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# A Gorenstein Ring with Larger Dilworth Number than Sperner Number

A counterexample is given to a conjecture of Ikeda by finding a class of Gorenstein rings of embedding dimension $3$ with larger Dilworth number than Sperner number. The Dilworth number of $A[Z/pZ\oplus Z/pZ]$ is computed when $A$ is an unramified principal Artin local ring.

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# Where to submit my proof that the set of real numbers can't be well ordered

## Recommended Posts

1 hour ago, discountbrains said:

I need to generalize my <* ordering. It needs to be an ordering on the whole set R because it has to be the possibility of any ordering. And, it certainly could be that a<*r<*c<* b<*....In fact, any number of elements not in T could be mixed up in there, but that doesnt mean there is a z in T with z≤*x for all x in T. I might write a typical ordering of mine as

....<* a<*r<*c<* b<*p<*... where r and p are not in R.

wtf, do you mean that Cantor rational number thing? 1    2      3      4 ...

1/2 2/2 3/2 ....

1/3 2/3 3/3....

.       .         .

This, of course, is where you start in the upper left corner and count back and forth diagonally showing thereś a one to one onto relationship to the natural numbers and doesnt apply to what im saying. Clearly this is a WO set. I dont know why u asked this. I was going to show what I think is a way to reorder any set of real numbers in a dense or çontinuous way, but will do that later. Im going to take a break from this now.This kinda gets to me after a while. If anyone has any comments or questions I will still read them.

> I need to generalize my <* ordering. It needs to be an ordering on the whole set R because it has to be the possibility of any ordering.

What does that mean, it has to be the "possibility of any ordering?" It's not even clear what you are trying to do. In order to make progress my suggestion to you would be to force yourself to write more clearly, one line at a time, and make sure each line makes sense. You don't have to write here, do it for yourself. Your ideas are jumbled because your prose is jumbled. That's why they make you write proofs in math class. Clear exposition leads to clear thinking.

> wtf, do you mean that Cantor rational number thing?

Yes, the fact that the natural numbers can be placed into bijection with the rationals.

>  I dont know why u asked this.

Because you have to walk before you can run. You are trying to investigate or understand the relation between the usual dense order on the reals, with the well-order on the reals. Now the reals are a very complicated set. There is a much simpler set we have lying around that has the same phenomenon. You should study it to try to gain insight into the analogous problem in the reals

Specifically, the usual order on the rationals is dense, just like the usual order on the reals is dense. Now you want to find a well-order. We know we can re-order the rationals so that they are well-ordered. This is a perfect analogy for what you are trying to do. That's why I suggested that you study it. It's so you can gain insight into the analogous problem for the reals: How to relate a well-order to a dense order on the same set.

It's a standard technique in math, when you are working on a hard problem, to look at simpler examples.

> I was going to show what I think is a way to reorder any set of real numbers in a dense or çontinuous way

But why? Isn't the usual order already dense or continuous? You don't have to work very hard to find a dense order on the reals, the usual order is dense.

> Im going to take a break from this now.This kinda gets to me after a while.

I hope you got your money's worth from the convo. You should definitely take another run at learning about the ordinals. The ordinals are all about well-ordered sets. If you are interested in well-orders you are by definition interested in the ordinal numbers, even if you don't know it.

I'll check this site from time to time in case you have more questions or comments. All the best, nice chatting with you.

By the way I have a really cool book recommendation for you. Infinity and the Mind by Rudy Rucker. https://www.amazon.com/Infinity-Mind-Philosophy-Infinite-Princeton/dp/0691121273

Edited by wtf


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# Show that $f$ is integrable using Archimedes-Riemann theorem

Define $$f(x)=\begin{cases} x, 2 \leq x \leq 3 \\ 2, 3 < x \leq 4 \end{cases}$$ Show that $$f$$ is integrable on $$[2,4]$$ using Archimedes-Riemann theorem (Hint: find an Archimedian sequence of partitions that all have 3 as a partition point).

My attempt: I tried to use the regular partition, but it skips the point 3, so I'm thinking of a partition with $$x_i-x_{i-1}=\frac{1}{n}.$$ Then $$m_i=\inf\{f(x)|x in [x_{i-1},x_i]\}=x_{i-1}, i =1,..., n, m_i = 2, i = n+1,...,2n.$$ $$M_i=\sup\{f(x)|x in [x_{i-1},x_i]\}=x_{i}, i =1,..., n, M_i = 2, i = n+1,...,2n.$$ Therefore, $$L(f,P_n)=\sum_{i=1}^{2n}m_i(x_i-x_{i-1})=2*\frac{1}{n}+(2+\frac{1}{n})*\frac{1}{n}+...+(2+\frac{n-1}{n})*\frac{1}{n}+2*\frac{1}{n}+...+2*\frac{1}{n}=\frac{9n-1}{2n}.$$ Similarly, $$U(f,P_n)=\sum_{i=1}^{2n}M_i(x_i-x_{i-1})=\frac{9n+3}{2n}.$$

$$\lim_{n \rightarrow \infty} L(f,P_n)=\lim_{n \rightarrow \infty} U(f,P_n)=\frac{9}{2},$$ so $$f$$ is integrable.

Is it correct? I noticed that what I've done actually is showing that $$x$$ is integrable on subinterval $$[2,3]$$ and 2 is integrable on $$[3,4]$$, so I'm kind of confused and I suppose this is not the right approach. Maybe I should've chosen another sequence of partitions... Correct me if I'm wrong.

• I've never heard of an "Archimedes-Riemann" theorem - I suspect the terminology is unique to your textbook. What exactly does it say? – jmerry Mar 3 at 6:11
• @jmerry . Ditto for me. Riemann and Archimedes were about 21 centuries apart. – DanielWainfleet Mar 3 at 7:36

I noticed that what I've done actually is showing that $$x$$ is integrable on subinterval $$[2,3]$$ and $$2$$ is integrable on $$[3,4]$$, so I'm kind of confused and I suppose this is not the right approach.
Theorem: Given $$a, a function $$f$$ is (Riemann) integrable on $$[a,c]$$ if and only if it is integrable on both $$[a,b]$$ and $$[b,c]$$, and $$\int_a^c f = \int_a^b f+\int_b^c f$$.
One quibble: $$M_{n+1}=\sup\{f(x)|x\in [x_n,x_{n+1}]\}$$ is equal to $$3$$, not $$2$$. After all, the lower endpoint of that interval is $$3$$, and $$f(3)=3$$. Fortunately, that only applies to a single subinterval, with a width that goes to zero.


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# Relation between a solution of Eikonal equation and curvature.

## Is there a relation between a solution of Eikonal equation and curvature?

Imagine someone hands you a local solution $\phi$ of Eikonal equation, $\left\| \nabla \phi \right\|_g = 1$, on Riemannian manifold $(M,g)$. Can you say me something about the curvature?

A more precise formulation:

Let $(M,g)$ be a two dimensional Riemannian manifold. Let $p\in M$ be a point and $v\in T_pM$ a direction at $p$. Now assume that we have a neighborhood $U$ of the point $p$, a geodesic $\gamma\subset U$ going through the point $p$ in the direction $v$ and a smooth solution $\phi$ to Eikonal equation, $\left\| \nabla \phi \right\|_g = 1$, on $U$ satisfying $\phi(x) = 0$ for $x \in \gamma$. Can we say something about the Riemann curvature tensor?

## Few observations:

Denote $u = \nabla \phi$. Then we know that $u \cdot v = 0$. Also by taking derivative of Eikonal equation we get $$\nabla^2 \phi \cdot \nabla \phi = 0.$$ The second derivative of $\phi$ has the following general form $$\nabla^2 \phi = a u\otimes u + b v \otimes v + \frac{c}{2}(u\otimes v + v \otimes u)$$ where $a,b,c$ are some scalars. Combining $\nabla^2 \phi \cdot \nabla \phi = 0$ and $u \cdot v = 0$ we deduce that $a=c=0$.

What can be say about $b$? Isn't it by any chance the Gaussian curvature?

Flat case: When the manifold is just $\mathbb{R}^2$ then the geodesics are just straight lines and the solution $\phi$ is just a affine function which has zero second derivatives.

Additional question: Can you generalize the question for manifolds with arbitrary dimension? The problem is I do not know how to set up the boundary condition of the Eikonal equation.

If $c$ is a geodesic of finite length, define $$X:=\{ \exp_{c(t)}\ \epsilon x(t)||t|\leq \epsilon_2\}$$ where $x$ is a unit parallel vector field along $c$ and $g(c'(t),x(t))=0$. Then define $$f(p)={\rm dist}\ (X,p)-\epsilon$$

Then $f|c=0$ and $|\nabla f|=1$ And since geodesic $c$ in smooth manifold is smooth, so $X,\ f$ are smooth. Hence we do not know curvature.

[Add] high dimension : If $U_\epsilon (c)$ is $\epsilon$-tubular neighborhood, then define smooth curve $\alpha$ in the boundary s.t. $\exp_{c(t)}\ \epsilon v(t)=\alpha(t)$ where $|v(t)|=\epsilon$.

So define $f(p)={\rm dist}\ (p,\alpha)-\epsilon$ so that it is a solution.

[Add2 - dimension 2 ] Consider a variation $F(t,s)=\exp_{c(t)}\ s \nabla f(c(t))$ Since $F_s(t,0)$ is a parallel along unit speed geodesic $c$, then \begin{align*} R(c',\nabla f,\nabla f,c')&= (\nabla_t\nabla_s F_s-\nabla_s\nabla_t F_s,c') \\&= -(\nabla_s\nabla_t F_s,c') =-(\nabla_t F_s,F_t)_s+(\nabla_t F_s,\nabla_s F_t) \\&=-(\nabla_t F_s,F_t)_s\end{align*}

• I see. You are right. – HK Lee Jun 15 '18 at 14:48
• @Rahul Dont you get $f(x,y)=|y-\epsilon | -\epsilon$? And you do not have problem with differentiability on $x$ axis. But I still do not understand the conclusion "Hence we do not know curvature." – tom Jun 15 '18 at 19:26
• Also you can obtain the solution in much more straight forward way. $f(p) = dist(c,p)$ on side of the geodesic and $f(p) = - dist(c,p)$ on the other side. Note that for this it is crucial that we are in 2d and we can talk about a side from a curve. – tom Jun 15 '18 at 19:30
• I see that your example is straight forward. And when we construct $f$, the curvature is not related. – HK Lee Jun 15 '18 at 23:50
• Ohh sorry, my question is not too clear in that regard. I'm suspecting that there might be a relation between second derivatives of $f$ and curvature. So is there one? – tom Jun 18 '18 at 23:16


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## Basic College Mathematics (9th Edition)

Published by Pearson

# Chapters 1-6 - Cumulative Review Exercises - Page 475: 44

140%

#### Work Step by Step

$\frac{252}{180}$=252$\div$180=1.4 1.4=$\frac{140}{100}$=140%

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# Equal perimeters of squares and right angled isosceles triangles

Consider a square ABCD having length l and breadth. Now start folding the sides AB and AC so that the figure becomes something like this

All the vertical and horizontal folds/stairs are equal in length.

The perimeter of the figure is equal to the the perimeter of the square.

As we increase the number of divisions, the length of each fold/stair decreases. Let the number of stairs/folds be n. As $n\rightarrow\infty$ the figure becomes a right angled isosceles triangle BCD. The perimeter of triangle BCD should be equal to the perimeter of square ABCD since as we increase the number of folds/stairs the perimeter remains the same.

Can anyone please correct me where I have gone wrong?

