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No, your condition doesn't imply Mahloness.
First, note that your first two conditions simply state that $M$ is inaccessible, and the third one gives that $M$ is limit of inaccessibles. Now consider the first cardinal $M_0$ which is inaccessible limit of inaccessibles. That is, every inaccessible below $M_0$ is not a limit of inaccessibles. Now consider $C$ consisting of limits of the set $I$ of inaccessible cardinals smaller than $M_0$. By choice of $M_0$, $C$ contains no inaccessibles below $M_0$, so if we show $C$ is a club below $M_0$, we will get that $M_0$ is not Mahlo.
$C$ is closed: this is true because the set of limit points of any set is closed. For the sake of completion, I will sketch the argument: suppose that each of $\alpha_i$ is a limit point of $I$, where $(\alpha_i)_{i<\kappa}$ is an increasing sequence of cardinals with limit $A$. Let $\beta<A$. This means that there is some $\alpha_i$ between $\beta$ and $A$. Since $\beta<\alpha_i$, there is some element of $I$ which is between $\beta$ and $\alpha_i$, hence between $\beta$ and $A$. So $A$ is a limit point of $I$, so $A\in C$. Hence $C$ is closed.
$C$ is unbounded below $M_0$: Let $\alpha<M_0$. Then $s(\alpha),s(s(\alpha)),s(s(s(\alpha))),...$ is an increasing sequence in $I$ of length $\omega$. Since cofinality of $M_0$ is (clearly) greater than $\omega$, limit of this sequence, which is an element of $C$, is between $\alpha$ and $M_0$. Hence $C$ is unbounded below $M_0$.
Indeed, the condition "inaccessible limit of inaccessibles" (or simply "regular limit of inaccessibles") is a condition known as 1-inaccessibility. It can be generalized: we can speak of 2-inaccessibles, which are regular limits of 1-inaccessibles, 3-inaccessibles which are regular limits of 2-inaccessibles and so on. This can be obviously extended to a successor ordinal, and for limit ordinal $\alpha$ (e.g. $\alpha=\omega$) we say that cardinal is $\alpha$-inaccessible if it's $\beta$-inacessible for $\beta<\alpha$ (this isn't circular definition, but rather it proceeds by transfinite induction).
For every ordinal $\alpha$, notion of $\alpha+1$-inaccessibility is strictly stronger than that of $\alpha$-inaccessibility. As can be shown, every Mahlo cardinal $\kappa$ is not only 1-inaccessible, 2-inaccessible, $\omega$-inaccessible, but even
$\kappa$-inaccessible. Notion of Mahloness is even stronger than that, but for that, I'd recommend reading this section on Wikipedia article. |
If we are given an arbitrary function $\vec{\textbf{a}}:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^k$, what would be an example of this? Say with $n=2, k=3$ or something along those lines?
Along the same line, what about a scalar function $a:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$? What would this look like?
My guess for $\vec{\textbf{a}}$ is something along the lines of $\vec{\textbf{a}}=\big<\vec{\textbf{x}}-\vec{\textbf{y}}, \vec{\textbf{x}}+\vec{\textbf{y}}\big>$ if $n=2, k=4$, but this seems strange. Having vectors inside of vectors. Does that make this a vector with 4 components? Also, this would be very limiting to what $k$ could be, without the dot product to make make entries of the vector be non-multiples of $n$.
I'm very unsure about the scalar function, and my only guess is to take a bunch of variables $x_i$ and $y_i$ and make a normal function with $2n$ scalar variables, but that seems wrong, because then it would just be $a:\mathbb{R}^{2n} \to \mathbb{R}$, and there would be no reason to have the cartesian product in there.
So, what would these functions look like? |
Is there a "simple" mathematical proof that is fully understandable by a 1st year university student that impressed you because it is beautiful?
closed as primarily opinion-based by Daniel W. Farlow, Najib Idrissi, user91500, LutzL, Jonas Meyer Apr 7 '15 at 3:40
Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.
Here's a cute and lovely theorem.
There exist two irrational numbers $x,y$ such that $x^y$ is rational.
Proof. If $x=y=\sqrt2$ is an example, then we are done; otherwise $\sqrt2^{\sqrt2}$ is irrational, in which case taking $x=\sqrt2^{\sqrt2}$ and $y=\sqrt2$ gives us: $$\left(\sqrt2^{\sqrt2}\right)^{\sqrt2}=\sqrt2^{\sqrt2\sqrt2}=\sqrt2^2=2.\qquad\square$$
(Nowadays, using the Gelfond–Schneider theorem we know that $\sqrt2^{\sqrt2}$ is irrational, and in fact transcendental. But the above proof, of course, doesn't care for that.)
How about the proof that
$$1^3+2^3+\cdots+n^3=\left(1+2+\cdots+n\right)^2$$
I remember being impressed by this identity and the proof can be given in a picture:
Edit: Substituted $\frac{n(n+1)}{2}=1+2+\cdots+n$ in response to comments.
Cantor's Diagonalization Argument, proof that there are infinite sets that can't be put one to one with the set of natural numbers, is frequently cited as a beautifully simple but powerful proof. Essentially, with a list of infinite sequences, a sequence formed from taking the diagonal numbers will not be in the list.
I would personally argue that the proof that $\sqrt 2$ is irrational is simple enough for a university student (probably simple enough for a high school student) and very pretty in its use of proof by contradiction!
Prove that if $n$ and $m$ can each be written as a sum of two perfect squares, so can their product $nm$.
Proof: Let $n = a^2+b^2$ and $m=c^2+d^2$ ($a, b, c, d \in\mathbb Z$). Then, there exists some $x,y\in\mathbb Z$ such that
$$x+iy = (a+ib)(c+id)$$
Taking the magnitudes of both sides are squaring gives
$$x^2+y^2 = (a^2+b^2)(c^2+d^2) = nm$$
I would go for the proof by contradiction of an infinite number of primes, which is fairly simple:
Assume that there is a finite number of primes. Let $G$ be the set of allprimes $P_1,P_2,...,P_n$. Compute $K = P_1 \times P_2 \times ... \times P_n + 1$. If $K$ is prime, then it is obviously notin $G$. Otherwise, noneof its prime factors are in $G$. Conclusion: $G$ is notthe set of allprimes.
I think I learned that both in high-school and at 1st year, so it might be a little too simple...
By the concavity of the $\sin$ function on the interval $\left[0,\frac{\pi}2\right]$ we deduce these inequalities: $$\frac{2}\pi x\le \sin x\le x,\quad \forall x\in\left[0,\frac\pi2\right].$$
The first player in Hex has a winning strategy.
There are no draws in hex, so one player must have a winning strategy. If player two has a winning strategy, player one can steal that strategy by placing the first stone in the center (additional pieces on the board never hurt your position) then using player two's strategy.
You cannot have two dice (with numbers $1$ to $6$) biased so that when you throw both, the sum is uniformly distributed in $\{2,3,\dots,12\}$.
For easier notation, we use the equivalent formulation "You cannot have two dice (with numbers $0$ to $5$) biased such that when you throw both, the sum is uniformly distributed in $\{0,1,\dots,10\}$."
Proof:Assume that such dice exist. Let $p_i$ be the probability that the first die gives an $i$ and $q_i$ be the probability that the second die gives an $i$. Let $p(x)=\sum_{i=0}^5 p_i x^i$ and $q(x)=\sum_{i=0}^5 q_i x^i$.
Let $r(x)=p(x)q(x) = \sum_{i=0}^{10} r_i x^i$. We find that $r_i = \sum_{j+k=i}p_jq_k$. But hey, this is also the probability that the sum of the two dice is $i$. Therefore, $$ r(x)=\frac{1}{11}(1+x+\dots+x^{10}). $$ Now $r(1)=1\neq0$, and for $x\neq1$, $$ r(x)=\frac{(x^{11}-1)}{11(x-1)}, $$ which clearly is nonzero when $x\neq 1$. Therefore $r$ does not have any real zeros.
But because $p$ and $q$ are $5$th degree polynomials, they must have zeros. Therefore, $r(x)=p(x)q(x)$ has a zero. A contradiction.
Given a square consisting of $2n \times 2n$ tiles, it is possible to cover this square with pieces that each cover $2$ adjacent tiles (like domino bricks). Now imagine, you remove two tiles, from two opposite corners of the original square. Prove that is is now no longer possible to cover the remaining area with domino bricks.
Proof:
Imagine that the square is a checkerboard. Each domino brick will cover two tiles of different colors. When you remove tiles from two opposite corners, you will remove two tiles with the
samecolor. Thus, it can no longer be possible to cover the remaining area.
(Well, it may be
too "simple." But you did not state that it had to be a university student of mathematics. This one might even work for liberal arts majors...)
One little-known gem at the intersection of geometry and number theory is Aubry's reflective generation of primitive Pythagorean triples, i.e. coprime naturals $\,(x,y,z)\,$with $\,x^2 + y^2 = z^2.\,$ Dividing by $\,z^2$ yields $\,(x/z)^2+(y/z)^2 = 1,\,$ so each triple corresponds to a rational point $(x/z,\,y/z)$ on the unit circle. Aubry showed that we can generate all such triples by a very simple geometrical process. Start with the trivial point $(0,-1)$. Draw a line to the point $\,P = (1,1).\,$ It intersects the circle in the
rational point $\,A = (4/5,3/5)\,$ yielding the triple $\,(3,4,5).\,$ Next reflect the point $\,A\,$ into the other quadrants by taking all possible signs of each component, i.e. $\,(\pm4/5,\pm3/5),\,$ yielding the inscribed rectangle below. As before, the line through $\,A_B = (-4/5,-3/5)\,$ and $P$ intersects the circle in $\,B = (12/13, 5/13),\,$ yielding the triple $\,(12,5,13).\,$ Similarly the points $\,A_C,\, A_D\,$ yield the triples $\,(20,21,29)\,$ and $\,(8,15,17),\,$ We can iterate this process with the new points $\,B,C,D\,$ doing the same we did for $\,A,\,$ obtaining further triples. Iterating this process generates the primitive triples as a ternary tree
$\qquad\qquad$
Descent in the tree is given by the formula
$$\begin{eqnarray} (x,y,z)\,\mapsto &&(x,y,z)-2(x\!+\!y\!-\!z)\,(1,1,1)\\ = &&(-x-2y+2z,\,-2x-y+2z,\,-2x-2y+3z)\end{eqnarray}$$
e.g. $\ (12,5,13)\mapsto (12,5,13)-8(1,1,1) = (-3,4,5),\ $ yielding $\,(4/5,3/5)\,$ when reflected into the first quadrant.
Ascent in the tree by inverting this map, combined with trivial sign-changing reflections:
$\quad\quad (-3,+4,5) \mapsto (-3,+4,5) - 2 \; (-3+4-5) \; (1,1,1) = ( 5,12,13)$
$\quad\quad (-3,-4,5) \mapsto (-3,-4,5) - 2 \; (-3-4-5) \; (1,1,1) = (21,20,29)$
$\quad\quad (+3,-4,5) \mapsto (+3,-4,5) - 2 \; (+3-4-5) \; (1,1,1) = (15,8,17)$
See my MathOverflow post for further discussion, including generalizations and references.
I like the proof that there are infinitely many Pythagorean triples.
Theorem:There are infinitely many integers $ x, y, z$ such that $$ x^2+y^2=z^2 $$ Proof:$$ (2ab)^2 + ( a^2-b^2)^2= ( a^2+b^2)^2 $$
One cannot cover a disk of diameter 100 with 99 strips of length 100 and width 1.
Proof: project the disk and the strips on a semi-sphere on top of the disk. The projection of each strip would have area at most 1/100th of the area of the semi-sphere.
If you have any set of 51 integers between $1$ and $100$, the set must contain some pair of integers where one number in the pair is a multiple of the other.
Proof: Suppose you have a set of $51$ integers between $1$ and $100$. If an integer is between $1$ and $100$, its largest odd divisor is one of the odd numbers between $1$ and $99$. There are only $50$ odd numbers between $1$ and $99$, so your $51$ integers can’t all have different largest odd divisors — there are only $50$ possibilities. So two of your integers (possibly more) have the same largest odd divisor. Call that odd number $d$. You can factor those two integers into prime factors, and each will factor as (some $2$’s)$\cdot d$. This is because if $d$ is the largest divisor of a number, the rest of its factorization can’t include any more odd numbers. Of your two numbers with largest odd factor $d$, the one with more $2$’s in its factorization is a multiple of the other one. (In fact, the multiple is a power of $2$.)
In general, let $S$ be the set of integers from $1$ up to some even number $2n$. If a subset of $S$ contains more than half the elements in $S$, the set must contain a pair of numbers where one is a multiple of the other. The proof is the same, but it’s easier to follow if you see it for a specific $n$ first.
The proof that an isosceles triangle ABC (with AC and AB having equal length) has equal angles ABC and BCA is quite nice:
Triangles ABC and ACB are (mirrored) congruent (since AB = AC, BC = CB, and CA = BA), so the corresponding angles ABC and (mirrored) ACB are equal.
This congruency argument is nicer than that of cutting the triangle up into two right-angled triangles.
Parity of sine and cosine functions using Euler's forumla:
$e^{-i\theta} = cos\ (-\theta) + i\ sin\ (-\theta)$
$e^{-i\theta} = \frac 1 {e^{i\theta}} = \frac 1 {cos\ \theta \ + \ i\ sin\ \theta} = \frac {cos\ \theta\ -\ i\ sin\ \theta} {cos^2\ \theta\ +\ sin^2\ \theta} = cos\ \theta\ -\ i\ sin\ \theta$
$cos\ (-\theta) +\ i\ sin\ (-\theta) = cos\ \theta\ +i\ (-sin\ \theta)$
Thus
$cos\ (-\theta) = cos\ \theta$
$sin\ (-\theta) = -\ sin\ \theta$
$\blacksquare$
The proof is actually just the first two lines.
I believe Gauss was tasked with finding the sum of all the integers from $1$ to $100$ in his very early schooling years. He tackled it quicker than his peers or his teacher could, $$\sum_{n=1}^{100}n=1+2+3+4 +\dots+100$$ $$=100+99+98+\dots+1$$ $$\therefore 2 \sum_{n=1}^{100}n=(100+1)+(99+2)+\dots+(1+100)$$ $$=\underbrace{101+101+101+\dots+101}_{100 \space times}$$ $$=101\cdot 100$$ $$\therefore \sum_{n=1}^{100}n=\frac{101\cdot 100}{2}$$ $$=5050.$$ Hence he showed that $$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}.$$
If $H$ is a subgroup of $(\mathbb{R},+)$ and $H\bigcap [-1,1]$ is finite and contains a positive element. Then, $H$ is cyclic.
Fermat's little theorem from noting that modulo a prime p we have for $a\neq 0$:
$$1\times2\times3\times\cdots\times (p-1) = (1\times a)\times(2\times a)\times(3\times a)\times\cdots\times \left((p-1)\times a\right)$$
Proposition (No universal set): There does not exists a set which contain all the sets (even itself)
Proof: Suppose to the contrary that exists such set exists. Let $X$ be the universal set, then one can construct by the axiom schema of specification the set
$$C=\{A\in X: A \notin A\}$$
of all sets in the universe which did not contain themselves. As $X$ is universal, clearly $C\in X$. But then $C\in C \iff C\notin C$, a contradiction.
Edit: Assuming that one is working in ZF (as almost everywhere :P)
(In particular this proof really impressed me too much the first time and also is very simple)
Most proofs concerning the Cantor Set are simple but amazing.
The total number of intervals in the set is zero.
It is uncountable.
Every number in the set can be represented in ternary using just
0 and
2. No number with a
1 in it (in ternary) appears in the set.
The Cantor set contains as many points as the interval from which it is taken, yet itself contains no interval of nonzero length. The irrational numbers have the same property, but the Cantor set has the additional property of being closed, so it is not even dense in any interval, unlike the irrational numbers which are dense in every interval.
The Menger sponge which is a 3d extension of the Cantor set
simultaneously exhibits an infinite surface area and encloses zero volume.
The derivation of first principle of differentiation is so amazing, easy, useful and simply outstanding in all aspects. I put it here:
Suppose we have a quantity $y$ whose value depends upon a single variable $x$, and is expressed by an equation defining $y$ as some specific function of $x$. This is represented as:
$y=f(x)$
This relationship can be visualized by drawing a graph of function $y = f (x)$ regarding $y$ and $x$ as Cartesian coordinates, as shown in Figure(a).
Consider the point $P$ on the curve $y = f (x)$ whose coordinates are $(x, y)$ and another point $Q$ where coordinates are $(x + Δx, y + Δy)$.
The slope of the line joining $P$ and $Q$ is given by:
$tanθ = \frac{Δy}{Δx} = \frac{(y + Δy ) − y}{Δx}$
Suppose now that the point $Q$ moves along the curve towards $P$.
In this process, $Δy$ and $Δx$ decrease and approach zero; though their ratio $\frac{Δy}{Δx}$ will not necessarily vanish.
What happens to the line $PQ$ as $Δy→0$, $Δx→0$? You can see that this line becomes a tangent to the curve at point $P$ as shown in Figure(b). This means that $tan θ$ approaches the slope of the tangent at $P$, denoted by $m$:
$m=lim_{Δx→0} \frac{Δy}{Δx} = lim_{Δx→0} \frac{(y+Δy)-y}{Δx}$
The limit of the ratio $Δy/Δx$ as $Δx$ approaches zero is called the derivative of $y$ with respect to $x$ and is written as $dy/dx$.
It represents the slope of the tangent line to the curve $y=f(x)$ at the point $(x, y)$.
Since $y = f (x)$ and $y + Δy = f (x + Δx)$, we can write the definition of the derivative as:
$\frac{dy}{dx}=\frac{d{f(x)}}{dx} = lim_{x→0} [\frac{f(x+Δx)-f(x)}{Δx}]$,
which is the required formula.
This proof that $n^{1/n} \to 1$ as integral $n \to \infty$:
By Bernoulli's inequality (which is $(1+x)^n \ge 1+nx$), $(1+n^{-1/2})^n \ge 1+n^{1/2} > n^{1/2} $. Raising both sides to the $2/n$ power, $n^{1/n} <(1+n^{-1/2})^2 = 1+2n^{-1/2}+n^{-1} < 1+3n^{-1/2} $.
Can a Chess Knight starting at any corner then move to touch every space on the board exactly once, ending in the opposite corner?
The solution turns out to be childishly simple. Every time the Knight moves (up two, over one), it will hop from a black space to a white space, or vice versa. Assuming the Knight starts on a black corner of the board, it will need to touch 63 other squares, 32 white and 31 black. To touch all of the squares, it would need to end on a white square, but the opposite corner is also black, making it impossible.
The Eigenvalues of a skew-Hermitian matrix are purely imaginary.
The Eigenvalue equation is $A\vec x = \lambda\vec x$, and forming the vector norm gives $$\lambda \|\vec x\| = \lambda\left<\vec x, \vec x\right> = \left<\lambda \vec x,\vec x\right> = \left<A\vec x,\vec x\right> = \left<\vec x, A^{T*}\vec x\right> = \left<\vec x, -A\vec x\right> = -\lambda^* \|\vec x\|$$ and since $\|\vec x\| > 0$, we can divide it from left and right side. The second to last step uses the definition of skew-Hermitian. Using the definition for Hermitian or Unitarian matrices instead yields corresponding statements about the Eigenvalues of those matrices.
I like the proof that not every real number can be written in the form $a e + b \pi$ for some integers $a$ and $b$. I know it's almost trivial in one way; but in another way it is kind of deep. |
You don’t even need an infinite number of monkeys! For any $\epsilon > 0$ and $k \geq l(\text{Hamlet})$, there is some number $N$ such that $N$ monkeys at typewriters, each typing for $k$ keystrokes, will produce a copy of Hamlet with probability greater than $1-\epsilon$. (This holds under some quite weak conditions on our model of monkey typing.)
This is an example of the general “soft analysis to hard analysis” principle, championed by Terry Tao among others: most any proof in analysis may be transformed into a proof of a quantitative statement such as the one above.
This can be made precise in some generality using various rather beautiful proof-theoretic methods, such as variants of Gödel’s Dialectica translation; lovely results along these lines have been obtained by e.g. Avigad, Gerhardy and Towsner. In this particular case, the bounds we get will of course depend on the model of monkey typing used.
For instance, if we assume that the keystrokes are independently uniformly distributed, if our Hamlet-recognition criterion is case-, punctuation- and whitespace-insensitive, then for the case $k = l(\text{Hamlet})$,
$$N = \left\lceil \frac{\log \epsilon}{\log \left(1 - \frac{1}{26^{l(\text{Hamlet})}}\right)}\right\rceil$$
will work. (The proof is an exercise for the reader.)
Project Gutenberg’s copy of Hamlet (first folio) weighs in at 117,496 alphanumeric characters. So if we want to produce Hamlet (first folio) with probability 1/2, in the minimal number of keystrokes, then by some quick slapdash estimating (rounding up a little to be on the safe side), something like $10^{170,000}$ monkeys should certainly suffice!
I guess empirical testing is out — ethical controls are so tricky. Anyone want to run some simulations? |
The
binomial theorem
is an important formula about the expansion of powers of sums. Its simplest version reads
<math>(x+y)^n=\sum_{k=0}^n{n \choose k}x^ky^{n-k}\quad\quad\quad(1)</math>
whenever
n
is any non-negative integer and the numbers
<math>{n \choose k}=\frac{n!}{k!(n-k)!}</math>
are the binomial coefficients
. This formula, and the triangular arrangement
of the binomial coefficients, are often attributed to Blaise Pascal
who described them in the 17th century
. It was however known long before to Chinese mathematicians.
The cases
n=2, n=3 and n=4 are the ones most commonly used:
( x + y) 2 = x 2 + 2 xy + y 2
( x + y) 3 = x 3 + 3 x + 3 2y xy + 2 y 3
( x + y) 4 = x 4 + 4 x + 6 3y x + 4 2y 2 xy + 3 y 4
Formula (1) is valid for all real or complex numbers
x and y, and more generally for any elements x and y of a ring as long as xy = yx.
Isaac Newton generalized the formula to other exponents by considering an infinite series:
<math>(x+y)^r=\sum_{k=0}^\infty {r \choose k} x^k y^{r-k}\quad\quad\quad(2)</math>
where
r can be any complex number (in particular r can be any real number, not necessarily positive and not necessarily an integer), and the coefficients are given by
<math>{r \choose k}=\frac{r(r-1)(r-2)\cdots(r-k+1)}{k!}</math>
(which in case
k
= 0 is a product of no numbers at all
and therefore equal to 1, and in case
k
= 1 is equal to
r
, as the additional factors (
r
- 1), etc., do not appear in that case).
The sum in (2) converges and the equality is true whenever the real or complex numbers
x and y are "close together" in the sense that the absolute value | x/y| is less than one.
The geometric series is a special case of (2) where we choose
y = 1 and r = -1.
Formula (2) is also valid for elements
x and y of a Banach algebra as long as xy = yx, y is invertible and || x/y|| < 1.
The binomial theorem can be stated by saying that the polynomial sequence
<math>\left\{\,x^k:k=0,1,2,\dots\,\right\}</math>
is of binomial type
.
All Wikipedia text is available under the terms of the GNU Free Documentation License |
\begin{align} \forall x\in(0,+\infty), \quad f(x) > 0 \tag 1 \\[10pt] \forall x\in ]0,+\infty[, \quad f(x) > 0 \tag 2 \\[10pt] \forall x\in {]}0,+\infty{[}, \quad f(x) > 0 \tag 3 \end{align} I was instructed from an early age to use the notation in line $(1)$ above. In France apparently the one in line $(3)$ prevails, or at least has some currency. I coded line $(3)$ like this:
\forall x\in {]}0,+\infty{[}, \quad f(x) > 0
Line $(2)$, on the other hand, is coded like this:
\forall x\in ]0,+\infty[, \quad f(x) > 0
In the conspicuous difference between the appearances of lines $(2)$ and $(3)$ we see that MathJax doesn't apply the spacing conventions that it applies well in line $(1)$. (Apparently to some users this is not conspicuous until the actual juxtaposition of lines $(2)$ and $(3)$ is seen.)
If one must use the French notation, how should that be done? Just the way I did in line $(3)$? Or is there a less crude way? |
Recall the definition of the Weyl Chambers: A
Weyl Chamber is a region of $V \setminus \bigcup_{\alpha \in \Phi} H_{\alpha}$, where $V$ is underlying Euclidean space, and $H_\alpha$ the hyperplane that is perpendicular to $\alpha$ the elements of root system $\Phi$ in $V$.
I want to show that:
The Weyl chambers are open, convex and connected.
My attempts:
A single hyperplane $H_\alpha$ separates the space $V$ into two half-spaces: positive half-space $H_\alpha^+=\{x \in V \mid (\alpha,x)>0\}$ and negative half-space $H_\alpha^-=\{x \in V \mid (\alpha,x)<0\}$. It is clear that $V \setminus H_\alpha = H_\alpha^+\cup H^-_\alpha.$ As the result, the Weyl chambers can be regarded as components of $$V \setminus \bigcup_{\alpha \in \Phi} H_{\alpha}= \bigcap_{\alpha \in \Phi} \left( V \setminus H_{\alpha} \right)= \bigcap_{\alpha \in \Phi} \left( H_\alpha^+\cup H_\alpha^- \right).$$ By the discussion above, a Weyl chamber is simply a $\textbf{intersection}$ of finite half-spaces. It is the case since one may write $$\left( H_\alpha^+\cup H_\alpha^- \right) \cap \left( H_\beta^+\cup H_\beta^- \right) =\left( H_\alpha^+\cap H_\beta^+ \right) \cup\left( H_\alpha^+\cap H_\beta^- \right) \cup \left( H_\alpha^-\cap H_\beta^+ \right) \cup\left( H_\alpha^-\cap H_\beta^- \right).$$ Therefore openness and convexity of the Weyl chambers are straightforward since half-spaces are open and convex, so are their intersections. Moreover, we would replace $V$ by $\Bbb R^{\dim V}$ since they are isomorphic. Now we can prove that the Weyl chamber is connected by showing that every convex set in $\Bbb R^{\dim V}$ is connected. And I'm sure that proof of this statement existed. So am I done? |
Ramification of prime numbers in number fields is a topic relevant to what I'm studying (arithmetic hyperbolic 3-manifolds), and many results from algebraic number theory are used there, however most of what I need comes from the case where the number field is quadratic. Now, eventually in this life I would like to master the theory for general number fields, but that is a very rich spectrum of ideas with many difficult theorems and many open problems. So for now, I'm really interested in just nailing down the generalizations one can make in the quadratic case. I've been working on finding more elementary proofs for facts that would require very advanced proofs in the general setting, in fact leaning toward proofs that rely on arithmetic calculations, because I find it can be very enlightening to see what kind of expressions pop up when you write the stuff out. I've made some progress but I think I could take it further with a little help (thanks in advance to any commenters or answerers).
I want to also mention there's an entry on math.stackexchange: https://math.stackexchange.com/questions/376590/vague-definitions-of-ramified-split-and-inert-in-a-quadratic-field with a nice answer that sheds some light on things, but there is more I want to see.
Probably the nicest result in this vein, which you can get by chasing down the definitions and looking at how things simplify for quadratics (which does take some work but the work is all relatively accessible algebra), is this:
Let $d\in\mathbb{Z}$ be squarefree, $p\in\mathbb{N}$ be prime $\Longrightarrow$ in the field $\mathbb{Q}(\sqrt{d})$:
$p$ is $\begin{cases} \text{inert} &; \quad p\neq 2 \;\&\; \Big(\frac{d}{p}\Big)=-1\text{, or } p=2 \;\&\; d\equiv_85\\ \text{split} &; \quad p\neq 2 \;\&\; \Big(\frac{d}{p}\Big)=1 \text{, or } p=2 \;\&\; d\equiv_81\\ \text{ramified} &; \quad p|d \text{, or } p=2 \;\&\; d\not\equiv_41 \end{cases}$.
This is great because you get a complete classification of what happens to any natural prime in any quadratic extension, but now I want to see what the prime factorizations of $p\mathcal{O}_d$ look like explicitly. It seems this should be derivable by the same methods but I keep getting stuck. As I said up top, I'm focusing on ramification for now, so let's just look at the 3rd case from the statement above.
First question:As mentioned in the link I gave, if $p|d$, then $p\mathcal{O}_d=(p,\sqrt{d})^2$. I wanted to see why this holds, and I found that (suppressing the calculations - I can show them if you like) $(p,\sqrt{d})^2=\big\{p\big(\alpha_1\alpha_3p+\alpha_2\alpha_4\frac{d}{p}+(\alpha_1\alpha_4+\alpha_2\alpha_3)\sqrt{d}\big)\mid\alpha_i\in\mathcal{O}_d\big\}$. This makes it clear it's contained in $(p)$, but is there a simple arithmetic way to see the reverse inclusion? Since $d$ is squarefree, we know $p$ and $\frac{d}{p}$ are coprime, so $\exists m,n\in\mathbb{Z}: mp+n\frac{d}{p}=1$, but how do we know we can find $\alpha_i\in\mathcal{O}_d: m=\alpha_1\alpha_3, n=\alpha_2\alpha_4,$ where $\alpha_1\alpha_4+\alpha_2\alpha_3=0$?
The next thing to do here is see what happens to 2. Like in the case for $p|d$, I'm comfortable with the classification of
when 2 ramifies, so let's go ahead and let $d\not\equiv_41$. But then the explicit factorization of $2\mathcal{O}_d$ remains ambiguous. Of course, for $d=-1$, we've got $(p)=(1+i)^2$, but it is not true in general that $(p)=(1+\sqrt{d})^2$, and I can't seem to find the appropriate cases to break it down into. I think it can be verified similarly to how the last one was, if you know the answer but... Second question:What is the prime factorization of $2\mathcal{O}_d$ when $d\not\equiv_41$, broken into the appropriate cases and in terms of $d$? I'd like to attempt verifying this by a more elementary computation like the last one.
One other thing to say about this is that sometimes writing $(p,\sqrt{d})$ is a little misleading because sometimes it's actually a principal ideal. Due to Stark, 1966, we have that there are only 9 $d$ values for which $\mathcal{O}_d$ is a PID, and which ones (note: 5 of them are Euclidean, implying PID, which can be done more easily algebraically - Stark used pretty advanced complex analysis), and clearly in those cases $\exists a\in\mathcal{O}_d:(p,\sqrt{d})=(a)$. Whether or not these ideals are principal turns out to have important consequences in the topological application I'm studying. So...
Third questionCan we say in general what this $a$ would be, casewise or all together in terms of $d$? Finally, are there non-PID $\mathcal{O}_d$s where $(p,\sqrt{d})$ is principal?
Also if anyone knows of a paper I can find where this is worked out that would be great. |
Suppose that we only have propositional variables and connectives. Suppose our rules of inference are detachment {C$\alpha$$\beta$, $\alpha$} $\vdash$ $\beta$, and uniform substitution. Suppose that we have a binary connective "C" and a unary connective "N". In 1953 C. A. Meredith found what is currently the shortest single axiom of classical C-N (Conditional-Negation) propositional calculus [p. 302 of A. N. Prior's Formal Logic indicates this and references Prior's paper " 'Single Axioms for the Systems (C, N), (C, O), and (A, N) of the Two-valued Propositional Calculus' JCS, vol. i, No. 3 (July 1953), pp. 155-64. Systems 1.5, 3.13, 6.13, 6.14.):
CCCCCpqCNrNsrtCCtpCsp.
Or using another prefix notation
→→→→→pq→¬r¬srt→→tp→sp.
So given the first notation, the formation rules are:
All lower case letters of the Latin alphabet, as well as any lower case letters of the Latin alphabet sub-scripted by Hindu-Arabic numerals are formulas. If xis a formula, then so is N x. If xand yare formulas, then so is C x y. The space between the xand the yin "C x y" is not necessary. Nothing else is a formula in this context.
To determine what is and what is not a formula the following suffices.
Assign -1 to all lower case letters, as well as all lower case letters which are sub-scripted. Assign 0 to N. Assign 1 to C. Sum the assigned numbers as you precede from left to right throughout any given string. A string will qualify as a formula if and only if it either starts and ends with -1, or if it starts with a member of {0, 1} and -1 is only reached at the spot corresponding to the last letter of the string, and that string ends with -1.
Now, via the above it turns out that any formula that starts with "C", has its antecedents corresponding to the longest subformulas which have a "0" corresponding to their last letter. So looking at the above formula, we have
C C C C C p q C N r N s r t C C t p C s p| | | | | | | | | | | | | | | | | | | | |1 2 3 4 5 4 3 4 4 3 3 2 1 0 1 2 1 0 1 0 -1
The longest subformula which ends with the first "t" is thus CCCCpqCNrNsrt. The longest subformula which ends with the second "p" is thus "Ctp", and "s" is the longest subformula corresponding to the third "0" spot. Thus, the main breaks of the formula can get located by the following "|" marks.
C|CCCCpqCNrNsrt|C|Ctp|C|s|p.
Furthermore, the main breaks of CCCCpqCNrNsrt is given by C|CCCpqCNrNsr|t, and so on.
Suppose we have a 2-valued model of C and N given by the following table:
C 0 1 N0 1 1 11 0 1 0
Does there exist any shorter single axiom, which under the rules above allows us to deduce all tautologies, and only those tautologies in the 2-valued model? Does there exist another single axiom of the same length?
If there exists a unique shortest axiom up to re-symbolization of variables and connectives, what is the shortest axiom (which is a tautology) that allows us to deduce a known axiom system for classical propositional calculus, such as the following axiom set? CpCqp. [p$\rightarrow$(q$\rightarrow$p)]. CCpCqrCCpqCpr. [(p$\rightarrow$(q$\rightarrow$r))$\rightarrow$((p$\rightarrow$q)$\rightarrow$(p$\rightarrow$r))]. CCNpNqCqp. [($\lnot$p$\rightarrow$$\lnot$q)$\rightarrow$(q$\rightarrow$p)]. If there exists more than one shortest axiom, what is the set of such shortest axioms up to re-symbolization of variables and connectives?
The length of an axiom is defined by the number of symbols it has when fully expressed as a well-formed formula in Polish/Lukasiewicz notation.
I've tried to find a way to compute all tautologies of a certain length in OTTER, but had no success. There has existed some similar work for shortest axioms in classical propositional calculus with the Sheffer Stroke, but in a private communication with one of the authors I've learned that they didn't do a search for the shortest axioms of the conditional-negation formulas for classical propositional calculus. |
I was wondering about important/famous mathematical constants, like $e$, $\pi$, $\gamma$, and obviously the golden ratio $\phi$. The first three ones are really well known, and there are lots of integrals and series whose results are simply those constants. For example:
$$ \pi = 2 e \int\limits_0^{+\infty} \frac{\cos(x)}{x^2+1}\ \text{d}x$$
$$ e = \sum_{k = 0}^{+\infty} \frac{1}{k!}$$
$$ \gamma = -\int\limits_{-\infty}^{+\infty} x\ e^{x - e^{x}}\ \text{d}x$$
Is there an interesting integral
* (or some series) whose result is simply $\phi$?
*
Interesting integral means that things like
$$\int\limits_0^{+\infty} e^{-\frac{x}{\phi}}\ \text{d}x$$
are not a good answer to my question. |
If you are interested in the case for the discrete Laplacian, check out the paper of Rigoli, Salvatori, and Vignati titled "Liouville properties on graphs" (DOI: 10.1112/S0025579300012031).
Among the results proven is the following:
Let $G$ be a graph and let $q$ be an arbitrary point in $G$. Let $u$ be a $p$-subharmonic function on $G$ for $p > 1$. Suppose that for all $R$ sufficiently large $$ \sup_{B_R(q)} u \lesssim \frac{(R\log R)^{(p-1)/p}}{|S_R(q)|^{1/p}} $$ and $$ |S_R(q)| \lesssim (R\log R)^{p-1} $$ where $S_R(q) = B_R(q) \setminus B_{R-1}(q) $ is the "sphere of the radius $R$", then $u$ is constant.
The requirement on the volume growth rate of balls of radius $R$ is typical: this is not just the case for graphs. Liouville theorems for non-compact, complete Riemannian manifolds are usually proven under the assumption of a lower bound on the Ricci curvature, which can be used to prove volume growth bounds on the Riemannian manifold (the simplest example being the Bishop-Gromov theorem). |
June 12th, 2018, 05:48 PM
# 1
Newbie
Joined: Apr 2018
From: Canada
Posts: 2
Thanks: 0
Given a velocity vector and a point, find the angle (urgent)
Hello! So I am doing this vectors as velocity question and I need help solving it. Basically, James Bond is driving 70.9km/h [N] and launches himself off a cliff. A helicopter is waiting for him 20m North and 15m West from him (at the same elevation). If a large gust of wind travelling 54km/h [W] hits him as he jumps from the car, what angle does he need to jump out of the car from in order to land in the helicopter?
So what I did so far was, I found the velocity vector that he is travelling if he jumps directly north from the car and that is:
sqrt((70.9)^2+(54)^2)
which gives you 89.12km/h [N37.3W]
I do not know how to figure the angle he should jump at to get to the point.
There isn't a vectors category and since in Canada the course is called "calculus and vectors" I just posted it here; sorry if it isn't supposed to be here.
Last edited by skipjack; June 13th, 2018 at 12:40 AM.
July 19th, 2018, 03:58 AM
# 2
Math Team
Joined: Jan 2015
From: Alabama
Posts: 3,264
Thanks: 902
The problem says "a large gust of wind travelling 54 km/h [W] hits him as he jumps from the car". Are we to assume that this velocity vector will be added to
his velocity? Wind does not work that way! If you were talking about a boat traveling on a river, then the boat is carried on the river so its velocity is added to the boat's. James Bond is not being "carried" by the wind. In order to answer the question, as asked, we would need to know a "coefficient of friction" between Bond and the wind.
Ignoring that error, take Bond's initial jump to make angle $\displaystyle \theta$ with north, angle $\displaystyle \phi$ with the horizontal plane. Then his initial velocity vector is $\displaystyle \left<v \sin(\theta), v \cos(\theta), v \cos(\phi)\right>$. Since the only force on him is -mg is the vertical direction (again, since we are not given a coefficient of friction we must assume the wind adds it velocity to his, not affecting his acceleration) we have, as his acceleration vector, $\displaystyle \vec{a}= \left<0, 0, -g\right>$. Integrating his velocity vector would be $\displaystyle \vec{v}= \left<C_1, C_2, -gt+ C_3\right>$. However, we are told that the wind (which, against all physical laws we are apparently supposed to assume
adds itself to his velocity) is [math]\left< 0, 54, 0\right> so we must have [math]\vec{v}= \left< C_1, C_2+ 54, -gt+ C_3\right> In order to determine the constants, we use his initial velocity vector: $\displaystyle \left<C_1, C_2, C_3\right>= \left<70.9 \sin(\theta), 70.9 \cos(\theta)+ 54, 70.9 \cos(\phi)\right>$.
Integrating again, his position vector is given by $\displaystyle \vec{x(t)}= \left<70.9 \sin(\theta)t+ D_1, 70.9 \cos(\theta)t+ 54t+ D_2, -\frac{g}{2}t^2+ 70.9 \cos(\phi)t+ D_3\right>$. Taking his initial position at the point where he launches the car to be (0, 0, 0), we have $\displaystyle \vec{v(0)}= \left<D_1, D_2, D_3\right>= <0, 0, 0>$ so his position, at time t, is given by $\displaystyle \vec{x(t)}= \left<70.9 \sin(\theta)t, 70.9 \cos(\theta)+ 54t , -\frac{g}{2}t^2+ v \cos(\phi)t\right>$.
Now, the helicopter is positioned at " 20m North and 15m West from him (at the same elevation)" or at $\displaystyle \left< 20, -15, 0\right>$ so that we must have, for
some t, $\displaystyle \left<70.9 \sin(\theta)t, 70.9 \cos(\theta)+ 54t, -\frac{g}{2}t^2+ 70.9 \cos(\phi)t\right>= \left< 20, -15, 0\right>$.
Solve the three equations $\displaystyle 70.9 \sin(\theta)t= 20$, $\displaystyle 70.9 \cos(\theta)+ 54t= -15$, $\displaystyle -\frac{g}{2}t^2+ 70.9 \cos(\phi)t= 0$ for $\displaystyle \theta$, $\displaystyle \phi$, and $\displaystyle t$.
Last edited by skipjack; July 19th, 2018 at 07:26 AM.
Tags angle, find, point, urgent, vector, velocity, velocity vector
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Revista Matemática Iberoamericana
Full-Text PDF (113 KB) | Metadata | Table of Contents | RMI summary
Volume 27, Issue 1, 2011, pp. 355–360 DOI: 10.4171/RMI/639
Published online: 2011-04-30
The Jet of an Interpolant on a Finite SetCharles Fefferman
[1]and Arie Israel [2](1) Princeton University, United States
(2) University of Texas at Austin, USA
We study functions $F \in C^m (\mathbb{R}^n)$ having norm less than a given constant $M$, and agreeing with a given function $f$ on a finite set $E$. Let $\Gamma_f (S,M)$ denote the convex set formed by taking the $(m-1)$-jets of all such $F$ at a given finite set $S \subset \mathbb{R}^n$. We provide an efficient algorithm to compute a convex polyhedron $\tilde{\Gamma}_f (S,M)$, such that $$ \Gamma_f (S,cM) \subset \tilde{\Gamma}_f (S,M) \subset \Gamma_f (S,CM), $ where $c$ and $C$ depend only on $m$ and $n$.
Keywords: Interpolation, jet, algorithm, Whitney extension theorem.
Fefferman Charles, Israel Arie: The Jet of an Interpolant on a Finite Set.
Rev. Mat. Iberoam. 27 (2011), 355-360. doi: 10.4171/RMI/639 |
$(X,d)$ be a metric space and $E \subset X$.
$\bar{E} = E \cup \{\text{limit points of $E$} \}$
$ int(E) = \{\text{collection of all interior points of $E$}\}$, where a point $e \in E$ is an interior point if there exists a neighborhood of $e$ contained in $E$. $\bar{E}$ denotes closure of $E$ and $int(E)$ denotes the interior of $E$. Then prove that
a) $\bar{E} = \bigcap_{E \subset F} F$, $F \subset X$ and $F$ is closed.
b) $int(E) = \bigcup_{V \subset E} V$, $V$ is open.
