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I'm right now learning about Monodromy from self-studying Rick Miranda's fantastic book "Algebraic Curves and Riemann surfaces". Today, I read about monodromy, and the monodromy representation of a holomorphic map between compact Riemann surfaces. I understand that we start by having a holomorphic map $F:X \rightarrow Y$, of degree d, where X and Y are Riemann surfaces, and then we remove the branch points from Y, and all the corresponding points in X mapping to them. Let $B=\{b_1,..,b_n\}$ be the branch points, $A=\{a_1,...,a_m\}$ the ramification points. So, fix a point $q \in V = Y-B$. We have that there are d preimages of q in $U=X-A$.
So for a specific branch point b, we choose some small open neighbourhood W of b so that $F^{-1}(W)$ gives a disjoint union $W_i$ of open neighbourhoods of the points mapping to b. Take some path from our basepoint $q$ to $q_0 \in W$, call this path $\alpha$. Choosing some small loop $\beta$ with basepoint q, with winding number 1 around b, and then considering $\alpha^{-1}\circ \beta \alpha$, gives a loop on V, based at q around b. We can now see that this loop only depends on $\beta$, in some sense.
Say that the points that maps to b has multiplicity $n_i,...,n_j$. Then we have that, according to local normal form, there are local coordinates $z_j$ on the open neighbourhoods from above, so that the map takes the form $z=z_j^{n_j}$. Now, we have that the loop around b, when we lift it up here, will simply yield a cyclic permutation of the preimages in the neighbourhood.
Now, my question is mostly:
How do I apply this concretely?Let us take an example (from Miranda's book) :"Let $f(z) = 4z^2(z-1)^2/(2z-1)^2$ define a holomorphic map of degree 4 from $P^1$ to itself. Show that there are three branch points, and that the three permutations in $S_4$ are $\rho_1=(12)(34)$, $\rho_2(13)(24)$ and $\rho_3=(14)(23)$ up to conjugacy."I can find the branch points, and I see that the multiplicity of the two points mapping to it has multiplicity 2, but I don't get how to rigorously show that the above are the associated permutations.
Hope I was clear, and sorry if I wasn't.
UPDATENow, rereading the question properly, maybe he doesn't want me to find the specific permutations, but just simply showing that they have that conjugacy class. I think that is the case. But I would still be curious of how to find the specific permutation that the monodromy induces. |
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. |
I am reading these lecture notes By Daniel Bump about Character Theory on Abelian Groups. If $G$ is a group, then $G^*$ denotes its characters, the set of homomorphisms $\pi : G \to \mathbb C^{\times}$, where $\mathbb C^{\times}$ denotes the multiplicative group of the complex numbers. At the end he introduces Fourier Analysis on Abelian groups and mentiones with regard to the isomorphism of $G$ and $G^{\ast}$.
However $G$ and $G^{\ast}$ may or may not be isomorphic. We have seen that they are isomorphic if $G$ is finite; or if $G = \mathbb R$ (the additive group) or $\mathbb Q_p$ (the additive group of $p$-adic numbers), then $G \cong G^{\ast}$. But if $G = \mathbb R / \mathbb Z$ then $G^{\ast} = \mathbb Z$, and it is in this setting that most people first encounter Fourier analysis.
The functions $e^{inx}$ for $ \in \mathbb Z$ are characters, i.e. homomorphic functions from $\mathbb R / \mathbb Z$ to $\mathbb C$ (or form $\mathbb R$ with the restriction of being periodic), and in analysis we are interested in what funtion $f : \mathbb R \to \mathbb C$ that are periodic (i.e. which could be identified with function on $\mathbb R/\mathbb Z$) have a series representation in $\{ e^{inx} \}_{n \in \mathbb Z}$ (or the trigonometric functions as it is sometimes presented, but then I guess it should be isomorphic to $\mathbb N\times\mathbb N$, i.e. the positive coefficient for the $\sin$- and $\cos$-terms).
As written on Wikipedia:Convergence of Fourier Series, if a periodic function $f : \mathbb R \to \mathbb C$ is of bounded variation, then its Fourier series converges everywhere. But here is my question, as said above, the set of
homomorphisms from $\mathbb R / \mathbb Z$ is isomorphic to $\mathbb Z$, but functions of bounded variation are not neccessariliy homomorphisms, i.e. they do not have to satisfy $f(x+y) = f(x)f(y)$, they are a more general class of functions. But when the Fourier Series is unique in this case, i.e. we have a bijection between the functions of bounded variation and elements from $\mathbb Z$ (i.e. the series), then we could not have an bijection between $\mathbb Z$ and the more restricted class of homomorphism, as some preimages of $\mathbb Z$ are not homomorphisms at all?
Hope someone could clarify in what sense every homomorphism $\mathbb R / \mathbb Z$ corresponds to an element from $\mathbb Z$, and what is their relation to the "ordinary" Fourier series everyone learns about in Analysis courses, which is not restricted to just homomorphic functions, but more general classes of periodic functions. |
How has the Taylor Rule performed? Matthew Martin6/29/2014 03:41:00 PM
Tweetable
The Taylor Rule is a theoretical equation that supposedly predicts what the Federal Reserve will set the nominal "fedfunds" interest rate at, depending on the inflation rate and unemployment rate at the time. Some version of it has reigned at the heart of all monetary macro models since the late 1980s. The equation looks like this: [$$]r=\pi+\alpha\left(\pi-\pi^*\right)+\beta\left(u^*-u\right)+\gamma[$$] where [$]r[$] is the fedfunds rate set by the Federal Reserve, [$]\pi[$] is the inflation rate, [$]\pi^*[$] is the target inflation rate, [$]u^*[$] is the estimate for "full employment" also known as the "non-accelerating inflation rate of unemployment" (NAIRU), [$]u[$] is the actual unemployment rate, and [$]\alpha,\beta,\gamma[$] are constants picked by the Fed. I found data for [$]r[$], [$]\pi[$], [$]u[$], and [$]u^*[$] on FRED. Recently, the Fed has announced an inflation target of 2 percent, though as best as we can tell this is what it has always been. Because there's no variation in [$]\pi^*[$] this leaves our model technically underidentified, but we can back out the estimates by rewriting the equation as [$$]r=\left(1+\alpha\right)\pi+\beta\left(u^*-u\right)+\gamma-\alpha\pi^*[$$] which can is straightforward to estimate with OLS. The Taylor Principle says that [$]\alpha\gt 0[$], and we would generally expect [$]\beta\gt 0[$], if the Fed desires to minimize unemployment as much as possible.
I've just run this regression 603 times, estimating Taylor Rules over every 5 year period from 1959 to present, on a monthly basis. I've extracted the coefficients from the OLS and plotted them below:I won't claim that my regressions are necessarily the most robust analysis, but it certainly seems that the notion of the Fed following any kind of Taylor rule is almost entirely apocryphal. In reality, the coefficients aren't even consistently
positiveover 5 year periods.
Or, to put this into terms John Taylor can understand, the Taylor Rule is not Lucas-robust. Not by a longshot.
I will leave it to my readers to interpret what the coefficients say about economic history of the United States. (For example, where's the easy money policy that caused thestagflation in the 1970s? Where's the tight money policy that caused the disinflation of the 1980s?)
As always, you're free to review my data (also need this) and my R script file (some editing required). |
There are many ways to do this; here’s the one that I prefer. Start with your recurrence:
$$s_n=s_{n-1}+2s_{n-2}\;.\tag{1}$$
This holds for $n\ge 2$, and in addition you have the initial conditions $s_0=0$ and $s_1=1$. If we assume that $s_n=0$ for all negative integers $n$, $(1)$ also works for $n=0$; it fails only for $n=1$, when it gives $s_1=0$ instead of $s_1=1$. I’ll fix this by adding a term:
$$s_n=s_{n-1}+2s_{n-2}+[n=1]\;,\tag{2}$$
where $[n=1]$ is an Iverson bracket whose value is $1$ if $n=1$ and $0$ otherwise. The modified recurrence $(2)$ gives the correct value for all $s_n$, again on the assumption that $s_n=0$ for $n<0$.
Now multiply $(2)$ by $x^n$ and sum over $n\ge 0$:
$$\begin{align*}\sum_{n\ge 0}s_nx^n&=\sum_{n\ge 0}\left(s_{n-1}+2s_{n-2}+[n=1]\right)x^n\\\\&=\sum_{n\ge 0}s_{n-1}x^n+2\sum_{n\ge 0}s_{n-2}x^n+\sum_{n\ge 0}[n=1]x^n\\\\&=x\sum_{n\ge 0}s_{n-1}x^{n-1}+2x^2\sum_{n\ge 0}s_{n-2}x^{n-2}+x\\\\&=x\sum_{n\ge 0}s_nx^n+2x^2\sum_{n\ge 0}s_nx^n+x\;.\end{align*}\tag{3}$$
Now let $g(x)$ be the generating function: by definition
$$g(x)=\sum_{n\ge 0}s_nx^n\;.$$
From $(3)$ we see that $g(x)=xg(x)+2x^2g(x)+x$, and solving this for $x$ yields
$$g(x)=\frac{x}{1-x-2x^2}\;.$$ |
Kernel Smoothed Intensity of Point Pattern
Compute a kernel smoothed intensity function from a point pattern.
Usage
# S3 method for pppdensity(x, sigma=NULL, …, weights=NULL, edge=TRUE, varcov=NULL, at="pixels", leaveoneout=TRUE, adjust=1, diggle=FALSE, se=FALSE, kernel="gaussian", scalekernel=is.character(kernel), positive=FALSE, verbose=TRUE)
Arguments x
Point pattern (object of class
"ppp").
sigma
Standard deviation of isotropic smoothing kernel. Either a numerical value, or a function that computes an appropriate value of
sigma.
weights
Optional weights to be attached to the points. A numeric vector, numeric matrix, an
expression, or a pixel image.
… edge
Logical value indicating whether to apply edge correction.
varcov
Variance-covariance matrix of anisotropic smoothing kernel. Incompatible with
sigma.
at
String specifying whether to compute the intensity values at a grid of pixel locations (
at="pixels") or only at the points of
x(
at="points").
leaveoneout
Logical value indicating whether to compute a leave-one-out estimator. Applicable only when
at="points".
adjust
Optional. Adjustment factor for the smoothing parameter.
diggle
Logical. If
TRUE, use the Jones-Diggle improved edge correction, which is more accurate but slower to compute than the default correction.
kernel
The smoothing kernel. A character string specifying the smoothing kernel (current options are
"gaussian",
"epanechnikov",
"quartic"or
"disc"), or a pixel image (object of class
"im") containing values of the kernel, or a
function(x,y)which yields values of the kernel.
scalekernel
Logical value. If
scalekernel=TRUE, then the kernel will be rescaled to the bandwidth determined by
sigmaand
varcov: this is the default behaviour when
kernelis a character string. If
scalekernel=FALSE, then
sigmaand
varcovwill be ignored: this is the default behaviour when
kernelis a function or a pixel image.
se
Logical value indicating whether to compute standard errors as well.
positive
Logical value indicating whether to force all density values to be positive numbers. Default is
FALSE.
verbose
Logical value indicating whether to issue warnings about numerical problems and conditions.
Details
This is a method for the generic function
density.
It computes a fixed-bandwidth kernel estimate (Diggle, 1985) of the intensity function of the point process that generated the point pattern
x.
By default it computes the convolution of the isotropic Gaussian kernel of standard deviation
sigma with point masses at each of the data points in
x. Anisotropic Gaussian kernels are also supported. Each point has unit weight, unless the argument
weights is given.
If
edge=TRUE, the intensity estimate is corrected for edge effect bias in one of two ways:
If
diggle=FALSE(the default) the intensity estimate is correted by dividing it by the convolution of the Gaussian kernel with the window of observation. This is the approach originally described in Diggle (1985). Thus the intensity value at a point \(u\) is $$ \hat\lambda(u) = e(u) \sum_i k(x_i - u) w_i $$ where \(k\) is the Gaussian smoothing kernel, \(e(u)\) is an edge correction factor, and \(w_i\) are the weights.
If
diggle=TRUEthen the code uses the improved edge correction described by Jones (1993) and Diggle (2010, equation 18.9). This has been shown to have better performance (Jones, 1993) but is slightly slower to compute. The intensity value at a point \(u\) is $$ \hat\lambda(u) = \sum_i k(x_i - u) w_i e(x_i) $$ where again \(k\) is the Gaussian smoothing kernel, \(e(x_i)\) is an edge correction factor, and \(w_i\) are the weights.
In both cases, the edge correction term \(e(u)\) is the reciprocal of the kernel mass inside the window: $$ \frac{1}{e(u)} = \int_W k(v-u) \, {\rm d}v $$ where \(W\) is the observation window.
The smoothing kernel is determined by the arguments
sigma,
varcov and
adjust.
if
sigmais a single numerical value, this is taken as the standard deviation of the isotropic Gaussian kernel.
alternatively
sigmamay be a function that computes an appropriate bandwidth for the isotropic Gaussian kernel from the data point pattern by calling
sigma(x). To perform automatic bandwidth selection using cross-validation, it is recommended to use the functions
bw.diggleor
bw.ppl.
The smoothing kernel may be chosen to be any Gaussian kernel, by giving the variance-covariance matrix
varcov. The arguments
sigmaand
varcovare incompatible.
Alternatively
sigmamay be a vector of length 2 giving the standard deviations of two independent Gaussian coordinates, thus equivalent to
varcov = diag(rep(sigma^2, 2)).
if neither
sigmanor
varcovis specified, an isotropic Gaussian kernel will be used, with a default value of
sigmacalculated by a simple rule of thumb that depends only on the size of the window.
The argument
adjustmakes it easy for the user to change the bandwidth specified by any of the rules above. The value of
sigmawill be multiplied by the factor
adjust. The matrix
varcovwill be multiplied by
adjust^2. To double the smoothing bandwidth, set
adjust=2.
If
at="pixels" (the default), intensity values are computed at every location \(u\) in a fine grid, and are returned as a pixel image. The point pattern is first discretised using
pixellate.ppp, then the intensity is computed using the Fast Fourier Transform. Accuracy depends on the pixel resolution and the discretisation rule. The pixel resolution is controlled by the arguments
… passed to
as.mask (specify the number of pixels by
dimyx or the pixel size by
eps). The discretisation rule is controlled by the arguments
… passed to
pixellate.ppp (the default rule is that each point is allocated to the nearest pixel centre; this can be modified using the arguments
fractional and
preserve).
If
at="points", the intensity values are computed to high accuracy at the points of
x only. Computation is performed by directly evaluating and summing the Gaussian kernel contributions without discretising the data. The result is a numeric vector giving the density values. The intensity value at a point \(x_i\) is (if
diggle=FALSE) $$ \hat\lambda(x_i) = e(x_i) \sum_j k(x_j - x_i) w_j $$ or (if
diggle=TRUE) $$ \hat\lambda(x_i) = \sum_j k(x_j - x_i) w_j e(x_j) $$ If
leaveoneout=TRUE (the default), then the sum in the equation is taken over all \(j\) not equal to \(i\), so that the intensity value at a data point is the sum of kernel contributions from all
other data points. If
leaveoneout=FALSE then the sum is taken over all \(j\), so that the intensity value at a data point includes a contribution from the same point.
If
weights is a matrix with more than one column, then the calculation is effectively repeated for each column of weights. The result is a list of images (if
at="pixels") or a matrix of numerical values (if
at="points").
The argument
weights can also be an
expression. It will be evaluated in the data frame
as.data.frame(x) to obtain a vector or matrix of weights. The expression may involve the symbols
x and
y representing the Cartesian coordinates, the symbol
marks representing the mark values if there is only one column of marks, and the names of the columns of marks if there are several columns.
The argument
weights can also be a pixel image (object of class
"im"). numerical weights for the data points will be extracted from this image (by looking up the pixel values at the locations of the data points in
x).
To perform spatial interpolation of values that were observed at the points of a point pattern, use
Smooth.ppp.
To compute a relative risk surface or probability map for two (or more) types of points, use
relrisk.
Value
By default, the result is a pixel image (object of class
"im"). Pixel values are estimated intensity values, expressed in “points per unit area”.
If
at="points", the result is a numeric vector of length equal to the number of points in
x. Values are estimated intensity values at the points of
x.
In either case, the return value has attributes
"sigma" and
"varcov" which report the smoothing bandwidth that was used.
If
weights is a matrix with more than one column, then the result is a list of images (if
at="pixels") or a matrix of numerical values (if
at="points").
If
se=TRUE, the result is a list with two elements named
estimate and
SE, each of the format described above.
Note
This function is often misunderstood.
The result of
density.ppp is not a spatial smoothing of the marks or weights attached to the point pattern. To perform spatial interpolation of values that were observed at the points of a point pattern, use
Smooth.ppp.
The result of
density.ppp is not a probability density. It is an estimate of the
intensity function of the point process that generated the point pattern data. Intensity is the expected number of random points per unit area. The units of intensity are “points per unit area”. Intensity is usually a function of spatial location, and it is this function which is estimated by
density.ppp. The integral of the intensity function over a spatial region gives the expected number of points falling in this region.
Inspecting an estimate of the intensity function is usually the first step in exploring a spatial point pattern dataset. For more explanation, see Baddeley, Rubak and Turner (2015) or Diggle (2003, 2010).
If you have two (or more) types of points, and you want a probability map or relative risk surface (the spatially-varying probability of a given type), use
relrisk.
Negative Values
Negative and zero values of the density estimate are possible when
at="pixels" because of numerical errors in finite-precision arithmetic.
By default,
density.ppp does not try to repair such errors. This would take more computation time and is not always needed. (Also it would not be appropriate if
weights include negative values.)
To ensure that the resulting density values are always positive, set
positive=TRUE.
References
Baddeley, A., Rubak, E. and Turner, R. (2015)
Spatial Point Patterns: Methodology and Applications with R. Chapman and Hall/CRC Press.
Diggle, P.J. (1985) A kernel method for smoothing point process data.
Applied Statistics (Journal of the Royal Statistical Society, Series C) 34 (1985) 138--147.
Diggle, P.J. (2003)
Statistical analysis of spatial point patterns, Second edition. Arnold.
Diggle, P.J. (2010) Nonparametric methods. Chapter 18, pp. 299--316 in A.E. Gelfand, P.J. Diggle, M. Fuentes and P. Guttorp (eds.)
Handbook of Spatial Statistics, CRC Press, Boca Raton, FL.
Jones, M.C. (1993) Simple boundary corrections for kernel density estimation.
Statistics and Computing 3, 135--146. See Also Aliases density.ppp Examples
# NOT RUN { if(interactive()) { opa <- par(mfrow=c(1,2)) plot(density(cells, 0.05)) plot(density(cells, 0.05, diggle=TRUE)) par(opa) v <- diag(c(0.05, 0.07)^2) plot(density(cells, varcov=v)) } # }# NOT RUN { Z <- density(cells, 0.05) Z <- density(cells, 0.05, diggle=TRUE) Z <- density(cells, 0.05, se=TRUE) Z <- density(cells, varcov=diag(c(0.05^2, 0.07^2))) Z <- density(cells, 0.05, weights=data.frame(a=1:42,b=42:1)) Z <- density(cells, 0.05, weights=expression(x)) # }# NOT RUN { # automatic bandwidth selection plot(density(cells, sigma=bw.diggle(cells))) # equivalent: plot(density(cells, bw.diggle)) # evaluate intensity at points density(cells, 0.05, at="points") plot(density(cells, sigma=0.4, kernel="epanechnikov")) # relative risk calculation by hand (see relrisk.ppp) lung <- split(chorley)$lung larynx <- split(chorley)$larynx D <- density(lung, sigma=2) plot(density(larynx, sigma=2, weights=1/D))# }
Documentation reproduced from package spatstat, version 1.55-1, License: GPL (>= 2) |
This vignette covers the use of functions
mdrical and
frrcal.
Incidence estimates from cross-sectional surveys using biomarkers for ‘recent infection’ require that the test for recent infection (usually an adapted diagnostic assay) be accurately characterised. The two critical parameters of test performance are the Mean Duration of Recent Infection (MDRI), denoted \(\Omega_T\), (with \(T\) the recency cutoff time), and False Recent Rate (FRR), denoted \(\beta_T\). The explicit time cutoff \(T\) was introduced by Kassanjee et al.
Epidemiology, 2012. 1 to differentiate between ‘true recent’ and ‘false recent’ results. Also see Kassanjee, McWalter, Welte. AIDS Research and Human Retroviruses, 2014. 2, which notes:
To lead to an informative estimator, this cut-off, though theoretically arbitrary, must be chosen to reflect the temporal dynamic range of the test for recent infection; i.e. at a time T post infection, the overwhelming majority of infected people should no longer be testing “recent”, and furthermore, T should not be larger than necessary to achieve this criterion.
3
MDRI is defined as the average time alive and returning a ‘recent’ result, while infected for times less than \(T\). FRR is defined as a cross sectional context specific proportion of subjects returning a ‘recent’ result while infected for longer than \(T\).
Test performance may be context-specific, and therefore, where available, local data should be used to calibrate tests. Often cross-sectional incidence surveys use a multi-step Recent Infection Testing Algorithm (RITA) and then the entire RITA must be appropriately calibrated. This may involve adapting MDRI estimates to account for the sensitivity of screening tests, or adapting FRR estimates based on weighted estimates for subpopulations such as treated individuals. Calibration should be performed using the same set of biomarkers used in a RITA, such as a viral load threshold to reduce false recency.
This package provides the function
mdrical to estimate MDRI for a given biomarker or set of biomarkers from a dataset of based on the test being applied to well-characterised specimens and subjects. That is, time since ‘infection’ should be well-known, as well as test result(s). While ‘infection time’ can be variously defined as refering to the exposure event, date of first detectability on an RNA assay, Western Blot seroconversion, etc., it should be consistently used. If the reference event used in test calibration differs from test conversion on the screening assay or algorithm that is used define someone as HIV-positive in a RITA, then the MDRI needs to be appropriately adapted to cater for this difference.
Function
mdrical estimates MDRI by fitting a model for the probability of testing ‘recent’ as a function of time since infection \(P_R(t)\). As an option, one of two functional forms (with their associated link functions) can be selected by the user. Fitting is performed using a generalised linear model (as implemented in the
glm2 package) to estimate parameters.
The linear binomial regression model takes the following form, with \(g()\) the link function
\[\begin{equation} g(P_R(t)) = f(t) \end{equation}\]
If the argument
functional_forms is specified with the value
"cloglog_linear", \(g()\) is the complementary log-log link function and \(\ln(t)\) as linear predictor of \(P_R(t)\), so that:
\[\begin{equation} \ln\left(-\ln(1-P_R(t)\right) = \beta_0 + \beta_1\ln(t) \end{equation}\]
If the argument
functional_forms is specified with the value
"logit_cubic", \(g()\) is the complementary log-log link function and the linear predictor of \(P_R(t)\) is a cubic polynomial in \(t\), so that:
\[\begin{equation} \ln{\left(\frac{P_R(t)}{1-P_R(t)}\right)} = \beta_0 + \beta_1t + \beta_2t^2 + \beta_3t^3 \end{equation}\]
In both cases, MDRI is the integral of \(P_R(t)\) from \(0\) to \(T\).
\[\begin{equation} \Omega_T = \int_{0}^{T} P_R(t)dt \end{equation}\]
The default behaviour is to implement both model forms if the argument
functional_forms is omitted.
Confidence intervals are computed by means of subject-level bootstrapping, as measurements from subjects with more than one measurement in the dataset cannot be considered indpendent observations. An MDRI estimate is then calculated using the resampled data. The number of bootstrap iterations is specified using the argument
n_bootstraps. We recommend 10,000 for reproducible confidence intervals and standard errors. To support subject level resampling, the subject identifier in the dataset must be specified using the
subid_var argument.
In addition to specifying the value of \(T\) (using the argument
recency_cutoff_time), an
inclusion_time_threshold is required, to
exclude data points beyond a certain time (post infection). This is to prevent falsely recent measurements from unduly affecting the fit between \(0\) and \(T\). This should typically be a value somewhat (but not too much) larger than \(T\).
To specify recency status, one can either supply a list of variables and thresholds (indicating in whether a result above or below the thresholds signifies recency) or specify the
recency_rule as
"binary_data", in which case a 1 indicates recency.
mdrical using the complementary log-log functional form and pre-classified data
Load the package in order to use it
The dataset
excalibdata contains example data from an evaluation of an assay measuring recency of infection. At an assay result of <10, the specimen is considered to be recently infected. It further contains viral load data, which is commonly used to reduce false recency. For example, when recency is defined as assay result <10 and viral load > 1000, the FRR is substantially lower (but the MDRI is also reduced).
The first example provides a variable that has pre-classified results, and uses only the complementary log-log functional form.
Note: To keep compute time reasonable during execution of the example code, only 1000 bootstraps are performed. To obtain reasonable standard errors and confidence intervals, 10,000 bootstraps are recommended.
## $MDRI## PE CI_LB CI_UB SE n_recent n_subjects## cloglog_linear 237.7463 213.0376 267.8011 13.7998 270 304## n_observations## cloglog_linear 708## ## $Models## $Models$cloglog_linear## ## Call: glm2::glm2(formula = (1 - recency_status) ~ 1 + I(log(time_since_eddi)), ## family = stats::binomial(link = "cloglog"), data = data, ## control = stats::glm.control(epsilon = tolerance, maxit = maxit, ## trace = FALSE))## ## Coefficients:## (Intercept) I(log(time_since_eddi)) ## -5.786 1.045 ## ## Degrees of Freedom: 707 Total (i.e. Null); 706 Residual## Null Deviance: 941.2 ## Residual Deviance: 714 AIC: 718## ## ## $Plots## $Plots$cloglog_linear
## ## ## $BSparms## NULL
mdrical using the logit cubic functional form and two independent thresholds on biomarkers
This example also specifies a vector of variables and a vector of paramaters to define recency, using the assay result and the viral load. The paramaters in the vector
c(10,0,1000,1) mean that recency is defined as an assay biomarker reading below 10 and a viral load reading above 1000.
Note: To keep compute time reasonable during execution of the example code, only 1000 bootstraps are performed. To obtain reasonable standard errors and confidence intervals, 10,000 bootstraps are recommended.
mdri <- mdrical(data=excalibdata, subid_var = "SubjectID", time_var = "DaysSinceEDDI", recency_cutoff_time = 730.5, inclusion_time_threshold = 800, functional_forms = c("logit_cubic"), recency_rule = "independent_thresholds", recency_vars = c("Result","VL"), recency_params = c(10,0,1000,1), n_bootstraps = 1000, alpha = 0.05, plot = TRUE)
## $MDRI## PE CI_LB CI_UB SE n_recent n_subjects## logit_cubic 259.2234 232.6863 289.6169 14.7569 270 295## n_observations## logit_cubic 644## ## $Models## $Models$logit_cubic## ## Call: glm2::glm2(formula = recency_status ~ 1 + I(time_since_eddi) + ## I(time_since_eddi^2) + I(time_since_eddi^3), family = stats::binomial(link = "logit"), ## data = data, control = stats::glm.control(epsilon = tolerance, ## maxit = maxit, trace = FALSE))## ## Coefficients:## (Intercept) I(time_since_eddi) I(time_since_eddi^2) ## 2.554e+00 -1.591e-02 2.184e-05 ## I(time_since_eddi^3) ## -1.471e-08 ## ## Degrees of Freedom: 643 Total (i.e. Null); 640 Residual## Null Deviance: 875.9 ## Residual Deviance: 635.3 AIC: 643.3## ## ## $Plots## $Plots$logit_cubic
## ## ## $BSparms## NULL
mdrical in which bootstraps are run in parallel
As above, but asking for both functional forms and parellelising the bootstrapping. In this case, the job is split over four cores.
Note: To make sure this vignette builds in a reasonable time, the example code, but not the output, is shown.
mdrical(data=excalibdata, subid_var = "SubjectID", time_var = "DaysSinceEDDI", recency_cutoff_time = 730.5, inclusion_time_threshold = 800, functional_forms = c("logit_cubic","cloglog_linear"), recency_rule = "independent_thresholds", recency_vars = c("Result","VL"), recency_params = c(10,0,1000,1), n_bootstraps = 10000, alpha = 0.05, plot = TRUE, parallel = TRUE, cores=4)
It is possible to output the fitting parameters for each bootstrap iteration, by using the switch
output_bs_parms = TRUE. This provides the user with the family of \(P_{R}(t)\) curves used to obtain the confidence intervals on the MDRI estimate, for potential further manipulation. Note that the functional form
cloglog_linear form has two parameters (labeled \(\beta_0\) and \(\beta_1\)), and the
logit_cubic form has four parameters (labeled \(\beta_0\) to \(\beta_3\)). In each case \(\beta_0\) denotes the intercept. In order to use these parameters to obtain predicted probabilities of testing recent for given times since infection, the complementary log-log and logit link functions must be “inverted”, so that the equations giving predicted probabilities are
\[\begin{equation} P_R(t) = \exp(-\exp(\beta_{0} + \beta_{1}\ln(t))) \end{equation}\]
and
\[\begin{equation} P_R(t) = \frac{1}{1 + \exp(-(\beta_{0} + \beta_{1}t + \beta_{2}t^2 + \beta_{3}t^3))} \end{equation}\]
respectively.
As a demonstration, we perform only 10 bootstrap iterations in the example below:
mdri <- mdrical(data=excalibdata, subid_var = "SubjectID", time_var = "DaysSinceEDDI", recency_cutoff_time = 730.5, inclusion_time_threshold = 800, functional_forms = c("cloglog_linear","logit_cubic"), recency_rule = "independent_thresholds", recency_vars = c("Result","VL"), recency_params = c(10,0,1000,1), n_bootstraps = 10, alpha = 0.05, plot = TRUE, output_bs_parms = TRUE)
## $cloglog_linear## ## Call: glm2::glm2(formula = (1 - recency_status) ~ 1 + I(log(time_since_eddi)), ## family = stats::binomial(link = "cloglog"), data = data, ## control = stats::glm.control(epsilon = tolerance, maxit = maxit, ## trace = FALSE))## ## Coefficients:## (Intercept) I(log(time_since_eddi)) ## -6.618 1.170 ## ## Degrees of Freedom: 643 Total (i.e. Null); 642 Residual## Null Deviance: 875.9 ## Residual Deviance: 637.5 AIC: 641.5## ## $logit_cubic## ## Call: glm2::glm2(formula = recency_status ~ 1 + I(time_since_eddi) + ## I(time_since_eddi^2) + I(time_since_eddi^3), family = stats::binomial(link = "logit"), ## data = data, control = stats::glm.control(epsilon = tolerance, ## maxit = maxit, trace = FALSE))## ## Coefficients:## (Intercept) I(time_since_eddi) I(time_since_eddi^2) ## 2.554e+00 -1.591e-02 2.184e-05 ## I(time_since_eddi^3) ## -1.471e-08 ## ## Degrees of Freedom: 643 Total (i.e. Null); 640 Residual## Null Deviance: 875.9 ## Residual Deviance: 635.3 AIC: 643.3
## $cloglog_linear## # A tibble: 10 x 2## beta0 beta1## <dbl> <dbl>## 1 -6.47 1.15## 2 -6.87 1.20## 3 -6.88 1.20## 4 -6.59 1.16## 5 -6.65 1.19## 6 -6.88 1.20## 7 -6.47 1.16## 8 -6.16 1.08## 9 -6.19 1.11## 10 -7.67 1.32## ## $logit_cubic## # A tibble: 10 x 4## beta0 beta1 beta2 beta3## <dbl> <dbl> <dbl> <dbl>## 1 2.08 -0.0118 0.0000116 -6.27e-9## 2 2.12 -0.0114 0.0000101 -6.46e-9## 3 2.15 -0.00937 0.000000345 4.78e-9## 4 2.69 -0.0182 0.0000264 -1.63e-8## 5 2.15 -0.0121 0.0000101 -1.97e-9## 6 2.95 -0.0188 0.0000260 -1.67e-8## 7 2.24 -0.0136 0.0000157 -9.86e-9## 8 2.11 -0.0132 0.0000223 -2.18e-8## 9 2.59 -0.0139 0.0000137 -7.22e-9## 10 2.08 -0.00754 -0.00000365 6.04e-9
frrcal
FRR is simply the binomially estimated probability of a
subject’s measurements post-\(T\) being ‘recent’ on the recency test. A binomial exact test is performed using
binom.test. All of a subject’s measurements post-\(T\) are evaluated and if the majority are recent, the subject is considered to have measured falsely recent. Inversely, if a majority are non-recent, the subject contributes a ‘true recent’ result. Each subject represents one trial. In the case that exactly half of a subject’s measurements are recent, they contribute 0.5 to the outcomes (which are rounded up to the nearest integer over all subjects).
This example calculates a false-recent rate, treating the data at subject level:
## FRRest LB UB alpha n_recent n_subjects n_observations## 0.0301 0.0131 0.0584 0.05 8 266 732
Kassanjee, R., McWalter, T.A., Baernighausen, T. and Welte, A. “A new general biomarker-based incidence estimator.” Epidemiology; 2012, 23(5): 721-728.↩
Kassanjee, R., McWalter, T.A. and Welte, A. “Short Communication: Defining Optimality of a Test for Recent Infection for HIV Incidence Surveillance.” AIDS Research and Human Retroviruses; 2014, 30(1): 45-49.↩
Kassanjee, R., McWalter, T.A., Baernighausen, T. and Welte, A. “A new general biomarker-based incidence estimator.” Epidemiology; 2012, 23(5): 721-728.↩ |
You have two different contexts of $A$-algebra in the literature. Both give an external definition of the situation $A\subset_{subring}B$. All depends on the context.
A) The first one (the most spread and your case) is that $B$ is a $A$-module with commutation of the multiplications i.e. for all $a\in A$ and $b_i\in B$, one has the identities (associativity w.r.t. to scaling) $$a(b_1b_2))=(ab_1)b_2=b_1(ab_2)$$
Here, "finitely generated as a $A$ algebra" means that there exists a finite set $F\subset B$ such that the smallest $B_1$ for which
$$A\cup F\subset_{subring}B_1\subset_{subring}B$$is precisely $B$.
From this, you see that
$B$ is finitely generated as a $A$-module (FGM) means that it exists $F=\{f_1,f_2,\cdots f_n\}\subset B$, such that for all $b\in B$ we have a decomposition $$b=\sum_{i=1}^n a_if_i$$with $a_i\in A$.
And
$B$ is finitely generated as a $A$-algebra (FGA) means that it exists $F=\{f_1,f_2,\cdots f_n\}\subset B$, such that for all $b\in B$ we have a decomposition $$b=\sum_{\alpha\in \mathbb{N}^F} a_\alpha F^\alpha$$where $\alpha$ is a (weight) mapping $F\to \mathbb{N}$ i.e. $\alpha(f_i)=\alpha_i$ and $F^\alpha=f_1^{\alpha_1}\cdots f_n^{\alpha_n}$ (multiindex notation) and $a_\alpha\in A$.
So (FGM) implies (FGA).
For the converse, you need to extend $F$ with the products of powers of the $f_i$, but remaining finite. There the condition that $B$ is integral has to be used. In view of [1], for all $i\in I$, one can write $$f_i^{d_i}=\sum_{k=0}^{d_i-1}a_k\,f_i^k$$this proves that every $F^\alpha$ can be written as a $A$-linear combination of the $F^\beta$ with $\beta_i< d_i$ for all $i$. But those $F^\beta$ are in finite number. So $B$ is (FGM).
B) In the second one, we have $B$ is a $A$-bimodule and one has, still for all $a\in A$ and for all $b_i\in B$ (associativity w.r.t. scalings) $$a(b_1b_2)=(ab_1)b_2\ ;\ b_1(ab_2)=(b_1a)b_2\ ;\ (b_1b_2)a=b_1(b_2a)$$[1] wikipedia page https://en.wikipedia.org/wiki/Integral_element |
Although I have not found an answer for my question I think that I have proved a weaker result that is enough for me.
Let $K_1 \subset \ \mbox{int}\ K_2 \subset K_2 \subset \ldots$ an exhaustion of $\Omega$ by compacts. For each $j\in\mathbb N$ we have that the set
$$\tilde K_{j} = \{(x,y) \in F : x \in K_j\ \mbox{or}\ y\in K_j\} \subset \Omega\times\Omega$$
is compact. Hence we can find $0<r_{j}$ such that
$$\tilde K_{j} + B(0,r_{j}) \subset \Omega\times\Omega,$$
where $B(0,r_{j})$ is the ball with center $0\in\mathbb R^{2n}$ and radius $r_{j}$. We are using the maximum norm here. We also can suppose that $r_j\longrightarrow 0$. We have
$$\overline{\tilde K_{j} + B(0,r_{j}/2)} \subset \Omega\times\Omega.$$
Let
$$A = \bigcup_{j\in \mathbb N}\tilde K_{j} + B(0,r_{j}/2).$$
It follows that $A \subset \Omega\times\Omega$ is a open set that contains $F$. Let $\{f,g\}$ be a partition of unity associated to the family $\{A,\Omega\setminus F\}$ (with $\mbox{supp} f \subset A$).
Claim: for every compact $K \subset \Omega$, the set
$$B_K=\{(x,y)\in \mbox{supp}f : x \in K\ \mbox{or}\ y \in K\}$$
is $\textit{relatively}$ compact in $\Omega$. We first observe that it is sufficient to prove the result when $K=K_j$ for some $j\in\mathbb N$. Notice that
$$B_{K_j} =\tilde K_{j} \cup \{(x,y) \in \mbox{supp}f \setminus F : x \in K_j \ \mbox{or}\ y \in K_j\}.$$
Since $\tilde K_{j}$ is compact, we have to prove that $\{(x,y) \in \mbox{supp}f \setminus F : x \in K_j \ \mbox{or}\ y \in K_j\}$ is relatively compact, or that this set is bounded and its closure (in $\mathbb R^n$) is a subset of $\Omega\times\Omega$.
(1) $\{(x,y) \in \mbox{supp}f \setminus F : x \in K_j \ \mbox{or}\ y \in K_j\}$ is bounded:
Suppose that this set is not bounded. Then we can find a sequence $(x_n,y_n)$ in this set such that $|(x_n,y_n)|$ tends to $\infty$ when $n$ tends to $\infty$. Since $x_n \in K_j$ or $y_n \in K_j$ for each $n$, we can suppose that $x_n \in K_j$ for all $n$ and that there exists $x_0 \in K_j$ such that $x_n \longrightarrow x_0$.
Now we recall that $(x_n,y_n) \in \mbox{supp}f \subset A$. Hence we can find $(z_n,w_n)\in \tilde K_{\ell_n}$ such that $|(x_n,y_n)-(z_n,w_n)|<r_{\ell_n}/2$.
Case 1: $\{\ell_n : n \in\mathbb N\}$ if finite. In this case we can suppose that there exists $\ell_0\in\mathbb N$ such that $(z_n,w_n) \in \tilde K_{\ell_0}$ and $|(x_n,y_n)-(z_n,w_n)| < r_{\ell_0}/2$ for all $n$. In particular,
$$|y_n| \le |(x_n,y_n)-(z_n,w_n)| + |w_n|,$$
which is bounded since $\tilde K_{\ell_0}$ is compact. We have reached a contradiction.
