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# Showing the Irreducibility of a Polynomial and “Splitting” the Field [duplicate]

This question already has an answer here:

The following problem is part of my Algebra problem set:

Let $\alpha= \sqrt[3]{2}+\omega$ ,where $\omega=e ^{2\pi i /3} = -\frac{1}{2} + \frac{\sqrt{-3}}{2}$ is a primitive cube root of $1$.

$\alpha$ is a zero of $f(x) = 9 + 9x +3x^3 + 6x^4 + 3x^5 + x^6$. Show that $f(x)$ is irreducible.

** Eisenstein's Criterion fails here I realize that Gauss's Lemma is the best method of proving this although I am unclear as to whether proving the irreducibility of a polynomial in a subfield is sufficient to prove that it is irreducible. **

Please Also note that we have not had any Galois Theory.

Show that $\mathbb{Q}(\alpha)$ is the smallest field which contains all three zeros of $x^3 -2$ (this is called the splitting field of $x^3 -2$).

Thank you in advance for your help!

-

## marked as duplicate by T. Bongers, user7530, tetori, Johannes Kloos, Dominic MichaelisNov 4 '13 at 6:09

excuse me... i was a bit confused.... –  Praphulla Koushik Oct 23 '13 at 9:31
Irreducible over $\Bbb Q$? –  leo Oct 23 '13 at 15:37
This is a partial duplicate only (the first part). –  Cameron Buie Oct 27 '13 at 16:14
Please, post only one question in one post. In particular, I suggest that you remove your first question from the post, as it has been previously answered, here. Posting several questions in the same post is discouraged and such questions may be closed, see meta. –  Cameron Buie Oct 27 '13 at 17:46
@user: Agreed. I was going to post the above comment, regarding that, but then I got sidetracked. –  Cameron Buie Oct 27 '13 at 17:48

We claim that the six degree polynomial you wrote down is irreducible by assuming that $\alpha = \sqrt[4]{2} + \omega$ is a root of it. We do this by contradiction. Suppose that it aint.

Claim 1: If $\alpha$ is not sextic over $\mathbb{Q}$ then it must be cubic over $\mathbb{Q}$.

As Paul said $\mathbb{Q}(\alpha)\subseteq \mathbb{Q}(\sqrt[3]{2},\omega)$ and since $[\mathbb{Q}(\sqrt[3]{2},\omega):\mathbb{Q}]=6$ this means $d=[\mathbb{Q}(\alpha):\mathbb{Q}]$ must divide $6$. Now look at the divisors of six. Clearly $d\not = 1$ as $\alpha$ is not a rational number (why not?), also $d\not = 6$ as we are assuming that $\alpha$ is not sextic over $\mathbb{Q}$. Therefore, $d=2\text{ or }3$.

But $d$ cannot be $2$, because if it was then it would mean $\alpha$ would satisfy a quadratic polynomial $X^2+aX+b$ with $a,b\in \mathbb{Q}$. Now if $\alpha$ is a root of this polynomial then so is its complex conjugate, now $\overline{\alpha} = \overline{\omega} + \sqrt[3]{2} = \omega^2 + \sqrt[3]{2}$. Recall that the sum of the roots of $X^2 + aX+b$ add up to $-a$ and so $\alpha + \overline{\alpha} = - a \in \mathbb{Q}$ but $\alpha + \overline{\alpha} = \omega + \omega^2 + 2\sqrt[3]{2} = -1 + 2\sqrt[3]{2}$ and this number is not rational. Thus, it must be the case that $\alpha$ is cubic over $\mathbb{Q}$.

Claim 2: $\omega \not \in \mathbb{Q}(\alpha)$

Mr. Paul basically showed this above. But here is another proof. If $\omega \in \mathbb{Q}(\alpha)$ then the degree of $\omega$ over $\mathbb{Q}$ would divide the degree of $\mathbb{Q}(\alpha)/\mathbb{Q}$ (why?), but the degree of $\omega$ over $\mathbb{Q}$ is two which does not divide three.

Claim 3: $\sqrt[3]{2}\not \in \mathbb{Q}(\alpha)$

If it was then $\omega = \alpha - \sqrt[3]{2} \in \mathbb{Q}(\alpha)$ contradicting Claim 2.

Claim 4: $\omega^2\sqrt[3]{2}\not \in \mathbb{Q}(\alpha)$

This is a bit computational. Suppose that $\omega^2\sqrt[3]{2}\in \mathbb{Q}(\alpha)$ then:

(i) $(\sqrt[3]{2}+\omega)\omega^2\sqrt[3]{2} = \omega^2\sqrt[3]{4} + \sqrt[3]{2} \in \mathbb{Q}(\alpha)$

(ii) $(\sqrt[3]{2}+\omega)(\omega^2\sqrt[3]{2})^2 = 2\omega + \omega^2\sqrt[3]{4} \in \mathbb{Q}(\alpha)$

Subtract (i) from (ii) to get $2\omega - \sqrt[3]{2}\in \mathbb{Q}(\alpha)$ but then $2\omega - \sqrt[3]{2} + (\omega + \sqrt[3]{2}) = 3\omega \in \mathbb{Q}(\alpha)$. This immediately would imply that $\omega \in \mathbb{Q}(\alpha)$ which contradicts Claim 2.

Claim 5: $\omega\sqrt[3]{2}\not \in \mathbb{Q}(\alpha)$.

This one is a similar computation but a bit trickier than Claim 4. Begin by assuming that $\omega\sqrt[3]{2} \in \mathbb{Q}(\alpha)$. First of all $\omega^2 = -1 - \omega$ (why?). Then, $(\sqrt[3]{2}+\omega)^2 = \sqrt[3]{4} + 2\sqrt[3]{2}\omega - 1 - \omega$. This implies that $\sqrt[3]{4} - \omega \in \mathbb{Q}(\alpha)$.