In (b) I am trying to show that LHS is contained in RHS and vice versa. LHS ⊂ RHS can be shown by considering a neighborhood N around a point x ∈E. N is an open set and N⊂E. So it is a subset of the union on the RHS. Now coming to other implication. Consider a point x in the RHS, so it must be element of some V⊂E. Would this V be a neighborhood? I know V is an open set but how do I ensure it is a neighborhood of $x$. My definition for neighborhood is as given in Rudin's Principles of Mathematical Analysis which states that a neighborhood of a point $p$ denoted by $Nr(p)$ consists of all points $q$ such that $d(p,q)<r$.
Please provide me with directions for part (a). |
Journal of the European Mathematical Society
Full-Text PDF (1398 KB) | Metadata | Table of Contents | JEMS summary
Volume 18, Issue 10, 2016, pp. 2315–2403 DOI: 10.4171/JEMS/642
Published online: 2016-09-19
Kirkwood gaps and diffusion along mean motion resonances in the restricted planar three-body problemJacques Féjoz
[1], Marcel Guardia [2], Vadim Kaloshin [3]and Pablo Roldán [4](1) IMCCE, Paris, France
(2) Universitat Politècnica de Catalunya, Barcelona, Spain
(3) University of Maryland, College Park, United States
(4) Escola Tècnica Superior d'Enginyeria Industrial de Barcelona, Spain
We study the dynamics of the restricted planar three-body problem near mean motion resonances, i.e. a resonance involving the Keplerian periods of the two lighter bodies revolving around the most massive one. This problem is often used to model Sun–Jupiter–asteroid systems. For the primaries (Sun and Jupiter), we pick a realistic mass ratio $\mu=10^{-3}$ and a small eccentricity $e_0>0$. The main result is a construction of a variety of Based on the work of Treschev, it is natural to conjecture that the time of diffusion for this problem is $\sim \frac{-\ln (\mu e_0)}{\mu^{3/2} e_0}$. We expect our instability mechanism to apply to realistic values of $e_0$ and we give heuristic arguments in its favor. If so, the applicability of Nekhoroshev theory to the three-body problem as well as the long time stability become questionable.
It is well known that, in the Asteroid Belt, located between the orbits of Mars and Jupiter, the distribution of asteroids has the so-called
non local diffusing orbits which show a drastic change of the osculating (instant) eccentricity of the asteroid, while the osculating semi major axis is kept almost constant. The proof relies on the careful analysis of the circular problem, which has a hyperbolic structure, but for which diffusion is prevented by KAM tori. In the proof we verify certain non-degeneracy conditions numerically.
Kirkwood gaps exactly at mean motion resonances of low order. Our mechanism gives a possible explanation of their existence. To relate the existence of Kirkwood gaps with Arnol'd diffusion, we also state a conjecture on its existence for a typical $\epsilon$-perturbation of the product of the pendulum and the rotator. Namely, we predict that a positive conditional measure of initial conditions concentrated in the main resonance exhibits Arnol’d diffusion on time scales $\frac{- \ln \epsilon }{\epsilon^{2}}$.
Based on the work of Treschev, it is natural to conjecture that the time of diffusion for this problem is $\sim \frac{-\ln (\mu e_0)}{\mu^{3/2} e_0}$. We expect our instability mechanism to apply to realistic values of $e_0$ and we give heuristic arguments in its favor. If so, the applicability of Nekhoroshev theory to the three-body problem as well as the long time stability become questionable.
It is well known that, in the Asteroid Belt, located between the orbits of Mars and Jupiter, the distribution of asteroids has the so-called
Keywords: Three-body problem, instability, resonance, hyperbolicity, Mather mechanism, Arnol’d diffusion, Solar System, Asteroid Belt, Kirkwood gap
Féjoz Jacques, Guardia Marcel, Kaloshin Vadim, Roldán Pablo: Kirkwood gaps and diffusion along mean motion resonances in the restricted planar three-body problem.
J. Eur. Math. Soc. 18 (2016), 2315-2403. doi: 10.4171/JEMS/642 |
Revista Matemática Iberoamericana
Full-Text PDF (215 KB) | Metadata | Table of Contents | RMI summary
Volume 24, Issue 3, 2008, pp. 1075–1095 DOI: 10.4171/RMI/567
Published online: 2008-12-31
Majorizing measures and proportional subsets of bounded orthonormal systemsOlivier Guédon
[1], Shahar Mendelson [2], Alain Pajor [3]and Nicole Tomczak-Jaegermann [4](1) Université Paris-Est, Marne-la-Vallée, France
(2) Technion - Israel Institute of Technology, Haifa, Israel
(3) Université Paris-Est, Marne-la-Vallée France
(4) University of Alberta, Edmonton, Canada
In this article we prove that for any orthonormal system $(\varphi_j)_{j=1}^n \subset L_2$ that is bounded in $L_{\infty}$, and any $1 < k < n$, there exists a subset $I$ of cardinality greater than $n-k$ such that on $\mathrm{span}\{\varphi_i\}_{i \in I}$, the $L_1$ norm and the $L_2$ norm are equivalent up to a factor $\mu (\log \mu)^{5/2}$, where $\mu = \sqrt{n/k} \sqrt{\log k}$. The proof is based on a new estimate of the supremum of an empirical process on the unit ball of a Banach space with a good modulus of convexity, via the use of majorizing measures.
Keywords: Empirical process, majorizing measure, orthonormal system
Guédon Olivier, Mendelson Shahar, Pajor Alain, Tomczak-Jaegermann Nicole: Majorizing measures and proportional subsets of bounded orthonormal systems.
Rev. Mat. Iberoam. 24 (2008), 1075-1095. doi: 10.4171/RMI/567 |
This is a question in the undergraduate-level textbook "Advanced Calculus" by Fitzpatrick.
Suppose that a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is twice differentiable such that for $\forall x$, $f'(x)\leq f(x)$, and $f(0)=0$. Then is $f$ the zero function?
The answer to this is not true as I was able to find a counter-example $f^*(x)= 1- e^x$. However we have only just learned about differentiation, the mean-value theorem and how to find extremes using 1st and 2nd derivatives, and we have only seen derivatives of polynomials so far, but I don't know how to disprove the above statement by using these.
(EDIT) For $1−e^x$ to be a valid counter-example, I need to "officially know" that the exponential function's derivative is equal to itself. But exponential functions are in the next chapter. Therefore unless I want to "cheat", I need to think of another function. |
I have to factor the polynomial at the field of 9 elements. For it I view the field $GF(3)[x]/(x^2+1)$. But if I view the field $GF(3)[x]/(x^2+x+2)$ I get another decomposition of this polynomial. So, why is it? And is it right to factor this polynomial over only one field?
closed as unclear what you're asking by user223391, GNUSupporter 8964民主女神 地下教會, B. Mehta, José Carlos Santos, Chris Godsil May 13 '18 at 0:01
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There should only be one decomposition
up to isomorphism, and I'm going to guess that your problem is a notational one.
Firstly, $GF(3)[x]/(x^2 + 1)$ and $GF(3)[x]/(x^2 + x + 2)$
are both fields of 9 elements, and they are isomorphic. But the variable $x$ doesn't play the same role in each. To keep the notation clean, I'm going to define some "generic" field of 9 elements, say $F$, and assume we have isomorphisms $GF(3)[x]/(x^2 + 1)\to F$ and $GF(3)[x]/(x^2 + x + 2)\to F$. Now: Let $\alpha$ be the image of $x$ in $F$ under the map $GF(3)[x] \to GF(3)[x]/(x^2 + 1) \to F$. Let $\beta$ be the image of $x$ in $F$ under the map $GF(3)[x] \to GF(3)[x]/(x^2 + x + 2) \to F$.
We can use this to write $F$ explicitly in two ways:
$F \cong GF(3)[\alpha]$, i.e. $F = \{0, 1, 2, \alpha, \alpha+1, \alpha+2, 2\alpha, 2\alpha+1, 2\alpha+2\}$ as a set, $F \cong GF(3)[\beta]$, i.e. $F = \{0, 1, 2, \beta, \beta+1, \beta+2, 2\beta, 2\beta+1, 2\beta+2\}$ as a set,
but $\alpha$ and $\beta$ are still
not the same thing, because the arithmetic works differently. After all, inside $F$, the following things are true: $\alpha^2 + 1 = 0$ (but $\beta^2 + 1 \neq 0$) $\beta^2 + \beta + 2 = 0$ (but $\alpha^2 + \alpha + 2 \neq 0$).
The issue is that we have two ways of
representing $F$ - as $GF(3)[\alpha]$ and $GF(3)[\beta]$ - and they're isomorphic, but the isomorphism is not just the map $\alpha\mapsto \beta$. Can you work out what it is?
Now, work out the decomposition of your polynomial in terms of $\alpha$, and separately in terms of $\beta$. What happens when you take the $\alpha$-decomposition and apply the isomorphism $GF(3)[\alpha]\to GF(3)[\beta]$ to it? |
In Blyth's book "Module Theory: An Approach to Linear Algebra" matrix theory is developed generally over a noncommutative ring $R$ with $1$. However, it seems that there is a mistake in way that an important property doesn't carry over to noncommutative rings, namely the isomorphic of a ring of $n\times n$ matrices with coefficients in $R$ and the endomorphism ring of a free module over $R$
Given an a homomorphism $\phi\colon M\to N$ of free $R$-modules with respective bases $(a_i)_m$ and $(b_i)_n$, the matrix $\mathrm{Mat}(\phi,(b_i)_n,(a_i)_m)$ of this homomorphism with respect to said bases is a matrix $(r_{ij})$ such that $r_{ij}$ is the unique element of $R$ so that $\phi(a_i) = s_{1i}b_1 + ... + r_{ij}b_j + ... + s_{ni}$.
One can prove that, for the respective bases, there is an isomorphism $\vartheta\colon\mathrm{Hom}_R(M,N)\to\mathrm{Mat}_{n\times n}(R), \phi \mapsto \mathrm{Mat}(\phi,(b_i)_n,(a_i)_m)$.
But am I right to assume that this is not a ring homomorphism? It seems modules over noncommutative rings lack the multiplicative property $$\mathrm{Mat}(\psi\circ\phi, (c_i)_p,(a_i)_n) = \mathrm{Mat}(\psi,(c_i)_p,(b_i)_m)\mathrm{Mat}(\phi,(b_i)_m,(a_i)_n)$$ for free modules $M,N,P$ with respective bases $(a_i)_n, (b_i)_m$ and $(c_i)_p$ and their homomorphisms $\phi\colon M\to N, \psi\colon N\to P$.
Am I right or there is a mistake there? I was doing all the matrix theory for commutative rings before now, but encountered a proof in the aforementioned book which uses (probably wrong) ring isomorphism.
Let $\mathrm{Mat}(\phi,(b_i)_m,(a_i)_n) = (r_{ij})$ and $\mathrm{Mat}(\psi,(c_i)_p, (b_i)_m) = (s_{ij})$. Then we have $\mathrm{Mat}(\psi,(c_i)_p, (b_i)_m)\mathrm{Mat}(\phi,(b_i)_m,(a_i)_n) = (t_{ik})$ where $t_{ik} = \sum_{j = 1}^m s_{ij}r_{jk}$. Also, $$(\psi\circ\phi)(a_i) = \psi\left(\sum_{i = 1}^n r_{ij}b_i\right) \\ = \sum_{i = 1}^n r_{ij}\psi(b_i) \\ = \sum_{i = 1}^n r_{ij}\left(\sum_{k = 1}^p s_{ki}c_k\right) \\ = \sum_{k = 1}^p \left(\sum_{i = 1}^n r_{ij}s_{ki}\right)c_k \\ \neq \sum_{k = 1}^p \left(\sum_{i = 1}^n s_{ki}r_{ij}\right)c_k $$ generally. |
So I have a Poisson distribution:
$ V \sim \operatorname{Po} \left({\rho v}\right) $
and I've calculated the maximum likelihood estimator $ \widehat{\rho} = \dfrac{\overline{v}}{v} $ from independent samples $ v_{1}...v_{n} $. How do I test whether $ \widehat{\rho} $ is unbiased?
Thanks |
Defining parameters
Level: \( N \) = \( 4000 = 2^{5} \cdot 5^{3} \) Weight: \( k \) = \( 1 \) Character orbit: \([\chi]\) = 4000.cf (of order \(50\) and degree \(20\)) Character conductor: \(\operatorname{cond}(\chi)\) = \( 500 \) Character field: \(\Q(\zeta_{50})\) Newforms: \( 0 \) Sturm bound: \(600\) Trace bound: \(0\) Dimensions
The following table gives the dimensions of various subspaces of \(M_{1}(4000, [\chi])\).
Total New Old Modular forms 200 0 200 Cusp forms 40 0 40 Eisenstein series 160 0 160
The following table gives the dimensions of subspaces with specified projective image type.
\(D_n\) \(A_4\) \(S_4\) \(A_5\) Dimension 0 0 0 0 Decomposition of \(S_{1}^{\mathrm{old}}(4000, [\chi])\) into lower level spaces
\( S_{1}^{\mathrm{old}}(4000, [\chi]) \cong \) \(S_{1}^{\mathrm{new}}(2000, [\chi])\)\(^{\oplus 2}\) |
Wolfram Alpha says that
$$\sum_{n=1}^{\infty} \frac{1}{n^2-3n+3} = 1 + \frac{\pi \tanh \left ( \frac{\sqrt{3}\pi}{2} \right )}{\sqrt{3}}$$
However I am unable to get it. It is fairly routine to prove that
$$\sum_{n=-\infty}^{\infty} \frac{1}{n^2-3n+3} = \frac{2\pi \tanh \left ( \frac{\sqrt{3}\pi}{2} \right )}{\sqrt{3}}$$
by using complex analysis ( contour integration ) but honestly I am stuck how to retrieve the original sum. Split up , the last sum gives:
\begin{align*} \sum_{n=-\infty}^{\infty} \frac{1}{n^2-3n+3} &= \sum_{n=-\infty}^{-1} \frac{1}{n^2-3n+3} + \frac{1}{3} + \sum_{n=1}^{\infty} \frac{1}{n^2-3n+3} \\ &=\frac{1}{3} +\sum_{n=1}^{\infty} \frac{1}{n^2+3n+3} + \sum_{n=1}^{\infty} \frac{1}{n^2-3n+3} \\ &=\frac{1}{3}+ \sum_{n=1}^{\infty} \left [ \frac{1}{n^2-3n+3} + \frac{1}{n^2+3n+3} \right ] \end{align*}
Am I overlooking something here?
P.S: Working with digamma on the other hand I am not getting the constant. I'm getting $\frac{1}{3}$ instead. |
MOP 1997: Let $Q$ be a quadrilateral whose side lengths are $a,b,c,d$ in that order. Show that the area of $Q$ does not exceed $(ac+bd)/2$.
My solution: Without loss of generality, let $a$ be the longest side. I first prove that the area cannot exceed (a
b+c d)/2. Drawing a diagonal, we can divide the quadrilateral into two triangles with sides are $a$ and $b$, and $c$ and $d$. The triangle with sides $a$ and $b$ can have maximum area if the angle between them is $\pi/2$. In this case, the area between them will be $\frac{1}{2}ab$. Similarly, the area of the other triangle can have a maximum of $\frac{1}{2}cd$. Adding them up, we get that the maximum area of the quadrilateral can be $\frac{1}{2}(ab+cd)$.
Now if this maximum is achieved, which means that the angle between $a$ and $b$, and $c$ and $d$ are both $\pi/2$,
by drawing a diagram, we can convince ourselves that $a=c$. This means that $a$ and $b$ are both the longest sides in the quadirlateral. Hence, by the rearrangement inequality, we have $(ab+cd)/2\leq (ac+bd)/2$. Hence, the area of the quadrilateral is less than or equal to $(ac+bd)/2$.
Is the reasoning given above correct? Does there exist a better proof? |
Does the series $\sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{n}}{\sqrt [3]{n+1}}}$ converge or diverge?
The series can be written as $\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n-1}}{\sqrt [3]{n}}}$, which remind me of the alternating harmonic series, wich is not convergent.
The terms are alternating, and therfor shuld the alternating series test be used. Have also checked the Leibniz cri.
Where do I go from here? |
Homework Statement: Henry Cavendish succeeded in measuring the value of the constant "G" way back in the late 1700s. His method was to put two known masses at a known distance and measure the attractive force between them; then he could use Newton's Law of Universal Gravitation to find "G"...
Homework Statement: The particle is moving in circular orbit such a way that the net force (F) is always towards the point p (point p is on the circumference of circle). Find the variation of force F with respect to r.i.e find the value of n in the expression F=kr^nHomework Equations: F=kr^n...
Nieuwenhuizen uses a method for calculating the propagator by decomposing the field ## h_{\mu\nu}, ## first into symmetric part ## \varphi_{\mu\nu} ## and antisymmetric part ## \psi_{\mu\nu} ##, and then by a spin decomposition using projector operators. Using this he writes the dynamical...
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I have a simulation I'm trying out (for fun). A self-gravitating ball of gas, in deep space. (The sim uses a fixed-time-step for each iteration.)I'd like to use Boyles Ideal gas law, the force of gravity, and energy as internal heat. (I don't want to touch enthalpy unless I don't realize...
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Consider a signal $$ x(t) = \cos(175\pi t) $$ which is sampled to produce discrete time signal $$ x[n] = x(nT_s) $$ The fundamental period of $x[n]$ is $$ N_0 = 7 $$
Given this, what is the smallest possible sample rate $T_s$? (Ans: 1.6327 ms).
I would assume that this is related to finding the Nyquist frequency. I was thinking:
Since, $$N_0 = 7 \implies f_0 = \frac17 \implies f_{\mathrm{Nyquist}} = 2 \frac17\implies T_s = \frac72 $$ However, this is obviously incorrect. I am not even using any information of the original signal. Any suggestions on what I could be doing wrong here? |
What is the proof, without leaving the Earth, and involving only basic physics, that the earth rotates around its axis?
By basic physics I mean the physics that the early physicists must've used to deduce that it rotates, not relativity.
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What is the proof, without leaving the Earth, and involving only basic physics, that the earth rotates around its axis?
By basic physics I mean the physics that the early physicists must've used to deduce that it rotates, not relativity.
Foucault pendulum. I don't know how the ancients did it, but it is surely pure classical mechanics.
The animation describes the motion of a Foucault Pendulum at a latitude of 30°N.
The Foucault pendulum is a great experiment which does demonstrate that the Earth is rotating, but it was only introduced in 1851. The Earth had been known to rotate for several centuries before that, probably stimulated by Copernicus and Galileo pushing the heliocentric model of the solar system during the 16th century.
A couple of decades before Faucalt's pendulum, the Coriolis effect was discovered. This effects (among other similarly large systems) hurricanes, causing them to rotate clockwise/anti clockwise depending on whether they're in the southern/northern hemisphere. It is an apparent force that appears in any rotating frame of reference (like a spinning planet). This again won't have helped early 'spinning-Earth' believers.
Early evidence that the Earth rotates was almost certainly the observation of the sun, planets and stars moving across the sky and then, with the help of telescopes, of the other planets also rotating. Of course this requires you to trust that the Earth is not the centre of the universe and so doesn't "prove" the Earth is spinning in the same way as observation of Foucault's pendulum or the Coriolis effect.
The idea of a "proof" in physics is a difficult one. A theory such as the 'spinning-Earth' will at first simply be presented to explain anomalous observations that current theories can't (such as why do the other planets and the sun move independently of the background stars if everything rotates together around the Earth?). It then gets tested by experiments inspired or predicted by the theory (if the Earth is spinning, what should we expect to observe?). If everything holds up, it's accepted as fact. Eventually, some bright spark realises he/she can demonstrate that the Earth is rotating with a clever little experiment (Foucault's pendulum/Coriolis effect/launching a rocket into space) and it is added to the mountain of evidence already accumulated on the subject.
I think the Foucault pendulum is the best answer, but for the sake of variety I'll add another very interesting one: the equatorial bulge affecting the figure of the Earth. This is the "pancaking" of the planet due to its rotation. You can measure the geometry of the Earth without leaving its surface, and find that it is bulging in accord with your expectations if that bulge were caused by rotation at the same rate as the one we observe relative to distant stars. As always, there's no such thing as "proof" in physics, but this is strong supporting evidence.
An indirect indication that the Earth rotates is the fact that the rotation varies over time. First of all, the orientation of the Earth's axis changes: long-term effects like precession and slow variations in the axial tilt, as well as small short-term variations like nutation. Precession was already known in the Ancient world (Hipparchus, Ptolemy,...) and the change in axial tilt was recognized by people like Fracastoro (in 1538). See the wiki pages for historic background.
The Earth's period of rotation also changes. First of all, the rotation is slowing down, caused by the tidal interaction between the Earth and the Moon: the length of the day increases by about 2 milliseconds per century. Edmond Halley was the first to notice that the orbital period of the Moon had changed compared with ancient records, and the effect was explained in the 18th and 19th century.
These days, we're able to measure the Earth's rotation very precisely, and we find that the rotation varies slightly from day to day. In particular, changes in the atmospheric winds and oceanic currents cause periodic fluctuations in the axial tilt and the period of rotation: an annual fluctuation with an amplitude of 0.34 milliseconds, a semiannual period with an amplitude of 0.29 milliseconds, 10-day fluctuations of the order of 0.1 milliseconds, fluctuations due to El Niño events, etc (see wikipedia, and also this post). Large earthquakes can also change the rotation period by a few microseconds, but these effects prove difficult to measure (see this article on the 2011 Japan earthquake).
Do these variations prove that the Earth rotates? Not directly, but I'd like to hear geocentrists explain how the entire universe can change its rotation, not only over thousands of years but even on a daily basis, when the causes of these variations can be traced back to events on Earth or its orbit.
Edit
Another proof that the Earth rotates, although it can only be measured with modern techniques:
aberration.
You're all familiar with this effect: suppose that you're standing in the rain, and there's no wind. Since the rain will fall vertically, you have to hold your umbrella straight up. But now start running. What happens? From your point of view, the rain will no longer fall vertically: you have to tilt your umbrella forward to keep your head dry. A similar phenomenon happens with light: the direction in which you see a beam of light depends on your velocity. This effect is called aberration.
For instance, suppose you have two stars, separated by an angle $\theta$ in a rest frame. If an observer is moving towards the star on the horizontal axis, then he will see the second star at an angle $\varphi$, instead of $\theta$. If the velocity of the observer changes, then the angle $\varphi$ changes, and the star will appear to 'wobble' with respect to stars at different positions.
The orbital velocity $v\approx 30\;\text{km/s}$ of the Earth around the Sun causes an
annual aberration: over the period of a year, every star and galaxy will appear to wobble on an ellipse, with a maximum displacement of $v/c\approx 20.5''$ around their mean position, regardless of their distance to the Earth (unlike parallax, which does depend on distance). It was first observed by James Bradley in 1725, and is direct proof that the Earth orbits the Sun.
But there's another, much smaller aberration: a
diurnal aberration, caused by the rotation of the Earth around its axis. The effect is greatest for an observer on the equator, who has an equatorial speed of $v = 0.465\;\text{km/s}$, while an observer on the poles will see no effect. For an observer on the equator, the position of every star wobbles on a daily basis, with a maximum displacement of $v/c\approx 0.32''$. It's an incredibly small effect, but it is measurable, and it has to be taken into account when doing high-precision astrometry. Moreover, it proves that the Earth does rotate.
Anything related to the Coriolis effect (some pretty pictures can be found in the link), i.e. even cannons will be (not precisely, rather seem) deflected because of the earth's rotation.
Measuring the geometry of the earth, we find that it has an equatorial bulge. We make no assumptions about the cause of the bulge, though it suggests already that the earth is rotating as @Mike has described.
We measure the acceleration due to gravity at the poles and on the equator. Most of the difference we find is accounted for by the bulge, but there is a $0.3\%$ discrepancy. The acceleration due to gravity is smaller on the equator than we expect. This discrepancy is nicely explained by assuming the earth is rotating. In fact, making this assumption we can make a rough calculation of the period of the earth.
We could go so far as to carefully measure the acceleration due to gravity along different lines of latitude and thus find the acceleration due to gravity as a function of latitude. We would find that these measurements were well described by a model in which the oblate earth were rotating with a period of about a day.
Let us check the claim that the earth is rotating about it's own axis. We may choose this axis to be the $z$-axis. The earth can be approximated by a sphere. Consider a pendulum living somewhere on the surface of the earth, initially swinging on a north-south line. The position of the bob is described a vector ${V} = (V^{\theta}, V^{\phi})$ living in the tangent space of this sphere. As the earth rotates, $V$ is parallely transported along the curved sketched out by the earth's rotation, call it $\gamma(t) = (\theta(t), \phi(t))$ (a circle). Then the equation obeyed by $V$ is given by, $$\nabla_{\dot{\gamma(t)}}V^{\alpha} = \dot{\theta}\nabla_{\theta}V^{\alpha} + \dot{\phi}\nabla_{\phi}V^{\alpha} = 0.$$ But since the earth only changes the $\phi$ angle, by rotating, $\dot{\gamma(t)} = (0, \dot{\phi})$ ($ = (0, 2\pi /24\text{hrs})$). The equation for parallel transport thus gives $$\nabla_\phi V^{\theta} = 0,$$ and $$\nabla_{\phi}V^{\phi} = 0.$$ These are simply the covariant derivatives, so that these equations are
$$\partial_{\phi}V^{\theta} + \Gamma^\theta_{\phi \mu}V^{\mu} = 0,$$ $$\partial_{\phi}V^{\phi} + \Gamma^\phi_{\phi \mu}V^{\mu} = 0.$$
An easy computation shows that the non-vanishing christoffel symbols of the sphere are $\Gamma^{\theta}_{\phi \phi} = -\text{sin}{\theta}\text{cos}{\theta},$ $\Gamma^{\phi}_{\phi \theta} = \text{cot}{\theta}.$ These equations become coupled equations as follows: $$\partial_{\phi}V^{\theta} -\text{sin}\theta \text{cos}\theta V^{\phi} = 0,$$ $$\partial_{\phi}V^{\phi} + \text{cot}\theta V^{\theta} = 0.$$ Uncoupling them by solving for $V^{\theta}$ in the second equation and plugging it into the first, we get, $$\partial^2_{\phi}V^{\phi} = -\text{sin}^2 \theta V^{\phi}.$$ This is easy to solve, and from this you get how much the pendulum should swing in the direction of $\phi$, in say a 24 hour period. Now compare with experiment.
The Sagnac effect can detect absolute rotation. This is used, for example, in the ring laser gyros used on modern aircraft as navigational devices. Not only is it possible to use this effect to prove that the earth rotates, its precision for detecting variations in the rotation is starting to become competitive with the most precise techniques: see http://www.wettzell.ifag.de/LKREISEL/G/LaserGyros.html
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The shear area is calculated as follows:
${\mathrm A}_{\mathrm y}\;=\;\frac{{\mathrm I}_{\mathrm z}^2}{\int_{\mathrm A^\ast}\left({\displaystyle\frac{{\mathrm S}_{\mathrm z}}{\mathrm t^\ast}}\right)^2\operatorname d\mathrm A^\ast}$
${\mathrm A}_{\mathrm z}\;=\;\frac{{\mathrm I}_{\mathrm y}^2}{\int_{\mathrm A^\ast}\left({\displaystyle\frac{{\mathrm S}_{\mathrm y}}{\mathrm t^\ast}}\right)^2\operatorname d\mathrm A^\ast}$
Where:I
zor I y: second moment of area in relation to the axis z or y S zor S y: first moment of area in relation to the axis z or y t*: effective element thickness for shear transfer
A*:
surface area based on effective shear thickness t*
The effective element thickness for shear transfer t* has a significant influence on the shear area. Therefore, the defined effective element thickness for shear transfer t* (Figure 1) of the elements should be checked.
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To prevent the error message (Figure 1) from appearing and the calculation to be started, it is necessary to assign a material to the corresponding point elements as follows:
Right-click the entire cross-section: 'Reduce Cross-Section into Elements', Figure 2 Selection of the corresponding elements or all point elements with the status ‘Insert’ by using the Views navigator, because it is only possible to define a material for them (in contrast to point elements with the ‘Remove’ status) Assign the corresponding material and confirm with OK, Figure 3 Deactivation of visibility Select all elements and right-click one of the selected elements: 'Create Cross-Section from Selected Elements', Figure 4
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If the results are not displayed exponentially (Figure 1) but as integers, the number of decimal places must be minimized, see Figure 2: Menu 'Options', 'Units and Decimal Places'.
Another option would be to change the unit from, for example, 'mm' to 'cm'.
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AnswerWhen determining the torsional moment of inertia, point elements are not considered for I tin the analytical equations. Therefore, it is possible to apply a correction factor for cross-sections with distinct fillets such as rolled cross-sections, to determine the effect of the point elements. Figure 01 shows recommendations for different types of rolled cross-sections. This correction factor can be adjusted in the calculation parameters. In addition, the torsional moment of inertia can also be calculated with the FE method, which also covers the effect of point elements and welds. This approach requires slightly more computing time than the analytical calculation.
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Production of $K*(892)^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$ =7 TeV
(Springer, 2012-10)
The production of K*(892)$^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$=7 TeV was measured by the ALICE experiment at the LHC. The yields and the transverse momentum spectra $d^2 N/dydp_T$ at midrapidity |y|<0.5 in ...
Transverse sphericity of primary charged particles in minimum bias proton-proton collisions at $\sqrt{s}$=0.9, 2.76 and 7 TeV
(Springer, 2012-09)
Measurements of the sphericity of primary charged particles in minimum bias proton--proton collisions at $\sqrt{s}$=0.9, 2.76 and 7 TeV with the ALICE detector at the LHC are presented. The observable is linearized to be ...
Pion, Kaon, and Proton Production in Central Pb--Pb Collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012-12)
In this Letter we report the first results on $\pi^\pm$, K$^\pm$, p and pbar production at mid-rapidity (|y|<0.5) in central Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV, measured by the ALICE experiment at the LHC. The ...
Measurement of prompt J/psi and beauty hadron production cross sections at mid-rapidity in pp collisions at root s=7 TeV
(Springer-verlag, 2012-11)
The ALICE experiment at the LHC has studied J/ψ production at mid-rapidity in pp collisions at s√=7 TeV through its electron pair decay on a data sample corresponding to an integrated luminosity Lint = 5.6 nb−1. The fraction ...
Suppression of high transverse momentum D mesons in central Pb--Pb collisions at $\sqrt{s_{NN}}=2.76$ TeV
(Springer, 2012-09)
The production of the prompt charm mesons $D^0$, $D^+$, $D^{*+}$, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at the LHC, at a centre-of-mass energy $\sqrt{s_{NN}}=2.76$ TeV per ...
J/$\psi$ suppression at forward rapidity in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012)
The ALICE experiment has measured the inclusive J/ψ production in Pb-Pb collisions at √sNN = 2.76 TeV down to pt = 0 in the rapidity range 2.5 < y < 4. A suppression of the inclusive J/ψ yield in Pb-Pb is observed with ...
Production of muons from heavy flavour decays at forward rapidity in pp and Pb-Pb collisions at $\sqrt {s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012)
The ALICE Collaboration has measured the inclusive production of muons from heavy flavour decays at forward rapidity, 2.5 < y < 4, in pp and Pb-Pb collisions at $\sqrt {s_{NN}}$ = 2.76 TeV. The pt-differential inclusive ...
Particle-yield modification in jet-like azimuthal dihadron correlations in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012-03)
The yield of charged particles associated with high-pT trigger particles (8 < pT < 15 GeV/c) is measured with the ALICE detector in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV relative to proton-proton collisions at the ...
Measurement of the Cross Section for Electromagnetic Dissociation with Neutron Emission in Pb-Pb Collisions at √sNN = 2.76 TeV
(American Physical Society, 2012-12)
The first measurement of neutron emission in electromagnetic dissociation of 208Pb nuclei at the LHC is presented. The measurement is performed using the neutron Zero Degree Calorimeters of the ALICE experiment, which ... |
Trying to evaluate this indefinite integral:
$$ \int (x^2 + 1)\cos2xdx$$
So far I have the following: $u=x^2 + 1 \Rightarrow du = 2xdx$ and $dv=\cos2x \Rightarrow v = \frac {\sin2x}{2}$. So the integral is equal to:
$$\int (x^2+1)\cos2xdx = (x^2+1)\frac{\sin2x}{2}-\int {\frac{\sin2x}{2}}2xdx$$
Next, I make another substitution for the integral on the right hand side; let $ u = x \Rightarrow du = dx$ and let $dv = \sin2x \Rightarrow v = \frac {-\cos2x}{2}$. Now I have the following:
$$\int (x^2+1)\cos2xdx = (x^2+1)\frac{\sin2x}{2}-\left (-\frac {x\cos2x}{2} - \int -\frac {cos2x}{2}dx\right)$$
Which after integrating becomes:
$$\int (x^2+1)\cos2xdx = (x^2+1)\frac{\sin2x}{2}-\left(-\frac {x\cos2x}{4} + \frac {\sin2x}{4}\right)$$
But when solving with the integrator on my calculator, I get a different answer (it looks like I am getting closer, but still off). What am I doing wrong here?? |
Let's say we have a convex objective function $f(\textbf{x})$, with $\textbf{x}\in R^n$ which we want to minimise under a set of constraints. The problem is that calculating $f$
exactly is not possible and only stochastic approximations are available, which are computably expensive.
Luckily the gradient $\nabla f$ can also be approximated such that we can use a stochastic gradient descent method:
$$\textbf{x}^{(k+1)}=\mathbf{x}^{(k)}-\alpha^{(k)} \nabla f(\textbf{x})$$where $\alpha^{(k)}$ is the step size and $\nabla f(\mathbf{x})$ is the
approximated gradient of $f(\mathbf{x})$. My question
This method is fast since it still converges even when the accuracy of the gradient is relatively low. But this will give an unconstrained solution, i.e., I will need that,
$$ x_j=0 \lor \alpha_j \le x_j \le \beta_j\quad \text{for all } \ \ j, \ \text{ and }\sum_j x_j \le 1 $$
where $\alpha_j$ and $\beta_j$ are minimum and maximum constraints such that $0<\alpha_j, \beta_j\le 1$.
Is there a way to use Mathematica's built in functions such as
FindMinimum[] or
Minimize[] to carry out a stochastic gradient descent (or any other method that will solve this problem).
Toy code
p = {.47, .53, .48, .31, .255, .27, .27, .311, .31, .466, .371, .304};b = {2.5, 2, 2.2, 3.5, 4.33, 4, 4, 3.4, 3.4, 2.2, 2.7, 3.3};z[o_, b_] := If[o == 1, b - 1, -1];n = Length[p];M = 10000;outcomes = Transpose[Table[RandomVariate[MultinomialDistribution[1,{p[[i]], 1 - p[[i]]}], M], {i, n}]];f[x_] := Block[{Y, R}, Mean[Reap[Do[Y = outcomes[[K]];R = 1 + Sum[z[Y[[i, 1]], b[[i]]] x[[i]], {i, n}]; Sow[-Log[R]];, {K, M}]][[2,1]]]] |
The coupled mode space NEGF method for computing transistor characteristics involves expanding the electronic wavefunction in a mode space basis
$$\Psi(x,y,z) = \sum_n\phi_n(x)\xi_n(y,z;x)$$
where $\xi_n(y,z;x)$ is the nth mode at position x (the x dimension being the direction of transport).
The method involves computing coupling between modes along the transport direction. E.g. coupling coefficients like
$$b_{mn}(x) = \langle\xi_m|\frac{\partial}{\partial x}|\xi_n\rangle$$
The numerical algorithm will divide the transistor into $X$ slices $\Delta x$ apart, and compute the modes on each slice. The issue is my numerical solver will sometimes compute $\xi_n$ as the nth eigenvalue, and sometimes compute $-\xi_n$, so phases might not be consistent across each slice. This changing phase will result in a large $\frac{\partial}{\partial x}|\xi_n\rangle$
My question is: Do the phases need to be consistent across each slice? If they do, how do NEGF algorithms ensure they are consistent? |
I'm trying to solve the fully coupled drift-diffusion system using Newton's Method. Although I eventually plan to potentially use a Jacobian-Free Newton-Krylov approach, this is still something that I want to implement and look into.
A bit of notation for the equations presented below:
$d_{ij}$ and $l_{ij}$ are control area/volume constants, not physical entities. The $i$ and $j$ indices are for the central point being considered ($i$), and all neighbouring points that are connected ($j$'s). $A_{\Omega_i}$ is the control area/volume associated to each vertex/node, and is also constant.
All quantities that are explicitly required during calculation are at node points. In the equations $B(\Delta_{ij}) \equiv B(\phi_i - \phi_j)$, the Bernoulli function, used as a smoothing method for the differences. This difference should
technically exist between node points, but that's for another time.
Here are the equations: $\phi$ is electrostatic potential, $n$ and $p$ are carrier concentrations. These are the independent variables to solve for. Everything else is either a constant, or a function of one of these variables.
$$\sum_{j\ne i} \frac{d_{ij}}{l_{ij}}(\phi_i - \phi_j) = -\frac{q}{\epsilon}(p_i - n_i - N_{D_i}^+ - N_{A_i}^-)A_{\Omega_i}$$
$$\frac{\partial{n_i}}{\partial t}A_{\Omega_i} = D_n \sum_{j \ne i} \left[\frac{d_{ij}}{l_{ij}} \{n_iB(\Delta_{ij}) - n_j B(-\Delta_{ij})\}\right] + (G_i - R_i)A_{\Omega_i}$$
$$\frac{\partial{p_i}}{\partial t}A_{\Omega_i} = D_p \sum_{j \ne i} \left[\frac{d_{ij}}{l_{ij}} \{p_iB(-\Delta_{ij}) - p_j B(\Delta_{ij})\}\right] + (G_i - R_i)A_{\Omega_i}$$
In residual form, and assuming equilibrium conditions so all time derivatives go to 0, these equations are:
$$\Phi = \sum_{j\ne i} \frac{d_{ij}}{l_{ij}}(\phi_i - \phi_j) +\frac{q}{\epsilon}(p_i - n_i - N_{D_i}^+ - N_{A_i}^-)A_{\Omega_i} = 0$$
$$N = D_n \sum_{j \ne i} \left[\frac{d_{ij}}{l_{ij}} \{n_iB(\Delta_{ij}) - n_j B(-\Delta_{ij})\}\right] + (G_i - R_i)A_{\Omega_i} = 0$$
$$P = D_p \sum_{j \ne i} \left[\frac{d_{ij}}{l_{ij}} \{p_iB(-\Delta_{ij}) - p_j B(\Delta_{ij})\}\right] + (G_i - R_i)A_{\Omega_i} = 0$$
Now, my question is whether I have these Jacobian elements correct. Here is the Jacobian element formula that I've tried to come up with. Only a few equations have been derived, since the remainining elements are just trivial variations of these 3 formulations. This post is mostly about getting more experienced eyes to look at these equations, and potentially see something that I haven't quite seen yet. Are these formulations for the Jacobian elements correct, or is there something wrong with them that I'm missing?
$$\frac{\partial\Phi}{\partial\phi_i} = \sum_{j\ne i} \frac{d_{ij}}{l_{ij}} +\frac{q}{\epsilon}(\frac{\partial p_i}{\partial \phi_i} - \frac{\partial n_i}{\partial \phi_i})A_{\Omega_i}$$
$$\frac{\partial{N}}{\partial \phi_i} = D_n \sum_{j \ne i} \left[\frac{d_{ij}}{l_{ij}} \left\{\frac{\partial{n_i}}{\partial \phi_i}B(\Delta_{ij}) + n_i \frac{\partial B(\Delta_{ij})}{\partial \phi_i})\right\}\right] + \frac{\partial(G_i - R_i)}{\partial \phi_i}A_{\Omega_i}$$
$$\frac{\partial N}{\partial n_i} = D_n \sum_{j \ne i} \left[\frac{d_{ij}}{l_{ij}} B(\Delta_{ij})\right] + \frac{\partial (G_i - R_i)}{\partial n_i}A_{\Omega_i}$$ |
Problem Statement: Let $A=\begin{bmatrix} 2 && 1 \\ 1 && 2 \end{bmatrix}$. Find an orthonormal basis for $\mathbb{C}^2$ with respect to the Hermitian form $\bar{x}^TAy$.
I am trying to figure out this proof, but not sure if what I am currently trying to do will help me. I was goin to start out by taking an arbitrary basis $B=\left\{v_1,v_2\right\}$ and write two arbitrary vectors in $\mathbb{C}^2$ as linear combinations of basis elements: $$v=x_1v_1+x_2v_2$$ $$w=y_1v_1+y_2v_2$$
Then break down the Hermitian form $$\langle v,w \rangle=\langle x_1v_1+x_2v_2, y_1v_1+y_2v_2 \rangle=\bar{x_1}\langle v_1, y_1v_1+y_2v_2 \rangle+\bar{x_2}\langle v_2, y_1v_1+y_2v_2 \rangle=\bar{x_1}y_1\langle v_1, v_1 \rangle+\bar{x_1}y_2\langle v_1, v_2 \rangle+\bar{x_2}y_1\langle v_2, v_1 \rangle+\bar{x_2}y_2\langle v_2, v_2 \rangle=\sum_{i=1,j=1}^2 \bar{x}_iy_j\langle v_i, v_j\rangle=\bar{x}^TAy$$
Then I want to use the fact that we're given $A$ in order to define $v_1,v_2$, but I am confused with one thing: In my notes, I have that $$a_{ij}=\langle v_i,v_j\rangle$$ but my notes also say for any orthonormal basis, that $$\langle v_i,v_j\rangle:=\begin{cases} 1\ \mathrm{if}\ i=j \\ 0\ \mathrm{if}\ i\neq j \end{cases}$$ which conflicts with the fact that $A=\begin{bmatrix} 2 && 1 \\ 1 && 2 \end{bmatrix}$ because that would mean that $$\langle v_i,v_j\rangle:=\begin{cases} 2\ \mathrm{if}\ i=j \\ 1\ \mathrm{if}\ i\neq j \end{cases}$$
Is there a more straightforward approach to this problem? |
Let $f:[a,b]\rightarrow \mathbb{R}$ be a positive and differentiable function. Prove: $$\frac{f(b)}{f(a)}=e^{(b-a)\frac{f'(c)}{f(c)}}$$
I tried to apply the MVT and make some algebraic manipulations, but got stuck.