Case 2: $\{\ell_n : n \in \mathbb N\}$ is infinity.
In this case we have
$$z_n \longrightarrow x_0 \ \mbox{and} \ |w_n| \longrightarrow \infty.$$
Since $z_n \longrightarrow x_0$ and $x_0 \in K_j \subset \mbox{int} K_{j+1}$ we can suppose that $z_n \in K_{j+1}$ for all $n$. Since $(z_n,w_n) \in F$ for all $n$ we have that
$$(z_n,w_n) \in \tilde K_{j+1}, \forall n \in \mathbb N,$$
which is compact. But this leads us to a contradiction with the fact that $|w_n|$ is unbounded. The proof of (1) is complete.
(2) Let $(x_0,y_0)$ be a point of the closure of $\{(x,y) \in \mbox{supp}f \setminus F : x \in K_j \ \mbox{or}\ y \in K_j\}$.
So $(x_0,y_0) = \lim (x_n,y_n)$, where $(x_n,y_n) \in \{(x,y) \in \mbox{supp}f \setminus F : x \in K_j \ \mbox{or}\ y \in K_j\}$ for all $n$. As before we can find $(z_n,w_n)\in \tilde K_{\ell_n}$ such that $|(x_n,y_n)-(z_n,w_n)|<r_{\ell_n}/2$.
Case 1: $\{\ell_n : n \in \mathbb N\}$ is finite.
We take $\ell_0$ such that $(z_n,w_n)\in \tilde K_{\ell_0}$ and $|(x_n,y_n)-(z_n,w_n)|<r_{\ell_0}/2$ for all $n$. Since $\tilde K_{\ell_0}$ is compact we can suppose that there exists $(z_0,w_0) \in \tilde K_{\ell_0}$ such that $|(z_n,w_n)-(z_0,w_0)|$ tends to $0$. Hence $|(x_0,y_0)-(z_0,w_0)| \le r_{\ell_0}/2$ and we have that $(x_0,y_0)\in\Omega\times\Omega$.
Case 2: $\{\ell_n : n \in \mathbb N\}$ is infinity.
In this case $(z_n,w_n)\longrightarrow (x_0,y_0)$. We can suppose that $x_n \in K_j$ for every $n$, and then $x_0 \in K_j$. Hence we can suppose that $z_n \in K_{j+1}$ for every $n$ and since $(z_n,w_n)\in F$, it follows that $(z_n,w_n) \in \tilde K_{j+1}$ for every $n$. Therefore $(x_0,y_0) \in \tilde K_{j+1} \subset \Omega\times \Omega$. |
You really should think of $\Delta$ as an $L^2$ self-adjoint, elliptic operator in it's own right irrespective of coordinates. $-\Delta$ has positive spectrum, with countable eigenvalues accumulating at $\infty$ etc. It's the average of the Hessian $\nabla^2 f$ in the sense that contracting the $(0, 2)$ tensor $\nabla^2 f$ with the metric $g$ produces a $(1, 1)$ tensor that you can trace (which is an averaging operation). In this sense there is no drift term.
On the other hand, there is a way think of the extra term $\mu$ coming from the connection. For a smooth function $f : M \to \mathbb{R}$, from the smooth structure alone, you can always form the differential $Df \in \Gamma^{\infty}(T^{\ast} M)$ as smooth section of the cotangent bundle. Without a connection, you can think of $Df : TM \to \mathbb{R}$ as a map between smooth manifolds, differentiate again, and this gives$$D^2 f : TTM \to \mathbb{R}$$
I used $D$ for the differential so as not to confuse this with the exterior derivative $d$ which would have $d^2 = 0$.
Now if you have a connection, you can split $TTM$ as$$TTM \simeq VTM \oplus HTM$$into vertical and horizontal sub-bundles. In this particular case both $VTM$ and $HTM$ are isomorphic to $TM$. $VTM$ is the kernel of the map $d\pi : TTM \to TM$ where $\pi: TM \to M$ is the bundle projection and this is isomorphic to $TM$. The connection allows you to "split" the map $d\pi : TTM \to TM$ obtaining an injective bundle morphism $TM \to TTM$ complementary to $VTM \simeq TM$ whose image I'll denoted $HTM$. The vertical bundle consists of elements of $TTM$ that are tangent to the fibres of $TM$ while the horizontal bundles consists of elements of $TTM$ that are tangent to the base $M$.
To help clarify, for a general vector bundle $\pi : E \to M$, the kernel $VE$ of $d\pi : TE \to TM$ is isomorphic to $E$ and a connection gives a splitting $TM \to TE$ whose image is denoted $HE$ and such that $TE \simeq VE \oplus HE \simeq E \oplus TM$.
Now, what this all has to do with "drift" is that under the identification $TTM \simeq TM \oplus TM$, the $g^{ij} \partial_{ij}$ term comes from $VE$ - it's the term tangent to the fibre and corresponds to $\partial_{ij} f = D^2 f$. The other term - the one with the connection coefficients $\Gamma$ corresponds to the part that is tangent to the base $M$.
In other words, the $g^{ij} \partial_{ij}$ part arises from differentiating while moving along the fibres of $TM$ and the $\Gamma$ part arises from moving along the base $M$.
Thus the drift is measuring the change in the fibres of the bundle $TM$ as measured by the connection as the basepoint $x \in M$ varies. |
This might be a somewhat open question, so any suggestion of improvement is welcome.
Suppose at time $t=0$, we have $N$ different assets whose weights are $w_1,\cdots,w_n$ ($\sum w_i = 1$), and they are all invested over the horizon $[0,T]$. Let $\{r_{i}^t\}_{i=1,\cdots,N}^{t=1,\cdots,T}$ be "returns" of asset $i$ over the interval $[t-1,t]$.
Question: I'm looking for a reasonable definition of "returns" that enables one to nicely represent/aggregate the total portfolio return from $\{r_{i}^t\}_{i=1,\cdots,N}^{t=1,\cdots,T}$.
At this point, I'm at the dilemma between choosing the ordinary return (something like $(S_{t+\Delta t}/S_t - 1)$) and the log return (something like $\log(S_{t+\Delta t}/S_t)$.) The two choices are kind of complementary in terms of the following:
The ordinary return enables easy linear aggregation across different
assets. For example, if we know all assets' weights $w_i$ and ordinary returns $r_i^t$ over year $t$, then the portfolio (ordinary) return over year $t$ is just a simple linear sum $$r_p^t=\sum_i w_ir_i^t$$ However, a very serious problem is that ordinary returns do NOT linearly add up over the timedimension ( note thatmy $T$ might be large so the naive linear approximation $\log(1+x)\approx x$ cannot apply). For example, for a certain asset $i$, if we know its ordinary return over each year $t$, then what is its total ordinary return over $[0,T]$? The answer is $r_i^{[0,T]}=\Pi_t (1+r_i^t) - 1$ which is quite ugly in terms of tractability (say, it's clearly impossible to compute the volatility of $r_i^{[0,T]}$ if only given those of $r_i^t$.).
The log return is more or less the contrary. It enables simple linear summation across the
timedimension: suppose the log returns of asset $i$ over year $t$ are $r_i^t$, then the total log return is simply $r_i^{[0,T]}:=\sum_t r_i^t$. However, it completely loses the compatibility with the assetdimension: given the log return $r_i$ of each asset $i$, the total log return of the portfolio would be a hideous expression $r_p = \log(\sum_i w_i \exp(r_i))$.
Is there any other version of "returns" that would hopefully allow simple aggregation (linear best) over
both the time and the asset dimension?
Thanks! |
Jupyter Notebook here: https://git.io/fjRjL PDF Article here: http://dx.doi.org/10.13140/RG.2.2.17472.58886 For an explicit compressible algorithm the maximum timestep one can take is dictated by both the advective and acoustic speeds in the flow. This is given by the famous CFL condition \begin{equation} V \frac{\delta t}{\Delta x} \leq 1 \end{equation} where $V$ is the maximum speed in the …
If you’ve worked in computational fluid dynamics, then you’re probably aware of the Taylor-Green vortex – at least the two-dimensional case. A simple google search will land you on this wikipedia page. The classic solution there is presented in the form \begin{equation} u = \cos x \sin y F(t);\quad v = -\sin x \cos y …
For Faster video speeds use: ffmpeg -i input.mov -filter:v “setpts=0.5*PTS” output.mov For Slower video speeds use: ffmpeg -i input.mov -filter:v “setpts=2*PTS” output.mov Note the factor multiplying PTS. If that factor is less than 1, then you get a faster video. The opposite otherwise. Thanks to: Modify Video Speed with ffmpeg
May all the best that time and chance have to offer be yours – Jeff Bendock: friend, mentor, and role model. Today (Aug 10) my wife and I went through an extraordinary experience. But what we experienced was not unique. We’re not the first or last people to experience it by any means. There was …
Here’s a neat trick I learned today to display all matplotlib plots as vector format rather than raster in Jupyter notebooks:
Convert and display in place: jupyter-nbconvert –to slides mynotebook.ipynb –reveal-prefix=reveal.js –post serve Convert to html: jupyter-nbconvert –to slides mynotebook.ipynb –reveal-prefix=reveal.js as always, try: jupyter-nbconvert –help
This issue appears to have showed up on Sierra. Here’s a simple fix: Edit your bash profile (emacs ~/.bash_profile) Add export BROWSER=open Close your shell or source it (source ~/.bash_profile) Things should get back to normal now. Ref: https://github.com/conda/conda/issues/5408
make sure you don’t fall into the trap of copying pointers in Python: import numpy as np a = np.zeros(2) print(“a = “, a) b = np.zeros(2) print(“b = “, b) b = a # this assignment is simply a pointer copy – b points to the same data pointed to by a a[0]=33.33 # …
First install a utility called highlight. It is available through macports and homebrew. sudo port install highlight Now that highlight is installed, you can “apply” it to a file and pipe it to the clipboard: highlight -O rtf MyCode.cpp | pbcopy Now go to Kenote and simply paste from clipboard (or command + v). Highlight …
Here are three ways to call a C++ classes’ operator () using a pointer: Say you have a C++ class name MyClass that overloads the operator “()”, something along the lines: I use method 2.
For those who have adopted inbox for their workflow, here’s how to add an event to your calendar. Your must be using google calendar & Chrom Download the google calendar extension and add it to Chrome. You will need to authorize your calendar. Back to inbox, select the text corresponding to your event (e.g. Thursday February 2, …
mkdir frames ffmpeg -i input -vf scale=320:-1:flags=lanczos,fps=10 frames/ffout%03d.png convert -loop 0 frames/ffout*.png -fuzz 20% -layers OptimizePlus output.gif Thanks to: http://superuser.com/questions/556029/how-do-i-convert-a-video-to-gif-using-ffmpeg-with-reasonable-quality
Observation When using a forward-Euler method for the time integration of the momentum equation for an inviscid-flow, it appears that the kinetic energy of the flow grows unbounded in time, regardless of the timestep size. Problem Statement Estimate the change in total kinetic energy when using forward-Euler to integrate the Euler momentum equations in a periodic box. Approach To …
Find and sort all files > 1Mb in size, in current directory and subdirectories find . -size +1M -exec du -h {} \; | sort -nr
I happened to be dealing with a bunch of mallocs that look like this double* f = (double*)malloc(sizeof(double)*nx*ny*nz); double* g = (double*)malloc(sizeof(double)*nx); double* h = (double*)malloc(sizeof(double)*nx*nz); I wanted to convert all of those those to use new, i.e. double* f = new double[nx*ny*nz] Using regex made all this possible search for: \(double\*\)malloc\(sizeof\(double\)\*(.*)\) replace with: new … |
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator |
For people who are real gluttons for punishment, here is all you would care to know about Atmospheric Escape Mechanics.
The article discusses many different mechanisms of gas losses. The easiest to model is the Thermal Escape mechanism (described in detail below). However, two other processes may contribute substantially to atmosphere loss in the absence of a magnetic field;
Pickup and electric field acceleration.
I've found no mathematical treatment of these last two but will describe them.
Pick-up
Pick-up is the process whereby hydrogen ions from the solar wind directly impinge upon the gas molecules of the atmosphere of planets with no or weak magnetic fields. When the ions impact molecules or atoms in the atmosphere, they impart momentum allowing them to escape. This is the dominant non-thermal loss mechanism for Mars' atmosphere and bodies with thin atmospheres.
Electric Field Acceleration
The dominant loss process for Venus' atmosphere is through electric
force field acceleration. As electrons are less massive than other
particles, they are more likely to escape from the top of Venus's
ionosphere.[3] As a result, a minor net positive charge develops. That
net positive charge, in turn, creates an electric field that can
accelerate other positive charges out of the atmosphere. As a result,
H+ ions are accelerated beyond escape velocity.
from the wikipedia article on Atmospheric Loss Mechanisms
This loss mechanism is the dominant non-thermal loss mechanism on bodies with thick atmospheres.
Thermal escape
In most cases, thermal escape is the dominant atmospheric loss mechanism.
As @HDE226868 posted, calculate surface temperature of the planet using solar luminosity, albedo, and distance from the sun. Then calculate the Vrms of the gases. Then compare to the body's escape velocity.
| Meas Calc
----------------------
Tmercury | 700 K 438 K
Tvenus | 735 K 185 K
Tearth | 313 K 254 K
Tmoon | 390 K 268 K
Tmars | 293 K 210 K
Tio | 130 K 95 K
Teuropa | 125 K 92 K
Tcallisto| 165 K 115 K
Ttitan | 93 K 85 K
Ttriton | 38 K 35 K
Tpluto | 55 K 36 K
NOTE1: the problem with this is it calculates the average surface temperature, whereas thermal escape mechanics rely most heavily on the highest / day side temperature. The shown measured temps are the "high temperatures".
NOTE2: I've got the albedo of these bodies but they tend to make the temperatures diverge further from the measured values than using HDE's approximation of 0.3 for an average albedo.
I use a ratio of 300 as the cut-off for gases. Values above this number indicate the planet can retain the gas for billions of years.
Next calculate the escape velocity of the body. These are the values that I get:
Escape
Body Velocity
Mercury = 4,250 m/s
Venus = 10,361 m/s
Earth = 11,178 m/s
Moon = 2,375 m/s
Mars = 5,021 m/s
Io = 2,560 m/s
Europa = 2,035 m/s
Callisto = 2,444 m/s
Titan = 2,641 m/s
Triton = 1,456 m/s
Pluto = 1,246 m/s
The formula to find lightest weight gas the planet can hold onto is as follows:
Vesc must be larger than some calculated multiple of Vrms (I based the form of this calculated multiple off the half-life formula)$$ ln\left (1 \times 10^{9} \div 9 \right )^2 = \frac{Vesc}{Vrms} $$
$$ ln\left (1 \times 10^{9} \div 9 \right ) = \left ( \frac{2GM \times m}{r \times 3RT} \right ) $$
$$ \large m = \frac{ln\left (1 \times 10^{9} \div 9 \right ) 3RTr}{2GM} $$
The 1e9 value is the number of years you want the gas to stick around, this represents 1 billion years. I believe the natural log portion of the equation to be my own empirical fit to the problem.
$ \large m $ - Molar mass of compound
$ R $ - Universal gas constant $ 8.3144621 \frac{J}{mol K} $
$ T $ - Temperature (K)
$ r $ - Planet's radius in meters (my statements above are wrong, it does play a factor)
$ G $ - Gravitational constant $ (6.67 \times 10^{-11} {N} \left (\frac{m}{kg} \right ) ^2 ) $
$ M $ - Planet's mass in kg
The gases each body can retain over geologic periods are:
Molar
Body Mass Gases
Mercury = 114 ~ Br2 + I2 only; all other gases escape; No ices
Venus = 20 ~ N2 and heavier
Earth = 9 ~ CH4 and heavier
Moon = 203 ~ I2 only; all other gases escape; No ices
Mars = 36 ~ F2 and heavier
Io = 58 ~ Kr + Cl2 only; all other gases escape; Ices of NH3, H2O, CO2, Br2, etc.
Europa = 89 ~ Kr + Cl2 only; all other gases escape; Ices of NH3, H2O, CO2, Br2, etc.
Callisto = 81 ~ Kr + Cl2 only; all other gases escape; Ices of NH3, H2O, CO2, Br2, etc.
Titan = 39 ~ N2 and heavier; Ices of CH4, NH3, H2O, CO2, O2, etc.
Triton = 53 ~ None; all gases escape; Ices of NH3, H2O, CO2, N2, O2, etc.
Pluto = 104 ~ None; all gases escape; Ices of NH3, H2O, CO2, N2, O2, etc.
Another twist to this is the fact that various molecules achieve much longer longevity when it is colder than their "snow line". Our solar system's snow line for water (the point at which it remains solid and doesn't evaporate/sublimate) occurs at the distance of our asteroid belt. Beyond this distance, solar system bodies can retain their ices.
Hypothetical Planets
It's been a long road but I think I finally have my answer. A simple swap between $ \large m $ and $ M $ generates the equation which determines what mass is required to retain a given gas if you use the assumptions below. First the equation:
$$ M = \frac{ln\left (1 \times 10^{9} \div 9 \right ) 3RTr}{2G \large m} $$
Now the assumptions 1. Replacement planets use the same density as Earth.
2. Replacement planets use the same albedo as Earth. 3. Planets need a Vesc/Vrms ratio of 400 to hold onto a gas for 4.5 billion years. 4. Planets need to retain gaseous water to maintain human habitability. 5. The daytime "hot" temp is 1.15x the temperature average. 6. Planets have a strong magnetic field so only thermal loss is important.
Which simplifies the equation to:
$$ M = \frac{1,200 \times 8.3144621 \frac{J}{mol K}Tr}{36 \times (6.67 \times 10^{-11} {N} \left (\frac{m}{kg} \right ) ^2)} $$
Then what I find is:
Orbit of Min Mass Vesc Surface G
Venus 0.55e 9,159 0.82
Earth 0.43e 8,437 0.75
Mars 0.32e 7,647 0.68
Remember the mass is the important thing, so the planet could possess a much lower density and, therefore, a much lower surface gravity should the world builder so desire.
Interestingly, if Mars were just about 3x its current mass, it
might have held onto a substantial atmosphere and been a pleasant place to live. |
Tagged: determinant of a matrix Problem 718
Let
\[ A= \begin{bmatrix} 8 & 1 & 6 \\ 3 & 5 & 7 \\ 4 & 9 & 2 \end{bmatrix} . \] Notice that $A$ contains every integer from $1$ to $9$ and that the sums of each row, column, and diagonal of $A$ are equal. Such a grid is sometimes called a magic square.
Compute the determinant of $A$.Add to solve later
Problem 686
In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not.
(a) The matrix $A$ is a $3 \times 3$ matrix with eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=0$.
Add to solve later
(b) The matrix $A$ is a $3 \times 3$ matrix with eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=-1$. Problem 582
A square matrix $A$ is called
nilpotent if some power of $A$ is the zero matrix. Namely, $A$ is nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix.
Suppose that $A$ is a nilpotent matrix and let $B$ be an invertible matrix of the same size as $A$.
Is the matrix $B-A$ invertible? If so prove it. Otherwise, give a counterexample. Problem 571
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.
There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold). The time limit was 55 minutes.
This post is Part 2 and contains Problem 4, 5, and 6.
Check out Part 1 and Part 3 for the rest of the exam problems.
Problem 4. Let \[\mathbf{a}_1=\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \mathbf{a}_2=\begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}, \mathbf{b}=\begin{bmatrix} 0 \\ a \\ 2 \end{bmatrix}.\]
Find all the values for $a$ so that the vector $\mathbf{b}$ is a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$.
Problem 5. Find the inverse matrix of \[A=\begin{bmatrix} 0 & 0 & 2 & 0 \\ 0 &1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix}\] if it exists. If you think there is no inverse matrix of $A$, then give a reason. Problem 6. Consider the system of linear equations \begin{align*} 3x_1+2x_2&=1\\ 5x_1+3x_2&=2. \end{align*} (a) Find the coefficient matrix $A$ of the system. (b) Find the inverse matrix of the coefficient matrix $A$. (c) Using the inverse matrix of $A$, find the solution of the system.
(
Linear Algebra Midterm Exam 1, the Ohio State University) Read solution Problem 546
Let $A$ be an $n\times n$ matrix.
The $(i, j)$
cofactor $C_{ij}$ of $A$ is defined to be \[C_{ij}=(-1)^{ij}\det(M_{ij}),\] where $M_{ij}$ is the $(i,j)$ minor matrix obtained from $A$ removing the $i$-th row and $j$-th column.
Then consider the $n\times n$ matrix $C=(C_{ij})$, and define the $n\times n$ matrix $\Adj(A)=C^{\trans}$.
The matrix $\Adj(A)$ is called the adjoint matrix of $A$.
When $A$ is invertible, then its inverse can be obtained by the formula
For each of the following matrices, determine whether it is invertible, and if so, then find the invertible matrix using the above formula.
(a) $A=\begin{bmatrix} 1 & 5 & 2 \\ 0 &-1 &2 \\ 0 & 0 & 1 \end{bmatrix}$. (b) $B=\begin{bmatrix} 1 & 0 & 2 \\ 0 &1 &4 \\ 3 & 0 & 1 \end{bmatrix}$. Problem 509
Using the numbers appearing in
\[\pi=3.1415926535897932384626433832795028841971693993751058209749\dots\] we construct the matrix \[A=\begin{bmatrix} 3 & 14 &1592& 65358\\ 97932& 38462643& 38& 32\\ 7950& 2& 8841& 9716\\ 939937510& 5820& 974& 9 \end{bmatrix}.\]
Prove that the matrix $A$ is nonsingular.Add to solve later
Problem 505
Let $A$ be a singular $2\times 2$ matrix such that $\tr(A)\neq -1$ and let $I$ be the $2\times 2$ identity matrix.
Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula: \[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\]
Using the formula, calculate the inverse matrix of $\begin{bmatrix}
2 & 1\\ 1& 2 \end{bmatrix}$. Problem 486
Determine whether there exists a nonsingular matrix $A$ if
\[A^4=ABA^2+2A^3,\] where $B$ is the following matrix. \[B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 2 & 1 & -4 \end{bmatrix}.\]
If such a nonsingular matrix $A$ exists, find the inverse matrix $A^{-1}$.
(
The Ohio State University, Linear Algebra Final Exam Problem) Read solution Eigenvalues of Orthogonal Matrices Have Length 1. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue Problem 419 (a) Let $A$ be a real orthogonal $n\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$.
Add to solve later
(b) Let $A$ be a real orthogonal $3\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an eigenvalue. |
Visiting addressNiels Henrik Abels hus Moltke Moes vei 35 (map)
0851 OSLO
Norway
Supported by the Foundation for Danish-Norwegian cooperation.
Christian Voigt (Glasgow) will give a talk with title: The structure of quantum permutation groups
Abstract: Quantum permutation groups, introduced by Wang, are a quantum analogue of permutation groups. These quantum groups have a surprisingly rich structure, and they appear naturally in a variety of contexts, including combinatorics, operator algebras, and free probability. In this talk I will give an introduction to these quantum groups, and review some results on their structure. I will then present a computation of the K-groups of the C*-algebras associated with quantum permutation groups, relying on methods from the Baum-Connes conjecture.
Alfons van Daele, University of Leuven (Belgium), will give a talk with title: Separability idempotents and quantum groupoids
Martijn Caspers (Münster) will give a talk with title: The Haagerup property for arbitrary von Neumann algebras
Abstract: The Haagerup property is an approximation property for both groups and operator algebras that has important applications in for example the Baum-Connes conjecture or von Neumann algebra theory. In this talk we show that the Haagerup property is an intrinsic invariant of an arbitrary von Neumann algebra. We also discuss stability properties of the Haagerup property under constructions as free products, graph products and crossed products. Finally we discuss alternative characterizations in terms of the existence of suitable quadratic forms.
Marco Matassa (UiO) will give a talk with title: Dirac Operators on Quantum Flag Manifolds
Abstract: I will review the paper "Dirac Operators on Quantum Flag Manifolds" by Ulrich Krähmer. The aim is to define Dirac operators on quantized irreducible flag manifolds. These will yield Hilbert space realizations of some distinguished covariant first-order differential calculi.
Adam Sørensen (UiO) will give a talk with title: Almost commuting matrices
Abstract: Two matrices A,B are said to almost commute if AB is close to BA (in a suitable norm). A question of Halmos, answered by Lin, asks if two almost commuting self-adjoint matrices are always close to two exactly commuting self-adjoint matrices. We will survey what is known about this and similar questions, and report on recent work with Loring concerning how the questions change if we look at real rather than complex matrices.
Abstract: We show that the discrete duals of the so called free orthogonal quantum groups have the completely contractive approximation property, analogous to the free groups. The proof relies on the structure of representation categories of these quantum groups, on the C*-algebraic structure of SUq(2), and on the free product techniques of Ricard and Xu. This talk is based on joint work with Kenny De Commer and Amaury Freslon.
Abstract: Independence has been introduced as a regularity property for pairs of commuting injective group endomorphisms of a discrete abelian group with finite cokernel by Joachim Cuntz and Anatoly Vershik. We discuss various characterisations of this regularity property and show how the statements need to be adjusted when removing the restrictions that the group has to be abelian and that the cokernels have to be finite. Somewhat surprisingly, this leads to the concept of *-commutativity. This property is defined for pairs of commuting self-maps of an arbitrary set. As an examples of *-commutativity, we explain a construction related to the Ledrappier shift and indicate how one obtains examples for independent group endomorphisms from this construction. If time permits, we will point out instances where the two notions have been readily used to obtain C*-algebraic results. Roughly speaking, both notions are designed to give rise to pairs of doubly commuting isometries, which significantly simplifies the analysis of the constructed C*-algebras. This is particularly useful when one tries to generalise results from the case of a single transformation to an action generated by finitely many transformations.
Abstract: We talk about independent resolutions for dynamical systems on totally disconnected spaces. Building on earlier work by Cuntz, Echterhoff and Li that allows one to compute the K-theory of totally disconnected systems that admit a so called independent invariant regular basis, we show how any totally disconnected dynamical system admits a resolution of such systems, which in some cases allows for K-theory computations. Based on work by me and X. Li.
Marco Matassa (UiO) will give a talk with title: On dimension and integration for spectral triples associated to quantum groups
Abstract: Abstract: I will discuss some aspects of the notions of spectral dimension and non-commutative integral in the context of modular spectral triples. I will focus on two examples: the modular spectral triple for SU_q(2) introduced by Kaad and Senior and the family of spectral triples for quantum projective spaces introduced by D'Andrea and Dąbrowski.
Jens Kaad (Trieste), will give a talk with title "Joint torsion line bundles of commuting operators"
Abstract:
In this talk I’ll associate a holomorphic line bundle to any commuting tuple of bounded operators on a Hilbert space. The transition functions for this bundle are given by the joint torsion which compares determinants of Fredholm complexes. The joint torsion is an invariant of the second algebraic K-group of the Calkin algebra (bounded operators modulo trace class operators). The main step is to prove that the transition functions for the joint torsion line bundle are indeed holomorphic. This is carried out by studying the Quillen-Freed holomorphic determinant line bundle over the space of Fredholm complexes. In particular I will construct a holomorphic section of a certain pull-back of this bundle. The talk is based on joint work with Ryszard Nest.
Magnus D. Norling will give a talk with title "Universal coefficient theorem in KK-theory". This presentation is part of the final act of the course on "The Baum-Connes conjecture and KK-theory".
Bas Jordans will give a talk with title "Higson's characterization of KK-theory". This presentation is part of the final act of the course on "The Baum-Connes conjecture and Kasparov's KK-theory".
Roberto Conti (Sapienza Università di Roma) will give a talk with title: Asymptotic morphisms in local quantum physics and study of some models
Abstract:
We discuss a notion of asymptotic morphisms that is suitable for a description of superselection sectors of a scaling limit theory. In some models, this leads to interesting questions about the explicit form of certain modular operators. (This talk is based on joint work with D. Guido and G. Morsella).
Magnus Landstad will give a talk with title: Quantum groups from almost matched pairs of groups - the groupoid approach
Abstract: If G is a locally compact group with two closed subgroups H,K s.t. G=HK, then (H,K) is called a matched pair of subgroups. The construction of a quantum group from such a pair goes back a long time. We shall look at the more general case where the subgroups are almost matched (the complement of HK in G has measure 0), then a groupoid approach to the construction is very useful and many formulas are obtained for free.
I shall start with explaining the concepts needed (quantum groups, groupoids, etc) and then how the groupoid is constructed. Finally we shall look at the special case where G has a compact open subgroup.
This is joint work with A. Van Daele.
Antoine Julien (NTNU) will give a talk with title: Tiling spaces, groupoids and K-theory
Abstract:
In this talk, I will describe how spaces, groupoids and C*-algebras can be associated with aperiodic tilings. In some cases, it is possible to describe the structure of the groupoid combinatorially in terms of augmented Bratteli diagrams. (joint work with Jean Savinien) Time permitting, I will expose a strategy for computing the K-theory of the tiling algebra in terms of the K-theory of AF-algebras (work in progress).
Takuya Takeishi, University of Tokyo, will give a talk with title: Bost-Connes system for local fields of characteristic zero
Abstract: The Bost-Connes system, which describes the relation between quantum statistical mechanics and class field theory, was first constructed by Bost and Connes for the rational field, and generalized for arbitrary number fields by the contribution of many researchers. In this talk, we will introduce a generalization of the Bost-Connes sysmtem for local fields of characteristic zero, and introduce some properties.
Judith Packer, University of Colorado (Boulder), will give a talk with title "Noncommutative solenoids and their projective modules"
Abstract: ``Noncommutative solenoids" are certain twisted group $C^*$-algebras, where the groups in question are countably infinitely generated; these algebras can also be generated as direct limits of rotation algebras. From examining the range of the trace of the $K_0$-groups of the noncommutative solenoids, their finitely generated projective modules can be constructed. We also discuss a way to construct Morita equivalence bimodules between noncommutative solenoids that goes back to work of M. Rieffel, with the new wrinkle of $p$-adic analysis appearing. This work is joint with F. Latr\'emoli\'ere.
Bas Jordans (UiO) will give a talk with title: Real dimensional spaces in noncommutative geometry
Abstract:
In noncommutative geometry geometric spaces are given by spectral triples. In this talk we consider a generalisation of these spectral triples to semifinite spectral triples. In analogy to the classical case it is possible to construct the product of two semifinite spectral triples. We will construct this product and derive properties thereof. Also we will describe for each z\in(0,\infty) a semifinite spectral triple which can be considered as having dimension z. As an application these "z-dimensional" semifinite triples will be used for two regularisation methods in physics.
Yusuke Isono from the University of Tokyo will give a talk with title: Strong solidity of II_1 factors of free quantum groups
Abstract:
We generalize Ozawa's bi-exactness to discrete quantum groups and give a new sufficient condition for strong solidity, which implies the absence of Cartan subalgebras. As a corollary, we prove that II_1 factors of free quantum groups are strongly solid. We also consider similar conditions on non-Kac type quantum groups, namely, non finite von Neumann algebras.
Erik Bédos will give a talk with title: On equivariant representations of C*-dynamical systems
Abstract:
Let \Sigma=(A, G, \alpha, \sigma) denote a unital discrete twisted C*-dynamical system. In our recent work with Roberto Conti (Rome), it has emerged that the so-called equivariant representations of \Sigma on Hilbert A-modules play an interesting role, complementing the one played by covariant representations. We will discuss some aspects of this notion and illustrate its usefulness in the study of the crossed products associated with \Sigma.
Adam Skalski (IMPAN) will give a talk with title: Closed quantum subgroups of locally compact quantum groups and some questions of noncommutative harmonic analysis (based on joint work with Matt Daws, Pawel Kasprzak and Piotr Soltan)
Abstract: The notion of a closed subgroup of a locally compact group is a very straightforward concept, often featuring in classical harmonic analysis. I will discuss the possible extensions of this notion to the quantum setting, focusing on the comparison of the two definitions proposed by S. Vaes and S.L. Woronowicz. I will describe some reformulations of these definitions and explain how they can beshown to be equivalent in many cases; I will also mention certain connections to other problems of quantum harmonic analysis.
Nicolai Stammeier, Westfälische Wilhelms-Universität Münster, will give a talk with title: Product Systems of Finite Type for Certain Algebraic Dynamics and their C*-algebras
Abstract: Let P be a lattice-ordered semigroup with unit acting on a discrete, abelian group G by injective endomorphisms with finite cokernel. Building on the work of Jeong Hee Hong, Nadia S. Larsen and Wojciech Szymanski on product systems of Hilbert bimodules and their KMS-states, one can associate a product system to this dynamical system that turns out to be of finite type. Imposing two additional conditions on the dynamics, namely independence of the endomorphisms for relatively prime elements in P and exactness, we derive presentations of the Nica-Toeplitz algebra and the Cuntz-Nica-Pimsner algebra. Moreover, the latter has lots of nice descriptions and is shown to be a unital UCT Kirchberg algebra.
Roberto Conti, Universita La Sapienza (Rome), Italy, will give a talk with title: The dark side of the Cuntz algebras
Abstract: We provide an overview of our recent work on the automorphism group of Cuntz algebras and discuss a number of open problems. (Joint work with J.H. Hong and W. Szymanski). |
Preludium
In chemistry we often use some techniques to factorize the (quite complicated) many-particle ground state of atoms and molecules into a series of one-particle states that have to fullfil certain requirements; examples are Hartree-Fock-Theory or Kohn-Sham-Density-Functional-Theory. These one-particle states are what we usually call orbitals or, when you differentiate them according to their spin, spin-orbitals. In the theories that I quoted you then have to solve a series of one-particle Schroedinger-equations which will give you certain
discrete energy levels that correspond to bound states of the system. Those energy levels are a fundamental property of the system, i.e. the eigenvalues of its Hamilton operator. You can't have energy levels in between or below. Now, the problem is that the orbitals and their energy levels depend on the number of electrons you have in the system, i.e. considering some species $\ce{A}$ you will find different orbitals at different energies for its anion $\ce{A-}$, neutral species $\ce{A}$ or cation $\ce{A+}$. That is so, because the electrons interact with one another and populating one orbital changes the potential which the other electrons feel. Main answer
But let's assume for a moment that the orbitals for $\ce{A-}$, $\ce{A}$ and $\ce{A+}$ are the same and only taking into account the electron-electron-repulsion of electrons occupying the same orbital: Starting from $\ce{A+}$ where we have all lower-lying orbitals doubly occupied and one orbital $\psi_n$ with energy $E_n$ singly occupied. Now, if you want add one more electron to this system to get to $\ce{A}$ you have to choose where to put this electron: You can either put it into $\psi_n$ or into some higher-lying orbital, like $\psi_{n+1}$ or $\psi_{n+2}$. If you put it into $\psi_n$, thus pairing the electrons up, you have to pay for that with the electron-electron-repulsion energy between the paired electrons (also known as spin pairing energy) $\Delta E_{\mathrm{pair}}$. So, compaired to $\ce{A+}$ the energy of your new system $\ce{A}_{\text{paired}}$ will be higher by an amount of $E_n + \Delta E_{\mathrm{pair}}$. If you instead put the new electron into $\psi_{n+1}$ you don't have to pair up any electrons, so you don't have to pay $\Delta E_{\mathrm{pair}}$. But the energy $E_{n+1}$ of orbital $\psi_{n+1}$ is higher than $E_n$. So, compaired to $\ce{A+}$ the energy of your new system $\ce{A}_{\text{unpaired}}$ will be higher by an amount of $E_{n+1}$. Now, the question: Which situation is better? For that you have to look at the energy difference between $\ce{A}_{\text{paired}}$ and $\ce{A}_{\text{unpaired}}$. This will be
\begin{equation} \Delta E = E(\ce{A}_{\text{paired}}) - E(\ce{A}_{\text{unpaired}}) = (E_n - E_{n+1}) + \Delta E_{\mathrm{pair}} \ .\end{equation}
If $\Delta E$ is negative then $\ce{A}_{\text{paired}}$ is lower in energy and thus energetically favored, i.e. pairing electrons is favored when $|(E_n - E_{n+1})| > \Delta E_{\mathrm{pair}}$. So, coming back to your original question: Typical values of the pairing energy are between 200 and 300 kJ/mol (or 2-3 eV) - for an example have a look here: there you can see that the pairing energy for carbon and nitrogen lies at about $20000 \, \mathrm{cm}^{-1}$ which is ca. 240 kJ/mol. The seperation between energy levels in atoms is usually much greater as can be seen in the table of ionization potentials given here (the differences between the ionization potentials of $\ce{s}$ and $\ce{p}$ valence orbitals are usually greater than 6 eV). Only in molecules and transition-metal complexes with frontier-orbitals that lie close in energy (or are degenerate) you have so called high-spin configurations where it is favorable not to pair up the electrons in a lower-lying orbital but instead to occupy the higher-lying orbital. So in conclusion, it really does "cost" energy to pair up electrons in a single orbital but it usually would cost a lot more energy to force an electron into a higher-lying orbital instead.
Finally, a word of caution: My answer oversimplifies things as I tried to point out in my preludium. There are other effects at work that might have a big influence in some cases. |
State the definition of the infimum of a set. For the set $$S = \left\{2(-1)^n+\frac{5}{n^2+2}: n \in \mathbb{N}^{+} \right\}$$
Find $\text{inf(S)}$ and prove your claim. Argue straight from the definition of infimum.
Let $A \subseteq X$, where $X$ is some ordered field. Then $\text{inf}(A)$ is $m \in X$ such that for any $x \in A$ we have $m \le x$ and for any number $k \in X$ such that $k \le x$ we have $k \le m$. But I'm unable to directly argue from it. Do you think what's presented below would get the marks, given this was from an exam?
Let $\displaystyle x_n = 2(-1)^n+\frac{5}{n^2+2}$, then $\displaystyle x_{2n} = 2+\frac{5}{(2n)^2+2}$ and $\displaystyle x_{2n+1} = \frac{5}{(2n+1)^2+2}-2$.
Let $j = 2n$ then $\displaystyle x_{j} = 2+\frac{5}{j^2+2}> 2+\frac{5}{(j+1)^2+2} = x_{j+1}.$ Thus $x_{j} > x_{j+1}.$ Hence $x_{j}$ is strictly decreasing from $\dfrac{11}{3}$ to $2$. Similarly, let $k = 2n+1$ then $\displaystyle x_k = \frac{5}{k^2+2}-2 >\frac{5}{(k+1)^2+2}-2$$ = x_{k+1}.$ Hence $x_k$ is strictly decreasing from $-\dfrac{1}{3}$ to $-2$. Hence $\min(-2, 2)<x_n < \max(11/3, -1/3)$ which implies that $-2<x_n < \dfrac{11}{3}$. Thus $\text{inf}(S) = -2.$ Does this count as arguing from definition?
If it doesn't I'd be grateful if someone could show me how to do this straight from the definition. |
What you have here is two blocks: one of mass $M$ and one of mass $+\infty$ and a spring of mass $m=0.$ So you have all the fun of zero mass and on top, now you have the fun of mass $+\infty$ too.
Newton's third law always holds. However, since the action-reaction pairs always apply to two different objects and only one of your objects has finite mass you don't use it very much. Or really need to worry about it at all if people have already told you how to figure it out. But let's see what happens if you track everything.
