## real analysis – formulate a mathematical solution to answer a multiple choice question

If I have a multiple choice question where $$int_ {e} ^ {2e} frac {x} {ln(x)} dx$$ is equal to:

a) 6.55…

b) 7.28…

c) 7.93…

d) 7.96…

e) 8.10…

f) 10.10…

the … means that there is integer number.

Find a method based on the ZF axiom to find the right answer?

I have already found a method but with the definition of a non-mathematical Z axiom.

who can solve any MCQ of this kind, poropser by geniuses of math for geniuses in math.

## solution verification – Are inverse images of separated presheaves separated?

$$newcommand{Sh}{operatorname{Sh}}$$
Let $$f: X to Y$$ be a continuous map, so that there is an induced geometric morphism $$f^* dashv f_* : Sh X to Sh Y$$, where
$$(f^*G)(U) = (f^dagger G)^a(U) = (operatorname{colim}_{V supseteq f(U)} G(V))^a$$
where $$cdot^a$$ denotes sheafification and $$f^dagger$$ denotes the inverse image presheaf.

Question. Is $$f^dagger G$$ separated when $$G$$ is separated?

My guess is yes. Here’s why.

First of all, we gotta get an hold on $$f^dagger G$$. My intuition is that sections $$s in f^dagger G(U)$$ are the same things as sections of $$G$$ defined on an open neighbourhood of $$f(U)$$, up to the equivalence defined by
$$(s, V_1) sim (t, V_2) quadtext{iff}quad exists V_1 cap V_2 supseteq W supseteq f(U), svert_W = tvert_W.$$
This is the same intuition behind the definition of stalk, which is indeed an inverse image.

Now, let $$U$$ be an open of $$X$$, and $${U_i}_{i in I}$$ an open cover of $$U$$. We want to check whether, given $$s, t in f^dagger G(U)$$, then $$svert_{U_i} = tvert_{U_i}$$ for all $$i in I$$ implies $$s = t$$.

Unpacking this, it means we are given $$s in G(V_1)$$ and $$t in G(V_2)$$, where $$V_1,V_2 supseteq f(U)$$, and we know that $$svert_{U_i} = tvert_{U_i}$$ for every $$i in I$$. Notice that restriction here is ‘trivial’, since $$V_1,V_2 supseteq f(U_i)$$ so $$s$$ and $$t$$ are valid elements of $$f^dagger G(U_i)$$ (up to equivalence, of course). What changes is the equivalence relation which is now larger, so when we say $$svert_{U_i} = tvert_{U_i}$$ we now mean
$$exists V_1 cap V_2 supseteq W_i supseteq f(U_i), svert_{W_i} = tvert_{W_i}.$$
Therefore we know $$s$$ and $$t$$ coincide on a neighbourhood of every $$f(U_i)$$. Let $$T = bigcup_{i in I}W_i$$. This is an open neighbourhood of $$f(U)$$ and it’s contained in $$V_1 cap V_2$$. We know $$G$$ is separated: thus from $$svert_{W_i} = tvert_{W_i}$$ for every $$i in I$$ we can conclude $$svert_T = tvert_T$$, which is enough to conclude $$s sim t$$ in $$f^dagger G(U)$$, proving the claim.

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OR

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Thanks for any help

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## node.js – Better solution instead of sending an image as binary through websocket for real time chat app

I’m building a real time chat application like Whatsapp. I have a websocket server with node+express, but I’m a bit confused on which flow I should use.

I’m considering sending the image as binary data through the websocket to the server, process it and store it in AWS s3, and then send the URL back to the user.

Another idea I have thought about is making have an endpoint to make a PUT request to the server, store that Image in S3, and then checking for the specific chatroom id in MongoDB, and then send the Url through websocket.

Can someone aware me on a better solution than what I currently have?

## Uniqueness of solution to heat equation when initial condition is a generalized function

Say $$u(t,x)$$ be a solution to the heat equation $$partial_t = partial_{xx} quad (t,x) in (0,T) times (-1,1)$$ subject to the initial/boundary conditions
$$u(0,x) = f(x), quad x in (-1,1), \ u(t,pm 1) = g^{pm}(t), quad t in (0,T),$$ with the usual compatibility conditions in corners: $$f(pm 1) = g^{pm}(0)$$. Suppose also that $$f$$ and $$g$$ are bounded and continuous. Then one can envoke the maximum principle or the energy method to prove that $$u$$ is the only solution.

What happens when $$f$$ or $$g$$ are unbounded? Say for instance, when $$f(x) = delta_0(x)$$ (point mass at zero) and $$g^pm equiv 0$$? This problem has a solution that can be easily represented as a series.

How does one go about proving uniqueness in such a situation?

In fact, come to think of it, how does one prove the uniqueness of the fundamental solution $$v(t,x) = exp {-x^2 / (4t)}/ sqrt{4 pi t}$$?

Is it some kind of weak uniqueness, where you show uniqueness of all classical solutions resulting from mollification of initial/boundary conditions? Is that the best one can do?

Any references would be deeply appreciated.

## complexity theory – State whether the language is in \$R\$, \$RE\$, etc. The intuition for the solution

The intuition is that finite languages are very simple in the sense that throwing a finite number of words from a language does not affect membership in $$text{R}$$ or in $$overline{text{RE}}$$, etc. Therefore, since $$L^{geq k}$$ is the result of throwing a finite number of words from $$L$$, we get that:

1. $$L(M)^{geq k} in text{RE}$$, and
2. $$overline{HP}^{geq k} notin text{RE}$$.