Any help appreciated.
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High-level idea: The reflex to use the MVT is a good one. Now, the issue is that you do not want to apply it directly to $f$, but to some other function $g$ related to $f$. In particular, the term $e^{b-a}$ strongly hints that you're going to take an exponential after that; and seeing $\frac{f'(c)}{f(c)}$ should prompt you to think of $(\ln f)' = \frac{f'}{f}$. These two together suggest we set $g=\ln f$, and apply the MVT to $g$: good thing, this is legitimate, as $f$ is assumed positive. So let's try that...
Let $g\colon [a,b]\to \mathbb{R}$ be defined by $g\stackrel{\rm def}{=} \ln f$. $g$ is well-defined, and differentiable, as $f$ is positive and differentiable.
Applying the Mean Value Theorem on $g$, we get that there exists $c\in(a,b)$ such that $$ g(b)-g(a) = g'(c) (b-a) $$ and, exponentiating both sides, we obtain $$ e^{g(b)-g(a)} = e^{g'(c)(b-a)}. $$ Recalling that $e^{g(b)-g(a)} = \frac{e^{g(b)}}{e^{g(a)}} = \frac{f(b)}{f(a)}$ and $g'(c) = \frac{f'(c)}{f(c)}$ concludes the proof.
Let $g(x)=\ln(f(x))$. According to the mean value theorem, $$ \frac{g(b)-g(a)}{b-a}=g^{\prime}(c)=\frac{f^{\prime}(c)}{f(c)} $$ for some $c\in(a,b)$. This implies that $$ \ln\Big(\frac{f(b)}{f(a)}\Big)=(b-a)\frac{f^{\prime}(c)}{f(c)} $$ hence $$ \frac{f(b)}{f(a)}=\exp\Big((b-a)\frac{f^{\prime}(c)}{f(c)}\Big) $$ |
If $\alpha(t)$ is a regular curve arbitrarily parametrized by $t$, not necessarily the arc-length of $\alpha$, and with nowhere vanishing curvature $\kappa_\alpha(t)$, the
involute of $\alpha(t)$ originating at the point $\alpha(t_0)$ is defined to be the curve
$\beta(t) = \alpha(t) - \dfrac{\dot \alpha(t)}{\vert \dot \alpha(t) \vert} \displaystyle \int_{t_0}^t \vert \dot \alpha(u) \vert \;du; \tag 1$
that is, $\beta(t)$ is the point $\alpha(t)$ translated by the distance
$\displaystyle \int_{t_0}^t \vert \dot \alpha(u) \vert \; du \tag 2$
in the direction of the unit vector
$-\dfrac{\dot \alpha(t)}{\vert \dot \alpha(t) \vert} \tag 3$
tangent to $\alpha(t)$ at $t$; the presence of the "$-$" sign in (1), (3) allows the analogy of the "stretched string", often used to describe the involute, to go forward: (1) describes, in mathematical terms, the free end of a taught string of length (2) having been "peeled off" of the
curve $\alpha(t)$ from the point $\alpha(t_0)$ in the direction of the point $\alpha(t)$; the presence of the "$-$" sign in (1) corresponds to the fact that as the string is peeled off towards $\alpha(t)$ from $\alpha(t_0)$, the distance of $\beta(t)$ from $\alpha(t)$ increases in the opposite direction.
These niceties of geometric visualization aside, equation (1) may be written in somewhat simpler terms, first
via the observation that
$\dfrac{\dot \alpha(t)}{\vert \dot \alpha(t) \vert} = T_\alpha(t) \tag 4$
is the
unit tangent vector field to $\alpha(t)$, and second by means of the fact that the integral (2) is in fact the arc-length along $\alpha(t)$ 'twixt $\alpha(t_0)$ and $\alpha(t)$:
$s - s_0 = \displaystyle \int_{t_0}^t \vert \dot \alpha(u) \vert \; du, \tag 5$
where $s_0$ is arbitrarily assigned to $\alpha(t_0)$ (relative to some unspecified reference point on $\alpha(t)$); by means of (4) and (5) we may write (1) in terms of arc-length parametrization along $\alpha(t)$ as
$\beta(s) = \alpha(s) - T_\alpha(s)(s - s_0). \tag 6$
We wish to find the
evolute of $\beta(s)$; to this end we find its tangent vector
$\dot \beta(s) = \dot \alpha(s) - \dot T_\alpha(s)(s - s_0) - T_\alpha(s)$$= T_\alpha(s) - \dot T_\alpha(s)(s - s_0) - T_\alpha(s) = -\kappa_\alpha(s)N_\alpha(s)(s - s_0), \tag 7$
where $\kappa_\alpha(s)$ and $N_\alpha(s)$ are the curvature and unit normal vector field to $\alpha(s)$; in deriving (7) we have called upon the Frenet-Serret equation
$\dot T_\alpha(s) = \kappa_\alpha(s) N_\alpha(s). \tag 8$
It should be borne in mind that though $s$ is the arc-length along $\alpha(s)$, it is not in general the arc-length $s'$ along $\beta(s)$; therefore we may not assume that
$\vert \dot \beta(s) \vert = 1, \tag 9$
where
$\dfrac{ds'(s)}{ds} = \vert \dot \beta(s) \vert, \tag{9.5}$
and to find the unit tangent vector $T_\beta(s)$ to $\beta(s)$ we must normalize (7); we have
$\vert \dot \beta(s) \vert = \vert -\kappa_\alpha(s) N_\alpha(s)(s - s_0) \vert = \kappa_\alpha(s)(s - s_0) ,\tag{10}$
since $\kappa_\alpha(s) > 0$ and we may take $s - s_0 > 0$ by assuming, without loss of generality, that $t > t_0$ in (5); then
$T_\beta(s) = \dfrac{\dot \beta(s)}{\vert \dot \beta(s) \vert} = \dfrac{-\kappa_\alpha(s)N_\alpha(s)(s - s_0)}{\kappa_\alpha(s)(s - s_0)} = -N_\alpha(s), \tag{11}$
from which we may compute the unit normal $N_\beta(s)$ to $\beta(s)$; indeed,
$\dot T_\beta(s) = -\dot N_\alpha(s) = \kappa_\alpha(s)T_\alpha(s), \tag{12}$
where we have used the Frenet-Serret equation complementing (8),
$\dot N_\alpha(s) = -\kappa_\alpha(s)T_\alpha(s); \tag{13}$
then in accord with (9.5),
$\kappa_\beta(s)N_\beta(s) = \dfrac{dT_\beta(s)}{ds'} = \dfrac{ds}{ds'}\dfrac{\dot T_\beta(s)}{ds} = \dfrac{\dot T_\beta(s)}{\vert \dot \beta(s) \vert} = \dfrac{\kappa_\alpha(s)T_\alpha(s)}{\kappa_\alpha(s)(s - s_0)} = \dfrac{T_\alpha(s)}{s - s_0}, \tag{14}$
from which
$\kappa_\beta(s) = \vert \kappa_\beta(s) \vert \vert N_\beta(s) \vert = \vert \kappa_\beta(s) N_\beta(s) \vert = \left \vert \dfrac{T_\alpha(s)}{s - s_0} \right \vert = \dfrac{\vert T_\alpha(s) \vert}{s - s_0} = \dfrac{1}{s - s_0}; \tag{15}$
thus (14) yields
$N_\beta(s) = T_\alpha(s), \tag{16}$
and
$\dfrac{1}{\kappa_\beta(s)} N_\beta(s) = (s - s_0)T_\alpha(s); \tag{17}$
the evolute of $\beta(s)$ is thus
$\beta(s) + \dfrac{1}{\kappa_\beta(s)}N_\beta(s)$$= \beta(s) + (s - s_0)T_\alpha(s) = \alpha(s) - (s - s_0)T_\alpha(s) + (s - s_0)T_\alpha(s) = \alpha(s), \tag{18}$
our original curve, parametrized now by its arc-length $s$. |
Inge Söderkvist (2009) has a nice write-up of solving the Rigid Body Movement Problem by singular value decomposition (SVD).
Suppose we are given 3D points $\{x_1,\ldots,x_n\}$ that after perturbation take positions $\{y_1,\ldots,y_n\}$ respectively. We seek a rigid "motion", i.e. a rotation $R$ and translation $d$ combined, applied to points $x_i$ that minimizes the sum of squares of "errors":
$$ \sum_{i=1}^n || Rx_i + d - y_i ||^2 $$
Think of the points $x_i,y_i$ as columns, and of rotation $R$ as a $3\times 3$ orthogonal matrix. Strictly speaking we require $\det(R)=1$ because as a continuous motion a rotation preserves the "orientation" of the points, i.e. will not involve reflecting them.
The first step is to subtract off the respective means $\overline{x},\overline{y}$ from points $x_i,y_i$, which has the effect of "eliminating" (for now) the unknown translation $d$. That is, the problem becomes:
$$ \min_{R\in SO(3)} || RA - B ||_F $$
where $A = [x_1 - \overline{x},\ldots,x_n - \overline{x}]$ and $B = [y_1 - \overline{y},\ldots,y_n - \overline{y}]$. Here $SO(3)$ denotes the special orthogonal group of rotation matrices in 3D we are allowed to choose $R$ from, and the matrix norm $F$ here denotes the Frobenius norm, i.e. the square root of the sum of squares of matrix entries (like a Euclidean norm, but on matrix entries).
Now $A,B$ are both $3\times n$ matrices. The minimization above is an orthogonal Procrustes problem allowing only rotation matrices. The solution $R$ is given by taking the singular value decomposition of the "covariance" matrix $C = BA^T$:
$$ C = U S V^T $$
where $U,V$ are orthogonal matrices and $S = \text{diag}(\sigma_1,\sigma_2,\sigma_3)$ is the diagonal matrix of singular values $\sigma_1 \ge \sigma_2 \ge \sigma_3 \ge 0$. Numerical linear algebra packages like Matlab and Octave will compute the SVD for you.
Once we have the SVD, define $R = U \text{diag}(1,1,\pm 1) V^T$ where the sign in the middle factor is chosen to make $\det(R) = 1$. Ordinarily a real world application will have $\det(UV^T) = 1$, and thus the sign chosen would be positive. If not, it means the best orthogonal (distance preserving) fit to the new points involves a reflection, and it suggests checking the data for mistakes.
Finally we define the translation $d = \overline{y} - R\overline{x}$. Done!
The variant of orthogonal Procrustes problems where only rotations are allowed is also the subject of another Wikipedia article on the Kabsch algorithm. Notation in the Wikipedia articles differs from ours in multiplying by $R$ on the right, rather than (as here) on the left. |
Solvable Leibniz Superalgebras Whose Nilradical Is A Lie Superalgebra Of Maximal Nilindex, 2019 National University of Uzbekistan; Institute of Mathematics, Uzbekistan Academy of Science
Solvable Leibniz Superalgebras Whose Nilradical Is A Lie Superalgebra Of Maximal Nilindex, Abror Khudoyberdiyev, Manuel Ladra, Khosiat Muratova Bulletin of National University of Uzbekistan: Mathematics and Natural Sciences
In this paper, we investigate solvable Leibniz superalgebras whose nilradical is a Lie superalgebra with maximal nilindex.It should be noted that Lie superalgebra with a maximal nilindex only exists in the variety of
Lie 2,m when mis odd. We give the classification of all solvable Leibniz superalgebras such that even part is a Lie algebra and nilradical is a Lie superalgebra with a maximal index of nilpotency.
Monoidal Supercategories And Superadjunction, 2019 University of Ottawa
Monoidal Supercategories And Superadjunction, Dene Lepine Rose-Hulman Undergraduate Mathematics Journal
We define the notion of superadjunction in the context of supercategories. In particular, we give definitions in terms of counit-unit superadjunctions and hom-space superadjunctions, and prove that these two definitions are equivalent. These results generalize well-known statements in the non-super setting. In the super setting, they formalize some notions that have recently appeared in the literature. We conclude with a brief discussion of superadjunction in the language of string diagrams.
Strengthening Relationships Between Neural Ideals And Receptive Fields, 2019 Texas A&M University
Strengthening Relationships Between Neural Ideals And Receptive Fields, Angelique Morvant Rose-Hulman Undergraduate Mathematics Journal
Neural codes are collections of binary vectors that represent the firing patterns of neurons. The information given by a neural code
C can be represented by its neural ideal J . In turn, the polynomials in C J can be used to determine the relationships among the receptive fields of the neurons. In a paper by Curto et al., three such relationships, known as the Type 1-3 relations, were linked to the neural ideal by three if-and-only-if statements. Later, Garcia et al. discovered the Type 4-6 relations. These new relations differed from the first three in that they were related ... C
Dissertation_Davis.Pdf, 2019 University of Kentucky
Dissertation_Davis.Pdf, Brian Davis brian davis
Enhanced Koszulity In Galois Cohomology, 2019 The University of Western Ontario
Enhanced Koszulity In Galois Cohomology, Marina Palaisti Electronic Thesis and Dissertation Repository
Despite their central role in Galois theory, absolute Galois groups remain rather mysterious; and one of the main problems of modern Galois theory is to characterize which profinite groups are realizable as absolute Galois groups over a prescribed field. Obtaining detailed knowledge of Galois cohomology is an important step to answering this problem. In our work we study various forms of enhanced Koszulity for quadratic algebras. Each has its own importance, but the common ground is that they all imply Koszulity. Applying this to Galois cohomology, we prove that, in all known cases of finitely generated pro-$p$-groups, Galois ...
Parametric Natura Morta, 2019 Independent researcher, Palermo, Italy
Parametric Natura Morta, Maria C. Mannone The STEAM Journal
Parametric equations can also be used to draw fruits, shells, and a cornucopia of a mathematical still life. Simple mathematics allows the creation of a variety of shapes and visual artworks, and it can also constitute a pedagogical tool for students.
Diagonal Sums Of Doubly Substochastic Matrices, 2019 Georgian Court University
Diagonal Sums Of Doubly Substochastic Matrices, Lei Cao, Zhi Chen, Xuefeng Duan, Selcuk Koyuncu, Huilan Li Electronic Journal of Linear Algebra
Let $\Omega_n$ denote the convex polytope of all $n\times n$ doubly stochastic matrices, and $\omega_{n}$ denote the convex polytope of all $n\times n$ doubly substochastic matrices. For a matrix $A\in\omega_n$, define the sub-defect of $A$ to be the smallest integer $k$ such that there exists an $(n+k)\times(n+k)$ doubly stochastic matrix containing $A$ as a submatrix. Let $\omega_{n,k}$ denote the subset of $\omega_n$ which contains all doubly substochastic matrices with sub-defect $k$. For $\pi$ a permutation of symmetric group of degree $n$, the sequence of elements $a_{1\pi(1 ...
In-Sphere Property And Reverse Inequalities For Matrix Means, 2019 Ton Duc Thang University
In-Sphere Property And Reverse Inequalities For Matrix Means, Trung Hoa Dinh, Tin-Yau Tam, Bich Khue T Vo Electronic Journal of Linear Algebra
The in-sphere property for matrix means is studied. It is proved that the matrix power mean satisfies in-sphere property with respect to the Hilbert-Schmidt norm. A new characterization of the matrix arithmetic mean is provided. Some reverse AGM inequalities involving unitarily invariant norms and operator monotone functions are also obtained.
Surjective Additive Rank-1 Preservers On Hessenberg Matrices, 2019 Walailak University
Surjective Additive Rank-1 Preservers On Hessenberg Matrices, Prathomjit Khachorncharoenkul, Sajee Pianskool Electronic Journal of Linear Algebra
Let $H_{n}(\mathbb{F})$ be the space of all $n\times n$ upper Hessenberg matrices over a field~$\mathbb{F}$, where $n$ is a positive integer greater than two. In this paper, surjective additive maps preserving rank-$1$ on $H_{n}(\mathbb{F})$ are characterized.
Solving The Sylvester Equation Ax-Xb=C When $\Sigma(A)\Cap\Sigma(B)\Neq\Emptyset$, 2019 Faculty of Sciences and Mathematics, University of Niš
Solving The Sylvester Equation Ax-Xb=C When $\Sigma(A)\Cap\Sigma(B)\Neq\Emptyset$, Nebojša Č. Dinčić Electronic Journal of Linear Algebra
The method for solving the Sylvester equation $AX-XB=C$ in complex matrix case, when $\sigma(A)\cap\sigma(B)\neq \emptyset$, by using Jordan normal form is given. Also, the approach via Schur decomposition is presented.
A Re-Emergent Analysis Of Early Algebraic Learning, 2019 Portland State University
A Re-Emergent Analysis Of Early Algebraic Learning, Steven Boyce, Diana Moss Extension Research
In this paper, we discuss a novel approach for collaborative retrospective analysis. One researcher was directly involved in a classroom teaching experiment, adopting an emergent perspective as an interpreter-witness of classroom interactions during a four-week algebra instructional unit with sixth-grade students. The other researcher experienced and analyzed the data in reverse chronological order. We describe how this re-emergent perspective revealed aspects of students’ early algebraic reasoning.
Resolution Of Conjectures Related To Lights Out! And Cartesian Products, 2019 University of Wyoming
Resolution Of Conjectures Related To Lights Out! And Cartesian Products, Bryan A. Curtis, Jonathan Earl, David Livingston, Bryan L. Shader Electronic Journal of Linear Algebra
Lights Out!\ is a game played on a $5 \times 5$ grid of lights, or more generally on a graph. Pressing lights on the grid allows the player to turn off neighboring lights. The goal of the game is to start with a given initial configuration of lit lights and reach a state where all lights are out. Two conjectures posed in a recently published paper about Lights Out!\ on Cartesian products of graphs are resolved.
A Note On Linear Preservers Of Semipositive And Minimally Semipositive Matrices, 2019 Indian Institute of Science, Bengaluru
A Note On Linear Preservers Of Semipositive And Minimally Semipositive Matrices, Projesh Nath Choudhury, Rajesh Kannan, K. C. Sivakumar Electronic Journal of Linear Algebra
Semipositive matrices (matrices that map at least one nonnegative vector to a positive vector) and minimally semipositive matrices (semipositive matrices whose no column-deleted submatrix is semipositive) are well studied in matrix theory. In this short note, the structure of linear maps which preserve the set of all semipositive/minimally semipositive matrices is studied. An open problem is solved, and some ambiguities in the article [J. Dorsey, T. Gannon, N. Jacobson, C.R. Johnson and M. Turnansky. Linear preservers of semi-positive matrices. {\em Linear and Multilinear Algebra}, 64:1853--1862, 2016.] are clarified.
Vector Cross Product Differential And Difference Equations In R^3 And In R^7, 2019 University of Beira Interior
Vector Cross Product Differential And Difference Equations In R^3 And In R^7, Patrícia D. Beites, Alejandro P. Nicolás, Paulo Saraiva, José Vitória Electronic Journal of Linear Algebra
Through a matrix approach of the $2$-fold vector cross product in $\mathbb{R}^3$ and in $\mathbb{R}^7$, some vector cross product differential and difference equations are studied. Either the classical theory or convenient Drazin inverses, of elements belonging to the class of index $1$ matrices, are applied.
Equivalence Of Classical And Quantum Codes, 2019 University of Kentucky
Equivalence Of Classical And Quantum Codes, Tefjol Pllaha Theses and Dissertations--Mathematics
In classical and quantum information theory there are different types of error-correcting codes being used. We study the equivalence of codes via a classification of their isometries. The isometries of various codes over Frobenius alphabets endowed with various weights typically have a rich and predictable structure. On the other hand, when the alphabet is not Frobenius the isometry group behaves unpredictably. We use character theory to develop a duality theory of partitions over Frobenius bimodules, which is then used to study the equivalence of codes. We also consider instances of codes over non-Frobenius alphabets and establish their isometry groups. Secondly ...
Upper Dimension And Bases Of Zero-Divisor Graphs Of Commutative Rings, 2019 University of Kashmir, Srinagar, India
Upper Dimension And Bases Of Zero-Divisor Graphs Of Commutative Rings, S. Pirzada, M. Aijaza, Shane P. Redmond EKU Faculty and Staff Scholarship
For a commutative ring R with non-zero zero divisor set Z∗(R), the zero divisor graph of R is Γ(R) with vertex set Z∗(R), where two distinct vertices x and y are adjacent if and only if x y = 0. The upper dimension and the resolving number of a zero divisor graph Γ(R) of some rings are determined. We provide certain classes of rings which have the same upper dimension and metric dimension and give an example of a ring for which these values do not coincide. Further, we obtain some bounds for the upper dimension in ...
An Invitation To Linear Algebra, 2019 CUNY Queensborough Community College
An Invitation To Linear Algebra, David N. Pham, Jonathon Funk, Wenjian Liu Open Educational Resources
This is an OER textbook on linear algebra.
Lattice Simplices: Sufficiently Complicated, 2019 University of Kentucky
Lattice Simplices: Sufficiently Complicated, Brian Davis Theses and Dissertations--Mathematics
Simplices are the "simplest" examples of polytopes, and yet they exhibit much of the rich and subtle combinatorics and commutative algebra of their more general cousins. In this way they are sufficiently complicated --- insights gained from their study can inform broader research in Ehrhart theory and associated fields.
In this dissertation we consider two previously unstudied properties of lattice simplices; one algebraic and one combinatorial. The first is the Poincar\'e series of the associated semigroup algebra, which is substantially more complicated than the Hilbert series of that same algebra. The second is the partial ordering of the elements of ...
The State Of Lexicodes And Ferrers Diagram Rank-Metric Codes, 2019 University of Kentucky
The State Of Lexicodes And Ferrers Diagram Rank-Metric Codes, Jared E. Antrobus Theses and Dissertations--Mathematics
In coding theory we wish to find as many codewords as possible, while simultaneously maintaining high distance between codewords to ease the detection and correction of errors. For linear codes, this translates to finding high-dimensional subspaces of a given metric space, where the induced distance between vectors stays above a specified minimum. In this work I describe the recent advances of this problem in the contexts of lexicodes and Ferrers diagram rank-metric codes.
In the first chapter, we study lexicodes. For a ring
R, we describe a lexicographic ordering of the left R-module R n. With this ordering we ... Polygonal Analogues To The Topological Tverberg And Van Kampen-Flores Theorems, Leah Leiner Senior Projects Spring 2019
Tverberg’s theorem states that any set of (q-1)(d+1)+1 points in d-dimensional Euclidean space can be partitioned into q subsets whose convex hulls intersect. This is topologically equivalent to saying any continuous map from a (q-1)(d+1)-dimensional simplex to d-dimensional Euclidean space has q disjoint faces whose images intersect, given that q is a prime power. These continuous functions have a Fourier decomposition, which admits a Tverberg partition when all of the Fourier coefficients, except the constant coefficient, are zero. We have been working with continuous functions where all of the Fourier coefficients except the ... |
(Short Writeup) Presentation of Symmetric Groups 1. Proving a set of elements are generators
If you can show that a set of elements generate all 2 cycles then they must generate $S_n$. This is because any $n$-cycles have a decomposition into 2-cycles. i.e. $$(a_1,a_2,a_3,\dots,a_n)=(a_1,a_2)(a_2,a_3,\dots,a_n)$$and we repeat the process on $(a_2,a_3,\dots,a_n)$. Then having all 2-cycles we can generate any longer cycles.
For your case, since $S_3$ has only three 2-cycles $\{(1,2),(1,3),(2,3)\}$ and you have picked $a=(1,2),b=(2,3)$, it suffices to show that you can get the last one.$$ab=(1,2)(2,3)=(1,2,3)$$$$aba=(1,2,3)(1,2)=(1,3)$$which implies that $\langle a,b\rangle$ does generates $S_3$.
Note that since you have chosen $a,b$ such that $a^2=b^2=1$, the only sequence you can generate is $abab\dots$. You can show that $baba\dots$ must generate the same sequence.
So if $\{a,b\}$ are really generators then you must be able to get $(1,3)$ by $abab\dots$. 2. Generating 2-cycles and Coxeter group
A simple way to generate all 2-cycles is by using generators of the form:$$t_k=(k,k+1), 1\leq k<n$$
Proof
: Suppose we can generate $(i,k)$ for some $i< k<n$. Then we have $(k,k+1)$ and:$$(k,k+1)(i,k)(k,k+1)=(k,k+1)(i,k,k+1)=(i,k+1)$$gives us $(i,k+1)$. Since we have $(i,i+1)$, we can generate any $(i,j), i<j\leq n$. Then since we have $(i,i+1)$ for $1\leq i<n$, we can generate all $(i,j)$ for $1\leq i<j\leq n$, which are all 2-cycles. $\square$
As a result, we expect a presentation using these $n-1$ generators. We can check some defining relations, which will turn out to be sufficient:\begin{align*}R_1: &t_k^2 =1, 1\leq k<n\\R_2: &(t_k t_{k+1})^3=1, 1\leq k<n-1\\R_3: &(t_jt_k)^2=1, 1\leq j<k-1<n-1\end{align*}In particular, $R_3$ represents the case where the two 2-cycles have no intersection.This type of group presentation is also known as a Coxeter group, i.e. a group with generators $\{s_i\}$ and relations are defined by:$$s_i^2=1 \text{ and }(s_is_j)^{m_{ij}}=1, m_{ij}\geq 2, i\neq j$$It is also the most common representation for symmetric groups.
For our presentation, we denote the generators as $g_i$. We can show that$$S_n\cong \langle g_1,g_2,\dots,g_{n-1}|R_1,R_2,R_3\rangle=G_n$$Where $R_1,R_2$ and $R_3$ are as before, but with $g_i$ in place of $t_i$.
3. Proof of Coxeter representation of for symmetric groups
This section describes a commonly used proof. Define a map$$\phi:G_n\to S_n$$$$g_i\mapsto t_i$$This is well defined since $R_1,R_2$ and $R_3$ are satisfied in $S_n$. The map is also surjective since the 2-cycles generate $S_n$. Therefore we have$$|G_n|\geq n!=|S_n|$$
Hence it suffices to show that$$|G_n|\leq n!=|S_n|$$so that $|G_n|=|S_n|$ and the map is 1-to-1 and we have an isomorphism.
This is true for $n=2$:$$G_2=\langle g_1|g_1^2=1\rangle\cong S_2$$such that $|G_2|\leq |S_2|$. Hence we assume $|G_{n-1}|\leq |S_{n-1}|=(n-1)!$ and prove the case for $n$. Define$$H=\langle g_2,\dots,g_{n-1}\rangle\subset G_n,$$which is a subgroup of $G_n$.The defining relations of $G_n$ on $H$ is equivalent to the case in $G_{n-1}$. Hence by induction we know that $|H|\leq (n-1)!$. Our goal is to show that$$[G_n:H]\leq n$$then using Lagrange's theorem for groups we have$$|G_n|=[G_n:H]|H|\leq n(n-1)!=n!$$The condition $[G_n:H]\leq n$ can be interpreted as at most $n$ cosets. Hence we try a natural approach to construct $n$ cosets as below:$$S=\{H_0=H, H_1=g_1H, H_2=g_2g_1H,\dots, H_{n-1}=g_{n-1}\dots g_2g_1H\}$$The goal now is to show that they are distinct.
The first observations are:$$g_iH_i=g_ig_ig_{i-1}\dots g_2g_1H=g_{i-1}\dots g_2g_1H=H_{i-1}$$$$g_iH_{i-1}=g_ig_{i-1}\dots g_2g_1H=H_i$$i.e. $g_i$ permutes $\{H_{i-1},H_i\}$. We can show that $g_i$ stabilizes all other $H_j$. In these cases, $|i-j|> 1$, so $i>j+1$ or $i<j-1$.
We will need to manipulate the relations. Recall that the relation$$(g_ig_j)^2=1, i< j-1$$ is equivalent to$$g_ig_j=g_jg_i,$$i.e. we can swap the elements.The other relation $(g_ig_{i+1})^3=1$ says that:$$(g_ig_{i+1})(g_ig_{i+1})(g_ig_{i+1})=1$$$$(g_ig_{i+1})(g_ig_{i+1})(g_ig_{i+1})(g_{i+1}g_ig_{i+1})=g_{i+1}g_ig_{i+1}$$$$g_ig_{i+1}g_i=g_{i+1}g_ig_{i+1}$$which allows us to do another kind of swapping. These two operations suffices to show $g_iH_j=H_j$.
First assume that $i> j+1$:\begin{align*}g_iH_j &=g_i(g_j\dots g_2g_1)H\\&=(g_j\dots g_2g_1)g_iH\\&= (g_j\dots g_2g_1)H\\&=H_j\end{align*}where we simply sequentially swap $g_i$ towards $H$, since none of the cycles intersect $g_i$.
Now assume that $i<j-1$:\begin{align*}g_iH_j&=\underline {g_i}g_j\dots g_{i+1}g_ig_{i-1}\dots g_2g_1H\\&= g_j\dots \underline{g_i}g_{i+1}g_ig_{i-1}\dots g_2g_1H\\&= g_j\dots g_{i+1}g_i \underline {g_{i+1}}g_{i-1}\dots g_2g_1H\\&= g_j\dots g_{i+1}g_ig_{i-1}\dots g_2g_1 \underline {g_{i+1}}H\\&= g_j\dots g_{i+1}g_ig_{i-1}\dots g_2g_1H\\&=H_j\end{align*}We did a shift followed by a triple-elements swap then by a shift. The step $g_{i+1}H=H$ is due to $g_{i+1}\in H$.
This shows that $G_n$ acts transitively on $S$, i.e. it preserves $S$. It is a theorem in group theory that in fact $S=G_n$ and there are exactly $n$ cosets, hence $[G_n:H]=n$ and we are done.
4. A Shorter Representation Using 2 Generators
Recall that the Coxeter group presentation for symmetric group is given by$$G_n=\langle g_1,g_2,\dots,g_{n-1}\;|\;R_1,R_2,R_3\rangle$$\begin{align*}R_1: &(g_i)^2=1, \;\;\;\;\;\;\;1\leq i\leq n-1\\R_2: &(g_ig_{i+1})^3=1, \;\; 1\leq i\leq n-1\\R_3: &(g_ig_j)^2=1,\;\;\;\; 1\leq i<j-1<n-1\end{align*}
We can obtain a shorter presentation by introducing a new generator $b=(1,2,\dots,n)$. By considering $a=g_1=(1,2)$ and $b$, we can show that they generate $g_2,\dots,g_{n-1}$. Therefore we can reduce the presentation to only 2 elements. We may then simplify the relations to involve only $a$ and $b$, the final result being$$G=\langle a,b\;|\;a^2=1,b^n=1,(ab)^{n-1}=1, (abab^{-1})^3=1, (ab^iab^{-i})^2=1,\;\;\;\; 2\leq i\leq n-2\rangle$$
5. Steps for Shorter Representation
We can add or delete generators and relations while preserving $G$, so that the new presentation of $G$ is structurally unchanged. i.e. $G\cong S_n$.
This procedure can be done through Tietze transformations, as mentioned by @Myself. It states that we can:
(1) add a relation if it can be derived from the current ones (2) remove a relation if it does not affect the group (3) add a generator if it can be generated by the group (4) remove a generator if it is expressible by other generators
5.1 Introducing new generator $b$ We can introduce a new generator $b$ by setting:$$b=g_1g_2\dots g_{n-1}$$then we adjust the presentation to\begin{align*}G &= \langle g_1,g_2,\dots, g_{n-1},b\;|\;R_1,R_2,R_3,R_4\rangle\\R_4 &: b=g_1g_2\dots g_{n-1}\end{align*}Let $g_1=a$, which we use interchangeably. The addition of $b$ allows us to derive new relations between $a,b$ and $g_i$:\begin{align*}bab^{-1}&=(g_1g_2\dots g_{n-1})g_1(g_{n-1}\dots g_2g_1) = (g_1g_2)g_1(g_2g_1)\\&=(g_2g_1g_2)(g_2g_1)\\&=g_2\\\text{Assume} &: b^{i-1}ab^{-(i-1)} = g_i\\\text{Induction step:}&\\b^iab^{-i}& =b(b^{i-1}ab^{-(i-1)})b^{-1}=bg_ib^{-1}=(g_1g_2\dots g_ig_{i+1}\dots g_{n-1})g_i(g_{n-1}\dots g_{i+1}g_i\dots g_2g_1)\\&= (g_1g_2\dots g_ig_{i+1})g_i(g_{i+1}g_i\dots g_2g_1) = (g_1g_2\dots g_{i-1})(g_{i+1}g_ig_{i+1})(g_{i+1}g_i\dots g_2g_1)\\&= (g_1g_2\dots g_{i-1})g_{i+1}(g_{i-1}\dots g_2g_1)=(g_1g_2\dots g_{i-1})(g_{i-1}\dots g_2g_1)g_{i+1}\\&= g_{i+1}\end{align*}Hence the induction shows that$$b^{i-1}ab^{-(i-1)}=g_i\text{ for }1\leq i\leq n-1.$$i.e. we can represent $g_i$ purely in terms of $a$ and $b$.Therefore we can remove all the generators and relations involving $g_2,g_3,\dots, g_{n-1}$ by this substitution.
5.2 Replacing relations $R_1$ The first consequence is $R_1$ is now redundant. Consider $ (g_{i+1})^2=1\in R_1$:$$(g_{i+1})^2=1\Longleftrightarrow (b^iab^{-i})^2=1\Longleftrightarrow 1=1$$which is trivial provided $(g_1)^2=a^2=1$ (also in $R_1$). Therefore we can keep $a^2=1$ and discard the rest, which turns $R_1$ into just $a^2=1$.
5.3 Replacing relations $R_2$ Now let us take any relation in $(g_ig_{i+1})^3=1\in R_2$:\begin{align*}(g_ig_{i+1})^3=1\longrightarrow ((b^{i-1}ab^{-(i-1)})(b^iab^{-i}))^3=1\\ (b^{i-1}abab^{-i})^3=1\\ (b^{i-1}iabab^{-i}) (b^{-1}iabab^{-i}) (b^{-1}iabab^{-i})=1\\ b^{i-1}abab^{-1}abab^{-1}abab^{-i}=1\\b^{-(i-1)}(b^{i-1}abab^{-1}abab^{-1}abab^{-i})(b^{i-1})=(b^{-(i-1)})(b^{i-1})\\abab^{-1}abab^{-1}abab^{-1}=1\\(abab^{-1})^3=1\end{align*}And therefore $R_2$ is reduced to a single relation: $(abab^{-1})^3=1$.
5.4 Replacing relations $R_3$ Our next step is to investigate $R_3$, with relations of the form$$R_3: (g_ig_j)^2=1, \;\;\;\;1\leq i\leq j-2\leq n-2$$\begin{align*}(g_ig_j)^2=1\longrightarrow ((b^{i-1}ab^{-(i-1)})(b^{j-1}ab^{-(j-1)}))^2=1\\(b^{i-1}ab^{-i+j}ab^{-(j-1)})^2=1\\(b^{i-1}ab^{-i+j}ab^{-(j-1)})(b^{i-1}ab^{-i+j}ab^{-(j-1)})=1\\b^{i-1}ab^{-i+j}ab^{-j+i}ab^{-i+j}ab^{-(j-1)}=1\\b^{-(i-1)}(b^{i-1}ab^{-i+j}ab^{-j+i}ab^{-i+j}ab^{-(j-1)})b^{i-1}=(b^{-(i-1)})(b^{i-1})\\ab^{-i+j}ab^{-j+i}ab^{-i+j}ab^{-j+i}=1\\(ab^{-i+j}ab^{-j+i})^2=1\\\end{align*}From the initial conditions, we observe that$$j-i\geq 2$$and when $i=1$ and $j-2=n-2$, we have$$j-i=(n-3)+1=n-2$$Therefore $2\leq j-i\leq n-2$. Replacing $j-i$ by $i$, we find the new $R_3$ as:$$R_3: (ab^iab^{-i})^2=1,\;\;\;\; 2\leq i\leq n-2$$
5.5 Replacing relations $R_4$ We are left with the last, new relation $R_4: b = g_1g_2\dots g_{n-1}$.
A simple substitution shows:$$b = g_1g_2\dots g_{n-1}=a(bab^{-1})(b^2ab^{-2})\dots(b^{n-2}ab^{-(n-2)}) = ababab\dots bab^{-(n-2)}$$$$b=(ab)^{n-1}b^{-n+1}$$$$b^n =(ab)^{n-1}$$It remains to show that $b^n=1=(ab)^{n-1}$. This is a bit tricky on $G$, so we use the isomorphism $G\cong S_n$ to deduce that $b=g_1g_2\dots g_{n-1}=(1,2,\dots n)$.Therefore $(ab)^{n-1}=b^n=(1,2,\dots,n)^n=1$ and we find:$$R_4: b^n=1, (ab)^{n-1}=1$$
5.6 Putting everything together Combining all the new relations:$$G=\langle a,b\;|\;a^2=1,b^n=1,(ab)^{n-1}=1, (abab^{-1})^3=1, (ab^iab^{-i})^2=1,\;\;\;\; 2\leq i\leq n-2\rangle$$where\begin{align*}R_1 &: a^2=1\\R_2 &: (abab^{-1})^3=1\\R_3 &: (ab^iab^{-i})^2=1,\;\;\;\; 2\leq i\leq n-2\\R_4 &: b^n=(ab)^{n-1}=1\end{align*}Note that we know that $G\cong S_n$ during all these substitutions, since the operations involved does not change $G$. This completes the presentation shown in section 4. |
Last year I made a post about the universal program, a Turing machine program $p$ that can in principle compute any desired function, if it is only run inside a suitable model of set theory or arithmetic. Specifically, there is a program $p$, such that for any function $f:\newcommand\N{\mathbb{N}}\N\to\N$, there is a model $M\models\text{PA}$ — or of $\text{ZFC}$, whatever theory you like — inside of which program $p$ on input $n$ gives output $f(n)$.
This theorem is related to a very interesting theorem of W. Hugh Woodin’s, which says that there is a program $e$ such that $\newcommand\PA{\text{PA}}\PA$ proves $e$ accepts only finitely many inputs, but such that for any finite set $A\subset\N$, there is a model of $\PA$ inside of which program $e$ accepts exactly the elements of $A$. Actually, Woodin’s theorem is a bit stronger than this in a way that I shall explain.
Victoria Gitman gave a very nice talk today on both of these theorems at the special session on Computability theory: Pushing the Boundaries at the AMS sectional meeting here in New York, which happens to be meeting right here in my east midtown neighborhood, a few blocks from my home.
What I realized this morning, while walking over to Vika’s talk, is that there is a very simple proof of the version of Woodin’s theorem stated above. The idea is closely related to an idea of Vadim Kosoy mentioned in my post last year. In hindsight, I see now that this idea is also essentially present in Woodin’s proof of his theorem, and indeed, I find it probable that Woodin had actually begun with this idea and then modified it in order to get the stronger version of his result that I shall discuss below.
But in the meantime, let me present the simple argument, since I find it to be very clear and the result still very surprising.
Theorem. There is a Turing machine program $e$, such that $\PA$ proves that $e$ accepts only finitely many inputs. For any particular finite set $A\subset\N$, there is a model $M\models\PA$ such that inside $M$, the program $e$ accepts all and only the elements of $A$. Indeed, for any set $A\subset\N$, including infinite sets, there is a model $M\models\PA$ such that inside $M$, program $e$ accepts $n$ if and only if $n\in A$. Proof. The program $e$ simply performs the following task: on any input $n$, search for a proof from $\PA$ of a statement of the form “program $e$ does not accept exactly the elements of $\{n_1,n_2,\ldots,n_k\}$.” Accept nothing until such a proof is found. For the first such proof that is found, accept $n$ if and only if $n$ is one of those $n_i$’s.
In short, the program $e$ searches for a proof that $e$ doesn’t accept exactly a certain finite set, and when such a proof is found, it accepts exactly the elements of this set anyway.
Clearly, $\PA$ proves that program $e$ accepts only a finite set, since either no such proof is ever found, in which case $e$ accepts nothing (and the empty set is finite), or else such a proof is found, in which case $e$ accepts only that particular finite set. So $\PA$ proves that $e$ accepts only finitely many inputs.
But meanwhile, assuming $\PA$ is consistent, then you cannot refute the assertion that program $e$ accepts exactly the elements of some particular finite set $A$, since if you could prove that from $\PA$, then program $e$ actually would accept exactly that set (for the shortest such proof), in which case this would also be provable, contradicting the consistency of $\PA$.
Since you cannot refute any particular finite set as the accepting set for $e$, it follows that it is consistent with $\PA$ that $e$ accepts any particular finite set $A$ that you like. So there is a model of $\PA$ in which $e$ accepts exactly the elements of $A$. This establishes statement (2).
Statement (3) now follows by a simple compactness argument. Namely, for any $A\subset\N$, let $T$ be the theory of $\PA$ together with the assertions that program $e$ accepts $n$, for any particular $n\in A$, and the assertions that program $e$ does not accept $n$, for $n\notin A$. Any finite subtheory of this theory is consistent, by statement (2), and so the whole theory is consistent. Any model of this theory realizes statement (3).
QED
One uses the Kleene recursion theorem to show the existence of the program $e$, which makes reference to $e$ in the description of what it does. Although this may look circular, it is a standard technique to use the recursion theorem to eliminate the circularity.
This theorem immediately implies the classical result of Mostowski and Kripke that there is an independent family of $\Pi^0_1$ assertions, since the assertions $n\notin W_e$ are exactly such a family.
The theorem also implies a strengthening of the universal program theorem that I proved last year. Indeed, the two theorems can be realized with the same program!
Theorem. There is a Turing machine program $e$ with the following properties: $\PA$ proves that $e$ computes a finite function; For any particular finite partial function $f$ on $\N$, there is a model $M\models\PA$ inside of which program $e$ computes exactly $f$. For any partial function $f:\N\to\N$, finite or infinite, there is a model $M\models\PA$ inside of which program $e$ on input $n$ computes exactly $f(n)$, meaning that $e$ halts on $n$ if and only if $f(n)\downarrow$ and in this case $\varphi_e(n)=f(n)$. Proof. The proof of statements (1) and (2) is just as in the earlier theorem. It is clear that $e$ computes a finite function, since either it computes the empty function, if no proof is found, or else it computes the finite function mentioned in the proof. And you cannot refute any particular finite function for $e$, since if you could, it would have exactly that behavior anyway, contradicting $\text{Con}(\PA)$. So statement (2) holds. But meanwhile, we can get statement (3) by a simple compactness argument. Namely, fix $f$ and let $T$ be the theory asserting $\PA$ plus all the assertions either that $\varphi_e(n)\uparrow$, if $n$ is not the domain of $f$, and $\varphi_e(n)=k$, if $f(n)=k$. Every finite subtheory of this theory is consistent, by statement (2), and so the whole theory is consistent. But any model of this theory exactly fulfills statement (3). QED
Woodin’s proof is more difficult than the arguments I have presented, but I realize now that this extra difficulty is because he is proving an extremely interesting and stronger form of the theorem, as follows.
