text:
# Bayesian estimation by sampling the prior?

Given a quadratic loss function, the Bayes estimator is given by $$\hat{\theta} = E[\theta|y] = \frac{\int_\Theta\theta p(y|\theta) p(\theta) d\theta}{\int_\Theta p(y|\theta) p(\theta) d\theta}$$ where $\theta\in\Theta$ is the parameter vector, $y$ is the data, $p(\theta)$ is the prior, $p(\theta|y)$ is the posterior, and $p(y|\theta)$ is the likelihood. A variety of numerical techniques (Rejection Sampling, MCMC methods, etc.) have been developed to draw from the posterior distribution when direct sampling from the posterior is unavailable. These techniques allow us to estimate $E[\theta|y]$ with $$E[\theta|y]\approx\frac{1}{n}\sum_{i=1}^n\theta_i$$ where $\theta_i$ is the $i^{th}$ draw from the posterior.

My question: If we are able to draw from the prior distribution, couldn't we just sample the prior to numerically approximate the integrals on the right hand side of the first equation? In other words, approximate $E[\theta|y]$ with$$E[\theta|y]\approx \frac{\frac{1}{n}\sum_{i=1}^n\theta_i p(y|\theta_i)}{\frac{1}{n}\sum_{i=1}^np(y|\theta_i)} = \frac{\sum_{i=1}^n\theta_i p(y|\theta_i)}{\sum_{i=1}^np(y|\theta_i)}$$ where $\theta_i$ is the $i^{th}$ draw from the prior. This would seem to be a more convenient and computationally efficient alternative to Rejection Sampling and MCMC techniques. I understand that in some cases we may not be able to draw from the prior, but in those cases why not just pick different priors? I haven't seen the method I'm proposing discussed in any textbooks or online, so I'm hoping someone can explain to me the folly of this approach.

• Numerical approximation is widely used compared to MCMC sampling. It is just a matter of choice really, although there are benefits and drawbacks to both. For people who aren't familiar with mathematical theories and concepts of numerical approximations, it is just simpler to understand the workings of MCMC algorithms as opposed to Newton methods and such. – Tony Hellmuth May 31 '18 at 2:48
• Do you know of any specific instances in which people have used this method? As I said, I have yet to come across any reference that makes use of this approach. I would also argue that the method I'm proposing is much easier to understand than MCMC techniques and it does not use any Newtonian methods. – Peter May 31 '18 at 16:14
• Yes certainly in hydrology, geo-spatial modelling and physics. There is a tendency to see computational power play a huge role since MCMC takes so freaking long to converge in say 10 parameter models and highly order differential equations when working with large, like massive datasets. Your answer below is certainly apt though, good find! – Tony Hellmuth May 31 '18 at 23:56

Alright, I think I found the answer to my question. The reason my approximation is rarely/never used is that it is generally a very poor approximation of the integrals. To see why, consider what happens when the likelihood function spikes somewhere in the tails of the prior (as will often be the case). The $\theta$'s surrounding the spike will rarely be sampled, resulting in a poor approximation of the likelihood function. See pages 25 and 26 of the document linked below for more information: https://engineering.purdue.edu/kak/Tutorials/MonteCarloInBayesian.pdf


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# Mock Objects Demystified

If you have ever tried writing any non trivial tests, mocks should not be a stranger to you. But what about some other “mock” like objects like stubs, spies and such? How are they different from each other? In this blog post, I will explain it as simple and easy to remember as possible.

Everything is a test double. Test double is just a general name for all mock like objects.

Then the simplest of them all is Dummy. A Dummy is an implementation of an interface that all methods return as close to nothing as possible. It don’t do nothing, it don’t return nothing, it’s just there.

We often use an dummy when we test a function and it requires some argument that is not related to what we want to test.

Next in the sequence of test doubles is stub, stub is a dummy, all functions in stub do nothing. The difference is that instead of return nothing, the functions return some predefined fixed value.

By return certain values, the stub drives the code to the path way that we want to test.

The next test double is the spy. The spy is a stub, it do nothing and return the value that are useful to the test. However, it remember certain interesting facts about the way it is called, and report the facts later to the test.

Sometimes you write a spy just to remember a function gets called, or how many times it gets called, or how many args are passed in, etc. But, a spy don’t do nothing, all the spy does is watch and remember.

At last, the mock. A mock is a spy, but it knows very specific how to check what are being spied. Usually it contains some verify…() function like verifyAlertTriggerOnThirdFailedAttempt(). So a true mock is a spy knows what to expect, and the test ask the mock if everything went as expected.

Finally there is the fake. A fake is not anything above. It is a simulator. The fake is very smart and complicated and it starts to look like the real deal. For example, a fake Authorizer may return User with id 1 if you pass in names that start with good, else it returns InvalidUser. But be caucious not to grow it too much or it will become a mess very quickly. And you probably want to have tests that test fakes. For unit tests fakes are really not that necessary, but for intergration tests, they can come quite handy. But in general, avoid fakes when you can.

That’s the end.

format: plain


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# Know the limits of an image after rotation

I have an image which I know the number of rows and the number of columns. I would like to calculate the limits of the new image after a rotation.

   BufferedImage myImage = ImageIO.read( new File( "D:\\Users...jpg" ) );
xmaxOrigine = myImage.getWidth() - 1;
ymaxOrigine = myImage.getHeight() - 1;
angle = 12;

p1X = (int) ( ( xmaxOrigine * cos ) - ( ymaxOrigine * sin ) );
p1Y = (int) ( ( xmaxOrigine * sin ) + ( ymaxOrigine * cos ) );

xmin_f = xmin_f < p1X ? xmin_f : p1X;
xmax_f = xmax_f < p1X ? p1X : xmax_f;
ymin_f = ymin_f < p1Y ? ymin_f : p1Y;
ymax_f = ymax_f < p1Y ? p1Y : ymax_f;

p2X = (int) ( ( 0  * cos ) - ( ymaxOrigine * sin ) );
p2Y = (int) ( ( 0 * sin ) + ( ymaxOrigine * cos ) );

xmin_f = xmin_f < p2X ? xmin_f : p2X;
xmax_f = xmax_f < p2X ? p2X : xmax_f;
ymin_f = ymin_f < p2Y ? ymin_f : p2Y;
ymax_f = ymax_f < p2Y ? p2Y : ymax_f;

p3X = (int) ( ( xmaxOrigine * cos ) - ( 0 * sin ) );
p3Y = (int) ( ( xmaxOrigine * sin ) + ( 0 * cos ) );

xmin_f = xmin_f < p3X ? xmin_f : p3X;
xmax_f = xmax_f < p3X ? p3X : xmax_f;
ymin_f = ymin_f < p3Y ? ymin_f : p3Y;
ymax_f = ymax_f < p3Y ? p3Y : ymax_f;

p4X = 0;
p4Y = 0;

xmin_f = xmin_f < p4X ? xmin_f : p4X;
xmax_f = xmax_f < p4X ? p4X : xmax_f;
ymin_f = ymin_f < p4Y ? ymin_f : p4Y;
ymax_f = ymax_f < p4Y ? p4Y : ymax_f;

widthFinal = xmax_f - xmin_f;
heightFinal = ymax_f - ymin_f;

As you can see I look for the xmin, xmax, ymin, ymax for each points. I would like to make this operation once if it is possible. Also, how can I condense this code?

• Where are the min and max variables declared and initialized? – JS1 Aug 19 '15 at 9:01

xmaxOrigine

So this looks like you are mixing a French word with English. Be careful of that. A native English reader will be mildly confused by this, as it looks similar to original which would make sense in that context. But the real problem is with someone for whom origine does not look like a word. For example, take native Mandarin speakers. Presumably they know how to look up English words that they don't understand in the English code. They won't however know that this is not an English word. So they won't find it in an English directory and won't know that they should look in a French dictionary.

Perhaps your context will have you overwhelmingly sharing this code with people who speak enough French to know what you mean. But if you do need to share with someone who doesn't speak French, that person may be confused by things like this.

    xmin_f = xmin_f < p1X ? xmin_f : p1X;
xmax_f = xmax_f < p1X ? p1X : xmax_f;
ymin_f = ymin_f < p1Y ? ymin_f : p1Y;
ymax_f = ymax_f < p1Y ? p1Y : ymax_f;

This would be easier to read if you used the natural language constructs for this.

    xmin_f = Math.min(xmin_f, p1X);
xmax_f = Math.max(p1X, xmax_f);
ymin_f = Math.min(ymin_f, p1Y);
ymax_f = Math.max(p1Y, ymax_f);

You also do this repeatedly. If these are object fields, you can make a method to do this for you.

private void expandBoundaries(int x, int y) {
xmin_f = Math.min(xmin_f, x);
xmax_f = Math.min(xmax_f, x);
ymin_f = Math.min(ymin_f, y);
ymax_f = Math.min(ymax_f, y);
}

If they are not object fields, you can make objects to track them. Then you'd use them like

    xRange.include(p1X);
yRange.include(p1Y);

and at the end

    widthFinal = xRange.getMax() - xRange.getMin();
heightFinal = yRange.getMax() - yRange.getMin();

or even

    widthFinal = xRange.getDifference();
heightFinal = yRange.getDifference();

with include defined as

public void include(T value) {
min = Math.min(min, value);
max = Math.max(max, value);
}

You can also save some multiplications by storing the values for later use:

    float rotatedX2X = xmaxOrigine * cos;
float rotatedY2X = ymaxOrigine * sin;
float rotatedX2Y = xmaxOrigine * sin;
float rotatedY2Y = ymaxOrigine * cos;

Then you could rewrite this code as

    BufferedImage myImage = ImageIO.read( new File( "D:\\Users...jpg" ) );
xmaxOrigine = myImage.getWidth() - 1;
ymaxOrigine = myImage.getHeight() - 1;

angle = 12;

float rotatedX2X = xmaxOrigine * cos;
float rotatedY2X = ymaxOrigine * sin;
float rotatedX2Y = xmaxOrigine * sin;
float rotatedY2Y = ymaxOrigine * cos;

xRange.include((int) (rotatedX2X - rotatedY2X));
yRange.include((int) (rotatedX2Y + rotatedY2Y));
xRange.include((int)-rotatedY2X);
yRange.include((int)rotatedY2Y);
xRange.include((int)rotatedX2X);
yRange.include((int)rotatedX2Y);
xRange.include(0);
yRange.include(0);

widthFinal = xRange.getDifference();
heightFinal = yRange.getDifference();

Note that this gets rid of p1X, p1Y, etc. They weren't used in the original code snippet. It's possible that they were used elsewhere. If so, you'd have to put them back.

Then your Range class:

public class Range<T extends Number> {
private T max;
private T min;

public Range(T min, T max) {
this.min = min;
this.max = max;
}

public void include(T value) {
min = Math.min(min, value);
max = Math.max(max, value);
}

public T getMin() {
return min;
}

public T getMax() {
return max;
}

public T getDifference() {
return max - min;
}
}

Note that you have to initialize the Range before using it. You don't show the initialization in the original code snippet, so I didn't show it here. It would look something like

    Range xRange = new Range(small, large);

I would start with a rotation matrix and think about, who you can combine all points to one matrix, so you "only" have to calculate one matrix multiplication.

