text:
### Semi-Supervised Learning by Augmented Distribution Alignment

In this work, we propose a simple yet effective semi-supervised learning approach called Augmented Distribution Alignment. We reveal that an essential sampling bias exists in semi-supervised learning due to the limited amount of labeled samples, which often leads to a considerable empirical distribution mismatch between labeled data and unlabeled data. To this end, we propose to align the empirical distributions of labeled and unlabeled data to alleviate the bias. On one hand, we adopt an adversarial training strategy to minimize the distribution distance between labeled and unlabeled data as inspired by domain adaptation works. On the other hand, to deal with the small sample size issue of labeled data, we also propose a simple interpolation strategy to generate pseudo training samples. Those two strategies can be easily implemented into existing deep neural networks. We demonstrate the effectiveness of our proposed approach on the benchmark SVHN and CIFAR10 datasets, on which we achieve new state-of-the-art error rates of $3.54\%$ and $10.09\%$, respectively. Our code will be available at \url{https://github.com/qinenergy/adanet}.

### Semi-supervised Protein Classification Using Cluster Kernels

A key issue in supervised protein classification is the representation of input sequencesof amino acids. Recent work using string kernels for protein datahas achieved state-of-the-art classification performance. However, suchrepresentations are based only on labeled data -- examples with known 3D structures, organized into structural classes -- while in practice, unlabeled data is far more plentiful.

### Rademacher Complexity Bounds for a Penalized Multiclass Semi-Supervised Algorithm

We propose Rademacher complexity bounds for multiclass classifiers trained with a two-step semi-supervised model. In the first step, the algorithm partitions the partially labeled data and then identifies dense clusters containing $\kappa$ predominant classes using the labeled training examples such that the proportion of their non-predominant classes is below a fixed threshold. In the second step, a classifier is trained by minimizing a margin empirical loss over the labeled training set and a penalization term measuring the disability of the learner to predict the $\kappa$ predominant classes of the identified clusters. The resulting data-dependent generalization error bound involves the margin distribution of the classifier, the stability of the clustering technique used in the first step and Rademacher complexity terms corresponding to partially labeled training data. Our theoretical result exhibit convergence rates extending those proposed in the literature for the binary case, and experimental results on different multiclass classification problems show empirical evidence that supports the theory.

### RSSL: Semi-supervised Learning in R

In this paper, we introduce a package for semi-supervised learning research in the R programming language called RSSL. We cover the purpose of the package, the methods it includes and comment on their use and implementation. We then show, using several code examples, how the package can be used to replicate well-known results from the semi-supervised learning literature.

### Word Sense Disambiguation with Semi-Supervised Learning

Current word sense disambiguation (WSD) systems based on supervised learning are still limited in that they do not work well for all words in a language. One of the main reasons is the lack of sufficient training data. In this paper, we investigate the use of unlabeled training data for WSD, in the framework of semi-supervised learning. Four semisupervised learning algorithms are evaluated on 29 nouns of Senseval-2 (SE2) English lexical sample task and SE2 English all-words task. Empirical results show that unlabeled data can bring significant improvement in WSD accuracy.

format: plain


text:
Matrices made negative semidefinite, but not simultaneously

Consider matrices $A \in \mathbb{R}^{n \times n}$, $B \in \mathbb{R}^{n \times m}$, such that $A$ has at least $1$ strictly positive eigenvalue.

Let $X_1, X_2 \in \mathbb{R}^{n \times n}$ such that $X_1, X_2 \succ 0$ (positive definite) and, respectively for each $i \in \{1,2\}$, $\exists Y_i \in \mathbb{R}^{m \times n}$ such that:

$$(A+B Y_i)^\top X_i + X_i (A+BY_i) \preccurlyeq 0 \ \text{ (negative semidefinite)}$$

(1) Find a case in which $\nexists Y \in \mathbb{R}^{m \times n}$ such that we simultaneously have

$$(A+B Y)^\top X_1 + X_1 (A+B Y) \preccurlyeq 0 \ \text{ and } \ (A + B Y)^\top X_2 + X_2 (A+BY) \preccurlyeq 0$$

(2) Find $x \in \mathbb{R}^n$ such that: $\nexists y \in \mathbb{R}^m$ such that we simultaneously have

$$x^\top X_1 (Ax+B y) \leq 0 \ \text{ and } \ x^\top X_2 (Ax+B y) \leq 0$$

-


text:
## Intermediate Algebra for College Students (7th Edition)

A quadratic equation can be represented as $px^2+qx+r=0$; where $p,q,r$ are real numbers and $p$ cannot be equal to zero, that is , $p \neq 0$


text:
# ISEE Middle-Level Math Practice Test Questions

Preparing your student for the ISEE Middle-Level Math test? Try these free ISEE Middle-Level Math Practice questions. Reviewing practice questions is the best way to brush up your student’s Math skills. Here, we walk you through solving 10 common ISEE Middle-Level Math practice problems covering the most important math concepts on the ISEE Middle-Level Math test.

These ISEE Middle-Level Math practice questions are designed to be similar to those found on the real ISEE Middle-Level Math test. They will assess your student’s level of preparation and will give you a better idea of what students need to study for the exam.

## 10 Sample ISEE Middle Level Math Practice Questions

1- What is the greatest common factor of 30 and 48?

☐A. 6

☐B. 7

☐C. 5

☐D. 8

2- Simplify.
$$2x^2+3y^5-x^2+2z^3-2y^2+2x^3-2y^5+5z^3$$

☐A. $$3x^2-2y^2+y^5+7z^3$$

☐B. $$x^2+2x^3-2y^2+y^5+7z^3$$

☐C. $$x^2+2x^3+3y^5+7z^3$$

☐D. $$x^2+2x^3-2y^2+5y^5+7z^3$$

3- The ratio of boys and girls in a class is $$4:7$$. If there are 44 students in the class, how many more boys should be enrolled to make the ratio $$1:1$$?

☐A. 8

☐B. 10

☐C. 12

☐D. 16

4- Solve for $$x: 4(x + 1) = 6(x − 4) + 20$$

☐A. 12

☐B. 4

☐C. 6.2

☐D. 5.5

5- A company pays its employer \$7000 plus $$2\%$$ of all sales profit. If $$x$$ is all sold profit, which of the following represents the employer’s revenue?

☐A. $$0.02x$$

☐B. $$0.98x-7000$$

☐C. $$0.02x+7000$$

☐D. $$0.98x+7000$$

6- Car A use 8-liter petrol per 100 kilometers; car B use 6-liter petrol per 100 kilometers. If both cars drive 250 kilometers, how much more petrol does car A use?

☐A. 5

☐B. 10

☐C. 15

☐D. 20

7- $$y=-4x-8$$

☐A. The quantity in Column A is greater.

☐B. The quantity in Column B is greater.

☐C. The two quantities are equal.

☐D. The relationship cannot be determined from the information given.

8-

☐A. The quantity in Column A is greater.

☐B. The quantity in Column B is greater.

☐C. The two quantities are equal.

☐D. The relationship cannot be determined from the information given.

9- The average age of Joe, Michelle, and Nicole is 32.

☐A. The quantity in Column A is greater.

☐B. The quantity in Column B is greater.

☐C. The two quantities are equal.

☐D. The relationship cannot be determined from the information given.

10-

☐A. The quantity in Column A is greater.

☐B. The quantity in Column B is greater.

☐C. The two quantities are equal.

☐D. The relationship cannot be determined from the information given.

## Best ISEE Middle Level Math Prep Resource for 2020

1- A
Prime factorizing of $$30=2×3×5$$
Prime factorizing of $$48=2×2×2×2×3$$
GCF$$=2×3=6$$

2- B
$$2x^2+3y^5-x^2+2z^3-2y^2+2x^3-2y^5+5z^3=$$
$$2x^2-x^2+2x^3-2y^2+3y^5-2y^5+2z^3+5z^3=$$
$$x^2+2x^3-2y^2+y^5+7z^3$$

3- C
The ratio of boy to girls is $$4:7$$. Therefore, there are 4 boys out of 11 students. To find the answer, first divide the total number of students by 11, then multiply the result by 4.
$$44 ÷ 11 = 4 ⇒ 4 × 4 = 16$$
There are 16 boys and $$28 (44 – 16)$$ girls. So, 12 more boys should be enrolled to make the ratio$$1:1$$

4- B
Simplify:
$$4(x + 1) = 6(x − 4) + 20$$
$$4x + 4 = 6x − 24 + 20$$
$$4x + 4 = 6x – 4$$
Subtract $$4x$$from both sides:
$$4 = 2x – 4$$
$$8 = 2x$$
$$4 = x$$

5- C
Employer’s revenue: $$0.2x+7000$$

6- A
Petrol of car A in $$250 \space km = \frac{(8×250)}{100} = 20$$
Petrol of car A in 250 km = $$\frac{(6×250)}{100} = 15$$
$$20-15=5$$

7- B
Column A: The value of $$x$$ when $$y = 12:$$
$$y=-4x-8→12=-4x-8→-4x=20→x=-5$$
Column $$B: -4$$
$$-4 \space is \space greater \space than \space -5.$$

8- A
Column A: Simplify.
$$\sqrt{36}+\sqrt{36}=6+6=12$$
12 is greater than $$\sqrt{72}$$ . $$(\sqrt{144}=12)$$

9- D
Column A: Based on information provided, we cannot find the average age of Joe and Michelle or average age of Michelle and Nicole.

10- A
Column A: Simplify.
$$\sqrt{(121-64)}=\sqrt{57}$$
Column B:
$$\sqrt{121}-\sqrt{64}=11-8=3$$
$$\sqrt{57}$$ is bigger than 3. $$(\sqrt 9 = 3)$$

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text:
# Many representations as a sum of three squares

Let $r_3(n) = \left|\{(a,b,c)\in {\mathbb Z}^3 :\, a^2+b^2+c^2=n \}\right|$. I am looking for the maximum asymptotic size of $r_3(n)$. That is, the maximum number of representations that a number can have as a sum of three squares.

Gauss proved that $r_3(n) = \frac{A\sqrt{n}}{\pi}\sum_{m-1}^\infty\left(\frac{-n}{m}\right) \frac{1}{m}$ (where $\left(\frac{a}{b}\right)$ is the Jacobi-Legendre symbol). Another similar-looking formula can be found, for example, in the bottom of the first page of this paper

A simple and non-constructive argument shows that there are infinitely many $n$'s for which $r_3(n) \ge c\sqrt{n}$ (for some constant $c$). Moreover, in an old paper of Erdős he mentions that $r_3(n) \ge c\sqrt{n}\log \log n$ can also be obtained. Unfortunately, Erdős does not provide any details and only states "By deep number theoretic results".

More specifically, my questions are (1) how can one obtain Erdős' bound? (2) What $n$'s achieve the above bounds? (3) Can one use the above formula for $r_3(n)$ to show this?

• In your lower bounds for $r_3(n)$, $n$ should be $\sqrt{n}$. See the responses below for more information. – GH from MO Sep 7 '15 at 22:58

The formula for $r_3(n)$ essentially connects this with a class number of an imaginary quadratic field, or (apart from the $\sqrt{n}$ scaling) with the value of an $L$-function at $1$. So your question may be reformulated as asking how large can $L(1,\chi_{d})$ be as $d$ runs over negative fundamental discriminants ($d$ is essentially $-n$). The distribution of these values has been extensively investigated (see for example Granville and Soundararajan where you'll find more references. To create large values of $L(1,\chi_d)$ you should find a $d$ with $\chi_d(p)=1$ for all small primes $p$ up to some point $y$. One can do this with $d$ of size about $\exp(y)$. Then for most such $d$ one will have $$L(1,\chi_d) \approx \prod_{p\le y} \Big(1-\frac{\chi_d(p)}{p} \Big)^{-1} \asymp \log \log |d|$$ by Mertens. Arguments of this type go back to Littlewood and Chowla (see the linked paper for references).