So first let's talk about infinite mass objects; basically they move at constant velocity regardless of what force you exert. Never let them collide. Luckily, they don't exist, so don't worry about it too much.
Next let's talk about zero mass Newtonian objects; if they feel a force imbalance they can instantly move in the direction of the force regardless of their current velocity and do so to adjust their location and velocity until they feel zero force. Their velocity at that point is indeterminate; however it often ends up similar to the parts they are next to unless there are multiple ways it can move or be and have no force on it. Never confuse a massless Newtonian object with a massless Relativistic object. The former can move at any speed, or even have an indeterminate speed, the later must move at the speed of light.
Now we can talk about objects with finite mass. They accelerate according to $F=ma$.
To talk about your problem, the first step is to be more explicit about the set-up. We can put them in space or we can out them on a level frictionless surface.
So the block of mass $M$ is moving to the right. To the right is a spring at rest of length $L$ and it is attached to a block of mass $+\infty$ on its right, which is also at rest.
Up until the block of mass $M$ touches the spring, there are no forces whatsoever. So, the blocks move at a steady velocity and the spring and the heavy block stay at rest.
Then it touches.
It must exert a force to move the spring out of its way. A real block would itself be like a very stiff spring; it would compress and that compression would be how it exerts a force on the spring. But since the spring is massless no force is required to make it move & so it can effortlessly move out of the way and since it is massless it will move until it feels zero force.
So let's describe the motion of the block of mass $M$. With the origin of time being when it first touches the spring and the origin of space being where it first touches the spring we have a position as a function of time $x(t)$ and $x(t)=vt$ for some fixed $v>0$ when $t\leq 0.$
This isn't very detailed. But it does show that it moved at constant velocity back when it had no force on it. And we can compute the acceleration on it after $t=0$ too. I'll use calculus since it was invented to do these physics problems. $Mx''(t)\hat x$ equals the force on the block of mass $M$. Super. Still not very detailed.
Now let's talk about the spring. It has lots of layers. A leftmost layer, a middle layer, a right most layer, and more in between. Each layer is in the way of the one to the left so when I say it can move at any speed since it is massless what happens in this case is that it compresses. You could even give each layer a bit of mass $dm$ and then let the mass go to zero after you set up your equations. That limit, after the fact, is actually how we figure out what massless Newtonian objects do.
So it compresses, starting at time $t=0.$ Remember that it compresses until it feels zero force. Let's find out the force. At time $t$, the left block is feeling a force $Mx''(t)\hat x$ so
by Newton's Third Law the spring is feeling a force due to the left block of $-Mx''(t)\hat x$ and it needs to compress until it feels zero force (because the massless parts have to move until they feel no force and because they are layers the motion is compression). So the mass compresses until the force it feels is zero and the force from the left block is $-Mx''(t)\hat x$ so it need to compress until it feels a force of $+Mx''(t)\hat x$ from the right block. And I say until, but since it is massless and Newtonian it happens instantly.
We could use
Newton's Third Law again to find the force the right block feels from the spring (which is equal and opposite to the force $+Mx''(t)\hat x$ the spring feels from the right block) and the right block feels a force of $-Mx''(t).$
So, the two blocks feel equal and opposite forces and the spring feels no net force. And the spring still accelerates (compresses) even though it feels no force because it is massless. You can think of it as having a super tiny mass and so it feels a super tiny force and so the two forces on it almost perfectly balance, so it can move and we are approximating the true force which is tiny with a zero force.
And every single force seems to be related to how the block is accelerating which seems a bit circular and vague. But basically all we've said about the spring is that it can exert forces and that it has very little, or no, mass. We now need to quantify how the spring exerts forces.
So here comes Hooke's Law. We say the spring exerts a force that is proportional to how much it is compressed. How long is the spring. At time $t=0$ it is the natural, rest length. Then being massless it stays out of the way of massive things by moving, and being made of layers so it gets in its own way it compresses. So the length is actually $L-x(t)$ so the force is proportional to $L-(L- x(t))$=$x(t).$ So $|\vec F|=|kx(t)|.$
So going back to the block, the force it felt was equal to $Mx''(t)\hat x$ which apparently really is equal in magnitude to $|kx(t)|$ now that we did the whole analysis. So, now we just need the signs. Positive $x$ goes to the right, so the force from the spring is outwards (also from Hooke's Law) so in the $-\hat x$ direction, so we get $Mx''(t)=-kx(t)$ for some positive $k.$
So, the forces come from the force law (in this case Hooke' Law). Newton's Third Law tells us the forces come in action-reaction pairs & so you can think of Hooke's Law as telling you how much force it takes to compress a spring or as telling you how much force a spring exerts when compressed. Newton says they are equal and opposite. We know that massless objects move or compress however they need to move until they feel no force (you can think of a slight imbalance making the very small mass move very quickly). And the infinitely massive objects simply don't move at all (you can think of them as moving a super tiny amount).
Edit responding to edit in question.
Absolutely nothing changes except one thing. The spring is no longer being pushed out of the block's way.
In particular the spring is still compressed, the spring still exerts a force to the left on the block, the block continues to feel a force from the spring just as it has since the spring first became compressed.
So let's step back. The spring got compressed because the block had velocity towards it and made contact. During the phase where it was getting more and more compressed it had to be compressed because a much more massive block was pushing it out of the way and so whatever forces they exchanged just allowed the much less massive spring to move however it needed to.
The same things happen now. But now at that moment when it is fully compressed the spring no longer is in the way so doesn't have to get out of the way. But the block still feels a force from the spring so it will accelerate to the left. But remember how the spring is pushing on the right block of mass $+\infty$ just as hard as it pushes on the block of mass $M$ on the left? Well if the block just took off to the left and the spring stayed there it would be pushed by the block on the right. So the block on the right is doing the pushing.
We could experimentally verify this if the block on the right was really another block of finite mass, with mass $m'$ with a pin keeping it stuck in place, and if the pin was pulled out and the block on the right was actually much less massive and the pin was keeping it in place before then indeed the mass $M$ block on the left could stay at rest as the very tiny mass $m$ spring and very tiny mass $m'$ right block shoot off to the right.This is because to the right block the left one is basically super massive. So there really is a force on the right block and the lack of motion of the right block and its force is what makes the spring push against the left block so hard it makes the left block move.
So the block on the right keeps pushing on the spring, so the very tiny mass spring is pushed with super acceleration to the left unless it touches the mass $M$ block on the left. So basically the spring is still forced (pun not intended) to be in contact with the block on the left, but now it is because of the block on the right, not because of the block on the left. But the same thing happens it adjusts its length at every moment until the length at that moment is just enough to have it push the two blocks just hard enough so they push back equal and opposite to each other (if the spring is massless, and if instead it has a tiny mass then nearly opposite). Here is where we use the third law again.
When does this stop? When the spring gets back to its natural length then it exerts no force on anything and nothing exerts any force on it. Since it is massless the velocity is actually indeterminate. Even if it is attached to the right block, things are still indeterminate. You can't predict everything when you over simplify. But there is
a solution, which is the spring and block having no force act on them and being momentarily at rest, just stay at rest. There is a form of the first law that asserts they do this. But of course we don't know the spring is at rest since it is massless so has indeterminate speed. But it is a solution. How about the block. It has zero force acting on it but has a velocity to the left so it just cruises on to the left at a steady velocity.
So the spring is the only thing we don't know about. But you could imagine that if just one air molecule bumps into it and it is massless then it will repeat the process, so it wiggles based in anything touching it. Zero mass objects are quite unstable, so saying where they are and what they are doing is probably a bit much to aim for.
So the mass compresses until the force it feels is zero and the force from the left block is $−Mx′′(t)\hat x$
so it need to compress until it feels a force of $+Mx′′(t)\hat x$
from the right block: I understood the statement but during that time when the right block imparts $+Mx′′(t)\hat x$ on the spring, would not the left block moved more towards right & the value of $-Mx′′(t)\hat x$ changes?
It takes no time under the limits of zero mass for the spring and infinite mass for the right block. Because the mass of the spring is zero, it instantly compresses to whatever length it needs to to experience zero total net force. And the right block never ever moves at all so it needs no time to compress itself to give an equal and opposite force. A real mass on the right would basically be a heavy and stiff spring itself and would take time to compress itself. But since it is infinite in mass it doesn't move so it is at its state of no compression (giving no force on others) and at its state of forcing others (as if it were compressed) ready to do neither at any time and thus needing no time to switch from one to the other.
The spring needs to change itself to itself give different forces, but since it is massless it can do instantly. The infinitely massive block doesn't need to change its length or speed to hive any amount of force or momentum or energy or to absorb any amount of force or momentum or energy.
I truly don't think studying zero mass and infinite mass objects is healthy. But at least it can teach you not to apply results beyond their domain of applicability. Infinite masses can have any amount of kinetic energy even though they never move. They can have any momentum even though they never move. They can compress and exert forces without compressing.
This is because a mass with a huge mass can have some finite energy and momentum while having a truly tiny velocity. And it can exert a force from its own compression with a truly tiny compression.
If you give every object a finite mass then a realistic model would depend on the speed of sound in the spring, not the literal speed of sound in air but the speed at which say a compression wave travels down a slinky. If the left mass hits faster than that speed then something totally different happens (it's a shock, which like speed of sound is a technical term) which is totally different than what otherwise would happen. And even when you hit at less than that speed, it still takes time before the right block starts to move. But this does require the spring to have a finite non-zero mass for the wave to take time to get from one side to the other. And for the mass on the right to move it needs a finite non-zero mass as well.
Since the spring compresses infinitely fast and the right block doesn't have to move it all happen instantly and everything is determined by the left block. |
This question already has an answer here:
I have encountered a problem regarding Heisenberg's uncertainty principle. Somewhere I found the expression
$$\Delta (x) \Delta (p) \geq \hbar$$
Whereas in some other places,
$$\Delta (x) \Delta(p) \geq \frac{h}{4\pi}$$
Here the second expression yields different expression than the first one. Why are the two expressions of uncertainty principle different? Which one is the correct form? |
If you’ve worked in computational fluid dynamics, then you’re probably aware of the Taylor-Green vortex – at least the two-dimensional case. A simple google search will land you on this wikipedia page. The classic solution there is presented in the form
\begin{equation} u = \cos x \sin y F(t);\quad v = -\sin x \cos y F(t);\quad F(t) = e^{-2\nu t} \end{equation}
There are also many variants on this form, some that exchange the sine and cosine such as McDermott and Almgren et. al – but the most interesting variation, is that presented by McDermott, and presents a non-stationary form (in space) – i.e. the vortices advect in space on an Eulerian computational grid.
There are four points that I want to address in this post:
It seems that the wrong paper is always cited when referring to the original work for this TG Vortex solution In light of the previous point, we should call the 2D solution the Taylor vortex and keep the Taylor-Green designation to the 3D initial condition How the heck do you obtain solutions that are non-stationary in space (Thanks to Lee Shunn for opening my mind about this) Finally, I want to actually derive the TG vortex solution and discuss it a bit further Point 1: We are citing the wrong paper
I am the first to admit – I have always cited the wrong paper when referring to the 2d TG vortex – namely:
Taylor, G. I., & Green, A. E. (1937). Mechanism of the production of small eddies from large ones. Proceedings of the Royal Society of London. Series A-Mathematical and Physical Sciences, 158(895), 499.
For the life of me I cannot figure out how to recover the 2D solution from that paper. The correct paper to cite for the 2D vortex is the following:
G.I. Taylor F.R.S. (1923) LXXV. On the decay of vortices in a viscous fluid, The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, 46:274, 671-674, DOI: 10.1080/14786442308634295
In light of the previous point, we should refer to the 2D solution as the Taylor vortex and reserve the Taylor-Green designation for the 3D case.
Point 3: How to obtain “advecting” solutions on an Eulerian grid?
Simple – Thanks to the mysterious art of Galilean transformations! Let’s say you find a solution for the velocity field $$\mathbf{u}(\mathbf{x},t)$$ in an inertial reference frame $(\mathbf{x},t)$ and now you want your solution in a moving frame $(\mathbf{x}’=\mathbf{x} – \mathbf {c} t, t)$, then, the solution in the moving frame, $\mathbf{u}’$ is
\begin{equation} \mathbf{u}’ =\mathbf{u}(\mathbf{x}-\mathbf{c} t) = \mathbf{u}(\mathbf{x}, t) – \mathbf{c} \end{equation} So that – in practice – where we implement our solutions on an Eulerian grid, we simply set \begin{equation} \mathbf{u}(\mathbf{x}, t) =\mathbf{c} + \mathbf{u}(\mathbf{x}-\mathbf{c} t) \end{equation} I owe it to Lee Shunn of Cascade Technologies for pointing that out to me in a private communication regarding one of his manufactured solutions for variable density flows. Point 4: How did Taylor do it?
Taylor is like Euler – not only virtue of their names rhyming – but also by virtue of their contributions to Fluid mechanics. His approach was based on a vorticity-streamfunction approach (heavily used in the rocket-motor stability analysis community – see Culick, Flandro, Majdalani, Saad) to eliminate the pressure. In two-dimensions, the only non-zero component of vorticity is
\begin{equation} \omega_z = \frac{\partial v}{\partial x} – \frac{\partial u}{\partial y} \end{equation} Now, using the streamfunction ($u=-\frac{\partial \psi}{\partial y}$, $v = \frac{\partial \psi}{\partial x}$) \begin{equation} \omega_z = \nabla^2 \psi \end{equation} Great so far. Now we write the vorticity transport equation \begin{equation} \frac{\partial\omega_z}{\partial t}+\mathbf{u}\cdot\nabla\omega_z=\nu\nabla^{2}\omega_z \end{equation} or \begin{equation} \left( \frac{\partial}{\partial t}+\mathbf{u}\cdot\nabla -\nu\nabla^{2}\right)\omega_z=0 \end{equation} Now Taylor makes a beautiful assumption: he sets $\omega_z = k \psi$. As I will explain later (maybe in the future), this is entire possible – this only means that lines of constant vorticity are also lines of constant streamfunction. More generally, this type of flow belongs to a general class of flows known as Generalized Beltrami flows where $\nabla\times\mathbf{u}\times\boldsymbol{\Omega} = 0$.
In 2d, this implies that the vorticity is an arbitrary function of $\psi$, i.e. $\omega_z = f(\psi)$. Taylor’s choice is simply $f(\psi) = k \psi$. This assumption leads to two simplifications:
We now have $\nabla ^2 \psi = k \psi$ The advection term in the vorticity equation disappears by virtue of the fact that $\mathbf{u}$ is perpendicular to $\nabla \psi$
The vorticity transport equation now reduces to
\begin{equation} \frac{\partial \psi}{\partial t} -\nu\psi=0 \end{equation} which leads to the general solution \begin{equation} \psi = F(x,y) e^{-k\nu t} \end{equation} where $F(x,y)$ is a solution to the vorticity equation (substitute $\psi = F e^{k\nu t}$ into the vorticity equation $\nabla^2 \psi = k \psi$) \begin{equation} \nabla F^2 = k F(x,y) \label{eq:f-equation} \end{equation} Generic solutions for $F$ are therefore of the sine, cosine, or exponential type. Taylor then proceeds to set a solution of the form \begin{equation} F(x,y) = A \cos (\pi \frac{x}{d}) \cos (\pi \frac{y}{d}) \end{equation} Substitution into \eqref{eq:f-equation} leads to \begin{equation} k = \frac{2\pi^2}{d^2} \end{equation} and the final solution is \begin{equation} \psi = A \cos (\pi \frac{x}{d}) \cos (\pi \frac{y}{d}) e^{-\frac{2\pi^2}{d^2} \nu t} \end{equation} This leads to the classical velocity field for the 2D Taylor vortex \begin{equation} u = A \frac{\pi}{d} \cos (\pi \frac{x}{d}) \sin (\pi \frac{y}{d})e^{-\frac{2\pi^2}{d^2} \nu t}; \\ v = – A \frac{\pi}{d} \sin (\pi \frac{x}{d}) \cos (\pi \frac{y}{d}) e^{-\frac{2\pi^2}{d^2} \nu t} \end{equation} or, more generally \begin{equation} \boxed{\psi = A \cos (\alpha x) \cos (\beta y) e^{-(\alpha^2 + \beta^2) \nu t}} \end{equation} \begin{equation} u = A \beta \cos (\alpha x) \sin (\beta y)e^{-(\alpha^2 + \beta^2) \nu t}; \\ v = – A \alpha\sin (\alpha x) \cos (\beta y)e^{-(\alpha^2 + \beta^2) \nu t} \end{equation} If implemented in an Eulerian computational code, this solution will just decay in time. To obtain advecting solutions, apply the Galilean transformation discussed above. This leads to solutions of the form \begin{equation} \boxed{ u = u_f + A \beta \cos [\alpha (x-u_f t)] \sin [\beta (y – v_f t)]e^{-(\alpha^2 + \beta^2) \nu t}; \\ v = v_f – A \alpha \sin [\alpha (x-u_f t)] \cos [\beta (y – v_f t)]e^{-(\alpha^2 + \beta^2) \nu t}} \end{equation}
Another interesting form (used by Almgren and McDermott) is
\begin{equation}
\boxed{\psi = -A \cos (\alpha x) \cos (\beta y) e^{-(\alpha^2 + \beta^2) \nu t}} \end{equation} \begin{equation} u = -A \beta \cos (\alpha x) \sin (\beta y)e^{-(\alpha^2 + \beta^2) \nu t}; \\ v = A \alpha\sin (\alpha x) \cos (\beta y)e^{-(\alpha^2 + \beta^2) \nu t} \end{equation}
or, for an advecting solution
\begin{equation}
u = u_f – A \beta \cos [\alpha (x-u_f t)] \sin [\beta (y – v_f t)]e^{-(\alpha^2 + \beta^2) \nu t}; \\ v = v_f + A \alpha \sin [\alpha (x-u_f t)] \cos [\beta (y – v_f t)]e^{-(\alpha^2 + \beta^2) \nu t} \end{equation} Generalized Beltrami Flows
Taylor’s vortex belongs a class of flows known as generalized Beltrami flows. These are flows where $\nabla\times\mathbf{u}\times\boldsymbol{\Omega} = 0$.
To be continued… |
I find myself needing to compute (or asymptotically estimate) the following sum over the $2^{S-1}$ compositions of $S$. I am hoping an expert in combinatorics (I am a computer scientist) will recognise this summation.
Let $B_{S,k}$ denote the set of compositions that have exactly $k$ parts, and define $T_{k}$ as,
$$T_{k}=\sum_{a \in B_{S,k}}\frac{a_{1}^{a_{2}}a_{2}^{a_{3}}...a_{k-1}^{a_{k}}}{a_{1}!a_{2}!...a_{k}!},\:\: \text{with}\:\: T_{1}=1.$$
The sum that I am interested in is $\sum_{k}T_{k}$.
Does an expression in terms of $S$ (either exact, or asymptotic) exist, or can it be found? |
Cadabra
Computer algebra system for field theory problems
Class to display expressions in a format that Mathematica can parse. Will throw an exception if a Cadabra Ex object cannot be understood by Mathematica. Can also convert expressions back to Cadabra notation.
#include <DisplayMMA.hh>
DisplayMMA (const Kernel &, const Ex &, bool use_unicode) void import (Ex &) Rewrite the output of mathematica back into a notation used by Cadabra. More... std::string preparse_import (const std::string &) Public Member Functions inherited from cadabra::DisplayBase DisplayBase (const Kernel &, const Ex &) void output (std::ostream &) void output (std::ostream &, Ex::iterator)
virtual bool needs_brackets (Ex::iterator it) override Determine if a node needs extra brackets around it. More...
bool use_unicode Protected Attributes inherited from cadabra::DisplayBase const Ex & tree const Kernel & kernel
void print_multiplier (std::ostream &, Ex::iterator) Output the expression to a mathematica-readable form. More... void print_opening_bracket (std::ostream &, str_node::bracket_t) void print_closing_bracket (std::ostream &, str_node::bracket_t) void print_parent_rel (std::ostream &, str_node::parent_rel_t, bool first) void print_children (std::ostream &, Ex::iterator, int skip=0) virtual void dispatch (std::ostream &, Ex::iterator) override For every object encountered, dispatch will figure out the most appropriate way to convert it into a LaTeX expression. More... void print_productlike (std::ostream &, Ex::iterator, const std::string &inbetween) Printing members for various standard constructions, e.g. More... void print_sumlike (std::ostream &, Ex::iterator) void print_fraclike (std::ostream &, Ex::iterator) void print_commalike (std::ostream &, Ex::iterator) void print_arrowlike (std::ostream &, Ex::iterator) void print_powlike (std::ostream &, Ex::iterator) void print_intlike (std::ostream &, Ex::iterator) void print_equalitylike (std::ostream &, Ex::iterator) void print_components (std::ostream &, Ex::iterator) void print_partial (std::ostream &str, Ex::iterator it) void print_matrix (std::ostream &str, Ex::iterator it) void print_other (std::ostream &str, Ex::iterator it) bool children_have_brackets (Ex::iterator ch) const
std::map< std::string, std::string > symmap Map from Cadabra symbols to Mathematica symbols. More... std::multimap< std::string, std::string > regex_map std::map< nset_t::iterator, Ex, nset_it_less > depsyms Map from symbols which have had dependencies added to an expression containing these dependencies. More...
private
overrideprivatevirtual
For every object encountered, dispatch will figure out the most appropriate way to convert it into a LaTeX expression.
This may be done by simply looking at the object's name (e.g.
\prod will print as a product) but may also involve looking up properties and deciding on the best course of action based on the attached properties.
Implements cadabra::DisplayBase.
void DisplayMMA::import ( Ex & ex )
Rewrite the output of mathematica back into a notation used by Cadabra.
This in particular involves converting 'Sin' and friends to
\sin and so on, as well as converting all the greek symbols. Currently only acts node-by-node, does not do anything complicated with trees.
overrideprotectedvirtual
Determine if a node needs extra brackets around it.
Uses context from the parent node if necessary. Has to be implemented in a derived class, because the answer depends on the printing method (e.g.
(a+b)/c needs brackets when printed like this, but does not need brackets when printed as
\frac{a+b}{c}).
Implements cadabra::DisplayBase.
std::string DisplayMMA::preparse_import ( const std::string & in )
private
private
private
private
private
private
private
private
private
private
Output the expression to a mathematica-readable form.
For symbols which cannot be parsed by mathematica, this can convert to an alternative, Such rewrites can then be undone by the 'import' member above.
private
private
private
private
private
private
Printing members for various standard constructions, e.g.
print as a list, or as a decorated symbol with super/subscripts etc. The names reflect the structure of the output, not necessarily the meaning or name of the object that is being printed.
private
Map from symbols which have had dependencies added to an expression containing these dependencies.
private
private
Map from Cadabra symbols to Mathematica symbols.
This is a bit tricky because MathLink does not pass
\[Alpha] and friends transparently. So we feed it UTF8 α and so on, but then we get
\[Alpha] back, and that needs to be regex-replaced before we feed it to our parser as ours does not swallow that kind of bracketing.
protected |
Motivation:
Given the roots of the quadratic $2x^2+6x+7=0$ find a quadratic with roots $\alpha^2-1$ and $\beta^2-1$
I was able to solve this problem in two ways:
Method 1:
Sum of the roots $\alpha+\beta=-\frac{b}{a}$
Product of roots $\alpha\beta=\frac{c}{a}$
Hence $\alpha+\beta=-3$ and $\alpha\beta=\frac{7}{2}$
We want an equation with roots $\alpha^2-1$ and $\beta^2-1$
The sum of the roots of the new quadratic will be $\alpha^2-1+\beta^2-1=\alpha^2+\beta^2-2$
The product of the roots of the new quadratic will be $(\alpha^2-1)(\beta^2-1)=\alpha^2\beta^2-(\alpha+\beta)+1$
We are able to compute $\alpha^2+\beta^2$ as it is $(\alpha+\beta)^2-2\alpha\beta$ and so the problem is solved.
Plugging the numbers gives $4u^2+45=0$
Method 2
Let $u=\alpha^2-1\implies\alpha=\sqrt{u+1}$ but we know that $\alpha$ solves the original equation so:
$$\begin{align}2\alpha^2+6\alpha+7&=0\\2(u+1)+6\sqrt{u+1}+7&=0\\\sqrt{u+1}&=\frac{-2u-9}{6}\\u+1&=\frac{1}{36}(-2u-9)^2\\36u+36&=4u^2+36u+81\\0&=4u^2+45\end{align}$$
Question:The first method clearly uses the values of $\alpha$ and $\beta$ but the second seemingly only requires $\alpha$. How is this possible? Sure, one man's $\alpha$ is another's $\beta$ and so you could relabel as the choice of $\alpha$ and $\beta $ is arbitrary. This is believable because of the symmetry involved in the new roots $\alpha^2-1$ looks much like a $\beta^2-1$ but I feel there must be more to this. Supposing one root of the new quadratic was $\alpha^2-1$ but the other was $\beta^3-2\beta$ or something worse? How would the second method know? This leads me to a more fundamental question.
Are there only certain functions of the roots of an old quadratic that can we can find a new quadratic in this way? I suppose we could consider $(x-f(\alpha)(x-g(\beta)=0$ where $f$ and $g$ are the functions of the old roots but then could we alway compute these numerically?
Thanks for taking the time to read this and for any contributions. |
Publications by Felix Parra Diaz
NUCLEAR FUSION
59 (2019) ARTN 112011
PLASMA PHYSICS AND CONTROLLED FUSION
60 (2018) ARTN 074004
NUCLEAR FUSION
58 (2018) ARTN 026003
JOURNAL OF PLASMA PHYSICS
84 (2018) ARTN 905840407
NUCLEAR FUSION
58 (2018) ARTN 124005
Plasma Physics and Controlled Fusion IOP Publishing (2018)
Using a kinetic model for the ions and adiabatic electrons, we solve a steady state, electron-repelling magnetic presheath in which a uniform magnetic field makes a small angle $\alpha \ll 1$ (in radians) with the wall. The presheath characteristic thickness is the typical ion gyroradius $\rho_{\text{i}}$. The Debye length $\lambda_{\text{D}}$ and the collisional mean free path of an ion $\lambda_{\text{mfp}}$ satisfy the ordering $\lambda_{\text{D}} \ll \rho_{\text{i}} \ll \alpha \lambda_{\text{mfp}}$, so a quasineutral and collisionless model is used. We assume that the electrostatic potential is a function only of distance from the wall, and it varies over the scale $\rho_{\text{i}}$. Using the expansion in $\alpha \ll 1$, we derive an analytical expression for the ion density that only depends on the ion distribution function at the entrance of the magnetic presheath and the electrostatic potential profile. Importantly, we have added the crucial contribution of the orbits in the region near the wall. By imposing the quasineutrality equation, we derive a condition that the ion distribution function must satisfy at the magnetic presheath entrance --- the kinetic equivalent of the Chodura condition. Using an ion distribution function at the entrance of the magnetic presheath that satisfies the kinetic Chodura condition, we find numerical solutions for the self-consistent electrostatic potential, ion density and flow across the magnetic presheath for several values of $\alpha$. Our numerical results also include the distribution of ion velocities at the Debye sheath entrance. We find that at small values of $\alpha$ there are substantially fewer ions travelling with a large normal component of the velocity into the wall.
PLASMA PHYSICS AND CONTROLLED FUSION
59 (2017) ARTN 065005
NUCLEAR FUSION
56 (2017) ARTN 016016
PLASMA PHYSICS AND CONTROLLED FUSION
59 (2017) ARTN 025015
NUCLEAR FUSION
57 (2017) ARTN 046008
PLASMA PHYSICS AND CONTROLLED FUSION
59 (2017) ARTN 055014 Ion-scale turbulence in MAST: anomalous transport, subcritical transitions, and comparison to BES measurements
PLASMA PHYSICS AND CONTROLLED FUSION
59 (2017) ARTN 114003 Turbulent momentum transport due to the beating between different tokamak flux surface shaping effects
PLASMA PHYSICS AND CONTROLLED FUSION
59 (2017) ARTN 024007
Physical review letters
118 (2017) 185002-
Measurements of the relaxation of a zonal electrostatic potential perturbation in a nonaxisymmetric magnetically confined plasma are presented. A sudden perturbation of the plasma equilibrium is induced by the injection of a cryogenic hydrogen pellet in the TJ-II stellarator, which is observed to be followed by a damped oscillation in the electrostatic potential. The waveform of the relaxation is consistent with theoretical calculations of zonal potential relaxation in a nonaxisymmetric magnetic geometry. The turbulent transport properties of a magnetic confinement configuration are expected to depend on the features of the collisionless damping of zonal flows, of which the present Letter is the first direct observation.
Overview of progress in European medium sized tokamaks towards an integrated plasma-edge/wall solution
NUCLEAR FUSION
57 (2017) ARTN 102014
PLASMA PHYSICS AND CONTROLLED FUSION
59 (2017) ARTN 034002
JOURNAL OF PLASMA PHYSICS
83 (2017) ARTN 905830206 |
Equilibrium Hydrolysis of Salts and Solubility and Buffer Solutions Degree of hydrolysis : K_{h} = \frac{C\alpha.C\alpha}{C - C\alpha} = \frac{C\alpha^{2}}{1 - \alpha} K h = Cα 2 , 1 − α ≅ 1 \alpha = \sqrt{\frac{K_{h}}{C}} pH of weak acid and strong base : [OH^{-}] = \sqrt{\frac{K_{w}.C}{K_{a}}} pOH = 7 - \frac{1}{2}pk_{a} - \frac{1}{2}logC pH = 14 - \left(7 - \frac{1}{2}pk_{a} - \frac{1}{2}logC\right) pH = 7 - \frac{1}{2}pk_{a} + \frac{1}{2}logC Salt of strong acid and weak base : NH_4^+ + H_{2}O \rightleftharpoons NH_{4}OH + H^{+} k_{h} = \frac{k_{w}}{k_{b}} \alpha = \sqrt{\frac{k_{h}}{C}} pH = 7 - \frac{1}{2}pk_{a} - \frac{1}{2}logC Salt of weak acid and weak base : k_{h} = \frac{k_{w}}{k_{a}k_{b}} k_{h} = \frac{C\alpha.C\alpha}{(C - \alpha) (C - \alpha)} = \alpha^{2} \alpha = \sqrt{k_{h}} pH = 7 + \frac{1}{2}pk_{a} - \frac{1}{2}pk_{a}
Types of buffers : Acidic buffer : Weak acid and its salt CH_{3}COOH + CH_{3}COONa HCN + KCN Basic buffer : Weak base and its salt NH_{3} + NH_{4}Cl NH_{4}OH + NH_{4}Cl Henderson equation pH of acidic buffer : pH = pk_{a} + log\left[\frac{salt}{Acid}\right] Basic buffer − Henderson equation pOH = pk_{b} + log\frac{\left[B^{+}\right]}{Base} pOH = pk_{b} + log\frac{N_{s}V_{s}}{N_{b}V_{b}} Buffer capacity (φ) : \tt \phi = \frac{no. of \ moles \ of \ SA \ or \ SB \ added \ to \ 1L \ of \ buffer \ solution}{change \ in \ pH(\triangle pH)} pH = pk a, [salt] = [Acid] pOH = Pk b, [salt] = [Base] Solubility product : A_{x}.B_{y} \rightleftharpoons xA^{y+} + yB^{x-} k_{sp} = [A^{y+}]^{x} [B^{x-}]^{y} Relation between solubility and solubility product. k_{sp} = (xs)^{x}.(ys)^{y} \Rightarrow x^{x}.y^{y}.s^{x + y} s = \sqrt{k_{sp}} s =\left(\frac{k_{sp}}{4}\right)^{1/3}; s =\left(\frac{k_{sp}}{27}\right)^{1/4}; s =\left(\frac{k_{sp}}{108}\right)^{1/5} View the Topic in this Video from 0:10 to 10:39
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1. For acidic buffer, pH = pK_{a} + log\frac{\left[salt\right]}{\left[acid\right]}
2. For basic buffer, pOH = pK_{b} + log\frac{\left[salt\right]}{\left[base\right]}
3. \tt Buffer \ capacity\left(\phi\right) = \frac{number \ of \ moles \ of \ acids \ or \ base \ added \ to \ 1 L \ of \ buffer}{change \ in \ pH}
4. For\ A_{x}B_{y} \rightleftharpoons xA^{y+} + y B^{x-}
K_{sp} = \left[A^{y+}\right]^{x} \left[B^{x-}\right]^{y} |
This question already has an answer here:
The task is:
Show that the equation $[A,B] :=AB-BA= iqI$, where $q\neq 0$ is a non-zero real constant and $I$ is the identity matrix cannot be satisfied by any finite dimensional Hermitian matrices $A,B$.
What I got so far:
If I set $A=B$ and $q=0$ than the right hand side is $[A,B]=[A,A]=0$ and the left hand side is $iqI=0$. This means the equation is satisfied in this case (counter-example).
Furhter I calculated $[A,B]= AB-BA = AB-(AB)^\dagger$ which I can use for the diagonal elements to show that $([A,B])_{ii} = \sum_k A_{ik}B_{ki}-A^\ast_{ik}B^\ast_{ki} = i \sum_k 2\, \Im(A_{ik}B_{ki})$ where I am stuck.
Do you have any idea how this statement can be proven even though I found a counter expample? |
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If the fractional part of 1x\dfrac{1}{x}x1 and x2x^2x2 are equal for some x∈(2,3)x \in (\sqrt{2},\sqrt{3})x∈(2,3), then find the value of x4−3x x^4-\dfrac{3}{x} x4−x3.
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Hypergraph isomorphic copy search¶
This module implements a code for the following problem:
INPUT:two hypergraphs \(H_1, H_2\)
OUTPUT:a copy of \(H_2\) in \(H_1\)
It is also possible to enumerate all such copies, and to require that such copies be induced copies. More formally:
A copy of \(H_2\) in \(H_1\) is an injection \(f:V(H_2)\mapsto V(H_1)\) such that for any set \(S_2\in E(H_2)\) we have \(f(S_2)\in E(H_1)\).
It is an
inducedcopy if no other set of \(E(H_1)\) is contained in \(f(V(H_2))\), i.e. \(|E(H_2)|=\{S:S\in E(H_1)\text{ and }S\subseteq f(V(H_2))\}\).
The functions implemented here lists all such injections. In particular, the number of copies of \(H\) in itself is equal to \(|Aut(H)|\).
The feature is available through
IncidenceStructure.isomorphic_substructures_iterator().
Implementation¶
A hypergraph is stored as a list of edges, each of which is a “dense” bitset over \(|V(H_1)|\) points. In particular, two sets of distinct cardinalities require the same memory space. A hypergraph is a C struct with the following fields:
n,m(
int) – number of points and edges.
limbs(
int) – number of 64-bits blocks per set.
set_space(
uint64_t *) – address of the memory used to store the sets.
sets(
uint64_t **) –
sets[i]points toward the
limbsblocks encoding set \(i\). Note also that
sets[i][limbs]is equal to the cardinality of
set[i], so that
setshas lenth
m*(limbs+1)*sizeof(uint64_t).
names(
int *) – associates an integer ‘name’ to each of the
npoints.
The operations used on this data structure are:
void permute(hypergraph * h, int n1, int n2)– exchanges points \(n1\) and \(n2\) in the data structure. Note that their names are also exchanged so that we still know which is which.
int induced_hypergraph(hypergraph * h1, int n, hypergraph * tmp1)– stores in
tmp1the hypergraph induced by the first \(n\) points, i.e. all sets \(S\) such that \(S\subseteq \{0,...,n-1\}\). The function returns the number of such sets.
void trace_hypergraph64(hypergraph * h, int n, hypergraph * tmp)– stores in
tmp1the trace of \(h\) on the first \(n\) points, i.e. all sets of the form \(S\cap \{0,...,n-1\}\).
Algorithm¶
We try all possible assignments of a representant \(r_i\in H_1\) for every \(i\in H_2\). When we have picked a representant for the first \(n<\) points \(\{0,...,n-1\}\subsetneq V(H_2)\), we check that:
The hypergraph induced by the (ordered) list \(0,...,n-1\) in \(H_2\) is equal to the one induced by \(r_0,...,r_{n-1}\) in \(H_1\). If \(S\subseteq \{0,...,n-1\}\) is contained in \(c\) sets of size \(k\) in \(H_2\), then \(\{r_i:i\in S\}\) is contained in \(\geq c\) sets of size \(k\) in \(H_1\). This is done by comparing the trace of the hypergraphs while remembering the original size of each set.
As we very often need to build the hypergraph obtained by the trace of the first \(n\) points (for all possible \(n\)), those hypergraphs are cached. The hypergraphs induced by the same points are handled similarly.
Limitations¶ Number of points For efficiency reason the implementation assumes that \(H_2\)has \(\leq 64\) points. Making this work for larger values means that calls to
qsort have to be replaced by calls to
qsort_r (i.e. to sort the edgesyou need to know the number of limbs per edge) and that induces a big slowdownfor small cases (~50% when this code was implemented). Also, 64 points for \(H_2\)is already very very big considering the problem at hand. Even \(|V(H_1)|> 64\)seems too much.
Vertex ordering The order of vertices in \(H_2\) has a huge influence on theperformance of the algorithm. If no set of \(H_2\) contains more that one of thefirst \(k<n\) points, then almost all partial assignments of representants arepossible for the first \(k\) points (though the degree of the vertices is takeninto account). For this reason it is best to pick an ordering such that thefirst vertices are contained in as many sets as possible together. A heuristicis implemented at
relabel_heuristic().
AUTHORS:
Nathann Cohen (November 2014, written in various airports between Nice and Chennai). Methods¶ class
sage.combinat.designs.subhypergraph_search.
SubHypergraphSearch¶
Bases:
object
relabel_heuristic()¶
Relabels \(H_2\) in order to make the algorithm faster.
Objective: we try to pick an ordering \(p_1,...,p_k\) of the points of \(H_2\) that maximizes the number of sets involving the first points in the ordering. One way to formalize the problems indicates that it may be NP-Hard (generalizes the max clique problem for graphs) so we do not try to solve it exactly: we just need a sufficiently good heuristic.
Assuming that the first points are \(p_1,...,p_k\), we determine \(p_{k+1}\) as the point \(x\) such that the number of sets \(S\) with \(x\in S\) and \(S\cap \{p_1,...,p_k\}\neq \emptyset\) is maximal. In case of ties, we take a point with maximum degree.
This function is called when an instance of
SubHypergraphSearchis created.
EXAMPLES:
sage: d = designs.projective_plane(3) sage: d.isomorphic_substructures_iterator(d).relabel_heuristic() |
Question:
A company uses a product that must be delivered by special trucks. As such, ordering (and delivery) costs are relatively high, at $3,600 per order. The product is packaged in four-liter containers. The cost of holding the product in storage is $90 per four-liter container per year. The annual demand for the product, which is constant over time, is 3,000 four-liter containers per year. The lead time from order placement to receipt is 16 days. The company operates 320 working days per year. Compute the optimal order quantity, the total minimum inventory cost, and the reorder point.