The rotation matrix is

new double[][] {
{ Math.cos(theta), -Math.sin(thera) },
{ Math.sin(theta), Math.cos(theta) }
}

And your matrix for the coordinates could be:

new double[][] {
{ x1, x2, ... },
{ y1, y2, ... }
}

You can find many examples how to calculate a matrix multiplication with Google. But you have to check for the right traversing over the arrays!

The result will be an array of rotated coordinates. Then you have to find the min and max of both rows by two for loops and you are done.

If you choose good names and extract some of the functionality in other methods (like creating the rotation matrix or the matrix multiplication, and so on) the code would be very short and nice to read.

It seems like you are coming from a C background. I primarily know the type suffixes like _f or _i for float and integer (you did only use the first one) only from C code. In Java you didn't do this. Most of the time you should use an IDE which can tell you the type of a variable.

Last thing I noticed: you cast a float to an integer. But this can lead to an off by one error:

int f = 0.6;
int a = (int) f; // is 0. Should be 1.
Int b = (int) f + 0.5; // is 1.

# Off by one?

At the top, you subtract off one from the width and height like this:

xmaxOrigine = myImage.getWidth() - 1;
ymaxOrigine = myImage.getHeight() - 1;

I think you did this because you are considering the rectangle to span between the center of pixels instead of the borders. But then at the end, you don't add one back in:

widthFinal = xmax_f - xmin_f;
heightFinal = ymax_f - ymin_f;

So if I had a 100x100 image, and rotated by 0 degrees, your function would return 99x99. I believe you should add one back in to return to the original units. You may also want to add another 0.5 to handle rounding:

widthFinal = xmax_f - xmin_f + 1.5f;
heightFinal = ymax_f - ymin_f + 1.5f;

Of course, I've made a few assumptions about what your program is intending to do. You should test your program yourself with various image sizes and angles to make sure your results match what you expect.


text:
# How do you simplify 3 (4 + 5s ) - 12 + ( - 3s )?

Jul 8, 2016

12s

#### Explanation:

Notice that there are three terms. We need to remove the brackets and then collect the like terms together.

$12 + 15 s - 12 - 3 s$

=$12 s$


text:
# Electronic – Program to make bode plot of transfer function

bode plot

For different purposes, I sometimes have to draw an bode plot. I first of would like to know if there is any piece of software which can draw them for me. So I gave in the transfer function and then it gives me the bode plot (both phase and the magnitude). This would save some time at occasions.

Ok, so now the real question. I have the following transfer function.

$$H(j\omega) = \frac{j \frac{\omega}{\omega_0} }{1 + 3j \frac{\omega}{\omega_0}}$$

The modulus can be calculated as follows (correct me if I am wrong):

$$|H(j\omega)| = 20 \log_{10}(p) – 20 \log_{10} \left(\sqrt{1+(3p)^2}\right)$$

(where \$p=\frac{\omega}{\omega_0} \$, with \$\omega_0=\frac{1}{500 \times 6.37 \times 10^{-7}} \$).

\$20 \log_{10}(p) \$, would be easy, but I am not sure how to calculate the second one? How do I calculate the (magnitude) bode plot of \$-20 \log_{10} \left(\sqrt{1+(3p)^2} \right) \$


text:
# Math Help - proof floor function

1. ## proof floor function

Can you help me to solve this please ?

2. Here are some examples.
$g^{ - 1} \left( {\{ 2\} } \right) = [2,3)\;\& {\kern 1pt} g^{ - 1} \left( {\{ - 1, - 2\} } \right) = [ - 2,0)\,$

Now you try.


text:
An invited review: neutron lifetime anomaly and mirror matter theory

It still feels like yesterday. Almost exactly four years ago, also around Chinese New Year, I finished my first paper (or to be exact, two) on the new mirror matter theory. Now I just finished my first invited review paper, which exactly details the original motivations on solving the puzzles of neutron lifetime in my first published paper. It feels like I just completed the circle. So many thanks for Dr. Ben Grinstein’s invitation. I’ve been trying to write a review on mirror matter theory and related experiments and observations for a long time. But it never came through. Ben’s invitation has really pushed me to finish this review paper earlier. It is not the full review paper I have imagined, but still a very important part of it. It focuses on the unique perspectives in the analysis of the neutron lifetime anomaly and the CKM unitarity issue, which have been mostly overlooked by the mainstream. It does not present the full picture and details of mirror matter theory. Instead, it gives the details of the phenomenological $n-n’$ oscillation model, and presents exactly how it can explain the above puzzles and how we can test its unique predictions in laboratory experiments.

Here is the paper: Neutron lifetime anomaly and mirror matter theory


text:
Definition 15.23.1. Let $R$ be a domain. We say an $R$-module $M$ is reflexive if the natural map

$j : M \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(M, R), R)$

which sends $m \in M$ to the map sending $\varphi \in \mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ to $\varphi (m) \in R$ is an isomorphism.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AUZ. Beware of the difference between the letter 'O' and the digit '0'.


text:
MathSciNet bibliographic data MR1409799 (97k:46074) 46L50 (46L10) Watanabe, Keiichi An application of orthoisomorphisms to non-commutative \$L\sp p\$$L\sp p$-isometries. Publ. Res. Inst. Math. Sci. 32 (1996), no. 3, 493–502. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
# Questions on Riemann Hypothesis prediction and explicit formula for $\hat{\Phi}(x)=\sum\limits_{n\le x}\frac{\phi(n)}{n}$

I've primarily been able to find information on the the summatory Euler totient function $$\Phi(x)$$ defined in (1) below, but I'm more interested in the $$\hat{\Phi}(x)$$ defined in (2) below.

(1) $$\quad\Phi(x)=\sum\limits_{n\le x}\phi(n)\,,\qquad\frac{\zeta(s-1)}{\zeta(s)}=\sum\limits_{n\le x}\frac{\phi(n)}{n^s}\quad\Re(s)>2$$

(2) $$\quad\hat{\Phi}(x)=\sum\limits_{n\le x}\frac{\phi(n)}{n}\,,\qquad\frac{\zeta(s)}{\zeta(s+1)}=\sum\limits_{n\le x}\frac{\phi(n)}{n^{s+1}}\quad\Re(s)>1$$

The following figure illustrates the $$\hat{\Phi}(x)$$ function defined in (2) above seems to exhibit a near-linear growth analogous the the first and second Chebyshev functions $$\vartheta(x)$$ and $$\psi(x)$$.

Figure (1): Illustration of formula (2) for $$\hat{\Phi}(x)$$

Question (1): What does the Riemann hypothesis predict for the asymptotic growth and error bound with respect to the $$\hat{\Phi}(x)$$ function defined in (2) above?

I've been told the explicit formula for $$\Phi(x)$$ defined in (3) below doesn't converge, but I'm wondering if there's one that converges for $$\hat{\Phi}(x)$$.

(3) $$\quad\Phi_o(x)=\frac{3\,x^2}{\pi^2}+\sum\limits_\rho\frac{x^\rho\,\zeta(\rho-1)}{\rho\,\zeta'(\rho)}+\frac{1}{6}+\sum\limits_{n}\frac{x^{-2\,n}\,\zeta(-2\,n-1)}{(-2\,n)\,\zeta'(-2\,n)}$$

Question (2): Is there a convergent explicit formula for the $$\hat{\Phi}(x)$$ function defined in (2) above and if so, what is the definition?

This question is motivated in part by the relationship between $$f(x)$$ and $$\hat{\Phi}(x)$$ illustrated in (6) below which leads to the formula for the Dirichlet eta function $$\eta(s)$$ defined in (7) below which I believe converges for $$\Re(s)>1$$. Note $$a_n$$ is 2-periodic (1, -3, 1, -3, ...).

(4) $$\quad f(x)=\sum\limits_{n\le x}(-1)^{n-1}\qquad\eta(s)=\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\,,\quad\Re(s)>0$$

(5) $$\quad\hat{\Phi}(x)=\sum\limits_{n\le x}\frac{\phi(n)}{n}\qquad\quad\frac{\zeta(s)}{\zeta(s+1)}=\sum\limits_{n=1}^\infty\frac{\phi(n)}{n^{s+1}}\,,\quad\Re(s)>1$$

(6) $$\quad f(x)=\sum\limits_{n\le x}\frac{a_n}{n}\hat{\Phi}\left(\frac{x}{n}\right)\,,\qquad a_n=(-1)^{n-1}((2 n-1)\bmod 4)$$

(7) $$\quad\eta(s)=\sum\limits_{n=1}^\infty\frac{a_n}{n^{s+1}}\sum\limits_{m=1}^\infty\frac{\phi(m)}{m}\,m^{-s}\,,\quad\Re(s)>1$$

Formula (7) above can be analytically continued as illustrated in formulas (8) and (9) below which confirm the correctness of relationship (6) above. In formulas (8) and (9) below the sum over $$m$$ in formula (7) above is analytically continued to $$\frac{\zeta(s)}{\zeta(s+1)}$$, and in formula (9) below the sum over $$n$$ in formula (7) above is also analytically continued to $$\left(1-2^{1-s}\right)\zeta(s+1)$$.

(8) $$\quad\eta(s)=\sum\limits_{n=1}^\infty\frac{a_n}{n^{s+1}}\frac{\zeta(s)}{\zeta(s+1)}\,,\quad\Re(s)>0$$

(9) $$\quad\eta(s)=\left(1-2^{1-s}\right)\zeta(s)$$

Note formula (8) above is equivalent to formula (10) below.

(10) $$\quad\zeta(s+1)=\frac{1}{1-2^{1-s}}\sum\limits_{n=1}^\infty\frac{a_n}{n^{s+1}}\,,\quad\Re(s)>0\land s\ne 2\,\pi\,i\,t\land t\in\mathbb{Z}$$

• No. Look at the residue theorem, at the estimates of $F(s)x^{-s}$ in each part, the problem of traversing the critical strip, the proofs of the explicit formula for $\sum_{p^k \le x} \log p$, look at those for $\sum_{n \le x} \mu(n)$ and their complication, see why it doesn't work for $\sum_{n \le x} \phi(n) n^{-k}$. – reuns Jul 1 at 3:11
• @reuns Ok, I'm disappointed there's no convergent explicit formula, but do these problems also rule out a Riemann hypothesis prediction? – Steven Clark Jul 1 at 3:48
• @StevenClark, the problem is that the error term for these functions is not dominated by the contribution from the zeroes, but rather an "arithmetic" part that is much bigger. As reuns mentioned, the issue is shifting the contour; one cannot bound the contours satisfactorily. See also this paper: sciencedirect.com/science/article/pii/S0022314X10001873 – Peter Humphries Jul 2 at 9:24
• @PeterHumphries The Nicolas criterion implies $\hat{\Phi}_n(N_k)-\hat{\Phi}_n(N_k-1)<\frac{\log(\log(N_k))}{e^{\gamma}}$ where $N_k$ is the product of the first $k$ primes is equivalent to the Riemann Hypothesis. I guess one can't do any better than that. – Steven Clark Jul 2 at 17:07


text:
# American Institute of Mathematical Sciences

ISSN:
1930-5346

eISSN:
1930-5338

All Issues

## Advances in Mathematics of Communications

Open Access Articles

2020, 14(3): i-i doi: 10.3934/amc.2020101 +[Abstract](588) +[HTML](180) +[PDF](91.27KB)
Abstract:
2020 doi: 10.3934/amc.2020073 +[Abstract](771) +[HTML](416) +[PDF](438.53KB)
Abstract:

Number theoretic public-key solutions, currently used in many applications worldwide, will be subject to various quantum attacks, making them less attractive for longer-term use. Certain group theoretic constructs are now showing promise in providing quantum-resistant cryptographic primitives, and may provide suitable alternatives for those looking to address known quantum attacks. In this paper, we introduce a new protocol called a Meta Key Agreement and Authentication Protocol (MKAAP) that has some characteristics of a public-key solution and some of a shared-key solution. Specifically, it has the deployment benefits of a public-key system, allowing two entities that have never met before to authenticate without requiring real-time access to a third-party, but does require secure provisioning of key material from a trusted key distribution system (similar to a symmetric system) prior to deployment. We then describe a specific MKAAP instance, the Ironwood MKAAP, discuss its security, and show how it resists certain quantum attacks such as Shor's algorithm or Grover's quantum search algorithm. We also show Ironwood implemented on several "internet of things" (IoT devices), measure its performance, and show how it performs significantly better than ECC using fewer device resources.