• You are perfectly correct, except when you say that the perimeter of the limit is the limit of the perimeters. – AlainD Jun 7 '14 at 16:34

The question Is Pi equal to 4? Does the same thing, using a circle inside of a square and slowly removing rectangles from the sides of the square and then using $circumference=2\pi r$ to say that $\pi$ is equal to 4. The answers there will probably answer this quite well.
If a sequence of paths converges (even uniformly) to a path $P$, that doesn't mean that the sequence of their lengths converges to the length of $P$.


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# ON THE AFFINE WEYL GROUP OF TYPE $\tilde{A}_{n-1}$ II

• Albar, Muhammad A. (Department of Mathematical Science Kfupm) ;
• Al-hamed, Maha A. (Department of Mathematical. College of Girls Dammam)
• Published : 1993.02.01

#### Abstract

In our paper [4] we showed that over bar $A_{n-1}$ is a split extension of (n-1) copies of Z by the symmetric group S$_{n}$ of degree n. We show in this paper how over bar $A_{n-1}$ appears naturally as a subgroup of the natural wreath product W=ZS$_{n}$.TEX> n/.


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# Difference between revisions of "2012 AMC 12A Problems/Problem 12"

## Problem

A square region $ABCD$ is externally tangent to the circle with equation $x^2+y^2=1$ at the point $(0,1)$ on the side $CD$. Vertices $A$ and $B$ are on the circle with equation $x^2+y^2=4$. What is the side length of this square?

$\textbf{(A)}\ \frac{\sqrt{10}+5}{10}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{5}\qquad\textbf{(C)}\ \frac{2\sqrt{2}}{3}\qquad\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}\qquad\textbf{(E)}\ \frac{9-\sqrt{17}}{5}$

## Solution 1

$[asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real a=1; real b=2; pair O=(0,0); pair A=(-(sqrt(19)-2)/5,1); pair B=((sqrt(19)-2)/5,1); pair C=((sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair D=(-(sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair E=(-(sqrt(19)-2)/5,0); path inner=Circle(O,a); path outer=Circle(O,b); draw(outer); draw(inner); draw(A--B--C--D--cycle); draw(O--D--E--cycle); label("A",D,NW); label("E",E,SW); label("O",O,SE); label("s+1",(D--E),W); label("\frac{s}{2}",(E--O),S); pair[] ps={A,B,C,D,E,O}; dot(ps); [/asy]$

The circles have radii of $1$ and $2$. Draw the triangle shown in the figure above and write expressions in terms of $s$ (length of the side of the square) for the sides of the triangle. Because $AO$ is the radius of the larger circle, which is equal to $2$, we can write the Pythagorean Theorem.

\begin{align*} \left( \frac{s}{2} \right) ^2 + (s+1)^2 &= 2^2\\ \frac14 s^2 + s^2 + 2s + 1 &= 4\\ \frac54 s^2 +2s - 3 &= 0\\ 5s^2 + 8s - 12 &=0 \end{align*}

$$s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \boxed{\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}}$$

## Solution 2

Using the diagram above, we look at the top-right vertex of the square. Let us call this point $(x,y)$. Then, we that since the square is symmetrical over the y-axis, that the y value is equal to $2x+1$, since we can multiply the x value(which is half of $s$) by two to get $s$, and we add one since the square lies one unit above the origin. Now, all we must do is find the intersection of the larger circle, $x^2 + y^2 = 4$, and the line $y=2x+1$. Substituting the second equation into the first, we get:

$5x^2 +4x -3 = 0$

Using the quadratic formula, we arrive with $x=\frac{-4 \pm 2\sqrt{19}}{10}$. However, recall that the x value is only one half of the side length. Multiplying this value by $2$, then, and using only the positive root(since the top right vertex of the square has a positive x value), we get:

$\frac{-4 + 2\sqrt{19}}{5} \Rightarrow \boxed{\textbf{(D)}}$

## Solution 3

Like the previous solutions, we can say that the square is symmetrical with respect to the y-axis.

Since CD passes through point (0,1), let C be (-x,1) and D be (x,1)

Point B is (-x, $sqrt(4-x^2)$) and Point A is (x,$sqrt(4-x^2)$). Now, we see that AB = 2x and BC = $sqrt(4-x^2)$-1

Setting those two equal, we get the same equation as the previous solutions: $5x^2 +4x -3 = 0$ and then solve the same way.

-Conantwiz2023


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#### Volume 15, issue 2 (2011)

On rational homology disk smoothings of valency $4$ surface singularities
 1 M Bhupal, A Stipsicz, Weighted homogeneous singularities and rational homology disk smoothings, Amer. J. Math. (to appear) arXiv:0902.2277 2 R Fintushel, R J Stern, Rational blowdowns of smooth 4–manifolds, J. Differential Geom. 46 (1997) 181 MR1484044 3 D Gay, A Stipsicz, On symplectic caps, from: "Perspectives in Analysis, Geometry and Topology", Progress in Mathematics, Birkhäuser (to appear) arXiv:0908.3774 4 G M Greuel, H A Hamm, Invarianten quasihomogener vollständiger Durchschnitte, Invent. Math. 49 (1978) 67 MR511096 5 G M Greuel, E Looijenga, The dimension of smoothing components, Duke Math. J. 52 (1985) 263 MR791301 6 G M Greuel, J Steenbrink, On the topology of smoothable singularities, from: "Singularities, Part 1 (Arcata, Calif., 1981)", Proc. Sympos. Pure Math. 40, Amer. Math. Soc. (1983) 535 MR713090 7 H A Hamm, Exotische Sphären als Umgebungsränder in speziellen komplexen Räumen, Math. Ann. 197 (1972) 44 MR0314073 8 T de Jong, D van Straten, Deformation theory of sandwiched singularities, Duke Math. J. 95 (1998) 451 MR1658768 9 J Kollár, Is there a topological Bogomolov–Miyaoka–Yau inequality?, Pure Appl. Math. Q. 4 (2008) 203 MR2400877 10 J Kollár, N I Shepherd-Barron, Threefolds and deformations of surface singularities, Invent. Math. 91 (1988) 299 MR922803 11 H B Laufer, Taut two-dimensional singularities, Math. Ann. 205 (1973) 131 MR0333238 12 Y Lee, J Park, A simply connected surface of general type with $p_g=0$ and $K^2=2$, Invent. Math. 170 (2007) 483 MR2357500 13 E Looijenga, Riemann–Roch and smoothings of singularities, Topology 25 (1986) 293 MR842426 14 E Looijenga, J Wahl, Quadratic functions and smoothing surface singularities, Topology 25 (1986) 261 MR842425 15 W D Neumann, An invariant of plumbed homology spheres, from: "Topology Symposium, Siegen 1979 (Proc. Sympos., Univ. Siegen, Siegen, 1979)", Lecture Notes in Math. 788, Springer (1980) 125 MR585657 16 W D Neumann, Abelian covers of quasihomogeneous surface singularities, from: "Singularities, Part 2 (Arcata, Calif., 1981)", Proc. Sympos. Pure Math. 40, Amer. Math. Soc. (1983) 233 MR713252 17 P Orlik, P Wagreich, Isolated singularities of algebraic surfaces with C$^{\ast}$ action, Ann. of Math. $(2)$ 93 (1971) 205 MR0284435 18 H Pinkham, Normal surface singularities with $C^*$ action, Math. Ann. 227 (1977) 183 MR0432636 19 H Pinkham, Deformations of normal surface singularities with $C^*$ action, Math. Ann. 232 (1978) 65 MR0498543 20 J Stevens, Deformations of singularities, Lecture Notes in Mathematics 1811, Springer (2003) MR1976016 21 A Stipsicz, On the $\bar\mu$–invariant of rational surface singularities, Proc. Amer. Math. Soc. 136 (2008) 3815 MR2425720 22 A Stipsicz, Z Szabó, J Wahl, Rational blowdowns and smoothings of surface singularities, J. Topol. 1 (2008) 477 MR2399141 23 J Wahl, Equisingular deformations of normal surface singularities. I, Ann. of Math. $(2)$ 104 (1976) 325 MR0422270 24 J Wahl, Smoothings of normal surface singularities, Topology 20 (1981) 219 MR608599 25 J A Wolf, Spaces of constant curvature, McGraw-Hill Book Co. (1967) MR0217740


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Let's say you have a library called libfoo located in /usr/local/lib and you want to link it with a node you're building called my_node. Put the following in your CMakeLists.txt:
# set the path to the library folder

(/usr/local/lib is probably in your environment path already, so the first line might be unnecessary)


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Notes on Trigonometric Ratios of Some Standard Angles | Grade 9 > Optional Mathematics > Trigonometry | KULLABS.COM

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Different angles have a different value with various trigonometric ratios. We shall consider 0°, 30°, 45° and 90° as the standard angles and we shall learn their values here. In this unit, we shall verify the values of 0°, 30°, 45° and 90° using geometrical proofs.

Trigonometrical Ratio of 45°

Let ABC be a right-angledisosceles trianglewhere $$\angle$$B = 90° and $$\angle$$A = $$\angle$$C = 45°

Also let BC = AC = $$\alpha$$.

We know, in right angled $$\triangle$$ABC

h2 = p2 + b2

or, (AC)2 = (AB)2 + (BC)2

or, (AC)2 = $$\alpha$$2 + $$\alpha$$2

or, (AC)2 = 2 $$\alpha$$2

∴ AC = $$\alpha$$$$\sqrt{2}$$

Taking $$\angle$$C as the reference angle, we get,

sin 45 = $$\frac{AB}{AC}$$= $$\frac{α}{α\sqrt{2}}$$ = $$\frac{1}{\sqrt{2}}$$

cos 45 = $$\frac{BC}{AC}$$= $$\frac{α}{α\sqrt{2}}$$ = $$\frac{1}{\sqrt{2}}$$

tan 45 = $$\frac{AB}{BC}$$= $$\frac{α}{α}$$ = 1

cosec 45 = $$\frac{AC}{AB}$$= $$\frac{α\sqrt{2}}{α}$$ = $$\sqrt{2}$$

sec 45 = $$\frac{AC}{AB}$$= $$\frac{α\sqrt{2}}{α}$$ =$$\sqrt{2}$$

cot 45 = $$\frac{BC}{AB}$$ = $$\frac{α}{α}$$ = 1

Let ABC be an equilateral triangle where$$\angle$$A = $$\angle$$B = $$\angle$$C = 60°

and AB = BC = CA = 2$$\alpha$$

Now, let's draw AD perpendicular to BC so that,

BD = DC = a and $$\angle$$BAD = $$\angle$$DAC = 30°

Now, in right angled $$\triangle$$ADC,

or, (AD)2 = (AC)2 + (DC)2

= (2α)2 - (α)2

= 4$$\alpha$$2 - $$\alpha$$2 = 3$$\alpha$$2

$$\therefore$$ AD = $$\alpha$$$$\sqrt{3}$$

Now, to find the trigonometric ratio for 60°, lets take $$\angle$$C = 60° as the reference angle, we get,

sin 60° = $$\frac{AD}{AC}$$ = $$\frac{\alpha\sqrt{3}}{2α}$$ = $$\frac{\sqrt{3}}{2}$$

cos 60° = $$\frac{DC}{AC}$$= $$\frac{α}{2α}$$ =$$\frac{1}{2}$$

tan 60° = $$\frac{AD}{DC}$$=$$\frac{\alpha\sqrt{3}}{2α}$$ = $$\sqrt{3}$$

cosec 60° = $$\frac{AC}{AD}$$= $$\frac{2α}{\alpha\sqrt{3}}$$ = $$\frac{2}{\sqrt{3}}$$

sec 60° = $$\frac{AC}{DC}$$= $$\frac{2α}{α}$$ = 2

cot 60° = $$\frac{DC}{AD}$$= $$\frac{α}{\alpha\sqrt{3}}$$ = $$\frac{1}{\sqrt{3}}$$

Now, to find out the trigonometric ratio for 30° , let's take $$\angle$$DAC = 30° as the reference angle, we get,

sin 30° =$$\frac{DC}{AC}$$= $$\frac{α}{2α}$$ =$$\frac{1}{2}$$

cos 30° = $$\frac{AD}{AC}$$ = $$\frac{\alpha\sqrt{3}}{2α}$$ = $$\frac{\sqrt{3}}{2}$$

tan 30° =$$\frac{DC}{AD}$$= $$\frac{α}{\alpha\sqrt{3}}$$ = $$\frac{1}{\sqrt{3}}$$

cosec 30° =$$\frac{AC}{DC}$$= $$\frac{2α}{α}$$ = 2

sec 30° = $$\frac{AC}{AD}$$= $$\frac{2α}{\alpha\sqrt{3}}$$ = $$\frac{2}{\sqrt{3}}$$

cot 30° =$$\frac{AD}{DC}$$=$$\frac{\alpha\sqrt{3}}{2α}$$ = $$\sqrt{3}$$

Trigonometrical Ratio of 0°

Let ABC be a right angled triangle where $$\angle$$B = 90° and $$\angle$$C = $$\theta$$.