• You beat me by one minute... – GH from MO Sep 7 '15 at 22:46
• @GH from MO: That's too bad -- one of us could have saved their energy. – Lucia Sep 7 '15 at 22:47
• No, it is fun! Let me upvote you... – GH from MO Sep 7 '15 at 22:52
• @AdamSheffer: Nothing wrong with what you wrote; but note that if $n=a^2+b^2+c^2$ then $n$ is the square of the distance. (i.e. you just have a scaling issue in what you wrote) – Lucia Sep 7 '15 at 23:25
• @AdamSheffer: In these arguments you can easily restrict $d$ to be in any progression that you want, and the result carries through. The constants in the bound may depend on the progressions. – Lucia Sep 9 '15 at 20:29

Let me restrict to the number of primitive representations $$r_3^\ast(n) = \left|\{(a,b,c)\in {\mathbb Z}^3 :\, a^2+b^2+c^2=n\ \text{and}\ \gcd(a,b,c)=1 \}\right|.$$ Note that $$r_3(n)$$ can be easily expressed from this quantity as $$r_3(n)=\sum_{d^2\mid n}r_3^\ast(n/d^2).$$ Clearly $$r_3^*(n)=0$$ when $$n\equiv 0,4,7\pmod{8}$$. For the remaining cases, it follows from the work of Gauss (1801) and Dirichlet (1839) on the class number that $$r_3^\ast(n) = \frac{24}{\pi}\,\sqrt{n}\,L\left(1,\left(\frac{D}{\cdot}\right)\right),$$ where $$D=\begin{cases} -4n,&n\equiv 1,2,5,6\pmod{8},\\ -n,&n\equiv 3\pmod{8}. \end{cases}$$ This tells us that the maximal (resp. minimal) order of $$r_3^*(n)$$ is determined by the maximal (resp. minimal) order of $$L\left(1,\left(\frac{D}{\cdot}\right)\right)$$, where $$D$$ depends on $$n$$ as above. Concerning the latter quantity, Littlewood (On the class-number of the corpus $$P(\sqrt{-k})$$, Proc. Lond. Math. Soc. (2) 27 (1928), 358-372) proved under GRH that $$(\log\log|D|)^{-1}\ll L\left(1,\left(\frac{D}{\cdot}\right)\right)\ll \log\log|D|,$$ and he also proved in the same article that $$L\left(1,\left(\frac{D}{\cdot}\right)\right)\gg\log\log|D|$$ holds for infinitely many $$D$$'s. This last bound was proved unconditionally by Walfisz (On the class-number of binary quadratic forms, Trav. Inst. Math. Tbilissi 11 (1942), 57-71), and was further explicated by Granville-Soundararajan (The distribution of values of $$L(1,\chi_d)$$, Geom. Funct. Anal. 13 (2003), 992-1028). In particular, see Theorem 5b on page 998 in Granville-Soundararajan's paper, which is also available as an arXiv preprint.

In short, there are infinitely many $$n$$'s such that $$r_3(n)\gg\sqrt{n}\log\log n$$, and this is sharp under GRH.

• Thanks a lot! From looking at Chowla's paper (repository.ias.ac.in/8839/1/8839.pdf) it seems to me that he derived the opposite direction. That is, that there are many $n$'s with small $L$ values. Does this somehow imply a similar result for large value? – Adam Sheffer Sep 8 '15 at 21:35
• @AdamSheffer: It seems I got confused with the references. It was Walfisz (1942) who proved that Littlewood's upper bound is sharp (apart from the constant), while Chowla (1947) proved that Littlewood's lower bound is sharp (apart from the constant). I will update my post with more precise references. (BTW I don't know a direct connection between the existence of small $L$-values and the existence of large $L$-values.) – GH from MO Sep 8 '15 at 21:59
• Now it all makes sense :) – Adam Sheffer Sep 9 '15 at 0:50
• @AdamSheffer: In fact Walfisz's paper concentrates on the negative fundamental discriminants. These are precisely the discriminants $D$ that arise from the square-free $n$'s with $n\equiv 1,2,3,5,6\pmod{8}$, see the precise formula for $D$ in my post. – GH from MO Sep 9 '15 at 20:40
• Thank you very much. Hopefully that's the end of these questions. – Adam Sheffer Sep 9 '15 at 20:51


text:
# Focus: Quantum Information Not Lost in Translation

Published March 30, 2012  |  Physics 5, 39 (2012)  |  DOI: 10.1103/Physics.5.39

#### Quantum Interface between an Electrical Circuit and a Single Atom

D. Kielpinski, D. Kafri, M. J. Woolley, G. J. Milburn, and J. M. Taylor

Published March 30, 2012

Physicists hoping to make an ultrapowerful quantum computer still haven’t settled on which quantum system to build it out of. In Physical Review Letters, a team proposes a way to faithfully transfer quantum information back and forth between two of the candidates—a single trapped ion and a superconducting circuit. If the scheme works out, it could allow researchers to explore hybrid technologies that combine the strengths of these two different quantum entities.

A simple quantum system can effectively be in two states simultaneously. When scaled up to many similar systems working in parallel, this property allows a huge amount of computational power, at least in theory. But there is not yet a clear winner among several possible building blocks to represent the individual quantum bits, or “qubits,” of quantum information. Each type of qubit has strengths and weaknesses, so researchers have recently been proposing hybrids. For example, an ion floating in an electromagnetic trap can retain its quantum state long enough to allow many sequential operations, but it’s complicated to engineer the required interactions between different ions. In contrast, the quantized electromagnetic oscillations in superconducting circuits are easier to manipulate using the complex wiring capabilities of modern microelectronics, but extending their lifetimes has been difficult. David Kielpinski of Griffith University in Nathan, Australia, and his collaborators now suggest a way to enmesh the quantum behavior of a circuit with that of an ion, to let each do what it does best.

Both the circuit and the ion reveal their quantum nature when a particular mode of oscillation is just barely excited. For the circuit, this mode is the familiar voltage and current oscillation of an $L\phantom{\rule{0}{0ex}}C$ circuit, whose frequency depends on its inductance ($L$) and its capacitance ($C$). For the ion, the researchers chose its relatively long-distance motion (tens of nanometers) back and forth inside an electromagnetic trap. The problem is that the typical oscillation frequencies are so different—the ion in the megahertz range and the circuit in the gigahertz range—so it’s hard to get them to communicate.

To bridge this thousandfold frequency gap, Kielpinski and his collaborators propose adding another electrical signal whose frequency precisely matches the difference. Mixing two signals to form a new signal at the sum or difference of their frequencies is widely used, for example, to encode audio signals for radio broadcasts. The challenge in this case is to preserve the delicate quantum information, which would be destroyed by simply irradiating an ion and circuit with a large microwave signal. To avoid this, the team proposes varying the parameters of the circuit at the required frequency, using a so-called bulk acoustic wave (BAW) to vibrate one of the plates of the capacitor in the $L\phantom{\rule{0}{0ex}}C$ circuit. Varying the capacitance would vary the resonant frequency of the circuit, and the combination signal would provide a megahertz-range electric field that would influence the ion’s motion.

The team calculates that, using existing technology and a well-designed circuit, experimenters could alter the capacitance by as much as $30$ percent, enough to generate a large signal that would match the ion frequency. Applying the BAW signal for just the right amount of time would then controllably swap the quantum states of the jiggling ion and the oscillating circuit.

If it works, one use of the new scheme would be to “let quantum information from superconducting circuits escape from the dilution refrigerator” that keeps them at ultralow temperatures, says Kielpinski. He imagines fiber-optic lines carrying quantum information between two distant circuits, with trapped ions as the intermediary technology between the optical and electronic signals. The team also proposes using the superconducting circuit to ferry quantum information between ions, perhaps allowing their interaction to be better controlled than if the two ions were placed in the same trap for a multi-ion calculation.

“An interface that connects different systems … can only help with the attempt to produce the optimal hardware,” says Steve Girvin of Yale University in New Haven, Connecticut. But he cautions that “some of these experiments are hard separately and hard-squared collectively.” Jose Aumentado of the National Institute of Standards and Technology in Boulder, Colorado, agrees that the proposal “looks really challenging, if it could be built.” But experiments on other systems have shown that quantum states can be swapped without adding noise, he says. “Parametric frequency conversion is super-important.”

–Don Monroe

Don Monroe is a freelance science writer in Murray Hill, New Jersey.


text:
Loops and guesses

I have to solve a system of equations in the following way: I first guess a value for a variable, let's call it y, calculate another variable z, which is a function of y, z(y), and then check if another condition was met on z. The simplest example that slightly resembles my problem looks like this:

a = 2;
b = 3;
y = 1; (*initial guess*)
z = x /. FindRoot[a*x + Log[x] - Log[y], {x, 0.01}];
FOCy = b*y + Log[y] + Log[z]

So I should keep doing this while FOCy is different than 0.

I usually use MATLAB for these kind of problem. There, I would discretize y on a grid and I would use a "shooting algorithm" (a fixed point using Intermediate Value Theorem). On Mathematica, I don't know what would be the recommended way to proceed.

Any thoughts?

Why not just include both equations in your FindRoot call:
a=2; b=3;


text:
The Mandelbrot set and Pi

22 views

Paul Melis

Jul 7, 1996, 3:00:00 AM7/7/96
to

Hello everybody,

when I was in a bookshop yesterday, I came across a book about chaos and
dynamics in which I found an interesting piece. It was about an article
that
was posted to USENET (I don't know if it was this group) in which the
following matter was discussed.

It appears that in the point -0.75+0i+epsilon (and also 0.25+0i+epsilon)
-
with epsilon being a fixed value - the number of iterations needed to
make
sure the point goes to infinity is quite interesting. See below (I hope I
remember this correctly) ...

epsilon | iterations needed
-------------------------------
0.1 | 3
0.01 | 31
0.001 | 314
0.0001 | 3141
0.00001 | 31415
. | .
. | .

The number of iterations needed appears to be some power of 10 times Pi,
with the decimals cut off.

?

Any feedback would be absolutely bananas :-)

Bye,

Paul
--
me...@cs.utwente.nl

Keith F. Lynch

Jul 8, 1996, 3:00:00 AM7/8/96
to

In article <31DFDB...@sms.utwente.nl>,
Paul Melis <p.m...@sms.utwente.nl> wrote:
> ... I came across a book ... in which I found an interesting piece.
> It was about an article that was posted to USENET ...

> It appears that in the point -0.75+0i+epsilon (and also

> 0.25+0i+epsilon) with epsilon being a fixed value - the number of

> iterations needed to make sure the point goes to infinity is quite
> interesting. See below (I hope I remember this correctly) ...

> epsilon | iterations needed
> -------------------------------
> 0.1 | 3
> 0.01 | 31
> 0.001 | 314

My initial reaction was that this was nonsense, since the point never
reaches infinity. And to "make sure" it will approach infinity, where
c is near 0.25, all that's necessary is that epsilon be more than zero.

But then I tried it for 0.25, seeing how many iterations it takes for
Z to exceed two. The results, in double precision C (about 15 decimal
digits), were suggestive. So I tried it again in MUMPS which has a
higher precision (about 20 decimal digits). Here were the results:

.1 9
.01 31
.001 98
.0001 313
.00001 992
.000001 3141
.0000001 9934
.00000001 31415
.000000001 99345
.0000000001 314158
.00000000001 993458
.000000000001 3141592
.0000000000001 9934590
.00000000000001 31416004

(The first number is epsilon. The second number is the number of
iterations it takes for Z = Z*Z + (0.25+epsilon) to exceed 2.0.)

Maybe those pi-like numbers are just a coincidence. But then again,
maybe not. The abberant last number can be explained away as due to
precision loss.

The value "2.0" isn't important. I would have gotten almost identical
numbers had I chosen 1.0, pi, or a trillion. Once Z leaves the vicinity
of 0.5, it takes off like an alien spaceship in _Independence Day_.

