@PM2Ring def theta(x): return sum(math.exp((log(p))^-1) for p in prime_range(1,floor(x)+1)) def psi(x): return sum(math.exp((log(1/p))^-1) for p in prime_range(1,floor(x)+1))
so basically it's useless because it just about halves the differences and doesn't effect the big O
@geocalc33 Here's a demo. We use a for loop to put the cumulative sums into a list. We could compute both sums in the same loop. This code takes a second or two to plot up to m=100000. Don't forget to comment out the print call. ;)
Speaking of prime random walks, here's an image I just made from the residues mod 5 of the primes <10,000,000. Start at (0, 0). Then if the next residue is 1 increase x, if it's 4 decrease x, if it's 2 increase y, if it's 3 decrease y. Count the times you visit each point, and convert that to a color. I do a log(1 + k) transform (where k is the visit count) to make the pattern more visible.
The recent kerfuffle about the Riemann hypothesis inspired me to write a little Sage / Python script to explore the graph of the cumulative Liouville function, A002819. There are a couple of graphs on Wikipedia, but they're pretty crude, and you can't zoom in.
Let $f,g:\Bbb R\to\Bbb R$ be nonnegative increasing (not necessarily strict) functions such that $f(x)\sim g(x)$ as $x\to\infty$. Then $\int_1^xf(t)\ dt\sim\int_1^xg(t)\ dt$ as $x\to\infty$?
Oh there is a known result that if one of the integral diverges then the statement holds.
@TedShifrin $f(x)\sim g(x)$ as $x\to\infty$ if $\lim_{x\to\infty}{f(x)\over g(x)} =1$.
Actually the problem is from complex analysis which is: if $\psi(x) = \sum_{p\leq x,p:\text{prime}}\lfloor {\log x\over\log p}\rfloor \log p$ and $\psi(x)\sim x$ as $x\to\infty$ then $\int_1^x\psi(u)\ du\sim{x^2/2}$ as $x\to\infty$.
So you're saying $F(x) = \int_1^x f(t)\ dt, G(t) = \int_1^xg(t)\ dt$ then by LH $${F(x)\over G(x)} = {f(x)\over g(x)}$$ and RHS $\to 1$ as $x\to\infty$ assumping $F,G\to\infty$ as $x\to\infty$.
Oh, i should have said $\nu (E)=\int_E \frac f{1-f} d\color{red}{\mu}$
It is true that $\mu<<\mu+\nu$ so there exists Radon Nikodym derivative $h$ so that $\mu(E)= \int h d(\mu+\nu)$. It is given that $\nu (E)=\int_E f d(\mu+\nu)$. Adding the two gives: $\mu+\nu(E)=\int_E (f+h) d(\mu+\nu)$ So by uniqueness of the Radon Nikodym derivative, I get $f+h= 1$ almost everywhere.
First, we prove a lemma:
Lemma: Let $(X, M, \mu)$ be a measure space. Let $f$ be a nonnegative measurable function on $(X, M)$. Define a measure $\nu$ on $(X, M)$ by $\nu(E) = \int_{E}fd\mu$ for $E \in M$. Then for any nonnegative, measurable function $F$ on $(X, M)$, we have
$$\int_{X}Fd\nu = ...
If $p:\Bbb R^{n+1}\setminus\{0\}\to S^n$ is a projection map $x\mapsto x/||x||$ and $\omega$ is a closed $n$-form on $\Bbb R^{n+1}\setminus\{0\}$ then is are any $\eta\in\Omega^n(S^n)$ such that $p^*\eta = \omega$?
@TedShifrin The actual question is different. If $\omega\in\Omega^n(\Bbb R^{n+1}\setminus\{0\})$ is a closed form and $M$ is a compact orientable manifold without boundary with a smooth map $f:M\to\Bbb R^{n+1}\setminus\{0\}$ then $\int_Mf^*\omega$ is an integer multiple of $\int_{S^n}\iota^*\omega$. If $p$ is the projection map I defined, then $\int_M(p\circ f)^*\omega = k\int_{S^n}(p\circ\iota)^*\omega$ for some $k\in\Bbb Z$ where $\iota$ is an inclusion map $S^n\to\Bbb R^{n+1}\setminus\{0\}$
One should say that f(x)= g(|x|), f is from R^n to R. So the integral I wrote simply becomes $\sigma (S^{n-1}) \int_0^\infty \color{red}{g(r)} r^{n-1} dr$.
There was a question in the exam, Using the CLT, show that $\lim\sqrt n P(X=n/2)=\sqrt{2/\pi}$ where $X$ is the number of heads when we flip a coin $n$ times. They also gave a hint that for a continuous function $\int_{-a}^a f(x)\sim 2af(x)$.
Here $n$ is even.