If $$\theta$$ tends to 0°

i.e $$\theta$$→0°, AC coincides with BC.

or, AC $$\approx$$BC

so, AC= BC = a (say)

Now, using pythagoras theorem

h2= p2+ b2

or, (AC)2=(AB)2+ (BC)2

or, a2 = (AB)2 + a2

or, (AB)2 = a2 - a2 = 0

$$\therefore$$ AB = 0

Now, taking right angled $$\triangle$$ABC

sin 0° = $$\frac{p}{h}$$ =$$\frac{AB}{AC}$$ =$$\frac{0}{a}$$ = 0

cos 0° =$$\frac{b}{h}$$ = $$\frac{BC}{AC}$$ = $$\frac{a}{a}$$ = 1

tan 0° =$$\frac{p}{b}$$ = $$\frac{AB}{BC}$$ = $$\frac{0}{a}$$ = 0

cosec 0° =$$\frac{h}{p}$$ = $$\frac{AC}{AB}$$ = $$\frac{a}{0}$$ = $$\infty$$

sec 0° =$$\frac{h}{b}$$ = $$\frac{AC}{BC}$$ = $$\frac{a}{a}$$ = 1

cot 0° =$$\frac{b}{p}$$ = $$\frac{BC}{AB}$$ = $$\frac{a}{0}$$ = $$\infty$$

Trigonometric Ratio of 90°

Let ABC be a right angled triangle where $$\angle$$B = 90° and $$\angle$$C = $$\theta$$ be the reference angle.

If $$\theta$$ tends to 90°

i.e $$\theta$$→90°, AC coincides with BC.

or, AC $$\approx$$AB

so, AC= AB = a (say)

Now, using pythagoras theorem

h2= p2+ b2

or, (AC)2=(AB)2+ (BC)2

or, a2 = (BC)2 + a2

or, (BC)2 = a2 - a2 = 0

$$\therefore$$ BC = 0

Now, taking right angled $$\triangle$$ABC

sin 90° = $$\frac{p}{h}$$ =$$\frac{AB}{AC}$$ =$$\frac{a}{a}$$ =1

cos 90° =$$\frac{b}{h}$$ = $$\frac{BC}{AC}$$ = $$\frac{0}{a}$$ = 0

tan 90° =$$\frac{p}{b}$$ = $$\frac{AB}{BC}$$ = $$\frac{a}{0}$$ =$$\infty$$

cosec 90° =$$\frac{h}{p}$$ = $$\frac{AC}{AB}$$ = $$\frac{a}{a}$$ =1

sec 90° =$$\frac{h}{b}$$ = $$\frac{AC}{BC}$$ = $$\frac{a}{0}$$ = $$\infty$$

cot 90° =$$\frac{b}{p}$$ = $$\frac{BC}{AB}$$ = $$\frac{0}{a}$$ =0

#### Complementary Angles

Two angles are Complementary when they add up to 90 degrees.

Let'a see the following example,

What is the angle added to 60° and 90°?

Let the angle be x.

Then, x + 60° = 90°

or, x = 90° - 60° = 30°

Hence, 30° and 60° when added together gives us 90°, So, 30° and 60° are called complements of each other.

The angles are said to be complementary if the sum of the angles is 90°.

Complementary Angles in Trigonometry

Let ABC be a right angledtriangle where $$\angle$$ABC = 90° and $$\angle$$ACB = $$\theta$$

Now, we know,

$$\angle$$ACB +$$\angle$$ABC +$$\angle$$BAC = 180° (sum of angles of a $$\triangle$$)

or, $$\theta$$ + 90° + $$\angle$$BAC = 180°

or, $$\angle$$BAC = 180° - 90° - $$\theta$$ = 90° - $$\theta$$

So, $$\angle$$ACB and$$\angle$$CAB are complementary angles.

Now, taking$$\angle$$A = 90° - $$\theta$$ as the reference angle, we get,

BC = perpendicular (p)

AC = hypotenuse (h)

AB = base (b)

So,

sin(90° - $$\theta$$) = $$\frac{p}{h}$$ =$$\frac{BC}{AC}$$ = cos$$\theta$$ (For reference angle $$\theta$$)

cos (90° - $$\theta$$) = $$\frac{b}{h}$$ =$$\frac{AB}{AC}$$ = sin$$\theta$$

tan(90° - $$\theta$$) =$$\frac{p}{h}$$ =$$\frac{BC}{AB}$$ = cot$$\theta$$

cot(90° - $$\theta$$) =$$\frac{b}{p}$$ =$$\frac{AB}{BC}$$ = tan$$\theta$$

sec(90° - $$\theta$$) =$$\frac{h}{b}$$ =$$\frac{AC}{AB}$$ = cosec$$\theta$$

cosec(90° - $$\theta$$) =$$\frac{h}{p}$$ =$$\frac{AB}{BC}$$ = sec$$\theta$$

Hence,

sin(90° - $$\theta$$) = cos$$\theta$$

cos(90° - $$\theta$$) = sin$$\theta$$

tan(90° - $$\theta$$) =cot$$\theta$$

cot(90° - $$\theta$$) =tan$$\theta$$

sec(90° - $$\theta$$) = cosec$$\theta$$

cosec(90° - $$\theta$$) =sec$$\theta$$

• h2 = p2 + b2
• Different angles have a different value with various trigonometric ratios. We shall consider 0°, 30°, 45° and 90° as the standard angles and we shall learn their values here. In this unit, we shall verify the values of 0°, 30°, 45° and 90° using geometrical proofs.
.

### Very Short Questions

Given that A = 30°

Then, LHS = sin3A

= sin3 × 30°

= sin90°

= 1

RHS = 3sinA - 4sin3A

= 3sin30° - 4(sin30°)3

= 3$$\frac{1}{2}$$ - 4($$\frac{1}{2}$$)3

= $$\frac{3}{2}$$ - 4 $$\frac{1}{8}$$

= $$\frac{3}{2}$$ - $$\frac{1}{2}$$

= $$\frac{2}{2}$$

= 1

$$\therefore$$ LHS = RHS verified.

Solution

LHS = cos 60

= $$\frac{1}{2}$$

RHS = cos230° - sin230°

($$\frac{\sqrt(3)}{2}$$)2 - ($$\frac{1}{2}$$)2

$$\frac{3}{4}$$ - $$\frac{1}{4}$$

$$\frac{2}{4}$$

$$\frac{1}{2}$$

$$\therefore$$ LHS = RHS proved.

Solution

sec2$$\frac{π}{4}$$ sec2$$\frac{π}{3}$$(cosec$$\frac{π}{6}$$ - cosec$$\frac{π}{2}$$)

= sec2 $$\frac{180°}{4}$$ × sec2$$\frac{180°}{3}$$ (cosec$$\frac{180°}{6}$$ - cosec$$\frac{180°}{2}$$)

= sec245° × sec260° (cosec30° - cosec90°)

= ($$\sqrt(2)$$)2 × (2)2 (2-1)

= 2×4×1

= 8

0%

Opposite
Same
Different

p2 + b2
p + b
b2 - p2
p2 - b2
• ### sin 45° = ______

(frac{sqrt{3}}{5})
(frac{sqrt{2}}{sqrt{3}})
(frac{1}{sqrt 2})
(frac{2}{1})
• ### cos 45° = ______

(frac{1}{sqrt 2})
(frac{sqrt{3}}{5})
(frac{sqrt{2}}{sqrt{3}})
(frac{2}{1})

1
2
0
(frac{1}{2})
• ### cosec 45° = ______

(sqrt{3})
(sqrt 2)
(frac{sqrt{2}}{sqrt{3}})
(frac{1}{2})
• ### sec 45° = ______

(sqrt 2)
(sqrt{3})
(frac{2}{3})
(frac{sqrt{3}}{2})
• ### cot 45° = ______

(sqrt{3})
0
1
(frac{sqrt{3}}{2})
• ### sin 60° = ______

1
(sqrt{2})
(frac{1}{2})
(frac{sqrt {3}}{2})

1
(frac{3}{2})
(frac{1}{2})

(frac{3}{2})
(frac{1}{2})
(sqrt3)
(sqrt{2})
• ### cosec 60° = ______

(frac{2}{sqrt {3}})
(frac{3}{2})
(frac{1}{2})
1

0
1
3
2
• ### cot 60° = ______

(frac{1}{2})
(frac{1}{sqrt 3})
(frac{2}{3})
(frac{sqrt{3}}{2})
• ### sin 30° = ______

(frac{1}{sqrt 2})
(frac{sqrt{3}}{2})
(frac{2}{3})
(frac{1}{2})
• ### sec 30° = ______

(frac{sqrt{3}}{2})
(frac{2}{sqrt 3})
(sqrt{3})
(frac{1}{sqrt{3}})
• ### tan 30° = ______

(frac{1}{sqrt 3})
(sqrt{2})
(frac{1}{2})
(frac{2}{sqrt{3}})
• ### cos 0° = _______

(sqrt{2})
1
0
(frac{2}{sqrt{3}})

1
(infty)
0
2

0
1
(infty)
(frac{1}{2})

2
0
1
(infty)
• ### tan 90° = ________

(frac{1}{2})
(infty)
(frac{2}{sqrt{3}})
(frac{1}{sqrt{3}})
• ### sec 90° = ________

1
0
(frac{2}{sqrt{3}})
(infty)


text:
Outlook: Colicity Inc. Units is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Hold
Time series to forecast n: 07 Feb 2023 for (n+1 year)
Methodology : Modular Neural Network (Emotional Trigger/Responses Analysis)

## Abstract

Colicity Inc. Units prediction model is evaluated with Modular Neural Network (Emotional Trigger/Responses Analysis) and Pearson Correlation1,2,3,4 and it is concluded that the COLIU stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Hold

## Key Points

1. Why do we need predictive models?
2. Which neural network is best for prediction?
3. What are the most successful trading algorithms?

## COLIU Target Price Prediction Modeling Methodology

We consider Colicity Inc. Units Decision Process with Modular Neural Network (Emotional Trigger/Responses Analysis) where A is the set of discrete actions of COLIU stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4

F(Pearson Correlation)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Emotional Trigger/Responses Analysis)) X S(n):→ (n+1 year) $∑ i = 1 n s i$

n:Time series to forecast

p:Price signals of COLIU stock

j:Nash equilibria (Neural Network)

k:Dominated move

a:Best response for target price

For further technical information as per how our model work we invite you to visit the article below:

How do AC Investment Research machine learning (predictive) algorithms actually work?