Here's the one line MUMPS program I used:

s e=1 f s e=e/10,z=0,c=0.25+e x "f j=1:1 q:z>2 s z=z*z+c" w e," ",j,!

> Does anybody know the original poster?

I did a Dejanews search, and found several articles on the topic that
were just over a year old. Some of them referred to an even earlier
article, supposedly from 1992. I can forward the 1995 articles if
you like.

I haven't looked at -0.75. But doesn't it converge on both (real)
sides of that number? I thought all real numbers from -2.0 to 0.25
were in the Mandelbrot set.
--
Keith Lynch, k...@access.digex.net
http://www.access.digex.net/~kfl/

G. A. Edgar

Jul 9, 1996, 3:00:00 AM7/9/96
to

> > Does anybody know the original poster?

That would be Dave Boll, 24 April 1991,
"Pi & the Mandelbrot Set", posted in sci.math.

> I did a Dejanews search, and found several articles on the topic that
> were just over a year old. Some of them referred to an even earlier
> article, supposedly from 1992. I can forward the 1995 articles if
> you like.
>
> I haven't looked at -0.75. But doesn't it converge on both (real)
> sides of that number? I thought all real numbers from -2.0 to 0.25
> were in the Mandelbrot set.

Correct. You need to approach from above (or below) the real axis.
--
Gerald A. Edgar ed...@math.ohio-state.edu

Jay R. Hill

Jul 9, 1996, 3:00:00 AM7/9/96
to k...@access5.digex.net, jh...@nosc.mil

k...@access5.digex.net (Keith F. Lynch) wrote:
>In article <31DFDB...@sms.utwente.nl>,
>Paul Melis <p.m...@sms.utwente.nl> wrote:
>> ... I came across a book ... in which I found an interesting piece.
>> It was about an article that was posted to USENET ...
>
>> It appears that in the point -0.75+0i+epsilon (and also
>> 0.25+0i+epsilon) with epsilon being a fixed value - the number of
>> iterations needed to make sure the point goes to infinity is quite
>> interesting. See below (I hope I remember this correctly) ...
>
>> epsilon | iterations needed
>> -------------------------------
>> 0.1 | 3
>> 0.01 | 31
>> 0.001 | 314
>
>My initial reaction was that this was nonsense, since the point never
>reaches infinity. And to "make sure" it will approach infinity, where
>c is near 0.25, all that's necessary is that epsilon be more than zero.
>
[snip]

>> Does anybody know the original poster?

Didn't the above mentioned book reproduce the original posting? I'll
have to check my copy at home.

>I haven't looked at -0.75. But doesn't it converge on both (real)
>sides of that number? I thought all real numbers from -2.0 to 0.25
>were in the Mandelbrot set.

Use z = (-0.75+i*eps).

Also try z = (-1.25+i*eps), z = -2 - eps.

Jay
--
int main(){float g,s,f,r,i;char*_="!/-,;<:!lnb/bh`r/ylqbAmmhI/S/x`K\n";int m,u,
e=0;_[32]++;for(;e<3919;){u=(256*(s=(r=.0325*(m=e%80)-2)*r+(i=.047*(e/80)-1.128
)*i)-96)*s+32*r<3?25:16+32*r+16*s<1?31:0;if(u==(s=f=0))do g=s*s-f*f+r;while((f=
2*s*f+i)*f+(s=g)*g<4&&++u<27);putchar(_[++e>3840&&m<25?31-m:m>78?32:u]^1);}}


text:
# Accounting Formulas

## What is an Accounting Formula?

The accounting equation or formula is contemplated to be the basis of the double-entry accounting method. The entirety of all the assets or belongings of a firm must be equivalent to the sum of all its records in the B/S. Based on this double-entry method, the accounting formula guarantees that the balance sheet persists ‘equalised’ and every entry obtained on the debit side must have an identical record (or coverage) on the credit side.

## What is the basic Accounting Equation?

The formula for accounting equation is obtained on the basic hypothesis that the equity owners have a claim on the entire assets of a firm post subtracting all the liabilities that is outstanding by the firm. This is depicted by the equation: Shareholders’ Equity = Assets – Liabilities.

 Important Accounting Formulas Assets = Liabilities + Equity $\frac{Current Assets}{Current Liabilities}= Current Ratio$ Income – Expenses = Net Income Beginning inventory value + Purchases of inventory – Ending inventory value = Cost of goods sold Sales – Cost of goods sold = Gross profit $\frac{Gross Profit}{Sales} = Gross Profit Margin$ $\frac{Fixed costs}{Sales \, price \, per \, unit-Variable \, cost \, per\, unit}=Break\, Even \, Point$ $Sales\, price \, per \, unit \, \times Break-even \, point \, in \, units = Breakeven\, point \, in \, Rupees$ $Inventory \, turnover \, ratio= \frac{Cost\, of \, Goods \, Sold}{Inventory}$ $Accounts \, Receivable \, Turnover \, Ratio = \frac{Sales\, on \, credit}{Accounts\, Receivable}$ $Total\, Asset\, Turnover= \frac{Sales}{Total\, Assets}$ $Debt\, to \, Equity \, Ratio =Debt – equity \, ratio = \frac{Total \, Liabilities}{Shareholder’s\, Equity}$ $Quick\, ratio= \frac{Current\, Assets-Inventory}{Current\, Liabilities}$ $Current\, Ratio= \frac{Current\, Assets}{Current\, Liabilities}$ $Return\, on\, Assets= \frac{Net\, Income}{Average\, Total\, Assets}$ $Return\, on\, Equity = \frac{Net\, Income}{Average\, Shareholder’s\, Equity}$ Beginning balance + net income – net losses – dividends = ending balance

## What is the comprehensive Accounting Equation?

The above mentioned is the concept, that is elucidated in detail about ‘Accounting Formulas’ for the Commerce students. To know more, stay tuned to BYJU’S.

You Might Also Like to Read: Useful Commerce Abbreviations

#### 1 Comment

1. Yashika

Easy way to learn


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function:RenderSidePart pageleftbodycaption pageleftbody sidenote PmWiki.WikiGroup-SideNote PmWiki.SideNote Site.SideNote

# Wiki Group

authors (basic)

PmWiki pages are organized into groups of related pages. This feature was added to PmWiki to allow authors to create their own wiki spaces of specialized content on their own, without having to become, or rely on, wiki administrators. See Pm's post to the pmwiki-users mailing list.

By default, page links are between pages of the same group; to create a link to a page in another group, add the name of the other group and a dot or slash to the page name. For example, links to Main/WikiSandbox could be written as:

 * [[Main.WikiSandbox]] * [[Main/WikiSandbox]] * [[(Main.Wiki)Sandbox]] * [[Main.WikiSandbox | link text]] * [[Main.WikiSandbox | +]]

To link to the default home page of a group, the name of the page can be omitted, like this:

 * [[Main.]] * [[Main/]]

## Creating groups

Creating a new group is as easy as creating new pages; simply edit an existing page to include a link to the new group's default home page (or any page in the new group) then click on the '?' to edit the page. As a rule, group names must start with a letter (but this can be changed by the wiki administrator by adding

$GroupPattern = '[[:upper:]\\d][\\w]*(?:-\\w+)*'; in config.php). For example, to make a default page in the group Foo, create a link to [[Foo/]] (or [[Foo.]]). To make a page called Bar in the group Foo, create a link to [[Foo/Bar]] and follow the link to edit that page. ## Groups in a standard PmWiki distribution • Main: The default group. On many wikis, it contains most of the author-contributed content. Main.HomePage and Main.WikiSandbox come pre-installed. • PmWiki: An edit-protected group that contains PmWiki documentation and help pages. • Site: Holds a variety of utility and configuration pages used by PmWiki, including SideBar, Search, Preferences, Templates, and AllRecentChanges. • SiteAdmin: Holds a number of password protected administration and configuration pages used by PmWiki, including ApprovedUrls, and Blocklist • To list all the groups in a site, try searching for "fmt=group". • To list all the pages in a group, try searching for "GroupName/". ## Special Pages in a Group By default, the Recent Changes page of each group shows only the pages that have changed within that group; the Site.All Recent Changes page shows all pages that have changed in all groups. Each group can also have Group Header or Group Footer pages that contain text to be automatically prepended or appended to every page in the group. A group can also have a Group Attributes page that defines attributes (read and edit passwords) shared by all pages within the group. Each page can also have its own individual read/edit password that overrides the group passwords (see Passwords). Finally, wiki administrators can set local customizations on a per-group basis--see Group Customizations. ## Group's default page The default "start page" for a group is a page whose name can be: 1. the same as the group (Foo/Foo) 2. HomePage (Foo/HomePage) 3. a name that the administrator has assigned to the {$DefaultName} variable in the configuration ([farm]config.php) file.

Note, on this site, the value of {$DefaultName} is HomePage and, thus, the default home page would be Foo/HomePage. You can usefully change the default search order for an entered page name by setting the variable$PagePathFmt in config.php, eg

$PagePathFmt = array('$Group.$1', '$1.$DefaultName', '$1.$1', '$DefaultGroup.$1', 'Profiles.$1');

where "$1" is the name of the page entered. If you are setting$DefaultName in order to make a start page for your groups, you will need to also define $PagePathFmt (see above) to get consistent use of this functionality. The simplest setting would be this:$PagePathFmt = array('$Group.$1', '$1.$DefaultName');

Note that the order of the definitions of these variables ($DefaultName and$PagePathFmt) is important - it must occur before any call to ResolvePageName() and it (therefore) it cannot occur in a per-page or per-group customization script.

As noted above, when linking to the default home page, authors can omit the page name and simply identify the group followed by a forward slash ([[Foo/]]).

Note the forward slash is required to ensure that the link unambiguously points to the identified group. If the slash is omitted, the link can end up being interpreted as pointing to an existing (or new) page in the current group (if the group, or its default home page, do not exist).

## Subgroups? Subpages?

No, PmWiki does not have subpages. Pm's reasons for not having subgroups are described at PmWiki:Hierarchical Groups, but it comes down to not having a good page linking syntax. If you create a link or pagename like [[A.B.C]] PmWiki doesn't think of "B.C" as being in group "A", it instead thinks of "C" as being in group "AB", which is a separate group from "A". Wiki administrators can look at Cookbook:Subgroup Markup and Cookbook:Include With Edit for recipes that may be of some help with developing subgroups or subpages.

## Restricting the creation of new groups

You can set PmWiki's $GroupPattern variable to only accept the group names you want to define. For example, to limit pages to the "PmWiki", "Main", "Profiles", and "Example" groups, add the following to local/config.php:$GroupPattern = '(?:Site|SiteAdmin|PmWiki|Main|Profiles|Example)';

With this setting, only the listed groups will be considered valid WikiGroups. You can add more groups to the list by placing additional group names separated by pipes (|).

See other solutions to this at Cookbook:Limit Wiki Groups and Cookbook:New Group Warning.

How can I get rid of the 'Main' group in urls for pages pointing to Main?

How can I limit the creation of new groups?

Why doesn't [[St. Giles and St. James]] work as a link? (It doesn't display anything.)

Because it contains periods, and destroys PmWiki's file structure, which saves pages as Group.PageName. Adding those periods disrupts this format. Links may only contain words. If you need a link precisely as shown, the page must be named eg StGilesAndStJames then you can use the (:title:) directive to have the page's title appear with periods (:title St. Giles and St. James:). (Although in US grammar the period is often omitted and in UK grammar the period must be omitted for contractions like St).

How can I delete a wiki group?

Normally you can't, as this requires an admin with server-side access to delete the file that makes up the group's RecentChanges page. But there is an option method of making it possible to delete RecentChanges pages from within the wiki if the admin enables the code found on Cookbook:RecentChanges Deletion.

How can I delete a wiki group's Group.RecentChanges page?