Theorem. (Woodin) There is a Turing machine program $e$ such that $\PA$ proves $e$ accepts at most a finite set, and for any finite set $A\subset\N$ there is a model $M\models\PA$ inside of which $e$ accepts exactly $A$. And furthermore, in any such $M$ and any finite $B\supset A$, there is an end-extension $M\subset_{end} N\models\PA$, such that in $N$, the program $e$ accepts exactly the elements of $B$.
This is a much more subtle claim, as well as philosophically interesting for the reasons that he dwells on.
The program I described above definitely does not achieve this stronger property, since my program $e$, once it finds the proof that $e$ does not accept exactly $A$, will accept exactly $A$, and this will continue to be true in all further end-extensions of the model, since that proof will continue to be the first one that is found. |
Short Version:
Let all lower latin variables mean strings, except $ n, m, l, r$ , which are indices and index bounds.
Let $ ab$ mean usual string concatenation of $ a$ to $ b$ .
Let $ t \leqslant s \iff \exists$ strings $ u,v$ such that $ utv = s$ .
Let $ \Sigma(s) = \{ a \leqslant s : |a| = 1\}$ .
Let $ V(n) = \{X_1, \dots, X_n\}$ be disjoint from $ \Sigma(s)$ always.
Let $ G(s, n) = \text{free-group}(V(n)\cup \Sigma(s))$
Let $ \text{pruned}(s, n) = \{$ grammars $ g$ for $ s$ on $ \leq n$ variables such that $ \forall X \in V(g)$ we have $ X \leqslant g^k(S)$ for some $ k \geq 0$ .
Define a
grammar $ g$ on $ n$ variables for $ s$to be a homomorphism of free groups: $ \alpha : G(s, n) \to G(s, n – 1)$ , such that $ \alpha$ is some mapping $ V(n) \to G(s, n-1)$ extended homomorphically, and identity on $ \Sigma(s)$ , and such that $ \alpha^k(S) = s$ for all $ k \geq n_g$ .
Let $ \text{SGP}(s, n) = \{\overline{x} \in G(s, n)^n :$ if $ \overline{x}\cdot \overline{g} = \overline{g’}$ then $ g’$ can be pruned such that an equivalent grammar resides in $ \text{pruned}(s, n)$ whenever $ g$ does $ \}$ .
From now on, identify $ G(s, n-1)$ with the first $ n-1$ components of $ G(s, n)$ .
Question is at the bottom of post.
Written with StackEdit.
Long Version:
A
grammar $ g$ of a string $ s$ over $ \Sigma$ is a string homomorphism from $ (\Sigma \cup V)^* \to (\Sigma \cup V\setminus \{S\})^*$ such that $ g^n(S) = s$ for all sufficiently large $ n$ , where $ V$ is a finite set of variable symbols, and $ S \in V$ is the designated start symbol which is defined to be present for any string $ s$ . A pruned grammar is one such that for all $ X \in V(g), $ the letter $ X$ appears as a substring to some iterate of $ g$ evaluated at $ S$ : $ X \leqslant g^k(S)$ for some $ k \geq 0$ . A pruned grammar just gets rid of unused rules.
Define the
size of a grammar $ g$ of $ s$ to be the sum of the lengths of the RHS of rules defininig $ g$ . Recall from context-free languages that a grammar can also be defined as a list of rules of the form $ X \to x$ . Formally we can express this as $ |g| := \sum_{X \in V(g)} |g(X)|$ .
In order to algebratize the problem, I didn’t find it beneficial to study string homomorphisms that are pruned grammars (the semigroup formed is $ (g \circ g’ = g’, \forall g,g’ \in \text{ the semigroup})$ . This is because composing two grammars like functions causes the first in the composition to hide the variables of the second, such that when the result is pruned you get the first operand as the result.
Instead, we’ll study transformations of the rules of grammars with $ n$ rules. Consider the direct product $ G^n$ , where $ G$ is the free group on $ \Sigma(g) = V(g) \cup \Sigma(s)$ , where the union is made disjoint by an unlimited supply of new variable symbols. Clearly if we assign a variable index to factor of the product, then we have an encoding of all posible grammars on $ n$ variables, a lot them being redundant and “very similar” to other grammars.
None the less, the smallest grammar problem should be statable as:
Find a grammar $ g$ of $ s$ such that $ |g|$ is minimized, or equivalently,
Find a vector $ \overline{g} \in G^n, n\geq 1$ such that $ |g|$ is minimized and $ g$ is a grammar for $ s$ .
Note the $ \geq n$ part. It implies that any naive algorithm would have to check all $ n = 1…|C(s)|$ where $ |C(s)| = O(|s|^2)$ and $ C(s)$ is the set of substrings of $ s$ . I.e. in the worst case assume you have to variablize approximately all substrings. This is not the case in practice or in theory, but it shouldn’t matter for this analysis.
Now we have to be careful about how we define $ |\cdot| : G \to \Bbb{Z}$ on a group, because it can literally be always equal to $ 1$ if not done right. Since $ G$ is generated by $ \Sigma(g)$ , we say that $ |x| = $ the length of a minimial run of letters in $ x_i \in \Sigma(g) \cup \Sigma(g)^{-1}$ . I.e. if $ x_1 \cdots, x_r = x$ . And we cannot do the same for $ r’ \lt r$ , then $ |x| := r$ .
Since the alphabet $ \Sigma(g)$ depends on the grammar, we have to introduce the alphabet that doesn’t change over $ \overline{g} \in G^n$ . That, we’ll just call $ \Sigma_n = \{X_1, \dots, X_n\}$ .
Now to define $ |\overline{x}|$ for $ \overline{x} \in G$ we need to take out rules that are easily seen to be not required or not helpful in making a grammar the smallest it can be. For instance if $ X_1 \to a \in \Sigma(s)$ , then $ |a| = 1$ and so any grammar would actually be larger than the equivalent grammar without it. Thus whenever $ t \in \Sigma_n \cup \Sigma$ we define $ |t|_n = 0$ . To get to the equivalent non-redundantly styled grammar, you just substitute in $ a$ for every occurence of $ X_1$ found on the RHS of other rules.
Define a
grammar symmetry from $ \overline{g}$ to be any element $ \sigma \in G^n$ such that when $ \sigma$ is applied componentwise on the left (or right) of $ \overline{g}$ , written $ \overline{h} = \sigma \overline{g}$ we have that $ h^k(S) = s$ for sufficiently large $ k$ is a preserved property.
We have $ G^n$ acting on $ G^n$ itself and a subset $ Y := \text{pruned-grammars}(n, s) = \{ \overline{g} : g$ is a pruned grammar for $ s\}$ . The subset $ H = \text{Stab}_{G^n}(Y)$ i.e. all group elements that stabilize pruned grammars for $ s$ together with their componentwise inverses, is a subgroup of $ G^n$ .
Proof. $ iY =Y, hY = Y \implies Y = h^{-1} Y \implies ih^{-1}Y = Y$ .
Let $ X := \text{smallest-grammars}(n, s)$ . Clearly $ X \subset Y$ since you can’t have extra junk in a smallest grammar of course.
Note: elements of $ X$ are precisely the smallest grammars one can achieve in $ n$ or less variables. The reason you can represent the lower variable counts in there as well, is that a component of $ \overline{g}$ can simply map to $ 1 \in \Sigma^*$ and it has length $ 0$ .
Lemma 0. $ H \times Y \to Y$ acts transitively.
Let $ g, h$ be two pruned grammars of $ s$ , each with $ n$ rules, or $ g,h \in Y$ . Consider the component of lowest index $ i$ such that $ g_i \neq h_i$ . Since we’re in a group, $ x g_i = h_i$ has a solution for all $ g_i, h_i \in \text{Stab}_{G^n}(Y)$ . Similarly we have that $ g_i = x^{-1} h_i$ . Do the same for each component, and call $ x_j$ the left multiplier to get from $ g_j$ to $ h_j$ . Then $ \overline{x} = (x_1, \dots, x_n)$ is such that $ \overline{x}\overline{g} = \overline{h}$ .
This is flawed, since $ \text{Stab}_{G^n}(Y)$ may not be writeable as a direct product.
Lemma -1. $ \text{Stab}_{G^n}(Y)$ is equal to a direct product of subgroups of $ G^n$ . Let $ K_i = \{ (1,1, \dots, x_i, \dots, 1, 1) : x_i$ , is in the $ i$ th position and $ (c_1, \dots, c_{i-1}, x_i, \dots, c_n) \in \text{Stab}_{G^n}(Y)$ for some $ c_j \in G\}$ . Then $ K_i$ forms a group under componentwise multiplicaiton.
Proof. We only need to show this for the $ i$ th component since the rest are all $ 1$ . Without loss of generality we only need to show that $ H_i = \{ x \in G : (c_1, \dots, c_{i-1}, x, \dots, c_n) \in \text{Stab}_{G^n}(Y)$ for some $ c_j \in G \}$ is a subgroup of $ G$ . Since $ \text{Stab}_{G^n}(Y)$ is in particular a subgroup of $ G^n$ any multiplication of its elements must be carried out componentwise, by definitino of direct product. Let $ x, y\in H_i$ . Then clearly $ (c_1 d_1, \dots, c_{i-1}d_{i-1}, xy^{-1}, \dots, c_n d_n) \in \text{Stab}_{G^n}(Y)$ , where similarly we can prove, using component-wiseness that $ y^{-1} \in H_i$ . $ \blacksquare$
That proves that $ \text{Stab}_{G^n}(Y)$ transitively acts on $ Y$ . Define $ H’ := \text{Stab}_{H}(X) = \{ \overline{x} \in H : \overline{x} X = X \}$ similarly, and we have a proof that $ H’ \times X \to X$ is a transitive group action.
Thus if we are given a grammar $ \overline{y} \in Y$ for $ s$ , say it might be approximately small, but not the smallest. Then there is a group element in $ H = \text{Stab}_{G^n}(Y)$ that will take $ \overline{y}$ to $ \overline{g}$ , a smallest grammar.
Thus the question can be phrased, compute $ \text{SGP}(s,n) = \text{argmin}_{\overline{x} \in H}\{|\overline{x} \cdot \overline{y}|\} \cdot \overline{y} =$ the set of smallest grammars of $ s$ on up to $ n$ variables, when given a pruned grammar $ y$ for $ s$ . You can go ahead and take $ y = \{ S \to s \}$ , with $ 1$ ‘s in the other components of $ \overline{y}$ .
Questions. For any smallest grammar $ g$ does there always exist two other smallest grammars $ h_1, h_2$ such that $ \overline{h_1}\overline{h_2} = \overline{g}$ in the direct product $ G^n$ ? In other words is there a recursive decomposition of the problem for each problem instance.
Written with StackEdit. |
Answer
In general, $\cos{(bx)} \ne b\cdot \cos{x}$ because they have different amplitudes and periods. Refer to the graph in the step-by-step part below.
Work Step by Step
RECALL: The function $y=a \cdot \cos{(bx)}$ has : amplitude = $|a|$ period = $\frac{2\pi}{b}$ Thus: The function $y=\cos{(3x)}$ has an amplitude of $|1|=1$ and a period of $\frac{2\pi}{3}$. The function $y=3\cos{x}$ has an amplitude of $|3|=3$ and a period of $\frac{2\pi}{1}=2\pi$. From the information above, it is obvious that the two functions are different from each other because they have different amplitudes and periods. Thus, it cannot be said that in general, $\cos{(bx)}=b \cdot \cos{x}$. Use a graphing utility to graph the two functions. (Refer to the attached image below for the graph, the green graph is $y=3\cos{x}$ while the red graph is $y=\cos{(3x)}$.) Use a graphing utility to graph the given functions. (Refer to the graph below.) Notice that the graphs are different. |
I wrote an entire blog post explaining the answers to 2.3 but Blogger decided to eat it. I don't want to redo those answers so here is 3.1:
For now on I will title my posts with the section number as well to help Google.
Question 3.1-1:Let $f(n)$ and $g(n)$be asymptotically non-negative functions. Using the basic definition of $\theta$-notation, prove that $\max(f(n) , g(n)) \in \theta(f(n) + g(n))$ .
CLRS defines $\theta$ as $\theta(g(n))= \{ f(n) :$ there exists some positive constants $c_1, c_2$, and $n_0,$ such that $0 \leq c_1g(n) \leq f(n) \leq c_2g(n)$ for all $n \geq n_0\}$ Essentially we must prove that there exists some $c_1$ and $c_2$ such that $c_1 \times (f(n) + g(n)) \leq \max(f(n), g(n)) \leq c_2 \times (f(n) + g(n))$ There are a variety of ways to do this but I will choose the easiest way I could think of. Based on the above equation we know that $\max(f(n), g(n)) \leq f(n) + g(n)$ (as f(n) and g(n) must both me non-negative) and we further know that $\max(f(n), g(n))$ can't be more than twice f(n)+g(n). What we have then are the following inequalities: $$\max(f(n), g(n)) \leq c_1 \times (f(n) + g(n))$$ and $$c_2 \times (f(n) + g(n)) \leq 2 \times \max(f(n), g(n))$$ Solving for $c_1$ we get 1 and for $c_2$ we get $\frac {1} {2}$
Question 3.1-2:Show for any real constants $a$ and $b$ where $b \gt 0$ that $(n+a)^b \in \theta(n^b)$ Because $a$ is a constant and the definition of $\theta$ is true after some $n_0$ adding $a$ to $n$ does not affect the definition and we simplify to $n^b \in \theta(n^b)$ which is trivially true Question 3.1-3:Explain why the statement "The running time of $A$ is at least $O(n^2)$," is meaningless.
I'm a little uncertain of this answer but I think this is what CLRS is getting at when we say a function $f(n)$ has a running time of $O(g(n))$ what we really mean is that $f(n)$ has an asymptotic upper bound of $g(n)$. This means that $f(n) \leq g(n)$ after some $n_0$. To say a function has a running time of at least g(n) seems to be saying that $f(n) \leq g(n) \And f(n) \geq g(n)$ which is a contradiction.
Question 3.1-4:Is $2^{n+1} = O(2^n)$? Is $2^{2n} = O(2^n)$?
$2^{n+1} = 2 \times 2^n$. which means that $2^{n+1} \leq c_1 \times 2^n$ after $n_0$ so we have our answer that $2^{n+1} \in o(2^n)$ Alternatively we could say that the two functions only differ by a constant coefficient and therefore the answer is yes.
There is no constant such that $2^{2n} = c \times 2^n$ and thefore $2^{2n} \notin O(2^n)$
Question 3.1-5:Prove that for any two functions $f(n)$ and $g(n)$, we have $f(n) \in \theta(g(n)) \iff f(n) \in O(g(n)) \And f(n) \in \Omega(g(n))$
This is an "if an only if" problem so we must prove this in two parts:
Firstly, if $f(n) \in O(g(n))$ then there exists some $c_1$ and $n_0$ such that $f(n) \leq c_1 \times g(n)$ after some $n_0$. Further if $f(n) \in Omega(g(n))$ then there exists some $c_2$ and $n_0$ such that $f(n) \geq c_2 \times g(n)$ after some $n_0$.
If we combine the above two statements (which come from the definitions of $\Omega$ and O) than we know that there exists some $c_1, c_2, and n_0,$ such that $c_1g(n) \leq f(n) \leq c_2g(n)$ for all $n \geq n_0\}$
We could do the same thing backward for the other direction: If $f(n) \in \theta(g(n))$ then we could split the above inequality and show that each of the individual statements are true.
Question 3.1-6:Prove that the running time of an algorithm is $\theta(g(n)) \iff$ its worst-case running time is $O(g(n))$ and its best case running time $\Omega(g(n))$. I'm going to try for an intuitive proof here instead of a mathematical one. If the worst case is asymptotically bound above in the worst case by a certain function and is asymptotically bound from below in the best case which means that the function is tightly bound by both those functions. f(n) never goes below some constant times g(n) and never goes above some constant times g(n). This is what we get from the above definition of $\theta(g(n)))$ A mathematical follows from question 3.1-5. Question 3.1-7:Prove that $o(g(n)) \cap \omega(g(n)) = \varnothing$
little o and little omega are defined as follows: \[o(g(n)) = \{ f(n) : \forall c > 0 \exists n_0 \text{such that } 0 \leq f(n) \leq c \times g(n) \forall n \gt n_0\] and \[\omega(g(n)) = \{ f(n) : \forall c > 0 \exists n_0 \text{such that } 0 \leq c \times g(n) \leq f(n) \forall n \gt n_0\]
In other words
$$f(n) \in o(g(n)) \iff \lim_{n \to \infty} \frac {f(n)} {g(n)} = 0$$ and $$f(n) \in \omega(g(n)) \iff \lim_{n \to \infty} \frac {f(n)} {g(n)} = \infty$$It is obvious that these can not be true at the same time. This would require that $0 = \infty$ |
Iterate of exponential Contents Integer and non-integer \(n\)
The most often are the first iteration of exponent, \(n=1\); \(\exp^1=\exp\)
and the minus first iteration, \(n=-1\); \(\exp^{-1} = \ln\).
Less often they appear with \(n = \pm 2\); \(\exp^2(z)=\exp(\exp(z))\), and \(\exp^{-2}(z)=\ln(\ln(z))\). Other values of number of iteration are not usual, and until year 2008, there was no regular way to evaluate iteration of exponential for any non–integer number \(n\) of iteration. However, with tetration tet, that is superfunction of exponent, and Arctetration ate, that is Abel function of exponent, the \(n\)th iteration can be expressed as follows:
\(\exp^n(z)=\mathrm{tet}(n+\mathrm{ate}(z))\)
Both, tet and ate are holomorphic functions; so, the representation above can be used for non-integer \(n\). The exponential can be iterated even complex number of times.
Iimplementation
Representation of \(\exp^n\) through function tet and ate defines the \(n\)th iterate of exponential for any complex number \(n\) of iterations.Methods for the evaluation are described in 2009 by D.Kouznetsov in Mathematics of Computation
[2], and the efficient C++ complex double implementation are described in 2010 in Vladikavkaz mathematical Journal in Russian; the English version is also loaded [3]. WIth known properties and the efficient implementation, functions tet, ate and non–integer ietrations of the exponent shouls be qualified as special functions; in computation, one can access them as if they would be elementary functions.The complex doube implementations of functions tet and ate are loaded to TORI, see fsexp.cin and fslog.cin; they run at various operational systems; at least under Linux and Macintosh. Reports of any problems with the use or the reproducible bugs should be appreciated.
Complex maps of the \(n\)th iteration of exponential, \(f=\exp^n(x+\mathrm i y)\) are shown in figures at right with lines \(u=\Re(f)\) and lines \(v=\Im(f)\) for various values \(n\) in the \(x\),\(y\) plane. As the function is real-holomorphic, the maps are symmetric; so the only upper half plane is shown in the figures.
Cut lines
While \(n\) is not integer, \(\exp^n(z)\) is holomorphic in the complex plane with two cut lines \(\Re(z)\le \Re(L)\), \(\Im(z)=\pm \Im(L)\), where \(L\approx 0.3+1.3 \mathrm i\) is fixed point of logarithm, id est, solution of equation
\(L=\ln(L)\).
In the figures at right, one of these cuts is seen; it is marked with dashed line. The additional levels \(\Re(L)\) for the real part of \(\exp^n\) and \(\Im(L)\) for the imaginary part are drown with thick green lines; of course, these lines cross each other at the branch point \(L\).
In addition, for negative number \(n\) of iterations (and, in particular, for \(n=-1\)), there is cut line along the negative part of the real axis, id est, from \(-\infty\) to \(\mathrm{tet}(-2-n)\).
Special function
Properties of the iteration of the exponential are described.
\(\exp^n(z)\) is holomorphic function with respect to \(z\) and with respect to \(n\). Properties of this function are analyzed and described. The efficient (fast and precise) algorithm for the evaluation is supplied with routines fsexp.cin and cslog.cin.
With achievements above, function \((n,z) \mapsto \exp^n(z)\) is qualified as special function. Designers of compilers and interpreters from the programming languages are invited to borrow the implementations of tetration and arctetration in order to provide the built-in function, that evaluates \(\exp^n(z)\) for complex values of \(n\) and \(z\). In particular, in Mathematica, there is already name for such a function; it should be called with Nest[Exp,z,n]. Up to year 2013, the built-in function Nest is implemented in such a way, that number \(n\) of iteration should be expressed with natural constant, positive integer number
[4]. Over-vice, the built-in function generates the error message instead of to perform the calculations and evaluations requested. With use of superfunctions and Abel functions, Nest could be implemented for more general case. References http://www.ams.org/journals/bull/1993-29-02/S0273-0979-1993-00432-4/S0273-0979-1993-00432-4.pdf Walter Bergweiler. Iteration of meromorphic functions. Bull. Amer. Math. Soc. 29 (1993), 151-188. http://www.ams.org/mcom/2009-78-267/S0025-5718-09-02188-7/home.html http://www.ils.uec.ac.jp/~dima/PAPERS/2009analuxpRepri.pdf http://mizugadro.mydns.jp/PAPERS/2009analuxpRepri.pdf D. Kouznetsov. Solution of \(F(x+1)=\exp(F(x))\) in complex \(z\)-plane. 78, (2009), 1647-1670 http://www.ils.uec.ac.jp.jp/~dima/PAPERS/2009vladie.pdf (English) http://mizugadro.mydns.jp/PAPERS/2010vladie.pdf (English) http://mizugadro.mydns.jp/PAPERS/2009vladir.pdf (Russian version) D.Kouznetsov. Superexponential as special function. Vladikavkaz Mathematical Journal, 2010, v.12, issue 2, p.31-45. http://reference.wolfram.com/mathematica/ref/Nest.html Nest, Wolfram Mathematica 9 Documentation center, 2013. |
فهرست مطالب Volume:7 Issue:4, 2018 تاریخ انتشار: 1397/02/29 تعداد عناوین: 5
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Pages 1-7Let $G$ be a groupý, ýwe say that $G$ satisfies the property $\mathcal{T}(\infty)$ provided thatý, ýevery infinite set of elements of $G$ contains elements $x\neq yý, ýz$ such that $[xý, ýyý, ýz]=1=[yý, ýzý, ýx]=[zý, ýxý, ýy]$ý. ýWe denote by $\mathcal{C}$ the class of all polycyclic groupsý, ý$\mathcal{S}$ the class of all soluble groupsý, ý$\mathcal{R}$ the class of all residually finite groupsý, ý$\mathcal{L}$ the class of all locally graded groupsý, ý$\mathcal{N}_2$ the class of all nilpotent group of class at most twoý, ýand $\mathcal{F}$ the class of all finite groupsý. ýIn this paperý, ýfirst we shall prove that if $G$ is a finitely generated locally graded groupý, ýthen $G$ satisfies $\mathcal{T}(\infty)$ if and only if $G/Z_2(G)$ is finiteý, ýand then we shall conclude that if $G$ is a finitely generated group in $\mathcal{T}(\infty)$ý, ýthený ý\[G\in\mathcal{L}\Leftrightarrow G\in\mathcal{R}\Leftrightarrow G\in\mathcal{S}\Leftrightarrow G\in\mathcal{C}\Leftrightarrow G\in\mathcal{N}_2\mathcal{F}.\]ý Keywords:?Finitely generated groups?, ?Residually finite groups?, ?Locally graded groups
Pages 9-16Let $G$ be a group and $Aut^{\Phi}(G)$ denote the group of all automorphisms of $G$ centralizing $G/\Phi(G)$ elementwiseý. ýIn this paperý, ýwe characterize the finite $p$-groups $G$ with cyclic Frattini subgroup for which $|Aut^{\Phi}(G):Inn(G)|=p$ý. Keywords:??Automorphism group?, ?Finite $p$, group?, ?Frattini subgroup?
Pages 17-26ýThe Magnus embedding of a free metabelian group induces the embedding of partially commutative metabelian group $S_\Gamma$ in a group of matrices $M_\Gamma$. Properties and the universal theory of the group $M_\Gamma$ are studied. Keywords:Partially commutative group, Metabeliah group, Universal theory, Equations in group
Pages 27-40In this paper we investigate the special automata over finite rank free groups and estimate asymptotic characteristics of sets they acceptý. ýWe show how one can decompose an arbitrary regular subset of a finite rank free group into disjoint union of sets accepted by special automata or special monoidsý. ýThese automata allow us to compute explicitly generating functionsý, ý$\lambda-$measures and Cesaro measure of thick monoidsý. ýAlso we improve the asymptotic classification of regular subsets in free groupsý. Keywords:free group, ?$\lambda, $measure, regular subset, special automaton, thick monoid
Pages 41-64ýýIn this paper we prove that the Maschke property holds for coprime actions on some important classes of $p$-groups likeý: ýmetacyclic $p$-groupsý, ý$p$-groups of $p$-rank two for $p>3$ and some weaker property holds in the case of regular $p$-groupsý. ýThe main focus will be the case of coprime actions on the iterated wreath product $P_n$ of cyclic groups of order $p$ý, ýi.eý. ýon Sylow $p$-subgroups of the symmetric groups $S_{p^n}$ý, ýwhere we also prove that a stronger form of the Maschke property holdsý. ýThese results contribute to a future possible classification of all $p$-groups with the Maschke propertyý. ýWe apply these results to describe which normal partition subgroups of $P_n$ have a complementý. ýIn the end we also describe abelian subgroups of $P_n$ of largest sizeý. Keywords:?Maschke's Theorem?, ?coprime action?, ?Sylow $p$-subgroup of symmetric group?, ?iterated wreath product?, ?uniserial action |
Electronic Journal of Probability Electron. J. Probab. Volume 22 (2017), paper no. 54, 43 pp. Muttalib–Borodin ensembles in random matrix theory — realisations and correlation functions Abstract
Muttalib–Borodin ensembles are characterised by the pair interaction term in the eigenvalue probability density function being of the form $\prod _{1 \le j < k \le N}(\lambda _k - \lambda _j) (\lambda _k^\theta - \lambda _j^\theta )$. We study the Laguerre and Jacobi versions of this model — so named by the form of the one-body interaction terms — and show that for $\theta \in \mathbb Z^+$ they can be realised as the eigenvalue PDF of certain random matrices with Gaussian entries. For general $\theta > 0$, realisations in terms of the eigenvalue PDF of ensembles involving triangular matrices are given. In the Laguerre case this is a recent result due to Cheliotis, although our derivation is different. We make use of a generalisation of a double contour integral formula for the correlation functions contained in a paper by Adler, van Moerbeke and Wang to analyse the global density (which we also analyse by studying characteristic polynomials), and the hard edge scaled correlation functions. For the global density functional equations for the corresponding resolvents are obtained; solving this gives the moments in terms of Fuss–Catalan numbers (Laguerre case — a known result) and particular binomial coefficients (Jacobi case). For $\theta \in \mathbb Z^+$ the Laguerre and Jacobi cases are closely related to the squared singular values for products of $\theta $ standard Gaussian random matrices, and truncations of unitary matrices, respectively. At the hard edge the double contour integral formulas provide a double contour integral form of the scaled correlation kernel obtained by Borodin in terms of Wright’s Bessel function.
Article information Source Electron. J. Probab., Volume 22 (2017), paper no. 54, 43 pp. Dates Received: 8 October 2016 Accepted: 26 April 2017 First available in Project Euclid: 23 June 2017 Permanent link to this document https://projecteuclid.org/euclid.ejp/1498183245 Digital Object Identifier doi:10.1214/17-EJP62 Mathematical Reviews number (MathSciNet) MR3666017 Zentralblatt MATH identifier 1366.60009 Citation
Forrester, Peter J.; Wang, Dong. Muttalib–Borodin ensembles in random matrix theory — realisations and correlation functions. Electron. J. Probab. 22 (2017), paper no. 54, 43 pp. doi:10.1214/17-EJP62. https://projecteuclid.org/euclid.ejp/1498183245 |
Last year I made a post about the universal program, a Turing machine program $p$ that can in principle compute any desired function, if it is only run inside a suitable model of set theory or arithmetic. Specifically, there is a program $p$, such that for any function $f:\newcommand\N{\mathbb{N}}\N\to\N$, there is a model $M\models\text{PA}$ — or of $\text{ZFC}$, whatever theory you like — inside of which program $p$ on input $n$ gives output $f(n)$.
This theorem is related to a very interesting theorem of W. Hugh Woodin’s, which says that there is a program $e$ such that $\newcommand\PA{\text{PA}}\PA$ proves $e$ accepts only finitely many inputs, but such that for any finite set $A\subset\N$, there is a model of $\PA$ inside of which program $e$ accepts exactly the elements of $A$. Actually, Woodin’s theorem is a bit stronger than this in a way that I shall explain.
Victoria Gitman gave a very nice talk today on both of these theorems at the special session on Computability theory: Pushing the Boundaries at the AMS sectional meeting here in New York, which happens to be meeting right here in my east midtown neighborhood, a few blocks from my home.
What I realized this morning, while walking over to Vika’s talk, is that there is a very simple proof of the version of Woodin’s theorem stated above. The idea is closely related to an idea of Vadim Kosoy mentioned in my post last year. In hindsight, I see now that this idea is also essentially present in Woodin’s proof of his theorem, and indeed, I find it probable that Woodin had actually begun with this idea and then modified it in order to get the stronger version of his result that I shall discuss below.
But in the meantime, let me present the simple argument, since I find it to be very clear and the result still very surprising.
Theorem. There is a Turing machine program $e$, such that $\PA$ proves that $e$ accepts only finitely many inputs. For any particular finite set $A\subset\N$, there is a model $M\models\PA$ such that inside $M$, the program $e$ accepts all and only the elements of $A$. Indeed, for any set $A\subset\N$, including infinite sets, there is a model $M\models\PA$ such that inside $M$, program $e$ accepts $n$ if and only if $n\in A$. Proof. The program $e$ simply performs the following task: on any input $n$, search for a proof from $\PA$ of a statement of the form “program $e$ does not accept exactly the elements of $\{n_1,n_2,\ldots,n_k\}$.” Accept nothing until such a proof is found. For the first such proof that is found, accept $n$ if and only if $n$ is one of those $n_i$’s.
In short, the program $e$ searches for a proof that $e$ doesn’t accept exactly a certain finite set, and when such a proof is found, it accepts exactly the elements of this set anyway.
Clearly, $\PA$ proves that program $e$ accepts only a finite set, since either no such proof is ever found, in which case $e$ accepts nothing (and the empty set is finite), or else such a proof is found, in which case $e$ accepts only that particular finite set. So $\PA$ proves that $e$ accepts only finitely many inputs.
But meanwhile, assuming $\PA$ is consistent, then you cannot refute the assertion that program $e$ accepts exactly the elements of some particular finite set $A$, since if you could prove that from $\PA$, then program $e$ actually would accept exactly that set (for the shortest such proof), in which case this would also be provable, contradicting the consistency of $\PA$.
Since you cannot refute any particular finite set as the accepting set for $e$, it follows that it is consistent with $\PA$ that $e$ accepts any particular finite set $A$ that you like. So there is a model of $\PA$ in which $e$ accepts exactly the elements of $A$. This establishes statement (2).
Statement (3) now follows by a simple compactness argument. Namely, for any $A\subset\N$, let $T$ be the theory of $\PA$ together with the assertions that program $e$ accepts $n$, for any particular $n\in A$, and the assertions that program $e$ does not accept $n$, for $n\notin A$. Any finite subtheory of this theory is consistent, by statement (2), and so the whole theory is consistent. Any model of this theory realizes statement (3).
QED
One uses the Kleene recursion theorem to show the existence of the program $e$, which makes reference to $e$ in the description of what it does. Although this may look circular, it is a standard technique to use the recursion theorem to eliminate the circularity.
This theorem immediately implies the classical result of Mostowski and Kripke that there is an independent family of $\Pi^0_1$ assertions, since the assertions $n\notin W_e$ are exactly such a family.
The theorem also implies a strengthening of the universal program theorem that I proved last year. Indeed, the two theorems can be realized with the same program!
Theorem. There is a Turing machine program $e$ with the following properties: $\PA$ proves that $e$ computes a finite function; For any particular finite partial function $f$ on $\N$, there is a model $M\models\PA$ inside of which program $e$ computes exactly $f$. For any partial function $f:\N\to\N$, finite or infinite, there is a model $M\models\PA$ inside of which program $e$ on input $n$ computes exactly $f(n)$, meaning that $e$ halts on $n$ if and only if $f(n)\downarrow$ and in this case $\varphi_e(n)=f(n)$. Proof. The proof of statements (1) and (2) is just as in the earlier theorem. It is clear that $e$ computes a finite function, since either it computes the empty function, if no proof is found, or else it computes the finite function mentioned in the proof. And you cannot refute any particular finite function for $e$, since if you could, it would have exactly that behavior anyway, contradicting $\text{Con}(\PA)$. So statement (2) holds. But meanwhile, we can get statement (3) by a simple compactness argument. Namely, fix $f$ and let $T$ be the theory asserting $\PA$ plus all the assertions either that $\varphi_e(n)\uparrow$, if $n$ is not the domain of $f$, and $\varphi_e(n)=k$, if $f(n)=k$. Every finite subtheory of this theory is consistent, by statement (2), and so the whole theory is consistent. But any model of this theory exactly fulfills statement (3). QED
Woodin’s proof is more difficult than the arguments I have presented, but I realize now that this extra difficulty is because he is proving an extremely interesting and stronger form of the theorem, as follows.
Theorem. (Woodin) There is a Turing machine program $e$ such that $\PA$ proves $e$ accepts at most a finite set, and for any finite set $A\subset\N$ there is a model $M\models\PA$ inside of which $e$ accepts exactly $A$. And furthermore, in any such $M$ and any finite $B\supset A$, there is an end-extension $M\subset_{end} N\models\PA$, such that in $N$, the program $e$ accepts exactly the elements of $B$.
This is a much more subtle claim, as well as philosophically interesting for the reasons that he dwells on.
The program I described above definitely does not achieve this stronger property, since my program $e$, once it finds the proof that $e$ does not accept exactly $A$, will accept exactly $A$, and this will continue to be true in all further end-extensions of the model, since that proof will continue to be the first one that is found. |
Since we know that the value of $e$, $i$, and $\pi$ are irrational reals, how about $$e^{i+\pi}\;?$$ Is it still irrational (that is, not a Gaussian rational)? The problem make me curious until now.
Building off of Clive's answer, if $e^\pi \cos 1$ and $e^\pi \sin 1$ were both rational numbers, then so would be their quotient $\tan 1$. But $\tan 1$ is irrational as a consequence of the Lindemann–Weierstrass theorem, because that theorem implies that $e^{2i}$ is transcendental, and because $e^{2i}=\dfrac{i-\tan 1}{\tan 1+i}$. Therefore $e^{\pi+i}$ cannot have both its real and imaginary parts rational.
If you mean
rational in the usual sense $-$ as a subset of the reals $-$ then $e^{i+\pi}$ is certainly not rational since it is not real. But the question of whether it's a Gaussian rational $-$ that is, its real and complex parts are rational $-$ may not be easy to prove. Notice that$$e^{i+\pi} = e^{\pi}\cos 1 + i e^{\pi}\sin 1$$
Given irrational numbers $\alpha$ and $\beta$ it's often quite hard to determine whether $\alpha^{\beta}$ and $\alpha \beta$ (and $\alpha + \beta$) are rational or irrational. For example, it is not currently known whether $e\pi$ is rational or irrational.
$e^i=\cos1+i\sin1$. (By Euler's Formula)
So $e^{i+\pi}=e^\pi\cos1+ie^\pi\sin1$, which is non-real and, of course, irrational. |
Some existence and concentration results for nonlinear Schrödinger equations
1.
Dipartimento di Matematica, Universita di Bari, via E. Orabona 4, 70125 Bari, Italy
$ i \h$$ \frac{\partial\psi}{\partial t}=-$ $\frac{ \h^2}{2m}\Delta \psi+V(x)\psi-\gamma|\psi|^{p-2}\psi,$ $\gamma>0,$ $ x\in\mathbb R^{N}$
where $\h$$ >0$, $p>2$, $\psi:\mathbb R^{N}\rightarrow\mathbb C,$ and the potential $V$ satisfies some symmetric properties. In particular the cases $N=2$ with $V$ radially symmetric and $N=3$ with $V$ having a cylindrical symmetry are discussed. Our main purpose is to study the asymptotic behaviour of such solutions in the semiclassical limit (i.e. as $\hbar \rightarrow 0^+$) when a concentration phenomenon around a point of $\mathbb R^N$ appears.
Mathematics Subject Classification:35J2. Citation:Teresa D'Aprile. Some existence and concentration results for nonlinear Schrödinger equations. Communications on Pure & Applied Analysis, 2002, 1 (4) : 457-474. doi: 10.3934/cpaa.2002.1.457
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System: the chemical reaction Surroundings: the environment which the chemical reaction is taking place in Latent heat: the heat released or absorbed by a chemical substance during a change of state
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Chemical Energy \(=\) Potential Energy (Stored Energy) \(+\) Kinetic Energy (Energy of Movement)
Enthalpy refers to the chemical energy or heat content of a substance. The exchange of heat between a system and its surroundings under constant pressure is referred to as the enthalpy change
Exothermic reactions, e.g. combustion release energy into the surroundings. Therefore, the energy of the reactants is higher than the products and \(\Delta H=H<0\)
Exothermic Reaction
Endothermic reactions, e.g. cold packs, absorb energy from the surroundings. Therefore, the energy of the products is higher than the reactants and \(\Delta H=H>0\)
Endothermic Reaction
Activation energy is the energy required to break the bonds of the reactants so that the reaction can proceed.
These chemical reactions are merely those which include the \(\Delta H\) value and must include states.
For example: \(CH_4(g)+O_2(g)\rightarrow CO_2(g)+H_2O(l)\,\,\Delta H=882kJ\,mol^{-1}\)
Specific heat capacity is the amount of energy needed to increase the temperature of a substance by \(1^oC\)
\(q=m\times c\times \Delta T\)
The heat released or absorbed by a chemical substance during a change of state
\(q=n\times L\)
The heat combustion of a fuel is defined as the enthalpy change that occurs when a fuel burns completely in oxygen. However, as many fuels are mixtures of chemicals that don’t have a specific molar mass, they cannot be expressed in kJ/mol.
\(q=n\times \Delta H_C\) |
I have the following exercise:
Consider the
heat equation$$u_t = k \Delta u \quad (*)$$ on a domain $\Omega$ with boundary conditions $u _{|\partial \Omega} =0$ and initial data $\phi(x)$. Suppose that $k$ is unknown, but it can be established by experiment that the total temperature in $\Omega$ decays like $\int_{\Omega} u(\cdot , t) \, dx = F(t)$, where $F$ is strictly monotonically decreasing.
Suppose also that the solution of the heat equation $$v_t = \Delta v \quad (**)$$ with the same boundary and initial conditions decays like $\int_{\Omega} v( \cdot , t) \, dx=G(t)$, where $G$ is strictly monotonically decreasing. Can you determine $k$ in terms of $F$ and $G$?
My attempt: we have that $$\int_{\Omega} u_t \, dx = \frac{d}{dt} \int_{\Omega} u \, dx = F'(t)$$and $$\int_{\Omega} v_t \, dx = \frac{d}{dt} \int_{\Omega} v \, dx = G'(t).$$Now, if we set $w(x,t) :=u(x,kt)$ we observe that $w$ is a solution for $(**)$. Then by uniqueness we have $v(x,t)=u(x,kt)$. Therefore $$\underbrace{\int_{\Omega} v_t \, dx }_{=G'(t)} = k \underbrace{\int_{\Omega} u_t (x,kt) \, dx}_{=F'(kt)}$$ and thus we can conclude $$k = \frac{G'(t)}{F'(kt)}>0.$$ The problem is that $F'$ seems to "depend" on $k$ itself, but I don't know if this is actually a problem... any comment about that?
Thanks in advance! |
$\textbf{The Problem:}$ Let $f:[a,b]\to\mathbb R$ be a continuous function. Prove that $f$ is Riemann integrable on $[a,b]$.
$\textbf{My Thoughts and Attempt:}$ Since $[a,b]$ is compact, $f$ is also uniformly continuous on it, so there is a $\delta(\varepsilon)>0$ such that for all $\varepsilon>0$ and all $x,y\in[a,b]$ with $\vert x-y\vert\leqslant\delta(\varepsilon)$ it follows that $\vert f(x)-f(y)\vert<\varepsilon.$ So let $\varepsilon>0$ be given and consider the partition $$\mathcal P=\{x_0=a,x_1=a+\delta(\varepsilon), x_2=a+2\delta(\varepsilon),\dots,a+(n-1)\delta(\varepsilon),x_n=b\},$$ where $n\in\mathbb N$ is such that $\vert x_j-x_{j-1}\vert\leqslant\delta(\varepsilon)$ implies $\color{red}{\vert f(x)-f(y)\vert<\displaystyle\frac{\varepsilon}{n}}$. Then the uniform continuity of $f$ guarantees that the values $$M_j=\sup\limits_{x\in[x_{j-1},x_j]}f(x)\quad\text{and}\quad m_j=\inf\limits_{x\in[x_{j-1},x_j]}f(x)$$ are attained and that $\color{red}{M_j-m_j<\displaystyle\frac{\varepsilon}{n}}$. So we see that \begin{align*} \mathcal U(f,\mathcal P)-\mathcal L(f,\mathcal P)&=\sum^{n}_{j=1}(x_{j}-x_{j-1})\cdot(M_j-m_j)\\ &\leqslant\delta(\varepsilon)\cdot\sum^{n}_{j=1}\frac{\varepsilon}{n}\\ &=\delta(\varepsilon)\cdot\varepsilon. \end{align*} And it follows that $f$ is Riemann integrable on $[a,b]$.
$\textbf{Note:}$ The notation $\delta(\varepsilon)$ means that $\delta$ can only depend on $\varepsilon.$
$\textbf{My Concern:}$ My concern is that I am not sure if the statements in $\color{red}{\text{red}}$ are correct. Also, if there are any other mistakes, please feel free to point them out.