2021, 15(2): 207-218 doi: 10.3934/amc.2020053 +[Abstract](1537) +[HTML](590) +[PDF](373.36KB)
Abstract:

We present a code-based public-key cryptosystem, in which we use Reed-Solomon codes over an extension field as secret codes and disguise it by considering its shortened expanded code over the base field. Considering shortened expanded codes provides a safeguard against distinguisher attacks based on the Schur product. Moreover, without using a cyclic or a quasi-cyclic structure we obtain a key size reduction of nearly \begin{document}$45 \%$\end{document} compared to the classic {McE}liece cryptosystem proposed by Bernstein et al.

2021, 15(1): 99-112 doi: 10.3934/amc.2020045 +[Abstract](1222) +[HTML](606) +[PDF](408.68KB)
Abstract:

We investigate a class of linear codes by choosing a proper defining set and determine their complete weight enumerators and weight enumerators. These codes have at most three weights and some of them are almost optimal so that they are suitable for applications in secret sharing schemes. This is a supplement of the results raised by Wang et al. (2017) and Kong et al. (2019).

2020, 14(2): 301-306 doi: 10.3934/amc.2020021 +[Abstract](1562) +[HTML](634) +[PDF](299.97KB)
Abstract:

McNie [8] is a code-based public key encryption scheme submitted to the NIST Post-Quantum Cryptography standardization [10] as a candidate. In this paper, we present Dual-Ouroboros, an improvement of McNie, which can be seen as a dual version of the Ouroboros-R protocol [1], another candidate to the NIST competition. This new improved protocol permits, first, to avoid an attack proposed by Gaborit [7] and second permits to benefit from a reduction security to a standard problem (as the original Ouroboros protocol).

2020, 14(2): 279-299 doi: 10.3934/amc.2020020 +[Abstract](1255) +[HTML](650) +[PDF](430.21KB)
Abstract:

This paper is concerned with the construction of algebraic-geometric (AG) codes defined from GGS curves. It is of significant use to describe bases for the Riemann-Roch spaces associated with some rational places, which enables us to study multi-point AG codes. Along this line, we characterize explicitly the Weierstrass semigroups and pure gaps by an exhaustive computation for the basis of Riemann-Roch spaces from GGS curves. In addition, we determine the floor of a certain type of divisor and investigate the properties of AG codes. Multi-point codes with excellent parameters are found, among which, a presented code with parameters \begin{document}$[216,190,\geqslant 18]$\end{document} over \begin{document}$\mathbb{F}_{64}$\end{document} yields a new record.

2020, 14(1): i-i doi: 10.3934/amc.2020012 +[Abstract](1800) +[HTML](479) +[PDF](212.1KB)
Abstract:
2019, 13(4): ⅰ-ⅱ doi: 10.3934/amc.2019033 +[Abstract](1510) +[HTML](402) +[PDF](96.27KB)
Abstract:
2017, 11(2): i-i doi: 10.3934/amc.201702i +[Abstract](2629) +[HTML](130) +[PDF](29.6KB)
Abstract:
2016, 10(3): i-ii doi: 10.3934/amc.201603i +[Abstract](1006) +[PDF](133.8KB)
Abstract:
The present issue of the Advances in Mathematics of Communications is dedicated to the conference

2016, 10(1): i-i doi: 10.3934/amc.2016.10.1i +[Abstract](2249) +[PDF](98.6KB)
Abstract:
The $4^{th}$International Castle Meeting on Coding Theory and its Applications (4ICMCTA) took place in the Palmela Castle, Portugal, on September 15--18, 2014. It was organized under the auspices of the Research & Development Center for Mathematics and Applications (CIDMA) from the University of Aveiro. Following in the spirit of the previous installments held at La Mota Castle, Spain, in 1999 and 2008, and at Cardona Castle, Spain, in 2011, the meeting has been a good opportunity for communicating new results, exchanging ideas, strengthening international cooperation, and introducing young researchers into the Coding Theory community.

2015, 9(1): 9-21 doi: 10.3934/amc.2015.9.9 +[Abstract](3721) +[PDF](424.1KB)
Abstract:
In this paper, a novel method for constructing complementary sequence set with zero correlation zone (ZCZ) is presented by interleaving and combining three orthogonal matrices. The constructed set can be divided into multiple sequence groups and each sequence group can be further divided into multiple sequence subgroups. In addition to ZCZ properties of sequences from the same sequence subgroup, sequences from different sequence groups are orthogonal to each other while sequences from different sequence subgroups within the same sequence group possess ideal cross-correlation properties, that is, the proposed ZCZ sequence set has inter-group orthogonal (IGO) and inter-subgroup complementary (ISC) properties. Compared with previous methods, the new construction can provide flexible choice for ZCZ width and set size, and the resultant sequences which are called IGO-ISC sequences in this paper can achieve the theoretical bound on the set size for the ZCZ width and sequence length.
2014, 8(4): i-ii doi: 10.3934/amc.2014.8.4i +[Abstract](2144) +[PDF](106.0KB)
Abstract:
Every second year, the GEOCRYPT conference brings together researchers from arithmetic geometry and cryptography. After Guadeloupe in 2009 and Corsica in 2011, French Polynesia was the host of its 2013 edition which took place in Punaauia, on the island of Tahiti, October 7-11.
The main topic of GEOCRYPT has always been the application of pure mathematical techniques to the safety and efficiency of modern communication systems, with particular interest in the fields of arithmetic and algebraic geometry over finite fields, algorithms for finite fields, error correcting codes, cryptology, boolean functions, discrete dynamical systems, and their interactions. The 2013 edition had the honour to feature six invited and eleven contributed talks from renowned international experts presenting strong results recently obtained on topics ranging from pure mathematics to cryptographic algorithms. The GEOCRYPT 2013 Program Committee carefully evaluated the submitted abstracts and selected the best contributions for presentation at the conference.

2013, 7(1): i-i doi: 10.3934/amc.2013.7.1i +[Abstract](2225) +[PDF](97.8KB)
Abstract:
Six years have passed since the founding of Advances in Mathematics of Communications, a journal devoted to all mathematical aspects of information and communications technology. As expressed in the editorial of the inaugural volume, communications technology is omnipresent in contemporary life. However, its ubiquity sometimes obscures the fact that the foundations of communications technology are genuinely mathematical, as are its methods, both analytical and constructive.

2011, 5(2): i-ii doi: 10.3934/amc.2011.5.2i +[Abstract](2230) +[PDF](127.0KB)
Abstract:
The present issue of the Advances in Mathematics of Communications has been dedicated to the conference

ALCOMA'10

(Algebraic Combinatorics and Applications, Designs and Codes)

which took place April 11-18, 2010 in the picturesque village of Thurnau near Bay\-reuth, Germany.
The purpose of this meeting was to bring together mathematicians from around the world to exchange recent results in all areas of Discrete Mathematics and Mathematics of Communications based on algebraic methods. Special emphasis was placed on the constructive theory of finite structures, in particular on combinatorial designs and codes. The conference was the third of its kind, after the inaugural meeting ALCOMA'99 in Gößweinstein, which was later followed by ALCOMA'05 at the same venue.

2010, 4(2): i-ii doi: 10.3934/amc.2010.4.2i +[Abstract](2037) +[PDF](42.7KB)
Abstract:
The Conference on Hyperellptic Curves, Discrete Logarithms, Encryption, etc. took place March 16-20, 2009 in Frutillar, Chile. This event derives its acronym CHiLE 2009 from its Spanish title

Curvas Hiperelĺpticas, Logaritmos discretos, Encriptación, etc.

The goal of this conference was to bring together researchers working on cryptographic aspects of algebraic geometry from all over the world, and especially from South America. It was hoped that this would help promote the development and expansion of this field of research in South America.

2007, 1(1): i-ii doi: 10.3934/amc.2007.1.1i +[Abstract](2533) +[PDF](53.2KB)
Abstract:
Almost 60 years have passed since Claude E.~Shannon's celebrated two-part article A Mathematical Theory of Communication appeared in July and October of 1948. With that paper, Shannon founded a discipline, whose wide reach could hardly have been foreseen at the time.
To commemorate the occasion of the 50th anniversary of the birth of Information Theory as a discipline, the IEEE Transactions on Information Theory published a special issue that included the following statement in its preface:
With communication engineering at the epicenter of the bombshell, the sensational aftermath of Shannon's paper soon reached Mathematics, Physics, Statistics, Computing, and Cryptology. Even Economics, Biology, Linguistics and other fields in the natural and social sciences felt the ripples of Shannon's new theory.

2019  Impact Factor: 0.734


text:
CONTINUITY OF APPROXIMATE POINT SPECTRUM ON THE ALGEBRA B(X)

Title & Authors
CONTINUITY OF APPROXIMATE POINT SPECTRUM ON THE ALGEBRA B(X)

Abstract
In this paper we provide a brief introduction to the continuity of approximate point spectrum on the algebra B(X), using basic properties of Fredholm operators and the SVEP condition. Also, we give an example showing that in general it not holds that if the spectrum is continuous an operator T, then for each $\small{{\lambda}{\in}{\sigma}_{s-F}(T){\setminus}\overline{{\rho}^{\pm}_{s-F}(T)}}$ and $\small{{\in}}$ > 0, the ball $\small{B({\lambda},{\in})}$ contains a component of $\small{{\sigma}_{s-F}(T)}$, contrary to what has been announced in [J. B. Conway and B. B. Morrel, Operators that are points of spectral continuity II, Integral Equations Operator Theory 4 (1981), 459-503] page 462.
Keywords
approximate point spectrum;continuity of the spectrum;
Language
English
Cited by
1.
WEYL'S THEOREM, TENSOR PRODUCT, FUGLEDE-PUTNAM THEOREM AND CONTINUITY SPECTRUM FOR k-QUASI CLASS An* OPERATO,;;

대한수학회지, 2014. vol.51. 5, pp.1089-1104
1.
WEYL'S THEOREM, TENSOR PRODUCT, FUGLEDE-PUTNAM THEOREM AND CONTINUITY SPECTRUM FOR k-QUASI CLASS An*OPERATO, Journal of the Korean Mathematical Society, 2014, 51, 5, 1089
References
1.
M. Ahues, A. Largillier, and B. V. Limaye, Spectral Computations for Bounded Operators, Chapman & Hall/CRC, 2001.