Normally you can't, as this requires an admin with server-side access to delete a file. But there is an optional method of making it possible to delete RecentChanges pages from within the wiki if the admin enables the code found on Cookbook:RecentChanges Deletion.

Can I delete a wiki group inside wiki.d folder on the server to eliminate the group?

Yes, if you delete all files named YourGroup.*, the pages from that group will be removed from the wiki. Note that the documentation (group PmWiki) and the site configuration (groups Site and SiteAdmin) that exist in the default installation, are located in wikilib.d and not in wiki.d, and some recipes provide files located in a wikilib.d subdirectory in the cookbook directory. (You shouldn't delete the groups Site and SiteAdmin, required for normal function.)

How can I list all pages in a WikiGroup?

In a wiki page use (:pagelist group=GroupName list=all:) or in a search box type GroupName/ list=all.


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# Cantor Set Homeomorphic to Inverse Limit Space

#### joypav

##### Active member
Problem:
For each positive integer $n$ let $X_n = \left\{1, 2, 3, ..., 2^n \right\}$ with the discrete topology and let $f_n : X_{n+1} → X_n$ be the function defined by:
$f_n(i) = i$ for $1 ≤ i ≤ 2^n$
$f_n(i) = 2^{n+1} − i + 1$ for $2^n < i ≤ 2^{n+1}$
Then
$X = lim_{←} \left\{X_i, f_i \right\}_{i=1}^{\infty}$
is homeomorphic to the Cantor set.

I've been trying to find a homeomorphism for this for way too long... anyone know what it would be?
I can take care of proving it's a homeomorphism once I have the function.


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# Flyclient on CKB

Flyclient is a superlight client design dedicated to PoW blockchains. With a few modifications to the blockchain data structure and an elegant interactive protocol, it allows a light client to verify the validity of the latest block with only a few hundred kilobytes of data transmission, rather than downloading all the block headers. Afterward, the client can verify any transaction in any block with a short proof.

In this post, I will describe how to apply the Flyclient design to Nervos CKB. Henceforth I will assume the readers have already read the Flyclient paper (the proofs are not necessary) or at least watched the talk. The rest of this post will be highly technical and thus contains absolutely no meaningful message to the general audience.

## Assumptions and Their Implications

### Two Crucial Assumptions

• A light client only looks for the heaviest chain and confirms that a certain transaction is embedded in this chain. The heaviest chain is, by assumption, syntactically valid. In other words, there is no need for a light client to verify whether all the transactions are valid in this chain.
• The light client connects to at least one honest full node.

Based on the second assumption, we immediately have that, when all full nodes send the light client the same latest block or a consistent history, the client only needs to sample blocks from one of them.

### Conflict Resolution

A point of discussion: what to do when two full nodes send us conflicting histories?

My suggestion is the following protocol:

1. Sample from both chains a series of blocks with the same accumulated difficulty.
2. Find the last common block in their replies, use it as the starting point for sampling, and repeat the first step.

Until we find an invalid block in one of the chains.

Set an upper bound on the times of this loop. Once the upper bound is reached, trust the heavier chain.

### What to do and What Not to do

Personally, I don’t see why we need non-interactive sampling. It’s an interactive protocol anyway.

A light client should save its latest block so that if it goes offline and comes back online, there is no need to sample from the genesis block. It can sample from the latest block before it went offline, if that block is still among the heaviest chain. Of course, a small upper bound should be applied to the local storage.

There are two challenges when implementing Flyclient on CKB. First, how to ensure a smooth transition to kick-off the MMR. Second, how to design the MMR data structure. We will focus on the second challenge for the rest of this post. The first challenge will be dealt with when some other things are settled (e.g., where to put the MMR root, what kind of fork).

## MMR Data Structure

Another point of discussion: do we want the number of blocks of all full MMR nodes to be powers of 2?

At this moment, I don’t think that is necessary. Please let me know if I am wrong. In Nervos CKB, every epoch contains roughly 300 to 1800 blocks. I think it is simpler if we have a certain level of MMR nodes that are dedicated to epoch roots. The benefit is, we can finish all operations regarding difficulty adjustment at the same level, which can simplify the algorithm and data structure design.

Now, what data field to put in the MMR node?

The main idea is, put all five parameters used in the difficulty adjustment in the MMR node so that we can reuse the code of calculating the new difficulty to verify whether the difficulty adjustment is correct!

\hat{H_{i-1}}: the adjusted hash rate of the i-1-th epoch

In the MMR, each leaf node is a block, and we term the level where each node represents an epoch epoch level.

Every MMR node contains:

• h: block/subtree hash—otherwise it is not a Merkle root.
• blockNum: (only applicable to non-leaf nodes) the number of leaf nodes in this subtree
• epochNum: (only applicable to nodes higher than epoch level) number of epochs, including incomplete ones
• w: the accumulated difficulty since the genesis block. Note that uncles do not contribute to the accumulated difficulty.
• deltaT: the time difference between the last block before this subtree (which is not part of this subtree), and the rightmost block in this subtree (which is part of this subtree). This value equals L_i for epoch-level nodes.
• firstEpochTarget, firstEpochDuration, firstEpochMainChainNum, firstEpochUncleNum：the target, duration, number of main chain blocks, and number of uncles of the first epoch in this subtree. These values equal T_i, L_i, C_{i,m}, C_{i,o}. Note that for all MMR nodes in an epoch, these four values only need to be transmitted once.
• nextTarget, nextMainChainNum: (only applicable when the last block of this subtree is the end of an epoch) the target and main chain block number of the next epoch (not in this subtree). These two values equal T_{i+1} and C_{i+1,m}.
• endAdjustedHashRate: (only applicable when the last block of this subtree is the end of an epoch) \hat{H_i} after the epoch ends.

Again, the benefit of this design is, we have all the inputs we need to re-execute the difficulty adjustment algorithm after every epoch.

## Verifying the Validity of Difficulty Adjustment in MMR

This part corresponds to page 18 at the end of Sect. 6.1. The goal is to verify that there is at least one valid difficulty transformation (execution of a series of difficulty adjustments) that is consistent with the current sampling. Once this is verified, many attacks become impossible.

Step 0. verify all leaf nodes’ information is consistent with their epoch-level node, and all epoch-level nodes’ difficulty adjustment results are correct.

Step 1. List all epoch-level MMR nodes. These nodes split the blockchain into many segments. The blockchain is valid if and only if all segments are valid. Now we focus on one segment.

Assuming there are n epochs in this segment, the Flyclient paper checks three groups of conditions targeting NC:

1. The difficulty change is between $\tau^n$and 1/\tau^n, where \tau is the dampening factor。
2. The upper and lower bounds of the accumulated difficulty w.
3. The upper and lower bounds of the total time.

Here is my solution:

The dampening factor.

In NC-Max, the adjusted hash rate estimation \hat{H_i} cannot change faster than \tau_1. So we compute \hat{H_i} for the first and last epoch of the segment, which allows us to check this condition.

The accumulated difficulty.

The total accumulated difficulty in epoch (i+1) is \hat{H_i}*L_{\rm ideal}. Among with, at least half is used on the main chain, at most all of them are used in the main chain. So the upper bound of the accumulated difficulty is to assume all of it is in the main chain; the lower bound is to assume only half of it is in the main chain.

Some further explanations on Equation 2 and 3 in the paper:

• If the ending difficulty is higher than the starting difficulty, then

• the upper bound of the total difficulty is, first spend (n+k)/2 epochs raising the difficulty, then spend (n-k)/2 epochs lowering the difficulty;
• the lower bound of the total difficulty is, first spend (n-k)/2 epochs lowering the difficulty, then spend (n+k)/2 epochs raising the difficulty;
• The other direction is very similar.

The total time.

The Flyclient paper didn’t say. Our solution is simple: compute the average epoch duration of each segment. If a certain segment’s average epoch duration deviates from the expectation too much (shorter than 3hr40min or longer than 4hr20min), then sample again from that segment, and verify everything in this suspicious segment.

6 Likes


text:
Language:   Search:   Contact
Zentralblatt MATH has released its new interface!
For an improved author identification, see the new author database of ZBMATH.

Query:
Fill in the form and click »Search«...
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Zbl 1157.46048
Mirmostafaee, Alireza Kamel; Moslehian, Mohammad Sal
(English)
[J] Result. Math. 52, No. 1-2, 161-177 (2008). ISSN 1422-6383; ISSN 0378-6218/e

The authors approximate a fuzzy almost quadratic function by a quadratic function in a fuzzy sense; this kind of fuzzy normed linear space was introduced by {\it T. Bag} and {\it S. K. Samanta} [J. Fuzzy Math. 11, 687--705 (2003; Zbl 1045.46048)].\par More precisely, they establish a fuzzy Hyers-Ulam-Rassias stability of the quadratic functional equation $f(x+y)+f(x-y)=2f(x)+2f(y)$. Their result can be regarded as a generalization of the stability phenomenon in the framework of normed linear spaces. They also prove a generalized version of fuzzy stability of the Pexiderized quadratic functional equation $f(x+y)+f(x-y)=2g(x)+2h(y)$.
[Congxin Wu (Harbin)]
MSC 2000:
*46S40 Fuzzy functional analysis
39B52 Functional equations for functions with more general domains
39B82 Stability, separation, extension, and related topics
26E50 Fuzzy real analysis
46S50 Functional analysis in probabilistic metric spaces

Keywords: fuzzy normed space; quadratic function; fuzzy almost quadratic function; Hyers-Ulam-Rassias stability

Citations: Zbl 1045.46048

Cited in: Zbl 1206.46065 Zbl 1201.46062

Highlights
Master Server


text:
# Tag Info

9

First, about teleportation, you say that you think quantum communication takes place in the protocol, but it doesn't. They only share an EPR pair they created together, hence the coordination and after, what Alice sends to Bob when communication takes place are classical bits, she sends the measured bits of the 2 qubits she has, so the only communication we ...

6

A necessary and sufficient condition on the unitary $B$ is that its columns all correspond to maximally entangled states. There also does not need to be any relationship between the two unitaries labeled $B$ and $B^{-1}$ in your figure: as long as you start with a maximally entangled state of systems 2 and 3, and then measure systems 1 and 2 with respect ...

6

The issue is that you are applying operations after measurement gates and this is currently not available on the real hardware. I think the hardware also does not support reset operations mid-way through a circuit at the moment. The best way forward is to keep running this on the simulator or try to find a different way of expressing the circuit such that ...

5

Suppose Alice wants to send Bob a 1000 bit message. To receive the message, Bob flips 1000 coins and writes down the results as 0s and 1s. About 50% of the random bits in the message that Bob generated are the same as in the intended message. Clearly this is an even better faster-than-light communication method than teleportation, because it succeeds 50% of ...

5

Your circuit does not measure $q_2$ qubit after teleportation; I guess that is why teleportation of $|1\rangle$ qubit is shown incorrectly.

5

Since your circuit is teleportation, $|C\rangle =|\varphi\rangle$ and since you measured $|11\rangle$ on $|AB\rangle$ the answer is $|ABC\rangle = |11\rangle|\varphi\rangle$. Now, let look why this is true. Firstly Hadamard and CNOT gate on second and third qubit prepares entangled Bell state \$|\beta_{00}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)...

4

Quantum teleportation has been suggested to be possible to occur in photosynthetic reaction centers. https://doi.org/10.1007/BF03166259 And it could in principle be occuring all the time through photochemistry in biology. Though to no real purpose, however, a related purposeful effect is the avian compass which(is hypothesized) to operate through quantum ...