## COLIU Stock Forecast (Buy or Sell) for (n+1 year)

Sample Set: Neural Network
Stock/Index: COLIU Colicity Inc. Units
Time series to forecast n: 07 Feb 2023 for (n+1 year)

According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Hold

X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)

Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)

Z axis (Grey to Black): *Technical Analysis%

## IFRS Reconciliation Adjustments for Colicity Inc. Units

1. Lifetime expected credit losses are generally expected to be recognised before a financial instrument becomes past due. Typically, credit risk increases significantly before a financial instrument becomes past due or other lagging borrower-specific factors (for example, a modification or restructuring) are observed. Consequently when reasonable and supportable information that is more forward-looking than past due information is available without undue cost or effort, it must be used to assess changes in credit risk.
2. An entity can rebut this presumption. However, it can do so only when it has reasonable and supportable information available that demonstrates that even if contractual payments become more than 30 days past due, this does not represent a significant increase in the credit risk of a financial instrument. For example when non-payment was an administrative oversight, instead of resulting from financial difficulty of the borrower, or the entity has access to historical evidence that demonstrates that there is no correlation between significant increases in the risk of a default occurring and financial assets on which payments are more than 30 days past due, but that evidence does identify such a correlation when payments are more than 60 days past due.
3. For the purpose of applying the requirement in paragraph 6.5.12 in order to determine whether the hedged future cash flows are expected to occur, an entity shall assume that the interest rate benchmark on which the hedged cash flows (contractually or non-contractually specified) are based is not altered as a result of interest rate benchmark reform.
4. An entity's business model refers to how an entity manages its financial assets in order to generate cash flows. That is, the entity's business model determines whether cash flows will result from collecting contractual cash flows, selling financial assets or both. Consequently, this assessment is not performed on the basis of scenarios that the entity does not reasonably expect to occur, such as so-called 'worst case' or 'stress case' scenarios. For example, if an entity expects that it will sell a particular portfolio of financial assets only in a stress case scenario, that scenario would not affect the entity's assessment of the business model for those assets if the entity reasonably expects that such a scenario will not occur. If cash flows are realised in a way that is different from the entity's expectations at the date that the entity assessed the business model (for example, if the entity sells more or fewer financial assets than it expected when it classified the assets), that does not give rise to a prior period error in the entity's financial statements (see IAS 8 Accounting Policies, Changes in Accounting Estimates and Errors) nor does it change the classification of the remaining financial assets held in that business model (ie those assets that the entity recognised in prior periods and still holds) as long as the entity considered all relevant information that was available at the time that it made the business model assessment.

*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.

## Conclusions

Colicity Inc. Units is assigned short-term Ba1 & long-term Ba1 estimated rating. Colicity Inc. Units prediction model is evaluated with Modular Neural Network (Emotional Trigger/Responses Analysis) and Pearson Correlation1,2,3,4 and it is concluded that the COLIU stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Hold

### COLIU Colicity Inc. Units Financial Analysis*

Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementBa3Caa2
Balance SheetBa3B1
Leverage RatiosCaa2Baa2
Cash FlowCaa2Caa2
Rates of Return and ProfitabilityB3B2

*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?

### Prediction Confidence Score

Trust metric by Neural Network: 91 out of 100 with 837 signals.

## References

1. Z. Wang, T. Schaul, M. Hessel, H. van Hasselt, M. Lanctot, and N. de Freitas. Dueling network architectures for deep reinforcement learning. In Proceedings of the International Conference on Machine Learning (ICML), pages 1995–2003, 2016.
2. Jacobs B, Donkers B, Fok D. 2014. Product Recommendations Based on Latent Purchase Motivations. Rotterdam, Neth.: ERIM
3. C. Claus and C. Boutilier. The dynamics of reinforcement learning in cooperative multiagent systems. In Proceedings of the Fifteenth National Conference on Artificial Intelligence and Tenth Innovative Applications of Artificial Intelligence Conference, AAAI 98, IAAI 98, July 26-30, 1998, Madison, Wisconsin, USA., pages 746–752, 1998.
4. J. N. Foerster, Y. M. Assael, N. de Freitas, and S. Whiteson. Learning to communicate with deep multi-agent reinforcement learning. In Advances in Neural Information Processing Systems 29: Annual Conference on Neural Information Processing Systems 2016, December 5-10, 2016, Barcelona, Spain, pages 2137–2145, 2016.
5. F. A. Oliehoek and C. Amato. A Concise Introduction to Decentralized POMDPs. SpringerBriefs in Intelligent Systems. Springer, 2016
6. Bai J. 2003. Inferential theory for factor models of large dimensions. Econometrica 71:135–71
7. K. Boda, J. Filar, Y. Lin, and L. Spanjers. Stochastic target hitting time and the problem of early retirement. Automatic Control, IEEE Transactions on, 49(3):409–419, 2004
Frequently Asked QuestionsQ: What is the prediction methodology for COLIU stock?
A: COLIU stock prediction methodology: We evaluate the prediction models Modular Neural Network (Emotional Trigger/Responses Analysis) and Pearson Correlation
Q: Is COLIU stock a buy or sell?
A: The dominant strategy among neural network is to Hold COLIU Stock.
Q: Is Colicity Inc. Units stock a good investment?
A: The consensus rating for Colicity Inc. Units is Hold and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of COLIU stock?
A: The consensus rating for COLIU is Hold.
Q: What is the prediction period for COLIU stock?
A: The prediction period for COLIU is (n+1 year)


text:
# Super-cobordisms

One can construct the $d$-dimensional bordism category by declaring the objects to be the $(d-1)$-dimensional compact manifolds without boundary and the morphisms the $d$-dimensional bordisms between them. Call it $\mathcal{Cob}_d$. It is well known that the connected components of the geometric realization of this category are in one-to-one correspondence with $\pi_d(MO)$, where $MO$ is the Thom spectrum for the orthogonal group. This is the classical Thom-Pontryagin theorem.

One can think of constructing a similar category with supermanifolds. Namely, $\mathcal{Cob}_{(d|k)}$ is the category whose objects are $(d-1|k)$-dimensional supermanifolds and the morphisms the $(d|k)$-dimensional bordisms.

Does anyone know of a Thom-Pontryagin type result for this category? Is there a spectrum $MO_{|k}$, whose homotopy groups recover the connected components of the geometric realization of $\mathcal{Cob}_{(d|k)}$?

• Just for completeness, there is a paper on that: Voronov and Zorich "Bordism theory and homotopy properties of supermanifolds" Feb 21 '15 at 15:19

There are a number of technical issues with making what you describe precise, for example: what precisely is a supermanifold with boundary? how can you glue/compose bordisms? etc. I am going to ignore these technicalities because I don't think it makes much of a difference to your question. Many of these technical issues have been addressed by other people, for example I would suggest looking at the work of Stolz-Teichner and the reference therein to see what sort of things people have tried to do.

In any case in the smooth category we have:

Batchelor's theorem: every $(d|k)$-dimensional supermanifold is (non-canonically!) of the form $\pi E$ for $E$ a rank $k$ vector bundle over a $d$-manifold. Moreover the isomorphism class of the vector bundle is uniquely determined by the super manifold.

Here $\pi E$ is the super manifold whose ring of functions is the global sections of the exterior algebra bundle $\wedge^* E^*$. So the morphisms from $\pi E$ to $\pi E'$ come from all algebra maps and are more than just the vector bundle maps (which correspond to homogeneous algebra maps).

If $(X, \mathcal{O}_X)$ is a super manifold, the vector bundle $E$ can be obtained by considering $\mathcal N / \mathcal N^2$ where $\mathcal N$ is the subsheaf of $\mathcal{O}_X$ of nilpotent elements.

So there are functors in both directions but there is no natural isomorphism from the identity functor on supermanifolds to $\pi(\mathcal N / \mathcal N^2)$. This is what the "non-canonical" means.

But we still get a bijection between isomorphism classes of supermanifolds and isomorphism classes of manifolds with vector bundles. Thus, unless you do something more fancy like work in families, when you pass to bordism classes you will just get the bordism group of $d$-manifolds equipped with rank $k$-vector bundles. This has two names:

$$\pi_d( MO \wedge BO(k)) = MO_d(BO(k))$$

either as the d-th homotopy group of the smash of the spectrum $MO$ and space $BO(k)$ (which is also another Thom spectrum) or equivalently as the d-th $MO$-homology group of $BO(k)$. So that identifies the bordism group.

I think it is a very interesting question whether the whole Pontryagin-Thom construction can be carried out inside the world of supermanifolds, but that is a different question.

• Thanks for the answer. I have a feeling that we are completely ignoring the automorphism information by this approach. If we thought of constructing some sort of a bordism $(\infty,n)$-category, it is interesting what's the resulting spectrum that comes from the geometric realization of this $(\infty,n)$-category. I assume these would be some sort of a filtration piece of $MO\wedge BO(k)$. Feb 19 '15 at 16:59
• My guess is that you would just get the filtration by $MTO(n) \wedge BO(k)$, as $n$-varies. Feb 20 '15 at 10:17
• How different are the automorphisms of $\pi E$ from those of $E$? Feb 20 '15 at 18:54
• Also as a side question: are the characteristic numbers of the supermanifolds formed by pairing the products of Stiefel-Whitney classes of the tangent bundle and those of the defining vector bundle $E$ with the fundamental class? Feb 22 '15 at 19:57
• Probably one can write something down, however, one will get something which depends on the bundles $E$ and $TM_0$ only and not really an invariant of the supermanifold. In fact, for any Lie supergroup $G$, isomorphism classes of principal $G$-bundles over the supermanifold $M$ are in one to one correspondence with isomorphism classes of principal $G_0$-bundles over $M_0$. This is so since $M_0 \rightarrow M$ and $G_0\rightarrow G$ are homotopy equivalences and the latter is in addition a morphism of groups. Feb 23 '15 at 10:52


text:
Torque question

Tags:
1. Feb 22, 2016

Cozma Alex

The definition of torque is τ= r x F

Let's consider a disk that is rotating around his axis of symmetry, we are usually used to calculate the torque applied by a tangential force at the disk using as r the radius of the disk, but can I define a torque of the force using as r the generic distance between a point external of the disk? In this case will be the torque by this point different by the one referred to the axis of rotation? In my opinion yes because it's the most logical thing

Can we define a torque in a non rotation situation? Think at a particle that is pulled by a force in a uniform accelerated motion, according to the definition of torque we can choose a point in space and get the distance between that point and the force and get a torque, is that right?

2. Feb 22, 2016

A.T.

Yes, the definition of torque says noting about rotation.

3. Feb 23, 2016

drvrm

one can define a torque from taking radius vector from the axis and the force vector applied at a point on the body- as its a vector product the torque will generate a rotation due to application of the force. so you can not say that its a non rotational system.

4. Feb 25, 2016

brotherStefan

The answer to the first question is "yes" -- this is what happens every time a wrench is used to turn a bolt or a nut; the force is applied at some distance external to the bolt head or nut.
The answer to the second question is, "it depends" -- the same force (F) applied to a longer lever-arm (r) will yield a proportionally greater torque. Said another way, a longer lever-arm allows the same torque to be developed with proportionately less force.


text:
# Math Help - Confusing Algebra in a Derivative

1. ## Confusing Algebra in a Derivative

Hi, I couldn't find anyone around my dorm that could help me so I thought I would try a forum. Thanks for any help I receive!

My problem is:

Find the derivative:
y = (3x^4 + 1)^4 * (x^3 + 4)

According to my book the answer is 3x^2(3x^4 + 1)^3 * (19x^4 + 64x + 1)

For some reason I never can get this solution.