2.
P. Aiena, Fredholm and Local Spectral Theory with Applications to Multipliers, Kluwer Acad., 2004.

3.
C. Apostol, L. A. Fialkow, D. A. Herrero, and D. Voiculescu, Approximation of Hilbert space operators. Vol. II, Res. Notes Math. 102, Pitman, Boston, 1984.

4.
L. Burlando, Continuity of spectrum and spectral radius in algebras of operators, Ann. Fac. Sci. Toulouse Math. 9 (1988), no. 1, 5-54.

5.
S. R. Caradus, W. E. Pfaffenberger, and B. Yood, Calkin Algebras and Algebras of Operators on Banach Spaces, Marcel Dekker, 1974.

6.
J. B. Conway and B. B. Morrel, Operators that are points of spectral continuity, Integral Equations Operator Theory 2 (1979), no. 2, 174-198.

7.
J. B. Conway and B. B. Morrel, Operators that are points of spectral continuity. II, Integral Equations Operator Theory 4 (1981), no. 4, 459-503.

8.
S. V. Djordjevic, The continuity of the essential approximative point spectrum, Facta Univ. Ser. Math. Inform. 10 (1995), 97-104.

9.
S. V. Djordjevic and Y. M. Han, Browder's theorems and spectral continuity, Glasg. Math. J. 42 (2000), no. 3, 479-486.

10.
S. V. Djordjevic and Y. M. Han, Operator matrices and spectral continuity, Glasg. Math. J. 43 (2001), no. 3, 487-490.

11.
B. P. Duggal, I. H. Jeon, and I. H. Kim, Continuity of the spectrum on a class of upper triangular operator matrices, J. Math. Anal. Appl. 370 (2010), 584-587.

12.
P. R. Halmos and G. Lumer, Square roots of operators. II, Proc. Amer. Math. Soc. 5 (1954), 589-595.

13.
R. E. Harte and W. Y. Lee, Another note on Weyl's theorem, Trans. Amer. Math. Soc. 349 (1997), no. 5, 2115-2124.

14.
J. D. Newburgh, The variation of Spectra, Duke Math. J. 18 (1951), 165-176.

15.
S. Sanchez-Perales and S. V. Djordjevic, Continuity of spectrum and approximate point spectrum on operator matrices, J. Math. Anal. Appl. 378 (2011), no. 1, 289-294.


text:
# Output format in MatrixForm

I have a correlation matrix, which I want to display in the following format:

• 2 decimal places (e.g., -0.56)
• number signs "+" and "-" (e.g., +0.76 or -0.34)

I've tried many different things, but I had no success so far... The code I'm using to generate the data is:

randomWalk[x_] := Accumulate[Prepend[RandomVariate[NormalDistribution[0, 1], x], 0]]
exchange = Table[Subscript[asset, i] = randomWalk[500], {i, 1, 5}];

Now the code I'm using to generate the correlation matrix:

N[Correlation[Transpose[exchange][[1 ;; 500, All]]], 2]//MatrixForm

However, the numbers generated are displayed with $MachinePrecision, and not 2 decimal places (although I've used N[#,2]). I've tried another code: MatrixForm[Round[Correlation[Transpose[exchange][[1 ;; 500, All]]], 0.01]]  In this case I get the correlation matrix with 2 decimal places, but it's still missing the "+" signal before the number. So I've tried this last code: NumberForm[Round[Correlation[Transpose[exchange][[1 ;; 500, All]]], 0.01],NumberSigns -> {"-", "+"}]//MatrixForm  In this last case I get the numbers correctly formatted, but I'm not able to put them in MatrixForm. Can someone give me a hint to solve this? • Why not NumberForm[MatrixForm[Round[...,NumbersSigns->{"-","+"}]? Apr 9 '13 at 16:00 • @ andre Thanks A LOT !!! It works perfectly now! – Rod Apr 9 '13 at 16:02 ## 2 Answers It is perfectly possible to wrap NumberForm around MatrixForm : NumberForm[MatrixForm[Round[...,NumbersSigns->{"-","+"}] does the job Of course there are many ways of doing this. Here's one: mat = RandomReal[{-1, 1}, {4, 4}] Table[NumberForm[mat[[i, j]], {3, 2}, NumberSigns -> {"-", "+"}], {i, 1, 4}, {j, 1, 4}] // MatrixForm  You get: • Imagine you have mat defined... try to use: NumberForm[mat, NumberSigns -> {"-", "+"}] // TableForm. It doesn't work... – Rod Apr 9 '13 at 14:47 • what's wrong with the output of the above, either Table or Map? Apr 9 '13 at 14:50 • You've generated numbers with 2 decimal places... try something with $MachinePrecision...
– Rod
Apr 9 '13 at 14:52
• mat has full real precision, it's the argument {3,2} that does the truncating to 3 significant digits plus two to the right of the decimal place. Apr 9 '13 at 14:54
• @RodLm What would be the problem there? You understand bill's matrix is just an example, right? We wouldn't like to display a 500x5 matrix here just to be closer to your matrix Apr 9 '13 at 15:37


text:
# Applets Directory

Henri Picciotto

Applets are small, single-purpose computer programs, usually available on the Web through a browser. For this site, I built the applets in GeoGebra, with the goal of making various mathematical ideas more visual and more interactive. However you cannot expect them to teach the concepts they illustrate! They are most effective if they are part of a multidimensional pedagogy that gets students to engage intellectually rather than watch passively.

You can use the applets online, and/or download the corresponding GeoGebra files and modify them to better fit your needs.

Some links below are to a single applet, others are to a launch page with links to several applets.

### Algebra

Changing a, b, c in y=ax^2+bx+c

Completing the Square

Complex Number Arithmetic Games

Constant Sums, Constant Products

Difference of Squares

Multiplying binomials

Off and On, understanding linear inequalities

Squaring a Binomial

Stairs, slope puzzles

Trinomial: Multiple Representations

Virtual Geoboard

Slices

Rectangles

### Functions

Doctor Dimension, exploring rate of change

Function Diagrams, a parallel axes representation

Sine Curve

Iterating Linear Functions:
with a table
with a 'cobweb' graph

Iterating f(x)=rx(1-x) (in two representations)

Bifurcation graph for f(x)=rx(1-x)

### Geometry

Area of a Circle

Dilation

Geometry of the Parabola

Isometries (Rigid Motions)

Only four kinds

Making Triangles

SSA

Pythagorean Theorem

Sum of the Angles in a Triangle

Symmetry with Pattern Blocks

Tri, Tri Again (inscribed circle iteration)

Virtual Circle Geoboard

Virtual Geoboard

Virtual Pentominoes

Virtual Tangrams

### Other

Parallelogram area

Pent, a two-person strategy game

Penumbra, some of the geometry of eclipses

Spirograph, exploring symmetric figures created by a circle in a circle

Virtual Grid Paper

As I create more applets, I'll try to remember to add links in this directory.


text:
# Calculating coefficient of restitution for a three-body system

I have been asked the following question:

Two billiard balls of the same size and mass are in contact on a billiard table. A third ball of the same mass and size strikes them symmetrically and remains at rest after the impact. Find the coefficient of restitution between the balls?

Given that I figure out the velocity of separation for the two billiard balls that were in touch, should I be considering the $$v_\textrm{sep}$$ for only one of the two balls (while it is clear that $$v_\textrm{app}$$ will be the same for both balls) when calculating the coefficient of restitution ($$e$$)?

If so, why?


text:
Listen to Your Favourite Programming Podcasts and Books on Audible - Here is a Free Trial - Click Here - Offer Limited Time Only
An Introduction to Sorting Algorithms

# An Introduction to Sorting Algorithms

## Marty Jacobs · Feb 24, 2017 00:00 · 1290 words · 7 minute read

Starting with some motivation on why we need to understand the best ways of sorting data. Consider the below quote, which was said 2 years ago… >World population projected to reach 9.7 billion by 2050. The current world population of 7.3 billion is expected to reach 8.5 billion by 2030, 9.7 billion in 2050 and 11.2 billion in 2100, according to a new UN DESA report, “World Population Prospects: The 2015 Revision”, launched today.Jul 29, 2015

hmm strange, this was also said 2 years ago.. >In 2015, the International Telecommunication Union estimated about 3.2 billion people, or almost half of the world’s population, would be online by the end of the year.

Yep, that’s an estimated 4.1 billion people in 2015 without internet. How could they live right? Well.. it’s going to be an interesting future when that gap is filled and those people are internet enabled. And it is very likely that it will become a reality, companies are working on this right now. Facebook’s Internet.org is just one of these companies aiming to connect the whole world with internet. So what happens when everyone is on the interwebs?

But what does this mean from a developer’s perspective? ..More data to sort through Zzzzz..

Luckily, sorting data is a very interesting problem to solve, and there isn’t always a one-size fits all solution. Sorting is really important as it makes data more readable and convenient for both users, and systems. When handling large datasets (millions and billions of files), a strategy needs to be formulated in order to solve sorting problems, while providing efficient results. Even for millions of items in a dataset, you would want to optimise your sort method for a more functional application. And you certainly don’t want to add extra time to your sorting method that’s for sure. There are algorithmic strategies that do pretty well against sorting large data sets, such as the divide and conquer strategy. Which is essentially, dividing the dataset into subsets, and conquering by sorting each subset and returning it to make a sorted dataset.

### There is no best algorithm

There are many go-to sorting algorithms that help you sort data in different ways, each with individual qualities and trade-offs. In order to choose the right algorithm for the job, we need to know more about the sort problem itself. Also the dataset properties need to be analysed, and understood in terms of how it needs to be sorted (lexicographical, numerical, alphabetical, customised ordering etc.).

Here are some questions to ask yourself when selecting the right one for you.

• How big is the data to be sorted? Can you fit it in main memory? Can you fit 12 in memory?
• Will the data input grow over time?
• What system resources are available?
• What is the data type of the data? (primitives, objects, both?)
• Can we fit the data in a database?
• Are we sorting just once?
• Is there a time/space constraints?

## Sorting algorithms

There is a large amount of implementations online, and some might not actually be optimal, so make sure you know exactly what the algorithm is doing. I won’t be using code in this post, but rather highlighting some pros and cons to help you narrow down some algorithm choices, by addressing the above questions. Also you might want to check out the difference between best case, worst case and average case performance to help make your decision. It is important if you need an algorithm to be consistent with your performance requirements. For example, if your algorithm has a best-case of O(n lg n) but a worst-case of (n^2) performance, and you might expect the amount of data input to increase ten-fold over the next year, then favouring worst-case running time will provide you with assurance that your algorithm will not lose its quality later.

### Stable sort vs. Unstable sort

A stable sort is a sort which guarantees that when two items are compared and determined to be the same, their positions in the set will not be compromised. Where as for unstable sort the positions of the items may change given the nature of the sort.

### Go big or go home

For large-sized datasets, we can use external sorting which is a class of algorithms that load chunks of data from secondary storage into main memory, then sorts upon it. The external merge sort is probably one of the most commonly used, as it loads the chunks of data into main memory (RAM) and sorts the chunks, merges them back together to then be written back to disk.