Then:

(iii) $(\sqrt[3]{4}-\omega)\omega\sqrt[3]{2} = 2\omega - \omega^2\sqrt[3]{2}$

(iv) $(\sqrt[3]{4}-\omega)(\omega\sqrt[3]{2})^2 = 2\omega^2\sqrt[3]{2} - \sqrt[3]{4}$

Add 2(iii) + (iv) to get that $4\omega - \sqrt[3]{4}\in \mathbb{Q}(\alpha)$ be we shown just above that $\sqrt[3]{4} - \omega \in \mathbb{Q}(\alpha)$ so when we add them we get $3\omega \in \mathbb{Q}(\alpha)$ which is a contradiction.

Claim 6: $\sqrt[3]{2}$ is cubic over $\mathbb{Q}(\alpha)$.

We know from Claim 3 that $\sqrt[3]{2}$ is not in the field and $\sqrt[3]{2}$ solves the polynomial $X^3-2$ so the degree of $\sqrt[3]{2}$ is either two or three. It cannot be two. Because if it degree two then $X^3-2$ must be reducible over $\mathbb{Q}(\alpha)$, in particular one of its roots: $\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2}$ must be contained in $\mathbb{Q}(\alpha)$ but Claim 3,4,5 show that is impossible. Thus $\sqrt[3]{2}$ must be cubic.

Claim 7: CONTRADICTION ARISES from assuming $\alpha$ was cubic.

Let $K=\mathbb{Q}(\alpha)$. By the above claims $K(\omega)/K$ is quadratic field and $K(\sqrt[3]{2})/K$ is a cubic field. Thus, $K(\omega,\sqrt[3]{2})/K$ is a sextic field which is a contradiction (do you see it?)*

*)Fact from Field Theory: If $F(\alpha)/F$ is degree $n$ and $F(\beta)/F$ is degree $m$ with $\gcd(n,m)=1$ then $F(\alpha,\beta)/F$ is degree $n\cdot m$.

-

Hint for the second part: Note that the splitting field of $x^3-2$ is exactly ${\mathbb Q}(\sqrt[3]{2},\omega)$. If you can show that $[{\mathbb Q}(\sqrt[3]{2},\omega) : {\mathbb Q}] = 6$, you should be done (why?).

For convinience consider: $[{\mathbb Q}(\sqrt[3]{2},\omega) : {\mathbb Q}] = [{\mathbb Q}(\sqrt[3]{2}):{\mathbb Q}] \times [{\mathbb Q}(\sqrt[3]{2},\omega) : {\mathbb Q}(\sqrt[3]{2})]$ and note that $\omega\not\in {\mathbb Q}(\sqrt[3]{2})$ (why?). :)

EDIT: Without Galois Theory, the first part looks very messy.

Here goes an idea: We are done, in view of the previous hint, if we can show ${\mathbb Q}(\alpha) = {\mathbb Q}(\sqrt[3]{2},\omega)$ (in fact, this should prove the second part directly).

Note that if $\omega\in {\mathbb Q}(\alpha)$, the so is $\sqrt[3]{2}$, and similarly if $\sqrt[3]{2}\in {\mathbb Q}(\alpha)$ then $\omega\in {\mathbb Q}(\alpha)$. Thus we will assume that $\omega\not\in {\mathbb Q}(\alpha)$ and $\sqrt[3]{2}\not\in {\mathbb Q}(\alpha)$, else we would be done.

We have, since ${\mathbb Q}(\alpha,\omega) = {\mathbb Q}(\sqrt[3]{2},\omega)$ (why?) : $$6 = [{\mathbb Q}(\alpha,\omega):{\mathbb Q}] =[{\mathbb Q}(\alpha,\omega):{\mathbb Q}(\alpha)]\times [{\mathbb Q}(\alpha):{\mathbb Q}]$$

Now use this and $\omega\not \in {\mathbb Q}(\alpha)$ to conclude that $[{\mathbb Q}(\alpha):{\mathbb Q}]=3$. Use this to prove that $2=[{\mathbb Q}(\alpha,\sqrt[3]{2}):{\mathbb Q}(\alpha)]$.

Now the minimal polynomial $m(x)$ for $\sqrt[3]{2}$ over ${\mathbb Q}(\alpha)[x]$ must divide $x^3-2 \in {\mathbb Q}(\alpha)[x]$ and $\deg m(x) = 2$.

Dicuss then according to what the 2 roots of $m(x)$ should be, bearing in mind that $m(x) | (x^3 - 2)$.

Finally, prove that if a possible choices produced a polynomial in ${\mathbb Q}(\alpha)[x]$, then either $\omega\in {\mathbb Q}(\alpha)$ or $\sqrt[3]{2}\in {\mathbb Q}(\alpha)$ contradicting our assumption.

-
At least two students from the same class asking the same question: math.stackexchange.com/questions/536985/… –  Alex B. Oct 23 '13 at 14:18
Thank you! However, we have NOT had any Galois Theory. Is is possible to answer the question without Galois Theory?? –  incompleteness Oct 23 '13 at 15:05
I am trying to prove the part about the irreducible polynomial, but withouth Galois Theory it is getting very messy. I might edit my post soon if I get somewhere. –  Pablo Rotondo Oct 23 '13 at 15:23
@incompleteness Edited the post. Very messy though, but this seems to prove (if correct) the second part directly. –  Pablo Rotondo Oct 23 '13 at 15:29


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# If $\overrightarrow{\alpha}$ and $\overrightarrow{\beta}$ are non zero and different vector such that $|\overrightarrow{\alpha} + \overrightarrow{\beta}| = |\overrightarrow{\beta} - \overrightarrow{\alpha}|$, then the angle between $\overrightarrow{\alpha}$ and $\overrightarrow{\beta}$ is :
( A ) $\overrightarrow{\pi}{6}$
( B ) $\overrightarrow{\pi}{3}$
( C ) $\overrightarrow{\pi}{4}$
( D ) $\overrightarrow{\pi}{2}$

format: plain


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Zum Hauptinhalt springen

# Inflation Schedule

Subject to change. Follow most recent economic discussions in the Solana forums: https://forums.solana.com

Validator-clients have two functional roles in the Solana network:

• Validate (vote) the current global state of their observed PoH.
• Be elected as ‘leader’ on a stake-weighted round-robin schedule during which time they are responsible for collecting outstanding transactions and incorporating them into their observed PoH, thus updating the global state of the network and providing chain continuity.

