVert_\infty$ is the sup-norm, right?
yes
awesome
No, it's the soup norm.
:P