Here are the 2 steps in the book that really confuse me:
3x^2(3x^4 + 1)^4 + 48x^3(x^3 + 4)(3x^4 + 1)^3
to this
3x^2(3x^4 + 1)^3[3x^4 + 1 + 16x(x^3 + 4)]

I get to the first step but I do not get the second step. I think I am just overlooking something simple, but I don't see it.

Thanks!!

2. y = (3x^4 + 1)^4 * (x^3 + 4)

To find the derivative, we're going to need to use a chain rule and a product rule.

y'= (3x^4 +1)^4 * d/dx (x^3 +4) + (x^3+4) * d/dx (3x^4 +1)^4
= (3x^4 +1)^4 * 3x^2 + (x^3+4) * 4(3x^4 +1)^3 * 12x
= 3x^2(3x^4 +1)^3 + 48x(x^3+4)(3x^4 +1)^3

From here, all you need to do is simplify the expression to get the textbook answer.

Oh, I see, didn't read your question thoroughly enough right away. Um, all you really should have to do is simplify the expression and then factor out a 3x^2(3x^4 + 1)^3 from the expression.

3. Originally Posted by stunirvana
Hi, I couldn't find anyone around my dorm that could help me so I thought I would try a forum. Thanks for any help I receive!

My problem is:

Find the derivative:
y = (3x^4 + 1)^4 * (x^3 + 4)

According to my book the answer is 3x^2(3x^4 + 1)^3 * (19x^4 + 64x + 1)

For some reason I never can get this solution.

Here are the 2 steps in the book that really confuse me:
3x^2(3x^4 + 1)^4 + 48x^3(x^3 + 4)(3x^4 + 1)^3
to this
3x^2(3x^4 + 1)^3[3x^4 + 1 + 16x(x^3 + 4)]

I get to the first step but I do not get the second step. I think I am just overlooking something simple, but I don't see it.

Thanks!!
$3x^2(3x^4 + 1)^4 + 48x^3(x^3 + 4)(3x^4 + 1)^3
$

$\textcolor{red}{3x^2(3x^4 + 1)^3} \cdot (3x^4+1) + \textcolor{red}{3x^2} \cdot 16x(x^3 + 4)\textcolor{red}{(3x^4 + 1)^3}$

note what factors the two terms have in common ...

both have the common factors $\textcolor{red}{3x^2}$ and $\textcolor{red}{(3x^4+1)^3}$ ... they are both factored out from each term ...

$\textcolor{red}{3x^2(3x^4+1)^3}[(3x^4+1) + 16x(x^3+4)]$

any clearer now?

4. I can get this. I think it should be:
Originally Posted by Math Major
y = (3x^4 + 1)^4 * (x^3 + 4)
y'= (3x^4 +1)^4 * d/dx (x^3 +4) + (x^3+4) * d/dx (3x^4 +1)^4
= (3x^4 +1)^4 * 3x^2 + (x^3+4) * 4(3x^4 +1)^3 * 12x^3
= 3x^2(3x^4 +1)^4 + 48x^3(x^3+4)(3x^4 +1)^3
Correct me if I am wrong above.

Its the simplification I'm having trouble with.

5. Originally Posted by skeeter

any clearer now?
So I just had to pull a $(3x^4+1)$ out of $(3x^+1)^4$ then pull them (I mean the two terms you highlighted) out to the front?

By the way, the math text thing is awesome!

6. Yes, you're correct. I just typed it too quickly.


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 45-45-90 Right Triangles

## Leg times sqrt(2) equals hypotenuse.

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45-45-90 Triangles
License: CC BY-NC 3.0

Audrey, has decided to make her classroom look more like a Math class by putting wall paper squares on the doors of the storage cabinets. She has chosen paper that shows squares, rectangles and triangles.

License: CC BY-NC 3.0

If the length of one of the legs of the 454590\begin{align*}45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*} triangles in the wallpaper is 8 inches in length, then how can Audrey figure out the length of the diagonal of the wallpaper square?

In this concept, you will learn to identify isosceles right triangles.

### Isosceles Triangle

A triangle that has two sides equal in length and two angles at the base of these sides equal in measure is called an isosceles triangle. If this concept is applied to a right triangle, then the two sides equal in length have to be the legs of the right triangle. Remember the longest side of the right triangle is called the hypotenuse and it is opposite the 90° angle. The two acute angles of a right triangle have a sum of 90° angle. If the legs are equal in length then the two acute angles of the triangle will each measure 45°. The following is an isosceles right triangle.

License: CC BY-NC 3.0

The angles of an isosceles right triangle will always have the measures of 454590\begin{align*} 45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*}. This triangle is often referred to as a 454590\begin{align*}45-45-90\end{align*} right triangle. The legs of an isosceles right triangle are equal in measure such that a=b\begin{align*}a = b\end{align*}. The length of the hypotenuse can be found using the Pythagorean Theorem.

First, determine the values of (a,b,c)\begin{align*}(a, b, c)\end{align*} for the Pythagorean Theorem.

abc===aa?\begin{align*}\begin{array}{rcl} a &=& a \\ b &=& a \\ c &=& ? \end{array}\end{align*}

Next, substitute the values for (a,b,c)\begin{align*}(a, b, c)\end{align*} into the Pythagorean Theorem.

c2c2==a2+b2(a)2+(a)2\begin{align*}\begin{array}{rcl} c^2 &=& a^2 + b ^2 \\ c^2 &=& (a) ^ 2 + (a)^2 \\ \end{array}\end{align*}

Next, perform the indicated squaring on the right side of the equation and simplify.

c2c2c2===(a)2+(a)2a2+a22a2\begin{align*}\begin{array}{rcl} c^2 &=& (a) ^ 2 + (a)^2 \\ c^2 &=& a^2 + a ^2 \\ c^2 &=& 2a^2 \\ \end{array}\end{align*}

Then, solve for the variable ‘c\begin{align*}c\end{align*}’ by taking the square root of both sides of the equation. Express the answer as a radical.

c2c2cc====2a22a22aa2\begin{align*}\begin{array}{rcl} c^2 &=& 2a^2 \\ \sqrt{c^2} &=& \sqrt{2} \cdot \sqrt{a^2} \\ c&=& \sqrt{2} \cdot a \\ c&=& a \sqrt{2} \\ \end{array}\end{align*}

The answer is a2.\begin{align*}a \sqrt{2}.\end{align*}

The length of the hypotenuse is a2 cm\begin{align*}a \sqrt{2} \ cm\end{align*}. Since a=b\begin{align*}a = b \end{align*} the length of the hypotenuse could be expressed as b2 cm\begin{align*}b \sqrt{2} \ cm\end{align*}.

The coefficient ‘a\begin{align*}a\end{align*}’ in front of 2\begin{align*}\sqrt{2} \end{align*} is the length of the legs of the given triangle. The length of the hypotenuse of an isosceles right triangle will always equal the product of the leg length and 2\begin{align*}\sqrt{2}\end{align*}.

If a 454590\begin{align*} 45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*} right triangle had a leg length of 5 inches then the length of the hypotenuse would be 5×2=52 in.\begin{align*}5 \times \sqrt{2} = 5 \sqrt{2}\ in.\end{align*}

### Examples

#### Example 1

Earlier, you were given a problem about Audrey and the wallpaper squares. She wants to figure out the length of the diagonal of the wallpaper square.

First, write down what you know.

aac===8 in= 8b = 8?\begin{align*}\begin{array}{rcl} a &=& 8 \ in = \ 8 \\ a &=& b \quad \ = \ 8 \\ c &=& ? \\ \end{array}\end{align*}

Next, write down an equation to show the relationship between the length of the hypotenuse and the length of a leg.

c=a2\begin{align*}c = a\sqrt{2}\end{align*}

Next, substitute what you know into the equation.

cc==a282\begin{align*}\begin{array}{rcl} c &=& a\sqrt{2} \\ c &=& 8\sqrt{2} \\ \end{array}\end{align*}

Remember there are four 454590\begin{align*}45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*} triangles along the diagonal of the wallpaper square.

Then, multiply the value of ‘c\begin{align*}c\end{align*}’ by the number of triangles.

ccc===824(82)322\begin{align*}\begin{array}{rcl} c &=& 8\sqrt{2} \\ c &=& 4(8\sqrt{2}) \\ c &=& 32 \sqrt{2} \\ \end{array} \end{align*}The answer is 322\begin{align*}32 \sqrt{2}\end{align*}.

The length of the diagonal of the entire wallpaper square is 322\begin{align*}32 \sqrt{2}\end{align*} inches.

#### Example 2

The length of the leg of the hypotenuse of a 454590\begin{align*} 45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*} triangle is 9 centimeters. What is the length of the hypotenuse?

First, write down what you know.

For the given 454590\begin{align*}45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*} triangle a=b=9\begin{align*}a = b = 9\end{align*}.

The length of the hypotenuse ‘c\begin{align*}c\end{align*}’ is a2\begin{align*}a\sqrt{2}\end{align*}.

Then, substitute ‘9’ for ‘a\begin{align*}a\end{align*}’ to find the length of the hypotenuse.

cc==a292\begin{align*}\begin{array}{rcl} c &=& a\sqrt{2} \\ c &=& 9\sqrt{2} \\ \end{array}\end{align*}

The answer is 92.\begin{align*}9 \sqrt{2}.\end{align*}

The length of the hypotenuse is 92 cm\begin{align*} 9 \sqrt{2} \ cm\end{align*}.

#### Example 3

For the given isosceles, right triangle, find the length of a leg.

License: CC BY-NC 3.0

First, write down what you know.

ACac===a = ? mb  = ? m10 m = 10\begin{align*}\begin{array}{rcl} AC &=& a \ = \ ? \ m \\ a &=& b \ \ = \ ? \ m \\ c &=& 10 \ m \ = \ 10 \\ \end{array}\end{align*}

Next, write down an equation to show the relationship between the length of the hypotenuse and the length of a leg.

c=AC2\begin{align*}c= AC \sqrt{2}\end{align*}Next, substitute what you know into the equation.

c10==AC2a2\begin{align*}\begin{array}{rcl} c&=& AC \sqrt{2} \\ 10&=& a \sqrt{2} \\ \end{array}\end{align*}

Next, isolate the variable ‘a\begin{align*}a\end{align*}’ by dividing both sides of the equation by 2\begin{align*}\sqrt{2}\end{align*} and simplify the equation.

\begin{align*}\begin{array}{rcl} 10 &=& a\sqrt{2} \\ \frac{10}{\sqrt{2}} &=& \frac{a{\cancel{\sqrt{2}}}} { \cancel{\sqrt{2}}} \\ \frac{10}{\sqrt{2}} &=& a \\ \end{array}\end{align*}

Next, multiply the left side of the denominator by \begin{align*}\frac{\sqrt{2}}{\sqrt{2}}\end{align*} to clear the radical from the denominator.

\begin{align*}\begin{array}{rcl} \frac{10}{\sqrt{2}} &=& a \\ \frac{10}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} &=& a \\ \frac{10 \sqrt{2}}{\sqrt{4}} &=& a \\ \end{array}\end{align*}

Then, simplify the left side of the equation to solve for the variable.

\begin{align*}\begin{array}{rcl} \frac{10 \sqrt{2}}{\sqrt{4}} &=& a \\ \frac{\overset{5} {\cancel{10}} \sqrt{2} }{{\cancel{2}}} &=& a \\ 5 \sqrt{2} &=& a \\ \end{array}\end{align*}

The answer is \begin{align*}5\sqrt{2}\end{align*}.

The length of a leg is \begin{align*}5 \sqrt{2}\end{align*} meters.

#### Example 4

The distance between the bases of a baseball diamond is 90 feet. If the umpire runs directly from home plate to second base, how far will he have to run? Round the answer to the nearest tenth.

First, draw and label a diagram to model the problem.