Validator-client rewards for these services are to be distributed at the end of each Solana epoch. As previously discussed, compensation for validator-clients is provided via a commission charged on the protocol-based annual inflation rate dispersed in proportion to the stake-weight of each validator-node (see below) along with leader-claimed transaction fees available during each leader rotation. I.e. during the time a given validator-client is elected as leader, it has the opportunity to keep a portion of each transaction fee, less a protocol-specified amount that is destroyed (see Validation-client State Transaction Fees).

The effective protocol-based annual staking yield (%) per epoch received by validation-clients is to be a function of:

• the current global inflation rate, derived from the pre-determined dis-inflationary issuance schedule (see Validation-client Economics)
• the fraction of staked SOLs out of the current total circulating supply,
• the commission charged by the validation service,
• the up-time/participation [% of available slots that validator had opportunity to vote on] of a given validator over the previous epoch.

The first factor is a function of protocol parameters only (i.e. independent of validator behavior in a given epoch) and results in an inflation schedule designed to incentivize early participation, provide clear monetary stability and provide optimal security in the network.

As a first step to understanding the impact of the Inflation Schedule on the Solana economy, we’ve simulated the upper and lower ranges of what token issuance over time might look like given the current ranges of Inflation Schedule parameters under study.

Specifically:

• Initial Inflation Rate: 7-9%
• Dis-inflation Rate: -14-16%
• Long-term Inflation Rate: 1-2%

Using these ranges to simulate a number of possible Inflation Schedules, we can explore inflation over time:

In the above graph, the average values of the range are identified to illustrate the contribution of each parameter. From these simulated Inflation Schedules, we can also project ranges for token issuance over time.

Finally we can estimate the Staked Yield on staked SOL, if we introduce an additional parameter, previously discussed, % of Staked SOL:

$\%~\text{SOL Staked} = \frac{\text{Total SOL Staked}}{\text{Total Current Supply}}$

In this case, because % of Staked SOL is a parameter that must be estimated (unlike the Inflation Schedule parameters), it is easier to use specific Inflation Schedule parameters and explore a range of % of Staked SOL. For the below example, we’ve chosen the middle of the parameter ranges explored above:

• Initial Inflation Rate: 8%
• Dis-inflation Rate: -15%
• Long-term Inflation Rate: 1.5%

The values of % of Staked SOL range from 60% - 90%, which we feel covers the likely range we expect to observe, based on feedback from the investor and validator communities as well as what is observed on comparable Proof-of-Stake protocols.

Again, the above shows an example Staked Yield that a staker might expect over time on the Solana network with the Inflation Schedule as specified. This is an idealized Staked Yield as it neglects validator uptime impact on rewards, validator commissions, potential yield throttling and potential slashing incidents. It additionally ignores that % of Staked SOL is dynamic by design - the economic incentives set up by this Inflation Schedule.

### Adjusted Staking Yield

A complete appraisal of earning potential from staking tokens should take into account staked Token Dilution and its impact on staking yield. For this, we define adjusted staking yield as the change in fractional token supply ownership of staked tokens due to the distribution of inflation issuance. I.e. the positive dilutive effects of inflation.

We can examine the adjusted staking yield as a function of the inflation rate and the percent of staked tokens on the network. We can see this plotted for various staking fractions here:


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# Newsmakers

Science  25 Apr 2008:
Vol. 320, Issue 5875, pp. 431
1. # IN PRINT

MISSING SEX? Carbon atoms can't have five bonds, chemist Alan Rosan thought as he pondered an illustration accompanying a New York Times review of a book about sex research. So Rosan, a professor at Drew University in Madison, New Jersey, e-mailed a dart to the paper wondering if the error represented the molecule's wanton desire “to form new, revealing bonding relationships” unheard of “in our more staid and prudish chemistry.”

But the joke was on him. The illustration actually spelled the word “sex,” pointed out an editor's note accompanying Rosan's letter and a similar complaint from a chemistry graduate student.

Rosan confesses to his mistake. “I tell my students all the time about the difference between looking and seeing. … In this case, I was guilty of not seeing,” says Rosan. However, he still thinks that the illustrator could have made the point without breaking the rules of chemistry.

2. # AWARDS

LET'S TRY IT. Alexander Varshavsky has thought of a new way to kill cancer cells, winning $1 million from a new prize that encourages people to share untested ideas in cancer research. Tumor cells often lose sections of DNA and pass the deletion to their daughter cells. Healthy tissues, however, still have the DNA. Varshavsky, a molecular biologist at the California Institute of Technology in Pasadena, wants to introduce a vector—a small piece of engineered DNA—throughout a patient's body. The vector would code for a toxin to kill cancer cells and for proteins that detect whether a cell contains the deletions. The vector would selfdestruct in healthy cells without the deletions. “I would be very surprised if this strategy ever works all the way to the bedside … in the shape that I suggested,” Varshavsky says, noting that the technical details are likely to evolve during actual implementation. Although the award—created by Joel Greenblatt and Robert Goldstein of Gotham Capital—is for personal use, Varshavsky says it will bolster his research, too. Varshavsky's idea was chosen from more than 500 submitted. A six-member scientific panel also awarded neurosurgeon and entrepreneur Mark Carol the$250,000 Ira Sohn Conference Foundation Prize in Pediatric Oncology for insights into radiation therapy.