Economic order quantity:
Economic order quantity is the level of unit to be ordered at a time. It provides minimum inventory cost, so it is considered as most economic level of quantity. It includes annual cost, ordering cost and carrying cost.
Answer and Explanation:
{eq}\textrm{Economic order Quantity }= \sqrt{\dfrac{2\times A\times O}{C}} {/eq}
A- Annual Demand, $ 3,000
O - Ordering cost, $ 3600
C0- Carrying cost $ 90
{eq}\textrm{Economic order Quantity }= \sqrt{\dfrac{2\times 3000\times 3600}{90}} {/eq}
= 489.8 or 490 four-liter
Total minimum inventory cost
Ordering cost 3000 /490 = 6 order( 6 * $3,600) = 21,600
Carrying cost{ (490/2)* $90} = $22,050
Total cost = $43,650
Reorder point.
= average daily unit sales * delivery lead time
Average daily sales = 3000/320 = 9.375
Lead time = 16 days
Reorder point = 9.375*16
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from Finance 101: Principles of FinanceChapter 19 / Lesson 8 |
I have the following group of equations:
\begin{align*} T' & =kn \\ b' & =\tau n\\ n' & =-\tau b-kT\end{align*}
I want to put these equations into one box and to put the description "Formule di Frenet" centered on the right.
In other words, I want to extend the following result valid for one equation:
\begin{equation*} \boxed{r:\alpha(t_0)+\Braket{\alpha'(t_0)}} \qquad \emph{retta tangente alla curva $\alpha$ nel punto $\alpha(t_0)$} \end{equation*} |
Let $R$ be a ring and let $n\in\mathbb{Z}$.
Given $a\in R$, I've seen $na$ defined as $$ na:=\begin{cases}0&\text{if }n=0,\\\underbrace{a+a+\cdots+a}_{n\text{ times}}&\text{if }n>0,\\\underbrace{-a-a-\cdots-a}_{|n|\text{ times}}&\text{if }n<0.\end{cases}\tag{1} $$ With this definition, I've even read that $$ (m+n)a=ma+na\tag{2} $$ and $$ (mn)1=(m1)(n1)\tag{3} $$ are obvious (and I do agree with this view).
Question:
(i)Isn't $(1)$ informal? If so, would $na$ be formally defined, for $n\geq0$, recursively via the equations \begin{align*} 0a&=0\tag{4}\\ (n+1)a&=na+a\tag{5} \end{align*} and for $n<0$ by $na:=-(-n)a$ ?
(ii)Would formal proofs of $(2)$ and $(3)$ be as given below?
I'm asking since I've never seen this done...
Proof of $(2)$: For fixed $m$, proceed by induction on $n$ for $n\geq-m$. Then, for $n<-m$,$$(m+n)a=-(-m-n)a=-((-m)a+(-n)a)=-(-m)a-(-n)a=ma+na$$ Proof of $(3)$: First show the result for $n\in\mathbb{Z}$ and $m\geq0$. To do so, prove by induction on $m$ for fixed $n$ that $(mn)1=m(n1)$ (making use of $(2)$) and then that this equals $(m1)(n1)$, by induction on $m$ again. Finally, for $m<0$,$$(mn)1=(-(-m)n)1=-((-m)n)1=-((-m)1)(n1)=(-(-m)1)(n1)=(m1)(n1)$$ |
Since Matt proved (1) and (2), I'll prove the rest.
(3)Let $RI^+(A)$ be the set of regular ideals of $A$.Clearly $RI^+(A)$ is an ordered commutative monoid with mulitiplications of ideals.Let $RI^+(B)$ be the set of ideals of $B$ which are relatively prime to $\mathfrak{f}$.$RI^+(B)$ is also an ordered commutative monoid.By (1) and (2), $RI^+(A)$ is canonically isomorphic to $RI^+(B)$ as an ordered commutative monoid.Since $B$ is a Dedekind domain, (3) follows immediately.
(4) follows immediately from (3) and the following lemma.
Lemma 1Let $P$ be a maximal ideal of $A$.$P$ is invertible if and only if $P$ is regular.
Proof:Suppose P is regular.By this, $A_P$ is integrally closed.Since $A_P$ is integrally closed, Noetherian and of dimension 1, it is a discrete valuation ring.Hence $PA_P$ is principal.Let $Q$ be a maximal ideal such that $Q \neq P$.Since $P$ is not contained in $Q$, $PA_Q = A_Q$.Hence $PA_Q$ is also principal.Since $A$ is Noetherian, $P$ is finitely generated over $A$.Hence P is invertible by this.
Suppose conversely P is invertible.By this, $PA_P$ is principal.Hence $A_P$ is a discrete valuation ring(e.g Atiyah-MacDonald).Hence $A_P$ is integrally closed.By this, $P$ is regular.
QED
Lemma 2Let $A$ be a commutative Noetherian ring.Let $I$ be a proper ideal of $A$ such that $dim A/I = 0$.Then $A/I$ is canonically isomorphic to $\prod_P A_P/IA_P$, where $P$ runs over all the maximal ideals of $A$ such that $I \subset P$.
This is well knowm.
Lemma 3Let $A$ be a Noetherian domain of dimension 1.Let $I$ be a non-zero proper ideal of $A$.Then $(A/I)^*$ is canonically isomorphic to $\bigoplus_{\mathfrak{p}} (A_\mathfrak{p}/IA_\mathfrak{p})^*$ as abelian groups, where $\mathfrak{p}$ runs over all the maximal ideals of $A$ such that $I \subset \mathfrak{p}$.
This follows immediately from
Lemma 2.
Lemma 4Let $A, K, B, \mathfrak{f}$ be as in the title question.Then $(B/\mathfrak{f})^*$ is canonically isomorphic to $\bigoplus_{\mathfrak{p}} (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$ as abelian groups, where $\mathfrak{p}$ runs over all the maximal ideal of $A$.
Proof:By
Lemma 3, $(B/\mathfrak{f})^*$ is canonically isomorphic to $\bigoplus_{\mathfrak{P}} (B_\mathfrak{P}/\mathfrak{f}B_\mathfrak{P})^*$ as an abelian group, where $\mathfrak{P}$ runs over all the maximal ideal of $B$ such that $\mathfrak{f} \subset \mathfrak{P}$.If $\mathfrak{p}$ is a regular prime ideal of $A$, $A_\mathfrak{p}$ is integrally closed by this.Hence $B_\mathfrak{p} = A_\mathfrak{p}$.Since $\mathfrak{f}A_\mathfrak{p} = A_\mathfrak{p}$, $B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p} = A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p} = 0$.Hence we only need to consider $\mathfrak{p}$ such that $\mathfrak{f} \subset \mathfrak{p}$.It's easy to see that $B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p}$ is canonically isomorphic to $\prod B_\mathfrak{P}/\mathfrak{f}B_\mathfrak{P}$, where $\mathfrak{P}$ runs over all the maximal ideals of $B$ lying over $\mathfrak{p}$.Hence $(B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$ is canonically isomorphic to $(\bigoplus B_\mathfrak{P}/\mathfrak{f}B_\mathfrak{P})^*$, where $\mathfrak{P}$ runs over all the maximal ideals of $B$ lying over $\mathfrak{p}$. QED
Lemma 5Let $B$ be an integral domain.Let $A$ be a subring of $B$ such that $B$ is integral over $A$.Let $I$ be an ideal of $A$.Let $\mathfrak{p}$ be a prime ideal of $A$ such that $I \subset \mathfrak{p}$.Let $B_\mathfrak{p}$ be the localization of $B$ with respect the multiplicative subset $A - \mathfrak{p}$. Let $f:B_\mathfrak{p} \rightarrow B_\mathfrak{p}/IB_\mathfrak{p}$ be the canonical homomorphism.$f$ induces a group homomorphism $g: (B_\mathfrak{p})^* \rightarrow (B_\mathfrak{p}/IB_\mathfrak{p})^*$.Then $g$ is surjective.
Proof:Since $A_\mathfrak{p}$ is a local ring and $B_\mathfrak{p}$ is integtral over $A_\mathfrak{p}$,every maximal ideal $\mathfrak{Q}$ of $B_\mathfrak{p}$ lies over $\mathfrak{p}A_\mathfrak{p}$.Hence $\mathfrak{Q}$ = $\mathfrak{P}B_\mathfrak{p}$, where $\mathfrak{P}$ is a maximal ideal of $B$ lying over $\mathfrak{p}$.Since $I \subset \mathfrak{p}$, $I \subset \mathfrak{P}$.Hence $IB_\mathfrak{p} \subset \mathfrak{Q}$.Let $x \in B_\mathfrak{p}$.Suppose $f(x)$ is invertible.Then $f(x)$ is not contained in any maximal ideal of $B_\mathfrak{p}/IB_\mathfrak{p}$.Suppose $x$ is not invertible.$x$ is contained in a maximal ideal of $B_\mathfrak{p}$.This is a contradiction.
QED
Lemma 6Let $A, K, B, \mathfrak{f}$ be as in the title question.Let $\mathfrak{p}$ be a prime ideal of $A$ such that $\mathfrak{f} \subset \mathfrak{p}$.Since $\mathfrak{f}$ is an ideal of both $A$ and $B$, $\mathfrak{f} = \mathfrak{f}A = \mathfrak{f}B$.Hence $\mathfrak{f}A_\mathfrak{p} = \mathfrak{f}B_\mathfrak{p}$.Hence $A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p} \subset B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p}$.Hence $(A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^* \subset (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$.We claim $(B_\mathfrak{p})^*/(A_\mathfrak{p})^*$ is isomorphic to $(B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*/(A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$.
Proof:By
lemma 5, $g: (B_\mathfrak{p})^* \rightarrow (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$is surjective.Let $\pi: (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^* \rightarrow (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*/(A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$ be the canonical homomorphism.Let $h: (B_\mathfrak{p})^* \rightarrow (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*/(A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$ be $\pi g$.Let $x \in (B_\mathfrak{p})^*$.Suppose $h(x) = 0$.Then $g(x) \in (A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$.Hence thers exists $y \in A_\mathfrak{p}$ such that $x \equiv y$ (mod $\mathfrak{f}B_\mathfrak{p}$).Since $\mathfrak{f}B_\mathfrak{p} = \mathfrak{f}A_\mathfrak{p}$, $x \in A_\mathfrak{p}$.Since $g(x) \in (A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$, $x \in (A_\mathfrak{p})^*$.Hence Ker$(h) = (A_\mathfrak{p})^*$. QED
(6)Let $A, K, B, \mathfrak{f}$ be as in the title question.There exists the following exact sequence of abelian groups.
$0 \rightarrow B^*/A^* \rightarrow (B/\mathfrak{f})^*/(A/\mathfrak{f})^* \rightarrow I(A)/P(A) \rightarrow I(B)/P(B) \rightarrow 0$
Proof:By this, there exists the following exact sequence of abelian groups.
$0 \rightarrow B^*/A^* \rightarrow \bigoplus_{\mathfrak{p}} (B_{\mathfrak{p}})^*/(A_{\mathfrak{p}})^* \rightarrow I(A)/P(A) \rightarrow I(B)/P(B) \rightarrow 0$
Here, $\mathfrak{p}$ runs over all the maximal ideals of $A$.
If $\mathfrak{p}$ is a regular prime ideal of $A$, $A_\mathfrak{p}$ is integrally closed.Hence $B_\mathfrak{p} = A_\mathfrak{p}$.Hence, in $\bigoplus_{\mathfrak{p}} (B_{\mathfrak{p}})^*/(A_{\mathfrak{p}})^*$,it suffices to consider only $\mathfrak{p}$ such that $\mathfrak{f} \subset \mathfrak{p}$.
By
Lemma 3, $(A/\mathfrak{f})^*$ is canonically isomorphic to $\bigoplus_\mathfrak{p} (A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$ as an abelian group, where $\mathfrak{p}$ runs over all the maximal ideals of $A$ such that $\mathfrak{f} \subset \mathfrak{p}$.
By
Lemma 4,$(B/\mathfrak{f})^*$ is canonically isomorphic to $\bigoplus_{\mathfrak{p}} (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$ as an abelian group, where $\mathfrak{p}$ runs over all the maximal ideal of $A$ such that $\mathfrak{f} \subset \mathfrak{p}$.
Now, by
Lemma 6, we are done. QED
Lemma 7Let $A, K, B, \mathfrak{f}$ be as in the title question.Let $\phi: I(A)/P(A) \rightarrow I(B)/P(B)$ be the canonical homomorphism.Let $C \in$ Ker($\phi$).Then $C$ contains an ideal of the form $A \cap \beta B$, where $\beta$ is an element of $B$ such that $\beta B + \mathfrak{f} = B$.
This follows immediately from
(6).
(5)Let $I$ be an invertible ideal of $A$.Since $B$ is a Dedekind domain, by this, there exist an ideal $\mathfrak{J}$ of $B$ and $\gamma \in K$ such that $\mathfrak{J} + \mathfrak{f} = B$and $IB = \mathfrak{J}\gamma$.Let $J = A \cap \mathfrak{J}$.By (2), $J$ is regular and $JB = \mathfrak{J}$.By (4), $J$ is invertible.Since $IB = \mathfrak{J}\gamma = J\gamma B$, $IJ^{-1}B = \gamma B$.By Lemma 7, there exists $\beta \in B$ such that $\beta B + \mathfrak{f} = B$ and $IJ^{-1} \equiv A \cap \beta B$ mod($P(A)$).Hence $I \equiv J(A \cap \beta B)$ mod($P(A)$).Since $J$ and $A \cap \beta B$ are regular, we are done. |
TL;DR: there is no mathematical certainty that every output value of common cryptographic hash functions is reachable, but for most that's overwhelmingly likely. A notable exception is double-SHA-256 (SHA256d) used in Bitcoin mining, where overwhelmingly likely there are some unreachable outputs.
For an idealized 256-bit hash, it becomes likely that every possible hash value $y$ is reached by hashing some $x$ when there are about $2^{264}$ possible values of $x$, including all 33-byte $x$; then odds that some $y$ is not covered become vanishingly small: less than one in $2^{b-264}$ if there are $2^b$ possible values of $x$; so while there is never a guarantee, it is practically certain that all $y$ are covered when we allow say 40-byte $x$: odds of the contrary are less than $2^{-56}$.
Argument: a hash can be idealized as a random function. The problem of if a random function covers all its image set is the coupon collector's problem. If the destination set has $n$ elements, the expected number of elements in the source set to fully cover the image set is $E(n)=n\log n+\gamma\,n+{1\over2}+O({1\over n})$ where $\gamma\approx0.577$ is the Euler–Mascheroni constant; and odds of not covering the input set after hashing $E(n)/\epsilon$ elements are less than $\epsilon$. Changing $n$ to $2^b$ for a $b$-bit hash, we get $E(2^b)\approx2^b(b\log(2)+\gamma)$ and $\begin{align}\log_2(E(2^b))&\approx b+\log_2\left(b+{\gamma\over\log(2)}\right)+\log_2(\log(2))\\&\approx b+\log_2(b+0.833)-0.528766\end{align}$
SHA-256 is a Merkle-Damgård hash using a compression function built per the Davies-Meyer construction. The problem of if all $y=\operatorname{SHA-256}(x)$ are reached reduces, up to 55 bytes hashed, to if all $\operatorname{Enc}_x(IV)$ are reached for a certain constant $IV$ and a certain block cipher $\operatorname{Enc}$ where $x$ is the key, and key scheduling encompasses the block padding. For an idealized block cipher the same analysis as for an idealized hash would hold, so we'd cover all values with overwhelming odds. SHA-256 uses an ARX block cipher and I see no reason why it would significantly differ from an idealized one, but that's a weak argument.
We can be more positive that we get full coverage when we allow two rounds (say, 100-byte hashes), because the problem is now if all $\operatorname{Enc'}_{x_1}(y_0)\boxplus y_0$ are reached where $x_1$ is the 36-byte second block, and $y_0$ is allowed to vary among the nearly-full set of 32-byte values reachable with the first round ( $\boxplus$ is 256-bit addition without carry across 32-bit boundaries). Still, there is no mathematical guarantee.
In Bitcoin mining, what's computed is $y=\operatorname{SHA-256}(\operatorname{SHA-256}(x))$. We are, for the outer hash, much below the coupon collector's bound and overwhelmingly above the birthday bound, therefore it is overwhelmingly likely that there are some values of $y$ not reached.
I second Dave Thomson's comment below. |
I want to find the distance between two points in spherical coordinates, so I want to express $||x-x'||$ where $x=(r,\theta, \phi)$ and $x' = (r', \theta',\phi')$ by the respective components. Is this possible? I just know that this is $\sqrt{r^2+r'^2-2rr'\cos(\theta- \theta')}$ if $\phi,\phi'$ is the same, but what is the most general distance?
The expression of the distance between two vectors in spherical coordinates provided in the other response is usually expressed in a more compact form that is not only easier to remember but is also ideal for capitalizing on certain symmetries when solving problems.
$$\begin{align} \|\mathbf{r}-\mathbf{r}^\prime\| &=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}\\ &=\sqrt{r^2+r'^2-2rr'\left[\color{red}{\sin(\theta)\sin(\theta')}\color{blue}{\cos(\phi)\cos(\phi')}+\color{red}{\sin(\theta)\sin(\theta')}\color{blue}{\sin(\phi)\sin(\phi')}+\cos(\theta)\cos(\theta')\right]}\\ &=\sqrt{r^2+r'^2-2rr'\left[\color{red}{\sin(\theta)\sin(\theta')}\color{blue}{\left(\cos(\phi)\cos(\phi')+\sin(\phi)\sin(\phi')\right)}+\cos(\theta)\cos(\theta')\right]}\\ &=\sqrt{r^2+r'^2-2rr'\left[\color{red}{\sin(\theta)\sin(\theta')}\color{blue}{\cos(\phi-\phi')}+\cos(\theta)\cos(\theta')\right]}.\\ \end{align}$$
This form makes it fairly transparent how azimuthal symmetry allows you to automatically eliminate some of the angular dependencies in certain integration problems. Another advantage of this form is that you now have at least two variables, namely $\phi$ and $\phi'$, that appear in the equation only once, which can make finding series expansions w.r.t. these variables a little less of a pain than the others.
You have simply to write it in Cartesian coordinates and change variables: $x=r\sin(\theta)\cos(\phi)$, $y=r\sin(\theta)\sin(\phi)$, $z=r\cos(\theta)$ $$\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}=$$$$=\sqrt{r^2+r'^2-2rr'\left[\sin(\theta)\sin(\theta')\cos(\phi)\cos(\phi')+\sin(\theta)\sin(\theta')\sin(\phi)\sin(\phi')+\cos(\theta)\cos(\theta')\right]}$$ But I don't see a way to really improve this mess.
Building on the answer from @[David H], I wrote the distance in a way that highlights the difference in angles: $$ ||\vec r_1 - \vec r_2|| = \sqrt{ {r_1}^2 + {r_2}^2 - 2\, r_1 r_2 \cos(\theta_1 - \theta_2) - 2\, r_1 r_2 \sin \theta_1 \sin \theta_2 \left( \cos(\phi_1 - \phi_2) - 1 \right) } $$ It highlights the contributions from the difference in polar angle $\theta$ and the difference in the azimuthal angle $\phi$, (third and fourth terms, respectively, under the square root symbol). Note the intuitive scaling of the azimuthal contribution by $\sin \theta_1 \sin \theta_2$. Of course, when the angular differences are both $0$, the distance reduces to $||r_1-r_2||$. |
I want to integrate a polynomial expression over a 4-node element in 3D. Several books on FEA cover the case where integrating is performed over an arbitrary flat 4-noned element. The usual procedure in this case is to find Jacobi matrix and use it's determinant to change the integration basis to the normalized one in which I have the simpler integration limits [-1;1] and the Gauss-Legendre quadrature technique is used easily.
In other words $\displaystyle\int_S f(x,y)\ \mathrm{d}x\,\mathrm{d}y\,$ is reduced to the form of $\displaystyle\int^{-1}_{1}\int^{-1}_{1} \tilde{f}(e,n)\ \left|\det(J)\right|\,\mathrm{d}e\,\mathrm{d}n$
But in 2D case I change the flat arbitrary element to the flat one but well-shaped square 2 by 2.
3D 4-noded element isn't flat in general but I suppose it still can be mapped with 2D coordinate system which is somehow related to cartesian coordinate system. I can't figure out how to express {x,y,z} in terms of {e,n} and what would be the size of the Jacobi matrix in this case (it's supposed to be square). |
M 3: a new muon missing momentum experiment to probe (g – 2) μ and dark matter at Fermilab Abstract
Here, new light, weakly-coupled particles are commonly invoked to address the persistent $$\sim 4\sigma$$ anomaly in $$(g-2)_\mu$$ and serve as mediators between dark and visible matter. If such particles couple predominantly to heavier generations and decay invisibly, much of their best-motivated parameter space is inaccessible with existing experimental techniques. In this paper, we present a new fixed-target, missing-momentum search strategy to probe invisibly decaying particles that couple preferentially to muons. In our setup, a relativistic muon beam impinges on a thick active target. The signal consists of events in which a muon loses a large fraction of its incident momentum inside the target without initiating any detectable electromagnetic or hadronic activity in downstream veto systems. We propose a two-phase experiment, M$^3$ (Muon Missing Momentum), based at Fermilab. Phase 1 with $$\sim 10^{10}$$ muons on target can test the remaining parameter space for which light invisibly-decaying particles can resolve the $$(g-2)_\mu$$ anomaly, while Phase 2 with $$\sim 10^{13}$$ muons on target can test much of the predictive parameter space over which sub-GeV dark matter achieves freeze-out via muon-philic forces, including gauged $$U(1)_{L_\mu - L_\tau}$$.
Authors: Princeton Univ., Princeton, NJ (United States) Fermi National Accelerator Lab. (FNAL), Batavia, IL (United States) Publication Date: Research Org.: Fermi National Accelerator Lab. (FNAL), Batavia, IL (United States) Sponsoring Org.: USDOE Office of Science (SC), High Energy Physics (HEP) (SC-25) OSTI Identifier: 1439466 Report Number(s): arXiv:1804.03144; FERMILAB-PUB-18-087-A Journal ID: ISSN 1029-8479; 1667037; TRN: US1900618 Grant/Contract Number: AC02-07CH11359 Resource Type: Journal Article: Accepted Manuscript Journal Name: Journal of High Energy Physics (Online) Additional Journal Information: Journal Volume: 2018; Journal Issue: 9; Journal ID: ISSN 1029-8479 Publisher: Springer Berlin Country of Publication: United States Language: English Subject: 79 ASTRONOMY AND ASTROPHYSICS; 46 INSTRUMENTATION RELATED TO NUCLEAR SCIENCE AND TECHNOLOGY; 72 PHYSICS OF ELEMENTARY PARTICLES AND FIELDS; Fixed target experiments Citation Formats
Kahn, Yonatan, Krnjaic, Gordan, Tran, Nhan, and Whitbeck, Andrew.
M3: a new muon missing momentum experiment to probe (g – 2)μ and dark matter at Fermilab. United States: N. p., 2018. Web. doi:10.1007/JHEP09(2018)153.
Kahn, Yonatan, Krnjaic, Gordan, Tran, Nhan, & Whitbeck, Andrew.
M3: a new muon missing momentum experiment to probe (g – 2)μ and dark matter at Fermilab. United States. doi:10.1007/JHEP09(2018)153.
Kahn, Yonatan, Krnjaic, Gordan, Tran, Nhan, and Whitbeck, Andrew. Wed . "M3: a new muon missing momentum experiment to probe (g – 2)μ and dark matter at Fermilab". United States. doi:10.1007/JHEP09(2018)153. https://www.osti.gov/servlets/purl/1439466.
@article{osti_1439466,
title = {M3: a new muon missing momentum experiment to probe (g – 2)μ and dark matter at Fermilab}, author = {Kahn, Yonatan and Krnjaic, Gordan and Tran, Nhan and Whitbeck, Andrew}, abstractNote = {Here, new light, weakly-coupled particles are commonly invoked to address the persistent $\sim 4\sigma$ anomaly in $(g-2)_\mu$ and serve as mediators between dark and visible matter. If such particles couple predominantly to heavier generations and decay invisibly, much of their best-motivated parameter space is inaccessible with existing experimental techniques. In this paper, we present a new fixed-target, missing-momentum search strategy to probe invisibly decaying particles that couple preferentially to muons. In our setup, a relativistic muon beam impinges on a thick active target. The signal consists of events in which a muon loses a large fraction of its incident momentum inside the target without initiating any detectable electromagnetic or hadronic activity in downstream veto systems. We propose a two-phase experiment, M$^3$ (Muon Missing Momentum), based at Fermilab. Phase 1 with $\sim 10^{10}$ muons on target can test the remaining parameter space for which light invisibly-decaying particles can resolve the $(g-2)_\mu$ anomaly, while Phase 2 with $\sim 10^{13}$ muons on target can test much of the predictive parameter space over which sub-GeV dark matter achieves freeze-out via muon-philic forces, including gauged $U(1)_{L_\mu - L_\tau}$.}, doi = {10.1007/JHEP09(2018)153}, journal = {Journal of High Energy Physics (Online)}, issn = {1029-8479}, number = 9, volume = 2018, place = {United States}, year = {2018}, month = {9} } Citation information provided by Web of Science
Web of Science
Figures / Tables: left) and vector ( right) forces that couple predominantly to muons. In both cases, a relativistic muon beam is incident on a fixed target and scatters coherently off a nucleus to produce the new particle as initial- ormore » |
tan(θ) calculates the tangent of the angle θ, which is defined by $\tan \theta=\frac{\sin \theta}{\cos \theta}$
The value returned depends on whether the calculator is in Radian or Degree mode. A full rotation around a circle is 2π radians, which is equal to 360°. The conversion from radians to degrees is angle*180/π and from degrees to radians is angle*π/180. The tan( command also works on a list of real numbers.
Since tangent is defined as the quotient of sine divided by cosine, it is undefined for any angle such that cos(θ)=0.
In radians:
tan(π/4) 1
In degrees:
tan(45) 1
Advanced Uses
tan(45°) 1
tan(π/4¹ ) 1
Error Conditions ERR:DATA TYPEis thrown if you supply a matrix or a complex argument. ERR:DOMAINis thrown if you supply an angle of π/2±nπ (in radians, where n is an integer) or 90±180n (in degrees, where n is an integer), or when the input is ≥1E12. Related Commands
. |
Theorem 11.4For any constant $\delta > 0$, every problem in NP has probabilistically checkable proofs of length poly($n$), where the verifier flips $O(\log n)$ coins and looks at three bits of the proof, with completeness $1-\delta$ and soundness $1/2+\delta$.
Moreover, the conditions under which Arthur accepts are extremely simple - namely, if the parity of these three bits is even.
This theorem has powerful consequences for inapproximability. If we write down all the triplets of bits that Arthur could look at, we get a system of linear equations mod 2. If Merlin takes his best shot at a proof, the probability that Arthur will accept is the largest fraction of these equations that can be simultaneously satisfied. Thus even if we know that this fraction is either at least $1-\delta$ or at most $1/2+\delta$, it is NP-hard to tell which. This implies that it is NP-hard to approximate Max-XORSAT within any constant factor better than $1/2$.
First, I don't think it can be as simple as just checking that the parity is even. Merlin could just make Arthur accept anything by handing him a proof of all zeros, and make him reject anything with a proof of all ones. Surely the parity must instead equal some value that Arthur computes along the way.
Further, It's not a priori clear to me that Arthur would select every triplet with equal likelihood. If that wasn't the case, wouldn't it be possible for a no-instance to yield a Max-XORSAT formula where more than $1/2+\delta$ of the clauses can be satisfied even though Arthur accepts with probability less than $1/2+\delta$?
Let's take a degenerate case as an example. Arthur asks Merlin to help him decide whether a given sequence of $n$ bits $\mathbf{b}$ is all ones - $\mathbf{b=1}$ for short. This problem is trivially in NP. And Arthur clearly doesn't need Merlin's help. But he can go out of his way to use it anyway. Here's what they do:
First Merlin gives Arthur a PCP $\mathbf{x} \in \{0,1\}^n$ (let's assume $n \gg 3$). Then Arthur checks if $\mathbf{b=1}.$ If it is, he checks that $x_1 \oplus x_2 \oplus x_3 = 1$ holds in Merlin's PCP. Otherwise, he checks either $x_1 \oplus x_2 \oplus x_3 = 1$ or $\overline{x_1} \oplus \overline{x_2} \oplus \overline{x_3} = 1$ (note that these can't both be satisfied) with equal probability 0.45. Or with probability 0.1 he checks $x_i \oplus x_j \oplus x_k = 1$ for a uniformly random triplet. Merlin can thus convince Arthur that $\mathbf{b=1}$ by giving him, for example, the proof consisting of all ones. But if $\mathbf{b\neq1}$ there's no proof that Arthur accepts with probability greater than 0.55. This gives PCP's with completeness $1-\delta$ (indeed we also have completeness 1) and soundness $1/2+\delta$ where $\delta = 0.05$. Now what's the Max-XORSAT formula corresponding to $\mathbf{b\neq1}$? Well, Arthur could look at any triplet $x_i, x_j, x_k$ as well as $\overline{x_1},\overline{x_2},\overline{x_3}$. So presumably it's the conjunction of all those triplets when XORed together. All but one of these clauses can be satisfied. So the fraction is very close to one. But that's nowhere near the probability that Arthur will actually accept.
PS: We need a PCP tag. |
The Debye model is a method developed by Peter Debye in 1912\(^{[7]}\) for estimating the phonon contribution to the specific heat (heat capacity) in a solid\(^{[1]}\). This model correctly explains the low temperature dependence of the heat capacity, which is proportional to \(T^3\) and also recovers the Dulong-Petit law at high temperatures. However, due to simplifying assumptions, its accuracy suffers at intermediate temperatures.
Introduction to Phonons
In 1912 Debye realized that, inconsistent with the Einstein model, low-energetic excitations of a solid material were not oscillations of a single atom\(^{[2]}\), but collective modes propagating through the material. Such vibrations can be considered to be sound waves, and their propagation speed is the speed of sound in the material\(^{[3]}\). Moreover, these modes only accept energy in discreet amounts.
Quantum theory uses the concepts of phonons, which are “quasi-particles” with definite energies and directions of motion, to treat the vibrations. The concept of phonon is analogous with photons of the electromagnetic wave. The relations between the energy of a phonon \(\varepsilon\), the angular frequency \(\omega\) and the wave vector \(\vec{q}\) are:
\[ \begin{align} \varepsilon=\hbar\omega \end{align} \label{1}\]
\[ \begin{align} \omega=\upsilon_{s}|\vec{q}| \end{align} \label{2}\]
where \(v_s\) is the velocity of the sound wave.
As a kind of Bosons, phonons obey Bose–Einstein statistics. The expectation number of bosons in a state with energy E is\(^{[6]}\):
\[ \begin{align} n_{(E)}=\dfrac{1}{e^{E/k_{B}T}-1}=\dfrac{1}{e^{\hbar\omega/k_{B}T}-1} \end{align} \label{3}\]
where \(k_B\)= 1.380 6504(24)×10
−23J/K is the Boltzmann constant. Debye frequency and Debye Temperature
Unlike electromagnetic radiation in a box, a phonon cannot have infinite frequency. Its frequency is bound by the medium of its propagation — the atomic lattice of the solid. If there are N primitive cells in the specimen, the total number of phonon modes are N. A cut-off frequency \({\omega}_D\), known as Debye frequency, is determined by the following manner\(^{[4]}\):
In the 3 dimensional reciprocal space, the volume for each allowed wave vector \(\vec{q}\) is:
\[ \begin{align} \left(\dfrac{2\pi}{L}\right)^{3}=\dfrac{8\pi^{3}}{V} \end{align} \label{4}\]
Since there is a cut-off wave vector \(q_{D}={\omega}_{D}/{\upsilon_{s}}\), all the modes are confined within a sphere with radius \(q_D\). Thus number of modes (not number of phonons) should be (5)
\[ \begin{align} N=\left(\dfrac{4}{3}\pi{q_{D}}^{3})/({\dfrac{8\pi^{3}}{V}}\right) \end{align} \label{5}\]
or
\[ \begin{align} q_{D}=\left(6\pi^{2}\dfrac{N}{V}\right)^{\frac{1}{3}} \end{align} \label{6}\]
\[ \begin{align} \omega_{D}=\upsilon_{s}\left(6\pi^{2}\dfrac{N}{V}\right)^{\frac{1}{3}} \end{align} \label{7}\]
Debye temperature \(T_{D}\) is defined as
\[\begin{align} T_{D}=\dfrac{\hbar\omega_{D}}{k_{B}}=\dfrac{\hbar\upsilon_{s}}{k_{B}}(6\pi^{2}\dfrac{N}{V})^{\frac{1}{3}} \end{align} \label{8}\]
The significance of this physical term will be discussed below. For elements in the same group, heavier atoms have lower Debye temperatures, simply because the velocity of sound decreases as the density increases.\(^{[4]}\)The Debye temperatures of several substances are listed in Table 1.\(^{[1]}\)
Aluminum 428K Iron 470K Silicon 645K Tungsten 400K Cadmium 209K Lead 105K Silver 225K Zinc 327K Chromium 630K Manganese 410K Tantalum 240K Carbon 2230K Copper 343.5K Nickel 450K Tin(white) 200K Ice 192K Gold 165K Platinum 240K Titanium 420K
Note
Heavier atoms have
lower Debye temperatures, because the velocity of sound decreases as the density increases. Derivation for Specific Heat
In the Debye approximation, the velocity of sound \(\upsilon_{s}\) is taken as constant for each polarization type, as it would be for a classical elastic continuum. According to equation (7), the density of states is:
\[ \begin{align} D_{(\omega)} =\dfrac{dN}{d\omega}=\dfrac{V\omega^{2}}{2\pi^{2}\upsilon_{s}^{3}} \end{align} \label{9}\]
Thus, thermal energy for each polarization type is given by\(^{[4]}\):
\[ \begin{align} U=\int d\omega D(\omega) n(\omega)\hbar\omega=\int_{0}^{\omega_{D}}d\omega\dfrac{V\omega^{2}}{2\pi^{2}\upsilon_{s}^{3}}\dfrac{\hbar\omega}{e^{\hbar\omega/k_{B}T}-1} \end{align} \label{10}\]
\[ \begin{align} U=\dfrac{3V\hbar}{2\pi^{2}\upsilon_{s}^{3}}\int_{0}^{\omega_{D}}d\omega\dfrac{\omega^{3}}{e^{\hbar\omega/k_{B}T}-1}=\dfrac{3Vk_{B}^{4}T^{4}}{2\pi^{2}\upsilon_{s}^{3}\hbar^{3}}\int_{0}^{x_{D}}dx\dfrac{x^{3}}{e^{x}-1}=9Nk_{B}T(\dfrac{T}{T_{D}})^{3}\int_{0}^{x_{D}}dx\dfrac{x^{3}}{e^{x}-1} \end{align} \label{11}\]
where \(x=\hbar\omega/k_{B}T\) and \(x_{D}\equiv T_{D}/T\).
The heat capacity is:
\[ \begin{align} C_{V}=\dfrac{\partial U}{\partial T}=9Nk_{B}(\dfrac{T}{T_{D}})^{3}\int_{0}^{x_{D}}dx\dfrac{x^{4}e^{x}}{(e^{x}-1)^{2}} \end{align} \label{12}\]
At the left of
Figure \(\PageIndex{1}\)\(^{[3]}\) below, the experimental results of specific heats of four substances are plotted as a function of temperature and they look very different. But if they are scaled to \(T/T_{D}\), they look very similar and are very close to the Debye theory. Figure \(\PageIndex{1}\) Specific Heats of Lead, Silver, Aluminum and Diamond High and Low Temperature Limits
The integral in Equation (12) cannot be evaluated in closed form. But the high and low temperature limits can be assessed.
High Temperature Limit
For the high temperature case where \(T\gg T_{D}\), the value of \(x\) is very small throughout the range of the integral. This justifies using the approximation to the exponential \(e^x \approx 1 + x\) and reduces equation (11) and (12) to
\[\begin{align} U=9Nk_{B}T \left(\dfrac{T}{T_{D}} \right)^{3}\int_{0}^{x_{D}}x^{2}dx=3Nk_{B}T \end{align} \label{14}\]
\[ \begin{equation} C_{V}=3Nk_{B} \end{equation} \label{13}\]
which is the classical
Dulong-Petit result.
When the temperature is above the Debye temperature, the heat capacity is very close to the classical value \(3Nk_{B} T\). For temperatures below the Debye temperature, quantum effects become important and \(C_v\) decreases to zero. Note that diamond, with a Debye temperature of 1860K, is a “quantum solid” at room temperature. \(^{[8]}\)
Low Temperature Limit
At very low temperature where \(T\ll T_{D}\), only long wavelength acoustic modes are thermally excited. These are just the modes that can be treated as elastic continuum with macroscopic elastic constants. The energy of those short wavelength modes are too high to be populated significantly at low temperatures. We may approximate \(x_D\equiv T_{D}/T\) to infinity and make use of the standard integral
\[ \begin{align} \int_{0}^{\infty}dx\dfrac{x^{3}}{e^{x}-1}=\dfrac{\pi^{4}}{15} \end{align} \label{15}\]
to obtain
\[ \begin{align} U=\dfrac{3\pi^{4}Nk_{B}T^{4}}{5T_{D}^{3}} \end{align} \label{16}\]
\[ \begin{align} C_{V}=\dfrac{12\pi^{4}Nk_{B}T^{3}}{5T_{D}^{3}}\cong324Nk_{B}\dfrac{T^{3}}{T_{D}^{3}} \end{align} \label{17}\]
Extension: Einstein-Debye Specific Heat
This \(T\) dependence of the specific heat at very low temperatures agrees with experiment for nonmetals. For metals the specific heat of highly mobile conduction electrons is approximated by Einstein Model, which is composed of single-frequency quantum harmonic oscillators. The temperature dependence of Einstein model is just T. It becomes significant at low temperatures and is combined with the above lattice specific heat in the Einstein-Debye specific heat\(^{[3]}\).
\[ \begin{align} C_{metal}=C_{electron}+C_{phonon}=\dfrac{\pi^{2}Nk^{2}}{2E_{f}}T+\dfrac{12\pi^{4}Nk_{B}}{5T_{D}^{3}}T^{3} \end{align} \label{18}\]
Finally, experiments suggest that amorphous materials do not follow the Debye \(T^3\) law even at temperatures below 0.01\(T_{D}\)\(^{[8]}\). There is more yet to be learned.
References http://en.wikipedia.org/wiki/Debye_model http://www.uam.es/personal_pdi/cienc...ivos/debye.pdf http://hyperphysics.phy-astr.gsu.edu...ds/phonon.html C. Kittel, introduction to solid state physics, pp.117-140. John Wiley & Sons, Inc., 1996 L. Landau and E. Lifshitz, Statistical Physics, Part, pp.191 -217. Elsevier, 1984 http://en.wikipedia.org/wiki/Bose%E2...ein_statistics Zur Theorie der spezifischen Waerm, Annalen der Physik (Leipzig) 39(4) http://www.phys.unsw.edu.au/~gary/SM3_6.pdf Contributors
ContribMSE5317 |
I am using
\boxed to place emphasis on important equations in an assignment I am typing up, and would like to give the final answer a bit of extra emphasis. I'm using
\begin{equation} ... \end{equation} to number each line, so that I can put in references in the text.