So the thing is that you have to write $P(X=n/2)=P(\frac{n-1}2<X<\frac{n+1}2)$ then you'll get the $\sqrt{2/\pi}$
What I did was $P(X=n/2)=P(\frac{n}2-\epsilon<X<\frac{n}2+\epsilon)$ where $\epsilon$ is a small real number $<1$. Then I found the end result in the form of $\epsilon$.
So changing the epsilon values we'll get different limits so limit shouldn't exist.
Why do we only take $\frac12$? I have seen other problems where they use $\epsilon=\frac12$.
Since binomial r.v. pmf is only concentrated at integers so it shouldn't matter what epsilon we take. RIght?
One (surprising?) thing I found: If $f:\Bbb C\to\Bbb C$ is a polynomial then we can extend this map to a continuous map $\bar{f}:\Bbb C_\infty\to\Bbb C_{\infty}$. The degree of $\bar{f}$ in a topological sense is equal to the degree of $f$ (polynomial degree).
I suspect I have found a typo in Spivak’s Calculus, third edition, on page 144. He writes “Then the Lemma implies that $f$ is $\epsilon$-good on $[a, a+\delta_0]$, so $a+\delta_0$ is in $A$…” I suspect he means $\alpha+\delta_0$ instead of $ a+\delta_0$.
@onepotatotwopotato Yes, it is known (and should appear in textbooks) that polynomials are meromorphic functions of the Riemann sphere, and the degree coincides with the index at infinity.
@Jakobian $X$ is a compact space, $C_p(X)$ is the space of continuous real valued functions $X\to\Bbb R$ with the pointwise convergence topology. I want to show that if $Y\subseteq C_p(X)$ is closed, then $Y$ is compact iff it is countably compact
It's fairly important in the study of compact subsets of function spaces and in the study of Rosenthal compacta. The main result is the Bourgain-Fremling-Talagrand dychotomy though
I don't understand a step in the proof, but it's toward the end, do you have time and are you interested in quickly going through the proof? (It's not long, less than a page in the book I'm looking at)
Alright. We assume that $Y$ is countably compact and we want to show that it is compact. Since countably compact means that every sequence has an accumulation point we can find, for every $x\in X$ a constant $M_x$ such that $|f(x)|\leq M_x$ for all $f\in Y$. This shows that $\overline{Y}$ (closure taken in $\Bbb R^X$) is compact, since it is closed in $\prod_{x\in X}[-M_x,M_x]$, so we have reduced to showing that in fact $\overline{Y}\subseteq C_p(X)$.
Suppose for a contradiction that there is a discontinuous $f\in\overline{Y}$. We can find $\varepsilon>0$ and $y\in X$ such that $Z=X\setminus f^{-1}(f(y)-\varepsilon,f(y)+\varepsilon)$ accumulates to $y$
Now this step is not 100% clear to me, but it is reasonable. Since $f$ is discontinuous we can find some $r\in\Bbb R$ and $\varepsilon>0$ such that $f^{-1}(B_\varepsilon(r))$ is not open, so $X\setminus f^{-1}(B_\varepsilon(r))$ is not closed, so it must accumulate to some point in its complement
But it's not clear to me how we get the same $y$ to be the center of the ball and the accumulation point at the same time
But anyway this doesn't seem crucial in the following so let's ignore this issue for a second
Alright so now the idea is to construct sequences of open nbhds of $y$, $\{U_n\}$, $\{x_n\}\subseteq Z$ and $\{f_n\}\subseteq Y$ such that, for all $n$:
0) $\overline{U_{n+1}}\subseteq U_n$
1) $|f_n(x)-f_n(y)|<\varepsilon/2^n$ for all $x\in U_n$
2) $x_n\in U_n\cap Z$ (that is $|f(x_n)-f(y)|\geq\varepsilon$
3) $|f_{n+1}(x_i)-f(y)|>\varepsilon/2$ for $i=1,\ldots,n$
4) $|f_n(y)-f(y)|<\varepsilon/2^n$
Here again let's just assume we have such sequences, this is not the issue
Now we let $x_\infty$ be any accumulation point of $\{x_n\}$, so in particular $x_\infty\in\bigcap\overline{U_i}$, and we let $S=\{x_n\}\cup\{x_\infty\}$. The restriction map $\Phi\colon C_p(X)\to C_p(S)$ is continuous, in particular $F=\Phi(Y)\subseteq C_p(Y)$ is countably compact (being the image of a countably compact set) and so it must actually be compact, since $\Bbb R^S$ is separable metric and countable compactness is the same as compactness for such spaces
Since $F$ is compact, we can find $g\in F\subseteq C_p(S)$ at which $\{f_n\}$ accumulates. But now we show that $g$ is discontinuous, giving the desired contradiction.