3. # MOVERS

BIOWARRIOR. Amicrobiologist with vaccine industry experience will head a new Department of Health and Human Services (HHS) agency aimed at helping companies combat bioterrorism. Robin Robinson, 52, will be the first director of the 1-year-old Biomedical Advanced Research and Development Authority (BARDA).

BARDA was created by Congress after companies complained about the hurdles to preparing drugs and vaccines for eligibility in BioShield, the federal biodefense procurement program (Science, 7 July 2006, p. 28). Robinson joined HHS in 2004 and has been “very successful” at leading the development and stockpiling of the first human vaccine against H5N1 influenza, says Brad Smith of the University of Pittsburgh Center for Biosecurity in Baltimore, Maryland. But BARDA's so far $100-million-a-year budget is a tiny fraction of the$3.4 billion a year that Smith estimates is needed.

4. # DEATHS

EYE FOR DETAIL. The father of modern chaos theory, meteorologist Edward Lorenz, died 16 April in his home in Cambridge, Massachusetts, at age 90. Lorenz had “plain old intelligence [and] was also an extremely persistent scientist. He would not settle for anything less than perfection in his work,” says meteorologist Kerry Emanuel of the Massachusetts Institute of Technology, where Lorenz spent his 60-year career.

That trait served Lorenz well in the early 1960s, when he got quite different results from two computer simulations of the weather because of an inadvertent, tiny difference in the simulated atmosphere's starting condition. A decade later, at the annual meeting of the American Association for the Advancement of Science (the publisher of Science), he gave a talk with a title that quickly entered popular culture: “Predictability: Does the Flap of a Butterfly's Wings in Brazil Set Off a Tornado in Texas?” Chaos and the atmosphere's exquisite sensitivity to initial conditions permeate meteorology today, from setting the limits of prediction to making the most of the computer's predictive powers.

5. # A LIFE IN SCIENCE

UNABASHED. Theoretical physicist John Wheeler, who died on 13 April at age 96, was insatiably curious and not afraid to look foolish, says William Unruh, a theorist at the University of British Columbia in Vancouver, Canada. Unruh recalls how Wheeler spent a year in unsuccessful pursuit of the idea that the uncertainties of quantum mechanics might be related to Gödel's incompleteness theorem, which says that, given a set of mathematically consistent axioms, there are true statements that can never be proved so. “He was like a little boy. He would jump into the stream and turn over the rocks, even if he got his pants wet,” says Unruh. That unselfconsciousness extended to Wheeler's personal life, Unruh adds. Wheeler loved the water but never learned to swim. So he would simply array himself in floats and go for a daily dip.

In a career spanning 7 decades, Wheeler helped explain nuclear fission, established general relativity as an essential part of astrophysics and cosmology, and pioneered the study of quantum gravity. A list of students reads like a who's who in gravitational theory.


text:
mersenneforum.org I found the next big prime number. How do I get my reward
 Register FAQ Search Today's Posts Mark Forums Read

2020-06-20, 17:46   #12
paulunderwood

Sep 2002
Database er0rr

3×52×72 Posts

Quote:
 Originally Posted by Parlak67 Please document how you found the factor.
I don't what software forumite Runtime Error used, but it is easy to check:

Code:
Mod(2,41094328366458894751937)^156635789-1
Mod(0, 41094328366458894751937)

Last fiddled with by paulunderwood on 2020-06-20 at 17:48

2020-06-20, 17:48   #13
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

10100000011012 Posts

Quote:
 Originally Posted by Parlak67 Please document how you found the factor.
You asked someone who did not find it, to document how he found it. After you made an extraordinary claim, with no documentation yet.
That seems ... odd.

Probably the factor finder used mfaktc or mfakto.

We do what we reasonably can to make such information available. Reference information blog
An Available software thread in the "New to GIMPS? Start Here" forum's Information and Answers subforum
And more, elsewhere.

Last fiddled with by kriesel on 2020-06-20 at 18:06

 2020-06-20, 18:14 #14 Uncwilly 6809 > 6502     """"""""""""""""""" Aug 2003 101×103 Posts 2·3·1,597 Posts Factors are absolute proof and the method of their discovery is basically irrelevant. They can be quickly verified to be true (a fraction of a second, IIRC). The PrimeNet server does that for all reported factors before they are entered into the database. And you have yet to state how you "knew" it to be prime. If you used any of the programs that I mentioned above, there was an error. Fixing any error in the program that produces false prime claims is important to us. Last fiddled with by Uncwilly on 2020-06-20 at 18:16
2020-06-20, 18:21   #15
Parlak67

Jun 2020

22×3 Posts

Quote:
 Originally Posted by kriesel Well done. This might be a a new speed record for empirically dispatching a dubious claim.
https://www.alpertron.com.ar/ECM.HTM

2020-06-20, 18:28   #16
Uncwilly
6809 > 6502

"""""""""""""""""""
Aug 2003
101×103 Posts

256E16 Posts

Quote:
 Originally Posted by Parlak67 https://www.alpertron.com.ar/ECM.HTM
That is a fine tool. But it is not capable of testing a Mersenne Prime of this size.

See the attached screen grab.
Attached Thumbnails

2020-06-20, 18:34   #17
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

3·5·17·37 Posts

We are looking at this from a wrong direction.
He simply found that 156635789 is prime. --

Quote:
 Originally Posted by Parlak67 156635789 you can look with the program you want?

2020-06-20, 18:34   #18
Parlak67

Jun 2020

22×3 Posts

Quote:
 Originally Posted by Uncwilly That is a fine tool. But it is not capable of testing a Mersenne Prime of this size. See the attached screen grab.
thanks for everything. I will start working again

2020-06-20, 18:34   #19
Runtime Error

Sep 2017
USA

5×47 Posts

I apologize for my earlier comment about IP addresses. There have been a few dubious claims recently that seem to be troll posts. Sorry for assuming.