There are some very nice looking boxes in the answers to Attractive Boxed Equations and framed equations, but both rely on colour, when I have a black and white printer. I was thinking of a double line around the final equation, but I'll take other answers if they are easier to code or look better.
I would love it if either the balanced boxes or the full width boxes were incorporated into the answer, but this is optional. It is a class assignment that the TA will look at for all of 10 minutes, not a publication.
Minimal Working Example:
\documentclass[letterpaper]{article}\usepackage{kpfonts}\usepackage{amssymb,mathtools}\begin{document}\begin{equation} \frac{M}{R} \propto R^2 \rho \end{equation}\begin{equation} \label{eqn:density} \boxed{ \rho \propto \frac{M}{R^3} } \end{equation}\end{document} |
Tagged: finite group If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later
Problem 455
Let $G$ be a finite group.
The centralizer of an element $a$ of $G$ is defined to be \[C_G(a)=\{g\in G \mid ga=ag\}.\]
A
conjugacy class is a set of the form \[\Cl(a)=\{bab^{-1} \mid b\in G\}\] for some $a\in G$. (a)Prove that the centralizer of an element of $a$ in $G$ is a subgroup of the group $G$.
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(b) Prove that the order (the number of elements) of every conjugacy class in $G$ divides the order of the group $G$. Problem 420
In this post, we study the
Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
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Problem. Let $G$ be a finite abelian group of order $n$. If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic to the cyclic group $Z_n=\Zmod{n}$ of order $n$. Problem 302
Let $R$ be a commutative ring with $1$ and let $G$ be a finite group with identity element $e$. Let $RG$ be the group ring. Then the map $\epsilon: RG \to R$ defined by
\[\epsilon(\sum_{i=1}^na_i g_i)=\sum_{i=1}^na_i,\] where $a_i\in R$ and $G=\{g_i\}_{i=1}^n$, is a ring homomorphism, called the augmentation map and the kernel of $\epsilon$ is called the augmentation ideal. (a) Prove that the augmentation ideal in the group ring $RG$ is generated by $\{g-e \mid g\in G\}$.
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(b) Prove that if $G=\langle g\rangle$ is a finite cyclic group generated by $g$, then the augmentation ideal is generated by $g-e$. Read solution |
Tagged: subspace Problem 709
Let $S=\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}\}$ where
\[ \mathbf{v}_{1}= \begin{bmatrix} 1 \\ 2 \\ 2 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{2}= \begin{bmatrix} 1 \\ 3 \\ 1 \\ 1 \end{bmatrix} ,\;\mathbf{v}_{3}= \begin{bmatrix} 1 \\ 5 \\ -1 \\ 5 \end{bmatrix} ,\;\mathbf{v}_{4}= \begin{bmatrix} 1 \\ 1 \\ 4 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{5}= \begin{bmatrix} 2 \\ 7 \\ 0 \\ 2 \end{bmatrix} .\] Find a basis for the span $\Span(S)$. Problem 706
Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set
\[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\] still a spanning set for $V$? If so, prove it. Otherwise, give a counterexample. Problem 663
Let $\R^2$ be the $x$-$y$-plane. Then $\R^2$ is a vector space. A line $\ell \subset \mathbb{R}^2$ with slope $m$ and $y$-intercept $b$ is defined by
\[ \ell = \{ (x, y) \in \mathbb{R}^2 \mid y = mx + b \} .\]
Prove that $\ell$ is a subspace of $\mathbb{R}^2$ if and only if $b = 0$.Add to solve later
Problem 659
Fix the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$, and let $\R^3$ be the vector space of $3 \times 1$ column vectors. Define
\[W = \{ \mathbf{v} \in \R^3 \mid \mathbf{b} \mathbf{v} = 0 \}.\] Prove that $W$ is a vector subspace of $\R^3$. Problem 658
Let $V$ be the vector space of $n \times n$ matrices with real coefficients, and define
\[ W = \{ \mathbf{v} \in V \mid \mathbf{v} \mathbf{w} = \mathbf{w} \mathbf{v} \mbox{ for all } \mathbf{w} \in V \}.\] The set $W$ is called the center of $V$.
Prove that $W$ is a subspace of $V$.Add to solve later
Problem 612
Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$. (a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.
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(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$. Problem 611
An $n\times n$ matrix $A$ is called
orthogonal if $A^{\trans}A=I$. Let $V$ be the vector space of all real $2\times 2$ matrices.
Consider the subset
\[W:=\{A\in V \mid \text{$A$ is an orthogonal matrix}\}.\] Prove or disprove that $W$ is a subspace of $V$. Problem 604
Let
\[A=\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 &1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 2 & 2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) Find a basis for the null space $\calN(A)$. (b) Find a basis of the range $\calR(A)$. (c) Find a basis of the row space for $A$.
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(The Ohio State University, Linear Algebra Midterm) Read solution Problem 601
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.
Let \[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix} a & b\\ c& -a \end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\] (a) Show that $W$ is a subspace of $V$. (b) Find a basis of $W$. (c) Find the dimension of $W$.
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(The Ohio State University, Linear Algebra Midterm) Read solution |
I'm trying to get an approximate answer to the following integral:
$\lim_{n \to \infty} \frac{n+1}{2} \int_1^{\ln(n)}xe^{-x}(1-\exp(-x - W(-xe^{-x})))^2 dx $
I'm currently just interested in if it converges to anything, so I want to just plug in a few values of $n$ and see what happens. My R 'code' looks like this:
>W <- function(x){(1001/2)*x*exp(-x)*(1-exp(-x-LambertW(-x*exp(-x))))^2}
>integrate(W, lower=1, upper=log(1000))
Unfortunately, R tells me that I have a non-finite function value. It's my understanding that LambertW calculates the principal branch of the function, which does not do anything strange for values between $-\frac 1e$ and $0$, which is where I am. I tried shifting the lower bound slightly up to avoid the branching point at $-\frac 1e$, but I got the same error message then. |
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Higher harmonic flow coefficients of identified hadrons in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Springer, 2016-09)
The elliptic, triangular, quadrangular and pentagonal anisotropic flow coefficients for $\pi^{\pm}$, $\mathrm{K}^{\pm}$ and p+$\overline{\mathrm{p}}$ in Pb-Pb collisions at $\sqrt{s_\mathrm{{NN}}} = 2.76$ TeV were measured ... |
Richard Feynman referred to Euler's Identity, $e^{i\pi} + 1 = 0$ as a "jewel."
I'm trying to demonstrate this jewel without recourse to a Taylor series.
Given $z = cos\theta + i sin\theta\; |\;|z| = 1$,
$$\frac{dz}{d\theta}= -sin\theta + icos\theta =i(isin\theta+cos\theta)=iz$$
Now, if I let $z=u(\theta)$, then, $$\frac{du}{d\theta}=iu(\theta)$$
Undoing my original derivative, $$\int iu(\theta) d\theta =u(\theta)+C$$ $$ \therefore z=u(\theta)=e^{i\theta}$$ which is the general case. Substituting $\pi =\theta$ for the special case, and invoking the original equation, we are left with $z=cos\pi + isin\pi =-1 = e^{i\pi}$ $$\therefore e^{i\pi}+1=0$$
When I first worked through this, the constant of integration disturbed me, like a nasty inclusion marring the jewel. But now, I think it's fair for me to excise it in the line, $\;\therefore z=u(\theta)=e^{i\theta}$
Is that correct? |
The Annals of Statistics Ann. Statist. Volume 18, Number 1 (1990), 429-442. Distribution Functions of Means of a Dirichlet Process Abstract
Let $\chi$ be a random probability measure chosen by a Dirichlet process on $(\mathbb{R}, \mathscr{B})$ with parameter $\alpha$ and such that $\int x\chi(dx)$ turns out to be a (finite) random variable. The main concern of this paper is the statement of a suitable expression for the distribution function of that random variable. Such an expression is deduced through an extension of a procedure based on the use of generalized Stieltjes transforms, originally proposed by the present authors in 1978.
Article information Source Ann. Statist., Volume 18, Number 1 (1990), 429-442. Dates First available in Project Euclid: 12 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aos/1176347509 Digital Object Identifier doi:10.1214/aos/1176347509 Mathematical Reviews number (MathSciNet) MR1041402 Zentralblatt MATH identifier 0706.62012 JSTOR links.jstor.org Subjects Primary: 62G99: None of the above, but in this section Secondary: 62E15: Exact distribution theory 60K99: None of the above, but in this section 44A15: Special transforms (Legendre, Hilbert, etc.) Citation
Cifarelli, Donato Michele; Regazzini, Eugenio. Distribution Functions of Means of a Dirichlet Process. Ann. Statist. 18 (1990), no. 1, 429--442. doi:10.1214/aos/1176347509. https://projecteuclid.org/euclid.aos/1176347509
Corrections See Correction: Donato Michele Cifarelli, Eugenio Regazzini. Correction: Distribution Functions of Means of a Dirichlet Process. Ann. Statist., Volume 22, Number 3 (1994), 1633--1634. |
In proximity theory, we use
\delta and
\ll as binary relations on the power set of a set X. How can one display NOT
\delta or NOT
\ll similar to how we can do
\not\in. Using
\not\delta and
\not\ll place the strike-through in bad positions over the
\delta and
\ll.
For NOT
\delta, I am currently using
\!\!\not\!\delta
For NOT
\ll, I am using
\hskip 0.4mm \not \hskip -0.4mm \ll
There are a few problems here. Sometimes the placement of the "not" over the "delta" is a little off but it varies throughout the pages. Also, at the end of the line, I may get the
\not appearing at the end of the current line and the
\delta appearing at the beginning of the next line. I have not noticed this issue with my defininition of NOT
\ll.
What is the best way to define these symbols? How can one make sure that the "not" is on the same line as the "delta"? |
Problem 676
Let $V$ be the vector space of $2 \times 2$ matrices with real entries, and $\mathrm{P}_3$ the vector space of real polynomials of degree 3 or less. Define the linear transformation $T : V \rightarrow \mathrm{P}_3$ by
\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = 2a + (b-d)x – (a+c)x^2 + (a+b-c-d)x^3.\]
Find the rank and nullity of $T$.Add to solve later
Problem 675
The space $C^{\infty} (\mathbb{R})$ is the vector space of real functions which are infinitely differentiable. Let $T : C^{\infty} (\mathbb{R}) \rightarrow \mathrm{P}_3$ be the map which takes $f \in C^{\infty}(\mathbb{R})$ to its third order Taylor polynomial, specifically defined by
\[ T(f)(x) = f(0) + f'(0) x + \frac{f^{\prime\prime}(0)}{2} x^2 + \frac{f^{\prime \prime \prime}(0)}{6} x^3.\] Here, $f’, f^{\prime\prime}$ and $f^{\prime \prime \prime}$ denote the first, second, and third derivatives of $f$, respectively.
Prove that $T$ is a linear transformation.Add to solve later
Problem 674
Let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis. Let $T : \mathrm{P}_4 \rightarrow \mathrm{P}_{4}$ be the map defined by, for $f \in \mathrm{P}_4$,
\[ T (f) (x) = f(x) – x – 1.\]
Determine if $T(x)$ is a linear transformation. If it is, find the matrix representation of $T$ relative to the standard basis of $\mathrm{P}_4$.Add to solve later
Problem 673
Let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the
standard basis.
Let $T : \mathrm{P}_3 \rightarrow \mathrm{P}_{5}$ be the map defined by, for $f \in \mathrm{P}_3$,
\[T (f) (x) = ( x^2 – 2) f(x).\]
Determine if $T(x)$ is a linear transformation. If it is, find the matrix representation of $T$ relative to the standard basis of $\mathrm{P}_3$ and $\mathrm{P}_{5}$.Add to solve later
Problem 672
For an integer $n > 0$, let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis.
Let $T : \mathrm{P}_n \rightarrow \mathrm{P}_{n+1}$ be the map defined by, for $f \in \mathrm{P}_n$,
\[T (f) (x) = x f(x).\]
Prove that $T$ is a linear transformation, and find its range and nullspace.Add to solve later
Problem 669 (a) Suppose that a $3\times 3$ system of linear equations is inconsistent. Is the coefficient matrix of the system nonsingular? (b) Suppose that a $3\times 3$ homogeneous system of linear equations has a solution $x_1=0, x_2=-3, x_3=5$. Is the coefficient matrix of the system nonsingular?
Add to solve later
(c) Let $A$ be a $4\times 4$ matrix and let \[\mathbf{v}=\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \text{ and } \mathbf{w}=\begin{bmatrix} 4 \\ 3 \\ 2 \\ 1 \end{bmatrix}.\] Suppose that we have $A\mathbf{v}=A\mathbf{w}$. Is the matrix $A$ nonsingular? Problem 668
Consider the system of differential equations
\begin{align*} \frac{\mathrm{d} x_1(t)}{\mathrm{d}t} & = 2 x_1(t) -x_2(t) -x_3(t)\\ \frac{\mathrm{d}x_2(t)}{\mathrm{d}t} & = -x_1(t)+2x_2(t) -x_3(t)\\ \frac{\mathrm{d}x_3(t)}{\mathrm{d}t} & = -x_1(t) -x_2(t) +2x_3(t) \end{align*} (a) Express the system in the matrix form. (b) Find the general solution of the system.
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(c) Find the solution of the system with the initial value $x_1=0, x_2=1, x_3=5$. Solve the Linear Dynamical System $\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =A\mathbf{x}$ by Diagonalization Problem 667 (a) Find all solutions of the linear dynamical system \[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}\mathbf{x},\] where $\mathbf{x}(t)=\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ is a function of the variable $t$.
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(b) Solve the linear dynamical system \[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}\mathbf{x}\] with the initial value $\mathbf{x}(0)=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$. Prove that $\{ 1 , 1 + x , (1 + x)^2 \}$ is a Basis for the Vector Space of Polynomials of Degree $2$ or Less Problem 665
Let $\mathbf{P}_2$ be the vector space of polynomials of degree $2$ or less.
(a) Prove that the set $\{ 1 , 1 + x , (1 + x)^2 \}$ is a basis for $\mathbf{P}_2$.
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(b) Write the polynomial $f(x) = 2 + 3x – x^2$ as a linear combination of the basis $\{ 1 , 1+x , (1+x)^2 \}$. Problem 663
Let $\R^2$ be the $x$-$y$-plane. Then $\R^2$ is a vector space. A line $\ell \subset \mathbb{R}^2$ with slope $m$ and $y$-intercept $b$ is defined by
\[ \ell = \{ (x, y) \in \mathbb{R}^2 \mid y = mx + b \} .\]
Prove that $\ell$ is a subspace of $\mathbb{R}^2$ if and only if $b = 0$.Add to solve later
Problem 659
Fix the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$, and let $\R^3$ be the vector space of $3 \times 1$ column vectors. Define
\[W = \{ \mathbf{v} \in \R^3 \mid \mathbf{b} \mathbf{v} = 0 \}.\] Prove that $W$ is a vector subspace of $\R^3$. Problem 658
Let $V$ be the vector space of $n \times n$ matrices with real coefficients, and define
\[ W = \{ \mathbf{v} \in V \mid \mathbf{v} \mathbf{w} = \mathbf{w} \mathbf{v} \mbox{ for all } \mathbf{w} \in V \}.\] The set $W$ is called the center of $V$.
Prove that $W$ is a subspace of $V$.Add to solve later |
I was reading Elad Hazan's book on Online Convex Optimization(http://ocobook.cs.princeton.edu/OCObook.pdf) and am facing difficulty understanding the proof given for the No regret algorithm for MAB (Pg 102-103). It would be great if someone can provide a clarification on this.
The No-regret (sub-linear) algorithm is given in Algorithm 17 (Pg 102), and proof for no regret is shown in lemma 6.1 . I will make the description as self-contained as possible, but a more detailed presentation can be found in pgs 102-103 in the book mentioned above.
Let $\mathcal{K}=\left[1, n \right]$ denote the set of experts(arms). Let $i_t \in \mathcal{K}$, denote the expert chosen by the algorithm in the $t^{th}$ round. Let $l_t(i_t)$, denote he loss function provided by the adversary in the $t^{th}$ round. Since we are dealing with the bandit setting, we do not know $l_t(i)$ for all $i \in \mathcal{K}\backslash i_t$. Further, the loss functions ($l_t$) are assumed to be bounded between 0 and 1. The way the algorithm works is, at each round we flip a coin, with bias $\delta$. If the outcome of the coin is heads, the algorithm chooses one of the actions i.e. $i_t$, and constructs a estimate of $l_t$ as follows: \begin{equation} \hat{l}_t = \begin{cases} \frac{n}{\delta} l_t(i_t), \text{ if } i = i_t\\ 0, \text{ otherwise} \end{cases} \end{equation}
If however, the outcome of the coin toss was tails, then it simply sets $\hat{l}_t = 0$.
It can be easily shown that by using the above scheme we have $E\left[ \hat{l}_t(i)\right] = l_t(i), \, \forall i \in \mathcal{K}$.
The regret as shown in the book is as follows. \begin{eqnarray} \label{Eq:1} E[regret_T] &=& E\left[\sum_{t=1}^{T} l_t(i_t) - \sum_{t=1}^{T}l_t(i^*)\right] \\ \label{Eq:2} & \leq& E\left[\sum_{t \not \in S_T} \hat{l}_t(i_t) - \sum_{t \not \in S_T} \hat{l}_t(i^*) + \sum_{t \in S_T} 1 \right] \end{eqnarray}
where $S_T \subseteq [1, T]$ denotes the round in which the coin toss was heads, and $i^* = \underset{i \in \mathcal{K}}{\text{arg min}} \sum_{t=1}^{T} l_t(i)$ . I am having a hard time showing why $E[regret_T] \leq E\left[\sum_{t \not \in S_T} \hat{l}(i_t) - \sum_{t \not \in S_t} \hat{l}_t(i^*) + \sum_{t \in S_T} 1 \right] $. The comment in the book is that $i^*$ is independent of $\hat{l}_t$, hence validity of inequality. I did not understand what that was supposed to mean.
My Attempt:
For my attempt, I will be using the some of the notation used in the proof of that book. We know that $\underset{i \in [1, \dotsc, n]}{\text{min}} \sum_{t=1}^T \hat{l}_t (i) \leq \sum_{t=1}^T \hat{l}_t (i), \, \forall i \in [1,n]$. Applying $E$ (expectation) on both sides we get, \begin{eqnarray} E\left[ \underset{i \in [1, \dotsc, n]}{\text{min}} \sum_{t=1}^T \hat{l}_t (i) \right] &\leq & E \left[\sum_{t=1}^T \hat{l}_t (i) \right], \, \forall i \in [1,n] \\ \implies E\left[ \underset{i \in [1, \dotsc, n]}{\text{min}} \sum_{t=1}^T \hat{l}_t (i) \right] &\leq & \underset{i \in [1, \dotsc, n]}{\text{min}} E \left[\sum_{t=1}^T \hat{l}_t (i) \right] \\ &=& \sum_{t=1}^T l_t (i^*) \end{eqnarray} , in the book it is easily shown that $E\left[\hat{l}(i)\right] = l(i)$. In view of the above inequality, we can show
\begin{eqnarray} E[regret_T] &=& E\left[\sum_{t=1}^{T} l(i_t) - \sum_{t=1}^{T}l_t(i^*)\right] \\ & \leq& E\left[ \sum_{t=1}^{T} \hat{l}(i_t) - \underset{i \in [1, \dotsc, n]}{\text{min}} \sum_{t=1}^T \hat{l}_t (i) \right] \end{eqnarray}
Clearly, I am making some mistakes in the attempt above. I will be grateful if someone can point me to those and help clarify the reasoning in the book. I say my attempt is incorrect because, in the way I have shown, I completely disregarded the set $S_T$. Without this set, the book shows the regret to be $\mathcal{O}(\sqrt{T})$, whereas with the $S_T$ set the regret is shown to be $\mathcal{O}(T^{\frac{3}{4}})$ |
In other words; Why can't I integrate the whole equation in one go like this?
$$\begin{align}f=\int df&=\int yz\,dx +\int xz\,dy+\int xy\,dz+\int a\,dz\\&=xyz+xyz+xyz+az+C\\&=3xyz + az +C\end{align}$$
This is strangely remarkably close to the correct answer, which is
$$f=xyz+az+C$$
I know that the differential $$df=yz\,dx+xz\,dy+(xy+a)\,dz\tag{a}$$ can be written as $$df=\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy+\frac{\partial f}{\partial z}\,dz\tag{b}$$
Matching equations $(\mathrm{a})$ and $(\mathrm{b})$ leads to $3$ more equations, namely: $$\frac{\partial f}{\partial x}=yz\tag{1}$$ $$\frac{\partial f}{\partial y}=xz\tag{2}$$ $$\frac{\partial f}{\partial z}=xy+a\tag{3}$$
Now integrating $(1)$, $(2)$, and $(3)$ with respect to their partial derivatives
$$f=xyz + P\quad\text{from} \quad(1)\tag{A}$$ $$f=xyz + Q\quad\text{from} \quad(2)\tag{B}$$ $$f=xyz + az+R\quad\text{from} \quad(3)\tag{C}$$ where $\mathrm{P}$, $\mathrm{Q}$ and $\mathrm{R}$ are constants of integration.
Now '
somehow' we decide that equation $(\mathrm{C})$ best describes the parent function $f$ and is therefore the function we desire:$$f=xyz + az+C$$with $R$ replacing $C$, since they are both constants.
So apart from the obvious "because it gives the correct answer", my question is as follows: Why do we have to integrate each term separately (independently) instead of the method I used at the beginning of this question (integrating the whole equation in one go)?
Also; What is the precise logic behind choosing $(\mathrm{C})$ to represent $f$ instead of $(\mathrm{A})$ or $(\mathrm{B})$?
Many thanks. |
Below is a visual proof (!) that $32.5 = 31.5$. How could that be?
(As noted in a comment and answer, this is known as the "Missing Square" puzzle.)
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It's an optical illusion - neither the first nor second set of blocks actually describes a triangle. The diagonal edge of the first is slightly concave and that of the second is slightly convex.
To see clearly, look at the gradients of the hypotenuses of the red and blue triangles - they're not 'similar'.
gradient of blue triangle hypotenuse = 2/5
gradient of red triangle hypotenuse = 3/8
Since these gradients are different, combining them in the ways shown in the diagram does not produce an overall straight (diagonal) line.
Overlay the two triangles to see the difference. This is known as the missing square puzzle.
Since this question has been bumped up, and some people find pictures helpful, may as well supply illustrations of the explanation.
The area of the red triangle is $\frac12(8)(3) = 12$ and that of the blue triangle is $\frac12(5)(2) = 5$. The area of the yellow and green regions are clearly 7 and 8 respectively, so the total coloured area is 12 + 5 + 7 + 8 = 32 which is less than 32.5, the area of a 13×5 triangle. Indeed, in the figure below you can see that the hypotenuses of the red and blue triangles "dip" below the actual hypotenuse of the large triangle; this accounts for the 0.5 difference in area.
Similarly, in the second figure, the hypotenuse of the red and blue triangles are above the actual hypotenuse, which accounts for the 0.5 difference in area again.
The difference between the two figures is a thin parallelogram with vertices at the endpoints of the hypotenuses in both figures (coloured pink below), which has area exactly 1. (If it looks much less, that's the nature of the optical illusion.)
Also, there are some links to similar figures at Wikipedia's article on this missing square puzzle.
Surprisingly, the general case remains to be explained.
Consider the recursion $F(n+2) = F(n+1) + F(n)$, with $F(1) = F(2) = 1$
producing the Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 ...
Let the blue triangle have horizontal and vertical sides of length $F(n+1)$ and $F(n-1)$, respectively.
Let the red triangle have horizontal and vertical sides of length $F(n+2)$ and $F(n)$, respectively. So that in the upper figure the two triangles create a rectangle with an area of $F(n)F(n+1)$ and in the lower figure the two triangles create a rectangle with an area of $F(n-1)F(n+2)$
By the Fibonacci identity $F(n-1)F(n+2) - F(n)F(n+1) = \pm1$
we can see that the two rectangles are bound to differ in area by exactly one unit.
As n increases the gradient/slope of each triangle tends to the square of the Golden Ratio $ \phi$,
one from above and one from below. Thus, as $\phi^2$ = 0.3819660112... 2/5 = 0.4 > $\phi^2$ 3/8 = 0.375 < $\phi^2$ 5/13 = 0.3846... > $\phi^2$ etc.
The red and blue triangles are not similar (the ratios of the sides are 3/8 = 0.375 and 2/5 = 0.4 respectively), so the "hypotenuse" of your big triangle is not a straight line.
No, this is not a proof of your statement. If you look very closely, you will see that the hypotenuse of the triangles aren't straight. You can also verify this algebraically by calculating the internal angles of the triangle using trigonometry.
The other way to demonstrate the non-straightness of the hypotenuse is in the fact that the red and blue triangles aren't similar, which can be seen by the difference in the ratios of their width and height. For the blue triangle, this is 2/5 and for the red triangle this is 3/8. As they are both right angled triangles, these ratios would be the same for similar triangles and the fact that they aren't means the interior angles are different.
I never looked at this, and thought that it was the same shape. You can't just change the number of angles in a shape and call it identical to the original. If you were to sketch only the outlines of each colored block on a blank piece of paper would the resulting "Shapes" be Identical or Even Similar? The answer is most definitely not. The illusion is looking at it on a graph VS in a Tangible form of any type, Imagine that you have 2 completely identical blocks of wood. You remove a identical portion as shown in figure 2, You didn't just make that hole appear out of nowhere because it can only be achieved by the removal of a block and thus a massive difference in the amounts of Angles laying within the "Shape" therefor making figure 2 nothing more than a combination of different shapes rather than being one definable shape of any type. I've yet to hear a better explanation than my own! |
Curve-shortening flow is the simplest example of a curvature flow. It moves each point on a plane curve \(\gamma\) in the inwards normal direction \(-\nu\) with speed proportional to the signed curvature \(k\) at that point, as described by the equation
$$ \frac{\partial \gamma}{\partial t} = - k \nu.$$
The name "curve-shortening" comes from the fact that the curve is always moving so as to decrease its length as efficiently as possible. The flow is the lowest-dimensional case of the
mean curvature flow: for example, the next case deforms surfaces so as to decrease their area efficiently.
For any initial curve that does not intersect itself, the flow will converge to a "round point" in finite time - this is the Gage-Hamilton-Grayson theorem.
New: You can make video recordings of this demonstration by following this link. Only works in Google Chrome (not mobile friendly).
The demonstration on this page scales time by a factor of \(1/\max |k|\) for each curve to avoid numerical errors.
Self-intersecting curves also jump over some cusps that are singularities of the smooth flow - this behaviour is qualitatively (and perhaps quantitatively) the same as that of the weak formulation of the flow.
This applet is open-source - it was developed by me and other contributors, and can be freely modified and redistributed by anyone. Check out the source code at GitHub.
« Home |
Bill Dubuque raised an excellent point here: Coping with *abstract* duplicate questions.
I suggest we use this question as a list of the generalized questions we create.
I suggest we categorize these abstract duplicates based on topic (please edit the question). Also please feel free to suggest a better way to list these.
Also, as per Jeff's recommendation, please tag these questions as faq.
Laws of signs (minus times minus is plus): Why is negative times negative = positive?
Order of operations in arithmetic: What is the standard interpretation of order of operations for the basic arithmetic operations?
Solving equations with multiple absolute values: What is the best way to solve an equation involving multiple absolute values?
Extraneous solutions to equations with a square root: Is there a name for this strange solution to a quadratic equation involving a square root?
Principal $n$-th roots:
$0! = 1$: Prove $0! = 1$ from first principles
Partial fraction decomposition of rational functions: Converting multiplying fractions to sum of fractions
Highest power of a prime $p$ dividing $N!$, number of zeros at the end of $N!$ and related questions: Highest power of a prime $p$ dividing $N!$
Solving $x^x=y$ for $x$: Is $x^x=y$ solvable for $x$?
What is the value of $0^0$? Zero to the zero power – is $0^0=1$?
Integrating polynomial and rational expressions of $\sin x$ and $\cos x$: Evaluating $\int P(\sin x, \cos x) \text{d}x$
Integration using partial fractions: Integration by partial fractions; how and why does it work?
Intuitive meaning of Euler's constant $e$: Intuitive Understanding of the constant "$e$"
Evaluating limits of the form $\lim_{x\to \infty} P(x)^{1/n}-x$ where $P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$ is a monic polynomial: Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$
Finding the limit of rational functions at infinity: Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q,P$ are polynomials
Divergence of the harmonic series: Why does the series $\sum_{n=1}^\infty\frac1n$ not converge?
Universal Chord Theorem: Universal Chord Theorem
Nested radical series: $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$
Derivative of a function expressed as $f(x)^{g(x)}$: Differentiation of $x^{\sqrt{x}}$, how?
Removable discontiuity: How can a function with a hole (removable discontinuity) equal a function with no hole?
Calculus Meets Geometry Volume of intersection between cylinders Two cylinders, same radius, orthogonal. This post is not particularly good but there are many existing duplicate-links. Note that this can be done without calculus. Two cylinders variation: different radii (orthogonal), non-orthogonal (same radius), and elliptic cylinders (essentially unsolved). Three cylinders: same radius and orthogonal. Number of permutations of $n$ where no number $i$ is in position $i$ How many equivalence relations on a set with 4 elements. How many ways can N elements be partitioned into subsets of size K? Seating arrangements of four men and three women around a circular table How to use stars and bars? How many different spanning trees of $K_n \setminus e$ are there? (or Spanning Trees of the Complete Graph minus an edge) Definition of Matrix Multiplication: (Maybe there should just be one canonical one?) On the determinant: Determinants of special matrices: Eigenvectors and Eigenvalues Gram-Schmidt Orthogonalization Prove that A + I is invertible if A is nilpotent A generalization for non-commutative rings Modular exponentiation: How do I compute $a^b\,\bmod c$ by hand? Solving the congruence $x^2\equiv1\pmod n$: Number of solutions of $x^2=1$ in $\mathbb{Z}/n\mathbb{Z}$ Can $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m$ are perfect squares? What is the period of the decimal expansion of $\frac mn$?
Geometric Series: Value of $\sum\limits_n x^n$
Summing series of the form $\sum_n (n+1) x^n$: How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$?
Finding the limit of rational functions at infinity: Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q,P$ are polynomials
Divergence of the harmonic series: Why does the series $\sum_{n=1}^\infty\frac1n$ not converge?
Nested radical series: $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$
Limit of exponential sequence and $n$ factorial: Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$.
There are different sizes of infinity: What Does it Really Mean to Have Different Kinds of Infinities?
Solving triangles: Solving Triangles (finding missing sides/angles given 3 sides/angles)
(Confusing) notation for inverse functions ($\sin^{-1}$ vs. $\arcsin$): $\arcsin$ written as $\sin^{-1}(x)$ |
I've been trying to get the limit of a function, but I don't know how.
The function is $\displaystyle{10^{n}\left(1 - \mathrm{e}^{\mathrm{i}t/10^{\,n}}\,\right)}$ and the solution says this converges to $-\mathrm{i}t$ as $n \to \infty$. The solution also told me to make use of the Euler's formula. I have no clue how they got to $-\mathrm{i}t$. Whenever I try to write $\mathrm{e}^{\mathrm{i}t/10^{n}} = \cos\left(t/10^n\right) + \mathrm{i}\,\sin\left(t/10^{n}\right)$ my function converges to $0$.
What did I do wrong ?.
Thank you! |
I am a bit confused about the following statement.
Let $f$ be a real function of a real variable and $x_0 \in \mathbb{R}$. If $f$ is differentiable in a neighborhood $I$ of $x_0$, $f'(x_0) = 0$ and $f'(x_0) > 0$ for all $x \in I \setminus \{x_0\}$, then $x_0$ is an inflection point.
Is this statement true? Reading this question: Prove that $f$ has an inflection point at zero if $f$ is a function that satisfies a given set of hypotheses, I think my statement is false, but I cannot give a counterexample.
Can anyone help me out?
EDIT. (Thanks to @Zestylemonzi's comment) My definition of inflection point is as follows:
$x_0$ is an inflection point if $f$ is differentiable at $x_0$ and there exists a neighborhood $J$ of $x_0$ such that the function $d(x) = f(x) − f(x_0) − f'(x_0) (x − x_0)$ has the same sign as $x−x_0$ for all $x \in J \setminus \{ x_0 \}$, or $d(x) = f(x) − f(x_0) − f'(x_0) (x − x_0)$ and $x−x_0$ have opposite signs for all $x \in J \setminus \{ x_0 \}$
I think that according to this definition my statement is true. I am aware of another possible definition of
inflection point:
$x_0$ is an inflection point if $f$ is differentiable at $x_0$ and there exists $\delta > 0$ such that $f$ is convex (respectively, concave) for $x \in (x_0 - \delta, x_0)$ and concave (respectively, convex) for $x \in (x_0, x_0 + \delta)$.
Is my statement true according to the second definition? |
I am on the way to proving that for $Y\subsetneq X$ closed subspace in a normed space $(X, \vert \vert \cdot \vert \vert)$ that:
$\exists \ell \in (\operatorname{span}(\{x_{0}\} \cup Y))^{*}$ where $\ell\vert_{Y}=0$ and $\ell(x_{0})={\operatorname{dist}(x_{0},Y)} $ and $\vert \vert \ell \vert \vert_{*}=1$. I used $\ell(\alpha x_{0})=\alpha {\operatorname{dist}(x_{0},Y)}$. It is rather simple to show that $\vert \vert \ell \vert \vert_{*}\leq1$ but on the issue of $\vert \vert \ell \vert \vert_{*}=1$, I have stumbled into trouble. When I look at the solutions, they do not make sense to me either.
The supposed solution: Let $(y_{n})_{n}\subseteq Y: \lim\limits_{n \to \infty} \vert\vert x_{0} - y_{n}\vert\vert={\operatorname{dist}(x_{0},Y)}$ , then it follows that $1=\ell (\frac{x_{0}-y_{n}}{\operatorname{dist}(x_{0},Y)})\leq\vert\vert \ell\vert\vert_{*}\vert\vert\frac{x_{0}-y_{n}}{\operatorname{dist}(x_{0},Y)}\vert\vert\Rightarrow \vert \vert \ell \vert \vert_{*}=1$.
Is this solution even correct, I do not understand where the first equality comes from and how would the entire inequality imply $\vert \vert \ell \vert \vert_{*}=1$. Appreciate the help. |
First, I am fresh new with the PGFPlotTable package! I have the following CSV file that I would like to plot in a LaTeX table:
truncatedtime;meanfinalfloatvalue;floatvaluedifference;datap2count;expecteddatap2count;performance2015-09-01 00:00:00;0.0375;0.0;48;48;1.02015-09-02 00:00:00;-0.247916666666667;0.0;48;48;1.02015-09-03 00:00:00;0.364583333333333;0.0;48;48;1.02015-09-04 00:00:00;0.397916666666667;0.0;48;48;1.02015-09-05 00:00:00;0.310416666666667;0.0;48;48;1.0[...]
Almost everything works fine with numeric columns, but I cannot compile LaTeX document when I include the first column. I have read the documentation and take a tour of TeX.SO related posts but I could not find any relevant solution.
My include code is the following:
\begin{table}[!ht]\centering\pgfplotstabletypeset[ col sep=semicolon, columns={truncatedtime,meanfinalfloatvalue,floatvaluedifference,datap2count,expecteddatap2count,performance}, columns/truncatedtime/.style={string type}, columns/meanfinalfloatvalue/.style={column name={$\bar{x}$},fixed,zerofill,precision=3}, columns/floatvaluedifference/.style={column name={$\Delta\bar{x}$},fixed,zerofill,precision=3}, columns/datap2count/.style={column name={$n_\mathrm{exp}$}}, columns/expecteddatap2count/.style={column name={$n_\mathrm{th}$}}, columns/performance/.style={column name={$\eta$},fixed,zerofill,precision=3}, header=has colnames, font=\footnotesize, dec sep align, fonts by sign={}{\color{red}}, every head row/.style={before row=\toprule,after row=\midrule}, every last row/.style={after row=\bottomrule}]{tables/Aggregate_day_dT_003_T1M003.csv}\caption{Aggregates (day) for Channel dT:003/T1M003 (MET)\protect\footnotemark}\label{tab:Aggregates_day_dT:003/T1M003 (MET)}\end{table}
This code leads to the following error when LaTeX try to parse it:
Package PGF Math Error: Could not parse input '2015-09-25 00:00:00' as a floating point number, sorry. The unreadable part was near '-09-25 00:00:00'..
It looks like
PGFPlotsTable does not notice the
string type style argument and try to cast it into float number which raise a conversion error.
What I have tried so far:
Enclose timestamp with quotes, does not work; Remove the
tableenvironment that wraps the include and all the features in it, it leads to the same error, it does not seem to have any effect;
Remove the
header=has colnamesargument which I think is not necessary as long as I name my columns, I get the same error
How must I force PGFPlotsTable to consider my column as text? What is going wrong with my code?
Update:
As suggested by
fmetz, I have enclosed my timestamps with curly-braces, now the CSV file has the following form, but it did not solve my problem:
truncatedtime;meanfinalfloatvalue;floatvaluedifference;datap2count;expecteddatap2count;performance{2015-09-01};0.0375;0.0;48;48;1.0{2015-09-02};-0.247916666666667;0.0;48;48;1.0{2015-09-03};0.364583333333333;0.0;48;48;1.0{2015-09-04};0.397916666666667;0.0;48;48;1.0{2015-09-05};0.310416666666667;0.0;48;48;1.0{2015-09-06};0.39375;0.0;48;48;1.0{2015-09-07};0.414583333333333;0.0;48;48;1.0{2015-09-08};0.375;0.0;48;48;1.0[...]
The error remains:
! Package PGF Math Error: Could not parse input '2015-09-06' as a floating point number, sorry. The unreadable part was near '-09-06'.. |
Research
I study the distribution of algebraic numbers, mathematical statistical physics and roots/eigenvalues of random polynomials/matrices.
Projects in Progress
1
The distribution of values of the non-archimedean absolute Vandermonde determinant and the non-archimedean Selberg integral (with Jeff Vaaler). The Mellin transform of the distribution function of the non-archimedean absolute Vandermonde (on the ring of integers of a local field) is related to a non-archimedean analog of the Selberg/Mehta integral. A recursion for this integral allows us to find an analytic continuation to a rational function on a cylindrical Riemann surface. Information about the poles of this rational function allow us to draw conclusions about the range of values of the non-archimedean absolute Vandermonde.
2
Non-archimedean electrostatics. The study of charged particles in a non-archimedean local field whose interaction energy is proportional to the log of the distance between particles, at fixed coldness $\beta$. The microcanonical, canonical and grand canonical ensembles are considered, and the partition function is related to the non-archimedean Selberg integral considered in 1. Probabilities of cylinder sets are explicitly computable in both the canonical and grand canonical ensembles.