4

The whole point of an EPR pair is that you cannot write (without losing some information) "This is what Alice has" and "This is what Bob has". Partial descriptions can be given using reduced density matrices. However, if you want to identify which bits Alice and Bob each have, we often use a notation like $$(|0\rangle_A|0\rangle_B+|1\rangle_A|1\rangle_B)/\... 3 Currently IBM Q does not support IF statement on real quantum processor. The IF can be used on simulator only. However, there is a theorem stating that quantum gates controlled by classical register can be replaced by those controlled by qubits before measurement. So, you can simply replace the statement measure q[1] -> c[1]; if (c == 1) x q[2]; by cx ... 3 I think faster than light communication is more stubborn than that :-) The principle of deferred measurement, to quote Nielsen and Chuang, page 186, is Measurements can always be moved from an intermediate stage of a quantum circuit to the end of the circuit; if the measurement results are used at any stage of the circuit then the classically controlled ... 3 As @Rammus points out this is just shorthand for taking the tensor product, rather than algebraically multiplying out as you may first assume. If we take the first term of \phi_2 and expand it out more explicitly with subscripts A for Alice and B for Bob: |0_A0_A\rangle(\alpha|0_B\rangle + \beta|1_B\rangle) = |0_A0_A\rangle \otimes (\alpha|0_B\... 3 Information can't be transmitted faster-than-light using quantum teleportation protocol, because as you correctly noticed quantum teleportation involves classical communication between sender and receiver. The paper's abstract is saying about implementing quantum teleportation using a particular technology (silicon integrated optics) for the first time. ... 3 There is a mistake in design of the first circuit. Both X and Z gate work when value 1 is in classical register. Gate X should work in case qubit q_2 is in state |1\rangle. Similarly Z acts when q_1 in state |1\rangle. Also you have to deal with state when both X and Z have to act. In your case you conditioned both X and Z on c==1. ... 3 1) In this step, you connect a teleported qubit with entangled qubits between Alice and Bob. This means, Bob now has an "access" to the teleported qubit. 2) Here you get some information about the teleported qubit and "partially colapse" Bob's qubit according to a state of the teleported qubit. 3) In this last step you bring information about the ... 3 Perhaps the following diagram helps to show what acts where. I think you've got a little muddled. 3 In the teleportation protocol, the two parties share the entangled Bell State and it is implemented via a CNOT gate between the state to be sent (suppose Alice is sending the state) and the part of entangled Bell state Alice has. The CNOT gate creates another entangled state whose measurement Alice will send to the second party, say it's Bob, to perform the ... 2 There's a mistake. It's incorrect even for a=0,b=0,U=I. In this case the correct formula is$$ |\phi\rangle \otimes |B_{00}\rangle = \frac{1}{2} \bigg(|B_{00}\rangle \otimes |\phi\rangle + |B_{01}\rangle \otimes Z|\phi\rangle + |B_{10}\rangle \otimes X|\phi\rangle + |B_{11}\rangle \otimes XZ|\phi\rangle\bigg) $$but their formula swaps X and Z in the ... 2 I think the code example closest to what you're looking for is task 1.4 of the Teleportation kata which gives the learner three qubits (the message qubit and the pair of qubits that Alice and Bob will entangle and share before teleportation) and asks them to transform the Bob's qubit into the message state. You'll notice that the test for this task doesn't ... 2 The top circuit uses classically controlled X and Z gates - gates that are conditioned on the measurement result. The bottom circuit uses quantum version of controlled X and Z gates - gates that are conditioned on the state of another qubit. Since each of the control qubits have been measured before applying CX and CZ gates, their state matches exactly the ... 2 The calculation of the resulting state of the described circuit: After applying the same circuit identities described in this answer (and here) to the connected question we will obtain a "simplified" circuit: Here are the links to the initial, first intermediate, second intermediate and final circuits presented in the Quirk. Whenever it was ... 2 Without that fixup, the state of Bob's qubit can be in any one of the states \alpha |0\rangle + \beta |1\rangle (the state we actually wanted to teleport) or \alpha |0\rangle - \beta |1\rangle, \beta |0\rangle + \alpha |1\rangle or \beta |0\rangle - \alpha |1\rangle (the states we don't want to end up with). X and Z gates are applied to fix the ... 2 It sounds like the source of your misunderstanding is the following Bob is able to harness the information of a qubit which contains more information than regular classical bits This statement is not true. There is a theorem known as Holevo's theorem which states that the maximum information that can be obtained from a single qubit is one bit. 1 About checking. I do not quite understand why you can't check how the GHZ-state was teleported in the most ordinary ways: in addition to measuring the state immediately after teleportation, as well as by YYX, YXY, XYY, XXX measurements, you can inverse your GHZ-state and make sure that all 0 are obtained, e.g. like as for this with the most usual GHZ-state ... 1 I think it is worth trying to understand the circuit that the authors really want to implement: Here they produce the Bell state they want to teleport onto the final two qubits, and cluster state (personally, I wouldn't call it a cluster state after they've added the two extra hadamards). Then they do two single-qubit teleportation protocols to make the ... 1 The question reads like quantum encryption is a special example of quantum teleportation, which is not the case. Teleportation solves the following problem: Alice wants to send Bob a qubit, but has no way to physically transport it without destroying it. However, Alice and Bob already share an entangled state, and they can communicate classically. They ... 1 Thanks to Martin Vesely! Now I modify method 1 by using the correct c values. I think I got the right answer. My goal is to see the 1st bit entangled with the 4th bit. As can be seen below, 1st and 4th bits always have the same values. So they are entangled. 1 I contacted @fran-cabrera from the IBMQ team and he could reproduce the bug! The problem is with the visualization of the transpiled circuit, not the execution (the result should be correct). The team is working on solving it and they expect to deploy a fix at the end of the week. I ran you example in Qiskit and it seems to work IBMQ.load_account() ... 1 If you are interested only in the measurements of the system that comprises only 2nd and 3rd qubits, then you could compute the reduced density matrix \rho_{23} on those qubits from the total density matrix \rho_{1234}=|\psi\rangle \langle\psi|, i.e.$$ \rho_{23} = \text{Tr}_{14}(\rho_{1234}) $$Density matrix carry all information that is need to ... 1 Judging from your computations, the Bell states are$$|B_{00}\rangle = |\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)|B_{10}\rangle = |\Phi^-\rangle = \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle)|B_{01}\rangle = |\Psi^+\rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)|B_{11}\rangle = |\Psi^-\rangle = \frac{1}{\sqrt{2}}(|01\...

1

"Assume that there is a setup with Bob which creates some sort of interference of the qubits he receives." I think you need to be more specific on this point. From Bob's perspective his qubit is simply a mixed state in the computational basis $$\rho_B=\frac{1}{2}|0\rangle\langle 0|+\frac{1}{2}|1\rangle\langle1|$$ so he will not be able to perceive any ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# Prove that for all real numbers x there is a y n which case x<y

1. Nov 5, 2012

### Andrax

1. The problem statement, all variables and given/known data
prove that
($\forall$x$\in$R) ($\exists$y$\in$R) : x < y

2. Relevant equations

3. The attempt at a solution
x < y
y= x+1
then x<x+1
which is correct but i'm kinda not sure of this answer...

2. Nov 5, 2012

### hedipaldi

y=x+1 satisfies x<y so what wrong with it?

3. Nov 5, 2012

### Andrax

just wanted to make sure someone actually confused me saying it dosen't


text:
## anonymous one year ago help help help

1. anonymous

???

2. anonymous

3. anonymous

Given , the side of the square as 36'' Hence, the circle has the diameter of 36'' Find the Radius of the circle by simply dividing the diameter by 2 i.e. Radius = diameter/2 Radius = 36/2 Radius = 18'' Now calculate the area of the Circle Use $Area = \pi \times r \times r$ Given the value of pi as 3.14 So, $Area = 3.14 \times 18 \times 18$ Now , 18 times 18 = 324 Hence, $Area = 3.14 \times 324$ WE , can take the same equation as $314 \times 324 \over 100$ equals area of Circle

4. anonymous

314 times 324 is equal to 101736 Now, divide 101736 by 100

5. anonymous

SO, we get 1017.36 which is A. Go through the procedure and ask doubts if any.

6. anonymous

Are you there @bubbleslove1234 ?

7. anonymous

thanks sooooooooooooooooo much......lol

8. anonymous

:)


text:
# Tuples

Tuples are tied using parentheses:

(a, b, c)

For a:A, b:B, and c:C, the type of such a tuple is (A, B, C). It is not possible to declare variables of tuple type, but tuples may be used to return multiple values from a function:

function f(x:X, y:Y) -> (A, B, C) {
a:A;
b:B;
c:C;
return (a, b, c);
}

To untie a tuple, use parentheses on the left:

a:A;
b:B;
c:C;
(a, b, c) <- f(x, y);


text:
# Module 8: Cluster Analysis¶

The following tutorial contains Python examples for solving classification problems. You should refer to Chapters 7 and 8 of the "Introduction to Data Mining" book to understand some of the concepts introduced in this tutorial. The notebook can be downloaded from http://www.cse.msu.edu/~ptan/dmbook/tutorials/tutorial8/tutorial8.ipynb.

Cluster analysis seeks to partition the input data into groups of closely related instances so that instances that belong to the same cluster are more similar to each other than to instances that belong to other clusters. In this tutorial, we will provide examples of using different clustering techniques provided by the scikit-learn library package.

Read the step-by-step instructions below carefully. To execute the code, click on the corresponding cell and press the SHIFT-ENTER keys simultaneously.

## 8.1 K-means Clustering¶

The k-means clustering algorithm represents each cluster by its corresponding cluster centroid. The algorithm would partition the input data into k disjoint clusters by iteratively applying the following two steps:

1. Form k clusters by assigning each instance to its nearest centroid.
2. Recompute the centroid of each cluster.

In this section, we perform k-means clustering on a toy example of movie ratings dataset. We first create the dataset as follows.

In [1]:
import pandas as pd

ratings = [['john',5,5,2,1],['mary',4,5,3,2],['bob',4,4,4,3],['lisa',2,2,4,5],['lee',1,2,3,4],['harry',2,1,5,5]]
titles = ['user','Jaws','Star Wars','Exorcist','Omen']
movies = pd.DataFrame(ratings,columns=titles)
movies

Out[1]:
user Jaws Star Wars Exorcist Omen
0 john 5 5 2 1
1 mary 4 5 3 2
2 bob 4 4 4 3
3 lisa 2 2 4 5
4 lee 1 2 3 4
5 harry 2 1 5 5

In this example dataset, the first 3 users liked action movies (Jaws and Star Wars) while the last 3 users enjoyed horror movies (Exorcist and Omen). Our goal is to apply k-means clustering on the users to identify groups of users with similar movie preferences.

The example below shows how to apply k-means clustering (with k=2) on the movie ratings data. We must remove the "user" column first before applying the clustering algorithm. The cluster assignment for each user is displayed as a dataframe object.

In [2]:
from sklearn import cluster

data = movies.drop('user',axis=1)
k_means = cluster.KMeans(n_clusters=2, max_iter=50, random_state=1)
k_means.fit(data)
labels = k_means.labels_
pd.DataFrame(labels, index=movies.user, columns=['Cluster ID'])

Out[2]:
Cluster ID
user
john 1
mary 1
bob 1
lisa 0
lee 0
harry 0

The k-means clustering algorithm assigns the first three users to one cluster and the last three users to the second cluster. The results are consistent with our expectation. We can also display the centroid for each of the two clusters.

In [3]:
centroids = k_means.cluster_centers_
pd.DataFrame(centroids,columns=data.columns)

Out[3]:
Jaws Star Wars Exorcist Omen
0 1.666667 1.666667 4.0 4.666667
1 4.333333 4.666667 3.0 2.000000

Observe that cluster 0 has higher ratings for the horror movies whereas cluster 1 has higher ratings for action movies. The cluster centroids can be applied to other users to determine their cluster assignments.