The book here just says that $g$ is discontinuous at $x_\infty$, since by construction $g(x_\infty)$ is not in the closure of $\{g(x_n)\}$, but I could definitely use some more details since I'm not seeing why is that the case
BTW the proof can be found as U.044 in Tkachuk: A Cp-Theory Problem Book.
I am unable to translate between the two notations quickly enough - but if nothing else, the final part of the proof (where continuity of $g$ leads to a contradiction) seems to be more detailed.
Hi everyone! I am trying to derive a formula for the 7-rough numbers, which are numbers relatively prime to 2, 3 and 5, or having only prime factors >= 7. I'd like to follow a reasoning similar to the one here (math.stackexchange.com/a/928114/719906), but I cannot make any sense out of the 7-rough sequence.
For sure the differences between adjacent numbers repeat with a pattern: [6, 4, 2, 4, 2, 4, 6, 2] repeated. Does this mean I should take the average value plus a constant as for the numbers relatively prime to 2 and 3? Or maybe a quadratic expression? And how should additional terms be included (based on odd or even, or other parameters I am not aware of)?
By the way, this is how the sequence of the 7-rough numbers looks like: 1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 77, 79, 83,...
Could maybe someone help me here in functional analysis:
This would be my question:
Let $X$ be a Banach space, $T\in B(X,X)=\{T:X\rightarrow X|~ T~\text{is linear and bounded operator}\}$ and denote by $\sigma(T):=\{\lambda \in \Bbb{C}: (\lambda I-T) ~\text{is not invertible}\}$ the spectrum of $T$.
Consider the map: $$\phi: \Bbb{C}\setminus \sigma(T)\rightarrow B(X,X): \lambda \mapsto (\lambda I-T)^{-1}$$ I want to check that $\phi$ is analytic. We did the following proof:
Fix $r>0$, consider $A_r$ of complex power series in one variable with convergence radius at least $r$ around $0$, we can think of those as the holomorphic functions from set $U_r:=\{z \in \Bbb C \mid |z|<r\}$ to $\Bbb C$. This is a $\Bbb C$-algebra. If $A$ is any banach algebra and $S \in A$ with $\|S\|<r$, then we get a ring homomorphism $A_r \to A, f \mapsto f(S)$
the low tech way of thinking about it in the case of this series is that all of the usual theorems for manipulating series still work (and in a banach algebra with much the same proofs).
but I don't get why B(X,X) needs to be a banach algebra for this, and firstly what is a banach algebra? Is it a set B(X,X) with a natural map $X\rightarrow B(X,X)$?
okay so we know that $B(X,X)$ is a banach space with the operator norm. We also have multiplication, just given by composition. We know that the operator norm satisfies $\|f \circ g \| \leq \|f\|\|g\|$, so we have a Banach algebra
many of those series manipulation arguments are implicitly using the cauchy criterion, which is why completeness is often assumed (and needed for many arguments to work)
an injective Lie group homorphism $G\rightarrow H$ between Lie groups of the same dimension with $G$ compact is always an embedding with clopen image, i.e. an isomorphism onto a collection of components of $H$. if $G=H$, the number of components is finite and since it's a homeomorphism invariant, it has to be all.
@Shaun You can certainly verify with generators and relations, but intuitively it's the automorphism group of $\Bbb P^1(\Bbb Z_3)$, which has 4 points.
I guess you should stop to verify that there are no transpositions in there.
@monoidal Yes, it's the number of connected components.
Ah, MSE is malfunctioning again. It's not wanting to post comments.
@TedShifrin Thank you. I'm looking for something more elementary than that though. I should have said so. I'm sorry. Here $PSL(n,F)$ is defined as the quotient of $SL(n,F)$ by its centre.
If you're not going to think about the projective geometry, then just work out generators and relations.
You should probably understand the basic situation with $\Bbb P^1(\Bbb C)$ and linear fractional transformations up to scalars (which again will be $PSL(2,\Bbb C)$).
@TedShifrin you can also show that $PGL(2,3) \cong S_4$ (via the action on $\Bbb P^1(\Bbb F_3)$, that's pretty direct) and then argue that $S_4$ has a unique subgroup of index 2
but the projective geometry really isn't harder than some basic linear algebra: $\Bbb P^1(\Bbb F_3)$ is the set of 1-d subspaces of $\Bbb F_3^2$
you don't even need homogenous coordinates for this I think
if $L$ is a one-dimensional subspace of $\Bbb F_3^2$ and $A \in \mathrm{GL}_2(\Bbb F_3)$, then $AL$ is again a one-dimenional subspace
this gives you an action of $\mathrm{GL}(2,3)$ on $\Bbb P^1(\Bbb F_3)$
and then you need to show that it factors over the center
@LukasHeger That's more in line with what I was looking for. Thank you. Again, I'll think it through in the meantime then ask for help later if needs be.