Thank you for your time. |
Isotomic Reciprocity What Might This Be About? Problem Tools
The main tool we'll use for solving the problem is the barycentricc coordinates. Much of what we'll need for the proof can be found in Paul Yiu's online book Introduction to the Geometry of the Triangle.
Let point $S$ have homogeneous barycentric coordinates $(x:y:z)$ with respect to $\Delta ABC.$ The cevian $AS$ hits side $BC$ at point $(0:y:z),$ etc. The isotomic conjugate $S'$ has coordinates $\displaystyle (\frac{1}{x}:\frac{1}{y}:\frac{1}{z}).$
A line through two points $S_{i}=(x_{i}:y_{i}:z_{i}),$ $i=1,2,$ is defined by the equation
$\left|\begin{array}{ccc} x & y & z\\ x_{1} & y_{1} & z_{1}\\ x_{2} & y_{2} & z_{2} \end{array}\right|=0.$
Proof
Assume $Q=(x:y:z)$ and $P=(u:v:w).$ Then $Q'=\displaystyle (\frac{1}{x}:\frac{1}{y}:\frac{1}{z})$ and $P'=\displaystyle (\frac{1}{u}:\frac{1}{v}:\frac{1}{w}).$ Thus $E=\displaystyle (\frac{1}{x}:0:\frac{1}{z})$ while $F=\displaystyle (\frac{1}{x}:\frac{1}{y}:0).$ Similarly, $V=\displaystyle (\frac{1}{u}:0:\frac{1}{w})$ and $W=\displaystyle (\frac{1}{u}:\frac{1}{v}:0).$
Point $Q=(x:y:z)$ lies on $VW$ only if
$\left|\begin{array}{ccc} x & y & z\\ \displaystyle\frac{1}{u} & 0 & \displaystyle\frac{1}{w}\\ \displaystyle\frac{1}{u} & \displaystyle\frac{1}{v} & 0\\ \end{array}\right|=0.$
The determinant equation evaluates to
$\displaystyle -\frac{x}{vw}+\frac{y}{uw}+\frac{z}{uv}=0,$
which is equivalent to
(*)
$-xu+yv+zw=0.$
This same equation is the condition for $P$ to lie on $EF.$
Now, to evaluate the cross-ratio $(VWMQ)=\displaystyle\frac{VM}{WM}:\frac{VQ}{WQ}$ we'll need to find those ratios. To this end, let's find $\alpha$ and $\beta$ such that $Q=\alpha V+\beta W.$ It will then follow that $\displaystyle\frac{VQ}{WQ}=\frac{\beta}{\alpha}.$ I rewrite $V=(w(v+u):0:u(v+u))$ and $W=(v(w+u):u(w+u):0)$ which insures that the pair of the homogeneous coordinates refers to the same frame of reference. $Q=\alpha V+\beta W$ then gives three equations:
$\begin{align} x&=\alpha w(v+u)+\beta v(w+u),\\ y&=\beta u(w+u),\\ z&=\alpha u(v+u). \end{align}$
From the last two $\displaystyle\alpha=\frac{z}{u(v+u)}$ and $\displaystyle\beta=\frac{y}{u(w+u)}.$ We need to check that these also satisfy the first equation:
$\begin{align}\displaystyle \alpha w(v+u)+\beta v(w+u)&=\frac{z}{u(v+u)}w(v+u)+\frac{y}{u(w+u)}v(w+u)\\ &=\frac{zw}{u}+\frac{yv}{u}\\ &=\frac{yv+zw}{u}\\ &=\frac{xu}{u}\\ &=x, \end{align}$
where we made use if the condition (*). Thus
$\displaystyle\frac{VQ}{WQ}=\frac{\beta}{\alpha}=\frac{y}{u(w+u)}\frac{u(v+u)}{z}=\frac{y(v+u)}{z(w+u)}.$
Point $M$ lies at the intersection of $VW$ and $BC$ whose equation is $x=0.$ Thus, from the determinant equation above $M=\displaystyle (0:\frac{1}{v}:-\frac{1}{w})=(0:w:-v).$ Like $Q,$ $M$ is a linear combination of $V$ and $W,$ say $M=\gamma V+\delta W.$ This again leads to three equations:
$\begin{align} 0&=\gamma w(v+u)+\delta v(w+u),\\ w&=\delta u(w+u),\\ -v&=\gamma u(v+u). \end{align}$
From the last two equations, $\displaystyle\gamma=-\frac{v}{u(v+u)}$ and $\displaystyle\delta=\frac{w}{u(w+u)}.$ The first equation is automatically satisfied. Thus we have
$\displaystyle\frac{VM}{WM}=-\frac{w(v+u)}{v(w+u)}.$
We conclude that
$\begin{align}\displaystyle (VWMQ)&=\frac{VM}{WM}:\frac{VQ}{WQ}\\ &=-\frac{w(v+u)}{v(w+u)}\cdot\frac{z(w+u)}{y(v+u)}\\ &=-\frac{wz}{vy}. \end{align}$
Due to the apparent symmetry $(EFNP)$ has the same value.
Note that $M$ and $U$ are harmonic conjugates with respect to the pair $B,C$ as are $N$ and $D.$ Acknowledgment
The above statement has been posted by Francisco Javier García Capitán at the CutTheKnotMath facebook page.
Barycenter and Barycentric Coordinates 3D Quadrilateral - a Coffin Problem Barycentric Coordinates Barycentric Coordinates: a Tool Barycentric Coordinates and Geometric Probability Ceva's Theorem Determinants, Area, and Barycentric Coordinates Maxwell Theorem via the Center of Gravity Bimedians in a Quadrilateral Simultaneous Generalization of the Theorems of Ceva and Menelaus Three glasses puzzle Van Obel Theorem and Barycentric Coordinates 1961 IMO, Problem 4. An exercise in barycentric coordinates Centroids in Polygon Center of Gravity and Motion of Material Points Isotomic Reciprocity An Affine Property of Barycenter Problem in Direct Similarity Circles in Barycentric Coordinates Barycenter of Cevian Triangle Concurrent Chords in a Circle, Equally Inclined
65617477 |
I've had no luck with this one. None of the convergence tests pop into mind.
I tried looking at it in this form $\sum \sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$ and apply Dirichlets test. I know that $\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n} \to 0$ but not sure if it's decreasing.
Regarding absolute convergence, I tried:
$$|\sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}|\geq \sin^2 n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}=$$
$$=\frac{1}{2}\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}-\frac{1}{2}\cos 2n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$$
But again I'm stuck with $\cos 2n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$.
Assuming it converges then I've shown that $\sum \sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$ doesn't converge absolutely. |
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∑n=1∞n45n \large \sum_{n=1}^\infty \dfrac{n^4}{5^n} n=1∑∞5nn4
The value of the infinite series above can be expressed in the form ab\frac{a}{b}ba, where aaa and bbb are positive coprime integers. Find a+ba+ba+b.
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I know this is sort of a broad question, but I'm having trouble getting a handle on the syntax for higher order logic, when going from first order logic. Basically I want to be able to do resolution proofs in higher order logic since it works so well for first order logic.
The syntax of first order logic I understand as this:
A first order
term is either a variable or a first order function of some arity. Functions of arity zero can be represented as constants. A first order function takes only terms as its arguments. There are an infinite number of first order element variables. There are an infinite number of free (constant) first order functions. Nothing else is a term.
A first order predicate takes only first order terms as arguments. Predicates that take no arguments can be represented as propositional constants.
A first order logical formula consists of predicates, and logical formula connected by logical connectives, including negation, conjunction, disjunction, and quantifiers. Quantifiers quantify element variables, either universal or existential. Logical connectives can connect to predicates or other logical formula. Nothing else is a logical formula.
Example first order statement: $\forall$x$\exists$y P(x,y) $\rightarrow$ Q(f(x), g(y))
Now, we can extend this to
second order logic by allowing quantification of functions and predicates/relations. Now first order function constants and predicate constants are no longer required to be free constants, but may be bound variables. Other than that it's identical to first order logic. The semantics may be more complicated, because the completeness and compactness theorem no longer apply in full semantics, and higher order proof calculus may be more complicated because higher order Skolemization is more complicated and may require the axiom of choice, but the syntax isn't much different. But quantified functions and relations in second order logic still take first order terms and return first order terms and truth values respectively.
Using our first order example as quantified second order logic:
$\forall$P$\exists$Q$\forall$f$\exists$g$\forall$x$\exists$y P(x,y) $\rightarrow$ $Q(f(x), g(y))$
This seems pretty straightforward if I want to do resolution proofs in second order logic. I can use most of the same proof calculus as first order logic, with new rules for unifying function variables and predicate variables.
But when we go higher orders the syntax seems more muddy. Do we have second order terms as different
types from first order terms? For instance I can have a third order function variable that takes a second order function variable as an argument... and have its domain be first order terms, or second order functions, of all sorts of different arities and signatures. Or a fourth order predicate that takes a third order function of a specific arity and a second order function, and a first order term. It looks to me like there's an explosion of syntactical complexity at third order logic and higher, but I haven't seen many explicit examples of third or higher order syntax.
In first and second order logic, as far as I understand, the only complexity of the signature of functions and relations is the arity; just how many arguments it takes. But it appears to me that the signature of higher order functions and relations is much more complicated, since higher order functions can take as arguments and have a domain more than just the first order terms.
I've heard higher order logic is similar to type theory, in that higher order terms are essentially different types, and higher order logic can be represented as many sorted first order logic. Can someone explain how higher order syntax is supposed to work? Or point me to a reference for formal syntax (or syntaxes?) for higher order logics? |
To find the $x$ coordinate of the point $P$ that divides the line segment $AB$ in ratio $m:n$ we can do this :
1) Projection of $AB$ along $x$ axis is $x_2-x_1$ 2) We must split this projection at a distance $\frac{m}{m+n}(x_2-x_1)$ from $x_1$ 3) Overall the $x$ coordinate of $P$ is $$x_1+\dfrac{m}{m+n}(x_2-x_1)$$
So far so good. But if I simplify above formula I get a nice looking result :
$$\dfrac{mx_2+nx_1}{m+n}$$
I'm wondering if this form has any nice interpretation, like, when $A$ is at the origin, the numerator becomes $mx_2$ and the result is easy to interpret - just $\frac{m}{m+n}$ of the length $x_2$. However when $x_1\ne 0$, we're adding $nx_1$ to the numerator. What does this quantity represent?
In short, do the expressions $mx_2$ and $nx_1$ represent any meaningful quantities?
EDIT :
If I rearrange the result as $$\dfrac{m}{m+n}x_2 + \dfrac{n}{m+n}x_1$$
it seems like a pair of meaningful quantities, but not sure how to interpret them. Help? |
Overview
First a few comments on absorbing boundaries: your second picture seems to imply that it's really the edges of your square which absorb the wave, but this is not quite true. Rather, one can consider your picture as in the following:
That is, you see only the excerpt of the full space in which the wave is considered as undisturbed. And outside of what you see is something happening. Most straightforwardly, that would be a grid large enough so that no reflections can occur. But depending on the parameters, this is numerically too costly.
For this case, one can apply "absorbing" techniques whose general idea is to make the solution vanish outside the region of interest, without changing the behavior inside (the outside boundary is here sketched as a circle, which might be more appropriate due to symmetry, but depending on your wave this doesn't have to be). In order to do so, a balance must be found between (i) having as little impact on the inner region as possible and (ii) an as small outer region as possible.
There are many different techniques to implement the absorbing which try to find such an appropriate balance, among them (complex) absorbing potential, complex exterior scaling, masking functions, etc. (see also the references in the comments).
Masking functions / absorbing potentials
As the OP asked for a simple method, I would suggest to try a masking function:for this you apply a given function $\text{mask}(x,y)$ to your solution at each time step which is smaller than in the outside region $R_\text{out}$. A suitable function could be
$$f_\text{mask}(x,y) = \tanh^2\left(\lambda (R - \sqrt{x^2+y^2})\right)$$
where $R$ denotes the maximum radius on the grid and the parameter $\lambda$ determines the width of the mask which is to be tuned appropriately. Many other choices are possible here as well, e.g. a suitably streched quarter wavelength of a cosine, etc.
There are basically two alternatives how the masking function is applied. The simple version is to simply multiply the solution $u(x,y)$ by $\text{mask}(x,y)$ at each timestep. More sophisticated, one can include the logarithm of the masking function to the partial differential equation,$$\frac{\partial u}{\partial t}=v + \log(f_\text{mask}(x,y)) u\\\frac{\partial v}{\partial t} = c^2\nabla^2u $$Why the log? Because in the operator one needs something which is zero inside, and smaller than zero outside. The latter version is very similar to the
(complex) absorbing potential technique.
Although there are more efficient methods available, imo nothing beats the conceptual simplicity of the masking approach.
EDIT: the previous applies to first-order-in-time PDEs, the kind of which I'm usually dealing with. For second order equations, represented as two first-order equations for the function and the temporal derivative, the absorbing potential must be applied only to the first equation corresponding to the function. The reason is that one does not want the derivative to be damped out, but the function value. |
I’d like to write about the situation that occurs in set theory when a forcing extension $V[G]=V[H]$ arises over a ground model $V$ in two different ways simultaneously, using generic filters over two different forcing notions $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$. The general fact, stated in theorem 1, is that in this case, the two forcing notions are actually isomorphic on a cone $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, with the isomorphism carrying the one generic filter to the other. In other words, below these respective conditions $b$ and $c$, the forcing notions and the respective generic filters are not actually different.
I have always assumed that this fact was part of the classical forcing folklore results, but
it doesn’t seem to be mentioned explicitly in the usual forcing literature (it appears as lemma 25.5 in Jech’s book), and so I am writing an account of it here. Victoria Gitman and I have need of it in a current joint project. (Bob Solovay mentions in the comments below that the result is due to him, and provides a possible 1975 reference.) Theorem 1. If $V[G]=V[H]$, where $G\subset \mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{B}\upharpoonright b$ is isomorphic to $\mathbb{C}\upharpoonright c$ by an isomorphism carrying $G$ to $H$.
The proof will also establish the following related result, concerning the situation where one extension is merely contained in the other.
Theorem 2. If $V[H]\subset V[G]$, where $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{C}\upharpoonright c$ is isomorphic to a complete subalgebra of $\mathbb{B}\upharpoonright b$.
By $\mathbb{B}\upharpoonright b$, where $b$ is a condition in $\mathbb{B}$ (that is, a nonzero element of $\mathbb{B}$), what I mean is the Boolean algebra consisting of the interval $[0,b]$ in $\mathbb{B}$, using relative complement $b-a$ as the negation of $a$. This is the complete Boolean algebra that arises when forcing with the conditions in $\mathbb{B}$ below $b$.
Proof: In order to prove theorem 2, let me assume at first only that $V[H]\subset V[G]$. It follows that $H=\dot H_G$ for some $\mathbb{B}$-name $\dot H$, and we may choose a condition $b\in G$ forcing that $\dot H$ is a $\check V$-generic filter on $\check{\mathbb{C}}$.
I claim that there is some $c\in H$ such that every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$. Note that every $d\in H$ has $[\![\check d\in\dot H]\!]\in G$ by the truth lemma, since $H=\dot H_G$, and so $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ for $d\in H$. If $c\in H$ forces that every $d$ in the generic filter has that property, then indeed every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ as claimed.
In other words, from the perspective of the $\mathbb{B}$ forcing, every $d\leq c$ has a nonzero possibility to be in $\dot H$.
Define $\pi:\mathbb{C}\upharpoonright c\to\mathbb{B}$ by $$\pi(d)=b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}.$$ Using the fact that $b$ forces that $\dot H$ is a filter, it is straightforward to verify that
$d\leq e\implies \pi(d)\leq\pi(e)$, since if $d\leq e$ and $d\in H$, then $e\in H$. $\pi(d\wedge e)=\pi(d)\wedge \pi(e)$, since $[\![\check d\in\dot H]\!]\wedge[\![\check e\in \dot H]\!]=[\![\check{(b\wedge e)}\in\dot H]\!]$. $\pi(d-e)=\pi(d)-\pi(e)$, since $[\![\check{(d-e)}\in\dot H]\!]=[\![\check d\in\dot H]\!]-[\![\check e\in\dot H]\!]$.
Thus, $\pi$ is a Boolean algebra embedding of $\mathbb{C}\upharpoonright c$ into $\mathbb{B}\upharpoonright\pi(c)$.
Let me argue that this embedding is a complete embedding. Suppose that $a=\bigvee A$ for some subset $A\subset\mathbb{C}\upharpoonright c$ with $A\in V$. Since $H$ is $V$-generic, it follows that $a\in H$ just in case $H$ meets $A$. Thus, $[\![\check a\in\dot H]\!]=[\![\exists x\in\check A\, x\in \dot H]\!]=\bigvee_{x\in A}[\![\check x\in\dot H]\!]$, and so $\pi(\bigvee A)=\bigvee_{x\in A}\pi(x)$, and so $\pi$ is complete, as desired. This proves theorem 2.
To prove theorem 1, let me now assume fully that $V[G]=V[H]$. In this case, there is a $\mathbb{C}$ name $\dot G$ for which $G=\dot G_H$. By strengthening $b$, we may assume without loss that $b$ also forces that, that is, that $b$ forces $\Gamma=\check{\dot G}_{\dot H}$, where $\Gamma$ is the canonical $\mathbb{B}$-name for the generic object, and $\check{\dot G}$ is the $\mathbb{B}$-name of the $\mathbb{C}$-name $\dot G$. Let us also strengthen $c$ to ensure that $c$ forces $\dot G$ is $\check V$-generic for $\check{\mathbb{C}}$. For $d\leq c$ define $\pi(d)=[\![\check d\in\dot H]\!]^{\mathbb{B}}$ as above, which provides a complete embedding of $\mathbb{C}\upharpoonright c$ to $\mathbb{B}\upharpoonright\pi(c)$. I shall now argue that this embedding is dense below $\pi(c)$. Suppose that $a\leq \pi(c)$ in $\mathbb{B}$. Since $a$ forces $\check a\in\Gamma$ and also $\check c\in\dot H$, it must also force that there is some $d\leq c$ in $\dot H$ that forces via $\mathbb{C}$ over $\check V$ that $\check a\in\dot G$. So there must really be some $d\leq c$ forcing $\check a\in\dot G$. So $\pi(d)$, which forces $\check d\in\dot H$, will also force $\check a\in\check{\dot G}_{\dot H}=\Gamma$, and so $\pi(d)\Vdash_{\mathbb{B}}\check a\in\Gamma$, which means $\pi(d)\leq a$ in ${\mathbb{B}}$. Thus, the range of $\pi$ on $\mathbb{C}\upharpoonright c$ is dense below $\pi(c)$, and so $\pi$ is a complete dense embedding of ${\mathbb{C}}\upharpoonright c$ to ${\mathbb{B}}\upharpoonright \pi(c)$. Since these are complete Boolean algebras, this means that $\pi$ is actually an isomorphism of $\mathbb{C}\upharpoonright c$ with $\mathbb{B}\upharpoonright \pi(c)$, as desired.
Finally, note that if $d\in H$ below $c$, then since $H=\dot H_G$, it follows that $[\![\check d\in\dot H]\!]\in G$, which is to say $\pi(d)\in G$, and so $\pi$ carries $H$ to $G$ on these cones. So $\pi^{-1}$ is the isomorphism stated in theorem 1.
QED
Finally, I note that one cannot get rid of the need to restrict to cones, since it could be that $\mathbb{B}$ and $\mathbb{C}$ are the lottery sums of a common forcing notion, giving rise to $V[G]=V[H]$, together with totally different non-isomorphic forcing notions below some other incompatible conditions. So we cannot expect to prove that $\mathbb{B}\cong\mathbb{C}$, and are content to get merely that $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, an isomorphism below respective conditions. |
This question already has an answer here:
I am reading Andy's lectures.
On page 22, one has the symplectic form for free electrodynamics as $$\Omega_{\Sigma}=-\frac{1}{e^2}\int_{\Sigma}\delta(*F)\wedge\delta A,\tag{1}$$ where $\Sigma$ is any Cauchy hypersurface (3-dimensional) and $\delta$ is the variation in the phase space. Subsequently, (1) gives the equation with indices as $$\Omega_{\Sigma}=-\frac{1}{e^2}\int_{\Sigma}d\Sigma^{\mu}\delta F_{\mu\nu}\wedge\delta A^{\nu},\tag{2}$$ where $d\Sigma^{\mu}$ is the induced mearsure times the unit norm vector to $\Sigma$.
Anybody know how to derive Eq.(2) from Eq.(1)? Further, what does it mean by "induced measure" here? Does it mean $\sqrt{|h|}dx^1dx^2dx^3$, where $h_{\mu\nu}$ is the induced metric? If it is indeed the case, then when the surface is null hypersurce $\mathcal{I}^{\pm}$, we have $h=0$. |
this is a mystery to me, despite having changed computers several times, despite the website rejecting the application, the very first sequence of numbers I entered into it's search window which returned the same prompt to submit them for publication appear every time, I mean ive got hundreds of them now, and it's still far too much rope to give a person like me sitting along in a bedroom the capacity to freely describe any such sequence and their meaning if there isn't any already there
my maturity levels are extremely variant in time, that's just way too much rope to give me considering its only me the pursuits matter to, who knows what kind of outlandish crap I might decide to spam in each of them
but still, the first one from well, almost a decade ago shows up as the default content in the search window
1,2,3,6,11,23,47,106,235
well, now there is a bunch of stuff about them pertaining to "trees" and "nodes" but that's what I mean by too much rope you cant just let a lunatic like me start inventing terminology as I go
oh well "what would cotton mathers do?" the chat room unanimously ponders lol
i see Secret had a comment to make, is it really a productive use of our time censoring something that is most likely not blatant hate speech? that's the only real thing that warrants censorship, even still, it has its value, in a civil society it will be ridiculed anyway?
or at least inform the room as to whom is the big brother doing the censoring? No? just suggestions trying to improve site functionality good sir relax im calm we are all calm
A104101 is a hilarious entry as a side note, I love that Neil had to chime in for the comment section after the big promotional message in the first part to point out the sequence is totally meaningless as far as mathematics is concerned just to save face for the websites integrity after plugging a tv series with a reference
But seriously @BalarkaSen, some of the most arrogant of people will attempt to play the most innocent of roles and accuse you of arrogance yourself in the most diplomatic way imaginable, if you still feel that your point is not being heard, persist until they give up the farce please
very general advice for any number of topics for someone like yourself sir
assuming gender because you should hate text based adam long ago if you were female or etc
if its false then I apologise for the statistical approach to human interaction
So after having found the polynomial $x^6-3x^4+3x^2-3$we can just apply Eisenstein to show that this is irreducible over Q and since it is monic, it follwos that this is the minimal polynomial of $\sqrt{1+\sqrt[3]{2}}$ over $\mathbb{Q}$ ? @MatheinBoulomenos
So, in Galois fields, if you have two particular elements you are multiplying, can you necessarily discern the result of the product without knowing the monic irreducible polynomial that is being used the generate the field?
(I will note that I might have my definitions incorrect. I am under the impression that a Galois field is a field of the form $\mathbb{Z}/p\mathbb{Z}[x]/(M(x))$ where $M(x)$ is a monic irreducible polynomial in $\mathbb{Z}/p\mathbb{Z}[x]$.)
(which is just the product of the integer and its conjugate)
Note that $\alpha = a + bi$ is a unit iff $N\alpha = 1$
You might like to learn some of the properties of $N$ first, because this is useful for discussing divisibility in these kinds of rings
(Plus I'm at work and am pretending I'm doing my job)
Anyway, particularly useful is the fact that if $\pi \in \Bbb Z[i]$ is such that $N(\pi)$ is a rational prime then $\pi$ is a Gaussian prime (easily proved using the fact that $N$ is totally multiplicative) and so, for example $5 \in \Bbb Z$ is prime, but $5 \in \Bbb Z[i]$ is not prime because it is the norm of $1 + 2i$ and this is not a unit.
@Alessandro in general if $\mathcal O_K$ is the ring of integers of $\Bbb Q(\alpha)$, then $\Delta(\mathcal O_K) [\mathcal O_K:\Bbb Z[\alpha]]^2=\Delta(\mathcal O_K)$, I'd suggest you read up on orders, the index of an order and discriminants for orders if you want to go into that rabbit hole
also note that if the minimal polynomial of $\alpha$ is $p$-Eisenstein, then $p$ doesn't divide $[\mathcal{O}_K:\Bbb Z[\alpha]]$
this together with the above formula is sometimes enough to show that $[\mathcal{O}_K:\Bbb Z[\alpha]]=1$, i.e. $\mathcal{O}_K=\Bbb Z[\alpha]$
the proof of the $p$-Eisenstein thing even starts with taking a $p$-Sylow subgroup of $\mathcal{O}_K/\Bbb Z[\alpha]$
(just as a quotient of additive groups, that quotient group is finite)
in particular, from what I've said, if the minimal polynomial of $\alpha$ wrt every prime that divides the discriminant of $\Bbb Z[\alpha]$ at least twice, then $\Bbb Z[\alpha]$ is a ring of integers
that sounds oddly specific, I know, but you can also work with the minimal polynomial of something like $1+\alpha$
there's an interpretation of the $p$-Eisenstein results in terms of local fields, too. If the minimal polynomial of $f$ is $p$-Eisenstein, then it is irreducible over $\Bbb Q_p$ as well. Now you can apply the Führerdiskriminantenproduktformel (yes, that's an accepted English terminus technicus)
@MatheinBoulomenos You once told me a group cohomology story that I forget, can you remind me again? Namely, suppose $P$ is a Sylow $p$-subgroup of a finite group $G$, then there's a covering map $BP \to BG$ which induces chain-level maps $p_\# : C_*(BP) \to C_*(BG)$ and $\tau_\# : C_*(BG) \to C_*(BP)$ (the transfer hom), with the corresponding maps in group cohomology $p : H^*(G) \to H^*(P)$ and $\tau : H^*(P) \to H^*(G)$, the restriction and corestriction respectively.
$\tau \circ p$ is multiplication by $|G : P|$, so if I work with $\Bbb F_p$ coefficients that's an injection. So $H^*(G)$ injects into $H^*(P)$. I should be able to say more, right? If $P$ is normal abelian, it should be an isomorphism. There might be easier arguments, but this is what pops to mind first:
By Schur-Zassenhaus theorem, $G = P \rtimes G/P$ and $G/P$ acts trivially on $P$ (the action is by inner auts, and $P$ doesn't have any), there is a fibration $BP \to BG \to B(G/P)$ whose monodromy is exactly this action induced on $H^*(P)$, which is trivial, so we run the Lyndon-Hochschild-Serre spectral sequence with coefficients in $\Bbb F_p$.
The $E^2$ page is essentially zero except the bottom row since $H^*(G/P; M) = 0$ if $M$ is an $\Bbb F_p$-module by order reasons and the whole bottom row is $H^*(P; \Bbb F_p)$. This means the spectral sequence degenerates at $E^2$, which gets us $H^*(G; \Bbb F_p) \cong H^*(P; \Bbb F_p)$.
@Secret that's a very lazy habit you should create a chat room for every purpose you can imagine take full advantage of the websites functionality as I do and leave the general purpose room for recommending art related to mathematics
@MatheinBoulomenos No worries, thanks in advance. Just to add the final punchline, what I wanted to ask is what's the general algorithm to recover $H^*(G)$ back from $H^*(P; \Bbb F_p)$'s where $P$ runs over Sylow $p$-subgroups of $G$?
Bacterial growth is the asexual reproduction, or cell division, of a bacterium into two daughter cells, in a process called binary fission. Providing no mutational event occurs, the resulting daughter cells are genetically identical to the original cell. Hence, bacterial growth occurs. Both daughter cells from the division do not necessarily survive. However, if the number surviving exceeds unity on average, the bacterial population undergoes exponential growth. The measurement of an exponential bacterial growth curve in batch culture was traditionally a part of the training of all microbiologists...
As a result, there does not exists a single group which lived long enough to belong to, and hence one continue to search for new group and activity
eventually, a social heat death occurred, where no groups will generate creativity and other activity anymore
Had this kind of thought when I noticed how many forums etc. have a golden age, and then died away, and at the more personal level, all people who first knew me generate a lot of activity, and then destined to die away and distant roughly every 3 years
Well i guess the lesson you need to learn here champ is online interaction isn't something that was inbuilt into the human emotional psyche in any natural sense, and maybe it's time you saw the value in saying hello to your next door neighbour
Or more likely, we will need to start recognising machines as a new species and interact with them accordingly
so covert operations AI may still exists, even as domestic AIs continue to become widespread
It seems more likely sentient AI will take similar roles as humans, and then humans will need to either keep up with them with cybernetics, or be eliminated by evolutionary forces
But neuroscientists and AI researchers speculate it is more likely that the two types of races are so different we end up complementing each other
that is, until their processing power become so strong that they can outdo human thinking
But, I am not worried of that scenario, because if the next step is a sentient AI evolution, then humans would know they will have to give way
However, the major issue right now in the AI industry is not we will be replaced by machines, but that we are making machines quite widespread without really understanding how they work, and they are still not reliable enough given the mistakes they still make by them and their human owners
That is, we have became over reliant on AI, and not putting enough attention on whether they have interpret the instructions correctly
That's an extraordinary amount of unreferenced rhetoric statements i could find anywhere on the internet! When my mother disapproves of my proposals for subjects of discussion, she prefers to simply hold up her hand in the air in my direction
for example i tried to explain to her that my inner heart chakras tell me that my spirit guide suggests that many females i have intercourse with are easily replaceable and this can be proven from historical statistical data, but she wont even let my spirit guide elaborate on that premise
i feel as if its an injustice to all child mans that have a compulsive need to lie to shallow women they meet and keep up a farce that they are either fully grown men (if sober) or an incredibly wealthy trust fund kid (if drunk) that's an important binary class dismissed
Chatroom troll: A person who types messages in a chatroom with the sole purpose to confuse or annoy.
I was just genuinely curious
How does a message like this come from someone who isn't trolling:
"for example i tried to explain to her that my inner heart chakras tell me that my spirit guide suggests that many ... with are easily replaceable and this can be proven from historical statistical data, but she wont even let my spirit guide elaborate on that premise"
3
Anyway feel free to continue, it just seems strange @Adam
I'm genuinely curious what makes you annoyed or confused yes I was joking in the line that you referenced but surely you cant assume me to be a simpleton of one definitive purpose that drives me each time I interact with another person? Does your mood or experiences vary from day to day? Mine too! so there may be particular moments that I fit your declared description, but only a simpleton would assume that to be the one and only facet of another's character wouldn't you agree?
So, there are some weakened forms of associativity. Such as flexibility ($(xy)x=x(yx)$) or "alternativity" ($(xy)x=x(yy)$, iirc). Tough, is there a place a person could look for an exploration of the way these properties inform the nature of the operation? (In particular, I'm trying to get a sense of how a "strictly flexible" operation would behave. Ie $a(bc)=(ab)c\iff a=c$)
@RyanUnger You're the guy to ask for this sort of thing I think:
If I want to, by hand, compute $\langle R(\partial_1,\partial_2)\partial_2,\partial_1\rangle$, then I just want to expand out $R(\partial_1,\partial_2)\partial_2$ in terms of the connection, then use linearity of $\langle -,-\rangle$ and then use Koszul's formula? Or there is a smarter way?
I realized today that the possible x inputs to Round(x^(1/2)) covers x^(1/2+epsilon). In other words we can always find an epsilon (small enough) such that x^(1/2) <> x^(1/2+epsilon) but at the same time have Round(x^(1/2))=Round(x^(1/2+epsilon)). Am I right?
We have the following Simpson method $$y^{n+2}-y^n=\frac{h}{3}\left (f^{n+2}+4f^{n+1}+f^n\right ), n=0, \ldots , N-2 \\ y^0, y^1 \text{ given } $$ Show that the method is implicit and state the stability definition of that method.
How can we show that the method is implicit? Do we have to try to solve $y^{n+2}$ as a function of $y^{n+1}$ ?
@anakhro an energy function of a graph is something studied in spectral graph theory. You set up an adjacency matrix for the graph, find the corresponding eigenvalues of the matrix and then sum the absolute values of the eigenvalues. The energy function of the graph is defined for simple graphs by this summation of the absolute values of the eigenvalues |
Revista Matemática Iberoamericana
Full-Text PDF (334 KB) | Metadata | Table of Contents | RMI summary
Volume 29, Issue 1, 2013, pp. 91–113 DOI: 10.4171/RMI/714
Published online: 2013-01-14
Monotonicity and comparison results for conformal invariantsAlbert Baernstein II
[1]and Alexander Yu. Solynin [2](1) Washington University, St. Louis, USA
(2) Texas Tech University, Lubbock, USA
Let $a_1,\dots,a_N$ be points on the unit circle $\mathbb{T}$ with $a_j=e^{i\theta_j}$, where $0=\theta_1\le\theta_2\le\dots\le \theta_N=2\pi$. Let $\Omega=\overline{\mathbb{C}}\setminus\{a_1,\dots,a_N\}$ and let $\Omega^*$ be $\overline{\mathbb{C}}$ with the $n$-th roots of unity removed. The maximal gap $\delta(\Omega)$ of $\Omega$ is defined by $\delta(\Omega)=\max\{\theta_{j+1}-\theta_j:\,0\le j\le N-1\}$. How should one choose $a_1,\dots,a_N$ subject to the condition $\delta(\Omega)\le 2\pi/n$ so that the Poincaré metric $\lambda_\Omega(0)$ of $\Omega$ at the origin is as small as possible? In this paper we answer this question by showing that $\lambda_\Omega(0)$ is minimal when $\Omega=\Omega^*$. Several similar problems on the extremal values of the harmonic measures and capacities are also discussed.
Keywords: Comparison theorem, hyperbolic metric, harmonic measure, capacity
Baernstein II Albert, Solynin Alexander: Monotonicity and comparison results for conformal invariants.
Rev. Mat. Iberoam. 29 (2013), 91-113. doi: 10.4171/RMI/714 |
I’d like to write about the situation that occurs in set theory when a forcing extension $V[G]=V[H]$ arises over a ground model $V$ in two different ways simultaneously, using generic filters over two different forcing notions $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$. The general fact, stated in theorem 1, is that in this case, the two forcing notions are actually isomorphic on a cone $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, with the isomorphism carrying the one generic filter to the other. In other words, below these respective conditions $b$ and $c$, the forcing notions and the respective generic filters are not actually different.
I have always assumed that this fact was part of the classical forcing folklore results, but
it doesn’t seem to be mentioned explicitly in the usual forcing literature (it appears as lemma 25.5 in Jech’s book), and so I am writing an account of it here. Victoria Gitman and I have need of it in a current joint project. (Bob Solovay mentions in the comments below that the result is due to him, and provides a possible 1975 reference.) Theorem 1. If $V[G]=V[H]$, where $G\subset \mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{B}\upharpoonright b$ is isomorphic to $\mathbb{C}\upharpoonright c$ by an isomorphism carrying $G$ to $H$.
The proof will also establish the following related result, concerning the situation where one extension is merely contained in the other.
Theorem 2. If $V[H]\subset V[G]$, where $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{C}\upharpoonright c$ is isomorphic to a complete subalgebra of $\mathbb{B}\upharpoonright b$.
By $\mathbb{B}\upharpoonright b$, where $b$ is a condition in $\mathbb{B}$ (that is, a nonzero element of $\mathbb{B}$), what I mean is the Boolean algebra consisting of the interval $[0,b]$ in $\mathbb{B}$, using relative complement $b-a$ as the negation of $a$. This is the complete Boolean algebra that arises when forcing with the conditions in $\mathbb{B}$ below $b$.
Proof: In order to prove theorem 2, let me assume at first only that $V[H]\subset V[G]$. It follows that $H=\dot H_G$ for some $\mathbb{B}$-name $\dot H$, and we may choose a condition $b\in G$ forcing that $\dot H$ is a $\check V$-generic filter on $\check{\mathbb{C}}$.
I claim that there is some $c\in H$ such that every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$. Note that every $d\in H$ has $[\![\check d\in\dot H]\!]\in G$ by the truth lemma, since $H=\dot H_G$, and so $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ for $d\in H$. If $c\in H$ forces that every $d$ in the generic filter has that property, then indeed every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ as claimed.
In other words, from the perspective of the $\mathbb{B}$ forcing, every $d\leq c$ has a nonzero possibility to be in $\dot H$.
Define $\pi:\mathbb{C}\upharpoonright c\to\mathbb{B}$ by $$\pi(d)=b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}.$$ Using the fact that $b$ forces that $\dot H$ is a filter, it is straightforward to verify that
$d\leq e\implies \pi(d)\leq\pi(e)$, since if $d\leq e$ and $d\in H$, then $e\in H$. $\pi(d\wedge e)=\pi(d)\wedge \pi(e)$, since $[\![\check d\in\dot H]\!]\wedge[\![\check e\in \dot H]\!]=[\![\check{(b\wedge e)}\in\dot H]\!]$. $\pi(d-e)=\pi(d)-\pi(e)$, since $[\![\check{(d-e)}\in\dot H]\!]=[\![\check d\in\dot H]\!]-[\![\check e\in\dot H]\!]$.
Thus, $\pi$ is a Boolean algebra embedding of $\mathbb{C}\upharpoonright c$ into $\mathbb{B}\upharpoonright\pi(c)$.
Let me argue that this embedding is a complete embedding. Suppose that $a=\bigvee A$ for some subset $A\subset\mathbb{C}\upharpoonright c$ with $A\in V$. Since $H$ is $V$-generic, it follows that $a\in H$ just in case $H$ meets $A$. Thus, $[\![\check a\in\dot H]\!]=[\![\exists x\in\check A\, x\in \dot H]\!]=\bigvee_{x\in A}[\![\check x\in\dot H]\!]$, and so $\pi(\bigvee A)=\bigvee_{x\in A}\pi(x)$, and so $\pi$ is complete, as desired. This proves theorem 2.
To prove theorem 1, let me now assume fully that $V[G]=V[H]$. In this case, there is a $\mathbb{C}$ name $\dot G$ for which $G=\dot G_H$. By strengthening $b$, we may assume without loss that $b$ also forces that, that is, that $b$ forces $\Gamma=\check{\dot G}_{\dot H}$, where $\Gamma$ is the canonical $\mathbb{B}$-name for the generic object, and $\check{\dot G}$ is the $\mathbb{B}$-name of the $\mathbb{C}$-name $\dot G$. Let us also strengthen $c$ to ensure that $c$ forces $\dot G$ is $\check V$-generic for $\check{\mathbb{C}}$. For $d\leq c$ define $\pi(d)=[\![\check d\in\dot H]\!]^{\mathbb{B}}$ as above, which provides a complete embedding of $\mathbb{C}\upharpoonright c$ to $\mathbb{B}\upharpoonright\pi(c)$. I shall now argue that this embedding is dense below $\pi(c)$. Suppose that $a\leq \pi(c)$ in $\mathbb{B}$. Since $a$ forces $\check a\in\Gamma$ and also $\check c\in\dot H$, it must also force that there is some $d\leq c$ in $\dot H$ that forces via $\mathbb{C}$ over $\check V$ that $\check a\in\dot G$. So there must really be some $d\leq c$ forcing $\check a\in\dot G$. So $\pi(d)$, which forces $\check d\in\dot H$, will also force $\check a\in\check{\dot G}_{\dot H}=\Gamma$, and so $\pi(d)\Vdash_{\mathbb{B}}\check a\in\Gamma$, which means $\pi(d)\leq a$ in ${\mathbb{B}}$. Thus, the range of $\pi$ on $\mathbb{C}\upharpoonright c$ is dense below $\pi(c)$, and so $\pi$ is a complete dense embedding of ${\mathbb{C}}\upharpoonright c$ to ${\mathbb{B}}\upharpoonright \pi(c)$. Since these are complete Boolean algebras, this means that $\pi$ is actually an isomorphism of $\mathbb{C}\upharpoonright c$ with $\mathbb{B}\upharpoonright \pi(c)$, as desired.
Finally, note that if $d\in H$ below $c$, then since $H=\dot H_G$, it follows that $[\![\check d\in\dot H]\!]\in G$, which is to say $\pi(d)\in G$, and so $\pi$ carries $H$ to $G$ on these cones. So $\pi^{-1}$ is the isomorphism stated in theorem 1.
QED
Finally, I note that one cannot get rid of the need to restrict to cones, since it could be that $\mathbb{B}$ and $\mathbb{C}$ are the lottery sums of a common forcing notion, giving rise to $V[G]=V[H]$, together with totally different non-isomorphic forcing notions below some other incompatible conditions. So we cannot expect to prove that $\mathbb{B}\cong\mathbb{C}$, and are content to get merely that $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, an isomorphism below respective conditions. |
Background
For a system consisting of two molecules (monomers or fragments are also used) X and Y, the binding energy is
$$\Delta E_{\text{bind}} = E^{\ce{XY}}(\ce{XY}) - [E^{\ce{X}}(\ce{X}) + E^{\ce{Y}}(\ce{Y})]\label{eq:sherrill-1} \tag{Sherrill 1}$$
where the letters in the parentheses refer to the atoms present in the calculation and the letters in the superscript refer to the (atomic orbital, AO) basis present in the calculation. The first term is the energy calculated for the combined X + Y complex (the dimer) with basis functions, and the next two terms are energy calculations for each isolated monomer with only their respective basis functions. The remainder of this discussion will make more sense if the complex geometry is used for each monomer, rather than the isolated fragment geometry.