Quote:
 Originally Posted by Parlak67 Please document how you found the factor.
Just to be clear, 156635789 is a prime number. If you wrote a program and found it on your own, good job! It just isn't very big compared to the size of today's known prime numbers.

However, the Mersenne number 2156635789-1 is not prime because it is divisible by 41094328366458894751937.

The software used: mfaktc version 0.21

Last fiddled with by Runtime Error on 2020-06-20 at 18:37

2020-06-20, 18:40   #20
Parlak67

Jun 2020

22×3 Posts

Quote:
 Originally Posted by Runtime Error I apologize for my earlier comment about IP addresses. There have been a few dubious claims recently that seem to be troll posts. Sorry for assuming. Just to be clear, 156635789 is a prime number. If you wrote a program and found it on your own, good job! It just isn't very big compared to the size of today's known prime numbers. However, the Mersenne number 2156635789-1 is not prime because it is divisible by 41094328366458894751937. The software used: mfaktc version 0.21
numbers

155654789
160456789
369452123

2020-06-20, 18:40   #21
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

3·5·17·37 Posts

Quote:
 Originally Posted by Runtime Error ... 2156635789-1 is not prime because it is divisible by 41094328366458894751937. The software used: mfaktc version 0.21
This reminds me of the Mr.Show sketch, "24 is the highest number."

...but wait! Runtime Error just said 41094328366458894751937, that is an enormous number!

2020-06-20, 18:45   #22
Parlak67

Jun 2020

22×3 Posts

Quote:
 Originally Posted by Batalov This reminds me of the Mr.Show sketch, "24 is the highest number." ...but wait! Runtime Error just said 41094328366458894751937, that is an enormous number!

My goal is not to make fun of someone here. Thank you for everything.

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 Eurasian Math. J.: Year: Volume: Issue: Page: Find

 Eurasian Math. J., 2020, Volume 11, Number 1, Pages 86–94 (Mi emj358)

An estimate of approximation of a matrix-valued function by an interpolation polynomial

V. G. Kurbatov, I. V. Kurbatova

Voronezh State University, 1 Universitetskaya Square, 394018 Voronezh, Russia

Abstract: Let $A$ be a square complex matrix; $z_1,…,z_n\in\mathbb{C}$ be (possibly repetitive) points of interpolation; $f$ be a function analytic in a neighborhood of the convex hull of the union of the spectrum of $A$ and the points $z_1,…,z_n$; and $p$ be the interpolation polynomial of $f$ constructed by the points $z_1,…,z_n$. It is proved that under these assumptions
$$||f(A)-p(A)||\leqslant \frac1{n!}\max_{t\in[0,1]\atop {\mu\in coż_1,z_2,…,z_n\}}}||\Omega(A)f^{(n)}((1-t)\mu\mathbf{1}+tA)||,$$
where $\Omega(z)=\prod_{k=1}^n(z-z_k)$ and the symbol $co$ means the convex hull.

Keywords and phrases: matrix function, polynomial interpolation, estimate.

 Funding Agency Grant Number Russian Foundation for Basic Research 19-01-00732_a Ministry of Education and Science of the Russian Federation 3.1761.2017/4.6 The first author was supported by the Ministry of Education and Science of the Russian Federation under state order No. 3.1761.2017/4.6. The second author was supported by the Russian Foundation for Basic Research under research project No. 19-01-00732 A.

DOI: https://doi.org/10.32523/2077-9879-2020-11-1-86-94

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References: PDF file   HTML file

MSC: 65F60, 97N50
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Citation: V. G. Kurbatov, I. V. Kurbatova, “An estimate of approximation of a matrix-valued function by an interpolation polynomial”, Eurasian Math. J., 11:1 (2020), 86–94

Citation in format AMSBIB
\Bibitem{KurKur20} \by V.~G.~Kurbatov, I.~V.~Kurbatova \paper An estimate of approximation of a matrix-valued function by an interpolation polynomial \jour Eurasian Math. J. \yr 2020 \vol 11 \issue 1 \pages 86--94 \mathnet{http://mi.mathnet.ru/emj358} \crossref{https://doi.org/10.32523/2077-9879-2020-11-1-86-94}


text:
# An entropy criterion for visualization

Posted on 21 Apr 2015
Tags: visualization, running, statistics, half-baked

Just throwing up a half-baked idea here. I can’t even confidently assert novelty since I still haven’t worked through relevant background like this.

In the running visualization 2-d histogram idea I’ve been playing with, I’ve been setting probably the most important free parameter arbitrarily. The (linear) grid size $$n$$ determines how coarse or fine the binning is, with the total number of bins being roughly $$n^2$$.1 From unsystematic exploration, a grid size of 3 is too coarse, a grid size of 200 is too fine, and a grid size appeared adequate enough to stop searching further.

There’s a philosophical, or at least methodological concern, that I need to address before moving on. If I’m asking “what grid size is best?”, I might really be asking or instead should be asking “what grid size is best for this particular task?”. Further, this task should have a concrete answer that is known to me, because otherwise how can I assess if the visualization is effective at helping a user complete this task?

I can sidestep this concern in two ways. For the first, I can ignore any pragmatic use of the visualization and just be concerned with aesthetics. A figure that looks nice or interesting is ceteris paribus2 to be treated as better than one that looks ugly or boring. For the second, the visualization can be treated as loss data reduction, with the amount of loss scaling with the coarseness of the grid. Task-specific visualizations can be approximated as operations on the reduced data. A data reduction that incurs low error for a range of tasks can then be treated as good without reference to any particular task.