License: CC BY-NC 3.0

First, write down what you know.

\begin{align*}\begin{array}{rcl} AC &=& a \ = \ ? \ m \\ a &=& b \ = \ ? \ m \\ c &=& 10 \ m \ = \ 10 \\ \end{array}\end{align*}

Next, write down an equation to show the relationship between the length of the hypotenuse and the length of a leg.

\begin{align*}C = AC \sqrt{2}\end{align*}

Next, substitute what you know into the equation.

\begin{align*}\begin{array}{rcl} c&=& AC \sqrt{2} \\ 10&=& a \sqrt{2} \\ \end{array}\end{align*}

Then, using the calculator, perform the operations indicated on the right side of the equation.

\begin{align*}\begin{array}{rcl} c&=& 90 \sqrt{2} \\ c &=& 90(1.41) \\ c&=& 126.9 \\ \end{array}\end{align*}The answer is 126.9.

The umpire runs 126.9 feet.

### Review

Find the length of the hypotenuse in each of the following \begin{align*}45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*} triangles.

1. \begin{align*}\text{Length of each leg} = 5\end{align*}

2. \begin{align*}\text{Length of each leg} = 4\end{align*}

3. \begin{align*}\text{Length of each leg} = 6\end{align*}

4. \begin{align*}\text{Length of each leg} = 3\end{align*}

5. \begin{align*}\text{Length of each leg} = 7\end{align*}

Use a calculator to figure out the approximate value of each hypotenuse. You may round to the nearest hundredth.

6. \begin{align*} 5\sqrt{2}\end{align*}

7. \begin{align*} 4\sqrt{2}\end{align*}

8. \begin{align*}6\sqrt{2}\end{align*}

9. \begin{align*} 3\sqrt{2}\end{align*}

10. \begin{align*}7\sqrt{2}\end{align*}

11. \begin{align*} 8\sqrt{2}\end{align*}

12. \begin{align*}10\sqrt{2}\end{align*}

13. \begin{align*} 13\sqrt{2}\end{align*}

14. \begin{align*}21\sqrt{2}\end{align*}

15. \begin{align*} 17\sqrt{2}\end{align*}

To see the Review answers, open this PDF file and look for section 7.11.

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### Vocabulary Language: English

TermDefinition
45-45-90 Theorem For any isosceles right triangle, if the legs are x units long, the hypotenuse is always x$\sqrt{2}$.
45-45-90 Triangle A 45-45-90 triangle is a special right triangle with angles of $45^\circ$, $45^\circ$, and $90^\circ$.
Hypotenuse The hypotenuse of a right triangle is the longest side of the right triangle. It is across from the right angle.
Isosceles Right Triangle An isosceles right triangle is a triangle with a ninety degree angle and exactly two sides that are the same length.
Legs of a Right Triangle The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle.
Radical The $\sqrt{}$, or square root, sign.


text:
# Why do farads multiplied by ohms produce a result that has a unit of seconds?

Why is the time constant (RC) measured in seconds even though the units are farads x ohms?

This is to fulfill my own curiosity as I haven't had much luck at finding the answer. I would be most grateful if someone could give me a solid answer or send me in the right direction.

• Because hours and minutes would be silly. What option is there besides seconds? – Majenko Nov 12 '14 at 23:05
• The second is the SI unit of time. What do you propose as an alternative? – The Photon Nov 12 '14 at 23:09
• Sorry I was asking why Farads multiplied by ohms gives seconds? – user37216 Nov 12 '14 at 23:37
• Unit cancellation is weird: what-if.xkcd.com/11 – Volker Siegel Nov 14 '14 at 0:39
• It's math. You just gotta take it on faith. – Robert Endl Jun 18 '17 at 23:18

It's the way the units work out.

Broken down to its form in SI units, a volt is

$$\mathrm{V = \frac{kg \cdot m^2}{A \cdot s^3}}$$

where A is amperes. So, when you divide by current to get ohms, you see that

$$\Omega = \mathrm{\frac{kg \cdot m^2}{A^2\cdot s^3}}$$

$$\mathrm{F=\frac{s^4 \cdot A^2}{m^2 \cdot kg}}$$

So when you multiply Ohms by Farads, you're left with seconds:

$$\Omega \cdot \mathrm{F = \frac{kg \cdot m^2}{A^2\cdot s^3} \cdot \frac{s^4 \cdot A^2}{m^2 \cdot kg} = s}$$

• Great answer, but IMO too far. I think it would have been enough to say $$\Omega \cdot \mathrm{F = \frac{V}{A} \cdot \frac{As}{V} = s}$$. – glglgl Nov 13 '14 at 18:45

Because seconds (s) is the fundamental unit of measurement of time. The other fundamental SI units are:
1. Meters (m) for distance
2. Kilograms (kg) for mass
3. Ampere (A) for current
4. Kelvin (K) for temperature
5. Mole (mol) for quantity
6. Candela (cd) for light intensity

All other units of measurement relevant to physics are derived from those seven fundamental units.

Resistance, represented in fundamental form, is $\frac{kg*m^2}{A^2*s^3}$
Likewise, Capacitance in fundamental form is $\frac{s^4*A^2}{m^2*kg}$
Therefore, $R*C = \frac{kg*m^2}{A^2*s^3}*\frac{s^4*A^2}{m^2*kg} = s$

If you charge the capacitor to some level and then connect it in parallel with the resistor, a current will begin to flow.

In reality this current will become smaller as the capacitor discharges (and the voltage across it therefore drops), but if we imagine that we somehow forced the current to stay at the initial magnitude through the resistor until the capacitor was fully discharged then it would take a certain amount of time until the capacitor was discharged to 0 V.

It turns out that this "certain amount of time" is the same no matter how much or little you charged the capacitor originally. (If you charge it more, there will be more charge to discharge, but the current will be proportionally higher because higher charge produces more voltage). This time is the product of the capacitance with the resistance -- or in other words your time constant.

And that is, intuitively, why the time constant has units of time.

(Alternatively, the time constant is how long it will take for the voltage to drop to $\frac1e$ of its original value, in the more realistic situation where we leave the system alone and let the current drop with the voltage as per Ohm's Law).

\begin{align} v&=IR \\ R&=\frac{v}{I} \\ &=\frac{1V}{1A} \\ I&=\frac{q}{t} \\ 1A &=\frac{1C}{1s} \\ C&=\frac{Q}{v} \\ &=\frac{1C}{1V} \end{align}

\begin{align} \text{.'. unit of } RC &=\frac{1V}{1A} \cdot \frac{1C}{1V} \\ &=\frac{1C}{1A} \\ &=\frac{1C}{1C/1s} \\ &=1s \\ \end{align}

Why is the time constant (RC) measured in seconds even though the units are farads x ohms?

Because farad is defined as charges held per unit of volt across the capacitor.

Charges are current times time. So farad is current x time over volt, or time over ohm.


text:
## Journals

### Many-body Chern number from statistical correlations of randomized measurements. (arXiv:2005.13543v1 [quant-ph])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

One of the main topological invariants that characterizes several topologically-ordered phases is the many-body Chern number (MBCN). Paradigmatic examples include several fractional quantum Hall phases, which are expected to be realized in different atomic and photonic quantum platforms in the near future. Experimental measurement and numerical computation of this invariant is conventionally based on the linear-response techniques which require having access to a family of states, as a function of an external parameter, which is not suitable for many quantum simulators. Here, we propose an ancilla-free experimental scheme for the measurement of this invariant, without requiring any knowledge of the Hamiltonian. Specifically, we use the statistical correlations of randomized measurements to infer the MBCN of a wavefunction. Remarkably, our results apply to disk-like geometries that are more amenable to current quantum simulator architectures.

Categories: Journals, Physics

### Hydrodynamics of non-integrable systems from relaxation-time approximation. (arXiv:2005.13546v1 [cond-mat.stat-mech])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

We develop a general kinetic theory framework to describe the hydrodynamics of strongly interacting, nonequilibrium quantum systems in which integrability is weakly broken, leaving a few residual conserved quantities. This framework is based on a generalized relaxation-time approximation; it gives a simple, but surprisingly accurate, prescription for computing nonequilibrium transport even in strongly interacting systems. This approximation reproduces the crossover from generalized to conventional hydrodynamics in interacting one-dimensional Bose gases with integrability-breaking perturbations, both with and without momentum conservation. It also predicts the hydrodynamics of chaotic quantum spin chains, in good agreement with matrix product operator calculations.

Categories: Journals, Physics

### Single-qubit rotations in parameterized quantum circuits. (arXiv:2005.13548v1 [quant-ph])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

With the advent of hybrid quantum classical algorithms using parameterized quantum circuits the question of how to optimize these algorithms and circuits emerges. In this paper we show that the number of single-qubit rotations in parameterized quantum circuits can be decreased without compromising the expressibility or entangling capability of the circuit. We also compare expressibility and entangling capability across different number of qubits in parameterized quantum circuits. We also consider a parameterized photonics quantum circuit, without any single-qubit rotations, which yields an expressibility and entangling capability comparable to the best regular parameterized quantum circuits.

Categories: Journals, Physics

### QMetrology from QCosmology: Study with Entangled Two Qubit Open Quantum System in De Sitter Space. (arXiv:2005.13555v1 [hep-th])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

In this paper, our prime objective is to apply the techniques of {\it parameter estimation theory} and the concept of {\it Quantum Metrology} in the form of {\it Fisher Information} to investigate the role of certain physical quantities in the open quantum dynamics of a two entangled qubit system under the Markovian approximation. There exist various physical parameters which characterize such system, but can not be treated as any quantum mechanical observable. It becomes imperative to do a detailed parameter estimation analysis to determine the physically consistent parameter space of such quantities. We apply both Classical Fisher Information (CFI) and Quantum Fisher Information (QFI) to correctly estimate these parameters, which play significant role to describe the out-of-equilibrium and the long range quantum entanglement phenomena of open quantum system. {\it Quantum Metrology}, compared to {\it classical parameter estimation theory}, plays a two-fold superior role, improving the precision and accuracy of parameter estimation. Additionally, in this paper, we present a new avenue in terms of {\it Quantum Metrology}, which beats the classical parameter estimation. We also present an interesting result of \textit{revival of out-of-equilibrium feature at the late time scales, arising due to the long-range quantum entanglement at early time scale and provide a physical interpretation for the same in terms of Bell's Inequality Violation in early time scale giving rise to non-locality.

Categories: Journals, Physics

### Persistent dark states in anisotropic central spin models. (arXiv:2005.13556v1 [cond-mat.str-el])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

Long-lived dark states, in which an experimentally accessible qubit is not in thermal equilibrium with a surrounding spin bath, are pervasive in solid-state systems. We explain the ubiquity of dark states in a large class of inhomogenous central spin models using the proximity to integrable lines with exact dark eigenstates. At numerically accessible sizes, dark states persist as eigenstates at large deviations from integrability, and the qubit retains memory of its initial polarization at long times. Although the eigenstates of the system are chaotic, exhibiting exponential sensitivity to small perturbations, they do not satisfy the eigenstate thermalization hypothesis. Rather, we predict long relaxation times that increase exponentially with system size. We propose that this intermediate chaotic but non-ergodic regime characterizes mesoscopic quantum dot and diamond defect systems, as we see no numerical tendency towards conventional thermalization with a finite relaxation time.

Categories: Journals, Physics

### Scaling up the Anderson transition in random-regular graphs. (arXiv:2005.13571v1 [cond-mat.dis-nn])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

We study the Anderson transition in lattices with the connectivity of a random-regular graph. Our results indicate that fractal dimensions are continuous across the transition, but a discontinuity occurs in their derivatives, implying the non-ergodicity of the metal near the Anderson transition. A critical exponent $\nu = 1.00 \pm0.02$ and critical disorder $W= 18.2\pm 0.1$ are found via a scaling approach. Our data support that the predictions of the relevant Gaussian Ensemble are only recovered at zero disorder.

