Long Exact Sequence in Group Cohomology Proof

enter image description here

How do you prove this? I have looked all around and am unable to find a real, in-depth proof.

algorithm analysis – Proof for time complexity of Insertion (k-proximate) Sort equals O(nk)

The following is the definition for Proximate Sorting given in my paper:

An array of distinct integers is k-proximate if every integer of the array is at most k places away
from its place in the array after being sorted, i.e., if the ith integer of the unsorted input array is the
jth largest integer contained in the array, then |i − j| ≤ k. In this problem, we will show how to
sort a k-proximate array faster than Θ(n log n).

The following is the proof for time complexity of k-proximate insertion sort:

To prove O(nk), we show that each of the n insertion sort rounds swap an
item left by at most O(k).

In the original ordering, entries that are ≥ 2k slots apart must already be ordered
correctly: indeed, if A(s) > A(t) but t − s ≥ 2k, there is no way to reverse the
order of these two items while moving each at most k slots. This means that for
each entry A(i) in the original order, fewer than 2k of the items A(0), . . . , A(i−1)
are less than A(i). Thus, on round i of insertion sort when A(i) is swapped into
place, fewer than 2k swaps are required, so round i requires O(k) time.

My problem with the proof is this line : “This means that for each entry A(i) in the original order, fewer than 2k of the items A(0), . . . , A(i−1) are less than A(i)”. The preceding statements just proved that t-s < 2k, otherwise the elements are sorted (as elements with distance greater than 2k cannot be swapped.) Isn’t the correct statement : “This means that for each entry A(i) in the original order, fewer than 2k of the items A(0), . . . , A(i−1) are greater than A(i)”?

Turn your free wifi into money with honeygain include payment proof – Other Money Making Opportunities

https://r.honeygain.money/BANTO3D44F
Paying a fix monthly payment for internet plan. Why not capitalize on the internet plan without paying a penny more and still earning passively.

Go a this link and install on your phone or laptop and let it run and generate traffic and earn passively

Payment proof
https://postimg.cc/PNHBvSXP

.

continuity – Baffling proof using function convexity

Let the function $f: mathbb{R} rightarrow mathbb{R}$ be convex, differentiable with derivative $f_x$ and Lipschitz continuous with constant $L.$ Then, for $a,b,c,d in mathbb{R}$ such that $a ge bge d $ and $ a ge cge d,$
begin{equation*}
begin{split}
& f(max{ b,c}) – f(a) + f(min{ b,c}) – f(d)\
& le f_x(min{ b,c})(b -d + c – a). \
end{split}
end{equation*}
Apparently this can be proven easily using $max{ b,c} – a le 0 $ and the convexity of $f,$ but I am stumped about how exactly that is done….almost seems like a mistake! Would really appreciate any pointers or tips.

For reference, this is from the proof of Lemma 3.2 in the paper: Boetius, Frederik, and Michael Kohlmann. “Connections between optimal stopping and singular stochastic control.” Stochastic Processes and their Applications 77.2 (1998): 253-281.

general topology – What are the caps in the proof of Poincare Conjecture and does the insertion of caps into initial manifold preserve homeomorphism?

Quote from Wikipedia article “Poincare Conjecture”:

“He wanted to cut the manifold at the singularities and paste in
caps (Question), and then run the Ricci flow again…
In essence, Perelman showed that all the strands that form can be
cut and capped (Question)…”

To my notice, the above “Question” is: the caps do not belong to the
original manifold; thus, there is no direct (one to one) correspondence between
the original manifold and the final sphere. In conclusion, this formulation
violates the Poincare Conjecture:

Every simply connected, closed 3-manifold is homeomorphic to the 3-sphere.

Should somebody explain this point in Wikipedia? The public deserves
to read a good article on Wikipedia.

lo.logic – Is law of excluded middle necessary in this proof?

I’m currently learning natural deduction and here is my question.

Is it possible to prove this
$vdash neg(P land Q)rightarrow (neg P lor neg Q)$
without referring to the law of excluded middle ?