3
Adèlic electrostatics and global zeta functions (with Joe Webster). The non-archimedean Selberg integral/canonical partition function are examples of Igusa zeta functions, and as such local Euler factors in a global zeta function. This global zeta function (the exact definition of which is yet to be determined) is also the partition function for a canonical electrostatic ensemble defined on the adèles of a number field. The archimedean local factors relate to the ordinary Selberg integral, the Mehta integral, and the partition function for the complex asymmetric $\beta$ ensemble. The dream would be a functional equation for the global zeta function via Fourier analysis on the idèles, though any analytic continuation would tell us something about the distribution of energies in the adèlic ensemble.
4
Pair correlation in circular ensembles when $\beta$ is an even square integer (with Nate Wells and Elisha Hulbert). This can be expressed in terms of a form in a grading of an exterior algebra, the coefficients of which are products of Vandermonde determinants in integers. Hopefully an understanding of the asymptotics of these coefficients will lead to scaling limits for the pair correlation function for an infinite family of coldnesses via hyperpfaffian/Berezin integral techniques. This would partially generalize the Pfaffian point process arising in COE and CSE. There is a lot of work to do, but there is hope.
5
Martingales in the Weil height Banach space (with Nathan Hunter). Allcock and Vaaler produce a Banach space in which $\overline{\mathbb Q}^{\times}/\mathrm{Tor}$ embeds densely in a co-dimension 1 subspace, the (Banach space) norm of which extends the logarithmic Weil height. Field extensions of the maximal abelian extension of $\mathbb Q$ correspond to $\sigma$-algebras, and towers of fields to filtrations. Elements in the Banach space (including those from $\overline{\mathbb Q}^{\times}/\mathrm{Tor}$) represent random variables, and the set up is ready for someone to come along and use martingale techniques—including the optional stopping time theorem—to tell us something about algebraic numbers. Instruction
I have three current PhD students and one current departmental Honors student. I have supervised two completed PhDs and six completed honors theses. You can find a list of current and completed PhD and honors students on my CV.
My teaching load has been reduced for the last five years (or so) due to an FTE release for serving on the Executive Council of United Academics. As President of United Academics, and Immediate Past President of the University Senate I am not teaching in the 2018 academic year. In AY2019, I am scheduled to teach a two-quarter sequence on mathematical statistical physics.
I take my teaching seriously. I prepare detailed lecture notes for most courses (exceptions being introductory courses, where my notes are better characterized as well-organized outlines). When practical and appropriate I use active learning techniques, mostly through supervised group work. I am a tough, but fair grader.
Service
Service encompasses pretty much everything that an academic does outside of teaching and research. This includes advising, serving on university and departmental committees, reviewing papers, writing letters of recommendation, organizing seminars and conferences, serving on professional boards, etc. The impossibility of doing it all allows academics to decide what types of service they are going specialize based on their interests and abilities.
I have spent the last three years heavily engaged in university level service. I currently serve as the president of United Academics of the University of Oregon, and I am the immediate-past president of the University Senate. Before that I was the Vice President of the Senate and the chair of the Committee on Committees. All of these roles are difficult and require a large investment of thought and energy. The reward for this hard work is a good understanding of how the university works, who to go to when issues need resolution, and who can be safely ignored.
I know what academic initiatives are underway, being involved in several of them. I am spearheading, with the new Core Education Council, the reform of general education at UO. I am working on the New Faculty Success Program—an onboarding program for new faculty—with the Office of the Provost and United Academics. I am currently on the Faculty Salary Equity Committee and its Executive Committee. I have been a bit player in many other projects and initiatives including student evaluation reform, the re-envisioning of the undergraduate multicultural requirement, and the creation of an expedited tenure process to allow the institution alacrity when recruiting imminent scholars. This list is incomplete.
Next year, with high probability, I will be the chair of the bargaining committee for the next collective bargaining agreement between United Academics and the University of Oregon (this assumes I am elected UA president). I will also be working with the Core Ed Council to potentially redefine the BA/BS distinction, with a personal focus on ensuring quantitative/data/information literacy is distributed throughout our undergraduate curriculum. I will also be working to help pilot (and hopefully scale) the Core Ed “Runways” (themed, cohorted clusters of gen ed courses) with the aspirational goal of having 100% of traditional undergraduates in a high-support, high-engagement, uniquely-Oregon first-year experience within the next 3-5 years.
As important as the service I
am doing, is the service I am not doing. I do little to no departmental service (though part of this derives from the CAS dean’s interpretation of the CBA) and I avoid non-required departmental functions (for reasons). I do routinely serve on academic committees for graduate/honors students, etc. I decline most requests to referee papers/grants applications, and serve on no editorial boards. The national organizations for which I am an officer are not mathematical organizations, but rather organizations dedicated to shared governance. Diversity & Equity
The two principles which drive my professional work are
truth and fairness.
I remember after a particularly troubling departmental vote, a senior colleague attempted to assuage my anger at the department by explaining that “the world is not fair.” I hate this argument because it removes responsibility from those participating in such decisions, and places blame instead on a stochastic universe. And, while there
is stochasticity in the universe, we should be working toward ameliorating inequities caused by chance, and in instances where we have agency, making decisions which do not compound them.
I do not think the department does a very good job at recognizing nor ameliorating inequities. Indeed, there are individuals, policies and procedures that negatively impact diversity. See my recent post Women & Men in Mathematics for examples.
My work on diversity and equity issues has been primarily through the University Senate and United Academics. As Vice-president of the UO Senate, I sat on the committee which vetted the Diversity Action Plans of academic units. I also worked on, or presided over several motions put forth by the University Senate which address equity, diversity and inclusion. Obviously, the work of the Senate involves many people, and in many instances I played only a bit part, but nonetheless I am proud to have supported/negotiated/presided over the following motions which have addressed diversity and equity issues on campus:
Implementing A System for the Continuous Improvement and Evaluation of Teaching Proposed Changes to Multicultural Requirement Resolution denouncing White Supremacy & Hate Speech on Campus Proposed Change to Admissions Policies Requiring Disclosure of Criminal and Disciplinary Hearing A Resolution in support of LGBTQAI Student Rights Declaring UO a Sanctuary Campus Reaffirming our Shared Values of Respect for Diversity, Equity, and Inclusion Student Sexual and gender-Based Harassment and Violence Complaint and Response Policy
Besides my work with the Senate, I have also participated in diversity activities through my role(s) with United Academics of the University of Oregon. United Academics supports both a Faculty of Color and LGBTQ* Caucus which help identify barriers and propose solutions to problems affecting those communities on campus. United Academics bargained a tenure-track faculty equity study, and I am currently serving on a university committee identifying salary inequities based on protected class and proposing remedies for them.
I have attended in innumerable rallies supporting social justice, and marched in countless marches. I flew to Washington D.C. to attend the March for Science. I’ve participated in workshops and trainings on diversity provided by the American Federation of Teachers, and the American Association of University Professors.
I recognize that I am not perfect. I cannot represent all communities nor emulate the diversity of thought on campus. I have occasionally used out-moded words and am generally terrible at using preferred pronouns (though I try). I recognize my short-comings and continually work to address them.
There are different tactics for turning advocacy into action, and individuals may disagree on their appropriateness and if/when escalation is called for. My general outlook is to work within a system to address inequities until it becomes clear that change is impossible from within. In such instances, if the moral imperative for change is sufficient then I work for change from without. This is my current strategy when tackling departmental diversity issues; I work with administrative units, the Senate and the union to put forth/support policies which minimize bias, discrimination and caprice in departmental decisions. I ensure that appropriate administrators know when I feel the department has fallen down on our institutional commitment to diversity, and I report incidents of bias, discrimination and harassment to the appropriate institutional offices (subject to the policy on Student Directed Reporters).
Fairness is as important to me as truth, and I look forward to the day where I can focus more of my time uncovering the latter instead of continually battling for the former. |
Let $p$ be a real number greater than $1$. It is well known (see Hall and Heyde's
Martingale limit theory and its applications, Theorem 2.10) that there exists a constant $C_p$ such that if $(X_i)_{i=1}^n$ is a real valued martingale difference with respect to the filtration $(\mathcal F_i)_{i=1}^n$ (that is, $(S_j:=\sum_{i=1}^jX_i)_{j=1}^n$ is a martingale with respect to this filtration), then $$\frac 1{C_p}\mathbb E\left(\sum_{i=1}^nX_i^2\right)^{p/2}\leqslant \mathbb E\left|\sum_{i=1}^nX_i\right|^p\leqslant C_p\mathbb E\left(\sum_{i=1}^nX_i^2\right)^{p/2}.$$Hence the $\mathbb L^p$ norm of the partial sum is controlled by those of the quadratic variation.
Now define for a real valued random variable $X$: $$\lVert X\rVert_{p,\infty}:=\left(\sup_{t\geqslant 0}t^p\mu\{|X|\geqslant t\}\right)^{1/p}.$$ This is equivalent to a norm (namely $N(X):=\sup_{\mu(A)>0}\mu(A)^{-1+1/p}\int_A|X|\mathrm d\mu$).
I would like to know whether there is a similar inequality to Rosenthal's one, that is, a control of $N(S_n)$ in terms of those of $N\left(\sqrt{\sum_{i=1}^nX_i^2}\right)$ plus maybe an other term. This seems to be a natural question which has probably been investigated, but I didn't manage to find a reference.
There are weak-$\mathbb L^p$ versions of Rosenthal's inequality for independent random variables, but I would like to see a reference dealing with an extension to martingale differences, namely:
Let $(\Omega,\mathcal F,\mu)$ be a probability space $p\gt 2$. Is there a constant $C_p$ such that if $n$ is an integer and $(X_j)_{1\leqslant j\leqslant n}$ is a martingale difference with respect to the filtration $(\mathcal F_j)_{1\leqslant j\leqslant n}$ with $\lVert X_j\rVert_{p,\infty}\lt\infty$, then $$C_p^{-1}\left\lVert \sqrt{\sum_{j=1}^nX_j^2}\right\rVert_{p,\infty}\leqslant \left\lVert \sum_{j=1}^nX_j\right\rVert_{p,\infty}\leqslant C_p\left\lVert \sqrt{\sum_{j=1}^nX_j^2}\right\rVert_{p,\infty}~? $$ |
All details in the question are for the case $p=2$ though I expect the answer shouldn't be that different for odd primes.
Adams showed (i think it was him) the following statement:
The element $h_j$ in the $1$-line of the ASS (of the sphere) persists (i.e. survives to $E_{\infty}$) iff the corresponding cohomology operation $Sq^{2^{j}}$ acts non-trivialy on the cohomology of some $2$-cell complex.
According to [Theorem $7.8.$] in Browder's article "The Kervaire Invariant of Framed Manifolds and its Generalization" which he assigns to Adams but which i was unable to find a proof of in the references he provides, The following statement is also true:
An element $h_ih_j$ in the 2-line persists andevaluates to 1 on a secondary cohomology operation $\psi$ iff $\psi$ acts non-trivially on the cohomology of some 2-cell complex.
Its an algebraic consequence of the minimal resolution of $\mathbb{F}_2$ over the steenrod algebra that the $s$-line in the ASS can be interpreted as
"relations (between relations)$^{s-1}$" and thus its natural to think of these as corresponding to some sort of $s$-order cohomology operations even if not in the most precise manner. The above statements are much stronger though as their consequences are about the stable homotopy groups of spheres.
Question 1:Is there a systematic connection between actions of $s$-order cohomology operations on the cohomology of $2$-cell complexes and persistence of elements in the $s$-line of the ASS? If so what is it?
In the end what i'm interested in is whether this can be used "systematically" to draw conclusions about differentials in Adams spectral sequences. I'll explain. It seems to me that the way Browder uses this theorem is by applying it to the $d_2$ in the ASS of a certain bordism spectrum he constructs to conclude that it vanishes at somewhete under certain conditions using the fact that he can calculate explicitly $d_1$ in the relevant ranges. I suspect that at least in the $E_2$-page the vanishing of the secondary cohomology operation associated to $d_1 \circ d_1=0$ always implies the vanishing of $d_2$ as the $d_2$ can be expressed as a certain Toda bracket which is "bounded from above" by the associated secondary cohomology operation. For higher differetials I have no idea...
Question 2:Is there a systematic technique lurking here for proving vanishing of $d_n$'s in Adams spectral sequences which exploits this interplay between differentials, Toda brackets, higher order cohomology operations, and the $E_2$ of the ASS of the sphere? If so what's the jist of it?
Finally let me add that i'm pretty convinced that there's an altogether different way to phrase this question which makes everything less confusing, comments about rephrasing and equivalent formulations (even if they are not full answers) would be very welcome. |
I got a full answer for my question on Jim Gatheral's book The Volatility Surface. I am going to try my luck again on another question on the same book. In Section The Decay of Skew Due to Jumps on page 63-64 , he claims that some $T^*$ that characterizes the decay time of the effect of jumps satisfies $$-(e^{\alpha+\frac{\delta^2}2}-1)\approx \sigma\sqrt{T^*}$$ where the occurrence of jumps is Poisson and the size is lognormally distributed with mean log-jump $\alpha$ and standard deviation $\delta$, and $\sigma$ is the volatility of the diffusion.
Does anyone have a reference to a somewhat rigorous justification of this claim?
@Quantuple made a suggestion in the comment section. While that makes intuitively plausible sense, I have been perplexed by how one exactly and rigorous describe the separate effects on implied volatility from diffusion and jumps that he alludes to and which Gatheral talks about in the paragraph right above that equation in question. Jump adds two effects on the implied volatility at the money in the long time asymptotics. One is the size, the other is the skew.
1) Size.
Is there a simple approximate addition formula to separate out the contributions from diffusion and jump?
2) Skew
Obviously, for time dependent deterministic diffusion volatility, the only cause of skew is from jumps, as there is no contribution to the at-the-money skew from diffusion. Actually, it is the ATM skew Gatheral is concerned about in this section. Since the formula above relates only sizes of jump and diffusion, I do not see what and how this relates to the skew in the long time asymptotics.
For a stochastic volatility model like Heston with jumps, how does the skew from the stochastic volatility relate and compare to that from the jump, in the long time asymptotics? |
Let $g$ be a Riemannian metric on the $d$-dimensional flat space $\mathbb R^d$, and consider the usual Lagrangian $$L(x, \dot x) = \tfrac 1 2 g_{ij}(x) \dot x^i \dot x^j.$$ Let $\hat g := \sqrt g$ denote the square root of the metric $g$, implicitly defined by the formula $\hat g_{ai} \hat g_{bj} \delta^{ab} = g_{ij}$, where $\delta^{ab}$ is the identity 2-tensor. I want to introduce phase-space-type coordinates $$u_i = \hat g_{ij} \dot x^j.$$ In the coordinates $(x,u)$, the metric on $u$ is just the Euclidean metric: $\langle u, u \rangle := \delta^{ij} u_i u_j$.
Let $x = x(t)$ denote a geodesic for the metric $g$, and define $u(t) := \hat g_{ij} \dot x^j$. These coordinates are convenient, because along the geodesic, $u(t)$ remains on the sphere of radius $|u(0)|$: $$\tfrac{d}{dt} \langle u, u \rangle = \tfrac{d}{dt} \delta^{ij} u_i u_j = \tfrac{d}{dt} g_{ij} \dot x^i \dot x^j = 0,$$ since geodesics are parametrized by unit speed. Conveniently, this means that $\langle u, \dot u \rangle \equiv 0$.
(You may wonder why don't I just use Hamiltonian phase-space coordinates $(x,p)$. In my research, I consider $g$ as a parameter, ranging over all possible Riemannian metrics on the plane $\mathbb R^2$. Hamiltonian coordinates have the nice property that for a fixed metric, the energy shells $\{ g^{ij}(x) p_i p_j = \mathrm{constant} \}$ are invariant under the geodesic flow. Unfortunately, these energy shells are not independent of the metric parameter $g$. In the coordinates $(x,u)$, on the other hand, the shells $\{ \langle u, u \rangle = \mathrm{constant} \}$ are just spheres in Euclidean space, and do not depend on $g$. In particular, it is important to me that these spherical shells are invariant under rotations in the phase space $\mathbb R^d \times \mathbb R^d$).
I want to calculate the geodesic equation in the coordinates $(x,u)$, particularly for the case that $d=2$. It is easy to see that $\dot x^j = \hat g^{ji} u_i$, where the superscripts denote the inverse of $\hat g$. When I calculate $\dot u$, though, I get a mess: $$\dot u_a = \big( \hat g_{ab,c} \hat g^{cj} \hat g^{bi} - \hat g_{ab} \Gamma_{uv}^b \hat g^{ui} \hat g^{vj} \big) u_i u_j,$$ where $\Gamma_{uv}^b$ are the Christoffel symbols for the metric $g$. I tried simplifying this expression, to no effect. There is plenty of symmetry around (e.g., $\langle u, \dot u \rangle = 0$), and I'm sure that the formula for $\dot u$ takes a much, much simpler form.
Question: Is there a simple expression in these coordinates for the evolution of $u(t)$?
Let me explain why the above expression is inadequate. For the metric $g$, let $U_g$ denote the vector field given by $U_g(x,u) = (u, \dot u)$ (where $\dot u$ is the expression above), so that solutions to the differential equation $(\dot x, \dot u) = U_g(x,u)$ are geodesics for the metric $g$. I need to calculate the (Euclidean) divergence $\operatorname{div} U$. I am pretty sure that in the end, $\operatorname{div} U$ can be expressed in some simple geometric quantities involving the metric (like the Riemannian divergence $\operatorname{div}_g$ of some vector field, scalar curvature $K_g$, etc.). For the messy $\dot u$ above, though, it is impossible for me to see what the true character of $\operatorname{div} U$ is. |
Chapter 9 Dynamic regression models
The time series models in the previous two chapters allow for the inclusion of information from past observations of a series, but not for the inclusion of other information that may also be relevant. For example, the effects of holidays, competitor activity, changes in the law, the wider economy, or other external variables, may explain some of the historical variation and may lead to more accurate forecasts. On the other hand, the regression models in Chapter 5 allow for the inclusion of a lot of relevant information from predictor variables, but do not allow for the subtle time series dynamics that can be handled with ARIMA models. In this chapter, we consider how to extend ARIMA models in order to allow other information to be included in the models.
In Chapter 5 we considered regression models of the form \[ y_t = \beta_0 + \beta_1 x_{1,t} + \dots + \beta_k x_{k,t} + \varepsilon_t, \] where \(y_t\) is a linear function of the \(k\) predictor variables (\(x_{1,t},\dots,x_{k,t}\)), and \(\varepsilon_t\) is usually assumed to be an uncorrelated error term (i.e., it is white noise). We considered tests such as the Breusch-Godfrey test for assessing whether the resulting residuals were significantly correlated.
In this chapter, we will allow the errors from a regression to contain autocorrelation. To emphasise this change in perspective, we will replace \(\varepsilon_t\) with \(\eta_t\) in the equation. The error series \(\eta_t\) is assumed to follow an ARIMA model. For example, if \(\eta_t\) follows an ARIMA(1,1,1) model, we can write \[\begin{align*} y_t &= \beta_0 + \beta_1 x_{1,t} + \dots + \beta_k x_{k,t} + \eta_t,\\ & (1-\phi_1B)(1-B)\eta_t = (1+\theta_1B)\varepsilon_t, \end{align*}\] where \(\varepsilon_t\) is a white noise series.
Notice that the model has two error terms here — the error from the regression model, which we denote by \(\eta_t\), and the error from the ARIMA model, which we denote by \(\varepsilon_t\). Only the ARIMA model errors are assumed to be white noise. |
(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.
Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.If\[A\begin{bmatrix}1 \\3 \\5\end{bmatrix}=B\begin{bmatrix}2 \\6 \\10\end{bmatrix},\]then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.
Let $A, B, C$ be $n\times n$ invertible matrices. When you simplify the expression\[C^{-1}(AB^{-1})^{-1}(CA^{-1})^{-1}C^2,\]which matrix do you get?(a) $A$(b) $C^{-1}A^{-1}BC^{-1}AC^2$(c) $B$(d) $C^2$(e) $C^{-1}BC$(f) $C$
Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less.Let\[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\]where\begin{align*}p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3.\end{align*}
(a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$.
(b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be a vector space and $B$ be a basis for $V$.Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$.Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form\[\begin{bmatrix}1 & 0 & 2 & 1 & 0 \\0 & 1 & 3 & 0 & 1 \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0\end{bmatrix}.\]
(a) What is the dimension of $V$?
(b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.Let\[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix}a & b\\c& -a\end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\]
(a) Show that $W$ is a subspace of $V$.
(b) Find a basis of $W$.
(c) Find the dimension of $W$.
(The Ohio State University, Linear Algebra Midterm)
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 3 and contains Problem 7, 8, and 9.Check out Part 1 and Part 2 for the rest of the exam problems.
Problem 7. Let $A=\begin{bmatrix}-3 & -4\\8& 9\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}-1 \\2\end{bmatrix}$.
(a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.
(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.
Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.
Problem 9.Determine whether each of the following sentences is true or false.
(a) There is a $3\times 3$ homogeneous system that has exactly three solutions.
(b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.
(c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.
(d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 2 and contains Problem 4, 5, and 6.Check out Part 1 and Part 3 for the rest of the exam problems.
Problem 4. Let\[\mathbf{a}_1=\begin{bmatrix}1 \\2 \\3\end{bmatrix}, \mathbf{a}_2=\begin{bmatrix}2 \\-1 \\4\end{bmatrix}, \mathbf{b}=\begin{bmatrix}0 \\a \\2\end{bmatrix}.\]
Find all the values for $a$ so that the vector $\mathbf{b}$ is a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$.
Problem 5.Find the inverse matrix of\[A=\begin{bmatrix}0 & 0 & 2 & 0 \\0 &1 & 0 & 0 \\1 & 0 & 0 & 0 \\1 & 0 & 0 & 1\end{bmatrix}\]if it exists. If you think there is no inverse matrix of $A$, then give a reason.
Problem 6.Consider the system of linear equations\begin{align*}3x_1+2x_2&=1\\5x_1+3x_2&=2.\end{align*}
(a) Find the coefficient matrix $A$ of the system.
(b) Find the inverse matrix of the coefficient matrix $A$.
(c) Using the inverse matrix of $A$, find the solution of the system.
(Linear Algebra Midterm Exam 1, the Ohio State University) |
Consider the following 3-SAT variant defined over the variables $x_1,\ldots,x_n$. In the $k$P$k$N-3SAT problem each variable $x_j$, $j \in [n]$, occurs exactly $k$ times as a positive literal in $\phi$, and exactly $k$ times as a negative literal in $\phi$, where $\phi$ is a 3-CNF formula. The problem is then to decide if such a formula has a satisfying assignment.
Is the $k$P$k$N-3SAT problem NP-complete?
In the $m$P$n$N-SAT problem each positive literal occurs exactly $m$ times in $\phi$, and each negative literal occurs exactly $m$ times in $\phi$, where $\phi$ is a CNF formula. It was shown in [1] that $2$P$1$N-SAT is NP-complete. This hints that the $k$P$k$N-3SAT problem is hard as well.
The $1$P$1$N-SAT is apparently easy, see a related question and answer here. Is $k$P$k$N-3SAT perhaps hard already for $k \geq 2$?
[1] Yoshinaka, Ryo. "Higher-order matching in the linear lambda calculus in the absence of constants is NP-complete." Term Rewriting and Applications. Springer Berlin Heidelberg, 2005. 235-249. |
Let's take into account a "
Basic Feedback System" (hereinafter BFS) block diagram first:
simulate this circuit – Schematic created using CircuitLab
We can write:
\$ V_{OUT}=A \cdot (V_{IN}+ \beta V_{OUT}) \$
Therefore the BFS overall gain:
$$ G= \frac{V_{OUT}}{V_{IN}}=\frac{A}{1- \beta A} \> \> \> \> (=\frac{1}{\frac{1}{A}- \beta})$$
if ( \$ 1- \beta A \$ ) → 0 , then G → \$ \infty \> \> \$ (the system becomes unstable)
so, for the stability of such a system it is required: \$ \> \> \beta A ≠ 1 \$
It shows that system stability depends on the \$ \beta \$A product - the
open loop gain (see the Nyquist stability criterion for instance for more details).
(For an ideal OpAmp with A → \$ \infty \> \> \$:\$ \> \> \> G= -\frac{1}{\beta}) \$
Now let's analyze those two cases in question: (starting with case 1; an inverting amplifier)
A)
simulate this circuit
\$ v_+ =0 \$
\$ V_{OUT}=A \cdot (v_+ - v_-)=-A \cdot v_- \$
=> \$ v_- = - \frac{V_{OUT}}{A} \$
\$ ( i_1 = ) \$\$ \frac{V_{IN}-v_-}{R_2} \$ = \$ \frac{v_--V_{OUT}}{R_F} \$ \$ (=i_2) \$
then:
\$ \frac{V_{IN}}{R_2}=v_- \cdot ( \frac{1}{R_2}+ \frac{1}{R_F})- \frac{V_{OUT}}{R_F} \$
Substituting now the above expression for \$ v_- \$, we obtain:
\$ \frac{V_{IN}}{R_2}=- \frac{V_{OUT}}{A} \cdot ( \frac{1}{R_2}+ \frac{1}{R_F})- \frac{V_{OUT}}{R_F} \$
and the overall gain is as follows:
$$ G= \frac{V_{OUT}}{V_{IN}}= \frac{(-1)}{ \frac{1}{A}(1+ \frac{R_2}{R_F})+ \frac{R_2}{R_F}} \> \> \> \> \> (1)$$
(Note that the denominator of this expression never can be 0! ; presuming A and both \$ R_2 \$ and \$ R_F \$ being positive, of course)
if A → \$ \infty \$ :
\$ G=- \frac{R_F}{R_2} \$
Comparing it now with the BFS:
\$ A'=-A \frac{R_F}{R_F+R_2} \$
\$ \beta = \frac{R_2}{R_F} \$
(here A' stands for /is analogical to/ the A in BFS)
Then:
\$ \beta A'=-A \frac{R_F}{R_F+R_2} \cdot \frac{R_2}{R_F}=-A \frac{R_2}{R_F+R_2}<0 \$ always (provided A>0, of course)
=> always* stable ( \$ \beta A' \$ ≠ 1)
*For "real" OpAmps this may not apply - under certain conditions (the phase angle between \$ V_{OUT} \$ and \$ (v_+ - v_-) \$ changes with rising frequency)
Continuing with the case 3 (positive feedback):
B)
simulate this circuit
\$ v_- =0 \$
\$ V_{OUT}=A \cdot (v_+ - v_-)=A \cdot v_+ \$
=> \$ v_+ = \frac{V_{OUT}}{A} \$
\$ (i_1=) \frac{V_{IN}-v_+}{R_2}= \frac{v_+-V_{OUT}}{R_F} (=i_2) \$
=> \$ \frac{V_{IN}}{R_2}=v_+ \cdot ( \frac{1}{R_2}+ \frac{1}{R_F})- \frac{V_{OUT}}{R_F} \$
Substituting now the above expression for \$ v_+ \$, we obtain:
\$ \frac{V_{IN}}{R_2}= \frac{V_{OUT}}{A} \cdot ( \frac{1}{R_2}+ \frac{1}{R_F})- \frac{V_{OUT}}{R_F} \$
and the overall gain is as follows:
$$G= \frac{V_{OUT}}{V_{IN}}= \frac{1}{ \frac{1}{A}(1+ \frac{R_2}{R_F})- \frac{R_2}{R_F}} \> \> \> \> \> (2)$$
(Note that the denominator in this case can be 0!)
if A → \$ \infty \$ :
\$ G=- \frac{R_F}{R_2} \$
Now, the limit values of the overall gain G (when A is approaching \$ \infty \$ ) are the same in both the cases A) and B):
$$G=-\frac{R_F}{R_2}$$
So it looks like it is the same at first sight...
BUT!
Comparing now the current case with the BFS:
\$ A'=A \frac{R_F}{R_F+R_2} \$
\$ \beta = \frac{R_2}{R_F} \$
(here A' again stands for /is analogical to/ the A in BFS)
\$ \beta A'=A \frac{R_F}{R_F+R_2} \cdot \frac{R_2}{R_F}=A \frac{R_2}{R_F+R_2}>0 \$,
so, if \$ \frac{R_F}{R_2}=(A-1) \$ then G → \$ \infty \$ => unstable!
The
exact expressions, ( 1) and ( 2), substantially differ one from another!I suppose their difference and its consequences are clearly evident from the analysis and the resulting formulas above. Due to usually very high value of A the stable case A) with negative feedback maintains, under the feedback influence, very low voltage between the Op Amp input terminal \$ v_+ \$, which is grounded, and the "live" input terminal \$ v_- \$. The latter is therefore at very low value (close to zero), that's why it is usually called virtual ground. (Maybe this "maintenance effect" is what you, sdarella, mean under the " stabilizer", am I right?) Unlike with the unstable case B), where the positive feedback leads to either oscillations or output saturation at \$ V_{OUT\_MAX} \$ or \$ V_{OUT\_MIN} \$, depending on the input conditions (see the case C) below).
C)
The case (3) with positive feedback can also be used but it works as a
comparator, with input voltage comparative levels \$ V_{IN\_LH} \$ and \$ V_{IN\_HL} \$ (i.e. input voltages at which the output voltage flips rapidly from a low level (L= \$ V_{OUT\_MIN} \$) to a high level (H= \$ V_{OUT\_MAX} \$) and vice versa, resp.). However, it is usually better to use "real" comparators made/intended right for this purpose.
simulate this circuit
we can write:
\$ \frac{V_{IN}-0}{R_2}= \frac{0-V_{OUT}}{R_F} \$
=> \$ V_{IN}=-\frac{R_2}{R_F}V_{OUT} \$ , (condition: \$ v_+ =0 \$ )
Provided the saturation values of \$ V_{OUT} \$ of the Op Amp are \$ V_{OUT\_MAX} \$and \$ V_{OUT\_MIN} \$ , we obtain the following:
for \$ V_{OUT\_MIN} (<0) \$:
$$V_{IN\_LH}=-\frac{R_2}{R_F} V_{OUT\_MIN} (>0)$$
and
for \$ V_{OUT\_MAX} (>0) \$:
$$V_{IN\_HL}=-\frac{R_2}{R_F} V_{OUT\_MAX} (<0)$$
(it's hysteresis is then \$ V_{HYST}=V_{IN\_LH}-V_{IN\_HL}=\frac{R_2}{R_F}(V_{OUT\_MAX}-V_{OUT\_MIN}) \$) |
The aim of this test case is to validate the following functions:
The simulation results of SimScale were compared to the analytical results derived from [Roark]. The meshes used in (A) and (B) were created with the parametrized-tetrahedralization-tool on the SimScale platform and then downloaded. Both meshes were then rotated and uploaded again. These were used in case (C) and (D). With this method we achieve the exact same mash for the rotated and non-rotated case.
The beam has a cross section A of 0.05 x 0.05 m
2 and a length = 1.0 m. Tool Type : CalculiX/Code_Aster Analysis Type : Static Mesh and Element types :
Case Mesh type Number of nodes Element type (A) linear tetrahedral 6598 3D isoparametric (B) quadratic tetrahedral 44494 3D isoparametric (C) linear tetrahedral 6598 3D isoparametric (D) quadratic tetrahedral 44494 3D isoparametric
Material:
Constraints:
Loads:
$$w_a l= V \rho g$$
$$w_a = 193.01175 \frac {N} {m}$$
$$I=\frac{bh^3}{12}=5.20833 \cdot 10^{-7} m^4$$
$$y(l) = -\frac{w_a l^4}{8 E I} = -2.2597 \cdot 10^{-4} m$$
Comparison of the displacement at the free end in the direction of the gravitation vector.
Case Tool Type [Roark] SimScale Error (A) CalculiX -2.2597E-4 -2.1340E-4 5.56% (B) CalculiX -2.2597E-4 -2.2559E-4 0.17% (C) CalculiX -2.2597E-4 -2.1340E-4 5.56% (D) CalculiX -2.2597E-4 -2.2560E-4 0.16% (A) Code_Aster -2.2597E-4 -2.1340E-4 5.56% (B) Code_Aster -2.2597E-4 -2.2559E-4 0.17% (C) Code_Aster -2.2597E-4 -2.1340E-4 5.56% (D) Code_Aster -2.2597E-4 -2.2560E-4 0.16%
[Roark] (1, 2, 3) (2011)”Roark’s Formulas For Stress And Strain, Eighth Edition”, W. C. Young, R. G. Budynas, A. M. Sadegh |
I'd like to compute the indefinite integral
$$\int \frac{dx}{(ax^2+bx+c)^n}, a,b,c \in \mathbb{R}$$
without resorting to the usual recursion relation method of solution
i.e. without using integration by parts.
But I'd also like to do it without simplifying $ax^2+bx+c$ into anything like $t^2+1$, because if I have to remember to use that simplification then I might as well just memorize the solution itself. I'd like to just work it out from first principles instead.
There are two possible methods.
a) differentiating with respect to a parameter (Hardy's method):
If I differentiate $\frac{1}{ax^2+bx+c}$, with respect to $c$, $n-1$ times we have
$${\frac { \left( -1 \right) ^{n-1} }{ \left( n-1 \right) !}}{\frac {\partial }{\partial c^{n-1}}}\frac{1}{ a{x}^{2}+bx+c } = \frac{1}{ \left(a{x}^{2}+bx+c \right)^n}$$
Then, if we know
$$\int \! \frac{1}{ a{x}^{2}+bx+c }{dx}=2\,\arctan \left( { \frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}} \right) {\frac {1}{\sqrt {4\,c a-{b}^{2}}}}$$
the answer is simply
$$\int \! \frac{1}{ \left(a{x}^{2}+bx+c \right)^n}{dx}= {\frac { \left( -1 \right) ^{n-1} }{ \left( n-1 \right) !}}{\frac {\partial }{\partial c^{n-1}}} \left( 2\,\arctan \left( {\frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}} \right) {\frac {1}{\sqrt {4\,ca-{b}^{2}}}} \right)$$
But computing $ \int \! \frac{1}{ a{x}^{2}+bx+c }{dx}=2\,\arctan \left( { \frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}} \right) {\frac {1}{\sqrt {4\,c a-{b}^{2}}}}$ requires using those ugly $t^2+1$ simplifications, so we need another method.
b) Partial fractions allowing for complex roots
For n = 1, we have
$$ \int \frac{dx}{ax^2+bx+c}=\frac{1}{a}\int \frac{dx}{(x-\alpha)(x-\beta)}=\frac{1}{a} \int \left( \frac{ \frac{1}{\alpha-\beta} }{x-\alpha} - \frac{\frac{1}{\alpha-\beta}}{x-\beta}\right)dx=\frac{1}{a} \frac{1}{\alpha-\beta} \ln\left|\frac{x-\alpha}{x-\beta}\right|$$
But the usual solution is given as
$$\int \! \frac{1}{ a{x}^{2}+bx+c }{dx}=2\,\arctan \left( { \frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}} \right) {\frac {1}{\sqrt {4\,c a-{b}^{2}}}}$$
This raises the question: is
$$\frac{1}{a} \frac{1}{\alpha-\beta} \ln\left|\frac{x-\alpha}{x-\beta}\right| = 2\,\arctan \left( { \frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}} \right) {\frac {1}{\sqrt {4\,c a-{b}^{2}}}}$$
true for both cases, i.e. when the quadratic has either two real, non-zero, roots or complex conjugate roots?
If it is true, could someone show me how to derive the arctan form from the logarithm form, in both real root and complex conjugate root cases?
(I looked in Bronstein's Symbolic Integration book and it looks like a very complicated matter to show these are equal, but I may be misunderstanding this)
Once the n = 1 case is solved, I believe the n > 1 case will be solved because
but this method does not look like it will give the arctan recursion relation solution
So, can someone help me integrate $\int \frac{dx}{(ax^2+bx+c)^n}, a,b,c \in \mathbb{R}$? |
The problem statement:
Measurement detects a position of a proton with accuracy of $\pm10pm$. How much is the position uncertainty $1s$ later? Assume the speed of a proton $v\ll c$.
What i understand:
I know that in general it holds that:
\begin{align} \Delta x \Delta p \geq \frac{\hbar}{2} \end{align}
This means i can calculate momentum uncertainty for the first measurement:
\begin{align} \Delta x_1\Delta p_1 &\geq \frac{\hbar}{2}\\ \Delta p_1 &\geq \frac{\hbar}{2\Delta x_1}\\ \Delta p_1 &\geq \frac{1.055\times10^{-34}Js}{2\cdot 10\times10^{-12}m \rlap{~\longleftarrow \substack{\text{should I insert 20pm}\\\text{instead of 10pm?}}}}\\ \Delta p_1 &\geq 5.275\times10^{-24} \frac{kg m}{s}\\ \Delta p_1 &\geq 9.845 keV/c \end{align}
Question:
Using the position uncertainty $\Delta x_1$ I calculated the momentum uncertainty in first measurement $\Delta p_1$.
How do I calculate the position uncertainty $\Delta x_2$ after $1s$? I am not sure what happens $1s$ later but is a momentum uncertainty conserved so that it holds $\Delta p_1 = \Delta p_2$? I know that the problem expects me to use the clasic relation $p=m_ev$ somehow but how? |
Does anyone have any tips/ideas/method for proving that a distribution is a member of the simple exponential family (SEF)?
Or is the process unique to each distribution?
For example, I am trying to show firstly that the Gamma distribution belongs to the exponential family:
\begin{align*} \rho(y|v,\alpha)&=\frac{\alpha^v}{\Gamma(v)}y^{v-1}e^{-\alpha y} \\ &=\exp[v\ln(\alpha)-\ln(\Gamma(v))+(v-1)\ln(y)-\alpha y] \end{align*}
But then I am not sure where to go from here to get it into the form
$$\exp[\frac{y\theta-b(\theta)}{\phi}+c(y,\phi)]$$
The obvious things to do is set $\theta=-\alpha$ but then $b(\theta)=v\ln(-\theta)$ which is still a function of $v$.
I'm just not sure of a general technique to pick the right $\theta$. |
What has one side and no edges? This isn't an impossible riddle but has an answer viz. the Klein Bottle.
Have you ever taken any two numbers, added the second to the first; written this down, added the result to the second number, written the result own, and so on?
If asked to find the area bounded by the parabola $y=x^2$ the $x$-axis and the line $x=a$, you would write, almost instinctively $$ \text{area } = \int_0^a x^2 dx = \frac{1}{3}a^3 $$
Soon after writing my previous letter to you, I noticed the error I had made in connection with numbers 7 and 9.
The folllowing people had sent correct solutions before the publication of this issue:
The game chosen for this issue of
Parabola is played by two people on a $10 \times 10$ chessboard.
Most candidates did not understand the question so ruled themselves out of consideration.
J221 Find a 2-digit number $AB$ such that $(AB)^2 - (BA)^2$ is a perfect square. |
Please help me to find the sum of $\sum\limits_{n=1}^\infty \left[ \frac{\left(\frac{3 - \sqrt 5}{2}\right)^n}{n^3}\right]$
Is there any special technique to solve this one ?
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Please help me to find the sum of $\sum\limits_{n=1}^\infty \left[ \frac{\left(\frac{3 - \sqrt 5}{2}\right)^n}{n^3}\right]$
Is there any special technique to solve this one ?