The criterion for grid size I’m fiddling with is based on maximizing entropy. In particular, for a range of grid sizes, a 2-d histogram is produced. Each region with at least one observation has a value giving the number of seconds I spent running in that region. Equivalently, this can be interpreted as color from white to blue. Ignoring the spatial structure, this can just be treated as a distribution over durations or colors. Because there is a limited capability to discriminate between similar colors (especially when they’re alpha-blended with the map), I bin the range of these values down to 10 categories. The entropy of this distribution is the quantity I seek to maximize.

It can be informally argued that in a 2-d histogram with high entropy as defined here, the region colors are conveying a lot of information about how much time was spent in the various regions.3

The choice of the number of duration bins itself has an effect on the outcome, and so must be chosen carefully. However, this choice may be made on the basis of the legibility of the color scale and without any direct reference to the running map.

Although the visualization at the grid size picked by this criterion doesn’t look strikingly optimal, it does at least look reasonable.

It would be conventional to state some open questions and ideas for future tests here, but so much about this idea is unsettled that pretty much any reasonable direction to go in appears open.

1. The “roughly” is due to details of exactly how hexbin packs a square region with hexagons.

2. Other things held constant, a foundational concept in causal inference.

3. On the other hand, this procedure ignores information about which areas of the map were ever visited versus those that were not. Attempts to incorporate this led to less compelling visualization, and have been put aside for now.


text:
## Saturday, March 10, 2012

### of human relations 3: drawings, drawings

[repeated: as i said, i'm looking for new ways to visualize human relations, feelings, real-life struggle as well as uplifting moments. looking back on centuries of art, i find myself surprised that there is so little art which addresses this in a way that i find provoking, uplifting, inspiring.

this gives new motivation to continue my investigation. i feel sure enough of the emotive strength of the drawings which come up in this sense. and this translates into paintings as well. but enough words, the images themselves should be stronger.']

self-portrait as a mother worried about her children (own work, 2012, 30 x 36 cm, click on the image for an enlargement)

rejection (own work, 2012, 21 x 20 cm, click on the image for an enlargement)

for the next image, also see the post on notre dame des anges:

seigneur notre retraite (in brown-red) (own work, 2011, 30 x 40 cm, click on the image for an enlargement)

another theme that crops up ever more frequently: the people who do not fit well into our wonderful' society. sure, they are the misfits, so to say. but who is the more crazy, he who accepts this crazy world, or he who does not fit in? if you have read my previous posts, you will know that i for one cannot simply answer that question. and my concern and sympathy are often with the misfits, and my anger is often directed at all the `respected' institutions which together comprise the status quo in our society.

woman out of kilter (own work, 2011, 20 x 28 cm, click on the image for an enlargement)

i think emotive expressive power comes from the eyes, the face, the body language...and the materials and techniques used. this is what i study on, laboriously, and much unappreciatedly...but the time of appreciation will come, i suppose, once enough people see the works and enough time has passed for them to compare this avenue of art to other avenues. anyway, if you see any art touching on the same themes, please let me know, since it would be inspiring to me to know other artists with a similar outlook.

detail of woman out of kilter (own work, 2011, 20 x 28 cm, click on the image for an enlargement)

taking care of the psychiatric patient iii (own work, 2011, 21 x 30 cm, click on the image for an enlargement)

grief and comfort iii (own work, 2011, 21 x 30 cm, click on the image for an enlargement)

self-portrait in orange unrest (own work, 2011, 21 x 30 cm, click on the image for an enlargement)

man, lost in thoughts and orange background (own work, 2011, 30 x 36 cm, click on the image for an enlargement)

st. frances posing (own work, 2011, 30 x 35 cm, click on the image for an enlargement)

woman in white bra (own work, 2012, 30 x 40 cm, click on the image for an enlargement)


text:
## anonymous one year ago Which of the following statements would best describe how to increase the work done by a heat engine? A. Lower the temperature of the hot reservoir. B. Raise the temperature of the exhaust. C. Lower the temperature of the exhaust. D. Raise the temperature of the cold reservoir. **Not sure!! thank you:)

1. Michele_Laino

the efficiency of a Thermal machine is given by the subsequent formula: $\eta = 1 - \frac{{{T_2}}}{{{T_1}}}$ where T_1 is the temperature of the hot reservoir and T_2 is the temperature of the cold reservoir

2. anonymous

okay! :) so to increase the work done, would we have to lower the temperature of the exhaust? :/

3. anonymous

i am not sure :/

4. Michele_Laino

yes! that's right! we have to lower the value of T_2. Please keep in mind that both T_1 and T_2 are absolute temperatures

5. anonymous

yay!! :) so choice C it is! thank you!!

6. Michele_Laino

yes! Option C thank you!! :)


text:
# Measurement of the production cross section for Z gamma to nu nu gamma in pp collisions at sqrt(s) = 7 TeV and limits on Z Z gamma and Z gamma gamma triple gauge boson couplings

Abstract : A measurement of the Z gamma to nu nu-bar gamma cross section in pp collisions at sqrt(s) = 7 TeV is presented, using data corresponding to an integrated luminosity of 5.0 inverse femtobarns collected with the CMS detector. This measurement is based on the observation of events with an imbalance of transverse energy in excess of 130 GeV and a single photon in the absolute pseudorapidity range abs(eta) < 1.4 with transverse energy above 145 GeV. The Z gamma to nu nu-bar gamma production cross section is measured to be 21.1 +/- 4.2 (stat.) +/- 4.3 (syst.) +/- 0.5 (lum.) fb, which agrees with the standard model prediction of 21.9 +/- 1.1 fb. The results are combined with the CMS measurement of Z gamma production in the l+ l- gamma final state (where l is an electron or a muon) to yield the most stringent limits to date on triple gauge boson couplings: abs(h[3,Z]) < 2.7E-3, abs(h[4,Z]) < 1.3E-5 for Z Z gamma and abs(h[3,gamma]) < 2.9E-3, abs(h[4,gamma]) < 1.5E-5 for Z gamma gamma couplings.
Document type :
Journal articles

http://hal.in2p3.fr/in2p3-00858232
Contributor : Sylvie Flores Connect in order to contact the contributor
Submitted on : Thursday, September 5, 2013 - 7:40:57 AM
Last modification on : Monday, December 13, 2021 - 9:15:13 AM