So the language $$L = { langle M rangle: exists k: L(M)^{geq k} = overline{HP}^{geq k}}$$ must be empty.

Formally speaking, we have the following claims:

Claim 1: For every non-trivial language $$A$$, and every finite strict subset $$Bsubsetneq A$$, it holds that $$A leq_m A setminus B$$.

Hint for the proof of claim 1: $$B$$ is finite and thus decidable. Therefore, given input $$x$$ for the reduction, we can check whether $$xin B$$. Then you can proceed easily. Think where to map inputs $$x$$ from $$B$$, and where to map inputs $$x$$ from $$overline{B}$$.

We also have the following similar claim which could be useful.

Claim 2: For every non-trivial language $$A$$, and every finite strict subset $$Bsubsetneq A$$, it holds that $$Asetminus B leq_m A$$.

Given the above claims, we’re done. Indeed, $$overline{HP} notin text{RE}$$, and for every $$k$$, $$overline{HP}^{geq k} = overline{HP}setminus {win overline{HP} : |w| < k}$$. That is, $$overline{HP}^{geq k}$$ equals a non-trivial infinite language minus a finite subset, and so by claim 1, $$overline{HP}^{geq k} notin text{RE}$$. Also, it holds that $$L(M)^{geq k} in RE$$ for every machine $$M$$ (this is easy, you can prove that directly using standard closure properties and the fact that $$L(M)^{geq k} = L(M)setminus { win L(M): |w| < k}$$. Alternatively, you can use claim 2 but you have to be careful regarding the edge cases where $$L(M)$$ is trivial, etc.). Therefore, it cannot be the case that there is a machine $$M$$ with $$L(M)^{geq k} = overline{HP}^{geq k}$$.

## solution verification – Proof of greatest element of a non-empty set

I am working from Elliott Mendelson’s “Number Systems and the Foundations of Analysis,” which I am thoroughly enjoying thus far.

The proof I am struggling with comes at the end of a section on order relations in the chapter on natural numbers:

By a greatest element of a set $$B subseteq P$$ we mean an object $$z$$ such that $$zin B$$ and $$(forall u)(uin B) implies u leq z)$$.

If $$emptyset neq A subseteq P$$ and $$A$$ is bounded above (that is, $$(exists w)(forall u)(u in A implies u leq w)$$), then $$A$$ has a greatest element. (Hint: Let $$B = {w: (forall u)(uin A implies u leq w)}$$. By hypothesis, $$B neq emptyset$$. Apply the least number principle.)

The hint basically prescribes how to satisfy the requirement of a greatest element possessing the property that there exists some $$z$$ such that $$(forall u)(uin B) implies u leq z)$$. In particular, define $$B$$ as given in the hint; since $$A$$ is bounded above, $$B$$ is non-empty. Since $$B$$ is, in turn, non-empty, it possesses a least element by the least number principle. This least element of $$B$$ is clearly the “$$z$$” referred to above, i.e., the greatest element of $$A$$. I have been struggling for quite a while with demonstrating that this “$$z$$” is indeed contained in $$A$$. I have attempted the method of proof by contradiction considering the alternative case that $$z notin A$$, but can’t seem to figure a means to arrive at a contradiction. I’ve also considered induction proofs as a means of reaching a contradiction, but likewise can’t reach any contradictions. Can anyone provide an extra tip or approach to explore, please?

## Why is the Kth Largest Element solution using a MinHeap O(N lgK) in complexity?

This is a rather well known solution to the $$k$$-th order statistic problem which requires us to find the $$k$$-th largest number in an unsorted array with $$n$$ elements where $$1 leq k leq n$$:

``````public int findKthLargest(int() nums, int k) {
PriorityQueue<Integer> heap = new PriorityQueue<Integer>();

for (int num: nums) {

if (heap.size() > k) {
heap.remove();
}
}

return heap.remove();
}
``````

A brief summary of this approach is that we maintain a min heap and keep polling from this heap each time its size exceeds $$k$$. This way, our final heap will contain $$k$$ elements, and these $$k$$ elements are guaranteed to be last $$k$$ elements in the sorted array (since we have been polling minimums). Within these last $$k$$ elements, the minimum element is guaranteed to be the $$k$$-th largest so we extract the min and return that as a result.

What I don’t particularly understand is why this problem is $$mathcal{O}(n lg{(k)})$$. For instance, after $$mathcal{O}(n)$$ creation of the heap, we do an extraction of the minimum $$n – k$$ times, which would require sifting up in the heap $$n – k$$ times and hence, shouldn’t this solution be $$mathcal{O}(n lg{(n – k)})$$?

Thanks!

## differential equations – Slow evaluation of derivative of Solution of NDSolve

I need to perform a Fourier Transform on the results of a parametric NDSolve function as well as on the derivative of one of the coefficients. Now I do this by evaluating the interpolaton function. This is awfully slow and I don’t know if there is any faster approach. Maybe you can help me out.

This is essentially my current approach with `c1` being a coefficient in my NDSolve solution `fermiD` and `c1'` its derivative :

``````x2 = Table[Evaluate[{c1'[2*Pi*0.001]
``````

In my code I loop it, because I need to evaluate at different frequencies. But I first tried looping one time:

``````For[i = 1, i < 1 + 1, i++, {m = 2*Pi*i*0.001, Print[m], x1 = Table[Evaluate[{c1[m]
``````