Another option that involves little coding is Google’s BigQuery, where you all you need to do is upload your dataset, and it utilises serverless technology saving you time setting up the infrastructure needed. You can just run an SQL (or SQL-like) query to sort your dataset. Easy as that.

### The mid line

If the dataset is just big enough to fit in RAM, then we can use an internal sort that might offer greater performance benefits. Another possibility we have to take into account is to identify if the data is mostly sorted already, and if that’s the case then you might want consider using a bubble sort or an insertion sort. I would suggest also to consider using a timsort, as it is a hybrid sort (combination of merge sort, and insertion sort) that runs at worst case O(n lg n). Python’s inbuilt sort() method utilises timsort.

### Rock bottom

The general rule is to use simple sorting algorithms on small datasets. Insertion sort is a good candidate for this as it promises good performance benefits, and is extremely simple to implement. If your dataset is really small then you should weigh up just using a standard library sort() method. For example, if your application is written in Java, then Collections.sort() is probably fast enough. Which leads to the next point..

### Benchmark testing

Time to dust off that profiler and put it to good use! Your algorithm is blazingly fast? but, how much space are you using? And vice versa. Benchmark testing helps you to understand and compare the performance of your algorithm.

### Parallel sorting

The best worst-case time complexity for a sorting algorithm is O(n lg n). But what if we could do better? What if we could split our sort algorithm into sub-problems, and solve those sub-problems simultaneously. It’s possible, and it can be done by solving each sub-problem on a seperate CPU core. Among other sort methods, the merge sort can be modified to utilise parallelism to give a speed increase, it is known as a Parallel merge sort. It is a parallel algorithm that provides sorting in O(lg n) worst-case running time. O(n lg n) / n = O(lg n). That is an incredible performance enhancement.

### As a final note

There are many sorting strategies out there, and it is about finding the best one for you. Another entirely likely possibility is the time required to sort may not be important at all. For instance, if you run into a problem where you only need to ever sort data once, then the choice of algorithm might not actually matter. Go grab a coffee and let that sort itself out. But generally speaking most of the time, sorting is performed routinely and it requires some analysis to help save time and space. On the bleeding edge, parallel sorting is being optimised in the field of high-performance computing (HPC) by using GPUs, as they have been proven worthy for general computing too!


text:
4.7k views

Consider the system, each consisting of $m$ linear equations in $n$ variables.

1. If $m < n$, then all such systems have a solution.
2. If $m > n$, then none of these systems has a solution.
3. If $m = n$, then there exists a system which has a solution.

Which one of the following is CORRECT?

1. $I, II$ and $III$ are true.
2. Only $II$ and $III$ are true.
3. Only $III$ is true.
4. None of them is true.

edited | 4.7k views
+3
$I.$ An underdetermined linear system has either no solution or infinitely many solutions.

The homogeneous (with all constant terms equal to zero) underdetermined linear system always has non-trivial solutions (in addition to the trivial solution where all the unknowns are zero).

$II.$ An overdetermined system is almost always inconsistent (it has no solution) when constructed with random coefficients. However, an overdetermined system will have solutions in some cases, for example if some equation occurs several times in the system, or if some equations are linear combination of the others.

The homogeneous case is always consistent (because there is a trivial, all-zero solution). There are two cases, depending on the number of linearly dependent equations: either there is just the trivial solution, or there is the trivial solution plus an infinite set of other solutions.

$III.$ If m=n, then there exists a system which has a solution.

$\therefore$ Ans: C

Correct answer => $C)$

why ?

$I)$ This is false. Consider a system with m < n, which are incosistent like

$a+b+c = 2$

$a+b+c = 3$

Here $m < n$ but no solution because of inconsistency !

$II)$ $m > n$ but no solution for none of system $=$$> What if this system of equations have 2 equations which are dependent ? ex =$$>$ $a+b = 2$

$2a + 2b = 4$

$a-b = 0$

Then $a = 1$, $b= 1$ is solutions . II) Is false.

$III)$ this is true, $M = 2$, $N = 2$

$a+b = 2$

$a-b = 0$

Then $m= 1$, $n= 1$ Now there exists system which has solution . III) is correct. Answer is $C$ !
by Boss (41.9k points)
edited
+2
but for the case of two parallel lines example y=x+5 and y=x+6 for these equations no solution so c should also be false.
+42

You need to read the statement III) Clearly. What you are deducing is incorrect ! If m = n , the there exists a system which has a solution

* there exists *

Counter example is used for disproving, for all, not * there exists *

+2
will the answer change for homogeneous equations?
+2

thank you .... it helped .. ur 'there exists'

+4
Statement 3) is " there exists a system which has a solution "  not  " there exists a solution "
0
ans should be option d ,    there may be cases where m=n  but those m equations be linearly dependent and results in no solutions so case iii will also be wrong
+1
@sushmita in case of homogeneous equeations when m>n the max rank will be n and this boild down to r<=n hence solution exists in case of m<n max rank is m and r<n infinite solutions exist and when m=n r<=n hence solution exists in all the three cases
0
for linear equations

a+b+c=2

a+b+c=3

a+b+c=5

system is inconsistent

so how option C is correct
0

If m>n, will ALL such systems always have a solution?

If m<n will NO such systems have a solution?

0

@Shubham101    the reason is again - there exists a system.  ;)

both are false.

0

Correct me if I'm wrong.

1. not all such but few systems have a solution because suppose in a 3d space(3 unknown) we're given 2 line eqn.It's underdetremined so we can't solve this type of equation.

2.It's overdetermined sysytem.suppose 3 line eqn are given in a 2d space.Now this 3 line may or may not be intersect at a same point.so none of this system has a solution is false.but few system has a solution is true.

3.It's perfectly determined.2 eqn. in 2d space but same concept as point 3, can be parallel or intersect at a point.So there exists a system which has a solution is true but for all it will be false statement.

0
can anyone explain with respect to the ranks why first two options are false for homogeneous system of equation they are right
0

@Kaluti I think for a system of linear homogenous equation all the 3 options are correct.

I think here we required rank(A | B) and rank(A) to conclude clearly . Becoz;

FOR HOMOGENEOUS SYSTEM :-

1) INCONSISTENT ;- Not possible

2) Unique solution :- Rank == No of variables

3) Many Solution :- Rank < No of variables.

FOR NON - HOMOGENEOUS SYSTEM :-

1) INCONSISTENT ;- when Rank(A|B) !=Rank (A)

2) Unique solution :- when Rank(A|B) =Rank (A) and Rank == No of variables

3) Many Solution :- when Rank(A|B) =Rank (A) and Rank < No of variables.

So It has m linear eqn so maxm rank can be = "m"

and no of variables are = "n"

and to conclude anything Further we need to assume Rank (A|B)=rank(A).

After assuming this only Case (III) can be true.

Becoz m can not be greater than n (Becoz if so , then we can not conclude anything) , hence (II) is false.

When m<n , then all such system has Many solution, Hence (I) is false..

by Active (1.5k points)
+2

i  think homogeneous systems always have at least one solution, namely, the case where all unknowns are equal to zero.  i.e.trivial solution to the homogeneous system, so whether m> n or m<n or m=n homogenous will have always a solution and can never be inconsistent.

Note: This isn't a direct answer to the question. I've covered all the possible cases I could. Don't read this if you want a straight answer to the given question.

The number of equations is $m$.

Hence, there can be a $m*m$ matrix.

Hence the maximum rank can be $m$.

Hence, $m$ is the upper bound of $r$.
We conclude that $r \leq m$ ALWAYS.

### Statement I

For homogeneous system of equations; if $m < n$, then $r < n$. Because $m$ is the upper bound of $r$. We'll have infinite solutions.

For non-homogeneous system of equations; maybe $r(A) \neq r(A|B)$. We'll have no solutions in this case.

So, all such systems have a solution? No. This statement is false.

### Statement II

$m > n$. The only thing we know is $r$ lies below $m$. It can be between $m$ and $n$. Or it can be lower than $n$.

ie

1. $m > r > n$ is possible.
2. $m > n > r$ is possible.

For case 2, $r<n$. It's good.

For case 1, $r > n$. Which is inconsistent and we'll have no solutions in case of non-homogeneous equations.

We always want $r \leq n$ be it any system of equations.

So, this statement is false, because we have a favourable case.

### Statement III

When $m = n$

$r$ could be $= m$, and hence, $r = n$.

$r$ could be $< m$, and hence $r < n$.

So, yeah, there exists favourable cases. This statement is correct.

Suppose $r$ is such that $r(A) \neq r(A|B)$. Then, we'll have no solution. But we need just one favourable case to make this statement True, and we did it.

Option C
ago by Loyal (6.9k points)

I & II are false already but similarly for III also

for the case of two parallel lines example y=x+5 and y=x+6 for these equations no solution so c should also be false.

Hence D is correct ans.

by (165 points)
+5

You need to read the statement III) Clearly. What you are deducing is incorrect ! If m = n , the there exists a system which has a solution

* there exists *

Counter example is used for disproving, for all, not * there exists *


text:
125 Answered Questions for the topic Scientific Notation

Scientific Notation Chemistry Mass Atoms

02/01/21

#### Allyl sulfide gives garlic its characteristic odor. Give the mass and number of carbon atoms.

a) The mass in grams of 2.62 mol of allyl sulfide. Answer in scientific notation.b) The number of carbon atoms in 72.8 g of allyl sulfide. Answer in scientific notation.

01/29/21

#### Convert the following into correct scientific notation:

I got the numbers right, but not the correct powers.a) 445.5 X 104 I got 4.455 X 10 to the? It is not power 4 and it is not power 2b) 7804.7 X 10-6 I got 7.8047 X 10 to the? It is not power -6 and... more
Scientific Notation Math Mathematics

09/28/20

#### Write a number in scientific notation...

Write a number in scientific notation that is between 0 and 1/3 and justify your response.
Scientific Notation Prealgebra

05/22/19

#### How do I figure this out?

If a phone has 10 gigabytes of hard drive space that means it has 10x109 bytes of space. A megabyte is 1x106. If our phone with 10 gigabytes of free storage downloads a game that takes 76... more
Scientific Notation Algebra 1

04/26/19

#### I need help with this worded question

On average, you breathe 24 000 times a day. How many breaths will you take in a year, assuming that it is not a leap year? Express your answer in scientific notation.
Scientific Notation Prealgebra Negative Exponents

03/25/19

#### Scientific notation negative exponents

Write the number in scientific notation 0.000001 7
Scientific Notation

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#### How many stacked pieces of paper would it take to reach the moon

I know that a sheet of paper is .1 milometers thick. I need to know how many sheets of paper stacked on top of each other would it take to be 1 kilometer thick. After that I need to know how many... more
Scientific Notation

10/22/18

#### how many sheetsof paper is needed to reach 1 kilometer thick

I know that a sheet of paper is .1 milometers thick. I need to know how many sheets of paper stacked on top of each other would it take to be 1 kilometer thick. After that I need to know how many... more
Scientific Notation

10/10/18

The maximum distance between Earth and sun is 1.5x10^8 km, and the minimum distance is 1.5x10^8 km. What is the average distance between Earth and the sun in scientific notation?
Scientific Notation

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#### The Earth is approximately 2 × 107 km from Venus at its closest approach. The Earth is also about 4 × 1013 km from Proxima Centauri, a star.