### Citation

S. Chatrchyan, M. Besançon, S. Choudhury, F. Couderc, M. Dejardin, et al.. Measurement of the production cross section for Z gamma to nu nu gamma in pp collisions at sqrt(s) = 7 TeV and limits on Z Z gamma and Z gamma gamma triple gauge boson couplings. Journal of High Energy Physics, Springer, 2013, 10(2013), pp.164. ⟨10.1007/JHEP10(2013)164⟩. ⟨in2p3-00858232⟩

### Metrics

Les métriques sont temporairement indisponibles


text:
Reconstructing Charge Distribution from Multipole Expansion

Let $$\rho$$ be a static, discrete or continuous charge distribution, and $$\phi(\mathbf{r})$$ the corresponding electric potential. We may expand $$\phi$$ in a multipole series, $$\phi(\mathbf{r}) = \frac{1}{r}\sum_{n \geq 0} \int \text{d}^3r' \rho(\mathbf{r}')\left(\frac{r'}{r}\right)^nP_n(\cos\theta).$$ My question is, if given the multipole expansion, one can determine the source $$\rho$$.

Evidently, this is possible if given the multipole expansion of two equally and oppositely charged point sources a distance $$a$$ apart. From the dipole moment, one can deduce $$a$$, and thus reconstruct the charge distribution. But is this true in general?

For a simple counterexample, consider two uniform solid spheres of charge, of different radii but with identical total charge $$Q$$: they will have vanishing multipole moments for all $$l\geq 1$$, and identical monopole moment, but the distributions are different.
• Fair enough. But, if we cannot obtain $\rho$ exactly, it still seems one can recover information about the geometry of the charge distribution. Is there a clear mapping from the multipole expansion to the geometry of the source generating the field? – fruitegg Sep 17 '19 at 16:58


text:
# Taking a closer look at Quantum gates and their operations

This post is a continuation of my earlier post ‘Exploring Quantum gate operations with QCSimulator’. Here I take a closer look at more quantum gates and their operations, besides implementing these new gates in my Quantum Computing simulator, the  QCSimulator in R.

Disclaimer: This article represents the author’s viewpoint only and doesn’t necessarily represent IBM’s positions, strategies or opinions

In  quantum circuits, gates  are unitary matrices which operate on 1,2 or 3 qubit systems which are represented as below

1 qubit
|0> = $\begin{pmatrix}1\\0\end{pmatrix}$ and |1> = $\begin{pmatrix}0\\1\end{pmatrix}$

2 qubits
|0> $\otimes$ |0> = $\begin{pmatrix}1\\ 0\\ 0\\0\end{pmatrix}$
|0> $\otimes$ |1> = $\begin{pmatrix}0\\ 1\\ 0\\0\end{pmatrix}$
|1> $\otimes$ |o> = $\begin{pmatrix}0\\ 0\\ 1\\0\end{pmatrix}$
|1> $\otimes$ |1> = $\begin{pmatrix}0\\ 0\\ 0\\1\end{pmatrix}$

3 qubits
|0> $\otimes$ |0> $\otimes$ |0> = $\begin{pmatrix}1\\ 0\\0\\ 0\\ 0\\0\\ 0\\0\end{pmatrix}$
|0> $\otimes$ |0> $\otimes$ |1> = $\begin{pmatrix}0\\ 1\\0\\ 0\\ 0\\0\\ 0\\0\end{pmatrix}$
|0> $\otimes$ |1> $\otimes$ |0> = $\begin{pmatrix}0\\ 0\\1\\ 0\\ 0\\0\\ 0\\0\end{pmatrix}$

|1> $\otimes$ |1> $\otimes$ |1> = $\begin{pmatrix}0\\ 0\\0\\ 0\\ 0\\0\\ 0\\1\end{pmatrix}$
Hence single qubit is represented as 2 x 1 matrix, 2 qubit as 4 x 1 matrix and 3 qubit as 8 x 1 matrix

1) Composing Quantum gates in series
When quantum gates are connected in a series. The overall effect of the these quantum gates in series is obtained my taking the dot product of the unitary gates in reverse. For e.g.

In the following picture the effect of the quantum gates A,B,C is the dot product of the gates taken reverse order
result = C . B . A

This overall action of the 3 quantum gates can be represented by a single ‘transfer’ matrix which is the dot product of the gates

If we had a Pauli X followed by a Hadamard gate the combined effect of these gates on the inputs can be deduced by constructing a truth table

 Input Pauli X – Output A’ Hadamard – Output B |0> |1> 1/√2(|0>  – |1>) |1> |0> 1/√2(|0>  + |1>)

Or

|0> -> 1/√2(|0>  – |1>)
|1> -> 1/√2(|0>  + |1>)
which is
$\begin{pmatrix}1\\0\end{pmatrix}$  ->1/√2 $\begin{pmatrix}1\\0\end{pmatrix}$$\begin{pmatrix}0\\1\end{pmatrix}$ = 1/√2  $\begin{pmatrix}1\\-1\end{pmatrix}$
$\begin{pmatrix}0\\1\end{pmatrix}$  ->1/√2 $\begin{pmatrix}1\\0\end{pmatrix}$ + $\begin{pmatrix}0\\1\end{pmatrix}$ = 1/√2  $\begin{pmatrix}1\\1\end{pmatrix}$
Therefore the ‘transfer’ matrix can be written as
T = 1/√2 $\begin{pmatrix}1 & 1\\ -1 & 1\end{pmatrix}$

2)Quantum gates in parallel
When quantum gates are in parallel then the composite effect of the gates can be obtained by taking the tensor product of the quantum gates.

