## [Get] LocalBitcoin Exploit, double your BTC balance! (Proof inside!) | NewProxyLists

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It works literally for any amount over 0.005 BTC. Personally tried this with 0.03 BTC deposit and it worked without problems.

Proof:

## Does the paper "Proof of Collatz's Conjecture" by Agelos Kratimenos contain valid proof?

The document "Proof of Collatz's conjecture" (link arXiv) by Agelos Kratimenos was submitted to arXiv a few days ago (4 Nov. 2019).

Is it really a valid proof?

I mean, I know that arXiv is a prestigious newspaper. Do they have peer reviews before an article is accepted? Or are submissions possible without that?

It would be amazing if it was really a proof of Collatz's conjecture.

## proof techniques – How to prove the Big-Theta recursive function without using repeated substitution, the main theorem or having the form closed?

I have a defined function: $$V (j, k)$$ or $$j, k in mathbb {N}$$ and $$t> 0 in mathbb {N}$$ and $$1 leq q leq j – 1$$. Note $$mathbb {N}$$ comprises $$0$$.

$$V (j, k) = start {case} tj & k leq 2 \ tk & j leq 2 \ tjk + V (q, k / 2) + T (j – q, k / 2) & j, k> 2 end {cases}$$

I am not allowed to use repeated substitution and supposed to prove it by induction. I can not use the master theorem because the recursive part is not in this form. Any ideas on how I can solve it with the given restrictions?

If I start with induction: I repair $$j, q$$ and enthrones $$k$$. So the basic case is $$k = 0$$. then $$V (j, 0) = tj$$. The question suggested that the function could be $$Theta (jk)$$ or maybe $$Theta (j ^ 2k ^ 2)$$ (but this is not necessarily one of the two).

I choose $$Theta (j, k)$$. In the basic case, that would mean that I should prove that $$tj = Theta (j, k)$$ when $$j = 0$$. However, when I start with Big-Oh, I should show that $$km leq mn = m cdot0 = 0$$ which currently I see not being possible.

I do not know if I misinterpreted the baseline scenario or if there is another approach.

## proof techniques – How to prove the Big-Oh recursive function without using repeated substitution, the main theorem or the closed form?

I have a function defined as $$V (j, k)$$ with two basic cases using $$j, k$$ and the recursive part has an extra variable $$q$$ that he uses too. As well, $$1 leq q leq j – 1$$. The recursive part has the form: $$jk + v (q, k / 2) + v (j – q, k / 2)$$I am not allowed to use repeated substitution and supposed to prove it by induction. I can not use the master theorem because the recursive part is not in this form. Any ideas on how I can solve it with the given restrictions?

## proof writing – how to find a derivative of the determining criterion

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## Theorem 2.1.5 in Durrett's probability theory and examples – does not include proof in the book

I read Durrett's Theory of Probability and Examples (4th ed.) And I have difficulty understanding the proof of Theorem 2.1.5:

assume $$mathcal {F_ {i, j}}$$, $$1 leq i leq n$$, $$1 leq j leq m (i)$$ are independent and leave $$mathcal {G_ {i}} = sigma ( bigcup_ {j} mathcal {F_ {i, j}})$$. then $$mathcal {G_ {1}}, …, mathcal {G_ {n}}$$ are independent.

Proof to Durrett:
Let $$mathcal {A_ {i}}$$ be the collection of sets of the form $$bigcap_ {j} A_ {i, j}$$ or $${A_ {i, j}} in mathcal {F_ {i, j}}$$. $$mathcal {A_ {i}}$$ is a $$pi$$-System that contains $$Omega$$ and contains $$bigcup_ {j} mathcal {F_ {i, j}}$$ So theorem 2.1.3 implies that $$sigma ( mathcal {A_ {i}}) = mathcal {G_ {i}}$$ are independent.

I do not understand two things in this proof:

i) why $$mathcal {A_ {i}}$$ contain $$bigcup_ {j} mathcal {F_ {i, j}}$$ ? I've tried to prove that an element of $$bigcup_ {j} mathcal {F_ {i, j}}$$ must be in $$mathcal {A_ {i}}$$but in vain for the moment.

ii) even if I prove it $$bigcup_ {j} mathcal {F_ {i, j}} subset mathcal {A_ {i}}$$ the fact then involves somehow that $$sigma ( mathcal {A_ {i}}) = sigma ( bigcup_ {j} mathcal {F_ {i, j}}) = mathcal {G_ {i}}$$ ?

It's been a while since I'm struggling with that and I'm running out of ideas to prove that or what Durrett is doing here. Any opinion would be very appreciated!