The Earth is approximately 2 × 107 km from Venus at its closest approach. The Earth is also about 4 × 1013 km from Proxima Centauri, a star. Which of the following is true? A. The distance from... more
Scientific Notation

09/19/18

Approximately 4x10^3 employees of a certain company average $30000 each year in salary. Write the total amount earned by all of the employees of this company in scientific notation. Scientific Notation Chemistry Algebra Significant Figures 09/04/18 #### Help with answering "Carry out each calculation. Report the answer using scientific notation and the proper number of significant figures." "Carry out each calculation. Report the answer using scientific notation and the proper number of significant figures." a. 41,390 × 0.725 b. 1,000/1.53 c. 23,000/0.5744 08/23/18 #### Carry out the following operations and express the answer with the appropriate number of significant figures: [(285.3x10^5)-(1.200x10^3)]x2.8954 How do I solve this problem with the scientific notation? And when it comes to the rules of significant figures, do I use the rule pertaining to subtraction or the one pertaining to... more Scientific Notation 08/21/18 #### explain the significance of the numbers, 2 megabytes, 32 megabytes and 1024 bytes . i need help. I am in year 9 Scientific Notation 08/12/18 #### Scientific notation A 5 pound bag of candy contains about 2500 pieces. If the company makes 4.7 million bags of candy per year, about how many pieces of candy are made per year? Scientific Notation 08/09/18 #### The density of a liquid is 8.03^4 x 10^-1 kilogram per liter. What is the mass (in kg) of liquid in a full 100,000 liter tank? Using Scientific Notation, what would be the answer to" density of a liquid is 8.03^4 x 10^-1 kilogram per liter. What is the mass (in kg) of liquid in a full 100,000 liter tank?" the ^are to... more Scientific Notation 07/19/18 #### A computer has a hard drive that stores approxamately 9*10^8 bytes. A high density 3.5 inch floppy disk holds approximately 1.4*10 bytes.. A computer has a hard drive that stores approxamately 9*10^8 bytes. A high density 3.5 inch floppy disk holds approximately 1.4*10^6 bytes. How many floppies are needed to back up the hard drive. Scientific Notation 05/11/18 #### Scientific notation A sample of microbes weighs 50 mg. There are approximately 2.5 * 10^6 microbes in a sample. What is the approximate mass of each microbe expressed using scientific Notation in grams. Show working... more Scientific Notation 04/10/18 #### Express answer in scientific notation In a state with a population of 2,000,000 the average citizen spends$6,000 on housing each year. What is the total spent on housing for the state? Express answer in scientific notation.
Scientific Notation

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#### . find the diameters of the earth in meters and write your answer in scientific notation

the circumference of a circle is 3.14 times greater than the diameters . find the diameters of the earth in meters and write your answer in scientific notation
Scientific Notation

04/03/18

#### the circumference of a circle is 3.14 times greater than the diameters .

the circumference of a circle is 3.14 times greater than the diameters . find the diameters of the earth in meters and write your answer in scientific notation?
Scientific Notation

04/02/18

#### Find the circumference of the earth

Find the circumference of the earth of the earth in meters and write your answer in scientific notation ?The circumference of the earth is approximately 40 075 km .1)Find the circumference of the... more
Scientific Notation Math Problem Math Help For College

03/08/18

#### scientific notation

Convert each number to scientific notation and multiply. Write your answer in scientific notation.(72,000,000)(400,000,000) Write 72,000.000 in scientific notation.72,000,000= Write 400,000,000... more

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text:
## Discrete Math helpppppppp!!!!!!!!!!!!11

1) Consider the infinite sequence A 1 , A 2 , A 3 , A 4 , A 5 , A 6 , A 7 , A 8 , A 9 , A 10 , â€¦ , which has A1 = 7 and satisfies Ak+1 = ( Ak ) + (14 k + 7) for all integers k ? 1.

(a) By taking k = 1, 2, 3, 4, and 5 in the recurrence relation above, compute the values of A2 , A3 , A4 , A5 , and A 6 (in that order).

(b) Prove (by induction): ? integers n ? 1, An = 7n^2

(2) Consider the infinite sequence B1 , B 2 , B 3 , B 4 , B 5 , B 6 , B 7 , B 8 , B 9 , B 10 , â€¦ , Which has B1 = 2 and satisfies Bk+1 = 3 (Bk ) + 2 for all integers k ? 1.

(a) By taking k = 1, 2, 3, 4, and 5 in the recurrence relation above, compute the values of B 2 , B 3 , B 4 , B5 , and B 6 (in that order).

(b) Prove (by induction): ? integers n ? 1, Bn = 3^n -1. (That is, prove that Bn is one less than the n-th power of 3.)


text:
Reference : UC Updatable Non-Hiding Committed Database with Efficient Zero-Knowledge Proofs
 Document type : E-prints/Working papers : First made available on ORBilu Discipline(s) : Engineering, computing & technology : Computer science Focus Areas : Computational Sciences To cite this reference: http://hdl.handle.net/10993/39421
 Title : UC Updatable Non-Hiding Committed Database with Efficient Zero-Knowledge Proofs Language : English Author, co-author : Rial Duran, Alfredo [University of Luxembourg > Interdisciplinary Centre for Security, Reliability and Trust (SNT) > >] Publication date : 2019 Number of pages : 25 Peer reviewed : No Keywords : [en] vector commitments ; universal composability ; modular design Abstract : [en] We define an ideal functionality $\Functionality_{\DB}$ and a protocol $\mathrm{\Pi_{\DB}}$ for an updatable non-hiding committed database ($\DB$). $\DB$ is described as the task of storing a database into a suitable data structure that allows you to efficiently prove in zero-knowledge (ZK) that a value is stored in the database at a certain position. The database is \emph{non-hiding} because both prover and verifier know its content. It is \emph{committed} in the sense that only ZK proofs about position-value pairs that are actually stored are possible. It is \emph{updatable} because its contents can be modified dynamically throughout the protocol execution. The $\DB$ task is used implicitly as building block of privacy-preserving protocols for e-commerce, smart billing and access control. In those protocols, this task is intertwined with others. Our functionality $\Functionality_{\DB}$ allows us to study constructions for this task in isolation. Furthermore, it allows us to improve modularity in protocol design, by using $\Functionality_{\DB}$ as building block of those protocols along with functionalities for other tasks. Our construction $\mathrm{\Pi_{\DB}}$ uses a non-hiding vector commitment (VC) scheme as building block. Thanks to the efficiency properties of non-hiding VC schemes, $\mathrm{\Pi_{\DB}}$ provides ZK proofs whose computation cost (after initialization) and whose size are both independent of the database size. Therefore, $\mathrm{\Pi_{\DB}}$ is suitable for large databases. Moreover, the database can be updated dynamically and very efficiently. Target : Researchers Permalink : http://hdl.handle.net/10993/39421 FnR project : FnR ; FNR11650748 > Alfredo Rial > SZK > Stateful Zero-Knowledge > 01/03/2018 > 28/02/2021 > 2017

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text:
# 40 CFR § 63.1412 - Continuous process vent applicability assessment procedures and methods.

(a) General. The provisions of this section provide procedures and methods for determining the applicability of the control requirements specified in § 63.1405(a) to continuous process vents.

(b) Sampling sites. Sampling sites shall be located as follows:

(1) Sampling site location. The sampling site for determining volumetric flow rate, regulated organic HAP concentration, total organic HAP, net heating value, and TRE index value, shall be after the final recovery device (if any recovery devices are present) but prior to the inlet of any control device that is present and prior to release to the atmosphere.

(2) Sampling site selection method. Method 1 or 1A of 40 CFR part 60, appendix A, as appropriate, shall be used for selection of the sampling site. No traverse site selection method is needed for process vents smaller than 0.33 foot (0.10 meter) in nominal inside diameter.

(c) Applicability assessment requirement. The organic HAP concentrations, volumetric flow rates, heating values, organic HAP emission rates, TRE index values, and engineering assessment control applicability assessment requirements are to be determined during maximum representative operating conditions for the process, except as provided in paragraph (d) of this section, or unless the Administrator specifies or approves alternate operating conditions. Operations during periods of malfunction shall not constitute representative conditions for the purpose of an applicability test.

(d) Exceptions. The owner or operator is not required to conduct a test that will cause any of the following situations:

(1) Causing damage to equipment;

(2) Necessitating that the owner or operator make a product that does not meet an existing specification for sale to a customer; or

(3) Necessitating that the owner or operator make a product in excess of demand.

(e) Organic HAP concentration. The organic HAP concentrations, used for TRE index value calculations in paragraph (j) of this section, shall be determined using the procedures specified in either § 63.1414(a) or by using the engineering assessment procedures in paragraph (k) of this section.

(f) Volumetric flow rate. The volumetric flow rate shall be determined using the procedures specified in § 63.1414(a), or by using the engineering assessment procedures in paragraph (k) of this section.

(g) Heating value. The net heating value shall be determined as specified in paragraphs (g)(1) and (2) of this section, or by using the engineering assessment procedures in paragraph (k) of this section.

(1) The net heating value of the continuous process vent shall be calculated using Equation 1:

${H}_{T}={K}_{1}\left(\sum _{j=1}^{n}{D}_{j}{H}_{j}\right)\phantom{\rule{0ex}{0ex}}\text{[Eq. 1]}$

Where:
HT = Net heating value of the sample, megaJoules per standard cubic meter, where the net enthalpy per mole of process vent is based on combustion at 25 °C and 760 millimeters of mercury, but the standard temperature for determining the volume corresponding to 1 mole is 20 °C, as in the definition of QS (process vent volumetric flow rate).
K1 = Constant, 1.740 × 10−7 (parts per million)−1 (gram-mole per standard cubic meter) (megaJoules per kilocalorie), where standard temperature for (gram-mole per standard cubic meter) is 20 °C.
Dj = Organic HAP concentration on a wet basis of compound j in parts per million, as measured by procedures indicated in paragraph (e) of this section. For process vents that pass through a final stream jet and are not condensed, the moisture is assumed to be 2.3 percent by volume.
Hj = Net heat of combustion of compound j, kilocalorie per gram-mole, based on combustion at 25 °C and 760 millimeters of mercury.

(2) The molar composition of the process vent (Dj) shall be determined using the methods specified in paragraphs (g)(2)(i) through (iii) of this section:

(i) The methods specified in § 63.1414(a) to measure the concentration of each organic compound.

(ii) American Society for Testing and Materials D1946-90 (Reapproved 1994) (incorporated by reference, see § 63.14) to measure the concentration of carbon monoxide and hydrogen.

(iii) Method 4 of 40 CFR part 60, appendix A to measure the moisture content of the stack gas.

(h) Organic HAP emission rate. The emission rate of organic HAP in the continuous process vent, as required by the TRE index value equation specified in paragraph (j) of this section, shall be calculated using Equation 2:

$E={K}_{2}\left(\sum _{j=1}^{n}{C}_{j}{M}_{j}\right)\phantom{\rule{0ex}{0ex}}\text{[Eq. 2]}$

Where:
E = Emission rate of organic HAP in the sample, kilograms per hour.
K2 = Constant, 2.494 × 10−6 (parts per million)−1 (gram-mole per standard cubic meter) (kilogram/gram) (minutes/hour), where standard temperature for (gram-mole per standard cubic meter) is 20 °C.
n = Number of components in the sample.
CJ = Organic HAP concentration on a dry basis of organic compound j in parts per million as determined by the methods specified in paragraph (e) of this section.
Mj = Molecular weight of organic compound j, gram/gram-mole.
QS = Continuous process vent flow rate, dry standard cubic meter per minute, at a temperature of 20 °C.