Categories: Journals, Physics

### Borromean states in discrete-time quantum walks. (arXiv:2005.13588v1 [quant-ph])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

We consider discrete-time quantum walks of many interacting particles. For a specific kind of pairwise interaction, we find peculiar multipartite bound states which fall apart if any subsystem is removed. This provides a conceptually simple physical model of structures known as Borromean rings. Interestingly, our approach highlights the role of entanglement in such systems. In order to form a Borromean bound state, the particles need to exhibit Greenberger-Horne-Zeillinger (GHZ) entanglement. Moreover, we discuss our findings in the context of formation of composite particles.

Categories: Journals, Physics

### Measurement-induced entanglement transitions in many-body localized systems. (arXiv:2005.13603v1 [quant-ph])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

The resilience of quantum entanglement to a classicality-inducing environment is tied to fundamental aspects of quantum many-body systems. The dynamics of entanglement has recently been studied in the context of measurement-induced entanglement transitions, where the steady-state entanglement collapses from a volume-law to an area-law at a critical measurement probability $p_{c}$. Interestingly, there is a distinction in the value of $p_{c}$ depending on how well the underlying unitary dynamics scramble quantum information. For strongly chaotic systems, $p_{c} > 0$, whereas for weakly chaotic systems, such as integrable models, $p_{c} = 0$. In this work, we investigate these measurement-induced entanglement transitions in a system where the underlying unitary dynamics are many-body localized (MBL). We demonstrate that the emergent integrability in an MBL system implies a qualitative difference in the nature of the measurement-induced transition depending on the measurement basis, with $p_{c} > 0$ when the measurement basis is scrambled and $p_{c} = 0$ when it is not. This feature is not found in Haar-random circuit models, where all local operators are scrambled in time. When the transition occurs at $p_{c} > 0$, we use finite-size scaling to obtain the critical exponent $\nu = 1.3(2)$, close to the value for 2+0D percolation. We also find a dynamical critical exponent of $z = 0.98(4)$ and logarithmic scaling of the R\'{e}nyi entropies at criticality, suggesting an underlying conformal symmetry at the critical point. This work further demonstrates how the nature of the measurement-induced entanglement transition depends on the scrambling nature of the underlying unitary dynamics. This leads to further questions on the control and simulation of entangled quantum states by measurements in open quantum systems.

Categories: Journals, Physics

### Calculating eigenvalues and eigenvectors of parameter-dependent hamiltonians using an adaptative wave operator method. (arXiv:2005.13611v1 [physics.comp-ph])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

We propose a wave operator method to calculate eigenvalues and eigenvectors of large parameter-dependent matrices, using an adaptative active subspace. We consider a hamiltonian which depends on external adjustable or adiabatic parameters, using adaptative projectors which follow the successive eigenspaces when the adjustable parameters are modified. The method can also handle non-hermitian hamiltonians. An iterative algorithm is derived and tested through comparisons with a standard wave operator algorithm using a fixed active space and with a standard block-Davidson method. The proposed approach is competitive, it converges within a few dozen iterations at constant memory cost. We first illustrate the abilities of the method on a 4-D coupled oscillator model hamiltonian. A more realistic application to molecular photodissociation under intense laser fields with varying intensity or frequency is also presented. Maps of photodissociation resonances of H${}_2^+$ in the vicinity of exceptional points are calculated as an illustrative example.

Categories: Journals, Physics

### Finite Quantum Instruments. (arXiv:2005.13642v1 [quant-ph])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

This article considers quantum systems described by a finite-dimensional complex Hilbert space $H$. We first define the concept of a finite observable on $H$. We then discuss ways of combining observables in terms of convex combinations, post-processing and sequential products. We also define complementary and coexistent observables. We then introduce finite instruments and their related compatible observables. The previous combinations and relations for observables are extended to instruments and their properties are compared. We present four types of instruments; namely, identity, trivial, L\"uders and Kraus instruments. These types are used to illustrate different ways that instruments can act. We next consider joint probabilities for observables and instruments. The article concludes with a discussion of measurement models and the instruments they measure.

Categories: Journals, Physics

### Exact spectral function of a Tonks-Girardeau gas in a lattice. (arXiv:2005.13646v1 [cond-mat.quant-gas])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

The single-particle spectral function of a strongly correlated system is an essential ingredient to describe its dynamics and transport properties. We develop a general method to calculate the exact spectral function of a strongly interacting one-dimensional Bose gas in the Tonks-Girardeau regime, valid for any type of confining potential, and apply it to bosons on a lattice to obtain the full spectral function, at all energy and momentum scales. We find that it displays three main singularity lines. The first two can be identified as the analogs of Lieb-I and Lieb-II modes of a uniform fluid; the third one, instead, is specifically due to the presence of the lattice. We show that the spectral function displays a power-law behaviour close to the Lieb-I and Lieb-II singularities, as predicted by the non-linear Luttinger liquid description, and obtain the exact exponents. In particular, the Lieb-II mode shows a divergence in the spectral function, differently from what happens in the dynamical structure factor, thus providing a route to probe it in experiments with ultracold atoms.

Categories: Journals, Physics

### Orthogonal Projections on Hyperplanes Intertwined With Unitaries. (arXiv:2005.13658v1 [quant-ph])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

Consider a sequence in a finite-dimensional complex (resp. real) vector space arising as the iterates of an arbitrary point under the composition of a unitary (resp. orthogonal) map with the orthogonal projection on the hyperplane orthogonal to the starting point. We show that, generically, the series of the squared norms of those points sums to the dimension of the underlying space. The exact formula for this series in non-generic cases is provided as well, along with the quantum-mechanical interpretation of this result.

Categories: Journals, Physics

### Rydberg spectrum of a single trapped Ca$^+$ ion: A Floquet analysis. (arXiv:2005.13659v1 [physics.atom-ph])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

We compute the Rydberg spectrum of a single Ca$^+$ ion in a Paul trap by incorporating various internal and external coupling terms of the ion to the trap in the Hamiltonian. The coupling terms include spin-orbit coupling in Ca$^+$, charge (electron and ionic core) coupling to the radio frequency and static fields, ion-electron coupling in the Paul trap, and ion center-of-mass coupling. The electronic Rydberg states are precisely described by a one-electron model potential for e$^-$+Ca$^{2+}$, and accurate eigenenergies, quantum defect parameters, and static and tensor polarizabilities for a number of excited Rydberg states are obtained. The time-periodic rf Hamiltonian is expanded in the Floquet basis, and the trapping-field-broadened Rydberg lines are compared with recent observations of Ca$^+(23P)$ and Ca$^+(52F)$ Rydberg lines.

Categories: Journals, Physics

### Extraction of many-body Chern number from a single wave function. (arXiv:2005.13677v1 [cond-mat.str-el])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

The quantized Hall conductivity of integer and fractional quantum Hall (IQH and FQH) states is directly related to a topological invariant, the many-body Chern number. The conventional calculation of this invariant in interacting systems requires a family of many-body wave functions parameterized by twist angles in order to calculate the Berry curvature. In this paper, we demonstrate how to extract the Chern number given a single many-body wave function, without knowledge of the Hamiltonian.

For FQH states, our method requires one additional integer invariant as input: the number of $2\pi$ flux quanta, $s$, that must be inserted to obtain a topologically trivial excitation. As we discuss, $s$ can be obtained in principle from the degenerate set of ground state wave functions on the torus, without knowledge of the Hamiltonian.

We perform extensive numerical simulations involving IQH and FQH states to validate these methods.

Categories: Journals, Physics

### Microscopic Models for Kitaev's Sixteenfold Way of Anyon Theories. (arXiv:2005.13683v1 [cond-mat.str-el])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

In two dimensions, the topological order described by $\mathbb{Z}_2$ gauge theory coupled to free or weakly interacting fermions with a nonzero spectral Chern number $\nu$ is classified by $\nu \; \mathrm{mod}\; 16$ as predicted by Kitaev [Ann. Phys. 321, 2 (2006)]. Here we provide a systematic and complete construction of microscopic models realizing this so-called sixteenfold way of anyon theories. These models are defined by $\Gamma$ matrices satisfying the Clifford algebra, enjoy a global $\mathrm{SO}(\nu)$ symmetry, and live on either square or honeycomb lattices depending on the parity of $\nu$. We show that all these models are exactly solvable by using a Majorana representation and characterize the topological order by calculating the topological spin of an anyonic quasiparticle and the ground-state degeneracy. The possible relevance of the $\nu=2$ and $\nu=3$ models to materials with Kugel-Khomskii-type spin-orbital interactions is discussed.

Categories: Journals, Physics

### Multipartite Gaussian Entanglement of Formation. (arXiv:2005.13733v1 [quant-ph])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

Entanglement of formation is a fundamental measure that quantifies the entanglement of bipartite quantum states. This measure has recently been extended into multipartite states taking the name $\alpha$-entanglement of formation. In this work, we follow an analogous multipartite extension for the Gaussian version of entanglement of formation, and focusing on the the finest partition of a multipartite Gaussian state we show this measure is fully additive and computable for 3-mode Gaussian states.

Categories: Journals, Physics

### Universal compressive characterization of quantum dynamics. (arXiv:2005.13776v1 [quant-ph])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

Recent quantum technologies utilize complex multidimensional processes that govern the dynamics of quantum systems. We develop an adaptive diagonal-element-probing compression technique that feasibly characterizes any unknown quantum processes using much fewer measurements compared to conventional methods. This technique utilizes compressive projective measurements that are generalizable to arbitrary number of subsystems. Both numerical analysis and experimental results with unitary gates demonstrate low measurement costs, of order $O(d^2)$ for $d$-dimensional systems, and robustness against statistical noise. Our work potentially paves the way for a reliable and highly compressive characterization of general quantum devices.

Categories: Journals, Physics

### Holographic Pseudo Entropy. (arXiv:2005.13801v1 [hep-th])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

We introduce a quantity, called pseudo entropy, as a generalization of entanglement entropy via post-selection. In the AdS/CFT correspondence, this quantity is dual to areas of minimal area surfaces in time-dependent Euclidean spaces which are asymptotically AdS. We study its basic properties and classifications in qubit systems. In specific examples, we provide a quantum information theoretic meaning of this new quantity as an averaged number of Bell pairs when the post-section is performed. We also present properties of the pseudo entropy for random states. We then calculate the pseudo entropy in the presence of local operator excitations for both the two dimensional free massless scalar CFT and two dimensional holographic CFTs. We find a general property in CFTs that the pseudo entropy is highly reduced when the local operators get closer to the boundary of the subsystem. We also compute the holographic pseudo entropy for a Janus solution, dual to an exactly marginal perturbation of a two dimensional CFT and find its agreement with a perturbative calculation in the dual CFT. We show the linearity property holds for holographic states, where the holographic pseudo entropy coincides with a weak value of the area operator. Finally, we propose a mixed state generalization of pseudo entropy and give its gravity dual.

Categories: Journals, Physics

### High-accuracy Ising machine using Kerr-nonlinear parametric oscillators with local four-body interactions. (arXiv:2005.13819v1 [quant-ph])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

A two-dimensional array of Kerr-nonlinear parametric oscillators (KPOs) with local four-body interactions is a promising candidate for realizing an Ising machine with all-to-all spin couplings, based on adiabatic quantum computation in the Lechner-Hauke-Zoller (LHZ) scheme. However its performance has been evaluated for only a few KPOs. By numerically simulating more KPOs, here we show that the performance can be dramatically improved by reducing inhomogeneity in photon numbers induced by the four-body interactions. The discrepancies of the photon numbers can be corrected by tuning the detunings of KPOs depending on their positions, without monitoring their states during adiabatic time evolution. This correction can be used independent of the number of KPOs in the LHZ scheme and thus can be applied to large-scale implementation.