In [4]:
import numpy as np

testData = np.array([[4,5,1,2],[3,2,4,4],[2,3,4,1],[3,2,3,3],[5,4,1,4]])
labels = k_means.predict(testData)
labels = labels.reshape(-1,1)
cols = movies.columns.tolist()
cols.append('Cluster ID')
newusers = pd.DataFrame(np.concatenate((usernames, testData, labels), axis=1),columns=cols)
newusers

Out[4]:
user Jaws Star Wars Exorcist Omen Cluster ID
0 paul 4 5 1 2 1
1 kim 3 2 4 4 0
2 liz 2 3 4 1 1
3 tom 3 2 3 3 0
4 bill 5 4 1 4 1

To determine the number of clusters in the data, we can apply k-means with varying number of clusters from 1 to 6 and compute their corresponding sum-of-squared errors (SSE) as shown in the example below. The "elbow" in the plot of SSE versus number of clusters can be used to estimate the number of clusters.

In [5]:
import matplotlib.pyplot as plt
%matplotlib inline

numClusters = [1,2,3,4,5,6]
SSE = []
for k in numClusters:
k_means = cluster.KMeans(n_clusters=k)
k_means.fit(data)
SSE.append(k_means.inertia_)

plt.plot(numClusters, SSE)
plt.xlabel('Number of Clusters')
plt.ylabel('SSE')

Out[5]:
Text(0,0.5,'SSE')

## 8.2 Hierarchical Clustering¶

This section demonstrates examples of applying hierarchical clustering to the vertebrate dataset used in Module 6 (Classification). Specifically, we illustrate the results of using 3 hierarchical clustering algorithms provided by the Python scipy library: (1) single link (MIN), (2) complete link (MAX), and (3) group average. Other hierarchical clustering algorithms provided by the library include centroid-based and Ward's method.

In [6]:
import pandas as pd

data

Out[6]:
Name Warm-blooded Gives Birth Aquatic Creature Aerial Creature Has Legs Hibernates Class
0 human 1 1 0 0 1 0 mammals
1 python 0 0 0 0 0 1 reptiles
2 salmon 0 0 1 0 0 0 fishes
3 whale 1 1 1 0 0 0 mammals
4 frog 0 0 1 0 1 1 amphibians
5 komodo 0 0 0 0 1 0 reptiles
6 bat 1 1 0 1 1 1 mammals
7 pigeon 1 0 0 1 1 0 birds
8 cat 1 1 0 0 1 0 mammals
9 leopard shark 0 1 1 0 0 0 fishes
10 turtle 0 0 1 0 1 0 reptiles
11 penguin 1 0 1 0 1 0 birds
12 porcupine 1 1 0 0 1 1 mammals
13 eel 0 0 1 0 0 0 fishes
14 salamander 0 0 1 0 1 1 amphibians
In [7]:
from scipy.cluster import hierarchy
import matplotlib.pyplot as plt
%matplotlib inline

names = data['Name']
Y = data['Class']
X = data.drop(['Name','Class'],axis=1)
dn = hierarchy.dendrogram(Z,labels=names.tolist(),orientation='right')

In [8]:
Z = hierarchy.linkage(X.as_matrix(), 'complete')
dn = hierarchy.dendrogram(Z,labels=names.tolist(),orientation='right')

### 8.3.3 Group Average¶

In [9]:
Z = hierarchy.linkage(X.as_matrix(), 'average')
dn = hierarchy.dendrogram(Z,labels=names.tolist(),orientation='right')

## 8.3 Density-Based Clustering¶

Density-based clustering identifies the individual clusters as high-density regions that are separated by regions of low density. DBScan is one of the most popular density based clustering algorithms. In DBScan, data points are classified into 3 types---core points, border points, and noise points---based on the density of their local neighborhood. The local neighborhood density is defined according to 2 parameters: radius of neighborhood size (eps) and minimum number of points in the neighborhood (min_samples).

For this approach, we will use a noisy, 2-dimensional dataset originally created by Karypis et al. [1] for evaluating their proposed CHAMELEON algorithm. The example code shown below will load and plot the distribution of the data.

In [10]:
import pandas as pd

data = pd.read_csv('chameleon.data', delimiter=' ', names=['x','y'])
data.plot.scatter(x='x',y='y')

Out[10]:
<matplotlib.axes._subplots.AxesSubplot at 0x124bc393a58>

We apply the DBScan clustering algorithm on the data by setting the neighborhood radius (eps) to 15.5 and minimum number of points (min_samples) to be 5. The clusters are assigned to IDs between 0 to 8 while the noise points are assigned to a cluster ID equals to -1.

In [11]:
from sklearn.cluster import DBSCAN

db = DBSCAN(eps=15.5, min_samples=5).fit(data)
labels = pd.DataFrame(db.labels_,columns=['Cluster ID'])
result = pd.concat((data,labels), axis=1)
result.plot.scatter(x='x',y='y',c='Cluster ID', colormap='jet')

Out[11]:
<matplotlib.axes._subplots.AxesSubplot at 0x124bc3374a8>

## 8.4 Spectral Clustering¶

One of the main limitations of the k-means clustering algorithm is its tendency to seek for globular-shaped clusters. Thus, it does not work when applied to datasets with arbitrary-shaped clusters or when the cluster centroids overlapped with one another. Spectral clustering can overcome this limitation by exploiting properties of the similarity graph to overcome such limitations. To illustrate this, consider the following two-dimensional datasets.

In [12]:
import pandas as pd

data1 = pd.read_csv('2d_data.txt', delimiter=' ', names=['x','y'])
data2 = pd.read_csv('elliptical.txt', delimiter=' ', names=['x','y'])

fig, (ax1,ax2) = plt.subplots(nrows=1, ncols=2, figsize=(12,5))
data1.plot.scatter(x='x',y='y',ax=ax1)
data2.plot.scatter(x='x',y='y',ax=ax2)

Out[12]:
<matplotlib.axes._subplots.AxesSubplot at 0x124bc560240>

Below, we demonstrate the results of applying k-means to the datasets (with k=2).

In [13]:
from sklearn import cluster

k_means = cluster.KMeans(n_clusters=2, max_iter=50, random_state=1)
k_means.fit(data1)
labels1 = pd.DataFrame(k_means.labels_,columns=['Cluster ID'])
result1 = pd.concat((data1,labels1), axis=1)

k_means2 = cluster.KMeans(n_clusters=2, max_iter=50, random_state=1)
k_means2.fit(data2)
labels2 = pd.DataFrame(k_means2.labels_,columns=['Cluster ID'])
result2 = pd.concat((data2,labels2), axis=1)

fig, (ax1,ax2) = plt.subplots(nrows=1, ncols=2, figsize=(12,5))
result1.plot.scatter(x='x',y='y',c='Cluster ID',colormap='jet',ax=ax1)
ax1.set_title('K-means Clustering')
result2.plot.scatter(x='x',y='y',c='Cluster ID',colormap='jet',ax=ax2)
ax2.set_title('K-means Clustering')

Out[13]:
Text(0.5,1,'K-means Clustering')

The plots above show the poor performance of k-means clustering. Next, we apply spectral clustering to the datasets. Spectral clustering converts the data into a similarity graph and applies the normalized cut graph partitioning algorithm to generate the clusters. In the example below, we use the Gaussian radial basis function as our affinity (similarity) measure. Users need to tune the kernel parameter (gamma) value in order to obtain the appropriate clusters for the given dataset.

In [14]:
from sklearn import cluster
import pandas as pd

spectral = cluster.SpectralClustering(n_clusters=2,random_state=1,affinity='rbf',gamma=5000)
spectral.fit(data1)
labels1 = pd.DataFrame(spectral.labels_,columns=['Cluster ID'])
result1 = pd.concat((data1,labels1), axis=1)

spectral2 = cluster.SpectralClustering(n_clusters=2,random_state=1,affinity='rbf',gamma=100)
spectral2.fit(data2)
labels2 = pd.DataFrame(spectral2.labels_,columns=['Cluster ID'])
result2 = pd.concat((data2,labels2), axis=1)

fig, (ax1,ax2) = plt.subplots(nrows=1, ncols=2, figsize=(12,5))
result1.plot.scatter(x='x',y='y',c='Cluster ID',colormap='jet',ax=ax1)
ax1.set_title('Spectral Clustering')
result2.plot.scatter(x='x',y='y',c='Cluster ID',colormap='jet',ax=ax2)
ax2.set_title('Spectral Clustering')

Out[14]:
Text(0.5,1,'Spectral Clustering')

## 3.5 Summary¶

This tutorial illustrates examples of using different Python's implementation of clustering algorithms. Algorithms such as k-means, spectral clustering, and DBScan are designed to create disjoint partitions of the data whereas the single-link, complete-link, and group average algorithms are designed to generate a hierarchy of cluster partitions.

References: [1] George Karypis, Eui-Hong Han, and Vipin Kumar. CHAMELEON: A Hierarchical Clustering Algorithm Using Dynamic Modeling. IEEE Computer 32(8): 68-75, 1999.


text:
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[; \frac{1}{2a - 3} - 5 = \frac{6a}{2a - 3} ;]
[; 1 - 5(2a - 3) = 6a ;]
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[; a = 1 ;]


text:
# Texas Go Math Grade 4 Lesson 1.3 Answer Key Compare and Order Numbers

Refer to our Texas Go Math Grade 4 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 4 Lesson 1.3 Answer Key Compare and Order Numbers.

## Texas Go Math Grade 4 Lesson 1.3 Answer Key Compare and Order Numbers

Unlock the Problem

Grand Canyon National Park in Arizona had 651,028 visitors in June and 665,188 visitors in July. In which month did the park have more visitors?

How many visitors were there in June?
It is given that
Grand Canyon National Park in Arizona had 651,028 visitors in June and 665,188 visitors in July
Hence, from the above,
We can conclude that
The number of visitors in June are: 651,028

How many visitors were there in July?
It is given that
Grand Canyon National Park in Arizona had 651,028 visitors in June and 665,188 visitors in July
Hence, from the above,
We can conclude that
The number of visitors in July are: 665,188

Share and Grow

Question 1.
Compare 145,158,190 and 145,185,170.
Write <, >, or = Use the place-value chart to help.

145,158,190 _________ 145,185,170
The given numbers are: 145,158,190 and 145,185,170
Now,
The given place-value chart for the given numbers is:

Now,
From the above place-value chart,
We can observe that
The Thousand’s period is greater in 145,185,170 than in 145,158,190
Hence, from the above,
We can conclude that

Compare. Write <, >, or =.

Question 2.
7,458,562,333 _________ 7,457,541,753
The given numbers are: 7,458,562,333 and 7,457,541,753
Now,
The representation of the given numbers in the place-value chart is:

Now,
From the above place-value charts,
We can observe that
The Million’s period is greater in 7,458,562,333 than in 7,457,541,753
Hence, from the above,
We can conclude that

Question 3.
56,991 _________ 52,880
The given numbers are: 56,991 and 52,880
Now,
The representation of the given numbers in the place-value chart is:

Now,
From the above place-value charts,
We can observe that
The Thousand’s period is greater in 56,991 than in 52,880
Hence, from the above,
We can conclude that

Question 4.
708,651,515 _________ 629,672,484
The given numbers are: 708,651,515 and 629,672,484
Now,
The representation of the given numbers in the place-value chart is:

Now,
From the above place-value charts,
We can observe that
The Million’s period is greater in 708,6511,515 than in 629,672,484
Hence, from the above,
We can conclude that

Question 5.
1,430,620 _________ 986,435
The given numbers are: 1,430,620 and 986,435
Now,
The representation of the given numbers in the place-value chart is:

Now,
From the above place-value charts,
We can observe that
The Million’s period is greater in 1,430,620 than in 986,435
Hence, from the above,
We can conclude that

Order from greatest to least

Question 6.
20,650; 21,150; 20,890
The given numbers are: 20,650. 21,150, and 20,890
Now,
The representation of the given numbers in the place-value chart is:

Now,
From the above place-value charts,
We can observe that
21,150 > 20,890 > 20,650
Hence, from the above,
We can conclude that
The order of the numbers from the greatest to the least is:
21,150 > 20,890 > 20,650

Problem Solving

H.O.T.
Algebra Write all of the digits that can replace each

Question 7.
567 < 5 5 < 582 ____________
Let the missing digit be x
Now,
The given pattern is:
567 < 5×5 < 582
Now,
From the above pattern,
We can observe that
The ten’s place-value in the given pattern is increasing
So,
The missing number should be: 7
Hence, from the above,
We can conclude that
The completed pattern is:
567 < 575 < 582

Question 8.
3,408 < 3, 30 < 3,540 ____________
Let the missing digit be x
Now,
The given pattern is:
3,408 < 3,x30 < 3,540
Now,
From the above pattern,
We can observe that
The hundred’s place-value in the given pattern is increasing and then becomes constant
So,
The missing number should be: 5
Hence, from the above,
We can conclude that
The completed pattern is:
3,408 < 3,530 < 3,540

Use the pictograph for 9-11.