(i) [Reserved]

(j) TRE index value. The owner or operator shall calculate the TRE index value of the continuous process vent using the equations and procedures in this paragraph, as applicable, and shall maintain records specified in § 63.1416(f).

(1) TRE index value equation. The equation for calculating the TRE index value is Equation 3:

$\mathrm{TRE}=1/{E}_{\mathrm{HAP}}*\left[A+B\left({Q}_{S}\right)+C\left({H}_{T}\right)\right]\phantom{\rule{0ex}{0ex}}\text{[Eq. 3]}$

Where:
TRE = TRE index value.
A, B, C = Coefficients presented in table 7 of this subpart.
EHAP = Emission rate of total organic HAP, kilograms per hour, as calculated according to paragraph (h) or (k) of this section.
QS = Continuous process vent volumetric flow rate, standard cubic meters per minute, at a standard temperature of 20 °C, as calculated according to paragraph (f) or (k) of this section.
HT = Continuous process vent net heating value, megaJoules per standard cubic meter, as calculated according to paragraph (g) or (k) of this section.

(2) TRE index calculation. The owner or operator of a continuous process vent shall calculate the TRE index value by using the equation and appropriate coefficients in Table 6 of this subpart. The owner or operator shall calculate the TRE index value for each control device scenario (i.e., flare, thermal incinerator with 0 percent recovery, thermal incinerator with 70 percent recovery). The lowest TRE index value is to be compared to the applicability criteria specified in § 63.1405(a).

(k) Engineering assessment. For purposes of TRE index value determinations, engineering assessments may be used to determine continuous process vent flow rate, net heating value, and total organic HAP emission rate for the representative operating condition expected to yield the lowest TRE index value. Engineering assessments shall meet the requirements of paragraphs (k)(1) through (4) of this section.

(1) If the TRE index value calculated using engineering assessment is greater than 4.0, the owner or operator is not required to perform the measurements specified in paragraphs (e) through (h) of this section.

(2) If the TRE index value calculated using engineering assessment is less than or equal to 4.0, the owner or operator is required either to perform the measurements specified in paragraphs (e) through (h) of this section for control applicability assessment or comply with the control requirements specified in § 63.1405(a).

(3) Engineering assessment includes, but is not limited to, the following examples:

(i) Previous test results, provided the tests are representative of current operating practices.

(ii) Bench-scale or pilot-scale test data representative of the process under representative operating conditions.

(iii) Maximum volumetric flow rate, organic HAP emission rate, organic HAP concentration, or net heating value limit specified or implied within a permit limit applicable to the continuous process vent.

(iv) Design analysis based on accepted chemical engineering principles, measurable process parameters, or physical or chemical laws or properties. Examples of analytical methods include, but are not limited to, the following:

(A) Use of material balances based on process stoichiometry to estimate maximum organic HAP concentrations;

(B) Estimation of maximum volumetric flow rate based on physical equipment design such as pump or blower capacities;

(C) Estimation of organic HAP concentrations based on saturation conditions; and

(D) Estimation of maximum expected net heating value based on the stream concentration of each organic compound.

[65 FR 3290, Jan. 20, 2000, as amended at 79 FR 60931, Oct. 8, 2014; 83 FR 51853, Oct. 15, 2018]


text:
When is it possible to arrive from $a$ to $b$ by this procedure?

Suppose that we take two natural numbers $a$ and $b$ with $a<b$.

Then, we may try to arrive at $b$ by starting at $a$ by this set of rules:

1) To arrive at $a_1$ which is such that we have $a< a_1 \leq b$ it is only allowed that we sum $a$ with one of its divisors different from $1$. If $a_1=b$ we are done.

2) If $a_1<b$ then to arrive at $a_2$ which is such that we have $a<a_1<a_2\leq b$ we perform step 1) with $a_1$.

3) Repeat 2) until we arrive at $a_{k(a,b)}=b$

As an example, take $a=12$ and $b=100$, we have $12 \to 16 \to 20 \to 40 \to 80 \to 100$.

As an example where $\gcd (a,b)=1$ take $a=15$ and $b=22$. Then we have $15 \to 18 \to 20 \to 22$.

As an example where $\gcd (a,b)=1$ and where we cannot arrive at $b$, take $a=18$ and $b=23$.

Of course, $k(a,b)$ can be multi-valued since for the example $a=12$ and $b=100$ we also have $12\to 24\to 48\to 96\to 100$. But this question is not about behaviour of $k(a,b)$ (although it could be that its behaviour has some awesome features).

If this procedure exists for some $a$ and $b$ and $a<b$ we may say that $b$ is reachable by $a$.

Can we find some condition(s) on $a$ and $b$ that is(are) both necessary and sufficient to guarantee that $b$ is reachable by $a$?

Since I am not sure did I explain this procedurally in a good way by 1), 2) and 3), I will describe here again the procedure in other words. It is very simple: choose natural numbers $a$ and $b$ with $a<b$. We are trying to arrive at $b$ by adding to $a$ some divisor of $a$ different than $1$. Then we arrive at $a_1=a+\alpha_1$ , where $\alpha_1 | a$. Then if we did not arrive at $b$ with that single step, we add to $a_1$ some divisor of $a_1$ different than $1$ to arrive at $a_2=a_1 + \alpha_2$, where $\alpha_2 | a_1$. We repeat this until we arrive at $b$, if possible. The question is: when is this possible?

• Do you have an example of a prime $b$ reachable by some $a<b$? Sep 26, 2017 at 18:10
• @SylvainJULIEN No, since such an example does not exist.
– user114642
Sep 26, 2017 at 18:27
• Does the following shorter description capture your procedure correctly? "For natural numbers $a$ and $b$ with $a < b$, we say that $b$ is one-step reachable from $a$ if $b - a$ is a divisor of $a$ other than $1$. We say that $b$ is $n$-step reachable from $a$ if there is a sequence $a = a_0, a_1, \dotsc, a_n = b$ in which $a_{i + 1}$ is one-step reachable from $a_i$ for all $0 \le i < n$." Then maybe you want to define $k(a, b)$ to be the set of all $n$ for which $b$ is $n$-step reachable from $a$, or to be the minimum of that set? Sep 26, 2017 at 18:55
• @LSpice I added description of the procedure in other words under the question raised in the text, you will not face any difficulties or ambiguities when reading it.
– user114642
Sep 26, 2017 at 18:57
• $$12 \to 15 \to 18 \to 20$$
– user114642
Sep 26, 2017 at 19:43

When $\gcd(a,b)>1$, then we can reach $b$ from $a$ by simply repeatedly adding $\gcd(a,b)$.

The least reachable number (LRN) from $a$ is adding its smallest prime divisor, and the greatest "arrivable" number (GAN) to $b$ is subtracting its smallest prime divisor. LRNs and GANs are always even (when defined). Since even numbers have $\gcd$ at least $2$, they are reachable when the difference between $a$ and $b$ is "large".

So the full statement is:

for $a<b$, $a$ can arrive at $b$ iff exactly one of the following holds:

1. $a$ can arrive at $b$ in $1$ step $\iff 1<b-a=\gcd(a,b)$
2. $a$ can arrive at $b$ in $2$ or more steps $\iff$ GAN of $b$ exists and is at least LRN of $a$

Sufficiency is constructive and necessity is obvious by definition of LRN and GAN.

• You can reword it to two exclusive conditions, amounting to your 1. and the last half of your 4. Gerhard "Brevity: The Soul Of Math?" Paseman, 2017.09.26. Sep 26, 2017 at 21:06
• 1 and last half of 4 weren't quite exclusive, but now they are. Sep 26, 2017 at 22:04

Let lpf stand for least prime factor. (Assuming gcd(a,b) is 1 and also both are larger than 1,) If b is at least as big as a plus lpf(a) plus lpf(b), then you can reach b from a. This is because you can almost always take steps of size 2 to reach the goal. When you can't take a step of size 2, then b must be at most lpf(a) plus lpf(b) plus a, and if b is reachable at all it must be at least that big.

Gerhard "Taking Small Steps Is Easier" Paseman, 2017.09.26.

• Of course, if gcd(a,b) is bigger than 1, then it is possible, especially if b-a is the least prime factor of both b and a. Gerhard "Also Leaves That To Reader" Paseman, 2017.09.26. Sep 26, 2017 at 19:58
• Further, one can always take a step to reach an even number, so an upper bound for minimal k(a,b) is 2 log_2 b . Gerhard "And Lower Bound Of Zero" Paseman, 2017.09.26. Sep 26, 2017 at 21:13
• Why $k(a,b)$ seems to be always divisor either of $a$ or $b$ or both?
– user114642
Sep 27, 2017 at 7:26
• Sorry, it is not always the case that it is so.
– user114642
Sep 27, 2017 at 7:31
• It does not need to be. Consider a and b squares of primes which differ by two. K can be 4 or 5, depending on how you count it. Gerhard "There Are Many Other Examples" Paseman, 2017.09.27. Sep 27, 2017 at 7:33


text:
My Math Forum Can you find side length of a triangle given three angles?

 Trigonometry Trigonometry Math Forum

 March 19th, 2018, 09:58 AM #1 Newbie   Joined: Jan 2018 From: Ontario Posts: 2 Thanks: 0 Can you find side length of a triangle given three angles? Can you mix dimensionless function or angles to find length in triangles? For example, two sides are composed of a distance of $0.85+0.4=1.25$ and at the same time $0.4=\cos\theta$and the base is $1$? For consecutive numbers or non-consecutive numbers $x March 19th, 2018, 02:22 PM #2 Global Moderator Joined: Dec 2006 Posts: 19,530 Thanks: 1750 Quote:  Originally Posted by Larrousse$\frac{\sin A}{\sin C}=a\frac{\sin B}{\sin C}=b\frac{\sin C}{\sin C}=c$If you are given the values of the angles A, B and C, the above equations give you the values of a, b and c in terms of the angles. If you aren't given the values of the angles A, B and C, the three equations just tell you that c = 1. Quote:  Originally Posted by Larrousse$\frac{h_c}{h_a}=a\frac{h_c}{h_b}=b\frac{h_c}{h_c}=c$The above three altitude quotient equations just tell you that c = 1.  March 19th, 2018, 03:00 PM #3 Newbie Joined: Jan 2018 From: Seattle, WA Posts: 20 Thanks: 6 Angles are insufficient to determine the size of a triangle. Thanks from v8archie March 19th, 2018, 03:36 PM #4 Newbie Joined: Jan 2018 From: Ontario Posts: 2 Thanks: 0 An angle has no measurement of units, and for that reason, it is difficult to describe the length of the side of the triangle or the unit doesn't matter. Attached Images  20180204_132123.jpg (80.4 KB, 5 views) Last edited by skipjack; March 19th, 2018 at 04:07 PM.  April 22nd, 2018, 05:11 PM #5 Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 For example, every equilateral triangle, whether its sides have length 1 cm or 1000 km, has its three angles the same with measure$\displaystyle \frac{\pi}{3}\$.