The counterpoise-corrected (CP-corrected) binding energy [1] to correct for basis set superposition error (BSSE) [2] is defined as
$$\Delta E_{\text{bind}}^{\text{CP}} = E^{\ce{XY}}(\ce{XY}) - [E^{\ce{XY}}(\ce{X}) + E^{\ce{XY}}(\ce{Y})]\label{eq:sherrill-3} \tag{Sherrill 3}$$
where the monomer calculations are now performed in the dimer/complex basis. Let's explicitly state how this works for the $E_{\ce{XY}}(\ce{X})$ term. The first molecule X contributes nuclei with charges, basis functions (AOs) centered on those nuclei, and electrons that will count to the final occupied molecular orbital (MO) index into the MO coefficient array. There is no reason why additional AOs that are not centered on atoms can't be added to a calculation. Depending on their spatial location, if they're close enough to have significant overlap, they may combine with atom-centered MOs, increasing the variational flexibility of the calculation and lowering the overall energy. Put another way, place the AOs that would correspond to molecule Y at their correct positions, but don't put the nuclei there, and don't consider the number of electrons they would contribute to the total number of occupied orbitals. This means that for the full electronic Hamiltonian
$$\hat{H}_{\text{elec}} = \hat{T}_{e} + \hat{V}_{eN} + \hat{V}_{ee}$$
calculating the electron-nuclear attraction $\hat{V}_{eN}$ term is now different. Considered explicitly in matrix form in the AO basis,
$$\begin{align*}V_{\mu\nu} &= \int \mathop{d\mathbf{r}_{i}} \chi_{\mu}(\mathbf{r}_{i}) \left( \sum_{A}^{N_{\text{atoms}}} \frac{Z_{A}}{|\mathbf{r}_{i} - \mathbf{R}_{A}|} \right) \chi_{\nu}(\mathbf{r}_{i}) \\&=\sum_{A}^{N_{\text{atoms}}} Z_{A} \left< \chi_{\mu} \middle| \frac{1}{r_{A}} \middle| \chi_{\nu} \right>\end{align*}$$
there are now fewer terms in the summation, since the nuclear charges from molecule Y are zero (the atoms just aren't there), but the number of $\mu\nu$ are the same as for the XY complex. This and the $\hat{T}_{e}, \hat{V}_{ee}$ terms aren't really mathematically or functionally different then; this is more to show where the additional basis functions enter, or to show where nuclei appear in the equations [3].
These atoms that don't have nuclei or electrons, only basis functions, are called
ghost atoms. Sometimes you also see the term ghost functions, ghost basis, or ghost {something} calculation. Adding the basis of monomer Y to make the full "dimer basis" means taking the monomer X and including basis functions at the nuclear positions for Y. Geometry optimization
Now to calculate the molecular gradient, that is, the derivative of the energy with respect to the $3N$ nuclear coordinates. This is the central quantity in any geometry optimization. For the sake of simplicity, consider a steepest descent-type update of the nuclear coordinates$$R_{A,x}^{(n+1)} = R_{A,x}^{(n)} - \alpha \frac{\partial E_{\text{total}}^{(n)}}{\partial R_{A,x}}\label{eq:steepest-descent} \tag{Steepest Descent}$$where $n$ is the optimization iteration number, $\alpha$ is some small step size with units [length
2][energy], and the last term is the derivative of the total (not just electronic) energy with respect to a change in atom $A$'s $x$-coordinate. Even Newton-Raphson-type updates with approximate Hessians (2nd derivative of the energy with respect to nuclear coordinates, rather than the 1st) need the gradient, so we must formulate it. Formulation of the energy
We're in a bit of trouble, because we want to replace $E_{\text{total}}$ in the gradient with $E_{\text{total}}^{\text{CP}}$, but all we have is $\Delta E_{\text{bind}}^{\text{CP}}$. The concept of CP correction can still be applied to a total energy, but the BSSE must be removed from each monomer. The BSSE correction itself for each monomer is$$\begin{split}E_{\text{BSSE}}(\ce{X}) &= E^{\ce{XY}}(\ce{X}) - E^{\ce{X}}(\ce{X}), \\E_{\text{BSSE}}(\ce{Y}) &= E^{\ce{XY}}(\ce{Y}) - E^{\ce{Y}}(\ce{Y}),\end{split}\label{eq:2}$$which, when subtracted from $\eqref{eq:sherrill-1}$, gives $\eqref{eq:sherrill-3}$. More correctly, considering that the geometry for each step is at the final cluster geometry and not the isolated geometry, the above is [4]$$\begin{split}E_{\text{BSSE}}(\ce{X}) &= E_{\ce{XY}}^{\ce{XY}}(\ce{X}) - E_{\ce{XY}}^{\ce{X}}(\ce{X}), \\E_{\text{BSSE}}(\ce{Y}) &= E_{\ce{XY}}^{\ce{XY}}(\ce{Y}) - E_{\ce{XY}}^{\ce{Y}}(\ce{Y}).\end{split}\label{eq:sherrill-10} \tag{Sherrill 10}$$
The CP-corrected
total energy is the full dimer energy with BSSE removed from each monomer is then$$\begin{split}E_{\text{tot}, \ce{\widetilde{XY}}}^{\text{CP}} &= E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY}) - E_{\text{BSSE}}(\ce{X}) - E_{\text{BSSE}}(\ce{Y}), \\&= E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY}) - \left[ E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{X}) - E_{\ce{\widetilde{XY}}}^{\ce{X}}(\ce{X}) \right] - \left[ E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{Y}) - E_{\ce{\widetilde{XY}}}^{\ce{Y}}(\ce{Y}) \right].\end{split}\label{eq:sherrill-15} \tag{Sherrill 15}$$Note that I have modified which geometry is used for each monomer in $\eqref{eq:sherrill-15}$. All monomers are calculated at the supermolecule geometry. This is convenient for two reasons: 1. We are only interested in removing the BSSE, not the effect of monomer deformation, and 2. a isolated monomer geometry without deformation doesn't make sense in the context of a geometry optimization. I also added the tilde to signify that the supermolecular/dimer geometry used may not be the final or minimum-energy geometry, as would be the case during a geometry optimization. We simply extract all structures consistently from a given geometry iteration. Perhaps $\ce{XY}(n)$ would be better notation. Formulation of the gradient
As Pedro correctly states, the differentiation operator is a linear operator. Because there are no products in $\eqref{eq:sherrill-15}$, the total gradient needed for $\eqref{eq:steepest-descent}$ will be a sum of gradients [5]:$$\frac{\partial E_{\text{tot}, \ce{\widetilde{XY}}}^{\text{CP}}}{\partial R_{A,x}} = \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY})}{\partial R_{A,x}} - \left[ \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{X})}{\partial R_{A,x}} - \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{X}}(\ce{X})}{\partial R_{A,x}} \right] - \left[ \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{Y})}{\partial R_{A,x}} - \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{Y}}(\ce{Y})}{\partial R_{A,x}} \right],$$so each step of a CP-corrected geometry optimization will require 5 gradient calculations rather than 1. Note that the nuclear gradient should be included for each term as well, which is a trivial calculation.
Extension to other molecular properties
Although not commonly done, counterpoise correction can be applied to any molecular property, not just energies or gradients. Simply replace $E$ or $\partial E/\partial R$ with the property of interest. For example, the CP-corrected polarizability $\alpha$ of two fragments is$$\alpha_{\text{tot}, \ce{\widetilde{XY}}}^{\text{CP}} = \alpha_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY}) - \left[ \alpha_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{X}) - \alpha_{\ce{\widetilde{XY}}}^{\ce{X}}(\ce{X}) \right] - \left[ \alpha_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{Y}) - \alpha_{\ce{\widetilde{XY}}}^{\ce{Y}}(\ce{Y}) \right]$$where I believe it now makes even less sense to have each individual fragment calculation not be at the cluster geometry. In papers that calculate CP-corrected properties, no mention is usually made of which geometry the individual calculations are performed at for this reason.
References Boys, S. Francis; Bernardi, F. The calculation of small molecular interactions by the differences of separate total energies. Some procedures with reduced errors. Mol. Phys. 1970, 19, 553-566. Sherrill, C. David. Counterpoise Correction and Basis Set Superposition Error. 2010, 1-6. One implementation note: Most common quantum chemistry packages should allow for the usage of ghost atoms in energy and gradient calculations. However, as Sherrill states, they do not properly allow for composing the full gradient expression to perform CP-corrected geometry optimizations. Gaussian can, and Psi4 may. For programs that can calculate gradients with ghost atoms, Cuby can be used to drive the calculation of CP-corrected geometries and frequencies. There is a typo in the Sherrill paper; the subscripts for all 4 energy terms should be $AB$, which here are $\ce{XY}$. Simon, S.; Bertran, J.; Sodupe, M. Effect of Counterpoise Correction on the Geometries and Vibrational Frequencies of Hydrogen Bonded Systems. J. Chem. Phys. A 2001, 105,, 4359-4364. |
Given a regular language $L$, then it is easy to prove that there is a constant $N$ such that is $\sigma \in L$, with $\lvert \sigma \rvert \ge N$ there exist strings $\alpha$, $\beta$ and $\gamma$ such that $\lvert \alpha \beta \rvert \le N$ and $\lvert \beta \rvert \ne \epsilon$, and for all $k$ it is $\alpha \beta^k \gamma \in L$. It is widely stated that the converse isn't true, but I haven't seen any clear example. Any suggestions? Clearly the proof that the offending language isn't regular has to use stronger methods than the typical "doesn't satisfy the pumping lemma". I'd be interested in simple examples, to present in introductory formal languages classes.
The language $\{ \$ a^nb^n \mid n \ge 1 \} \cup \{ \$^kw \mid k\neq 1, w\in \{a,b\}^* \}$ seems to be simple. The second part is regular (and can be pumped). The first part is nonregular, but can be pumped "into" the second part by choosing $\$$ to pump.
(added) Of course, this can be generalized to $\$L \cup \{ \$^k \mid k\neq 1 \} \cdot \{a,b\}^*$ for
any $L\subseteq \{a,b\}^*$. Sometimes the formulation is in the "if ... then ..." style: if $w$ starts with a single $\$$ then it is of the form. That I personally find less intuitive.
As noted by @vonbrand the (possibly) non-regular part of the language is isolated by intersecting with $\$\{a,b\}^*$. This can be separately tested using the pumping lemma if needed. |
Setup (complete, but all very standard):
My problem is how to best calculate the cumulative integral of a function which comes out of Spectral Collocation with a chebyshev basis. Take some function $f : [0, \bar{z}] \to \mathbb{R}$ approximated by $$ f(z) \approx \sum_{n=0}^{N-1}d_n T_n(z),\, z \in [0, \bar{z}] $$ where $T_n(z)$ is a basis of Chebyshev polynomials adapted to the $[0, \bar{z}]$ domain. Denote the vectors of coefficients as $d\in \mathbb{R}^N$. Calculate the Chebyshev polynomial roots (adapted to the $[0, \bar{z}]$ domain) and define them as, $$ \vec{z}_{\mathrm{int}} \equiv \{z_1,\ldots z_{N}\}\in \mathbb{R}^{N} $$
And the complete set of nodes including boundary values as (with $z_0 \equiv 0$ and $z_{N+1} \equiv \bar{z}$ as $\vec{z}\equiv \{0, z_1,\ldots z_N, \bar{z}\}\in \mathbb{R}^{N+2}$ Now, define the basis matrices as \begin{align} B &\equiv \begin{bmatrix} T_0(z_0)& \ldots & T_{N-1}(z_0)\\ \ldots & & \ldots\\ T_0(z_{N+1}) & \ldots & T_{N-1}(z_{N+1}) \end{bmatrix} \in \mathbb{R}^{(N+2)\times N}\\ B' &\equiv \begin{bmatrix} T_0'(z_0)& \ldots & T'_{N-1}(z_0)\\ \ldots & & \ldots\\ T_0'(z_{N+1}) & \ldots & T_{N-1}'(z_{N+1}) \end{bmatrix} \in \mathbb{R}^{(N+2)\times N} \end{align} Then, given the coefficient matrix $d$, you can find $f$ or the derivative $f'$ at every point in $\vec{z}$ with \begin{align} \vec{f} &\equiv \{f(z_n)\}_{n=0}^{N+1} = B \cdot d\in \mathbb{R}^{N+2}\\ \vec{f}' &\equiv \{f'(z_n)\}_{n=0}^{N+1} = B' \cdot d\in \mathbb{R}^{N+2}\\ \end{align} with the function at the interior nodes as $\vec{f}_{\mathrm{int}} \equiv \vec{f}(1:N) \in \mathbb{R}^N$.
Finally, we can find the weighting vector for Chebyshev-Gauss quadrature (https://en.wikipedia.org/wiki/Gaussian_quadrature) on the exact same roots $\vec{z}_{\mathrm{int}}$, and call it $\omega \in \mathbb{R}^N$. With this, we can approximate integrals for some $g(z)$ with $g(\vec{z}_{\mathrm{int}}) \equiv \{g(z) | z \in \vec{z}_{\mathrm{int}}\}$ $$ \int_0^{\bar{z}} g(z) d z \approx \omega \cdot g(\vec{z}_{\mathrm{int}}) $$ (note that this quadrature scheme does not use the endpoints).
My Problem:Define the cumulative function,$$F(z) \equiv \int_0^z f(\tilde{z}) d\tilde{z}$$If all I cared about was $F(\bar{z})$, then I have a nice approximation,$$F(\bar{z}) \approx \omega \cdot \vec{f}_{\mathrm{int}}$$But what an approximation for the cumulative integral, $F(z)$ for all $z \in \vec{z}$ given only the above? Current Solution:Note that as I am not able to evaluate $f(z)$ at other points, I cannot naively use Gauss-Chebyshev quadrature as the quadrature roots depend on the domain of integration. My, grossly imprecise, method is to use trapezoidal integration at the unevenly spaced chebyshev roots. Or,$$\Delta \vec{z} = \{\vec{z}(n) - \vec{z}(n-1)\}_{n=1}^{N+1}$$Then, to calculate the integral to one of the nodes,$$\int_0^{z_{n-1}}f(\hat{z}) d \hat{z} \approx \frac{1}{2}\Delta \vec{z}(0:n) \cdot (\vec{f}(0:n-1) + \vec{f}(1:n))$$I can even get fancier and calculate this all in one step,
\begin{align} \Omega \equiv \frac{1}{2}\begin{bmatrix}0 &0 & 0 &0 & \ldots & 0 \\ \Delta \vec{z}(0:1) & & 0 & 0 & \ldots & 0 \\ \Delta \vec{z}(0:2) & & & 0 & \ldots & 0 \\ \ldots & & & & & \ldots \\ \Delta \vec{z}(0:N)& & & & & \end{bmatrix}\in \mathbb{R}^{(N+2)\times (N+1)} \end{align}
Which gives the complete set of integrals as, $$ \vec{F} =\Omega \cdot (\vec{f}(0:N) + \vec{f}(1:N+1))\in\mathbb{R}^{N+2} $$ This is especially useful for me because I am solving a spectral collocation method, so having auto-differentiation within the calculation of the residual makes the problem solvable.
Are there better approaches?I am hoping that there is a more precise way of calculating these partial integrals. For example, I found https://math.stackexchange.com/questions/344073/integrating-non-uniform-grid-data-from-an-accelerometer which gives some ideas for a non-uniform Simpson's Rule, but it seems intractable to get a quadratic-form anything like my $\Omega$ above. Has anyone done that work, or is there a method I am forgetting about? |
I’d like to write about the situation that occurs in set theory when a forcing extension $V[G]=V[H]$ arises over a ground model $V$ in two different ways simultaneously, using generic filters over two different forcing notions $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$. The general fact, stated in theorem 1, is that in this case, the two forcing notions are actually isomorphic on a cone $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, with the isomorphism carrying the one generic filter to the other. In other words, below these respective conditions $b$ and $c$, the forcing notions and the respective generic filters are not actually different.
I have always assumed that this fact was part of the classical forcing folklore results, but
it doesn’t seem to be mentioned explicitly in the usual forcing literature (it appears as lemma 25.5 in Jech’s book), and so I am writing an account of it here. Victoria Gitman and I have need of it in a current joint project. (Bob Solovay mentions in the comments below that the result is due to him, and provides a possible 1975 reference.) Theorem 1. If $V[G]=V[H]$, where $G\subset \mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{B}\upharpoonright b$ is isomorphic to $\mathbb{C}\upharpoonright c$ by an isomorphism carrying $G$ to $H$.
The proof will also establish the following related result, concerning the situation where one extension is merely contained in the other.
Theorem 2. If $V[H]\subset V[G]$, where $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{C}\upharpoonright c$ is isomorphic to a complete subalgebra of $\mathbb{B}\upharpoonright b$.
By $\mathbb{B}\upharpoonright b$, where $b$ is a condition in $\mathbb{B}$ (that is, a nonzero element of $\mathbb{B}$), what I mean is the Boolean algebra consisting of the interval $[0,b]$ in $\mathbb{B}$, using relative complement $b-a$ as the negation of $a$. This is the complete Boolean algebra that arises when forcing with the conditions in $\mathbb{B}$ below $b$.
Proof: In order to prove theorem 2, let me assume at first only that $V[H]\subset V[G]$. It follows that $H=\dot H_G$ for some $\mathbb{B}$-name $\dot H$, and we may choose a condition $b\in G$ forcing that $\dot H$ is a $\check V$-generic filter on $\check{\mathbb{C}}$.
I claim that there is some $c\in H$ such that every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$. Note that every $d\in H$ has $[\![\check d\in\dot H]\!]\in G$ by the truth lemma, since $H=\dot H_G$, and so $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ for $d\in H$. If $c\in H$ forces that every $d$ in the generic filter has that property, then indeed every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ as claimed.
In other words, from the perspective of the $\mathbb{B}$ forcing, every $d\leq c$ has a nonzero possibility to be in $\dot H$.
Define $\pi:\mathbb{C}\upharpoonright c\to\mathbb{B}$ by $$\pi(d)=b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}.$$ Using the fact that $b$ forces that $\dot H$ is a filter, it is straightforward to verify that
$d\leq e\implies \pi(d)\leq\pi(e)$, since if $d\leq e$ and $d\in H$, then $e\in H$. $\pi(d\wedge e)=\pi(d)\wedge \pi(e)$, since $[\![\check d\in\dot H]\!]\wedge[\![\check e\in \dot H]\!]=[\![\check{(b\wedge e)}\in\dot H]\!]$. $\pi(d-e)=\pi(d)-\pi(e)$, since $[\![\check{(d-e)}\in\dot H]\!]=[\![\check d\in\dot H]\!]-[\![\check e\in\dot H]\!]$.
Thus, $\pi$ is a Boolean algebra embedding of $\mathbb{C}\upharpoonright c$ into $\mathbb{B}\upharpoonright\pi(c)$.
Let me argue that this embedding is a complete embedding. Suppose that $a=\bigvee A$ for some subset $A\subset\mathbb{C}\upharpoonright c$ with $A\in V$. Since $H$ is $V$-generic, it follows that $a\in H$ just in case $H$ meets $A$. Thus, $[\![\check a\in\dot H]\!]=[\![\exists x\in\check A\, x\in \dot H]\!]=\bigvee_{x\in A}[\![\check x\in\dot H]\!]$, and so $\pi(\bigvee A)=\bigvee_{x\in A}\pi(x)$, and so $\pi$ is complete, as desired. This proves theorem 2.
To prove theorem 1, let me now assume fully that $V[G]=V[H]$. In this case, there is a $\mathbb{C}$ name $\dot G$ for which $G=\dot G_H$. By strengthening $b$, we may assume without loss that $b$ also forces that, that is, that $b$ forces $\Gamma=\check{\dot G}_{\dot H}$, where $\Gamma$ is the canonical $\mathbb{B}$-name for the generic object, and $\check{\dot G}$ is the $\mathbb{B}$-name of the $\mathbb{C}$-name $\dot G$. Let us also strengthen $c$ to ensure that $c$ forces $\dot G$ is $\check V$-generic for $\check{\mathbb{C}}$. For $d\leq c$ define $\pi(d)=[\![\check d\in\dot H]\!]^{\mathbb{B}}$ as above, which provides a complete embedding of $\mathbb{C}\upharpoonright c$ to $\mathbb{B}\upharpoonright\pi(c)$. I shall now argue that this embedding is dense below $\pi(c)$. Suppose that $a\leq \pi(c)$ in $\mathbb{B}$. Since $a$ forces $\check a\in\Gamma$ and also $\check c\in\dot H$, it must also force that there is some $d\leq c$ in $\dot H$ that forces via $\mathbb{C}$ over $\check V$ that $\check a\in\dot G$. So there must really be some $d\leq c$ forcing $\check a\in\dot G$. So $\pi(d)$, which forces $\check d\in\dot H$, will also force $\check a\in\check{\dot G}_{\dot H}=\Gamma$, and so $\pi(d)\Vdash_{\mathbb{B}}\check a\in\Gamma$, which means $\pi(d)\leq a$ in ${\mathbb{B}}$. Thus, the range of $\pi$ on $\mathbb{C}\upharpoonright c$ is dense below $\pi(c)$, and so $\pi$ is a complete dense embedding of ${\mathbb{C}}\upharpoonright c$ to ${\mathbb{B}}\upharpoonright \pi(c)$. Since these are complete Boolean algebras, this means that $\pi$ is actually an isomorphism of $\mathbb{C}\upharpoonright c$ with $\mathbb{B}\upharpoonright \pi(c)$, as desired.
Finally, note that if $d\in H$ below $c$, then since $H=\dot H_G$, it follows that $[\![\check d\in\dot H]\!]\in G$, which is to say $\pi(d)\in G$, and so $\pi$ carries $H$ to $G$ on these cones. So $\pi^{-1}$ is the isomorphism stated in theorem 1.
QED
Finally, I note that one cannot get rid of the need to restrict to cones, since it could be that $\mathbb{B}$ and $\mathbb{C}$ are the lottery sums of a common forcing notion, giving rise to $V[G]=V[H]$, together with totally different non-isomorphic forcing notions below some other incompatible conditions. So we cannot expect to prove that $\mathbb{B}\cong\mathbb{C}$, and are content to get merely that $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, an isomorphism below respective conditions. |
Research talks;Number Theory
For a non-principal Dirichlet character $\chi$ modulo $q$, the classical Pólya-Vinogradov inequality asserts that
$M (\chi) := \underset{x}{max}$$| \sum_{n \leq x}$$\chi(n)| = O (\sqrt{q} log$ $q)$. This was improved to $\sqrt{q} log$ $log$ $q$ by Montgomery and Vaughan, assuming the Generalized Riemann hypothesis GRH. For quadratic characters, this is known to be optimal, owing to an unconditional omega result due to Paley. In this talk, we shall present recent results on higher order character sums. In the first part, we discuss even order characters, in which case we obtain optimal omega results for $M(\chi)$, extending and refining Paley's construction. The second part, joint with Alexander Mangerel, will be devoted to the more interesting case of odd order characters, where we build on previous works of Granville and Soundararajan and of Goldmakher to provide further improvements of the Pólya-Vinogradov and Montgomery-Vaughan bounds in this case. In particular, assuming GRH, we are able to determine the order of magnitude of the maximum of $M(\chi)$, when $\chi$ has odd order $g \geq 3$ and conductor $q$, up to a power of $log_4 q$ (where $log_4$ is the fourth iterated logarithm).
For a non-principal Dirichlet character $\chi$ modulo $q$, the classical Pólya-Vinogradov inequality asserts that $M (\chi) := \underset{x}{max}$$| \sum_{n \leq x}$$\chi(n)| = O (\sqrt{q} log$ $q)$. This was improved to $\sqrt{q} log$ $log$ $q$ by Montgomery and Vaughan, assuming the Generalized Riemann hypothesis GRH. For quadratic characters, this is known to be optimal, owing to an unconditional omega result due to Paley. In this talk, we ...
11L40 ; 11N37 ; 11N13 ; 11M06
... Lire [+] |
Convolutions from a DSP perspective
I'm a bit late to this but still would like to share my perspective and insights. My background is theoretical physics and digital signal processing. In particular I studied wavelets and convolutions are almost in my backbone ;)
The way people in the deep learning community talk about convolutions was also confusing to me. From my perspective what seems to be missing is a proper separation of concerns. I will explain the deep learning convolutions using some DSP tools.
Disclaimer
My explanations will be a bit hand-wavy and not mathematical rigorous in order to get the main points across.
Definitions
Let's define a few things first. I limit my discussion to one dimensional (the extension to more dimension is straight forward) infinite (so we don't need to mess with boundaries) sequences $x_n = \{x_n\}_{n=-\infty}^{\infty} = \{\dots, x_{-1}, x_{0}, x_{1}, \dots \}$.
A pure (discrete) convolution between two sequences $y_n$ and $x_n$ is defined as
$$ (y * x)_n = \sum_{k=-\infty}^{\infty} y_{n-k} x_k $$
If we write this in terms of matrix vector operations it looks like this (assuming a simple kernel $\mathbf{q} = (q_0,q_1,q_2)$ and vector $\mathbf{x} = (x_0, x_1, x_2, x_3)^T$):
$$ \mathbf{q} * \mathbf{x} = \left( \begin{array}{cccc} q_1 & q_0 & 0 & 0 \\ q_2 & q_1 & q_0 & 0 \\ 0 & q_2 & q_1 & q_0 \\ 0 & 0 & q_2 & q_1 \\ \end{array} \right) \left( \begin{array}{cccc} x_0 \\ x_1 \\ x_2 \\ x_3 \end{array} \right) $$
Let's introduce the down- and up-sampling operators, $\downarrow$ and $\uparrow$, respectively. Downsampling by factor $k \in \mathbb{N}$ is removing all samples except every k-th one:
$$ \downarrow_k\!x_n = x_{nk} $$
And upsampling by factor $k$ is interleaving $k-1$ zeros between the samples:
$$ \uparrow_k\!x_n = \left \{ \begin{array}{ll} x_{n/k} & n/k \in \mathbb{Z} \\ 0 & \text{otherwise} \end{array} \right.$$
E.g. we have for $k=3$:
$$ \downarrow_3\!\{ \dots, x_0, x_1, x_2, x_3, x_4, x_5, x_6, \dots \} = \{ \dots, x_0, x_3, x_6, \dots \} $$$$ \uparrow_3\!\{ \dots, x_0, x_1, x_2, \dots \} = \{ \dots x_0, 0, 0, x_1, 0, 0, x_2, 0, 0, \dots \} $$
or written in terms of matrix operations (here $k=2$):
$$ \downarrow_2\!x = \left( \begin{array}{cc} x_0 \\ x_2 \end{array} \right) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{cccc} x_0 \\ x_1 \\ x_2 \\ x_3 \end{array} \right) $$
and
$$ \uparrow_2\!x = \left( \begin{array}{cccc} x_0 \\ 0 \\ x_1 \\ 0 \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{array} \right) \left( \begin{array}{cc} x_0 \\ x_1 \end{array} \right) $$
As one can already see, the down- and up-sample operators are mutually transposed, i.e. $\uparrow_k = \downarrow_k^T$.
Deep Learning Convolutions by Parts
Let's look at the typical convolutions used in deep learning and how we write them. Given some kernel $\mathbf{q}$ and vector $\mathbf{x}$ we have the following:
a strided convolution with stride $k$ is $\downarrow_k\!(\mathbf{q} * \mathbf{x})$, a dilated convolution with factor $k$ is $(\uparrow_k\!\mathbf{q}) * \mathbf{x}$, a transposed convolution with stride $k$ is $ \mathbf{q} * (\uparrow_k\!\mathbf{x})$
Let's rearrange the transposed convolution a bit:$$ \mathbf{q} * (\uparrow_k\!\mathbf{x}) \; = \; \mathbf{q} * (\downarrow_k^T\!\mathbf{x}) \; = \; (\uparrow_k\!(\mathbf{q}*)^T)^T\mathbf{x}$$
In this notation $(\mathbf{q}*)$ must be read as an operator, i.e. it abstracts convolving something with kernel $\mathbf{q}$.Or written in matrix operations (example):
$$ \begin{align} \mathbf{q} * (\uparrow_k\!\mathbf{x}) & = \left( \begin{array}{cccc} q_1 & q_0 & 0 & 0 \\ q_2 & q_1 & q_0 & 0 \\ 0 & q_2 & q_1 & q_0 \\ 0 & 0 & q_2 & q_1 \\ \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{array} \right) \left( \begin{array}{c} x_0\\ x_1\\ \end{array} \right) \\ & = \left( \begin{array}{cccc} q_1 & q_2 & 0 & 0 \\ q_0 & q_1 & q_2 & 0 \\ 0 & q_0 & q_1 & q_2 \\ 0 & 0 & q_0 & q_1 \\ \end{array} \right)^T \left( \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ \end{array} \right)^T \left( \begin{array}{c} x_0\\ x_1\\ \end{array} \right) \\ & = \left( \left( \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ \end{array} \right) \left( \begin{array}{cccc} q_1 & q_2 & 0 & 0 \\ q_0 & q_1 & q_2 & 0 \\ 0 & q_0 & q_1 & q_2 \\ 0 & 0 & q_0 & q_1 \\ \end{array} \right) \right)^T \left( \begin{array}{c} x_0\\ x_1\\ \end{array} \right) \\ & = (\uparrow_k\!(\mathbf{q}*)^T)^T\mathbf{x} \end{align}$$
As one can see the is the transposed operation, thus, the name.
Connection to Nearest Neighbor Upsampling
Another common approach found in convolutional networks is upsampling with some built-in form of interpolation. Let's take upsampling by factor 2 with a simple repeat interpolation.This can be written as $\uparrow_2\!(1\;1) * \mathbf{x}$. If we also add a learnable kernel $\mathbf{q}$ to this we have $\uparrow_2\!(1\;1) * \mathbf{q} * \mathbf{x}$. The convolutions can be combined, e.g. for $\mathbf{q}=(q_0\;q_1\;q_2)$, we have $$(1\;1) * \mathbf{q} = (q_0\;\;q_0\!\!+\!q_1\;\;q_1\!\!+\!q_2\;\;q_2),$$
i.e. we can replace a repeat upsampler with factor 2 and a convolution with a kernel of size 3 by a transposed convolution with kernel size 4. This transposed convolution has the same "interpolation capacity" but would be able to learn better matching interpolations.
Conclusions and Final Remarks
I hope I could clarify some common convolutions found in deep learning a bit by taking them apart in the fundamental operations.
I didn't cover pooling here. But this is just a nonlinear downsampler and can be treated within this notation as well. |
How would i find the limit as $\lim\limits_{x\to3}\frac{4x(x-3)}{|x-3|}$? that is the absolute value of x-3 in the denominator. I thought my professor told my class that we were able to omit the absolute value sign for whatever reason. If that is true can you tell me why? Thanks!
You can find limits from each side. When taking the limit from the right, $x \gt 3$, so you can delete the absolute value signs. $$\lim_{x \to 3^+}\frac{4x(x-3)}{|x-3|}=\lim_{x \to 3^+}\frac{4x(x-3)}{x-3}=\lim_{x \to 3^+}4x=12$$ From the left, $x \lt 3$, so you must replace $|x-3|$ with $3-x$. $$\lim_{x \to 3^-}\frac{4x(x-3)}{|x-3|}=\lim_{x \to 3^-}\frac{4x(x-3)}{3-x}=\lim_{x \to 3^-}-4x=-12$$ As the left and right limits disagree, there is not a single limit.
I guess you have to be told about the right and left limits: $$ r = \lim\limits_{x\to3+0}\frac{4x(x-3)}{|x-3|} = \lim\limits_{x\to3+0}\frac{4x(x-3)}{x-3} = 12 $$ $$ l = \lim\limits_{x\to3-0}\frac{4x(x-3)}{|x-3|} = \lim\limits_{x\to3-0}\frac{4x(x-3)}{-(x-3)} = -12 $$
The limit exists iff $r=l$ and in that case it is equal to each of them. That's not your case though. |
Revista Matemática Iberoamericana
Full-Text PDF (498 KB) | Metadata | Table of Contents | RMI summary
Volume 29, Issue 1, 2013, pp. 183–236 DOI: 10.4171/RMI/718
Published online: 2013-01-14
Hardy spaces and regularity for the inhomogeneous Dirichlet and Neumann problemsXuan Thinh Duong
[1], Steve Hofmann [2], Dorina Mitrea [3], Marius Mitrea [4]and Lixin Yan [5](1) Macquarie University, Sydney, Australia
(2) University of Missouri, Columbia, USA
(3) University of Missouri, Columbia, United States
(4) University of Missouri, Columbia, USA
(5) Zhongshan University, Guangzhou, China
This article has three aims. First, we study Hardy spaces, $h^p_L(\Omega)$, associated with an operator $L$ which is either the Dirichlet Laplacian $\Delta_{D}$ or the Neumann Laplacian $\Delta_{N}$ on a bounded Lipschitz domain $\Omega$ in ${\mathbb{R}}^n$, for $0 < p \leq 1$. We obtain equivalent characterizations of these function spaces in terms of maximal functions and atomic decompositions. Second, we establish regularity results for the Green operators, regarded as the inverses of the Dirichlet and Neumann Laplacians, in the context of Hardy spaces associated with these operators on a bounded semiconvex domain $\Omega$ in ${\mathbb{R}}^n$. Third, we study relations between the Hardy spaces associated with operators and the standard Hardy spaces $h^p_r(\Omega)$ and $h^p_z(\Omega)$, then establish regularity of the Green operators for the Dirichlet problem on a bounded semiconvex domain $\Omega$ in ${\mathbb{R}}^n$, and for the Neumann problem on a bounded convex domain $\Omega$ in ${\mathbb{R}}^n$, in the context of the standard Hardy spaces $h^p_r(\Omega)$ and $h^p_z(\Omega)$. This gives a new solution to the conjecture made by D.-C. Chang, S. Krantz and E. M. Stein regarding the regularity of Green operators for the Dirichlet and Neumann problems on $h^p_r(\Omega)$ and $h^p_z(\Omega)$, respectively, for all $\frac{n}{n+1} < p\leq 1$.
Keywords: Hardy space, heat semigroup, atom, inhomogeneous Dirichlet and Neumann problems, Green operator, semiconvex domain, convex domain
Duong Xuan Thinh, Hofmann Steve, Mitrea Dorina, Mitrea Marius, Yan Lixin: Hardy spaces and regularity for the inhomogeneous Dirichlet and Neumann problems.
Rev. Mat. Iberoam. 29 (2013), 183-236. doi: 10.4171/RMI/718 |
Defining parameters
Level: \( N \) = \( 5077 \) Weight: \( k \) = \( 2 \) Nonzero newspaces: \( 16 \) Sturm bound: \(4295988\) Dimensions
The following table gives the dimensions of various subspaces of \(M_{2}(\Gamma_1(5077))\).
Total New Old Modular forms 1076535 1076535 0 Cusp forms 1071460 1071460 0 Eisenstein series 5075 5075 0 Decomposition of \(S_{2}^{\mathrm{new}}(\Gamma_1(5077))\)
We only show spaces with even parity, since no modular forms exist when this condition is not satisfied. Within each space \( S_k^{\mathrm{new}}(N, \chi) \) we list the newforms together with their dimension.
Label \(\chi\) Newforms Dimension \(\chi\) degree 5077.2.a \(\chi_{5077}(1, \cdot)\) 5077.2.a.a 1 1 5077.2.a.b 205 5077.2.a.c 216 5077.2.b \(\chi_{5077}(5076, \cdot)\) n/a 422 1 5077.2.c \(\chi_{5077}(1629, \cdot)\) n/a 842 2 5077.2.e \(\chi_{5077}(1630, \cdot)\) n/a 844 2 5077.2.f \(\chi_{5077}(360, \cdot)\) n/a 2526 6 5077.2.h \(\chi_{5077}(2402, \cdot)\) n/a 2532 6 5077.2.i \(\chi_{5077}(382, \cdot)\) n/a 7596 18 5077.2.k \(\chi_{5077}(321, \cdot)\) n/a 19366 46 5077.2.l \(\chi_{5077}(21, \cdot)\) n/a 7614 18 5077.2.m \(\chi_{5077}(25, \cdot)\) n/a 19412 46 5077.2.o \(\chi_{5077}(89, \cdot)\) n/a 38732 92 5077.2.q \(\chi_{5077}(22, \cdot)\) n/a 38824 92 5077.2.r \(\chi_{5077}(9, \cdot)\) n/a 116196 276 5077.2.t \(\chi_{5077}(3, \cdot)\) n/a 116472 276 5077.2.u \(\chi_{5077}(7, \cdot)\) n/a 349416 828 5077.2.w \(\chi_{5077}(4, \cdot)\) n/a 350244 828
"n/a" means that newforms for that character have not been added to the database yet |
This is the canonical Arrow-Pratt "portfolio" model. Couple of points on terminology:
For a function $u$, we define the
risk aversion function by $r_u(x):=-\frac{u''(x)}{u'(x)}$. In your utility function, $r_u(x) = \lambda$; hence, it is a constant absolute risk aversion utility and $\lambda$ is the "coefficient of risk aversion," not the "risk coefficient aversion".
The two points in time, $t_0,t_1$ can be seen as "beginning of period" and "end of period", where "period" is here the time interval $[t_0,t_1]$. This may be important: you don't need a dynamic approach as was suggested by some people. The guy in your problem allocates $a$ to the risky asset and $w_0-a$ to the riskless, over a time interval included between $t_0$ and $t_1$.
Your problem is the basic, canonical portfolio choice model with utility over final wealth. The guy in your problem just consumes what he has in the end of the time period. This is also important to bear in mind.
It's "negative exponential", not "exponential negative".
Rewrite $w(a)$ for final wealth (end of period, or at $t_1$); it depends on $a$, i.e. the part of $w_0$ that is invested in the risky asset. Your problem is:
$$\max_a \;E[U(w(a)] = \max_a \;E[-e^{-\lambda(x-r_f)}]$$
Let $\chi = x-r_f$, i.e. the
excess return of the risky asset (relative to the risk-free). Denote its distribution function by $dF(\chi)$ and hence
$$ \max_a \;E[-e^{-\lambda\chi}] = \max_a \;\int -e^{-\lambda z} dF(z)$$
Let $a^* = \arg\max_a \;E[U(w(a))]$. The following condition should hold in order for the (interior) optimum $a^*$ of this function to be bounded (note the redundancy in what I wrote just now):
Assumption (I) The values of the excess return random variable $\chi = x - r_f$ alternate in sign, i.e. $\chi$ takes values $\underline{\chi}\leq 0 \leq \overline{\chi}$ with positive probability.
If $\chi$ was positive
almost surely, then $a$ is unbounded precisely because the objective is unbounded, as you very well understood from the beginning. Hence, Assumption (I) should be retained.
Trust your intuition - your professor is wrong.
Addendum:
if $a^*\rightarrow \infty$, i.e. if the optimal solution is unbounded, then the
derivative of the expected utility evaluated at the optimal solution is zero - and since this doesn't make any sense, you have to rephrase it as
$$ \lim_{a\rightarrow\infty} E\left[\frac{d}{da}U(w(a)) \right] = 0 $$
Now $U$ is concave. Hence, in order for $a^* \rightarrow \infty$ not to be a critical point, you have to have
$$ \lim_{a\rightarrow\infty} E\left[\frac{d}{da}U(w(a)) \right] <0 $$
and not positive. Replace the parametric form, take the derivative, and you will find a (strict) inequality relating the distribution function and the marginal utility at the limits.
And since this is supposed to be a hint and not a homework helpdesk, I have to stop here :) Already, you were right in your original answer, but you have to
prove it as well. |
My model has 3 independent parameters $\{\rho, \alpha, \beta\}$ (polar coordinates), and a set of observables $\{Q_i\}$ and $\{T_{ij}\}$ where $i=1,2,...,642$ and $j=1,2,...,Q_i$ (if $Q_i=0$, there is no $T_{ij}$).
From $\{\rho, \alpha, \beta\}$, a set of intermediate (spherical harmonic) coefficients $q^m_l(\rho, \alpha, \beta)$ and $t^m_1(\rho, \alpha, \beta)$ where $l=1,2,3~m=-l,...,-1,0,1,...,l$ are obtained:
A rotational transformation $R_l^m(\alpha, \beta)$ (trigonometric polynomials in $(\alpha, \beta)$):
$q^m_l(\rho, \alpha, \beta) = q^0_l(\rho, 0, 0) R^m_l(\alpha, \beta)$
$t^m_1(\rho, \alpha, \beta) = t^0_1(\rho, 0, 0) R^m_l(\alpha, \beta)$
$q^0_l(\rho, 0, 0)$ and $t^0_l(\rho, 0, 0)$ are calibrated empirical functions.
$Q_i$ obeys Poisson distribution[a],
$Q_i \sim \text{Pois}(\sum_{l=1}^3\sum_{m=-l}^lq^m_l(\rho, \alpha, \beta)S^m_l(\theta_i, \phi_i))$, where $S^m_l(\theta_i, \phi_i)$ is the value of real spherical harimonic function at $i$.
$T_{ij}$ obeys shifted exponential and is treated with quantile regression[b] at 1% (empirical) percentile,
$T_{ij} \sim \text{Quant}_{0.01}(\sum_{l=1}^3\sum_{m=-l}^lq^m_l(\rho, \alpha, \beta)S^m_l(\theta_i, \phi_i))$
I would like to do inference, and I can see 2 options.
I have millions of observations, and am concerned about the feasibility of this approach on a small cluster (less than 1000 processor cores).
Ensemble of predictors: given an observation $\{Q_i, T_{ij}\}$, fit $\hat{q}^m_l$ and $\hat{t}^m_l$. For each $l$ fit $\rho, \alpha, \beta$ by $q$ or $t$. With 6 predictors $(\hat{\rho}_l^{t/q}, \hat{\alpha}_l^{t/q}, \hat{\beta}_l^{t/q})$, what is the best way to make a combined predictor? I know boosting for classifiers, is there a continuous counterpart?
Am I going to lose accuracy compared to hierarchical Bayesian? How can I estimate that loss?
Footnotes:
a. actually quasi-Poisson, where the variance is proportional to mean. |
I’d like to write about the situation that occurs in set theory when a forcing extension $V[G]=V[H]$ arises over a ground model $V$ in two different ways simultaneously, using generic filters over two different forcing notions $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$. The general fact, stated in theorem 1, is that in this case, the two forcing notions are actually isomorphic on a cone $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, with the isomorphism carrying the one generic filter to the other. In other words, below these respective conditions $b$ and $c$, the forcing notions and the respective generic filters are not actually different.
I have always assumed that this fact was part of the classical forcing folklore results, but
it doesn’t seem to be mentioned explicitly in the usual forcing literature (it appears as lemma 25.5 in Jech’s book), and so I am writing an account of it here. Victoria Gitman and I have need of it in a current joint project. (Bob Solovay mentions in the comments below that the result is due to him, and provides a possible 1975 reference.) Theorem 1. If $V[G]=V[H]$, where $G\subset \mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{B}\upharpoonright b$ is isomorphic to $\mathbb{C}\upharpoonright c$ by an isomorphism carrying $G$ to $H$.