## automata – NFA DFA proof that for any normal language such as L in REG, this implies that s (L) IN REG

$$varphi_ {1} (L) = left {w in Sigma ^ {*} | text {there is a} alpha in Sigma ^ {*} text {with} | alpha | = | w | text {and} alpha w in L right }$$
$$begin {array} {l} { text {Proof. We want to show that} varphi_ {1} (L) text {for all languages} L in operatorname {REG} text {is normal. Be there} L in operatorname {REG} text {random. Because} L text {is normal}} \ { text {there is a DFA} M text {with L} (M) = L text {We are building now} M text {a} lambda text {NFA} {} operatorname {with L} left (M ^ { prime} right) = varphi_ {1} (L). text {For this, we describe}} \ { text {the working method} M ^ { prime} text {formal the condition space of} M ^ { prime} text {is (of design) that of} M. text {The calculation}} \ { text {of} M ^ { prime} text {works (conceptually) as follows: instead of a stone at the beginning of the calculation , place}} q_ {0} text {,}} end {array}$$
$$begin {array} {l} { text {we place 3 stones (1 white, 1 red and 1 blue) on the states of} M. text {We place the blue}} \ { text { on} q_ {0} text {, and white on a step det. guessed} q_ {i} text {(both stones at the same condition). }} \ { text {The calculation of} M ^ { prime} text {at entry} w in Sigma ^ {*} text {functions (of design) as follows:

Could someone help me with that? Searched for regular expressions DFA + NFA + Conversion, but still do not know how to solve this problem.

## gr.group theory – Equivalence of harmonic measures on hyperbolic groups: a basic proof?

Consider a group of Gromov-hyperbolic $$Gamma$$ and let $$mu$$ to be a measure of probability finely supported on $$Gamma$$. Suppose the support of $$mu$$ generates $$Gamma$$ in a semi-group, in other words, a random walk $$X_n$$ driven by $$mu$$ can visit the whole group $$Gamma$$.

Fact: Random walking $$X_n$$ almost surely converges to a point on the border of Gromov $$partial X$$ of $$X$$. Let $$nu$$ to be the output measure on $$partial X$$. Then, $$( partial X, nu)$$ is a model for the so-called Poisson limit. Measurement $$nu$$ is called the harmonic measure (as far as $$mu$$).

Now consider the opposite measure $$check { mu}$$ Defined by $$check { mu} (g) = mu (g ^ {- 1})$$ and let $$check { nu}$$ to be the corresponding harmonic measure.

Question : Are $$nu$$ and $$check { nu}$$ equivalent?

It seems that the answer is positive. In the document Harmonic versus Quasi-Formal Measurements for Hyperbolic Groups, Blachère, Haïssinsky and Mathieu prove that $$h = lv$$ on a hyperbolic group if and only if the harmonic measure $$nu$$ and the Patterson-Sullivan measure $$rho$$ are equivalent. Right here, $$h$$, is the asymptotic entropy, $$l$$ is the asymptotic drift and $$v$$ is the growth in volume. Very roughly, the proof goes like this.

First, the two measures are ergodic and are therefore equivalent or singular. The Hausdorff dimension of $$nu$$ is $$h / l$$ and the Hausdorff dimension of $$rho$$ is $$v$$. Using the Lebesgue Differentiation Theorem, they show that they are equivalent if and only if they have the same Hausdorff dimension. This part is very technical, the basic ingredients being the shadow lemmas for the Patterson-Sullivan measure and for the harmonic measure.

Since the lemmas for harmonic measurements $$nu$$ and $$check { nu}$$ hold and since $$nu$$ and $$check { nu}$$ have the same dimension of Hausdorff, I think we can adapt their evidence to show that $$nu$$ and $$check { nu}$$ are equivalent.

However, I would like to prove a similar result in a broader context and this strategy seems too complicated and too technical. So I wonder if there is a much simpler proof.

Note that you really need to use hyperbolicity somewhere. For example, consider a non-centered probability measure $$mu$$ sure $$mathbb {Z} ^ d$$, C is
$$p = sum_ {x in mathbb {Z} ^ d} x mu (x) neq 0.$$
Then, $$p$$ it's called drifting the random walk. According to Ney and Spitzer's theorem, Martin's limit of the random walk is a sphere: a sequence of points $$x_n$$ converges to a point on the boundary if and only if $$x_n$$ goes to infinity and $$x_n$$ converge in direction, that is to say $$x_n / | x_n |$$ converges. The harmonic measure is based on the direction given by $$p$$. In particular, the harmonic measure associated with $$check { nu}$$ is supported on $$-p$$ and so $$nu$$ and $$check { nu}$$ are singular.

## Proof Verification – Given several sets of start and end numbers, how to validate any level crossing?

Let's say you have the following numbers of start and end. How to validate that the ranges do not cross?

1 – 4
2 – 6
3 – 5

Obviously, each set interferes with another. 2 is between 1 and 4, while the range of 3-5 is between 2-6.

You will browse the list in a loop, comparing them to all the others.
My initial thought was

Current = The whole you will compare to others
Other = any of the other sets

CurrentStartNumber> OtherFinalNumber OR
CurrentFinalNumber <OtherStartNumber

This one succeeds, when it should not, when the current set is between another (that is to say 3-5 vs 2-6).

Honestly, I did not know which label to use. Everything did not seem to match, so took the most appropriate ..