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text:
We're wrapping up the release of CommandBox 3.9.0 which is available now as a release candidate.  I'd like to get as many eyes as possible to ensure everything is working good before we finalize the release.

# Installation

The easiest way to get ahold of the release candidate is to run the following command.  It will provide you with the URL to download:

CommandBox> upgrade --latest

The pre-release binaries are located on our artifacts server here:

http://integration.stg.ortussolutions.com/artifacts/ortussolutions/commandbox/3.9.0-rc/

# What's New

We've closed 36 tickets in this release and here's a rundown of the most important stuff.

## UNC Network path support

This has been pretty big for Windows users who access files on their servers over a UNC network path like \\server-name\foo\bar.  You can now cd into paths starting with \\, perform file operations like cat against those paths, etc.  Note backslashes need escaped in the CommandBox shell.

CommandBox> cd \\\\server-name/share
CommandBox> cd \\\\192.168.123.105/share

This was pretty straightforward since Java already supports this, but I had to change a lot of core path handling to make sure the correct slashes were preserved.  This needs a fair amount of testing to make sure we nailed it down for good.  If your network share requires permissions, you'll need to have saved those in Windows Explorer already  or execute a "net use" OS command from the shell.

When running a task from the CLI, the user will be automatically prompted if they don't supply all the required args.  This is just like commands work now.

CommandBox> task run myTask
Please enter required field "Foo": _

We also fixed several bugs with passing positional parameters and flag to task runners.

# positional
# named
# Use a Flag
CommandBox> task run taskFile=taskFileName target=targetName :param1=value1 :param2=value2 --:param3

This has been a long time coming, but if you want to work on a CommandBox module now, you don't have to keep copying files over to your CommandBox installation just to test. Instead, just run this from the root of your module's repo:

CommandBox> package link

This will symlink (works on Windows and *nix) your module into the core CLI's modules folder and reload the shell so you can immediately start testing.  When you're done, just run package unlink.  If you'd like to use this same feature, but to link a ColdBox module's repo over to a test application so you can test it without making a copy, you can pass in the path to the remote modules folder you'd like to link to.

CommandBox> package link /path/to/test/app/modules

This is a little easier than using your OS's native symlink commands and can be used in recipes that will work across operating systems!

## Everything Else (mostly)

The rest of the changes don't really need a dedicated section but they're worth mentioning, so I've put them in this tidy list :)  If you'd like the ticket numbers, you can get them out of this full list of tickets in JIRA: https://ortussolutions.atlassian.net/secure/ReleaseNote.jspa?projectId=11000&version=20400

• box install returns failing exit code if a package install fails. This helps builds fail correctly if things go wrong
• When prompted to type in something like a missing parameter, your answer is no longer added to the command history
• The --debug flag works correctly when starting a server from your OS shell like \$> box server start --debug
• box.json dependencies are stored with forward slashes so Mac and Windows devs stop fighting over which file to commit
• config set no longer prints out the value to avoid leaking secrets in build script output.
• Updating a package now uninstalls the previous version first to ensure a fresh start since the new version may have removed files.
• If you're setting CommandBox behind an AJP proxy, we exposed the flag and ports for that as first class citizens of server.json.
• Visual markers for private packages in the forgebox search command.
• The package init command creates a valid slug for private packages in the @user/slug format.
• Command parameters defaults work on all the aliases for a command now.
• New --local flag to server list to show all servers that have been started in the current working directory
• If for some reason you want to supply some ad-hoc JVM args to the actual CLI process, you can create a new environment var called BOX_JAVA_PROPS="foo=bar;brad=wood"
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# Even plus odd functions

1. Jan 9, 2015

### {~}

I have been looking at my old calculus textbook because to my dismay I seem to have forgotten most of the calculus I learned. I am given 3 cases of $(f+g)(x)$.

Case 1 both f and g are even:
I know $f(x) = f(-x)$ and $g(x)=g(-x)$ for the domain of the function. I can reason by substitution that
$f(x)+g(x)=f(-x)+g(-x)$
$(f+g)(x)=(f+g)(-x)$
$(f+g)(x)$ is even. So far so good.

Case 2 both f and g are odd:
I found that if $f(x)=-g(x)+c$ then
$(f+g)(x)=c$ which is even.
Otherwise I think that $(f+g)(x)$ would be odd though I don't know how to assert that.

Case 3 f is even and g is odd:
I think that other than is special case where one or both of our functions are zero for all x in the domain $(f+g)(x)$ would neither be even or odd. I don't know how to prove this.

I know that the notions of even an odd is defined in terms of sets rather than algebraically like I did here. I think if I understood sets better I might have more of a handle on this, I don't know. Hints?

2. Jan 9, 2015

### DarthMatter

If $f$ and $g$ are odd:

$(f+g)(-x)=f(-x)+g(-x)=-f(x)-g(x)=-(f+g)(x)$.

3. Jan 9, 2015

### {~}

I had something similar, but why does this look like it's always true when it isn't? Or is my special case wrong?

4. Jan 9, 2015

### DarthMatter

What special case? $-g(x)+c$ is not odd in general.

5. Jan 9, 2015

### {~}

Thanks for pointing out my error. That made things less confusing.

6. Jan 9, 2015

### jbunniii

Suppose that $f$ is even and $g$ is odd.

If $f+g$ is even then $f(x) + g(x) = f(-x) + g(-x)$ for all $x$. Since $f(x) = f(-x)$, it follows that $g(x) = g(-x)$. But $g$ is odd, so also $g(x) = -g(-x)$. Therefore, $g(-x) = -g(-x)$, or equivalently $2g(-x) = 0$. This is true for all $x$, so $g$ is the zero function.

If $f+g$ is odd then $f(x) + g(x) = -f(-x) - g(-x)$ for all $x$. Since $g(x) = -g(-x)$, this is equivalent to $f(x) = -f(-x)$. But $f$ is even, so also $f(x) = f(-x)$. Therefore, $f(-x) = -f(-x)$, or $2f(-x) = 0$. This is true for all $x$, so $f$ is the zero function.

Last edited: Jan 9, 2015
7. Jan 11, 2015

### FactChecker

In Case 3 you can not say anything about the sum of an even + odd function. Any function, h(x), can be represented as the sum of an even function and an odd function. So you can not state any conclusions about h.
( Define even(x) = (h(x) + h(-x))/2 and odd(x) = (h(x) - h(-x))/2. Prove that h(x) = even(x) + odd(x), that even(x) is even, and that odd(x) is odd.)

Last edited: Jan 11, 2015


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# Applications Using the Pythagorean Theorem

## Word problems using the Pythagorean Theorem

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Practice Applications Using the Pythagorean Theorem
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Applications Using the Pythagorean Theorem

What if a fireman needed to rescue a cat from a tree? The cat is 40 feet up from the ground. The fireman places a ladder 30 feet away from the base of the tree? How tall does the ladder need to be for him to reach the cat? After completing this Concept, you'll be able to solve real-world applications like this one using the Pythagorean Theorem and its converse.

### Guidance

The Pythagorean Theorem and its converse have many applications for finding lengths and distances.

#### Example A

Maria has a rectangular cookie sheet that measures . Find the length of the diagonal of the cookie sheet.

Solution

Draw a sketch:

Define variables: Let length of the diagonal.

Write a formula: Use the Pythagorean Theorem:

Solve the equation:

Check: and . The solution checks out.

#### Example B

Find the area of the shaded region in the following diagram:

Solution

Draw the diagonal of the square in the figure:

Notice that the diagonal of the square is also the diameter of the circle.

Define variables: Let diameter of the circle.

Write the formula: Use the Pythagorean Theorem: .

Solve the equation:

The diameter of the circle is , therefore the radius .

Area of a circle formula: .

The area of the shaded region is therefore .

#### Example C

In a right triangle, one leg is twice as long as the other and the perimeter is 28. What are the measures of the sides of the triangle?

Solution

Make a sketch and define variables:

Let: length of the short leg

length of the long leg

length of the hypotenuse

Write formulas:

The sides of the triangle are related in two different ways.

The perimeter is 28, so

The triangle is a right triangle, so the measures of the sides must satisfy the Pythagorean Theorem:

Solve the equation:

Plug the value of we just obtained into the perimeter equation:

The short leg is:

The long leg is:

The hypotenuse is:

Check: The legs of the triangle should satisfy the Pythagorean Theorem:

The results are approximately the same.

The perimeter of the triangle should be 28:

Watch this video for help with the Examples above.

### Vocabulary

• The Pythagorean Theorem is a statement of how the lengths of the sides of a right triangle are related to each other. A right triangle is one that contains a 90 degree angle. The side of the triangle opposite the 90 degree angle is called the hypotenuse and the sides of the triangle adjacent to the 90 degree angle are called the legs.

• If we let and represent the legs of the right triangle and represent the hypotenuse then the Pythagorean Theorem can be stated as:

In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. That is: .

### Guided Practice

Mike is loading a moving van by walking up a ramp. The ramp is 10 feet long and the bed of the van is 2.5 feet above the ground. How far does the ramp extend past the back of the van?

Solution

Make a sketch:

Define variables: Let how far the ramp extends past the back of the van.

Write a formula: Use the Pythagorean Theorem:

Solve the equation:

Check by plugging the result in the Pythagorean Theorem:

. So the ramp is 10 feet long. The answer checks out.

### Practice

1. In order to make a ramp that is high and covers of ground, how long must the ramp be?
2. A regulation baseball diamond is a square with 90 feet between bases. How far is second base from home plate?
3. Emanuel has a cardboard box that measures .
1. What is the length of the diagonal across the bottom of the box?
2. What is the length of the diagonal from a bottom corner to the opposite top corner?
4. Samuel places a ladder against his house. The base of the ladder is 6 feet from the house and the ladder is 10 feet long.
1. How high above the ground does the ladder touch the wall of the house?
2. If the edge of the roof is 10 feet off the ground and sticks out 1.5 feet beyond the wall, how far is it from the edge of the roof to the top of the ladder?
5. Find the area of the triangle below if the area of a triangle is defined as :
6. Instead of walking along the two sides of a rectangular field, Mario decided to cut across the diagonal. He thus saves a distance that is half of the long side of the field.
1. Find the length of the long side of the field given that the short side is 123 feet.
2. Find the length of the diagonal.
7. Marcus sails due north and Sandra sails due east from the same starting point. In two hours Marcus’ boat is 35 miles from the starting point and Sandra’s boat is 28 miles from the starting point.
1. How far are the boats from each other?
2. Sandra then sails 21 miles due north while Marcus stays put. How far is Sandra from the original starting point?
3. How far is Sandra from Marcus now?
8. Determine the area of the circle below. (Hint: the hypotenuse of the triangle is the diameter of the circle.)
9. A rectangle's length is longer than its width and if the diagonal has a length of , what are the lengths of the sides of the rectangle?
10. For an isosceles triangle with sides of the length given, find the length of hypotenuse:
1. 1
2. 2
3. 3

### Vocabulary Language: English

converse

converse

If a conditional statement is $p \rightarrow q$ (if $p$, then $q$), then the converse is $q \rightarrow p$ (if $q$, then $p$. Note that the converse of a statement is not true just because the original statement is true.