More precisely, using only the following set of inference rules. These rules are being introduced in the book logic: the laws of truth Page 410.
I assume these rules are complete and have tried a long time , however, still cannot come up with a correct derivation without referring to the law of excluded middle which is not included in the following rules.
enter image description here

dg.differential geometry – Schoen and Yau’s proof of the higher dimensional positive mass theorem

In April 2017 Schoen and Yau posted on the arxiv their solution of the time-symmetric positive mass theorem in all dimensions, which has been a significant conjecture since the 70s. As of now, July 2020, it hasn’t appeared in a journal. Is there any consensus in the geometric analysis community on the veracity of the proof?

https://arxiv.org/abs/1704.05490

proof explanation – Find the cardinality of {{{1, 4}, a, b, {{3, 4}}, {ø}}}?

So I processed to solve this: set within a set {{{1, 4}, a, b, {{3, 4}}, {ø}}}
by doing the following, finding the cardinality of {{1, 4}, a, b, {{3, 4}}, {ø}} which has 5 elements in the set and then afterwards finding the cardinality of {5} which is 1 element. Therefore {{{1, 4}, a, b, {{3, 4}}, {ø}}} = 1.
Is it correct or I have made a mistake?

algorithms – Difficulty in understanding the proof of the lemma : “Matroids exhibit the optimal-substructure property”

I was going through the text “Introduction to Algorithms” by Cormen et. al. where I came across a lemma in which I could not understand a vital step in the proof. Before going into the lemma I give a brief description of the possible prerequisites for the lemma.


Let $M=(S,ell)$ be a weighted matroid where $S$ is the ground set and $ell$ is the family of subsets of $S$ called the independent subsets of $S$. Let $w:Srightarrowmathbb{R}$ be the corresponding weight function ($w$ is strictly positive).

Let us have an algorithm which finds an optimal subset of $M$ using greedy method as:

$text{G}{scriptstyle{text{REEDY}}}(M,w):$

$1quad Aleftarrowphi$

$2quad text{sort $S(M)$ into monotonically decreasing order by weight $w$}$

$3quad text{for each $xin S(M)$, taken in monotonically decreasing order by weight $w(x)$}$

$4quadquad text{do if $Acup{x} in ell(M)$}$

$5quadquadquadtext{then $Aleftarrow Acup {x}$}$

$6quad text{return $A$}$


I was having a problem in understanding a step in the proof of the lemma below.

Lemma: (Matroids exhibit the optimal-substructure property)

Let $x$ be the first element of $S$ chosen by $text{G}{scriptstyle{text{REEDY}}}$ for the weighted matroid $M = (S, ell)$. The remaining problem of finding a maximum-weight independent subset containing $x$ reduces to finding a maximum-weight independent subset of the weighted matroid $M’ = (S’, ell’)$, where

$S’ = {yin S:{x,y}in ell}$ ,

$ell’ = {В subseteq S – {x} : В cup {x} in ell}$ ,

and the weight function for $M’$ is the weight function for $M$, restricted to $S’$. (We call $M’$ the contraction of $M$ by the element $x$.)

Proof:

  1. If $A$ is any maximum-weight independent subset of $M$ containing $x$, then $A’ = A – {x}$ is an independent subset of $M’$.

  2. Conversely, any independent subset $A’$ of $M’$ yields an independent subset $A = A’cup{x}$ of $M$.

  3. We have in both cases $w(A) = w(A’) + w(x)$.

  4. Since we have in both cases that $w(A) = w(A’) + w(x)$, a maximum-weight solution in $M$ containing $x$ yields a maximum-weight solution in $M’$, and vice versa.


I could understand $(1),(2),(3)$. But I could not get how the line $(4)$ was arrived in the proof from $(1),(2),(3)$, especially the part in bold-italics. Could anyone please make it clear to me?

automata – validation of a pumping lemma proof for regular languages

i have the following regular expression:

image of the regular expression

of course i could think of a world like w=a^(m+2)b^(m+2)c^(2m+3) and continue with the proof BUT
i was just wondering, because L is made up of a union of two expressions, is it valid to split L into
L1=a^(i)b^(j)c^(2j-1)| i<=j & i,j>0
L2=a^(i)b^(j)c^(2j-1)| i>=2j-1 & i,j>0

show for each L1 and L2 that the pummping lemma does not work on them (so for l1 i just show that for lets say each k i is not smaller or equal to j, and for l2 i show that i is not always bigger or equal to 2j-1 for each k)

by that i show that l1 and l2 are not regular which means that the union of the two will also be not regular.. is this corrent?

thank you.