The proof will also establish the following related result, concerning the situation where one extension is merely contained in the other.
Theorem 2. If $V[H]\subset V[G]$, where $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{C}\upharpoonright c$ is isomorphic to a complete subalgebra of $\mathbb{B}\upharpoonright b$.
By $\mathbb{B}\upharpoonright b$, where $b$ is a condition in $\mathbb{B}$ (that is, a nonzero element of $\mathbb{B}$), what I mean is the Boolean algebra consisting of the interval $[0,b]$ in $\mathbb{B}$, using relative complement $b-a$ as the negation of $a$. This is the complete Boolean algebra that arises when forcing with the conditions in $\mathbb{B}$ below $b$.
Proof: In order to prove theorem 2, let me assume at first only that $V[H]\subset V[G]$. It follows that $H=\dot H_G$ for some $\mathbb{B}$-name $\dot H$, and we may choose a condition $b\in G$ forcing that $\dot H$ is a $\check V$-generic filter on $\check{\mathbb{C}}$.
I claim that there is some $c\in H$ such that every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$. Note that every $d\in H$ has $[\![\check d\in\dot H]\!]\in G$ by the truth lemma, since $H=\dot H_G$, and so $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ for $d\in H$. If $c\in H$ forces that every $d$ in the generic filter has that property, then indeed every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ as claimed.
In other words, from the perspective of the $\mathbb{B}$ forcing, every $d\leq c$ has a nonzero possibility to be in $\dot H$.
Define $\pi:\mathbb{C}\upharpoonright c\to\mathbb{B}$ by $$\pi(d)=b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}.$$ Using the fact that $b$ forces that $\dot H$ is a filter, it is straightforward to verify that
$d\leq e\implies \pi(d)\leq\pi(e)$, since if $d\leq e$ and $d\in H$, then $e\in H$. $\pi(d\wedge e)=\pi(d)\wedge \pi(e)$, since $[\![\check d\in\dot H]\!]\wedge[\![\check e\in \dot H]\!]=[\![\check{(b\wedge e)}\in\dot H]\!]$. $\pi(d-e)=\pi(d)-\pi(e)$, since $[\![\check{(d-e)}\in\dot H]\!]=[\![\check d\in\dot H]\!]-[\![\check e\in\dot H]\!]$.
Thus, $\pi$ is a Boolean algebra embedding of $\mathbb{C}\upharpoonright c$ into $\mathbb{B}\upharpoonright\pi(c)$.
Let me argue that this embedding is a complete embedding. Suppose that $a=\bigvee A$ for some subset $A\subset\mathbb{C}\upharpoonright c$ with $A\in V$. Since $H$ is $V$-generic, it follows that $a\in H$ just in case $H$ meets $A$. Thus, $[\![\check a\in\dot H]\!]=[\![\exists x\in\check A\, x\in \dot H]\!]=\bigvee_{x\in A}[\![\check x\in\dot H]\!]$, and so $\pi(\bigvee A)=\bigvee_{x\in A}\pi(x)$, and so $\pi$ is complete, as desired. This proves theorem 2.
To prove theorem 1, let me now assume fully that $V[G]=V[H]$. In this case, there is a $\mathbb{C}$ name $\dot G$ for which $G=\dot G_H$. By strengthening $b$, we may assume without loss that $b$ also forces that, that is, that $b$ forces $\Gamma=\check{\dot G}_{\dot H}$, where $\Gamma$ is the canonical $\mathbb{B}$-name for the generic object, and $\check{\dot G}$ is the $\mathbb{B}$-name of the $\mathbb{C}$-name $\dot G$. Let us also strengthen $c$ to ensure that $c$ forces $\dot G$ is $\check V$-generic for $\check{\mathbb{C}}$. For $d\leq c$ define $\pi(d)=[\![\check d\in\dot H]\!]^{\mathbb{B}}$ as above, which provides a complete embedding of $\mathbb{C}\upharpoonright c$ to $\mathbb{B}\upharpoonright\pi(c)$. I shall now argue that this embedding is dense below $\pi(c)$. Suppose that $a\leq \pi(c)$ in $\mathbb{B}$. Since $a$ forces $\check a\in\Gamma$ and also $\check c\in\dot H$, it must also force that there is some $d\leq c$ in $\dot H$ that forces via $\mathbb{C}$ over $\check V$ that $\check a\in\dot G$. So there must really be some $d\leq c$ forcing $\check a\in\dot G$. So $\pi(d)$, which forces $\check d\in\dot H$, will also force $\check a\in\check{\dot G}_{\dot H}=\Gamma$, and so $\pi(d)\Vdash_{\mathbb{B}}\check a\in\Gamma$, which means $\pi(d)\leq a$ in ${\mathbb{B}}$. Thus, the range of $\pi$ on $\mathbb{C}\upharpoonright c$ is dense below $\pi(c)$, and so $\pi$ is a complete dense embedding of ${\mathbb{C}}\upharpoonright c$ to ${\mathbb{B}}\upharpoonright \pi(c)$. Since these are complete Boolean algebras, this means that $\pi$ is actually an isomorphism of $\mathbb{C}\upharpoonright c$ with $\mathbb{B}\upharpoonright \pi(c)$, as desired.
Finally, note that if $d\in H$ below $c$, then since $H=\dot H_G$, it follows that $[\![\check d\in\dot H]\!]\in G$, which is to say $\pi(d)\in G$, and so $\pi$ carries $H$ to $G$ on these cones. So $\pi^{-1}$ is the isomorphism stated in theorem 1.
QED
Finally, I note that one cannot get rid of the need to restrict to cones, since it could be that $\mathbb{B}$ and $\mathbb{C}$ are the lottery sums of a common forcing notion, giving rise to $V[G]=V[H]$, together with totally different non-isomorphic forcing notions below some other incompatible conditions. So we cannot expect to prove that $\mathbb{B}\cong\mathbb{C}$, and are content to get merely that $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, an isomorphism below respective conditions. |
You say you want to understand how $\lambda$ and $\eta$ affect the cost function.
If you hold the weights $w$ fixed, the equation for $C$ tells you how $\lambda$ affects the cost function, and $\eta$ doesn't affect the cost function; it only affects the sequence of steps taken by gradient descent.
But the tricky thing is that the final weights $w$ depend on $\lambda,\eta$. If we choose $\lambda,\eta$ and then use gradient descent to train a set of weights, the final weights $w$ will depend on $\lambda$ and $\eta$.
So, we can think of the final weights $w$ as actually being a function of $\lambda,eta$. Similarly, the final cost $C$ is actually a function of $w,\lambda,eta$, and since $w$ in turn depends on $\lambda,\eta$, this means that the final cost $C$ is really a function of the two variables $\lambda,\eta$. Suppose we write it as a function, to make the dependence clearer: $C(\lambda,\eta)$.
Now we want minimize $C(\lambda,\eta)$. If you want to minimize this using gradient descent, you need to be able to compute the partial derivatives ${\partial \over \partial \lambda} C(\lambda,\eta)$ and ${\partial \over \partial \eta} C(\lambda,\eta)$. Unfortunately, it is not at all clear how to write down an analytical expression for those partial derivatives. The tricky bit is that it is not at all clear how the final weights $w$ depend on $\lambda,\eta$ -- we have no nice way to write down an analytical expression for that. So, we can differentiate to get
$${\partial \over \partial \lambda} C(\lambda,\eta) = {\partial \over \partial \lambda} C_0(\lambda,\eta) + {\lambda \over 2n} \sum_i 2 w_i {\partial w_i \over \partial \lambda} + {1 \over 2n} \sum_i w_i^2,$$
but how do we compute ${\partial w_i \over \partial \lambda}$? And how do we compute ${\partial \over \partial \lambda} C_0(\lambda,\eta)$? It's not clear, as we don't have an analytical expression for the final weights $w_i$ as a function of $\lambda$; the only way we have to compute the weights is to run the gradient descent algorithm to completion.
So that's the challenge with using gradient descent to optimize $\lambda,\eta$, and one reason why people often use grid search instead.
I'm not saying applying gradient descent is impossible, but one must apply other black-box methods to compute the gradients, and there are other reasons why this might fail due to multiple local minima. Usually grid search is easy enough to apply and thus that is what is used. |
The mod 4 behaviour of total Lie algebra cohomology 34 Downloads Abstract.
We show that if
L is a unimodular Lie algebra over a field of characteristic \(\ne 2\), then the dimension \(\sigma\)( L) of the total cohomology of L is a multiple of 4 when \(\dim(L)\not\equiv 3\) (mod 4). However, contrary to a claim by Deninger and Singhof, we give an example of a rational nilpotent algebra L of dimension 15 with \(\sigma(L)\not\equiv 0\) (mod 4). Over fields of characteristic 2, we completely classify those algebras L with \(\sigma(L)\not\equiv 0\) (mod 4). KeywordsAlgebra Cohomology Nilpotent Algebra Total Cohomology Preview
Unable to display preview. Download preview PDF. |
Let there be the random variables $X$, $Y$, and $Z$. Let all the bivariate PDFs $f_{X, Y}$, $f_{X, Z}$, and $f_{Y, Z}$ be known.
Can we write the unknown trivariate PDF $f_{X, Y, Z}$ in terms of the known bivariate PDFs?
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Here is an example, obtained by tweaking a 2D counterexample: Let $(X,Y,Z)$ be such that $$ f_{X,Y,Z}(x,y,z)= \begin{cases} 2 \phi(x)\phi(y)\phi(z) & xyz>0\\ 0 & \text{otherwise} \end{cases} $$ where $\phi$ is the standard 1D Gaussian pdf. Then the bivariates $f_{X,Y}, f_{Y,Z}, f_{Z,X}$ are all standard 2D Gaussians, but of course $(X,Y,Z)$ is not Gaussian. Now both the standard 3D Gaussian and this $f$ give the same distribution of bivariates.
Let $f(x,y,z)=1$ when $x,y,z\in [0,1]$, and be zero otherwise. Then $f_{X,Y}$ is uniform on the unit square, and same for $f_{X,Z}$ and $f_{Y,Z}$. On the other hand, let $$ g(x,y,z)=1+\sin(2\pi (x+y+z)) \hspace{1cm}\text{for }x,y,z\in [0,1] $$ Then $g_{X,Y},g_{X,Z}$ and $g_{Y,Z}$ are also uniform on the unit square. |
There are several equivalent ways of implementing Dirichlet boundary conditions with the finite element method. I'll give a brief overview, but you'll probably need more details, which can be found in the book Understanding and Implementing the Finite Element Method.
It may be easier to analyze this for the Poisson problem $-\nabla^2u = f$ than for the heat equation; the techniques are the same in both cases, and if you understand them for the steady-state problem then it shouldn't be a big leap to adapt them to the time-dependent one. For the initial value, that amounts to setting $u^0_i = \mathbf{u}_0(x_i)$. You'll then use this as part of the right-hand side in the linear system you've written down when you go to solve for the next time-step $\mathbf{u}^1$. The boundary conditions on $\partial\Omega$ are trickier.
One way to implement Dirichlet boundary conditions is to alter the matrices $A$ and $B$ and the load vector in order to guarantee that the boundary conditions are satisfied exactly. Suppose that $i$ is some unknown of the linear system corresponding to a Dirichlet boundary point; the modification you have to make is to zero out row $i$ of $B$, set row $i$ of $A$ to be the corresponding row in the identity matrix and set $\mathbf{l}_i$ equal to $g(x_i)$.
To illustrate this approach, suppose you had a $3 \times 3$ chunk of the linear system that looked like this:
$\left[\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}\right]\left[\begin{matrix} u_1 \\ u_2 \\ u_3\end{matrix}\right]= \left[\begin{matrix} l_1 \\ l_2 \\ l_3\end{matrix}\right]$
and that $u_2$ is a degree of freedom corresponding to a boundary node. The modified system would look like this:
$\left[\begin{matrix} a_{11} & a_{12} & a_{13} \\ 0 & 1 & 0 \\ a_{31} & a_{32} & a_{33} \end{matrix}\right]\left[\begin{matrix} u_1 \\ u_2 \\ u_3\end{matrix}\right]= \left[\begin{matrix} l_1 \\ g_2 \\ l_3\end{matrix}\right]$
You can see how the modified system isn't symmetric anymore. Nonetheless, $u_1$ and $u_3$ are still coupled to $u_2$. The asymmetry of the system isn't usually an issue in practice provided that your initial guess for $u$ exactly satisfies the Dirichlet boundary conditions; try throwing the problem into a naive CG solver and see what happens. (Can you come up with a reason why this might be so?)
Another approach is to reduce the system size. Instead of solving for all of the coefficients $u_i$ in the expansion $\mathbf{u} = \sum_iu_i\phi_i$, you'll only solve for the coefficients corresponding to interior nodes. This means extracting sub-matrices out of $A$ and $B$ for those unknowns, and adding an extra component to the load vector $\mathbf{l}$ consisting of the effective forcing on the interior from boundary conditions. Referring back to the example matrices from before, we're left with the following reduced system for $u_1$ and $u_3$:
$\left[\begin{matrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{matrix}\right]\left[\begin{matrix} u_1 \\ u_3\end{matrix}\right] =\left[\begin{matrix} l_1 - a_{12}g_2 \\ l_3 - a_{32}g_2\end{matrix}\right]$
Changing the linear system or deflating it are, from what I've observed, the most common approaches. You can also add on boundary conditions later as a constraint, or you can use penalty methods, but these are less common. Of the two, I tend to favor the first approach. |
9 0 Einstein-scalar field action --> Einstein-scalar field equations
Dear friends,
Just a small question I do not know how to derive.
From the Einstein-scalar field action defined by
[tex]S\left( {g,\psi } \right) = \int_{} {\left( {R(g) - \frac{1}{2}\left| {\nabla \psi } \right|_g^2 - V\left( \psi \right)} \right)d{\eta _g}}[/tex]
one gets the so-called Einstein-scalar field equations given by
[tex]{\rm Eins}_{\alpha \beta} = {\nabla _\alpha }\psi {\nabla _\beta }\psi - \frac{1}{2}{g_{\alpha \beta }}{\nabla _\mu }\psi {\nabla ^\mu }\psi - {g_{\alpha \beta }}V(\psi ).[/tex]
My question is how to derive such equations. It seems that we need to take derivative.... but how? Thanks. |
this is a mystery to me, despite having changed computers several times, despite the website rejecting the application, the very first sequence of numbers I entered into it's search window which returned the same prompt to submit them for publication appear every time, I mean ive got hundreds of them now, and it's still far too much rope to give a person like me sitting along in a bedroom the capacity to freely describe any such sequence and their meaning if there isn't any already there
my maturity levels are extremely variant in time, that's just way too much rope to give me considering its only me the pursuits matter to, who knows what kind of outlandish crap I might decide to spam in each of them
but still, the first one from well, almost a decade ago shows up as the default content in the search window
1,2,3,6,11,23,47,106,235
well, now there is a bunch of stuff about them pertaining to "trees" and "nodes" but that's what I mean by too much rope you cant just let a lunatic like me start inventing terminology as I go
oh well "what would cotton mathers do?" the chat room unanimously ponders lol
i see Secret had a comment to make, is it really a productive use of our time censoring something that is most likely not blatant hate speech? that's the only real thing that warrants censorship, even still, it has its value, in a civil society it will be ridiculed anyway?
or at least inform the room as to whom is the big brother doing the censoring? No? just suggestions trying to improve site functionality good sir relax im calm we are all calm
A104101 is a hilarious entry as a side note, I love that Neil had to chime in for the comment section after the big promotional message in the first part to point out the sequence is totally meaningless as far as mathematics is concerned just to save face for the websites integrity after plugging a tv series with a reference
But seriously @BalarkaSen, some of the most arrogant of people will attempt to play the most innocent of roles and accuse you of arrogance yourself in the most diplomatic way imaginable, if you still feel that your point is not being heard, persist until they give up the farce please
very general advice for any number of topics for someone like yourself sir
assuming gender because you should hate text based adam long ago if you were female or etc
if its false then I apologise for the statistical approach to human interaction
So after having found the polynomial $x^6-3x^4+3x^2-3$we can just apply Eisenstein to show that this is irreducible over Q and since it is monic, it follwos that this is the minimal polynomial of $\sqrt{1+\sqrt[3]{2}}$ over $\mathbb{Q}$ ? @MatheinBoulomenos
So, in Galois fields, if you have two particular elements you are multiplying, can you necessarily discern the result of the product without knowing the monic irreducible polynomial that is being used the generate the field?
(I will note that I might have my definitions incorrect. I am under the impression that a Galois field is a field of the form $\mathbb{Z}/p\mathbb{Z}[x]/(M(x))$ where $M(x)$ is a monic irreducible polynomial in $\mathbb{Z}/p\mathbb{Z}[x]$.)
(which is just the product of the integer and its conjugate)
Note that $\alpha = a + bi$ is a unit iff $N\alpha = 1$
You might like to learn some of the properties of $N$ first, because this is useful for discussing divisibility in these kinds of rings
(Plus I'm at work and am pretending I'm doing my job)
Anyway, particularly useful is the fact that if $\pi \in \Bbb Z[i]$ is such that $N(\pi)$ is a rational prime then $\pi$ is a Gaussian prime (easily proved using the fact that $N$ is totally multiplicative) and so, for example $5 \in \Bbb Z$ is prime, but $5 \in \Bbb Z[i]$ is not prime because it is the norm of $1 + 2i$ and this is not a unit.
@Alessandro in general if $\mathcal O_K$ is the ring of integers of $\Bbb Q(\alpha)$, then $\Delta(\mathcal O_K) [\mathcal O_K:\Bbb Z[\alpha]]^2=\Delta(\mathcal O_K)$, I'd suggest you read up on orders, the index of an order and discriminants for orders if you want to go into that rabbit hole
also note that if the minimal polynomial of $\alpha$ is $p$-Eisenstein, then $p$ doesn't divide $[\mathcal{O}_K:\Bbb Z[\alpha]]$
this together with the above formula is sometimes enough to show that $[\mathcal{O}_K:\Bbb Z[\alpha]]=1$, i.e. $\mathcal{O}_K=\Bbb Z[\alpha]$
the proof of the $p$-Eisenstein thing even starts with taking a $p$-Sylow subgroup of $\mathcal{O}_K/\Bbb Z[\alpha]$
(just as a quotient of additive groups, that quotient group is finite)
in particular, from what I've said, if the minimal polynomial of $\alpha$ wrt every prime that divides the discriminant of $\Bbb Z[\alpha]$ at least twice, then $\Bbb Z[\alpha]$ is a ring of integers
that sounds oddly specific, I know, but you can also work with the minimal polynomial of something like $1+\alpha$
there's an interpretation of the $p$-Eisenstein results in terms of local fields, too. If the minimal polynomial of $f$ is $p$-Eisenstein, then it is irreducible over $\Bbb Q_p$ as well. Now you can apply the Führerdiskriminantenproduktformel (yes, that's an accepted English terminus technicus)
@MatheinBoulomenos You once told me a group cohomology story that I forget, can you remind me again? Namely, suppose $P$ is a Sylow $p$-subgroup of a finite group $G$, then there's a covering map $BP \to BG$ which induces chain-level maps $p_\# : C_*(BP) \to C_*(BG)$ and $\tau_\# : C_*(BG) \to C_*(BP)$ (the transfer hom), with the corresponding maps in group cohomology $p : H^*(G) \to H^*(P)$ and $\tau : H^*(P) \to H^*(G)$, the restriction and corestriction respectively.
$\tau \circ p$ is multiplication by $|G : P|$, so if I work with $\Bbb F_p$ coefficients that's an injection. So $H^*(G)$ injects into $H^*(P)$. I should be able to say more, right? If $P$ is normal abelian, it should be an isomorphism. There might be easier arguments, but this is what pops to mind first:
By Schur-Zassenhaus theorem, $G = P \rtimes G/P$ and $G/P$ acts trivially on $P$ (the action is by inner auts, and $P$ doesn't have any), there is a fibration $BP \to BG \to B(G/P)$ whose monodromy is exactly this action induced on $H^*(P)$, which is trivial, so we run the Lyndon-Hochschild-Serre spectral sequence with coefficients in $\Bbb F_p$.
The $E^2$ page is essentially zero except the bottom row since $H^*(G/P; M) = 0$ if $M$ is an $\Bbb F_p$-module by order reasons and the whole bottom row is $H^*(P; \Bbb F_p)$. This means the spectral sequence degenerates at $E^2$, which gets us $H^*(G; \Bbb F_p) \cong H^*(P; \Bbb F_p)$.
@Secret that's a very lazy habit you should create a chat room for every purpose you can imagine take full advantage of the websites functionality as I do and leave the general purpose room for recommending art related to mathematics
@MatheinBoulomenos No worries, thanks in advance. Just to add the final punchline, what I wanted to ask is what's the general algorithm to recover $H^*(G)$ back from $H^*(P; \Bbb F_p)$'s where $P$ runs over Sylow $p$-subgroups of $G$?
Bacterial growth is the asexual reproduction, or cell division, of a bacterium into two daughter cells, in a process called binary fission. Providing no mutational event occurs, the resulting daughter cells are genetically identical to the original cell. Hence, bacterial growth occurs. Both daughter cells from the division do not necessarily survive. However, if the number surviving exceeds unity on average, the bacterial population undergoes exponential growth. The measurement of an exponential bacterial growth curve in batch culture was traditionally a part of the training of all microbiologists...
As a result, there does not exists a single group which lived long enough to belong to, and hence one continue to search for new group and activity
eventually, a social heat death occurred, where no groups will generate creativity and other activity anymore
Had this kind of thought when I noticed how many forums etc. have a golden age, and then died away, and at the more personal level, all people who first knew me generate a lot of activity, and then destined to die away and distant roughly every 3 years
Well i guess the lesson you need to learn here champ is online interaction isn't something that was inbuilt into the human emotional psyche in any natural sense, and maybe it's time you saw the value in saying hello to your next door neighbour
Or more likely, we will need to start recognising machines as a new species and interact with them accordingly
so covert operations AI may still exists, even as domestic AIs continue to become widespread
It seems more likely sentient AI will take similar roles as humans, and then humans will need to either keep up with them with cybernetics, or be eliminated by evolutionary forces
But neuroscientists and AI researchers speculate it is more likely that the two types of races are so different we end up complementing each other
that is, until their processing power become so strong that they can outdo human thinking
But, I am not worried of that scenario, because if the next step is a sentient AI evolution, then humans would know they will have to give way
However, the major issue right now in the AI industry is not we will be replaced by machines, but that we are making machines quite widespread without really understanding how they work, and they are still not reliable enough given the mistakes they still make by them and their human owners
That is, we have became over reliant on AI, and not putting enough attention on whether they have interpret the instructions correctly
That's an extraordinary amount of unreferenced rhetoric statements i could find anywhere on the internet! When my mother disapproves of my proposals for subjects of discussion, she prefers to simply hold up her hand in the air in my direction
for example i tried to explain to her that my inner heart chakras tell me that my spirit guide suggests that many females i have intercourse with are easily replaceable and this can be proven from historical statistical data, but she wont even let my spirit guide elaborate on that premise
i feel as if its an injustice to all child mans that have a compulsive need to lie to shallow women they meet and keep up a farce that they are either fully grown men (if sober) or an incredibly wealthy trust fund kid (if drunk) that's an important binary class dismissed
Chatroom troll: A person who types messages in a chatroom with the sole purpose to confuse or annoy.
I was just genuinely curious
How does a message like this come from someone who isn't trolling:
"for example i tried to explain to her that my inner heart chakras tell me that my spirit guide suggests that many ... with are easily replaceable and this can be proven from historical statistical data, but she wont even let my spirit guide elaborate on that premise"
3
Anyway feel free to continue, it just seems strange @Adam
I'm genuinely curious what makes you annoyed or confused yes I was joking in the line that you referenced but surely you cant assume me to be a simpleton of one definitive purpose that drives me each time I interact with another person? Does your mood or experiences vary from day to day? Mine too! so there may be particular moments that I fit your declared description, but only a simpleton would assume that to be the one and only facet of another's character wouldn't you agree?
So, there are some weakened forms of associativity. Such as flexibility ($(xy)x=x(yx)$) or "alternativity" ($(xy)x=x(yy)$, iirc). Tough, is there a place a person could look for an exploration of the way these properties inform the nature of the operation? (In particular, I'm trying to get a sense of how a "strictly flexible" operation would behave. Ie $a(bc)=(ab)c\iff a=c$)
@RyanUnger You're the guy to ask for this sort of thing I think:
If I want to, by hand, compute $\langle R(\partial_1,\partial_2)\partial_2,\partial_1\rangle$, then I just want to expand out $R(\partial_1,\partial_2)\partial_2$ in terms of the connection, then use linearity of $\langle -,-\rangle$ and then use Koszul's formula? Or there is a smarter way?
I realized today that the possible x inputs to Round(x^(1/2)) covers x^(1/2+epsilon). In other words we can always find an epsilon (small enough) such that x^(1/2) <> x^(1/2+epsilon) but at the same time have Round(x^(1/2))=Round(x^(1/2+epsilon)). Am I right?
We have the following Simpson method $$y^{n+2}-y^n=\frac{h}{3}\left (f^{n+2}+4f^{n+1}+f^n\right ), n=0, \ldots , N-2 \\ y^0, y^1 \text{ given } $$ Show that the method is implicit and state the stability definition of that method.
How can we show that the method is implicit? Do we have to try to solve $y^{n+2}$ as a function of $y^{n+1}$ ?
@anakhro an energy function of a graph is something studied in spectral graph theory. You set up an adjacency matrix for the graph, find the corresponding eigenvalues of the matrix and then sum the absolute values of the eigenvalues. The energy function of the graph is defined for simple graphs by this summation of the absolute values of the eigenvalues |
CGAL 4.14.1 - dD Range and Segment Trees
#include <CGAL/Segment_tree_d.h>
A \( d\)-dimensional segment tree stores \( d\)-dimensional intervals and can be used to find all intervals that enclose, partially overlap, or contain a query interval, which may be a point.
Implementation
A \( d\)-dimensional segment tree is constructed in \( {O}(n\log n^d)\) time. An inverse range query is performed in time \( {O}(k+{\log}^d n )\), where \( k\) is the number of reported intervals. The tree uses \( {O}(n\log n^d)\) storage.
typedef unspecified_type Data container
Data.
typedef unspecified_type Window container
Window.
Segment_tree_d< Data, Window, Traits > s (Tree_base< Data, Window > sublayer_tree) A segment tree is defined, such that the subtree of each vertex is of the same type prototype
sublayer_tree is. More...
bool make_tree (In_it first, In_it last) The tree is constructed according to the data items in the sequence between the element pointed by iterator
first and iterator
last. More...
OutputIterator window_query (Window win, OutputIterator result)
win \( =[a_1,b_1),\ldots, [a_d,b_d)\), \( a_i,b_i\in T_i\), \( 1\le i\le d\). More...
OutputIterator enclosing_query (Window win, OutputIterator result) All elements that enclose the associated \( d\)-dimensional interval of
win are placed in the associated sequence container of
OutputIterator and returns an output iterator that points to the last location the function wrote to.
bool is_valid () The tree structure is checked. More... bool is_inside (Window win, Data object) returns
true, if the interval of
object is contained in the interval of
win,
false otherwise.
bool is_anchor () returns false.
bool CGAL::Segment_tree_d< Data, Window, Traits >::is_valid ( )
The tree structure is checked.
For each vertex either the sublayer tree is a tree anchor, or it stores a (possibly empty) list of data items. In the first case, the sublayer tree of the vertex is checked on being valid. In the second case, each data item is checked weather it contains the associated interval of the vertex and does not contain the associated interval of the parent vertex or not.
true is returned if the tree structure is valid,
false otherwise.
bool CGAL::Segment_tree_d< Data, Window, Traits >::make_tree ( In_it first, In_it last )
The tree is constructed according to the data items in the sequence between the element pointed by iterator
first and iterator
last.
true is returned. Otherwise, nothing is done but a
CGAL warning is given and
false returned.
Segment_tree_d<Data, Window, Traits> CGAL::Segment_tree_d< Data, Window, Traits >::s ( Tree_base< Data, Window > sublayer_tree )
A segment tree is defined, such that the subtree of each vertex is of the same type prototype
sublayer_tree is.
We assume that the dimension of the tree is \( d\). This means, that
sublayer_tree is a prototype of a \( d-1\)-dimensional tree. All data items of the \( d\)-dimensional segment tree have container type
Data. The query window of the tree has container type
Window.
Traits provides access to the corresponding data slots of container
Data and
Window for the \( d\)-th dimension. The traits class
Traits must at least provide all functions and type definitions described, for example, in the reference page for
tree_point_traits. The template class described there is fully generic and should fulfill the most requirements one can have. In order to generate a one-dimensional segment tree instantiate
Tree_anchor<Data, Window> sublayer_tree with the same template parameters
Data and
Window
Segment_tree_d is defined. In order to construct a two-dimensional segment tree, create
Segment_tree_d with a one-dimensional
Segment_tree_d with the corresponding
Traits of the first dimension.
Traits::Data==Data and
Traits::Window==Window.
OutputIterator CGAL::Segment_tree_d< Data, Window, Traits >::window_query ( Window win, OutputIterator result )
win \( =[a_1,b_1),\ldots, [a_d,b_d)\), \( a_i,b_i\in T_i\), \( 1\le i\le d\).
All elements that intersect the associated \( d\)-dimensional interval of
win are placed in the associated sequence container of
OutputIterator and returns an output iterator that points to the last location the function wrote to. In order to perform an inverse range query, a range query of \( \epsilon\) width has to be performed. |
Here is a direct proof of exactness at $B \otimes_R N$. Summary: it's very annoying to prove things in detail using the generators-and-relations definition of tensor products. Even in this long-winded version, significant detail is omitted. I would be curious to know if anyone has typed something like this into Coq.
This is usually treated using the universal property of the tensor product, as in the other answers, or in Dummit and Foote, or left as an exercise. In my opinion, leaving this as an exercise is mostly about not wanting to write this much detail. I have written this out because I can't find another source that does so.
First we state two clarifying lemmas:
Lemma 1 If $f \colon S \rightarrow T$ is a map of sets, then the kernel of the induced map of free $R$-modules $f \colon R[S] \rightarrow R[T]$ is generated by $\{s-s' \in R[S] \mid f(s) = f(s')\}$.
Lemma 2 If $$X \xrightarrow{f} Y \xrightarrow{g} Z \rightarrow 0$$ are maps of $R$-modules and $\mathrm{Ker}(f)$ is generated by $\{x_i\}$, $\mathrm{Ker} (g)$ is generated by elements $\{y_j\}$, and we have $x'_j \in X$ such that $f(x'_j) = y_j$, then $\mathrm{Ker} (gf)$ is generated by $\{x_i\} \cup \{x'_j\}$.
To prove that $\mathrm{Ker}(\beta \otimes \mathrm{Id}) \subset \mathrm{Im}(\alpha \otimes \mathrm{Id})$ we work with the definition of the tensor products as quotients of the free modules $R[B \times M]$ and $R[C \times M]$. If $(\beta \otimes \mathrm{Id} )(\sum m_i \otimes n_i) = 0 \in C \otimes_R M$, then the composition $$R[B \times M] \xrightarrow{ \beta \times \mathrm{Id}} R[C \times M] \xrightarrow{\pi} C \otimes_R M$$ sends $\sum (m_i ,n_i)$ to zero, where $\pi$ is the quotient map defining the tensor product.
Our goal is to show that any such element $\sum (m_i ,n_i)$ can be expressed ($\star$) as a sum $\sum_j (\alpha(a_j),m_j)$ plus a linear combination of tensor-product relation elements in $R[B \times M]$ (there are 4 types). This is equivalent to showing that $\sum m_i \otimes n_i$ is in the image of $\alpha \otimes \mathrm{Id}$.
That is to say, we want to show that the kernel of the composition $\pi \circ (\beta \otimes \mathrm{Id})$ is generated by 5 types of elements. Lemma 1 tells us that the kernel of $\beta \times \mathrm{Id}$ is generated ($\triangle$) by $\{(b,m)-(b',m) \in R[B \times M] \mid \beta(b) = \beta(b')\}$. The kernel of $\pi$ is generated by tensor-product relation elements by definition. Further, any tensor-product relation element in $R[C \times M]$ is the image of a tensor-product relation element in $R[B \times M]$, because $ \beta $ is onto.
We can write the annoying expression$$(b,m)-(b',m) = (b-b', m) - [(b+(-b'),m)-(b,m)-(-b',m)]+[(-1)(b',m)-(-b',m)]$$Since $\beta(b-b') = 0$, $b-b' = \alpha(a)$, so the first term on the RHS is an element in the image of $\alpha \times \mathrm{Id}$ and the other two terms are tensor product relations terms for $B \otimes_R M$.
By Lemma 2, we know a set of generators of $\mathrm{Ker}(\pi \circ (\beta \otimes \mathrm{Id}))$. We want to show that all of these can be expressed as described above ($\star$). The generators arising from tensor-product relation elements of $R[C \times M]$ are tensor-product relation elements in $R[B \times M]$, so there is nothing to prove. The other type of generator $\triangle$ is also expressible in the form $\star$, because of the annoying expression above. |
The term you are reducing is $(K_y \Omega)$ where $K_y$ is the constant function $\lambda x. y$ (it always returns $y$, ignoring its argument) and $\Omega = (\lambda x. (x \, x) \: \lambda x. (x \, x))$ is a non-terminating term. In some sense $\Omega$ is the ultimate non-terminating term: it beta-reduces to itself, i.e. $\Omega \to \Omega$. (Make sure to work it out on paper at least once in your life.)
Applicative order reduction must reduce the argument of the function to a normal form, before it can evaluate the top-level redex. Since the argument $\Omega$ has no normal form, applicative order reduction loops infinitely. More generally, for any term $M$, $M \Omega \to M \Omega$, and this is the reduction chosen by the applicative order strategy.
Normal order reduction starts by reducing the top-level redex (because the function $K_y$ is already in normal form). Since $K_y$ ignores its argument, $(K_y \Omega) \to y$ in one step. More generally, $K_y N \to y$ for any term $N$, and this is the reduction chosen by the normal order strategy.
This case illustrates a more general phenomenon: applicative order reduction only ever finds a normal form if the term is strongly normalizing, whereas normal order reduction always finds the normal form if there is one. This happens because applicative order always evaluates fully arguments first, and so misses the opportunity for an argument to turn out to be unused; whereas normal order evaluates arguments as late as possible, and so always wins if the argument turns out to be unused.
(The flip side is that applicative order tends to be faster in practice, because it's relatively rare for an argument to be unused; whereas it's common for an argument to be used multiple times, and under applicative order the argument is only evaluated once. Normal order evaluates the argument as often as it's used, be it 0, 1 or many times.) |
Let me illustrate the issue with an simplified example. Suppose we want to solve the following problem with finite difference method (FDM):
$$\frac{\partial u(t,x)}{\partial t}=\frac{\partial^2 u(t,x)}{\partial x^2}$$ $$u(0,x)=x(1-x),\ u(t,0)=0$$ $$t\in[0,1],\ x\in[0,1] $$
One boundary condition (b.c.) is missing, but let's ignore it for a while and discretize the equation, initial condition (i.c.) and b.c. with difference formula. Here we use 2nd order difference formula. Clearly if we only use
$$f' (x_i)\simeq \frac{f (x_{i}+h)-f (x_{i}-h)}{2 h}$$ $$f'' (x_i)\simeq\frac{f (x_{i}-h)-2 f (x_i)+f (x_{i}+h)}{h^2}$$
we cannot generate equation for $t=1,\ x=1$, so we need one-sided formula (If you're not familiar with one-sided formula, start from page 6 of this book ):
$$f' (x_n)\simeq \frac{f (x_{n}-2h)-4 f (x_{n}-h)+3 f (x_n)}{2 h}$$ $$f'' (x_n)\simeq \frac{-f (x_{n}-3h)+4 f (x_{n}-2h)-5f (x_{n}-h)+2 f (x_{n})}{h^2}$$
OK, now here comes the problem: we surprisingly find that, even if b.c. is not enough, we still obtain a closed algebraic equation system!
One may suspect the system will be kind of ill-posed e.g. "This system won't be solvable!" or "This system will have infinite many solutions!" etc., but once again, surprisingly, further check shows the system determines a unique solution and seems to be stable i.e. with smaller grid size or higher order difference formula, the solution doesn't have significant change.
How to explain this solution? If a hidden b.c. has been imposed, what's it?
At the very least, if the solution is really nothing more than numeric error, I'd like to know the
exact cause of the error.
Here's the source code for solving the problem above in
Mathematica. I've used
pdetoae for the generation of difference formula:
eq = D[u[t, x], t] == D[u[t, x], x, x];ic = u[0, x] == x (1 - x);bc = u[t, 0] == 0;points = 25;grid = tgrid = Array[# &, points, {0, 1}];ptoafunc = pdetoae[u[t, x], {tgrid, grid}, 2];ae = Rest /@ (ptoafunc[eq] // Rest);aeic = ptoafunc[ic] // Rest;aebc = ptoafunc[bc];var = Outer[u, tgrid, grid] // Flatten;{b, m} = CoefficientArrays[{ae, aeic, aebc} // Flatten, var];solarray = Partition[LinearSolve[N@m, -b], points];solfunc = ListInterpolation[solarray, {tgrid, grid}];style = Style[#, 16] &;Plot3D[solfunc[t, x], {t, 0, 1}, {x, 0, 1}, AxesLabel -> style /@ {"t", "x"}]
Plot[solfunc[t, #] & /@ grid // Evaluate, {t, 0, 1}]
The following is the norm of solution $\sqrt{\int_0^1 u(t,x)^2 \, dx}$:
norm[t_?NumericQ, order_: 2, func_: solfunc] := Power[NIntegrate[func[t, x]^order, {x, 0, 1}, Method -> {Automatic, "SymbolicProcessing" -> 0}], 1/order]Plot[norm[t], {t, 0, 1}, AxesLabel -> {t, norm}, PlotRange -> All]
Remark
Notice that the value of i.c. is used at the corner $t=0,\ x=1$.
LinearSolveis not a iterative solver but a symbolic solver for linear algebraic equation system.
Also, it's actually enough to reproduce the issue by discretizing in $x$ direction only, and the resulting ordinary differential equation (ODE) system can be solved
symbolically. The following is the corresponding Mathematica code. I've used
pdetoode for the generation of ODE system:
eq = D[u[t, x], t] == D[u[t, x], x, x];ic = u[0, x] == x (1 - x);bc = u[t, 0] == 0;points = 4;grid = Array[# &, points, {0, 1}];ptoofunc = pdetoode[u[t, x], t, grid, 2];ode = ptoofunc[eq] // Rest;odeic = ptoofunc[ic] // Rest;odebc = ptoofunc[bc];var = u /@ grid;asol = DSolve[{ode, odeic, odebc}, var[t] // Through, t] // Simplify // First
$$ \begin{array}{l} u(0)(t)=0 \\ u\left(\frac{1}{3}\right)(t)=\frac{1}{18} \left(-18 t+e^{-18 t}+3\right) \\ u\left(\frac{2}{3}\right)(t)=\frac{2}{9}-2 t \\ u(1)(t)=-3 t-\frac{e^{-18 t}}{18}+\frac{1}{18} \\ \end{array} $$
style = Style[#, 16] &;ParametricPlot3D[ Flatten@{t, #} & /@ ({grid, asol[[All, -1]]}\[Transpose]) // Evaluate, {t, 0, 1}, BoxRatios -> {1, 1, 0.4}, AxesLabel -> style /@ {"t", "x", "u"}]
I only used 4 grid points, or else the symbolic calculation will be too slow, but as you can see the result agrees well with the FDM solution.
For completeness, here are implementations in some other programming languages. Notice I've only discretized $x$ in the following.
Maple
points := 24: y[0](t):=0: h := 1/points: eq := seq((D(y[i]))(t) = (y[i-1](t)-2*y[i](t)+y[i+1](t))/h^2, i = 1 .. points-1), (D(y[points]))(t) = (-y[points-3](t)+4*y[points-2](t)-5*y[points-1](t)+2*y[points](t))/h^2: ic := seq(y[points*x](0) = -x^2+x, x = h .. 1, h): var := [seq([t, y[i](t)], i = 1 .. points)]: p := dsolve([eq, ic], numeric, range = 0 .. 1): with(plots): odeplot(p, var, size = [default, .618]);
Notice
Maple can also calculate the symbolic solution of the sytem with the following line (remember to choose a small
points first):
dsolve([eq, ic]);
The solution is consistent with that of
Mathematica. Octave
The code isn't tested in, but should be compatible with MATLAB.
Notice $u(t,0)=0$ isn't plotted in the resulting graph.
points=24;dx=1/points;one=ones(points,1);mat1=spdiags([one -2*one one],[-1 0 1],points-1,points);mat2=zeros(1,points);mat2(points-3:points)=[-1 4 -5 2];mat=[mat1;mat2]/dx^2;span=dx:dx:1;[T,Y]=ode45(@(t,y) mat*y,[0,1],span.*(1-span));plot(T,Y)
Python
from numpy import zeros,linspacefrom scipy.integrate import odeintpoints=24;h=1/points;def func(y,t0): dydt=zeros(points+1) dydt[0]=0 for i in range(1,points): dydt[i]=(y[i-1]-2*y[i]+y[i+1])/h**2 dydt[points]=(-y[points-3]+4*y[points-2]-5*y[points-1]+2*y[points])/h**2 return dydty0=[x*h-x*x*h*h for x in range(0,points+1)]t=linspace(0,1,200)sol=odeint(func,y0,t)from matplotlib.pyplot import plot,showplot(t,sol)show()
As you can see, all these implementations lead to the same solution.
Background Information
OK, let me explain a bit about why I insist on finding the meaning of this strange difference scheme. Actually what I've reproduced in this question is the behavior of
Mathematica function
NDSolve when insufficient b.c. is added to it for solving time dependent PDE. For more information, check this post. As I wrote there:
I know the best countermeasure is to add the missing boundary condition, I'm just curious. Anyway, if the output with insufficient boundary condition is completely meaningless, why doesn't
NDSolvesimply stop calculating and return the input?