This question appears to be off-topic. The users who voted to close gave this specific reason:
I don't know anything about the polylogarithm function myself, so if I were forced to solve the problem I would use an elementary technique. Here, through a series of differentiations and multiplication by $x$, I can deduce an integral equation for a function with the given series expansion:
Assume $f(x) = \sum_{n=1}^{\infty} {x^n \over n^3}$. Then we want $f\left({3-\sqrt{5} \over 2} \right)$. This series is dominated by the geometric series so it must converge absolutely at the given point (which has modulus less than 1). Now, term-by-term we have
$$ x{d\over dx} \left[x {d\over dx} \left[ x{d\over dx} \left[{x^n \over n^3} \right] \right] \right] = x {d\over dx} \left[ x{d\over dx} \left[{x^n \over n^2} \right] \right] = ... = x^n, $$ so $$ \sum_{n=1}^{\infty} x{d\over dx} \left[x {d\over dx} \left[ x{d\over dx} \left[{x^n \over n^3} \right] \right] \right] = \sum_{n=1}^{\infty} x^n = {x\over 1-x}.$$ Moving the sum through the derivatives, we find $x(x(xf'(x))')' = {x\over 1-x}$. Solving this differential equation gives an expression for $f$ which can be used to evaluate the sum.
Practically speaking, considering the terms in $$S_n=\sum_{i=1}^n \frac{a^n}{n^3}$$ with $a=\frac{1}{2} \left(3-\sqrt{5}\right)$, you can notice that $a$ is rather small ($\approx 0.382$) which means that the numerator of the fraction $a^n$ will decrease quite fast while the denominator $n^3$ will increase quite fast; this makes that each term will be significantly smaller than the previous.
Let us compute the partial sums and get for $6$ significant figures $$S_1=0.381966$$ $$S_2=0.400203$$ $$S_3=0.402267$$ $$S_4=0.402600$$ $$S_5=0.402665$$ $$S_6=0.402679$$ $$S_7=0.402683$$ $$S_8=0.402684$$ Now, the question is : how many terms $k$ have to be added in order to reach an accuracy such that $$\frac{a^k}{k^3} \leq \epsilon$$
The answer is given by the solution of $a^k =k^3 \epsilon$ which can be expressed in terms of Lambert function (another special function) $$k=-\frac{3 }{\log (a)}W\left(\frac{1}{3 \sqrt[3]{-\frac{\epsilon }{\log ^3(a)}}}\right)$$ which can seem very complex. However, very good approxations exist for Lambert function such as $$W(x)\approx L_1-L_2+\frac{L_2}{L_1}$$ where $L_1=\log(x)$ and $L_2=\log(L_1)$.
Applied to the case $a=\frac{1}{2} \left(3-\sqrt{5}\right)$ and $\epsilon=10^{-6}$, this would give for the value of the argument of Lambert function $x=32.0808$ from which $W(x)=2.58319$ and then $k=8.05213$ which is what we saw ealier.
If you want to change the tolerance to $\epsilon=10^{-p}$, quick and dirty fit would show that the number of terms to be added is approximately given by $$k=0.0148695 p^2+1.67776 p-2.80166$$
According to Maple, $$ \text{polylog}\left(3, \dfrac{3-\sqrt{5}}{2}\right) = \dfrac{4}{5} \zeta(3) + \dfrac{\pi^2}{15} \ln \left(\dfrac{3-\sqrt{5}}{2}\right) - \dfrac{1}{12} \ln\left(\dfrac{3-\sqrt{5}}{2}\right)^3 $$
I don't know where it gets this rather remarkable identity. |
I know no general method to approach a question when the inverse functions are given in exponents, but I guess that in general such questions hide simple ideas under a cumbersome form with artificially chosen functions and coefficients.. For instance, in this question the inequality holds for all positive $k$. We can show this as follows.
Since $\operatorname{arcsin} x+\operatorname{arccos} x=\pi/2$ for each $x$,
$$12^{\operatorname{arcsin} x}+12^{\operatorname{arccos} x}+12^{\operatorname{arctan} x}>$$ $$12^{\operatorname{arcsin} x}+12^{\operatorname{arccos} x}\ge \mbox{ (by AM-GM) }$$ $$12^{(\operatorname{arcsin} x+\operatorname{arccos} x)/2}=2\cdot 12^{\pi/4}\approx 14.080.$$
On the other hand, the derivative of the function $3k^{\pi/k}$ is $\frac{3\pi k^{\pi/k}}{k^2}(1-\ln k)$. Therefore the function increases when $0<k<e$, attains its maximum $3e^{\pi/e}\approx 9.529$ at $k=e$, and decreases when $k>e$.
PS. The following graph (drawn by Mathcad) suggests that the minimum value of a function
$f(x)=12^{\operatorname{arcsin} x}+12^{\operatorname{arccos} x}+12^{\operatorname{arctan} x}$ is about $18.5237$ and is attained when $x$ is a solution of a transcendental equation $f’(x)=0$, that is when
$$(12^{\operatorname{arccos} x}-12^{\operatorname{arcsin} x})(1+x^2)=12^{\operatorname{arctan} x}\sqrt{1-x^2}.$$ |
(This answer does not construct a model of ZFC - Extensionality. Instead, it discusses the purpose of the axiom of extensionality and its relationship to first order logic with equality.)
The point of the axiom of extensionality is to avoid the situation where the way in which the set is defined, not just its members, affects which sets contain the set. So to make extensionality fail you will just need a model in which there are two sets $A,B$ which have the same elements but are not equal.
The easiest way to do this is to take any set $A$, make a copy of it, color one copy red, and color the other copy blue. Declare that any set that was a member of $A$ is a member of both red-$A$ and blue-$A$, and that any set which contained $A$ contains both red-$A$ and blue-$A$. Then the resulting thing will still be a model of ZFC in the language with just $\in$ (without $=$ yet).
The point of the axiom of extensionality is then to say that this situation does not occur: there are not two sets which are somehow different despite having all the same elements.
In fact, a modification of the axiom of extensionality lets us define $=$ in terms of $\in$, by declaring that two sets will be regarded as equal if they have all the same members. To make this work, we rewrite the axiom of extensionality without $=$ as:
$$(\forall x)(x \in A \Leftrightarrow x \in B) \Rightarrow (\forall y)(A \in y \Leftrightarrow B \in y)$$Then we can
define$$A = B \Leftrightarrow (\forall x)(x \in A \Leftrightarrow x \in B)$$and we can prove the substitution axioms for equality in terms of the equality-free axiom of extensionality and the definition of $=$.
This can be done in
any model of ZFC in the language with just $\in$ (satisfying the modified version of extensionality) to obtain a model of ZFC in the language ($\in$, $=$), including the usual axiom of extensionality. But here $=$ may not be interpreted as the true equality relation, for example if the original model had red sets and blue sets. If we want the $=$ symbol to be interpreted as actual equality, we have to mod out by the equivalence relation induced by the interpretation of $=$ in our model. In the setting where $=$ is treated as a logical symbol which must be interpreted as true equality, the point of the axiom of extensionality is to ensure that this modding-out has already been performed: if we would define to sets to be "equal" in this indirect way, then they are already equal. |
This question is an exact duplicate of:
Suppose we have a bivariate normal (X, Y), whose mean is zero, and variance 1, and correlation \rho. Is there an easy formula for P(X < x | Y < y ) ?
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This question is an exact duplicate of:
Suppose we have a bivariate normal (X, Y), whose mean is zero, and variance 1, and correlation \rho. Is there an easy formula for P(X < x | Y < y ) ?
This question was marked as an exact duplicate of an existing question.
Sure:
$$ \int_{-\infty}^x \frac{\int_{-\infty}^{y} f_{X,Y}(u,v) dv}{ \int_{-\infty}^{y}f_{Y}(v) dv} du $$
Didn't even require normality. |
I'm currently studying atomic term symbols. I wanted to try it on a simple atomic carbon with the electron configuration $1s^22s^22p^2$.
I know, that only open-shell electrons are involved in the term symbol classification, so that leaves us working with solely 2 electrons in $2p$ sub-shell.
First of all, we have ${6}\choose{2}$$= \frac{6!}{2!(6-2)!}=15$ possible microstates for the $1s^22s^22p^2$ configuration.
And, if I understand it correctly, we can categorize them into several "subsets" labeled with the corresponding term symbols.
The atomic term symbol is defined as ${}^{2S+1}L_J$.
In our case, the possible values of S, L and J are following: $$\max(M_L) = \max(L) = +2 \Rightarrow L = 0,1,2$$ $$\max(M_S) = \max(S) = +1 \Rightarrow S = 0, 1$$ $$J = L+S, L+S-1, \ldots, |L-S| = 3,2,1,0$$
Considering possible values of $S$, multiplicity can be $1$ or $3$, i.e. singlet or triplet. Values of $L$ enable symbols $S,P,D$.
So, our presumed set of states is ${}^3D, {}^1D, {}^3P, {}^1P, {}^3S, {}^1S$.
${}^3D$ "contains" $S=1$ and $L=2$. In that situation, both electrons would have to be spin-up and positioned together in the orbital with $m_l = +1$. That is impossible, as it contradicts Pauli exlusion principle. I.e. the state ${}^3D$ does not exist.
${}^1D$ is possible ($L=2,1,0, S=0$), so we can compute the number of corresponding microstates $N$ with the formula $N = (2L+1)(2S+1) = 5\cdot 1 = 5$.
Using the same formula, ${}^3P$ is going to contain 9 microstates.
With the "substraction method" described in this video, we can arrive to the conclusion, that both ${}^1P$ and ${}^3S$ won't exist.
And finally, ${}^1S$ will contain the last one state.
Question Now I know the possible term symbols, but I'm not sure, which microstates belong to them specifically.
It's clear, that microstates 13 and 15 will belong to ${}^1D$, but how can I determine it for the other states? ${}^1D$ should contain 3 more microstates with $S=0$ and $m_L$ being equal to -1, 0 and +1, but there are multiple candidates for every configuration.
So how could I distinguish between, e.g. microstates 7 and 9 or 8, 11 and 14?
I don't understand this point, as even $J=2$ for ${}^1D$, so it can't be used to distinguish among the "similar" microstates and to assign them properly. |
I've been fascinated by the simplicity of the driver circuit based in the YX8018 in the prominent solar garden lights, they were able to simplify the circuit a lot in a simple boost topology by just running at fixed frequency and using the LED as the output rectifier. I was interested in building my own circuit using a 555 timer driving a small transistor, just to learn more about the topology, but in my research I was not able to find a equation, or at least a simpler explanation between the frequency, input voltage, and inductor value to be able to calculate the output current to the LED. Could someone help me understand how I can calculate the output current given these parameters?
The closer I got to an answer was by reading the SLVA372C App Note, in the equations section, the inductor ripple current:
$$ \Delta Il = \dfrac{Vin \times D}{f \times L} $$
Plugging the values I tested on my solar light's circuit:
$$ \Delta Il = \dfrac{1.5 \times 0.5}{200000 \times 330 \times 10^{-6}} = 0.0107 $$
But when I measured the current to the LED using my DMM I got 6.3mA. |
I want to give an example that came up within my master thesis: integration of internal functions over convex domains in non-standard analysis.
The set of hyperreals ${^*\mathbb R}$ is an ordered field extending the reals which contains infinitesimal ($|x|<\tfrac 1n$ for all $n\in\mathbb N$) and unlimted numbers ($|x|>n$ for all $n\in\mathbb N$). An internal function ${^*f:{}^*\mathbb R \to {}^*\mathbb R}$ arises as a extension of a sequence $f_n$ of ordinary function to unlimited numbers. As an example
$$f_\epsilon(x) = \frac{1}{\sqrt{2\pi}\epsilon}\exp(-\frac{1}{2}\frac{x^2}{\epsilon^2})$$
can be extended to infinitesimal $\epsilon>0$ whereupon it acts like a the dirac delta, in the sense that for any compactly supported, smooth test function $\varphi$,
$$ \int f_\epsilon(x) \varphi(x) dx \simeq \varphi(0)$$
where $\simeq$ means that the quantities only differ by an infinitesimal amount. This type of integral is easier to rigorize because we integrate over a compact domain. However one would like to integrate over more general convex domains in $^*\mathbb R$.
Problem: Develop a theory that allows the evaluation of integrals of the form $\int_C {^*f(x)} dx$, where $^*f$ is internal and $C\subset{^*\mathbb R}$ is convex.
There are some obvious problems. For example what should $\int_{\mathbb I} 1 dx$ be? Here $\mathbb I$ is the set of infinitesimals. There is no largest infinitesimal, yet at the same time the integral is 'obviously' bounded by any positive non-infinitesimal number.
A way out of this dilemma is, instead of letting the integral be a map onto $^*\mathbb R$, to consider a sort of fuzzy integral that maps a function onto an external number $\alpha \in \mathbb E$. An external number is nothing but a pair $(\alpha_0,\mathcal N)$, where $\alpha_0\in{^*\mathbb R}$ is the
value of the integral and $\mathcal N\subset {}^*\mathbb R$ is sort of it's uncertainty. In the above case we'd have $\alpha_0 = 0$ and $\mathcal N = \mathbb I$.
Here the set $\mathcal N$ is always a so called
neutrix, that is a convex additive subgroup of ${^*\mathbb R}$. These sets can have very peculiar properties. This theory of integration is called external integration and was developed by I. van den Berg and Fouad Koudjeti.
Neutrices, external numbers, and external calculus (unfortunately paywalled)
Some more background on external numbers can be found here
One of the the issues with this theory is however that it is rather complicated and only can handle non-negative functions. I believe that it should be possible to develop a 'neat' nonstandard integration theory that addresses these issues.
Finally I want to give a nice example of how one can intuitively work with nonstandard integration which is also from the cited book but translated into a more approachable language.
We use a nonstandard version of Laplace's method used to prove Stirling's formula $n!\sim \sqrt{2\pi n} (\frac ne)^n$: By the gamma function representation, $n! = \int_0^\infty t^n e^{-t}dx$. Instead of $n$, we insert an unlimited hypernatural number $N$. Note that the integrand has a unique global maximum at $t=N$. Doing a coordinate transformation $x=t/N-1$ we obtain
$$N! = \int_{-1}^\infty e^{-N(x+1)}N^{N+1} (x+1)^N dx = N^{N+1}e^{-N}\int_{-1}^\infty e^{-N(x-\log(1+x))}dx $$
Here note that the integrand is infinitesimal whenever $x-\log(1+x)$ is not infinitesimal itself. In fact the only non-negligible contribution can occur when $-N(x-\log(1+x)) \sim -\tfrac 12 Nx^2 +O(x^3)$ is not infinitesimal, which is the case when $x\in \frac{1}{\sqrt N} \mathbb L$, where $\mathbb L$ is the set of limited numbers. ($|x|<n$ for some $n\in\mathbb N$). Hence we can restrict the integral to this set. By doing so we make a slight error in the form of $N! = (1+\epsilon)\cdot\ldots$ with $\epsilon$ infinitesimal.
In fact an equivalence relation $\asymp$ is given on the hyperreals by $x\asymp y \iff x \in y(1+\epsilon)$ for some infinitesimal $\epsilon$.
$$ N! \asymp N^{N+1} e^{-N} \int_{\frac{1}{\sqrt N }\mathbb L} e^{-N(x-\log(1+x))}dx $$
Since this set is contained within the set of infinitesimals, we can replace $x-\log(1+x)$ by its first order Taylor approximation and only do a small error again:
$$ N! \asymp N^{N+1} e^{-N} \int_{\frac{1}{\sqrt N }\mathbb L} e^{-\tfrac 12 Nx^2}dx $$
Now we can to a retransformation $x = \frac{1}{\sqrt N}y$ leading to
$$ N! \asymp N^{N} e^{-N} \sqrt N \int_{\mathbb L} e^{-\tfrac 12 y^2}dy = \sqrt{2\pi N} N^N e^{-N} $$
which proves Stirlings formula as we have shown that for all unlimited $N$ there exists an infinitesimal $\epsilon$ such that
$$ \frac{N!}{\sqrt{2\pi N} (N/e)^N} = 1 + \epsilon$$ |
"Abstract: We prove, once and for all, that people who don't use superspace are really out of it. This includes QCDers, who always either wave their hands or gamble with lettuce (Monte Zuma calculations). Besides, all nonsupersymmetric theories have divergences which lead to problems with things like renormalons, instantons, anomalons, and other phenomenons. Also, they can't hide from gravity forever."
Can a gravitational field possess momentum? A gravitational wave can certainly possess momentum just like a light wave has momentum, but we generally think of a gravitational field as a static object, like an electrostatic field.
"You have to understand that compared to other professions such as programming or engineering, ethical standards in academia are in the gutter. I have worked with many different kinds of people in my life, in the U.S. and in Japan. I have only encountered one group more corrupt than academic scientists: the mafia members who ran Las Vegas hotels where I used to install computer equipment. "
so Ive got a small bottle that I filled up with salt. I put it on the scale and it's mass is 83g. I've also got a jup of water that has 500g of water. I put the bottle in the jug and it sank to the bottom. I have to figure out how much salt to take out of the bottle such that the weight force of the bottle equals the buoyancy force.
For the buoyancy do I: density of water * volume of water displaced * gravity acceleration?
so: mass of bottle * gravity = volume of water displaced * density of water * gravity?
@EmilioPisanty The measurement operators than I suggested in the comments of the post are fine but I additionally would like to control the width of the Poisson Distribution (much like we can do for the normal distribution using variance). Do you know that this can be achieved while still maintaining the completeness condition $$\int A^{\dagger}_{C}A_CdC = 1$$?
As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including:ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room.An altern...
You're always welcome to ask. One of the reasons I hang around in the chat room is because I'm happy to answer this sort of question. Obviously I'm sometimes busy doing other stuff, but if I have the spare time I'm always happy to answer.
Though as it happens I have to go now - lunch time! :-)
@JohnRennie It's possible to do it using the energy method. Just we need to carefully write down the potential function which is $U(r)=\frac{1}{2}\frac{mg}{R}r^2$ with zero point at the center of the earth.
Anonymous
Also I don't particularly like this SHM problem because it causes a lot of misconceptions. The motion is SHM only under particular conditions :P
I see with concern the close queue has not shrunk considerably in the last week and is still at 73 items. This may be an effect of increased traffic but not increased reviewing or something else, I'm not sure
Not sure about that, but the converse is certainly false :P
Derrida has received a lot of criticism from the experts on the fields he tried to comment on
I personally do not know much about postmodernist philosophy, so I shall not comment on it myself
I do have strong affirmative opinions on textual interpretation, made disjoint from authoritorial intent, however, which is a central part of Deconstruction theory. But I think that dates back to Heidegger.
I can see why a man of that generation would be leaned towards that idea. I do too. |
Integration Exercises on indefinite and definite integration of basic algebraic and trigonometric functions.
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The following identities may also prove useful:$$\sin^2x = \frac{1}{2} - \frac{1}{2} \cos 2x \text{ and } \cos^2x = \frac{1}{2} + \frac{1}{2} \cos 2x$$
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Let $P=\{x \in \mathbb{R}^n \mid Ax=b, x\geq 0\}$ be a nonempty convex polyhedron (not bounded).
Show that $P$ is bounded (i.e., it is a polytope) if and only if the linear inequality $Ax=0, \,\, x\geq 0$ has trivial solution $x=0$ only.
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Let $P=\{x \in \mathbb{R}^n \mid Ax=b, x\geq 0\}$ be a nonempty convex polyhedron (not bounded).
Show that $P$ is bounded (i.e., it is a polytope) if and only if the linear inequality $Ax=0, \,\, x\geq 0$ has trivial solution $x=0$ only.
This question appears to be off-topic. The users who voted to close gave this specific reason:
If there exists some nontrivial solution $d$ to $Ax=0,x\geqslant 0$, then for any $x\in P$ we have for any $\varepsilon>0$ $$A(x+\varepsilon d)=Ax+\varepsilon Ad=Ax+0=Ax=b,$$ so that $x+\varepsilon d\in P$. It follows that $P$ is not bounded.
Suppose $P$ is not bounded. Then there are $x_n$ with $\|x_n\| \ge n$ such that $Ax_n =b, x_n \ge 0$.
Let $x'_n = {1 \over \|x_n\|} x_n$, and note that $A x'_n \to 0$ and $x'_n \ge 0$.
For some subsequence we have $x'_{n_k} \to x$ for some unit norm $x$. We see that $Ax = 0$ and $x \ge 0$, and so there is a non trivial solution. |
Consider a sequence of $n$ independent Bernoulli trials drawn from a list of biases $p_1,p_2,...,p_n\in[0,1]$, respectively. We set the random variable $X$ to be the sum of these trials. On wikipedia, the distribution of $X$ is called the Poisson binomial distribution. We define the sample mean and sample variance of our list of Bernoulli biases as $$ \bar{p}=\frac{1}{N}\sum_{i=1}^n p_i $$ and $$ \sigma_p^2 =\frac{1}{N}\sum_{i=1}^N(p_i-\bar{p}) =\frac{1}{N}\sum_{i=1}^N p_i^2 - \bar{p}^2. $$
Since the trials are independent, it is easy to compute that $$ \mathbb{E}[X] = \sum_{i=1}^n p_i = N\bar{p} $$ and \begin{align*} \mathbb{Var}[X] &= \sum_{i=1}^n p_i(1-p_i) \\ &= N\bar{p} - N(\sigma_p^2+\bar{p}^2) \\ &= N\bar{p}(1-\bar{p}) - N\sigma_p^2. \end{align*}
The expectation value of $X$ is not surprising. Also, when $\sigma_p^2=0$ we must have $\bar{p}=p_1=\cdots=p_n$ and so $X$ is binomially distributed, which matches $\mathbb{Var}[X]$ computed above.
My confusion is this: why does the variance of $X$ go
down as the sample variance $\sigma_p^2$ goes up (with $\bar{p}$ and $N$ fixed)? I find this very counter-intuitive, and would appreciate an explanation. I would expect with a greater variance of biases, there would be a broader distribution of possible sums of the result... |
I have been looking at the modified Cholesky decomposition suggested by the following paper: Schnabel and Eskow, A Revised Modified Cholesky Factorization Algorithm,
SIAM J. Optim. 9, pp. 1135-1148 (14 pages).
The paper talks about an implementation of the Cholesky decomposition, modified by the fact that when the matrix is not positive definite, a diagonal perturbation is added before the decomposition takes place. The algorithm given in the paper (Algorithm 1) suggests that it finds factorization such that $LL^\top = A + E, E \ge 0$. But, when implemented as stated in the paper, what I get is a lower triangular factor matrix $L$, such that: $$P\cdot(L\cdot L^\top)\cdot P^\top = A + E$$ where, $P$ is a permutation matrix.
This $L$ matrix is not the same as when using the normal Cholesky factorization for a PD matrix, where $A = L_1L_1^\top$, say.
Now, I am using it in optimization context, specifically in trust region methods, where this decomposition is followed by inverting $L$ (to compute the trust region step), so it would be helpful to have factors in lower triangular form. Is there a way to get back $L_1$ (the original Cholesky factor) for a positive definite matrix from the $P$ and $L$ matrix obtained from the modified Cholesky factorization? I am a bit surprised that the algorithm misstates what it would produce, so maybe I am missing some step here.
Related threads I have found so far:
The second thread says that the relationship is not possible to find for any given set of matrix (not necessarily for a modified Cholesky decomposition as mentioned here). Not sure if the answer still holds in this case as well.
Example:Consider the following $4\times4$ matrix which is PD.$$A = \begin{bmatrix}6 & 3 & 4 & 8 \\ 3 & 6 & 5 & 1 \\ 4 & 5 & 10 & 7 \\ 8 & 1 & 7 & 25 \end{bmatrix}$$The vanilla Cholesky factor $L_1$, such that $L_1L_1^\top=A$ is:$$L_1 = \begin{bmatrix}2.4495 & 0 & 0 & 0 \\ 1.2247 & 2.1213 & 0 & 0 \\ 1.6330 & 1.4142 & 2.3094 & 0 \\ 3.2660 & -1.4142 & 1.5877 & 3.1325 \end{bmatrix}$$Now, if I perform the modified Cholesky, I get:$$L=\begin{bmatrix}5 & 0 & 0 & 0 \\ 1.4 & 2.83549 & 0 & 0 \\ 0.2 & 1.66462 & 1.78579 & 0 \\ 1.6 & 0.62070 & 0.92215 & 1.48471 \end{bmatrix}$$and$$P=\begin{bmatrix}0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}$$Such that, $P\cdot (L\cdot L^\top)\cdot P^\top = A$. Of course, since $A$ is PD, $E=0$. |
The Cartesian product of two topological spaces \(M\)and \(N\) can be thought of as the set of all ordered pairs selected from \(M\) and \(N\). (There's more to the definition than this, but this is what's pertinent for our purposes.) For example, the Cartesian product of two open/closed intervals is an open/closed rectangle. This means that if you're going to tell me that AdS
n has the same topology as \(\mathbb{R}^n \times S^m\), I should always be able to pick a point from the Euclidean space and a point from the m-sphere, and you can show me the point in AdS n that corresponds to that ordered pair.
With that in mind: if you write your constraint as
\((x^0)^2 + (x^1)^2 = \alpha^2 + \vec{x}^2\),
then it's not too hard to see that this defines a circle in the \(x^0\text{-}x^1\) plane
for any value of \(\vec{x}\), with radius \(\sqrt{\alpha^2 + \vec{x}^2}\). This means that if I pick a point on the circle \(S^1\), and pick a point in the Euclidean space \(\mathbb{R}^{n-1}\), then there will always be a unique point in AdS n corresponding to that choice.
On the other hand, if you write
\(\vec{x}^2 = (x^0)^2 + (x^1)^2 - \alpha^2\),
then this does not define an (n-2)-sphere for all possible choices of \(x^0\) and \(x^1\), since the right-hand side can become negative. Thus, your proposed map from ordered pairs in \(\mathbb{R}^2\) and \(S^{n-2}\) to AdS
n doesn't work, since I can choose points from the base spaces that have no corresponding point in AdS n (namely, any point from \(\mathbb{R}^2\) that has \((x^0)^2 + (x^1)^2 < \alpha^2\), and any point at all from \(S^{n-2}\).)
Of course, this flaw in your second argument doesn't prove on its own that AdS
n isn't homeomorphic to \(\mathbb{R}^2 \times S^{n-2}\)—merely that this map isn't a homeomorphism. But since we already found another argument that AdS n was homeomorphic to \(\mathbb{R}^{n-1} \times S^1\), and since these two product spaces aren't homeomorphic to each other (they're topologically distinct, so they can't be), we can conclude that AdS n isn't homeomorphic to \(\mathbb{R}^2 \times S^{n-2}\). (Unless, of course, \(n = 3\).) |
I am trying to calculate the beta weighted delta and gamma for a portfolio of options of different underlying stocks, but I can't seem to find the correct formula.
Can someone point me to it or a book that contains it?
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I've started thinking about this, too. My gedanken conclusion turned out to be too simple once I found what I was after: http://www.investment-and-finance.net/derivatives/o/option-beta.html, which I've confirmed in Black & Scholes (1973) p10 (eq 15).
In short: $$ \beta_{\text{option}} = \frac{S\cdot\Delta}{O}{\beta_S} $$ where $S$ is the underlying price ($x$ in the B&S paper), $O$ is the option price ($w_1$ in B&S), $\Delta$ is the usual $\partial{O}/\partial{S}$, and $\beta_S$ is $\beta$ for the underlying.
Regarding your question, you'd just have to re-arrange this to use an empirically measured option $\beta$, and differentiate for Gamma. I'm not sure that gets you where you want to go, though. |
The aim of this test case is to validate the following parameters of compressible steady-state turbulent flow through a de Laval Nozzle subsonic and supersonic flow regimes:
The
rhoSimpleFoam solver is used for the subsonic case, while for the supersonic case the sonicFoam solver is employed. The k−ω model was used to model turbulence. Simulaton results of SimScale were compared to analytical results obtained from methods elucidated in [1]. The mesh was created locally using the blockMesh tool and then imported on to the SimScale platform.
A typical configuration of the de Laval nozzle with a non-smooth throat was chosen as the geometry (see Table 1 for coordinates). Since the nozzle is axisymmetric, it was modeled as an angular slice of the complete geometry, with a wedge angle of 18 degrees (see Fig.1.).
Point x y z A 0 0 0 B 0 5.5434 35 C 0 -5.5434 35 D 68.68 0 0 E 68.68 3.1677 20 F 68.68 -3.1677 20 G 238.8 6.3354 40 H 238.8 -6.3354 40
A non-uniformly-spaced hexahedral mesh was generated using the blockMesh tool (see Fig.2.). Flow near the nozzle wall was resolved using inflation of y+=30 for the subsonic and 300 for the supersonic case. In order to keep the flow two-dimensional, the mesh was designed to have only one layer in the y direction.
Tool Type : OPENFOAM® Analysis Type : Compressible Steady-state (Turbulent) Mesh and Element types :
Mesh type Cells in x Cells in y Cells in z Number of nodes Type blockMesh 150 1 100 15000 2D hex Fluid:
Table 3 encapsulates the properties of fluids used in the subsonic and supersonic case simulations. The need for using a different fluid for supersonic case arises from the courant number restriction.
Case m(g/mol) cp(J/kgK) mu(N/ms) Pr Subsonic 28.9 1005 1.79×10−5 1 Supersonic 11640.3 2.5 1.8×10−5 1 Boundary Conditions:
Parameter Inlet (ABC) Outlet (GHI) Wall (AEHIFC) Wedges (AGHEBA + AGIFCA) Velocity 7.58 ms−1 Zero Gradient 0.0 ms−1 Wedge Pressure Zero Gradient 1.3×105 Nm−2 Zero Gradient Wedge Temperature 300 K Zero Gradient Zero Gradient Wedge k 0.862 m2/s2 Zero Gradient Wall Function Wedge ω 484.269 s−1 Zero Gradient Wall Function Wedge αt 0(calculated) 0(calculated) Wall Function Wedge μt 0(calculated) 0(calculated) Wall Function Wedge
Parameter Inlet (ABC) Outlet (GHI) Wall (AEHIFC) Wedges (AGHEBA + AGIFCA) Velocity Zero Gradient Zero Gradient 0.0 ms−1 Wedge Pressure 1.2999 Nm−2 Wave Transmissive 0.0296 Nm−2 Zero Gradient Wedge Temperature 1.0388 K Zero Gradient Zero Gradient Wedge k 3.75×10−5 m2/s2 Zero Gradient Wall Function Wedge ω 0.144 s−1 Zero Gradient Wall Function Wedge αt 0(calculated) 0(calculated) Wall Function Wedge μt 0(calculated) 0(calculated) Wall Function Wedge
Results for the subsonic cases are calculated from Bernoulli’s equation and the ideal gas equation as follows:
Subsonic flow does not see a significant temperature rise. So the speed of sound remains almost constant and can be calculated as:
\[c = \sqrt{\frac{\gamma RT}{m}}\]
Here, γ , T, m and R represent the specific heat ratio, temperature, and molecular weight of the fluid, and the universal gas constant respectively. The Mach number can then be calculated as:
\[M = \frac{A_{in}U_{in}}{Ac}\]
Assuming a stagnation pressure of Pstag=0.1301 MPa, the static pressure can be computed as:
\[P_{stat} = P_{stag} – \frac{1}{2}\rho u^2\]
The relation between nozzle cross-section area A and Mach number M is what governs flow characteristics in supersonic flow through a de Laval nozzle [1]:
\[\frac{A}{A_t} = \frac{1}{M}\left[\frac{2}{\gamma + 1}\left(1+\frac{\gamma – 1}{2}M^2\right)\right]^{\frac{\gamma + 1}{2(\gamma – 1)}}\]
Using the known area ratio, the mach number variation is calculated by solving the above equation. The pressure can be calculated using [1]:
\[p = p_0\left(1+\frac{\gamma -1}{2}M^2\right)^{\frac{\gamma}{1-\gamma}}\]
A comparison of the Mach number and pressure variation in the nozzle obtained with SimScale with analytical results is given in Fig.3A and 3B for subsonic and Fig.4A and 4B for supersonic flow.
Fig.3. Visualization of Mach number and pressure (A, B) along the nozzle for subsonic flow
Fig.4. Visualization of Mach number and pressure (A, B) along the nozzle for supersonic flow
The deviation from analytical results exists because the latter is calculated with a one-dimensional hypothesis. Thus, all parameters are assumed to be uniform in the radial direction. Fig.5. shows that this is in fact not the case – there exists radial variation in flow variables. This is one reason why some deviation is seen between the two.
Fig.5. Contours of velocity and temperature in the nozzle. Clearly, there exists radial variation in these parameters.
[1] (1, 2, 3) H. W. Leipmann and A. Roshko: Elements of Gasdynamics. *Courier Corporation (1957)*
This offering is not approved or endorsed by OpenCFD Limited, producer and distributor of the OpenFOAM software and owner of the OPENFOAM® and OpenCFD® trade marks. OPENFOAM® is a registered trade mark of OpenCFD Limited, producer and distributor of the OpenFOAM software. |
I'm not sure about your calculation but Matlab yields the correct result. In this problem, the common approach is to use
Routh-Hurwitz criterion and search for a row of zeros that yields the possibility for imaginary axis roots. For convert the system to the closed-loop transfer function, hence $$\frac{K}{s^4 + 10s^3 + 88s^2 + 256s + K}$$
The Routh table is
$$\begin{matrix}s^4 &&&& 1 &&&& 88 &&&& K \\s^3 &&&& 10 &&&& 256 \\s^2 &&&& 62.4 &&&& K \\s^1 &&&& \frac{15974.4-10K}{62.4} \\s^0 &&&& K \end{matrix}$$
The \$ s^1 \$ row is the only row that can yield a row of zeros. From the preceding row, we obtain
$$\begin{align}&15974.4 - 10K = 0 \\K &= \frac{15974.4}{10} = 1597.44\end{align}$$
Now we take a look at the row above \$s^1\$ and construct the following polynomial, hence
$$\begin{align}62.4 s^2 + K &= 0 \\62.4 s^2 + 1597.44 &= 0 \\ s^2 &= \frac{-1597.44}{62.4} \\s_{1,2} &= \pm j \sqrt{25.6} \\s_{1,2} &= \pm j 5.0596 \\\end{align}$$
The root locus crosses the imaginary axis at \$\pm j5.0596\$ at the gain \$K=1597.44\$. Consequently, the gain \$K\$ must be less than 1597.44 for the system to be stable. |
Are there any known upper bounds on: $$\#\left\{\text{hyperbolic knots }K\subseteq S^3\middle|\operatorname{Vol}(S^3\setminus K)<M\right\}$$ ? I expect this grows at least exponentially in $M$, and I am interested to know whether there also exists an exponential upper bound. If we restrict to alternating knots, then work of Lackenby is relevant.
Here is an expansion of Ian's answer.
Dehn surgery on a hyperbolic knot or link generically gives another hyperbolic manifold. This follows from the Dehn surgery theorem; see Theorem 5.8.2 in chapter five of Thurston's book. Furthermore, the new manifold has smaller hyperbolic volume than the original. See Theorem 6.5.6 in chapter six. Some further discussion and references can be found in Ian's answer here.
So: fix a hyperbolic link $L$ in the three-sphere with an unknotted component $C$. Surgering $L$ along $C$ gives an infinite family of hyperbolic links, all with volume less than that of $L$. For explicit pictures of this, see the last few pages of chapter five of Thurston's book. (Actually, the pictures go in the opposite direction -- he explains how the Whitehead link is the "geometric limit" of a sequence of twist knots.)
This question is so close to two interesting open questions, I thought it might warrant a second answer.
First, as Ian Agol and Sam Need have explained, once the volume $M$ gets above $v_8 \approx 3.66386237671..$ the volume of regular ideal hyperbolic octahedron (aka the volume of the Whitehead link), then there are infinitely many knots in your set. However, it is known that for a fixed volume $v$ the set of hyperbolic manifolds with volume $<v$ can be obtained by filling finitely many hyperbolic manifolds. To give a flavor of this idea, all the twist knots can be obtained by dehn filling one component of the Whitehead link. Furthermore, the Whitehead link is known to be one of the smallest volume (orientable) two-cusped manifolds so there can only be finitely many (orientable) one-cusped manifold for a fixed $\epsilon >0$, there can be only finitely many hyperbolic knot complements (or more hyperbolic generally orientable hyperbolic manifolds) that have volume at most $v_8 -\epsilon$. Finding this number would involve classifying manifolds in this set which do not come from filling the Whitehead link.
Second, one could ask: for a specific volume, how many hyperbolic manifolds have this volume? Hodgson & Masai investigated this question and found examples of closed manifolds that are distinguished by their volume and link complements that share a volume with at least $f(v)$ other manifolds, where $f(v)$ is a function that grows exponentially in $v$. More relevant to your question, Millichamp followed up on this work by considering (amongst a number of other things) lower bounds on how many hyperbolic knot complements have specific given volumes. |
6.2 Moving averages
The classical method of time series decomposition originated in the 1920s and was widely used until the 1950s. It still forms the basis of many time series decomposition methods, so it is important to understand how it works. The first step in a classical decomposition is to use a moving average method to estimate the trend-cycle, so we begin by discussing moving averages.
Moving average smoothing
A moving average of order \(m\) can be written as\[\begin{equation} \hat{T}_{t} = \frac{1}{m} \sum_{j=-k}^k y_{t+j}, \tag{6.1}\end{equation}\]where \(m=2k+1\). That is, the estimate of the trend-cycle at time \(t\) is obtained by averaging values of the time series within \(k\) periods of \(t\). Observations that are nearby in time are also likely to be close in value. Therefore, the average eliminates some of the randomness in the data, leaving a smooth trend-cycle component. We call this an
\(m\)-MA, meaning a moving average of order \(m\).
For example, consider Figure 6.4 which shows the volume of electricity sold to residential customers in South Australia each year from 1989 to 2008 (hot water sales have been excluded). The data are also shown in Table 6.1.
Year Sales (GWh) 5-MA 1989 2354.34 1990 2379.71 1991 2318.52 2381.53 1992 2468.99 2424.56 1993 2386.09 2463.76 1994 2569.47 2552.60 1995 2575.72 2627.70 1996 2762.72 2750.62 1997 2844.50 2858.35 1998 3000.70 3014.70 1999 3108.10 3077.30 2000 3357.50 3144.52 2001 3075.70 3188.70 2002 3180.60 3202.32 2003 3221.60 3216.94 2004 3176.20 3307.30 2005 3430.60 3398.75 2006 3527.48 3485.43 2007 3637.89 2008 3655.00
In the last column of this table, a moving average of order 5 is shown, providing an estimate of the trend-cycle. The first value in this column is the average of the first five observations (1989–1993); the second value in the 5-MA column is the average of the values for 1990–1994; and so on. Each value in the 5-MA column is the average of the observations in the five year window centred on the corresponding year. In the notation of Equation (6.1), column 5-MA contains the values of \(\hat{T}_{t}\) with \(k=2\) and \(m=2k+1=5\). This is easily computed using
There are no values for either the first two years or the last two years, because we do not have two observations on either side. Later we will use more sophisticated methods of trend-cycle estimation which do allow estimates near the endpoints.