Question 9.
In which month shown did the Grand Canyon National Park have about 7,500 tent campers?
The given pictograph is:

Now,
From the above pictograph,
We can observe that

Hence, from the above,
We can conclude that
September month shows that the Grand Canyon National Park has about 7,500 tent campers

Question 10.
Which months had more than 10,000 tent campers?
The given pictograph is:

Now,
From the above pictograph,
We can observe that

Hence, from the above,
We can conclude that
July and August months had more than 10,000 tent campers

Question 11.
What if during the month of October, the park had 22,500 tent campers? How many symbols would be placed on the pictograph for October?
It is given that
During the month of October, the park had 22,500 tent campers
Now,
The given pictograph is:

So,
From the above pictograph,
We can observe that
The number of symbols that would be placed on the pictograph for October = $$\frac{22,500}{5,000}$$
= 4.5
Hence, from the above,
We can conclude that
The number of symbols that would be placed on the pictograph for October is: 4 full symbols + $$\frac{1}{2}$$ symbol

Question 12.
H.O.T. What’s the Question?
Compare: 17,643,251; 17,633,512; and 17,633,893. The answer is 17,633,512,
The given numbers are: 17,643,251, 17,633,512, and 17,633,893
Now,
The representation of the given numbers in the place-value chart is:

Now,
From the above place-value chart,
We can observe that
The order of the numbers from the greatest to the least is:
17,643,251 > 17,633,893 > 17,633,512
Hence, from the above,
We can conclude that
The question required for this problem is:
What is the least number?

Question 13.
Multi-Step What’s the Error?
Max said that 36,594,145 is less than 5,980,251 because 3 is less than 5. Describe Max’s error and give the correct answer.
It is given that
Max said that 36,594,145 is less than 5,980,251 because 3 is less than 5
Now,
The given numbers is: 36,594,145 and 5,980,251
Now,
The representation of the given numbers in the place-value chart is:

Now,
From the above place-value chart,
We can observe that
The Million’s period is greater i.e., 3 > 0
Hence, from the above,
We can conclude that
Max’s error is: Misplacement of digits in the place-value chart
The correct answer is:
36,594,145 > 5,980,251

Fill in the bubble completely to show your answer.

Question 14.
The hotter a pepper, the higher the number of units it is rated. The Scotch bonnet pepper isn’t as hot as a Fatali pepper. Which could be the rating of the Scotch bonnet pepper?

(A) 325,500 units
(B) 400,000 units
(C) 330,000 units
(D) 300,000 units
It is given that
The hotter a pepper, the higher the number of units it is rated. The Scotch bonnet pepper isn’t as hot as a Fatali pepper
Now,
The given rating of Fatali pepper is: 325,000 units
Now,
According to the given information,
The rating of Scotch bonnet pepper will be less than the rating of Fatali pepper i.e., less than 325,000 units
Hence, from the above,
We can conclude that
The rating of the Scotch bonnet pepper is:

Question 15.
Which statement about 3,584,271 and 3,564,092 is true?
(A) 3,584,271 < 3,564,092
(B) 3,584,271 > 3,564,092
(C) 3,584,271 = 3,564,092
(D) 3,564,092 > 3,584,271
The given numbers are: 3,584,271 and 3,564,092
Now,
The representation of the given numbers in the place-value chart is:

So,
From the above place-value chart,
We can observe that
3,584,271 > 3,564,092
Hence, from the above,
We can conclude that
The statement about the given numbers that is true is:

Question 16.
Multi-Step The attendance for 3 baseball games was 38,454 people; 38,198 people; and 36,572 people. Which shows the number of people who attended each game written in order from least to greatest?
(A) 36,572 < 38,198 < 38,454
(B) 38,454 < 38,198 < 36,572
(C) 36,572 < 38,454 < 38,198
(D) 38,454 <36,572 <38,198
It is given that
The attendance for 3 baseball games was 38,454 people; 38,198 people; and 36,572 people
Now,
The representation of the given numbers in the place-value chart is:

So,
From the above place-value chart,
We can observe that
36,572 < 38,198 < 38,454
Hence, from the above,
We can conclude that
The number of people who attended each game written in order from the least to the greatest is:

TEXAS Test Prep

Question 17.
Zachary’s school set a goal of collecting 12,155 cans of food each day. In the first 3 days they collected 12,250 cans; 10,505 cans; and 12,434 cans. Which total was less than their daily goal?
(A) 12,434
(B) 12,250
(C) 12,155
(D) 10,505
It is given that
Zachary’s school set a goal of collecting 12,155 cans of food each day. In the first 3 days they collected 12,250 cans; 10,505 cans; and 12,434 cans
Now,
According to the given information,
The comparison of numbers is:
12,155 < 12,250
12,155 > 10,505
12,155 < 12,434
Hence, from the above,
We can conclude that
The total that was less than their daily goal is:

### Texas Go Math Grade 4 Lesson 1.3 Homework and Practice Answer Key

Compare. Write <, >, or =.

Question 1.
$6,462 __________$6,471
The given numbers are: 6,462 and 6,471
Now,
The representation of the given numbers in the place-value chart is:

Now,
From the above place-value chart,
We can observe that
Ten’s place is greater in 6,462 than in 6,471
Hence, from the above,
We can conclude that

Question 2.
453,536 __________ 451,626
The given numbers are: 453,536 and 451,626
Now,
The representation of the given numbers in the place-value chart is:

Now,
From the above place-value chart,
We can observe that
Thousand’s place is greater in 453,536 than in 451,626
Hence, from the above,
We can conclude that

Order from greatest to least.

Question 3.
87,262; 78,529; 78,811
The given numbers are: 87,262, 78,529, and 78,811
Now,
The representation of the given numbers in the place-value chart is:

Now,
From the above place-value chart,
We can observe that
87,262 > 78,811 > 78,529
Hence, from the above,
We can conclude that
The order of the numbers from the greatest to the least is:
87,262 > 78,811 > 78,529

Question 4.
125,162; 212,225; 128,361
The given numbers are: 125,162, 212,225, and 128,361
Now,
The representation of the given numbers in the place-value chart is:

Now,
From the above place-value chart,
We can observe that
212,225 > 128,361 > 125,182
Hence, from the above,
We can conclude that
The order of the numbers from the greatest to the least is:
212,225 > 128,361 > 125,182

Write all of the digits that can replace each .
Question 5.
735 < 7 4 < 761
Let the missing digit be: x
Now,
The given pattern is:
735 < 7×4 < 761
Now,
From the above pattern,
We can observe that
Ten’s place-value is increasing
So,
The missing number should be: 4 (or) 5
Hence, from the above,
We can conclude that
735 < 744 < 761 (or) 735 < 754 < 761

Question 6.
2,307 < 2 55 < 2,739
Let the missing digit be: x
Now,
The given pattern is:
2,307 < 2,x55 < 2,739
Now,
From the above pattern,
We can observe that
Hundred’s place-value is increasing
So,
The missing number should be: 4 (or) 5 (or) 6
Hence, from the above,
We can conclude that
2,307 < 2,455 < 2,739 (or) 2,307 < 2,555 < 2,739 (or) 2,307 < 2,655 < 2,739

Problem Solving

Use the pictograph for 7-8.

Question 7.
During which months did more than 10,000 people attend the circus?
The given pictograph is:

Now,
From the above pictograph,
We can observe that

Hence, from the above,
We can conclude that
During June and July months, more than 10,000 people attended the circus

Question 8.
List the months in order from greatest attendance to least attendance.
The given pictograph is:

Now,
From the above pictograph,
We can observe that

Hence, from the above,
We can conclude that
The order of the months in order from the greatest attendance to the least attendance is:
July > June > August > May

Lesson Check

Fill in the bubble completely to show your answer.

Question 9.
Leah’s car has 156,261 miles on the odometer. Casey’s car has 165,002 miles on the odometer. Which statement correctly compares the two mileages?
(A)156,261 < 165,002
(B) 165,002 = 156,261
(C) 156,261> 165,002
(D) 165,002 < 156,261
It is given that
Leah’s car has 156,261 miles on the odometer. Casey’s car has 165,002 miles on the odometer
Now,
The given numbers are: 156,261 and 165,002
Now,
The representation of the given numbers in the place-value chart is:

Now,
From the above place-value chart,
We can observe that
165,002 > 156,261
Hene, from the above,
We can conclude that
The statement that correctly compares the two mileages is:

Question 10.
Which number is greater than 53,614,293?
(A) 35,828,781
(B) 53,641,192
(C) 53,416,998
(D) 50,967,299
The given number is: 53,614,293
Hence, from the above,
We can conclude that
The number that is greater than 534,614,293 is:

Question 11.
Which number makes this statement true?
42,607 > ____________ >40,352
(A) 40,255
(B) 38,976
(C) 42,875
(D) 42,410
Let the missing number be: x
Now
The given pattern is:
42,607 > x > 40,352
Now,
From the above pattern,
We can observe that
x < 42,607 and x > 40,352
Hence, from the above,
We can conclude that
The number that makes the given statement true is:

Question 12.
Which digits can replace the to make a true statement?
6,456 < 6, 12 < 6,788
(A) 5, 6
(B) 4, 5, 6
(C) 5, 6, 7
(D) 4, 5, 6, 7
Let the missing digit be: x
Now,
The given pattern is:
6,456 < 6,x12 < 6,788
Now,
From the above pattern,
We can observe that
The Hundred’s place-value is increasing
So,
The missing digit should be: 5 (or) 6
Hence, from the above,
We can conclude that
The digits that can replace the to make a true statement are:

Question 13.
Multi-Step At Ted’s Used Cars, the sales staff set a goal of $25,500 in sales each week. The sales for three weeks were$28,288; $25,369; and$25,876. Which total did not meet the goal?
(A) $25,500 (B)$28,288
(C) $25,369 (D)$25,876
It is given that
At Ted’s Used Cars, the sales staff set a goal of $25,500 in sales each week. The sales for three weeks were$28,288; $25,369; and$25,876
Now,
The given numbers are: 28,288, 25,369, and 25,876
Now,
Compare the given numbers with 25,500
So,
25,500 < 28,888
25,500 > 25,369
25,200 < 25,876
Hence, from the above,
We can conclude that
The total that did not meet the goal is:

Question 14.
Multi-Step Over three years a factory produced 3,516,235; 2,965,005 and 3,261,636 cans of soup. Which lists the numbers in order from least to greatest?
(A) 2,965,005; 3,261,636; 3,516,235
(B) 3,516,235; 3,261,636; 2,965,005
(C) 2,965,005; 3,516,235; 3,261,636
(D) 3,261,636; 2,965,005; 3,516,235
It is given that
Over three years a factory produced 3,516,235; 2,965,005 and 3,261,636 cans of soup
Now,
The given numbers are: 3,516,235, 2,965,005, and 3,261,636
Now,
The representation of the given numbers in the place-value chart is:

Now,
From the above place-value chart,
We can observe that
3,516,235 > 3,261,636 > 2,965,005
Hence, from the above,
We can conclude that
The order of the numbers from the least to the greatest is:

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text:
0
Research Papers

# Full State Feedback Control of Steam Temperature in a Once-Through Direct Steam Generation Receiver Powered by a Paraboloidal Dish

[+] Author and Article Information
José I. Zapata

Solar Thermal Group,
Research School of Engineering
Australian National University,
Canberra 0200, Australia
e-mail: jose.zapata@anu.edu.au

The Hartman–Grobman theorem establishes that under certain conditions, the linear approximation has the same qualitative behavior as the nonlinear model near the operating point [15].