So, perhaps we'll have to admit the solution is really meaningless in the end, but I think this should be the last thing to do. |
this is a mystery to me, despite having changed computers several times, despite the website rejecting the application, the very first sequence of numbers I entered into it's search window which returned the same prompt to submit them for publication appear every time, I mean ive got hundreds of them now, and it's still far too much rope to give a person like me sitting along in a bedroom the capacity to freely describe any such sequence and their meaning if there isn't any already there
my maturity levels are extremely variant in time, that's just way too much rope to give me considering its only me the pursuits matter to, who knows what kind of outlandish crap I might decide to spam in each of them
but still, the first one from well, almost a decade ago shows up as the default content in the search window
1,2,3,6,11,23,47,106,235
well, now there is a bunch of stuff about them pertaining to "trees" and "nodes" but that's what I mean by too much rope you cant just let a lunatic like me start inventing terminology as I go
oh well "what would cotton mathers do?" the chat room unanimously ponders lol
i see Secret had a comment to make, is it really a productive use of our time censoring something that is most likely not blatant hate speech? that's the only real thing that warrants censorship, even still, it has its value, in a civil society it will be ridiculed anyway?
or at least inform the room as to whom is the big brother doing the censoring? No? just suggestions trying to improve site functionality good sir relax im calm we are all calm
A104101 is a hilarious entry as a side note, I love that Neil had to chime in for the comment section after the big promotional message in the first part to point out the sequence is totally meaningless as far as mathematics is concerned just to save face for the websites integrity after plugging a tv series with a reference
But seriously @BalarkaSen, some of the most arrogant of people will attempt to play the most innocent of roles and accuse you of arrogance yourself in the most diplomatic way imaginable, if you still feel that your point is not being heard, persist until they give up the farce please
very general advice for any number of topics for someone like yourself sir
assuming gender because you should hate text based adam long ago if you were female or etc
if its false then I apologise for the statistical approach to human interaction
So after having found the polynomial $x^6-3x^4+3x^2-3$we can just apply Eisenstein to show that this is irreducible over Q and since it is monic, it follwos that this is the minimal polynomial of $\sqrt{1+\sqrt[3]{2}}$ over $\mathbb{Q}$ ? @MatheinBoulomenos
So, in Galois fields, if you have two particular elements you are multiplying, can you necessarily discern the result of the product without knowing the monic irreducible polynomial that is being used the generate the field?
(I will note that I might have my definitions incorrect. I am under the impression that a Galois field is a field of the form $\mathbb{Z}/p\mathbb{Z}[x]/(M(x))$ where $M(x)$ is a monic irreducible polynomial in $\mathbb{Z}/p\mathbb{Z}[x]$.)
(which is just the product of the integer and its conjugate)
Note that $\alpha = a + bi$ is a unit iff $N\alpha = 1$
You might like to learn some of the properties of $N$ first, because this is useful for discussing divisibility in these kinds of rings
(Plus I'm at work and am pretending I'm doing my job)
Anyway, particularly useful is the fact that if $\pi \in \Bbb Z[i]$ is such that $N(\pi)$ is a rational prime then $\pi$ is a Gaussian prime (easily proved using the fact that $N$ is totally multiplicative) and so, for example $5 \in \Bbb Z$ is prime, but $5 \in \Bbb Z[i]$ is not prime because it is the norm of $1 + 2i$ and this is not a unit.
@Alessandro in general if $\mathcal O_K$ is the ring of integers of $\Bbb Q(\alpha)$, then $\Delta(\mathcal O_K) [\mathcal O_K:\Bbb Z[\alpha]]^2=\Delta(\mathcal O_K)$, I'd suggest you read up on orders, the index of an order and discriminants for orders if you want to go into that rabbit hole
also note that if the minimal polynomial of $\alpha$ is $p$-Eisenstein, then $p$ doesn't divide $[\mathcal{O}_K:\Bbb Z[\alpha]]$
this together with the above formula is sometimes enough to show that $[\mathcal{O}_K:\Bbb Z[\alpha]]=1$, i.e. $\mathcal{O}_K=\Bbb Z[\alpha]$
the proof of the $p$-Eisenstein thing even starts with taking a $p$-Sylow subgroup of $\mathcal{O}_K/\Bbb Z[\alpha]$
(just as a quotient of additive groups, that quotient group is finite)
in particular, from what I've said, if the minimal polynomial of $\alpha$ wrt every prime that divides the discriminant of $\Bbb Z[\alpha]$ at least twice, then $\Bbb Z[\alpha]$ is a ring of integers
that sounds oddly specific, I know, but you can also work with the minimal polynomial of something like $1+\alpha$
there's an interpretation of the $p$-Eisenstein results in terms of local fields, too. If the minimal polynomial of $f$ is $p$-Eisenstein, then it is irreducible over $\Bbb Q_p$ as well. Now you can apply the Führerdiskriminantenproduktformel (yes, that's an accepted English terminus technicus)
@MatheinBoulomenos You once told me a group cohomology story that I forget, can you remind me again? Namely, suppose $P$ is a Sylow $p$-subgroup of a finite group $G$, then there's a covering map $BP \to BG$ which induces chain-level maps $p_\# : C_*(BP) \to C_*(BG)$ and $\tau_\# : C_*(BG) \to C_*(BP)$ (the transfer hom), with the corresponding maps in group cohomology $p : H^*(G) \to H^*(P)$ and $\tau : H^*(P) \to H^*(G)$, the restriction and corestriction respectively.
$\tau \circ p$ is multiplication by $|G : P|$, so if I work with $\Bbb F_p$ coefficients that's an injection. So $H^*(G)$ injects into $H^*(P)$. I should be able to say more, right? If $P$ is normal abelian, it should be an isomorphism. There might be easier arguments, but this is what pops to mind first:
By Schur-Zassenhaus theorem, $G = P \rtimes G/P$ and $G/P$ acts trivially on $P$ (the action is by inner auts, and $P$ doesn't have any), there is a fibration $BP \to BG \to B(G/P)$ whose monodromy is exactly this action induced on $H^*(P)$, which is trivial, so we run the Lyndon-Hochschild-Serre spectral sequence with coefficients in $\Bbb F_p$.
The $E^2$ page is essentially zero except the bottom row since $H^*(G/P; M) = 0$ if $M$ is an $\Bbb F_p$-module by order reasons and the whole bottom row is $H^*(P; \Bbb F_p)$. This means the spectral sequence degenerates at $E^2$, which gets us $H^*(G; \Bbb F_p) \cong H^*(P; \Bbb F_p)$.
@Secret that's a very lazy habit you should create a chat room for every purpose you can imagine take full advantage of the websites functionality as I do and leave the general purpose room for recommending art related to mathematics
@MatheinBoulomenos No worries, thanks in advance. Just to add the final punchline, what I wanted to ask is what's the general algorithm to recover $H^*(G)$ back from $H^*(P; \Bbb F_p)$'s where $P$ runs over Sylow $p$-subgroups of $G$?
Bacterial growth is the asexual reproduction, or cell division, of a bacterium into two daughter cells, in a process called binary fission. Providing no mutational event occurs, the resulting daughter cells are genetically identical to the original cell. Hence, bacterial growth occurs. Both daughter cells from the division do not necessarily survive. However, if the number surviving exceeds unity on average, the bacterial population undergoes exponential growth. The measurement of an exponential bacterial growth curve in batch culture was traditionally a part of the training of all microbiologists...
As a result, there does not exists a single group which lived long enough to belong to, and hence one continue to search for new group and activity
eventually, a social heat death occurred, where no groups will generate creativity and other activity anymore
Had this kind of thought when I noticed how many forums etc. have a golden age, and then died away, and at the more personal level, all people who first knew me generate a lot of activity, and then destined to die away and distant roughly every 3 years
Well i guess the lesson you need to learn here champ is online interaction isn't something that was inbuilt into the human emotional psyche in any natural sense, and maybe it's time you saw the value in saying hello to your next door neighbour
Or more likely, we will need to start recognising machines as a new species and interact with them accordingly
so covert operations AI may still exists, even as domestic AIs continue to become widespread
It seems more likely sentient AI will take similar roles as humans, and then humans will need to either keep up with them with cybernetics, or be eliminated by evolutionary forces
But neuroscientists and AI researchers speculate it is more likely that the two types of races are so different we end up complementing each other
that is, until their processing power become so strong that they can outdo human thinking
But, I am not worried of that scenario, because if the next step is a sentient AI evolution, then humans would know they will have to give way
However, the major issue right now in the AI industry is not we will be replaced by machines, but that we are making machines quite widespread without really understanding how they work, and they are still not reliable enough given the mistakes they still make by them and their human owners
That is, we have became over reliant on AI, and not putting enough attention on whether they have interpret the instructions correctly
That's an extraordinary amount of unreferenced rhetoric statements i could find anywhere on the internet! When my mother disapproves of my proposals for subjects of discussion, she prefers to simply hold up her hand in the air in my direction
for example i tried to explain to her that my inner heart chakras tell me that my spirit guide suggests that many females i have intercourse with are easily replaceable and this can be proven from historical statistical data, but she wont even let my spirit guide elaborate on that premise
i feel as if its an injustice to all child mans that have a compulsive need to lie to shallow women they meet and keep up a farce that they are either fully grown men (if sober) or an incredibly wealthy trust fund kid (if drunk) that's an important binary class dismissed
Chatroom troll: A person who types messages in a chatroom with the sole purpose to confuse or annoy.
I was just genuinely curious
How does a message like this come from someone who isn't trolling:
"for example i tried to explain to her that my inner heart chakras tell me that my spirit guide suggests that many ... with are easily replaceable and this can be proven from historical statistical data, but she wont even let my spirit guide elaborate on that premise"
3
Anyway feel free to continue, it just seems strange @Adam
I'm genuinely curious what makes you annoyed or confused yes I was joking in the line that you referenced but surely you cant assume me to be a simpleton of one definitive purpose that drives me each time I interact with another person? Does your mood or experiences vary from day to day? Mine too! so there may be particular moments that I fit your declared description, but only a simpleton would assume that to be the one and only facet of another's character wouldn't you agree?
So, there are some weakened forms of associativity. Such as flexibility ($(xy)x=x(yx)$) or "alternativity" ($(xy)x=x(yy)$, iirc). Tough, is there a place a person could look for an exploration of the way these properties inform the nature of the operation? (In particular, I'm trying to get a sense of how a "strictly flexible" operation would behave. Ie $a(bc)=(ab)c\iff a=c$)
@RyanUnger You're the guy to ask for this sort of thing I think:
If I want to, by hand, compute $\langle R(\partial_1,\partial_2)\partial_2,\partial_1\rangle$, then I just want to expand out $R(\partial_1,\partial_2)\partial_2$ in terms of the connection, then use linearity of $\langle -,-\rangle$ and then use Koszul's formula? Or there is a smarter way?
I realized today that the possible x inputs to Round(x^(1/2)) covers x^(1/2+epsilon). In other words we can always find an epsilon (small enough) such that x^(1/2) <> x^(1/2+epsilon) but at the same time have Round(x^(1/2))=Round(x^(1/2+epsilon)). Am I right?
We have the following Simpson method $$y^{n+2}-y^n=\frac{h}{3}\left (f^{n+2}+4f^{n+1}+f^n\right ), n=0, \ldots , N-2 \\ y^0, y^1 \text{ given } $$ Show that the method is implicit and state the stability definition of that method.
How can we show that the method is implicit? Do we have to try to solve $y^{n+2}$ as a function of $y^{n+1}$ ?
@anakhro an energy function of a graph is something studied in spectral graph theory. You set up an adjacency matrix for the graph, find the corresponding eigenvalues of the matrix and then sum the absolute values of the eigenvalues. The energy function of the graph is defined for simple graphs by this summation of the absolute values of the eigenvalues |
Let $\chi$ be a Dirichlet character mod 4. Show $\sum_{n=1}^\infty \chi(n)\phi(n)n^{-s} = \frac{L(\chi,s-1)}{L(\chi,s)}$ and $\sum_{n=1}^\infty \chi(n)d(n)n^{-s}=L(\chi,s)^2$. ($\phi$ is the Euler totient function and $d(n)$ is the number of divisors of $n$.)
First, is this true just for characters mod 4 and not true in general? I'm not sure what specific properties about characters mod 4 I should use besides that $\chi(n)=0$ for $n$ even.
I took the log of both sides and tried to use the following: $$L(\chi,s)=\prod_{p \text{ prime}}\frac{1}{1-\frac{\chi(p)}{p^s}}$$
$$\log L(\chi,s)=\sum_{p \text{ prime}}\sum_{n=1}^\infty \frac{\chi(p)^n}{np^{ns}}$$
$\chi$ and $\phi$ are multiplicative, so we can express $$\sum_{n=1}^\infty \chi(n)\phi(n)n^{-s}$$ as the Euler product $$\prod_p(1+\frac{\chi(p)\phi(p)}{p^s}+\frac{\chi(p^2)\phi(p^2)}{p^{2s}}+\cdots).$$
Manipulating things are not quite working. Any help would be appreciated. |
However, it appears to me that there's another very simple way to do inference directly on the average return and standard deviation implied by the Sharpe. I'm wondering (1) if this approach is valid, or whether I'm overlooking something; (2) if there's some literature on this and similar approaches.
Given a strategy with (required / targeted / IS-based) daily Sharpe ratio $SR = \frac{\mu}{\sigma}$, with returns statistics $\mu$ (mean) and $\sigma$ (standard deviation).
The question is how long of an OS period the strategy needs at minimum to establish statistically significant returns.
Assuming iid returns (big assumption), the sample mean of the daily returns in the OS period after $n$ days can be modelled with (probably more appropriate to use t distribution though) $$\hat\mu \sim \mathcal{N}\left(\mu, \frac{\sigma^2}{n}\right)$$.
We define statistical significance at level $\alpha$ as the $z_{1-\alpha}$ score (one-sided test) from zero (i.e., the null is that $\mu=0$): $$\hat\mu \geq z_{1-\alpha} \frac{\sigma}{\sqrt{n}}$$.
With the definition of the daily Sharpe $SR = \frac{\mu}{\sigma}$ and substituting $\mu=\hat\mu$ this becomes:
$$n \geq \left(\frac{z_{1-\alpha}}{SR}\right)^2$$
Intuitively this makes sense: A lower Sharpe, or higher significance level increases the minimum sampling period. For instance, at $\alpha=90\%$ significance we obtain for a strategy with an annual Sharpe of 1: $n \geq (1.28 * \sqrt{252} / 1)^2 \approx 413$ days, but at $\alpha=95\%$ already $n \geq 682$ days. |
I struggle with this exercice.
If $f: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ is differentiable and $g: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ is given by $g(x_1, x_2) = f(x^2_1- x^2_2, 2x_1x_2)$. Show that $g$ is also differentiable and compute $\frac{\partial{g}}{\partial{x_1}}$ and $\frac{\partial{g}}{\partial{x_2}}$ in terms of $\frac{\partial{f}}{\partial{x_1}}$ and $\frac{\partial{f}}{\partial{x_2}}$
Clearly, this is a compound function so we have to use the chain rule somewhere.
As we know : if $g = f \circ h$ then $Dg(a)[v] = Df(h(a))[Dh(x)[v]]$ where $a$ is an interior point of $\mathbb{R}^3$ and $v \in \mathbb{R}^3$ is a direction vector.
In class, teacher showed this :
Let $x = (x_1, x_2)$ with $x_1 = (x \cdot e_1)$ and $x_2 = (x \cdot e_2)$
$g = f \circ h$ with $h(x) = ((x \cdot e_1)^2 - (x \cdot e_2)^2)e_1 + 2(x \cdot e_1)(x \cdot e_2)e_2$
So, $\frac{\partial{g}}{\partial{x_1}} = Dg(x)[e_1] = Df(h(x))[Dh(x)[e_1]]$
From here, I don't really know what to do. I don't understand the use of a basis $(e_1, e_2)$ in differentiation and how to derive a dot product with a vector of this basis.
Thanks for the help. |
Is there any comprehensive list (books, online, ...) of rules for manipulating infinite series (partial sums) to find convergence of a sum? Often authors use some "trick" to compute an infinite series. Following this trick is always a disclaimer, such as "adding infinite sequences is not the same as adding discrete values so the usual rules of algebra will not work."
Huh? Then how am I supposed to learn what I can and cannot do when all I have been shown is a trick that works in a particular case? For example,
\begin{align*} \sum_{n=1}^{\infty}\frac{1}{n(n+1)}\rightarrow S_{n}&=\sum_{k=1}^{n}\frac{1}{k(k+1)}\\\ &=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{n(n+1)}\\\ &=\; \sum_{k=1}^{n}(\frac{1}{k}-\frac{1}{k+1})\\ &=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots+(\frac{1}{n}-\frac{1}{n+1})\\\ &=1-\frac{1}{n+1}\rightarrow \lim_{n\rightarrow \infty}\left(1-\frac{1}{n+1}\right)=1-0=1. \end{align*} OK, I see the author used partial fractions expansion followed by grouping like-terms. However, elsewhere I'll find that grouping like-terms will lead you to the wrong answer. Such is the case with this infinite sum:
$$ S_{n}=1-1+1-1+1+\cdots \stackrel{?}{\Longrightarrow} (1-1)+(1-1)+\cdots=0 $$ $$ S_{n}=1-1+1-1+1+\cdots\stackrel{?}{\Longrightarrow} 1+(-1+1)+(-1+1)+\cdots=1 $$ ...So does this mean infinite series are not associative? Or are they? What are the invariant properties of infinite series that can be confidently used when manipulating an infinite series?
FOLLOW UP:Thanks for the great answers so far, yet they have led me to question the practicality of writing in summation form. Is it not more useful to simply write the infinite sum as its corresponding infinite partial sum sequence?
E.g.: Why is the fourier series written as an infinite sum? I realize it can be used to create a continuous analog for many types of discontinuous functions, and yet I feel I gain nothing when I write out the fourier series to say, a solution to a partial differential equation, because Idk how to evaluate the output to a particular input since its defined by an infinite sum of sin's and cosines. (Besides finding the partial sum sequence) Is there not a more direct way? |
Assume we use FEM with piecewise linear finite elements to discretize the BVP over $\omega = (0,1)$: $-u''+ bu' + u = 2x$, $u(0) = u(1) = 0$ for parameter $b\in R$. Given a mesh $T = \left\{x_i\right\}_{i=0}^{N}$ where $x_i = 1 - (1-\frac{i}{N})^{\beta}$ for some $\beta > 1$. Using principle of error equilibration, find a value of $\beta$.
My attempt: I tried to look up the literature for the mathematical definition of "error equilibrium in FEM," but I failed to find it. Could someone please help with a useful reference?
On the other hand, I have a question on my discretization method: as we know, multiplying both sides of the PDE by a test function $v\in H_{0}^{1}(w)$. Then we would have the following variational formulation:
$\int_{0}^{1} (u'v' - buv' + uv) = \int_{0}^{1} 2xv$ for every $v\in H_{0}^{1}(w)$
Now, define $B[u,v] = \int_{0}^{1} (u'v' - buv' + uv)$ and $F(v) = 2xv$. Let $\left\{\phi_{i}\right\}_{i=0}^{N-2}$ be the basis of the finite-dimensional subspace $V_h\subset H_{0}^{1}(w)$ over the given mesh $T$. Now, we can express $u(x) = \sum_{i=0}^{N-2} \phi_{i}(x)u_i$ where $\phi_{i}(x)$ are piecewise linear functions defined over the mesh $T$.
My questions:
(1) Due to the boundary conditions: $u(0) = u(1) = 0$, would the first basis function (i.e, $\phi_0(x)$) be a piecewise hat function with vertices $(x_0, 0), (x_1, 1)$ and $(x_2,0)$, and the last basis function (i.e, $\phi_N(x)$) are piecewise hat functions with vertices $(x_{N-2}, 0), (x_{N-1}, 1)$ and $(x_N,0)$? Usually, for Dirichlet condition, $\phi_0(x)$ are with vertices $(x_0, 1), (x_1,0)$ and $(x_0, 0)$, but then in that case, $u(x) = \sum_{i=1}^{N} \phi_{i}(x)u_i + \alpha \phi_{0}(x)$ where $\alpha = u(0)$, and $u(1) = 0$.
(2) Is working with $v\in H_{0}^{1}(w)$ the right setting in this case? Usually, for one-dimensional problem like this, we are only allowed to let either $v(0) = 0$ or $v(1) = 0$, but in this case as I could see that the information from the given data at $x=1$ and $x=0$ are "useless," I decided to set $v(1) = v(0) = 0$. This resulted in the slight change in the set of basis functions $\left\{\phi_{i}\right\}_{i=0}^{N}$.
(3) Assume the BCs now are $u'(0) = 1$ and $u(1) = 1$, could I use the same set of test function as described in (1) above, but adding two elements: $\phi_{0}(x) = $ linear function between 2 points $(0,1)$ and $(x_1, 0)$, while $\phi_{N}(x) = $ linear function connecting $(x_{N-1},0)$ and $(x_N, 1).$ |
Linear Regression and Likelihood
The linear estimator $y$ is
As usual, we have redefined our data to get rid of the intercept $\beta^0$.
In ordinary linear models, we find the error being the difference between the target $\hat y$ and the estimator $y$
which is required to have a minimun absolute value.
We could use least squares to solve the problem. However, instead of using a deterministic estimator $\beta^m X_m^{\phantom{m}n}$, we assume a Gaussian random estimator
where we have used the knowledge of linear regression, that the mean of the estimator should be a linear model $\beta^m X_m^{\phantom{m}n}$. The likelihood becomes
It is not surprising that requiring the maximum likelihood will lead to the same result as least squares due to log takes out exponential.
Bayesian Linear Model
Applying Bayes’ theorem to this problem,
Since ${\color{red}P(\hat y^n)}$ doesn’t depend on the parameters and is a constant, we will ignore it for the sake of optimization.
We will assume a least information model for $P([X_m^{\phantom{m}n}, \beta^m])$, that is
Our posterior becomes
This is nothing but Ridge loss with coefficient $\lambda$, where |
Moreover, what happens if we require the monoid to be
commutative?
(I believe that my question with "semigroup" instead of "monoid" has the same answer – removing or adding the identity element shouldn't change so much).
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Moreover, what happens if we require the monoid to be
commutative?
(I believe that my question with "semigroup" instead of "monoid" has the same answer – removing or adding the identity element shouldn't change so much).
Let $G$ be an arbitrary group. We construct an edge-colured graph $(V,E,C,\Gamma)$ having $G$ as automorphism group: As set of vertices, we take $V=G$, as set of edges, we take $E=\{\,g\to h: g,h\in G, g\ne h\,\}$ (i.e., we are dealing with thge complete directed graph on $G$). As set of colours, we take $C=G\setminus\{1\}$, and let the colouring $\Gamma\colon E\to C$ be given as $\Gamma(g\to h)=hg^{-1}$.
Claim. Let $G$ ba any group and $(V,E,C,\Gamma)$ as described above. Then $\operatorname{Aut}(V,E,C,\Gamma)\cong G$. Proof. If $a\in G$, then $\phi_a\colon V\to V$, $g\mapsto ga$ induces an automorphism. Indeed, if $g\to h$ is an edge then so is $ga\to ha$, and we have $\Gamma(ga\to ha)=ha(ga)^{-1}=hg^{-1}=\Gamma(g\to h)$.As $\phi_a(1)=a$, we see that $a\mapsto \phi_a$ is a monomorphism $G\to \operatorname{Aut}(V,E,C,\Gamma)$.On the other hand, let $\phi$ be an arbitrary automorphism of $(V,E,C,\Gamma)$ and let $a=\phi(1)$. Then $\phi=\phi_a$ because for any $g\ne 1$, the edge $a\to \phi(g)$ must have the same colour as $1\to g$, hence $\phi(g)a^{-1}=g$, $\phi(g)=ga$. $\square$
Let $(V,E,C,\Gamma)$ be an edge-coloured directed graph, where wlog. $C$ is an ordinal. We define an (un-coloured) directed graph $(V',E')$ as follows. Let $V'$ consist of
Let $E'$ consist of
Claim. Let $(V',E')$ be obtained as described from $(V,E,C,\Gamma)$. we have $\operatorname{Aut}(V'E')\cong\operatorname{Aut}(V,E,C,\Gamma)$. Proof.Given an automorphism $\phi$ of $(V,E,C,\Gamma)$, we can readily find a corresponding automorphism of $(V',E')$ by mapping vertices $$\begin{align}v&\mapsto \phi(v),\\ \hat v&\mapsto \widehat{\phi(v)},\\ (\beta,v,w)&\mapsto (\beta,\phi(v),\phi(w)).\end{align}$$This gives us a monomorphism $\operatorname{Aut}(V,E,C,\Gamma)\to \operatorname{Aut}(V'E')$. We need to show that it is onto.Vertices of the second type ("$\hat v$") are characterized by the fact that they have no successors. Vertices of the first type ("$v$") are characterized by the fact that they have a successor of the first type. Thus also vertices of the third type are characterized.An automorphism $\phi$ of $(V',E')$ must respect types.Moreover, vertices of the third type with ordinal $0$ are characterized by being of third type and having a type one predecessor, so this must also be respected by an automorphism.Assume $\phi((0,v,w))=(0,a,b)$ where $v\to w$ is an edge in $E$ of colur $c$. Then $\phi(v)=a$ by looking at the unique predecessors and $\phi(\hat v)=\hat a$ by their type 1 successors.I claim that $\phi((\beta,v,w))=(\beta,a,b)$ for all $\beta\le c$. Indeed, if $\beta>0$ and the claim holds for all smaller ordinals, then $\phi((\beta,v,w))$ is determined among all type 3 vertices by the fact that precisely all $(\gamma,a,b)$ with $\gamma<\beta$ are predecessors.After that, note that $w$ is the unique successor of $(c,v,w)$, hence $(c,a,b)$ must have a sucessor of second type; this must be $b$ and therefore $\Gamma(a\to b)=c$ and $\phi(w)=b$. We conclude that $\phi$ induces a permutation of $V$ and is uniquely determined by that permutation, and if $v\to w$ is an edge in $E$ then $\phi(v)\to \phi(w)$ is also an edge and is of the same colour. In other words, our monomorphism $\operatorname{Aut}(V,E,C,\Gamma)\to \operatorname{Aut}(V'E')$ is in fact an isomorphism. $\square$
Next, $(V,E)$ be a directed graph. We define an undirected graph $(V',E')$ as follows.
Let $V'=V\cup (E\times\{1,2,3,4\}$, i.e., $V'$ consists of
and let $E'$ consist of
Claim. With $(V',E')$ obtained from $(V,E)$, we have$\operatorname{Aut}(V,E)\cong \operatorname{Aut}(V',E')$. Proof.As before, from an automorphism $\phi$ of $(V,E)$, we readily obtain an automoprhism of $(V','E')$ that maps $v\mapsto \phi(v)$ and $(v,w)_i\mapsto (\phi(v),\phi(w))_i$., thereby obtaining a monomoprhism $\operatorname{Aut}(V,E)\to\operatorname{Aut}(V',E')$.
Now let $\phi$ be any automophism of $(V',E')$. Observe that vertices of the form $(v,w)_1$ are precisely those that are of degree $2$ and with an edge between their neighbours. These two neighbours have precisely one other common neighbour, namely the corrsponding $(v,w)_3$. Exactly one of these common neighbours, namely $(v,w)_2$, has a neigbour of degree $1$, which is $(v,w)_4$; the other common neighbour must be $v$. And the only neighbour of $(v,w)_3$ not yet mentioned is $w$. We conclude that $\phi$ must map $(v,w)_1$ to some $(a,b)_1$ and that then $\phi(v)=a$, $\phi(w)=b$, $\phi((v,w)_i)=(a,b)_i$. In particular $\phi$ induces a permutation of $V\subset V'$ and such that $(V,E)$ has an edge $\phi(v)\to\phi(w)$ whenever $v\to w$ is an edge. In other words, our monomorphism $\operatorname{Aut}(V,E)\to\operatorname{Aut}(V',E')$ is an isomorphism. $\square$
Finally, given an undirected graph $(V,E)$, we construct a commutative semigroup with the same automorphism group: Let $S=V\cup E\cup\{0\}$ and declare
Any automorphism of the graph gives rise to an automorphism of $S$. On the other hand, any automorphism of $S$ must respect vertices (idempotent elements) and edges (other non-zero elements)products) and vertex-edge incidence (vertex and edge with non-zero product)
(This essentially repeats the construction given in the accepted answer to one of the referenced similar questions).
This shows the final
Claim. Let $(V,E)$ be an undirected graph. Then there exists a commutative monoid $S$ such that $\operatorname{Aut}(V,E)\cong\operatorname{Aut}(S)$. $\square$
Combining all of the above, we see that for every (finite or inifnite) group $G$, there exists a commutative monoid $S$ such that $$\operatorname{Aut}(S)\cong G. $$ |
In the proof that $L^p(\mu)$ is complete for $p\in[1,\infty]$ (as done in
Saxe, Theorem 3.21 or in Folland, Theorem 6.6, the latter of which is outlined here) we make use of the following completeness criterion:
Lemma:Let $(X,\|\cdot\|_X)$ be a normed vector space and consider a sequence $(x_n)_{n\in\mathbb N}$ in $X$. If $\sum_{i=1}^\infty x_i$ converges whenever $\sum_{i=1}^\infty \|x_i\|_X$ converges, then $(X,\|\cdot\|_X)$ is complete.
In order to show completeness, we start with a Cauchy sequence $(f_k)_{k\in\mathbb N}$ in $L^p(\mu)$ and we aim to show,
with the help of the above result, that $(f_k)_{k\in\mathbb N}$ converges in $L^p(\mu)$. In particular, we suppose that Cauchy $(f_k)_{k\in\mathbb N}$ is such that $\sum_{i=1}^\infty \|f_k\|_p$ converges. The authors then go to show that the corresponding series converges, which yields the result.
What I don't understand is why
is this sufficient to show that any Cauchy sequence of elements converges in $L^p(\mu)$? In considering those Cauchy sequences $(f_k)_{k\in\mathbb N}$ which satisfy that $\sum_{i=1}^\infty \|f_k\|_p$ converges, a re we not restricting our attention to a select few Cauchy sequences? Why is focusing on this subset of Cauchy sequences enough?
-- -- -- --
I had thought, perhaps, that there was a relation between Cauchy sequences and their corresponding series' converging absolutely, but that has been indicated to not be true here. |
It is clearly enough to show that an infinite dimensional vector space $V$ has smaller dimension that its dual $V^*$.
Let $B$ be a basis of $V$, let $\mathcal P(B)$ be the set of its subsets, and for each $A\in\mathcal P(B)$ let $\chi_A\in V^*$ be the unique functional on $V$ such that the restriction $\chi_A|_B$ is the characteristic function of $A$. This gives us a map $\chi:A\in\mathcal P(B)\mapsto\chi_A\in V^*$.
Now a complete infinite boolean algebra $\mathcal B$ contains an independent subset $X$ such that $|X|=|\mathcal B|$---here, that $X$ be independent means that whenever $n,m\geq0$ and $x_1,\dots,x_n,y_1,\dots,y_m\in X$ we have $x_1\cdots x_n\overline y_1\cdots\overline y_n\neq0$. (This is true in this generality according to [Balcar, B.; Franěk, F. Independent families in complete Boolean algebras. Trans. Amer. Math. Soc. 274 (1982), no. 2, 607--618. MR0675069], but when $\mathcal B=\mathcal P(Z)$ is the algebra of subsets of an infinite set $Z$, this is a classical theorem of [Fichtenholz, G. M; Kantorovich L. V. Sur les opérations linéaires dans l'espace des fonctions bornées. Studia Math. 5 (1934) 69--98.] and [Hausdorff, F. Über zwei Sätze von G. Fichtenholz und L. Kantorovich. Studia Math. 6 (1936) 18--19])
If $X$ is such an independent subset of $\mathcal P(B)$ (which is a complete infinite boolean algebra), then $\chi(X)$ is a linearly independent subset of $V^*$, as one can easily check. It follows that the dimension of $V^*$ is at least $|X|=|\mathcal P(B)|$, which is strictly larger than $|B|$.
Later: The proof of the existence of an independent subset is not hard; it is given, for example, in this notes by J. D. Monk as Theorem 8.9. In any case, I think this proof is pretty because it captures precisely the intuition (or, rather, my intuition) of why this is true. I have not seen the paper by Fichtenhold and Kantorovich (I'd love to get a copy!) but judging from its title one sees that they were doing similar things... |
Difference between revisions of "Lower attic"
From Cantor's Attic
Line 12: Line 12:
* the [[Bachmann-Howard]] ordinal
* the [[Bachmann-Howard]] ordinal
* [[admissible]] ordinals and [[admissible#relativized_admissible | relativized Church-Kleene $\omega_1^x$]]
* [[admissible]] ordinals and [[admissible#relativized_admissible | relativized Church-Kleene $\omega_1^x$]]
−
*
+
* [[Church-Kleene omega_1 | $\omega_1^{ck}$]], the supremum of the computable ordinals
* the [[Feferman-Schütte]] ordinal [[Feferman–Schütte | $\Gamma_0$]]
* the [[Feferman-Schütte]] ordinal [[Feferman–Schütte | $\Gamma_0$]]
* [[epsilon naught | $\epsilon_0$]] and the hierarchy of [[epsilon naught#epsilon_numbers | $\epsilon_\alpha$ numbers]]
* [[epsilon naught | $\epsilon_0$]] and the hierarchy of [[epsilon naught#epsilon_numbers | $\epsilon_\alpha$ numbers]]
Revision as of 08:06, 30 December 2011
Welcome to the lower attic, where the countably infinite ordinals climb ever higher, one upon another, in an eternal self-similar reflecting ascent.
$\omega_1$, the first uncountable ordinal, and the other uncountable cardinals of the middle attic stable ordinals The ordinals of infinite time Turing machines, including the Bachmann-Howard ordinal admissible ordinals and relativized Church-Kleene $\omega_1^x$ Church-Kleene $\omega_1^{ck}$, the supremum of the computable ordinals the Feferman-Schütte ordinal $\Gamma_0$ $\epsilon_0$ and the hierarchy of $\epsilon_\alpha$ numbers the small countable ordinals, such as $\omega,\omega+1,\ldots,\omega\cdot 2,\ldots,\omega^2,\ldots,\omega^\omega,\ldots,\omega^{\omega^\omega},\ldots$ up to $\epsilon_0$ Hilbert's hotel $\omega$, the smallest infinity down to the parlour, where large finite numbers dream |
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Change to browse by: References & Citations Bookmark(what is this?) Mathematics > Probability Title: A Characterization of the Normal Distribution by the Independence of a Pair of Random Vectors
(Submitted on 1 Jan 2016)
Abstract: Kagan and Shalaevski 1967 have shown that if the random variables $X_1,\dots,X_n$ are independent and identically distributed and the distribution of $\sum_{i=1}^n(X_i+a_i)^2$ $a_i\in \mathbb{R}$ depends only on $\sum_{i=1}^na_i^2$ , then each $X_i$ follows the normal distribution $N(0, \sigma)$. Cook 1971 generalized this result replacing independence of all $X_i$ by the independence of $(X_1,\dots, X_m) \textrm{ and } (X_{m+1},\dots,X_n )$ and removing the requirement that $X_i$ have the same distribution. In this paper, we will give other characterizations of the normal distribution which are formulated in a similar spirit. Submission historyFrom: Wiktor Ejsmont [view email] [v1]Fri, 1 Jan 2016 12:49:26 GMT (10kb) |
Numerical solutions of viscoelastic bending wave equations with two term time kernels by Runge-Kutta convolution quadrature
Department of Mathematics, Hunan Normal University, Changsha 410081, Hunan, China
$u_{t}(x,~t)-\int_{0}^{t}[\beta_{1}(t-s)\,u_{xx}(x,~s) - \beta_{2}(t-s)\,u_{xxxx}(x,~s)]ds = f(x,~t),$
$ 0<x<1,~ 0<t\leq T $
$ \beta_{1}(t) $
$ \beta_{2}(t) $
$ (0,~\infty) $ Mathematics Subject Classification:Primary: 45K05, 65J08; Secondary: 65D3. Citation:Da Xu. Numerical solutions of viscoelastic bending wave equations with two term time kernels by Runge-Kutta convolution quadrature. Discrete & Continuous Dynamical Systems - B, 2017, 22 (6) : 2389-2416. doi: 10.3934/dcdsb.2017122
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show all references
References:
[1] [2] [3] [4] [5] [6] [7]
H. S. Carslaw and J. C. Jaeger,
Conduction of Heat in Solids, 2 edition, Charendon Press, Oxford, 1959.
Google Scholar
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E. Hairer, S. P. Nϕrsett and G. Wanner,
[13]
E. Hairer and G. Wanner,
[14] [15] [16] [17] [18] [19] [20]
F. Liu, M. M. Meerschaert, R. J. McCough, P. Zhuang and Q. Liu,
Numerical methods for solving the multi-term time-fractional wave-diffusion equation,
[21] [22] [23] [24]
W. McLean and V. Thomée,
Maximum-norm error analysis of a numerical solution via Laplace transformation and quadrature of a fractional order evolution equation,
[25] [26] [27] [28] [29] [30] [31] [32]
A. K. Pani, G. Fairweather and R. I. Fernandes,
Alternating direction implicit orthogonal spline collocation methods for an evolution equation with a positive-type memory term,
[33]
J. Prüss,
[34]
M. Renardy, W. J. Hrusa and J. A. Nohel,
[35] [36] [37]
J. Tang and D. Xu,
The global behavior of finite difference-spatial spectral collocation methods for a partial integro-differential equation with a weakly singular kernel,
[38]
D. V. Widder,
The Laplace Transform, Princeton University Press, Princeton, NJ., 1946.
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Uniform
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H. Ye, F. Liu, I. Turner, V. Anh and K. Burrage,
Series expansion solutions for the multiterm time and space fractional partial differential equations in two-and three-dimensions,
K Rate 2 - 4 8 2 - 8 Theory
K Rate 2 - 4 8 2 - 8 Theory
K Rate 2 - 4 8 2 - 8
K Rate 2 - 4 8 2 - 8
K Rate 2 - 4 8 2 - 8 Theory
K Rate 2 - 4 8 2 - 8 Theory
K Rate 2 - 4 8 2 - 8
K Rate 2 - 4 8 2 - 8
K Rate 2 - 4 8 16 2 - 8 2 - 16 4 - 16
K Rate 2 - 4 8 16 2 - 8 2 - 16 4 - 16
K Rate 2 - 4 8 2 - 8
K Rate 2 - 4 8 2 - 8
K Rate 4 - 8 16 4 - 16
K Rate 4 - 8 16 4 - 16
K Rate 2 - 4 8 2 - 8
K Rate 2 - 4 8 2 - 8
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what does it mean by becoming extensional in the first place?
The axiom of extensionality relates to what it means for two functions to be equal. Specifically, extensionality says:
$f = g \iff \forall x \ldotp f(x) = g(x)$
That is, functions are equal if they map equal inputs to equal outputs. By this definition, quicksort and mergesort are equal, even if they don't have the same implementations, because they behave the same
as functions.
How does it become extensional
What's missing is the rule of definitional equality for functions. It usually looks like this:
$\frac{\Gamma, (x : U) \vdash (f x) = (g x):V}{\Gamma \vdash f = g: (x : U) \to V}\text{(Fun-DefEq)}$
That is, two functions are definitionally equal when they produce equal results
when applied to an abstract variable. This is similar in spirit to the way we typecheck polymorphic functions: you make sure it holds for all values by making sure it holds for an abstract value.
We get extensionality when we combine the two: if two functions always produce the same result, we should be able to find some equality proof $P$ such that $\Gamma,(x: U) \vdash P:Id_V(f x, g x)$ i.e. the proof that the two functions always produce the same result. But, if we combine this with the rule $\text{(Id-DefEq)}$, then any time two functions are extensionally equal (i.e. we can find the proof term $P$, then they are also
definitionally equal.
This is in stark contrast to an intensional system, where two functions are equal if and only if their bodies are
syntactically identical. So mergesort and quicksort are intensionally different, but extensionally the same.
The $\text{(Id-DefEq)}$ means that extensional equality is baked into the type system: if you have a type constructor $T : ((x : U) \to V) \to \mathsf{Set}$, then you can use a value of type $T\ f$ in a context expecting $T\ g$ if $f$ and $g$ map equal inputs to equal outputs. Again this is not true in an intensional system, where $f$ and $g$ might be incompatible if they're syntactically different.
Does the above mean that we purposefully drop the proof that M and N are equal and just consider them to be equal definitionally (like a presumption)?
It's even a bit stronger than that. It's saying that $M$ and $N$ are definitionally equal any time
there exists some proof that they are propositionally equal. So on the one hand, if you have a propositional proof that two values are equal, you can forget that proof and say that they are definitionally equal. But on the other hand, if you are trying to prove that two values are definitionally equal (as a dependent type checking algorithm would), then you cannot say that they are not equal unless you are certain that no proof $P$ exists. This is why it is undecidable. |
In general, one extracts a manifold invariant from a TQFT by interpreting the closed manifold as a bordism from the empty set to the empty set. The TQFT sends this bordism to a homomorphism of the ground field, which is a number. Such invariants are always multiplicative under disjoint union, this is a consequence of the TQFT being a
monoidal functor:$$\mathcal{Z}(M_1 \sqcup M_2) = \mathcal{Z}(M_1) \otimes \mathcal{Z}(M_2) = \mathcal{Z}(M_1) \cdot \mathcal{Z}(M_2)$$
Some TQFTs, like the Crane-Yetter invariant (but not, say, the Turaev-Viro model) give manifold invariants that are multiplicative under connected sum $\#$.
One way to see this is to notice that they can be defined (for connected manifolds) with Kirby calculus: Given a manifold, choose a handle decomposition and consider its link diagram. The diagram is then labelled with morphisms from a ribbon fusion category and the whole diagram is evaluated as a morphism from the monoidal identity to itself, again a number. Now the evaluation of the disjoint union of two link diagrams must then give the product of the evaluations of the respective diagrams, since a ribbon fusion category is monoidal. But the disjoint union of two link diagrams of manifolds $M_1$ and $M_2$ is the link diagram of the connected sum $M_1 \# M_2$!
Which leads me to believe that this multiplicativity secretly comes from monoidality of some functor again. Is there a category of bordisms where the monoidal product of morphisms (=bordisms) is the connected sum, and not the disjoint union? Are such TQFTs actually monoidal functors from this bordism category to $\mathrm{Vect}$?
A related, noncategorical question was asked here: Monoid structure of oriented manifolds with connect sum |
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