To see what the trend-cycle estimate looks like, we plot it along with the original data in Figure 6.5.
Notice that the trend-cycle (in red) is smoother than the original data and captures the main movement of the time series without all of the minor fluctuations. The order of the moving average determines the smoothness of the trend-cycle estimate. In general, a larger order means a smoother curve. Figure 6.6 shows the effect of changing the order of the moving average for the residential electricity sales data.
Simple moving averages such as these are usually of an odd order (e.g., 3, 5, 7, etc.). This is so they are symmetric: in a moving average of order \(m=2k+1\), the middle observation, and \(k\) observations on either side, are averaged. But if \(m\) was even, it would no longer be symmetric.
Moving averages of moving averages
It is possible to apply a moving average to a moving average. One reason for doing this is to make an even-order moving average symmetric.
For example, we might take a moving average of order 4, and then apply another moving average of order 2 to the results. In the following table, this has been done for the first few years of the Australian quarterly beer production data.
Year Quarter Observation 4-MA 2x4-MA 1992 Q1 443 1992 Q2 410 451.25 1992 Q3 420 448.75 450.00 1992 Q4 532 451.50 450.12 1993 Q1 433 449.00 450.25 1993 Q2 421 444.00 446.50 1993 Q3 410 448.00 446.00 1993 Q4 512 438.00 443.00 1994 Q1 449 441.25 439.62 1994 Q2 381 446.00 443.62 1994 Q3 423 440.25 443.12 1994 Q4 531 447.00 443.62 1995 Q1 426 445.25 446.12 1995 Q2 408 442.50 443.88 1995 Q3 416 438.25 440.38 1995 Q4 520 435.75 437.00 1996 Q1 409 431.25 433.50 1996 Q2 398 428.00 429.62 1996 Q3 398 433.75 430.88 1996 Q4 507 433.75 433.75
The notation “\(2\times4\)-MA” in the last column means a 4-MA followed by a 2-MA. The values in the last column are obtained by taking a moving average of order 2 of the values in the previous column. For example, the first two values in the 4-MA column are 451.25=(443+410+420+532)/4 and 448.75=(410+420+532+433)/4. The first value in the 2x4-MA column is the average of these two: 450.00=(451.25+448.75)/2.
When a 2-MA follows a moving average of an even order (such as 4), it is called a “centred moving average of order 4”. This is because the results are now symmetric. To see that this is the case, we can write the \(2\times4\)-MA as follows:\[\begin{align*} \hat{T}_{t} &= \frac{1}{2}\Big[ \frac{1}{4} (y_{t-2}+y_{t-1}+y_{t}+y_{t+1}) + \frac{1}{4} (y_{t-1}+y_{t}+y_{t+1}+y_{t+2})\Big] \\ &= \frac{1}{8}y_{t-2}+\frac14y_{t-1} + \frac14y_{t}+\frac14y_{t+1}+\frac18y_{t+2}.\end{align*}\]It is now a weighted average of observations that is symmetric. By default, the
ma() function in R will return a centred moving average for even orders (unless
center=FALSE is specified).
Other combinations of moving averages are also possible. For example, a \(3\times3\)-MA is often used, and consists of a moving average of order 3 followed by another moving average of order 3. In general, an even order MA should be followed by an even order MA to make it symmetric. Similarly, an odd order MA should be followed by an odd order MA.
Estimating the trend-cycle with seasonal data
The most common use of centred moving averages is for estimating the trend-cycle from seasonal data. Consider the \(2\times4\)-MA: \[ \hat{T}_{t} = \frac{1}{8}y_{t-2} + \frac14y_{t-1} + \frac14y_{t} + \frac14y_{t+1} + \frac18y_{t+2}. \] When applied to quarterly data, each quarter of the year is given equal weight as the first and last terms apply to the same quarter in consecutive years. Consequently, the seasonal variation will be averaged out and the resulting values of \(\hat{T}_t\) will have little or no seasonal variation remaining. A similar effect would be obtained using a \(2\times 8\)-MA or a \(2\times 12\)-MA to quarterly data.
In general, a \(2\times m\)-MA is equivalent to a weighted moving average of order \(m+1\) where all observations take the weight \(1/m\), except for the first and last terms which take weights \(1/(2m)\). So, if the seasonal period is even and of order \(m\), we use a \(2\times m\)-MA to estimate the trend-cycle. If the seasonal period is odd and of order \(m\), we use a \(m\)-MA to estimate the trend-cycle. For example, a \(2\times 12\)-MA can be used to estimate the trend-cycle of monthly data and a 7-MA can be used to estimate the trend-cycle of daily data with a weekly seasonality.
Other choices for the order of the MA will usually result in trend-cycle estimates being contaminated by the seasonality in the data.
Example: Electrical equipment manufacturing
Figure 6.7 shows a \(2\times12\)-MA applied to the electrical equipment orders index. Notice that the smooth line shows no seasonality; it is almost the same as the trend-cycle shown in Figure 6.1, which was estimated using a much more sophisticated method than a moving average. Any other choice for the order of the moving average (except for 24, 36, etc.) would have resulted in a smooth line that showed some seasonal fluctuations.
Weighted moving averages
Combinations of moving averages result in weighted moving averages. For example, the \(2\times4\)-MA discussed above is equivalent to a weighted 5-MA with weights given by \(\left[\frac{1}{8},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{8}\right]\). In general, a weighted \(m\)-MA can be written as \[ \hat{T}_t = \sum_{j=-k}^k a_j y_{t+j}, \] where \(k=(m-1)/2\), and the weights are given by \(\left[a_{-k},\dots,a_k\right]\). It is important that the weights all sum to one and that they are symmetric so that \(a_j = a_{-j}\). The simple \(m\)-MA is a special case where all of the weights are equal to \(1/m\).
A major advantage of weighted moving averages is that they yield a smoother estimate of the trend-cycle. Instead of observations entering and leaving the calculation at full weight, their weights slowly increase and then slowly decrease, resulting in a smoother curve. |
The aim of this test case is to validate the following functions:
The simulation results of SimScale were compared to the analytical results in [Roark]. The meshes used in (A) and (B) were created with the parametrized-tetrahedralization-tool on the SimScale platform and the meshes used in (C) and (D) were meshed with Salome.
The bar has a cross section of 0.05 x 0.05 m
2 and a length of l = 1.0 m. Tool Type : Calculix, Code_Aster Analysis Type : Thermomechanical Mesh and Element types :
Case Mesh type Number of nodes Element type Tool type (A) linear tetrahedral 84 3D isoparametric Calculix (B) quadratic tetrahedral 369 3D isoparametric Calculix (C) linear hexahedral 369 3D isoparametric Calculix (D) quadratic hexahedral 1221 3D isoparametric Calculix (E) linear tetrahedral 84 3D isoparametric Code_Aster (F) quadratic tetrahedral 369 3D isoparametric Code_Aster (G) linear hexahedral 369 3D isoparametric Code_Aster (H) quadratic hexahedral 1221 3D isoparametric Code_Aster
Material:
Constraints (Boundary Conditions):
Loads:
$$\sigma = \Delta T \gamma E$$
The equation used to solve the problem is derived in [Roark]. Inserting the values described in the previous chapter results in the unit stress σ = 24.6 Mpa in the whole beam.
Comparison of the unit stress σ in the beam obtained with SimScale with the results derived from the equations presented in [Roark].
Case [Roark] SimScale Error (%) (A) 24.6 24.6 0.00% (B) 24.6 24.6 0.00% (C) 24.6 24.599 0.004% (D) 24.6 24.599-24.601 ± 0.004% (E) 24.6 24.6 0.00% (F) 24.6 24.6 0.00% (G) 24.6 24.6 0.00% (H) 24.6 24.6 0.00%
Note: The worst results was obtained with mesh (D). This was due to an interpolation error and is shown in the picture below.
[Roark] (1, 2, 3, 4) (2011)”Roark’s Formulas For Stress And Strain, Eighth Edition”, W. C. Young, R. G. Budynas, A. M. Sadegh |
Tagged: determinant Problem 548
An $n\times n$ matrix $A$ is said to be
invertible if there exists an $n\times n$ matrix $B$ such that $AB=I$, and $BA=I$,
where $I$ is the $n\times n$ identity matrix.
If such a matrix $B$ exists, then it is known to be unique and called the
inverse matrix of $A$, denoted by $A^{-1}$.
In this problem, we prove that if $B$ satisfies the first condition, then it automatically satisfies the second condition.
So if we know $AB=I$, then we can conclude that $B=A^{-1}$.
Let $A$ and $B$ be $n\times n$ matrices.
Suppose that we have $AB=I$, where $I$ is the $n \times n$ identity matrix.
Prove that $BA=I$, and hence $A^{-1}=B$.Add to solve later
Problem 452
Let $A$ be an $n\times n$ complex matrix.
Let $S$ be an invertible matrix. (a) If $SAS^{-1}=\lambda A$ for some complex number $\lambda$, then prove that either $\lambda^n=1$ or $A$ is a singular matrix. (b) If $n$ is odd and $SAS^{-1}=-A$, then prove that $0$ is an eigenvalue of $A$.
Add to solve later
(c) Suppose that all the eigenvalues of $A$ are integers and $\det(A) > 0$. If $n$ is odd and $SAS^{-1}=A^{-1}$, then prove that $1$ is an eigenvalue of $A$. Problem 438
Determine whether each of the following statements is True or False.
(a) If $A$ and $B$ are $n \times n$ matrices, and $P$ is an invertible $n \times n$ matrix such that $A=PBP^{-1}$, then $\det(A)=\det(B)$. (b) If the characteristic polynomial of an $n \times n$ matrix $A$ is \[p(\lambda)=(\lambda-1)^n+2,\] then $A$ is invertible. (c) If $A^2$ is an invertible $n\times n$ matrix, then $A^3$ is also invertible. (d) If $A$ is a $3\times 3$ matrix such that $\det(A)=7$, then $\det(2A^{\trans}A^{-1})=2$. (e) If $\mathbf{v}$ is an eigenvector of an $n \times n$ matrix $A$ with corresponding eigenvalue $\lambda_1$, and if $\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_2$, then $\mathbf{v}+\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_1+\lambda_2$.
(
Stanford University, Linear Algebra Exam Problem) Read solution Eigenvalues of Orthogonal Matrices Have Length 1. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue Problem 419 (a) Let $A$ be a real orthogonal $n\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$.
Add to solve later
(b) Let $A$ be a real orthogonal $3\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an eigenvalue. Problem 391 (a) Is the matrix $A=\begin{bmatrix} 1 & 2\\ 0& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 1& 2 \end{bmatrix}$? (b) Is the matrix $A=\begin{bmatrix} 0 & 1\\ 5& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ 4& 3 \end{bmatrix}$? (c) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 0& 2 \end{bmatrix}$?
Add to solve later
(d) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ -1& 4 \end{bmatrix}$? Problem 389 (a) A $2 \times 2$ matrix $A$ satisfies $\tr(A^2)=5$ and $\tr(A)=3$. Find $\det(A)$. (b) A $2 \times 2$ matrix has two parallel columns and $\tr(A)=5$. Find $\tr(A^2)$. (c) A $2\times 2$ matrix $A$ has $\det(A)=5$ and positive integer eigenvalues. What is the trace of $A$?
(
Harvard University, Linear Algebra Exam Problem) Problem 374
Let \[A=\begin{bmatrix}
a_0 & a_1 & \dots & a_{n-2} &a_{n-1} \\ a_{n-1} & a_0 & \dots & a_{n-3} & a_{n-2} \\ a_{n-2} & a_{n-1} & \dots & a_{n-4} & a_{n-3} \\ \vdots & \vdots & \dots & \vdots & \vdots \\ a_{2} & a_3 & \dots & a_{0} & a_{1}\\ a_{1} & a_2 & \dots & a_{n-1} & a_{0} \end{bmatrix}\] be a complex $n \times n$ matrix. Such a matrix is called circulant matrix. Then prove that the determinant of the circulant matrix $A$ is given by \[\det(A)=\prod_{k=0}^{n-1}(a_0+a_1\zeta^k+a_2 \zeta^{2k}+\cdots+a_{n-1}\zeta^{k(n-1)}),\] where $\zeta=e^{2 \pi i/n}$ is a primitive $n$-th root of unity. Problem 363 (a) Find all the eigenvalues and eigenvectors of the matrix \[A=\begin{bmatrix} 3 & -2\\ 6& -4 \end{bmatrix}.\]
Add to solve later
(b) Let \[A=\begin{bmatrix} 1 & 0 & 3 \\ 4 &5 &6 \\ 7 & 0 & 9 \end{bmatrix} \text{ and } B=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 &0 \\ 0 & 0 & 4 \end{bmatrix}.\] Then find the value of \[\det(A^2B^{-1}A^{-2}B^2).\] (For part (b) without computation, you may assume that $A$ and $B$ are invertible matrices.) Problem 338
Each of the following sets are not a subspace of the specified vector space. For each set, give a reason why it is not a subspace.
(1) \[S_1=\left \{\, \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \in \R^3 \quad \middle | \quad x_1\geq 0 \,\right \}\] in the vector space $\R^3$. (2)\[S_2=\left \{\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \in \R^3 \quad \middle | \quad x_1-4x_2+5x_3=2 \,\right \}\] in the vector space $\R^3$.
(3)\[S_3=\left \{\, \begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2 \quad \middle | \quad y=x^2 \quad \,\right \}\] in the vector space $\R^2$.
(4)Let $P_4$ be the vector space of all polynomials of degree $4$ or less with real coefficients.
\[S_4=\{ f(x)\in P_4 \mid f(1) \text{ is an integer}\}\] in the vector space $P_4$.
(5)\[S_5=\{ f(x)\in P_4 \mid f(1) \text{ is a rational number}\}\] in the vector space $P_4$. (6)Let $M_{2 \times 2}$ be the vector space of all $2\times 2$ real matrices.
\[S_6=\{ A\in M_{2\times 2} \mid \det(A) \neq 0\} \] in the vector space $M_{2\times 2}$.
(7)\[S_7=\{ A\in M_{2\times 2} \mid \det(A)=0\} \] in the vector space $M_{2\times 2}$.
(
Linear Algebra Exam Problem, the Ohio State University) (8)Let $C[-1, 1]$ be the vector space of all real continuous functions defined on the interval $[a, b]$.
\[S_8=\{ f(x)\in C[-2,2] \mid f(-1)f(1)=0\} \] in the vector space $C[-2, 2]$.
(9)\[S_9=\{ f(x) \in C[-1, 1] \mid f(x)\geq 0 \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$. (10)Let $C^2[a, b]$ be the vector space of all real-valued functions $f(x)$ defined on $[a, b]$, where $f(x), f'(x)$, and $f^{\prime\prime}(x)$ are continuous on $[a, b]$. Here $f'(x), f^{\prime\prime}(x)$ are the first and second derivative of $f(x)$.
\[S_{10}=\{ f(x) \in C^2[-1, 1] \mid f^{\prime\prime}(x)+f(x)=\sin(x) \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$.
(11)Let $S_{11}$ be the set of real polynomials of degree exactly $k$, where $k \geq 1$ is an integer, in the vector space $P_k$. (12)Let $V$ be a vector space and $W \subset V$ a vector subspace. Define the subset $S_{12}$ to be the complementof $W$,
\[ V \setminus W = \{ \mathbf{v} \in V \mid \mathbf{v} \not\in W \}.\] Add to solve later |
During October (2017) I will write a program per day for some well-known numerical methods in both Python and Julia. It is intended to be an exercise then don't expect the code to be good enough for real use. Also, I should mention that I have almost no experience with Julia, so it probably won't be idiomatic Julia but more Python-like Julia.
Finite Difference: Eigenproblems
Today we have a Finite difference method to compute the vibration modes of a beam. The equation of interest is
with
Following are the codes.
Python from __future__ import division, print_function import numpy as np from scipy.linalg import eigh as eigsh import matplotlib.pyplot as plt def beam_FDM_eigs(L, N): x = np.linspace(0, L, N) dx = x[1] - x[0] stiff = np.diag(6*np.ones(N - 2)) -\ np.diag(4*np.ones(N - 3), -1) - np.diag(4*np.ones(N - 3), 1) +\ np.diag(1*np.ones(N - 4), 2) + np.diag(1*np.ones(N - 4), -2) vals, vecs = eigsh(stiff/dx**4) return vals, vecs, x N = 1001 nvals = 20 nvecs = 4 vals, vecs, x = beam_FDM_eigs(1.0, N) #%% Plotting num = np.linspace(1, nvals, nvals) plt.rcParams["mathtext.fontset"] = "cm" plt.figure(figsize=(8, 3)) plt.subplot(1, 2, 1) plt.plot(num, np.sqrt(vals[0:nvals]), "o") plt.xlabel(r"$N$") plt.ylabel(r"$\omega\sqrt{\frac{\lambda}{EI}}$") plt.subplot(1, 2 ,2) for k in range(nvecs): vec = np.zeros(N) vec[1:-1] = vecs[:, k] plt.plot(x, vec, label=r'$n=%i$'%(k+1)) plt.xlabel(r"$x$") plt.ylabel(r"$w$") plt.legend(ncol=2, framealpha=0.8, loc=1) plt.tight_layout() plt.show() Julia using PyPlot function beam_FDM_eigs(L, N) x = linspace(0, L, N) dx = x[2] - x[1] stiff = diagm(6*ones(N - 2)) - diagm(4*ones(N - 3), -1) - diagm(4*ones(N - 3), 1) + diagm(1*ones(N - 4), 2) + diagm(1*ones(N - 4), -2) vals, vecs = eig(stiff/dx^4) return vals, vecs, x end N = 1001 nvals = 20 nvecs = 4 vals, vecs, x = beam_FDM_eigs(1.0, N) #%% Plotting num = 1:nvals # Missing line for setting the math font figure(figsize=(8, 3)) subplot(1, 2, 1) plot(num, sqrt.(vals[1:nvals]), "o") xlabel(L"$N$") ylabel(L"$\omega\sqrt{\frac{\lambda}{EI}}$") subplot(1, 2 ,2) for k in 1:nvecs vec = zeros(N) vec[2:end-1] = vecs[:, k] plot(x, vec, label="n=$(k)") end xlabel(L"$x$") ylabel(L"$w$") legend(ncol=2, framealpha=0.8, loc=1) tight_layout() show()
Both have (almost) the same result, as follows
Comparison Python/Julia
Regarding number of lines we have: 40 in Python and 39 in Julia. The comparisonin execution time is done with
%timeit magic command in IPython and
@benchmark in Julia.
For Python:
with result
For Julia:
with result
BenchmarkTools.Trial: memory estimate: 99.42 MiB allocs estimate: 55 -------------- minimum time: 665.152 ms (17.14% GC) median time: 775.441 ms (21.76% GC) mean time: 853.401 ms (16.86% GC) maximum time: 1.236 s (15.68% GC) -------------- samples: 6 evals/sample: 1
In this case, we can say that the Python code is roughly 4 times faster than Julia. |
TABLE OF CONTENTS 1. /barBar_BarHypKaon 1.1. barBar_BarHypKaon/enable 1.2. barBar_BarHypKaon/parameter_set 1.3. barBar_BarHypKaon/a 1.4. barBar_BarHypKaon/BB_BYK 1.5. barBar_BarHypKaon/barBar_barBarMeson_strange 1.6. barBar_BarHypKaon/xs_param 1.7. barBar_BarHypKaon/get_Channels_BYK 1.8. barBar_BarHypKaon/Get_Charge 1.9. barBar_BarHypKaon/Get_ID 1.10. barBar_BarHypKaon/get_ChannelParameters /barBar_BarHypKaon [ Modules ]
NAME
module
barBar_BarHypKaon
PURPOSE
Administrates the calculation of cross sections for strangeness production from BB->BYK collisions (with B=N,Delta and Y=Lambda,Sigma,Lambda*,Sigma*). The properties of the final-state channels are calculated as well.
NOTES
All isospin channels are included. Ref.: K. Tsushima et al., PRC59 (1999) 369. The original Tsushima model was extended and updated in arXiv:1404.7011. Useful for reactions near and above threshold (up to sqrt(s)=3.6 GeV), not appropriate for sqrt(s) far below threshold (cross section does not fit well existing data & perturbative treatment of kaon production necessary). barBar_BarHypKaon/enable [ Global module-variables ]
SOURCE
logical, save ::
enable = .true.
PURPOSE
Enable the production of BB -> B Hyperon Kaon channels. B=Nucleon^{0,1},Delta^{-,0.+,++}; Hyperon=Lambda^{0},Sigma^{0,-,+}; Kaon=K^{+,0}
barBar_BarHypKaon/parameter_set [ Global module-variables ]
SOURCE
integer, save :: parameter_set= 2
PURPOSE
Select a particular parameter set for BB->BYK collisions. Possible values:
1 = original Tsushima model: Tsushima et al., PRC59 (1999) 369 2 = extended/adjusted model, fitted to HADES data: Agakishiev et al., arXiv:1404.7011 3 = custom parameters based on Tsushima values (as given by the array 'a' in the jobcard; those values not given in the jobcard are adopted from Tsushima, i.e. parameter set 1) 4 = custom parameters based on HADES values (as given by the array 'a' in the jobcard; those values not given in the jobcard are adopted from HADES, i.e. parameter set 2) barBar_BarHypKaon/a [ Global module-variables ]
SOURCE
real, dimension(1:Nch), save :: a= -1.
PURPOSE
This array contains the "
a" parameters (in microbarn) for the 30 primary channels, see: Tsushima et al., PRC59 (1999) 369, table III Agakishiev et al., arXiv:1404.7011, chapter IV
Note: The values given in the jobcard are only used for parameter_set = 3 and 4.
barBar_BarHypKaon/BB_BYK [ Namelists ]
NAME
NAMELIST
BB_BYK
PURPOSE
Includes the switches:
barBar_BarHypKaon/barBar_barBarMeson_strange [ Functions ]
NAME
function
barBar_barBarMeson_strange (srts, teilchenIN) result (sigma_BYK)
PURPOSE
XSections for strangeness production from BB->BYK collisions.
INPUTS
real :: srts -- sqrt(s) of the incoming channel type(particle),dimension(1:2) :: teilchenIN -- the incoming particles
OUTPUT
real, dimension(1:18) :: sigma_BYK -- cross section for each channel (mb)
NOTES
This routine calculates also the properties of the finalstates, needed for the construction of the Type(PreEvent) variable. B stands for nucleon or a \Delta or a N* resonance. Other resonances not included. N* resonances are trated as nucleons. In RMF mode sqrt(s*) already contains the self energy of the incoming channel. See its definition in generateFinalState in master_2Body.f90-module variable "srtS_XS". barBar_BarHypKaon/xs_param [ Functions ]
NAME
real function
xs_param (s, j)
PURPOSE
Calculate the parametrized cross section for the primary channels (without isospin factor).
INPUTS
real, intent(in) :: s --- Mandelstam s in GeV**2 integer, intent(in) :: j --- channel number
OUTPUT
Cross section in mb. barBar_BarHypKaon/get_Channels_BYK [ Functions ]
NAME
function
get_Channels_BYK (InChan) result (parts)
PURPOSE
Defines the elements of the type(PreEvent) function
INPUTS
integer, intent(in) :: InChan -- Number of final channel
OUTPUT
type(preEvent), dimension(1:3) :: parts
barBar_BarHypKaon/Get_Charge [ Functions ]
NAME
function
Get_Charge
PURPOSE
Convert the variables (between GiBUU & Munich-RBUU) related to the isospin states.
INPUTS
NOTES
Only for nucleons, Delta's, N*'s, hyperons (Lambda,Sigma) and kaons (K^{+,0}).
barBar_BarHypKaon/Get_ID [ Functions ]
NAME
Integer Function
Get_ID (species, iso)
PURPOSE
Convert identification number (ID) between GiBUU and Munich-RBUU.
INPUTS
NOTES
Only for nucleons, Delta's, N*'s, hyperons (Lambda,Sigma) and kaons (K^{+,0}).
barBar_BarHypKaon/get_ChannelParameters [ Subroutines ]
NAME
subroutine
get_ChannelParameters
PURPOSE
Returns the number of final states & their properties.
INPUTS
integer :: iso1 -- charge of incoming particle Nr 1 integer :: iso2 -- charge of incoming particle Nr 2 integer :: isot -- total charge of incoming channel
OUTPUT
integer :: nexit -- Number of final states for BB-->BYK^{+} integer :: iexit -- Relation to table (see above) integer :: fac -- isospin factor for each final state in BYK^{+} integer :: isohf -- isospin-state of hyperon Y in BYK^{+} integer :: isobf -- isospin-state of the baryon B in BYK^{+} integer :: nexit0 -- Number of final states for BB-->BYK^{0} integer :: iexit0 -- Relation to table (see above) integer :: fac0 -- isospin factor for each final state in BYK^{0} integer :: isohf0 -- isospin-state of hyperon Y in BYK^{0} integer :: isobf0 -- isospin-state of the baryon B in BYK^{0}
NOTES
Notation: B : p,n,\DELTA^{-},\DELTA^{0},\DELTA^{+},\DELTA^{++} Y : \Lambda^{0},\Sigma^{-},\Sigma^{0},\Sigma^{+} K : Kaon^{+},Kaon^{0} This is a partial copy of the Munich-RBUU code. For this reason we use here - and only here - local variables for the different isospin-states of the considered particle species, which may differ from those used in the GiBUU code. Please, DO NOT USE THE FOLLOWING PRESCRIPTIONS OUTSIDE OF THIS ROUTINE!!! Relation between local variables and baryon/meson isospin states, used in this routine: PROTONS = 1 NEUTRONS = 0 \DELTA^{-} = 10 \DELTA^{0} = 11 \DELTA^{+} = 12 \DELTA^{++} = 13 N*^{0} = 20 N*^{+} = 21 \Lambda^{0} = 0 \Sigma^{-} = 1 \Sigma^{0} = 2 \Sigma^{+} = 3 Kaon^{+} = 1 Kaon^{0} = 0 |
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Mixture A is 15 percent alcohol, and mixture B is 50 percent[#permalink]
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Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?
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The Question: Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?[/color] A. 1.5 B. 1.7 C. 2.3 D. 2.5 E. 3.0
You can treat this question as wighted average problem: \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(0.3=\frac{0.15*A+0.5(4-A)}{4}\) --> \(A\approx{2.3}\).
Answer: C.
Also you can notice that as wighted average (alcohol share) of final mixture (30%) is closer to that of mixture A (15%) than to that of mixture B (50%) then there should be more of mixture A in final solution of 4 gallons than of mixture B, so answer choices A and B are out right away. Plus, if in final mixture there were equal amounts of mixtures A and B then the final solution would have (15%+50%)/2=32.5% of alcohol, and as 32.5% is a little more than 30% (actual concentration) then there should be a little more of mixture A than mixture B in 4 gallons, answer choice C fits best.
Re: Mixture Problem (What is the best way to approach this?)[#permalink]
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14 Jan 2011, 07:37
2
1
Hi, It is always better to approach this problem as a simple word problem rather than a Problem of Mixtures. Word problems should be framed in to simple equations. Here let us assume mixture A is x gallons and mixture B is Y gallons.
Number of gallons of alcohol in A = 0.15x (Mixture A is 15 percent alcohol) Number of gallons of alcohol in B = 0.5Y(mixture B is 50 percent alcohol) Number of gallons of alcohol in a mixture formed by mixing A and B i.e in a mixture of x+y gallons =0.3(x+y)
Hence equating the alcohol content we get the equation 0.15x+0.5y=0.3(x+y) We also have x+y=4(two Mixtures are poured together to create a 4-gallon mixture) Here you can solve the two equations to get X, else substitute the choices. While substitution it is always preferable to start substituting from a median of options(Here 2.3 is median among the set of answers 1.5,1.7,2.3,2.5,3.0). This approach helps us to eliminate other two options based on the result we get by substituting median value, by either moving up or down.Here 2.3 satisfies our equation. Hope this helps.
Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent[#permalink]
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14 Jul 2013, 00:19
Nobody wrote:
Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?[/color]
A. 1.5 B. 1.7 C. 2.3 D. 2.5 E. 3.0
All the good methods are already outlined above: Another way of doing this problem, without the need to write down anything:
Assume that both Mixture A and Mixture B are in equal proportion \(\to\) The total amount of alcohol would be 50 % of 2 gallons+15% of 2 gallons \(\to\)1+0.3 = 1.3 gallons. As per the problem, we need only 1.2 gallons. Thus, to account for this extra 0.1 gallons, and we can safely assume that the mixture which has more percentage of alcohol would be lesser, and Mixture A will be a little more than 2 gallons
Now, if we are confused between 2.3 or 2.5, we can always back calculate :
For eg, assuming that 2.5 is the correct answer, amount of Mixture B = 4-2.5 = 1.5 gallons \(\to\) 50% of this is 0.75 gallons.
Now, we know that 10% of 2.5 gallons can be calculated in a jiffy and it equals 0.25. Thus, 15% would be 10%+5% \(\to\) 0.25+0.125 = 0.375 and the net alcohol content stands at 1.125 which is less than 1.2 gallons.Hence, we have to increase the alcohol content a little, and this can be done by increasing amount of Mixture B. Thus, the correct answer is C._________________
Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent[#permalink]
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05 Jul 2015, 05:22
Bunuel wrote:
Nobody wrote:
The Question: Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?[/color] A. 1.5 B. 1.7 C. 2.3 D. 2.5 E. 3.0
You can treat this question as wighted average problem: \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(0.3=\frac{0.15*A+0.5(4-A)}{4}\) --> \(A\approx{2.3}\).
Answer: C.
Also you can notice that as wighted average (alcohol share) of final mixture (30%) is closer to that of mixture A (15%) than to that of mixture B (50%) then there should be more of mixture A in final solution of 4 gallons than of mixture B, so answer choices A and B are out right away. Plus, if in final mixture there were equal amounts of mixtures A and B then the final solution would have (15%+50%)/2=32.5% of alcohol, and as 32.5% is a little more than 30% (actual concentration) then there should be a little more of mixture A than mixture B in 4 gallons, answer choice C fits best.
Hope it's clear.
Why can't we use this formula :
0,15A + 0,5B = 0,3 A + B = 4
and then solve for A ?
I have difficulties understanding when to use: 0,15A + 0,5B = 0,3 (A+B) and when to use the formula above.
Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent[#permalink]
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05 Jul 2015, 05:33
LaxAvenger wrote:
Bunuel wrote:
Nobody wrote:
The Question: Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?[/color] A. 1.5 B. 1.7 C. 2.3 D. 2.5 E. 3.0
You can treat this question as wighted average problem: \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(0.3=\frac{0.15*A+0.5(4-A)}{4}\) --> \(A\approx{2.3}\).
Answer: C.
Also you can notice that as wighted average (alcohol share) of final mixture (30%) is closer to that of mixture A (15%) than to that of mixture B (50%) then there should be more of mixture A in final solution of 4 gallons than of mixture B, so answer choices A and B are out right away. Plus, if in final mixture there were equal amounts of mixtures A and B then the final solution would have (15%+50%)/2=32.5% of alcohol, and as 32.5% is a little more than 30% (actual concentration) then there should be a little more of mixture A than mixture B in 4 gallons, answer choice C fits best.
Hope it's clear.
Why can't we use this formula :
0,15A + 0,5B = 0,3 A + B = 4
and then solve for A ?
I have difficulties understanding when to use: 0,15A + 0,5B = 0,3 (A+B) and when to use the formula above.
Hi lax, it is this very formula that has been used earlier too.. your two eqs are: \(0.15A + 0.5B = 0.3(A+B)\) A + B = 4 or B=4-A.... substitue value of B and you will get \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(0.3=\frac{0.15*A+0.5(4-A)}{4}\)_________________
Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent[#permalink]
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05 Jul 2015, 06:10
Why can't we use this formula :
0,15A + 0,5B = 0,3 A + B = 4
and then solve for A ?
I have difficulties understanding when to use: 0,15A + 0,5B = 0,3 (A+B) and when to use the formula above.[/quote]
Hi lax, it is this very formula that has been used earlier too.. your two eqs are: \(0.15A + 0.5B = 0.3(A+B)\) A + B = 4 or B=4-A.... substitue value of B and you will get \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(0.3=\frac{0.15*A+0.5(4-A)}{4}\)[/quote]
If I solve the eq 0,15A + 0,5B = 0,3 A + B = 4
I end up getting: -1.7 = -0.35A -> A = 4.8
so I do not understand why for some problems this formula gives correct results and for some we have to use another ..
Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent[#permalink]
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05 Jul 2015, 06:52
LaxAvenger wrote:
Why can't we use this formula :
0,15A + 0,5B = 0,3 A + B = 4
and then solve for A ?
I have difficulties understanding when to use: 0,15A + 0,5B = 0,3 (A+B) and when to use the formula above.
Hi lax, it is this very formula that has been used earlier too.. your two eqs are: \(0.15A + 0.5B = 0.3(A+B)\) A + B = 4 or B=4-A.... substitue value of B and you will get \(weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}\) --> \(0.3=\frac{0.15*A+0.5(4-A)}{4}\)[/quote]
If I solve the eq 0,15A + 0,5B = 0,3 A + B = 4
I end up getting: -1.7 = -0.35A -> A = 4.8
so I do not understand why for some problems this formula gives correct results and for some we have to use another ..[/quote]
Hi LaxAvenger, where you are going wrong is in the formation of equation.. 0,15A + 0,5B = 0,3.... 0.3 of what? 0.3 of final mix that is equal to A+B.. so the eq will become 0,15A + 0,5B = 0,3(A+B) now try and you will get the correct answer.._________________
Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent[#permalink]
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12 Apr 2017, 10:10
Nobody wrote:
Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?
A. 1.5 B. 1.7 C. 2.3 D. 2.5 E. 3.0
.15A+.5B=.3(A+B) A/B=4/3 A/(A+B)=4/7 4/7*4 gallons=2.3 gallons C
Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent[#permalink]
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16 Jan 2019, 05:35
Nobody wrote:
Mixture A is 15 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 4-gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?
A. 1.5 B. 1.7 C. 2.3 D. 2.5 E. 3.0
Let´s see an "alternate" way to apply the ALLIGATION method:
Re: Mixture A is 15 percent alcohol, and mixture B is 50 percent[#permalink]
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16 Jan 2019, 12:24
Hi All,
We're told that Mixture A is 15 percent alcohol and mixture B is 50 percent alcohol and when a certain amount of each is poured together to create a 4-gallon mixture, the result contains 30 percent alcohol. We're asked approximately how many gallons of mixture A are in the mixture. This "mixture" question can be approached in a number of different ways, including by TESTing THE ANSWERS.
To start, IF we had an EQUAL amount of each Mixture, then the average of the 4-gallon mixture would be (15 + 50)/2 = 65/2 = 32.5% alcohol. This result is TOO HIGH though; the actual mixture is 30%. This means that we'll need a little more of Mixture A... meaning that a little more than 2 gallons (of the 4-gallon total) will be Mixture A. Looking at the answer choices, it's highly likely that the correct answer is Answer C.
We can actually check to see if either Answer C is a match or if Answer D is "too small." We're looking for an answer that would lead to (4)(.3) = 1.2 gallons of alcohol. Answer D is a little 'easier' to work with, so let's TEST that Answer....
Answer D: 2.5 gallons
IF.... we have 2.5 gallons of Mixture A, then we have 4 - 2.5 = 1.5 gallons of Mixture B... (2.5)(.15) + (1.5)(.5) = .375 + .75 = 1.125 gallons of Alcohol. This answer is TOO SMALL (we need there to be 1.2 gallons of alcohol). To get more alcohol we need LESS of Mixture A than we have here. There's only one answer that fits... |
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Hi,
I reinstalled Cadabra, but now the previous notebooks no longer work. It looks like something breaks at the "Depends" property. I use "Depends" on indices like
{\dot{#}, \bar{#}}::Symbol;
{\alpha, \beta, \gamma, \delta}::Indices(chiral, position=fixed);
{\dalpha, \dbeta, \dgamma, \ddelta}::Indices(antichiral, position=fixed);
followed by
\theta{#}::Depends{\alpha, \beta, \gamma, \delta, \dalpha, \dbeta, \dgamma, \ddelta};
which gives me an error like
RuntimeError: Depends: \prod lacks property Coordinate, Derivative, Accent or Indices.
In 2.x, make sure to write dependence on a derivative as A::Depends(\partial{#}), note the '{#}'.
I didn't get to the place where I had the problem with the other "accented" symbols, but I will try to see if now it works or not. |
I'm struggling with understanding how context free languages can be closed under union but are not closed under intersection. I was wondering if there was a simple proof or example demonstrating that CFLs are not closed under intersection.
Let us assume 2 CFLs L1 and L2 and their corresponding grammars be S1 and S2 respectively. It is very straightforward to see that the union of the two, represented by the new grammar as:
S -> S1 | S2
is also a CFG, as the rule of being context-free is still not violated.Context-free grammar
But to prove that they are not closed under intersection, I'll provide an example.
Let L1 and L2 respectively be:
L1 = { a
nb nc m | n, m >= 0 }
L2 = { a
mb nc n | n, m >= 0 }
It is not hard to see that:
L, which is L1 ∩ L2:
L = { a
nb nc n | n, m >= 0 } is not CFL ( a PDA cannot be created ).
Hope this explains.
Complementing the counterexample in the other answer, let me also what mentions when you try to construct a machine accepting the intersection of two context-free languages. Recall that a context-free language can be accepted by a pushdown automaton. Hence we can try to simulate two pushdown automata operating simultaneously. Since each PDA requires a stack, altogether we need to use two stacks. However, using two stacks we can simulate a Turing machine (exercise), so in general two stacks cannot be implemented in a single stack. Now it might be the case that in this particular case this reduction in the number of stacks is possible, but this hope is shattered by the example given by Akash Mahapatra.
Similar considerations apply for the
quotient of two context-free languages $L_1/L_2 = \{ x \in \Sigma^* : \exists y \in L_2 \, xy \in L_1 \}$. In this case one can show that every r.e. language can be represented as the quotient of two context-free languages. Hence this particular way of using two stacks is as general as possible. The same doesn’t hold for intersection, since the intersection of any (finite) number of context-free languages is always recursive, indeed belongs to the much smaller class LOGCFL. |
We know if every language in $EXP$ has polysize circuits, then $P\neq NP$ and $EXP=PSPACE=\Sigma_2^P\cap\Pi_2^P$.
If every language in $PSPACE$ has polysize circuits, then does it give anything (note if it gave does $P\neq PSPACE$ then this would give $P\neq PSPACE$ unconditionally since $P\subseteq P/poly$ and $P=PSPACE\implies PSPACE\subseteq P/poly\implies P\neq PSPACE$ a contradiction)?
If every language in $PSPACE$ is in $BPP$ then $BPP=\Sigma_2^P\cap\Pi_2^P=PSPACE\neq EXP\not\subseteq P/poly$.
If every language in $PSPACE$ is in $RNC$ then $NC\neq P=PSPACE=RNC=BPP=\Sigma_2^P\cap\Pi_2^P\neq EXP$ (note $NC=PSPACE$ is not possible unconditionally).
If every language in $PSPACE$ is in $BPP$ or is in $RNC$ do they give anything beyond? |
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