The subscript 0 indicates a specific value for a variable or sets of variables.

The system will converge to the equilibrium point $x0,y0$ when subject to $u0,v0$.

Other studies employing moving-boundary formulations of two-phase flow heat exchangers have obtained further model reductions that tend to preserve only the dominant behavior of the system [20,21].

Typical receiver heat losses in operation are approximately 45–60 kW [25].

Contributed by the Solar Energy Division of ASME for publication in the JOURNAL OF SOLAR ENERGY ENGINEERING: INCLUDING WIND ENERGY AND BUILDING ENERGY CONSERVATION. Manuscript received July 13, 2014; final manuscript received October 14, 2014; published online November 17, 2014. Assoc. Editor: Markus Eck.

J. Sol. Energy Eng 137(2), 021017 (Apr 01, 2015) (10 pages) Paper No: SOL-14-1202; doi: 10.1115/1.4028919 History: Received July 13, 2014; Revised October 14, 2014; Online November 17, 2014

<>

## References

Birnbaum, J., Feldhoff, J. F., Fichtner, M., Hirsch, T., Jöcker, M., Pitz-Paal, R., and Zimmermann, G., 2011, “Steam Temperature Stability in a Direct Steam Generation Solar Power Plant,” Sol. Energy, 85(4), pp. 660–668.
Valenzuela, L., Zarza, E., Berenguel, M., and Camacho, E., 2004, “Direct Steam Generation in Solar Boilers,” IEEE Control Syst. Mag., 24(2), pp. 15–29.
Camacho, E., Rubio, F., Berenguel, M., and Valenzuela, L., 2007, “A Survey on Control Schemes for Distributed Solar Collector Fields. Part I: Modeling and Basic Control Approaches,” Sol. Energy, 81(10), pp. 1240–1251.
Valenzuela, L., Zarza, E., Berenguel, M., and Camacho, E. F., 2005, “Control Concepts for Direct Steam Generation in Parabolic Troughs,” Sol. Energy, 78(2), pp. 301–311.
Eck, M., and Hirsch, T., 2007, “Dynamics and Control of Parabolic Trough Collector Loops With Direct Steam Generation,” Sol. Energy, 81(2), pp. 268–279.
Schlipf, D., Hanel, L., and Maier, H., 2012, “Model Based Controller Design for a Steam Drum in Linear Fresnel CSP-Plant Using Direct Evaporation,” Proceedings of the 18th SolarPACES Conference, Sept. 11–14.
Maffezzoni, C., and Parigi, F., 1982. “Dynamic Analysis and Control of a Solar Power Plant—II. Control System Design and Simulation,” Sol. Energy, 28(2), pp. 117–128.
Rawlings, J. B., and Mayne, D. Q., 2009, Model Predictive Control: Theory and Design, Nob Hill Publishing, Madison, WI.
Zapata, J. I., Pye, J., and Lovegrove, K., 2013, “A Transient Model for the Heat Exchange in a Solar Thermal Once Through Cavity Receiver,” Sol. Energy, 93(0), pp. 280–293.
Diaz, G., 2007, “Controllability of Cross-Flow Two-Phase Heat Exchangers,” Int. J. Heat Mass Transfer, 50(23–24), pp. 4559–4567.
Li, B., Jain, N., and Alleyne, A. G., 2012, “LMI Control Design for Nonlinear Vapor Compression Cycle Systems,” ASME Paper No. DSCC2012-MOVIC2012-886.
Lovegrove, K., Burgess, G., and Pye, J., 2011, “A New 500 m2 Paraboloidal Dish Solar Concentrator,” Sol. Energy, 85(4), pp. 620–626.
Kaneff, S., 1999. “Viable Distributed Dish/Central Plant Solar Power: Status, New Developments, Potential,” J. Phys. IV France, 9, pp. Pr3–195–Pr3–200.
Zapata, J., 2014, “A Modified Extended Kalman Filter to Estimate the State of the SG4 Receiver at the Australia National University,” Proceedings of the Australian Solar Council Conference Melbourne, Australia, May 8–9.
Teschl, G., 2012, Ordinary Differential Equations and Dynamical Systems, Vol. 140, AMS Bookstore, Providence, RI.
Åstrom, K. J., and Murray, R. M., 2008, Feedback Systems, Princeton University Press, Princeton, NJ.
LeVeque, R. J., 2007, Finite Difference Methods for Ordinary and Partial Differential Equations: Steady-State and Time-Dependent Problems, SIAM, Philadelphia, PA.
Hoagg, J., and Bernstein, D., 2007, “Nonminimum-Phase Zeros – Much to do About Nothing – Classical Control Revisited Part II,” IEEE Control Syst., 27(3), pp. 45–57.
Hairer, E., and Wanner, G., 1999, “Stiff Differential Equations Solved by Radau Methods,” J. Comput. Appl. Math., 111(1–2), pp. 93–111.
He, X.-D., Liu, S., Asada, H. H., and Itoh, H., 1999, “Multivariable Control of Vapor Compression Systems,” Proceedings of the ASHRAE Winter Meeting, Vol. 105, ASHRAE, Atlanta, GA, Conference Code: 55431.
Rasmussen, B., and Alleyne, A., 2006, “Dynamic Modeling and Advanced Control of Air Conditioning and Refrigeration Systems,” Air Conditioning and Refrigeration Center, College of Engineering, University of Illinois at Urbana-Champaign, Technical Report No. TR-244.
Franklin, G. F., Powell, J. D., and Emami-Naeini, A., 2010, Feedback Control of Dynamics Systems, 6th ed., Pearson, Upper Saddle River, NJ.
Hodel, A., Tenison, R. B., Clem, D. A., and Ingram, J. E., 1996, “The Octave Control Systems Toolbox: A MATLAB TM-Like CACSD Environment,” Proceedings of the IEEE International Symposium on Computer-Aided Control System Design, Dearborn, MI, Sept. 15–18, pp. 386–391.
Varga, A., 1981, “A Schur Method for Pole Assignment,” IEEE Trans. Autom. Control, AC-26(2), pp. 517–519.
Burgess, G., Lovegrove, K., Mackie, S., Zapata, J., and Pye, J., 2011, “Direct Steam Generation Using the SG4 500 m2 Paraboloidal Dish Concentrator,” Proceedings of the 17th SolarPACES Conference, Granada, Spain, Sept. 15–18.
Wittenmark, B., 1989, “Integrators, Nonlinearities, and Anti-Reset Windup for Different Control Structures,” American Control Conference, Pittsburgh, PA, June 21–23, pp. 1679–1683.
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## Figures

Fig. 1

SG4 steam generation system diagram

Fig. 2

Signal diagram for the SG4 receiver outlet temperature controller. Thick arrows represent vector signals and thin arrows represent scalar signals.

Fig. 3

Complex plane diagram with dominant pole original and desired locations. Inset shows location of zero b7 and the pole/zero cancellation effect of the controller

Fig. 4

Simulated performance of the receiver outlet temperature controller, when transitioning from manual to automatic feed-water mass flow settings, and maintaining temperature during variations in DNI. Vertical dotted lines indicate the period where the temperature controller is active. (a) DNI. (b) Feed-water and outlet mass flow. (c) Receiver outlet temperature, with dotted lines to indicate 500 ± 15 °C. (d) Average receiver pressure. (e) Cumulative length of fluid regions, with respect to tube length (horizontal dotted line).

Fig. 5

Simulated performance of the receiver outlet temperature controller, when rejecting a triple drop to zero in DNI, akin a cloud disturbance. Simulations include the controlled without the AW and FF terms to show their effect on the controller. Simulations also include a PI controller with kp = 8×10-5 and ki = 2.22×10-7 for comparison. (a) DNI. (b) Feed-water and outlet mass flow. (c) Receiver outlet temperature, with dotted lines to indicate 500 ± 15 °C. (d) Average receiver pressure. (e) Cumulative length of fluid regions, with respect to tube length (horizontal dotted line).

Fig. 6

Unsuccessful attempts at experimental temperature control on 25 October 2013. Vertical dotted lines indicate periods when the temperature controller is engaged. (a) DNI. (b) Feed-water mass flow. (c) Receiver outlet temperature. Horizontal dotted lines indicate 500 ± 15 °C. (d) Inlet, outlet, and average receiver pressure. (e) Cumulative length of fluid regions, with respect to tube length (horizontal dotted line).

Fig. 7

Successful experimental test of temperature control on 30 October 2013. A vertical dotted line indicates when the temperature controller was engaged. (a) DNI. (b) Feed-water mass flow. (c) Receiver outlet temperature. Horizontal dotted lines indicate 500 ± 15 °C. (d) Inlet, outlet, and average receiver pressure. (e) Cumulative length of fluid regions, with respect to tube length (horizontal dotted line).

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## Murray : the surname Murray

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### Extract from the top surnames nationwide

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Question

# Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius $$10 m ,$$ mass $$100 kg$$ is filled with helium at $$1.66$$ bar at $$27^o C ($$ Density of air $$= 1.2 kg m^{-3}$$ and $$R = 0.83$$ bar $$dm^3 K^{-1} mol^{-1} ) .$$

A
1138.12 Kg
B
3811.12 Kg
C
1381.12 Kg
D
3112.38 kg

Solution

## The correct option is D $$3811.12$$ KgPayload of the ballon = mass of the displaced air- mass of the ballonRadius of the ballon, $$r=10m$$Mass of the ballon, $$m=100kg$$Therefore volume of the ballon $$=4.3nr^3=4/3\times 22/7\times (10)^3=4190.5\, m^3$$Now volume of the displaced air $$=4190.5\,m^3$$Given :Density of air $$=1.2\, kg\,m^{-3}$$Therefore, the mass of the displaced air $$=4190.5\times 1.2=5028.6\,kg$$Let $$w$$ be the mass of helium gas filled into the ballon, then $$PV=(w/m)RT$$OR $$w=PVM/RT$$$$=(1.66\times 4190.5\times 10^3\times 4)/(0.083\times 300)$$$$=1117\,kg$$ (approx)Total mass of the balloon filled with He $$=1117+100=1217\,kg$$Therefore payload of the balloon $$=5028.6-1217=3811.6\,kg$$Hence,m the pay load of the balloon is $$3811.6\,kg$$Therefore, the correct option is B.Chemistry

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# How do solve x^2+21>10x and write the answer as a inequality and interval notation?

Nov 6, 2016

Rewrite as a quadratic equation, solve, and use test points to see in what interval the solution lies.

${x}^{2} + 21 = 10 x$

${x}^{2} - 10 x + 21 = 0$

$\left(x - 7\right) \left(x - 3\right) = 0$

$x = 7 \mathmr{and} 3$

TEST POINT 1: $x = 2$

2^2 + 21 >^? 10(2)

25 > 20" "color(green)(√)

TEST POINT 2: $x = 4$

4^2 + 21 >^? 10(4)

$37 {>}^{\emptyset} 40 \text{ } \textcolor{red}{\times}$

TEST POINT 3: $x = 8$

8^2 + 21>^?10(8)

85 > 80" "color(green)(√)

So, the solution is $x < 3 \mathmr{and} x > 7$.

In interval notation, this would be $\left(- \infty , 3\right)$ and $\left(7 , \infty\right)$.

Hopefully this helps!