Categories: Journals, Physics

### Observation of exceptional point in a PT broken non-Hermitian system simulated using a quantum circuit. (arXiv:2005.13828v1 [quant-ph])

arXiv.org: Quantum Physics - Fri, 2020-05-29 14:52

Exceptional points (EPs), the degeneracy point of non-Hermitian systems, have recently attracted great attention after its ability to greatly enhance the sensitivity of micro-cavities is demonstrated experimentally. Unlike the usual degeneracies in Hermitian systems, at EPs, both the eigenenergies and eigenvectors coalesce. Although some of the exotic properties and potential applications of EPs are explored, the range of EPs studied is largely limited by the experimental capability. Some of the systems, e.g. with higher-order EPs, are hard to achieve with conventional simulations. Here we propose an extendable method to simulate non-Hermitian systems on the quantum circuits, where a wide range of EPs can be studied. The system is inherently parity-time (PT) broken due to the non-symmetric controlling effects and post-selection. Inspired by the quantum Zeno effect, the circuit structure grantees the success rate of the post-selection. A sample circuit is implemented in a quantum programming framework, and the phase transition at EP is demonstrated. Considering the scalable and flexible nature of quantum circuits, our model is capable of simulating large scale systems with higher-order EPs. We believe this work may lead to broader applications of quantum computers and provide a tool to the studies for non-Hermitian systems and the associated EPs.

Categories: Journals, Physics


text:
# Control Mechanisms¶

## Overview¶

A ControlMechanism is a subclass of ModulatoryMechanism that is restricted to using only ControlSignals and not GatingSignals. Accordingly, its constructor has a control_signals argument in place of a modulatory_signals argument. It also lacks any attributes related to gating. In all other respects it is identical to its parent class, ModulatoryMechanism.

Examples

The following example creates a ControlMechanism by specifying its objective_mechanism using a constructor that specifies the OutputStates to be monitored by its objective_mechanism and the function used to evaluated these:

>>> import psyneulink as pnl
>>> my_transfer_mech_A = pnl.TransferMechanism(name="Transfer Mech A")
>>> my_DDM = pnl.DDM(name="My DDM")
>>> my_transfer_mech_B = pnl.TransferMechanism(function=pnl.Logistic,
...                                            name="Transfer Mech B")

>>> my_control_mech = pnl.ControlMechanism(
...                          objective_mechanism=pnl.ObjectiveMechanism(monitor=[(my_transfer_mech_A, 2, 1),
...                                                                               my_DDM.output_states[pnl.RESPONSE_TIME]],
...                                                                     name="Objective Mechanism"),
...                          function=pnl.LinearCombination(operation=pnl.PRODUCT),
...                          control_signals=[(pnl.THRESHOLD, my_DDM),
...                                           (pnl.GAIN, my_transfer_mech_B)],
...                          name="My Control Mech")

This creates an ObjectiveMechanism for the ControlMechanism that monitors the primary OutputState of my_Transfer_mech_A and the RESPONSE_TIME OutputState of my_DDM; its function first multiplies the former by 2 before, then takes product of their values and passes the result as the input to the ControlMechanism. The ControlMechanism’s function uses this value to determine the allocation for its ControlSignals, that control the value of the threshold parameter of my_DDM and the gain parameter of the Logistic Function for my_transfer_mech_B.

The following example specifies the same set of OutputStates for the ObjectiveMechanism, by assigning them directly to the objective_mechanism argument:

>>> my_control_mech = pnl.ControlMechanism(
...                             objective_mechanism=[(my_transfer_mech_A, 2, 1),
...                                                  my_DDM.output_states[pnl.RESPONSE_TIME]],
...                             control_signals=[(pnl.THRESHOLD, my_DDM),
...                                              (pnl.GAIN, my_transfer_mech_B)])
...

Note that, while this form is more succinct, it precludes specifying the ObjectiveMechanism’s function. Therefore, the values of the monitored OutputStates will be added (the default) rather than multiplied.

The ObjectiveMechanism can also be created on its own, and then referenced in the constructor for the ControlMechanism:

>>> my_obj_mech = pnl.ObjectiveMechanism(monitored_output_states=[(my_transfer_mech_A, 2, 1),
...                                                               my_DDM.output_states[pnl.RESPONSE_TIME]],
...                                      function=pnl.LinearCombination(operation=pnl.PRODUCT))

>>> my_control_mech = pnl.ControlMechanism(
...                        objective_mechanism=my_obj_mech,
...                        control_signals=[(pnl.THRESHOLD, my_DDM),
...                                         (pnl.GAIN, my_transfer_mech_B)])

Here, as in the first example, the constructor for the ObjectiveMechanism can be used to specify its function, as well as the OutputState that it monitors.

See Specifying Control for a System for examples of how a ControlMechanism, the OutputStates its objective_mechanism, and its control_signals can be specified for a System.

## Class Reference¶

class psyneulink.core.components.mechanisms.adaptive.control.controlmechanism.ControlMechanism(system=None monitor_for_control=None, objective_mechanism=None, function=Linear, control_signals=None, modulation=ModulationParam.MULTIPLICATIVE combine_costs=np.sum, compute_reconfiguration_cost=None, compute_net_outcome=lambda x, y:x-y, params=None, name=None, prefs=None)

Subclass of AdaptiveMechanism that modulates the parameter(s) of one or more Component(s).

Parameters: system (System or bool : default None) – specifies the System to which the ControlMechanism should be assigned as its controller. monitor_for_control (List[OutputState or Mechanism] : default None) – specifies the OutputStates to be monitored by the ObjectiveMechanism specified in the objective_mechanism argument; for any Mechanisms specified, their primary OutputState are used. objective_mechanism (ObjectiveMechanism or List[OutputState specification] : default None) – specifies either an ObjectiveMechanism to use for the ControlMechanism, or a list of the OutputStates it should monitor; if a list of OutputState specifications is used, a default ObjectiveMechanism is created and the list is passed to its monitored_output_states argument. function (TransferFunction : default Linear(slope=1, intercept=0)) – specifies function used to combine values of monitored OutputStates. control_signals (ControlSignal specification or List[ControlSignal specification, ..]) – specifies the parameters to be controlled by the ControlMechanism; a ControlSignal is created for each (see Specifying ControlSignals for details of specification). modulation (ModulationParam : ModulationParam.MULTIPLICATIVE) – specifies the default form of modulation used by the ControlMechanism’s ControlSignals, unless they are individually specified. combine_costs (Function, function or method : default np.sum) – specifies function used to combine the cost of the ControlMechanism’s control_signals; must take a list or 1d array of scalar values as its argument and return a list or array with a single scalar value. compute_reconfiguration_cost (Function, function or method : default None) – specifies function used to compute the ControlMechanism’s reconfiguration_cost; must take a list or 2d array containing two lists or 1d arrays, both with the same shape as the ControlMechanism’s control_allocation attribute, and return a scalar value. compute_net_outcome (Function, function or method : default lambda outcome, cost: outcome-cost) – function used to combine the values of its outcome and costs attributes; must take two 1d arrays (outcome and cost) with scalar values as its arguments and return an array with a single scalar value. params (Dict[param keyword: param value] : default None) – a parameter dictionary that can be used to specify the parameters for the Mechanism, parameters for its function, and/or a custom function and its parameters. Values specified for parameters in the dictionary override any assigned to those parameters in arguments of the constructor. name (str : default see name) – specifies the name of the ControlMechanism. prefs (PreferenceSet or specification dict : default Mechanism.classPreferences) – specifies the PreferenceSet for the ControlMechanism; see prefs for details.
system

System_Base – The System for which the ControlMechanism is a controller. Note that this is distinct from a Mechanism’s systems attribute, which lists all of the Systems to which a Mechanisms belongs – a ControlMechanism can belong to but not be the controller of a System.

objective_mechanism

ObjectiveMechanismObjectiveMechanism that monitors and evaluates the values specified in the ControlMechanism’s objective_mechanism argument, and transmits the result to the ControlMechanism’s OUTCOME input_state.

monitor_for_control

List[OutputState] – each item is an OutputState monitored by the ObjectiveMechanism listed in the ControlMechanism’s objective_mechanism attribute; it is the same as that ObjectiveMechanism’s monitored_output_states attribute (see Monitor for specification). The value of the OutputStates in the list are used by the ObjectiveMechanism to generate the ControlMechanism’s input.

monitored_output_states_weights_and_exponents

List[Tuple(float, float)] – each tuple in the list contains the weight and exponent associated with a corresponding OutputState specified in monitor_for_control; these are the same as those in the monitored_output_states_weights_and_exponents attribute of the objective_mechanism, and are used by the ObjectiveMechanism’s function to parametrize the contribution made to its output by each of the values that it monitors (see ObjectiveMechanism Function).

outcome

1d array – the value of the ControlMechanism’s primary InputState, which receives its Projection from the OUTCOME OutputState of its objective_mechanism.

function

TransferFunction : default Linear(slope=1, intercept=0) – determines how the value s of the OutputStates specified in the monitor_for_control argument of the ControlMechanism’s constructor are used to generate its control_allocation.

control_allocation

2d array – each item is the value assigned as the allocation for the corresponding ControlSignal listed in the control_signals attribute; the control_allocation is the same as the ControlMechanism’s value attribute).

control_signals

ContentAddressableList[ControlSignal] – list of the ControlSignals for the ControlMechanism, including any inherited from a system for which it is a controller (same as ControlMechanism’s output_states attribute); each sends a ControlProjection to the ParameterState for the parameter it controls

compute_reconfiguration_cost

Function, function or method – function used to compute the ControlMechanism’s reconfiguration_cost; result is a scalar value representing the difference — defined by the function — between the values of the ControlMechanism’s current and last control_alloction, that can be accessed by reconfiguration_cost attribute.

costs

list – current costs for the ControlMechanism’s control_signals, computed for each using its compute_costs method.

combine_costs

Function, function or method – function used to combine the cost of its control_signals; result is an array with a scalar value that can be accessed by combined_costs.

Note

This function is distinct from the combine_costs_function of a ControlSignal. The latter combines the different costs for an individual ControlSignal to yield its overall cost; the ControlMechanism’s combine_costs function combines those costs for its control_signals.

combined_costs

1d array – result of the ControlMechanism’s combine_costs function.

compute_net_outcome

Function, function or method – function used to combine the values of its outcome and costs attributes; result is an array with a scalar value that can be accessed by the the net_outcome attribute.

net_outcome

1d array – result of the ControlMechanism’s compute_net_outcome function.

control_projections

List[ControlProjection] – list of ControlProjections, one for each ControlSignal in control_signals.

modulation

ModulationParam – the default form of modulation used by the ControlMechanism’s ControlSignals, unless they are individually specified.

name

str – the name of the ControlMechanism; if it is not specified in the name argument of the constructor, a default is assigned by MechanismRegistry (see Naming for conventions used for default and duplicate names).

prefs

PreferenceSet or specification dict – the PreferenceSet for the ControlMechanism; if it is not specified in the prefs argument of the constructor, a default is assigned using classPreferences defined in __init__.py (see PreferenceSet for details).

assign_as_controller(system: psyneulink.core.components.shellclasses.System_Base, context=<ContextFlags.COMMAND_LINE: 512>)

Assign ControlMechanism as controller for a System.

system must be a System for which the ControlMechanism should be assigned as the controller. If the specified System already has a controller, it will be replaced by the current one, and the current one will inherit any ControlSignals previously specified for the old controller or the System itself. If the current one is already the controller for another System, it will be disabled for that System.

outputStateTypes

alias of ControlSignal