If we consider the combined action of a Pauli X gate and a Hadamard gate in parallel

 A B A’ B’ |0> |0> |1> 1/√2(|0>  + |1>) |0> |1> |1> 1/√2(|0>  – |1>) |1> |0> |0> 1/√2(|0>  + |1>) |1> |1> |0> 1/√2(|0>  – |1>)

Or

|00> => |1> $\otimes$ 1/√2(|0>  + |1>) = 1/√2 (|10> + |11>)
|01> => |1> $\otimes$ 1/√2(|0>  – |1>) = 1/√2 (|10> – |11>)
|10> => |0> $\otimes$ 1/√2(|0>  + |1>) = 1/√2 (|00> + |01>)
|11> => |0> $\otimes$ 1/√2(|0>  – |1>) = 1/√2 (|10> – |11>)

|00> = $\begin{pmatrix}1\\ 0\\ 0\\0\end{pmatrix}$ =>1/√2$\begin{pmatrix} 0\\ 0\\ 1\\ 1\end{pmatrix}$
|01> = $\begin{pmatrix}0\\ 1\\ 0\\0\end{pmatrix}$ =>1/√2$\begin{pmatrix} 0\\ 0\\ 1\\ -1\end{pmatrix}$
|10> = $\begin{pmatrix}0\\ 0\\ 1\\0\end{pmatrix}$ =>1/√2$\begin{pmatrix} 1\\ 0\\ 1\\ -1\end{pmatrix}$
|11> = $\begin{pmatrix}0\\ 0\\ 0\\1\end{pmatrix}$ =>1/√2$\begin{pmatrix} 1\\ 0\\ -1\\ -1\end{pmatrix}$

Here are more Quantum gates
a) Rotation gate
$U = \begin{pmatrix}cos\theta & -sin\theta\\ sin\theta & cos\theta\end{pmatrix}$

b) Toffoli gate
The Toffoli gate flips the 3rd qubit if the 1st and 2nd qubit are |1>

 Toffoli gate Input Output |000> |000> |001> |001> |010> |010> |011> |011> |100> |100> |101> |101> |110> |111> |111> |110>

c) Fredkin gate
The Fredkin gate swaps the 2nd and 3rd qubits if the 1st qubit is |1>

 Fredkin gate Input Output |000> |000> |001> |001> |010> |010> |011> |011> |100> |100> |101> |110> |110> |101> |111> |111>

d) Controlled U gate
A controlled U gate can be represented as
controlled U = $\begin{pmatrix}1 & 0 & 0 & 0\\ 0 &1 &0 & 0\\ 0 &0 &u11 &u12 \\ 0 & 0 &u21 &u22 \end{pmatrix}$   – (A)
where U = $\begin{pmatrix}u11 &u12 \\ u21 & u22\end{pmatrix}$

e) Controlled Pauli gates
Controlled Pauli gates are created based on the following identities. The CNOT gate is a controlled Pauli X gate where controlled U is a Pauli X gate and matches the CNOT unitary matrix. Pauli gates can be constructed using

a) H x X x H = Z    &
H x H = I

b) S x X x S1
S x S1 = I

the controlled Pauli X, Y , Z are contructed using the CNOT for the controlled X in the above identities
In general a controlled Pauli gate can be created as below

f) CPauliX
Here C is the 2 x2  Identity matrix. Simulating this in my QCSimulator
CPauliX I=matrix(c(1,0,0,1),nrow=2,ncol=2) # Compute 1st composite a = TensorProd(I,I) b = CNOT2_01(a) # Compute 1st composite c = TensorProd(I,I) #Take dot product d = DotProduct(c,b) #Take dot product with qubit e = DotProduct(d,q) e }

Implementing the above with I, S, H gives Pauli X, Y and Z as seen below

library(QCSimulator)
I4=matrix(c(1,0,0,0,
0,1,0,0,
0,0,1,0,
0,0,0,1),nrow=4,ncol=4)

#Controlled Pauli X
CPauliX(I4)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    0    1
## [4,]    0    0    1    0
#Controlled Pauli Y
CPauliY(I4)
##      [,1] [,2] [,3] [,4]
## [1,] 1+0i 0+0i 0+0i 0+0i
## [2,] 0+0i 1+0i 0+0i 0+0i
## [3,] 0+0i 0+0i 0+0i 0-1i
## [4,] 0+0i 0+0i 0+1i 0+0i
#Controlled Pauli Z
CPauliZ(I4)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0   -1

g) CSWAP gate

q00=matrix(c(1,0,0,0),nrow=4,ncol=1)
q01=matrix(c(0,1,0,0),nrow=4,ncol=1)
q10=matrix(c(0,0,1,0),nrow=4,ncol=1)
q11=matrix(c(0,0,0,1),nrow=4,ncol=1)
CSWAP(q00)
##      [,1]
## [1,]    1
## [2,]    0
## [3,]    0
## [4,]    0
#Swap qubits
CSWAP(q01)
##      [,1]
## [1,]    0
## [2,]    0
## [3,]    1
## [4,]    0
#Swap qubits
CSWAP(q10)
##      [,1]
## [1,]    0
## [2,]    1
## [3,]    0
## [4,]    0
CSWAP(q11)
##      [,1]
## [1,]    0
## [2,]    0
## [3,]    0
## [4,]    1

h) Toffoli state
The Toffoli state creates a 3 qubit entangled state 1/2(|000> + |001> + |100> + |111>)

Simulating the Toffoli state in IBM Quantum Experience we get

h) Implementation of Toffoli state in QCSimulator

#ToffoliState
# Computation of the Toffoli State
H=1/sqrt(2) * matrix(c(1,1,1,-1),nrow=2,ncol=2)
I=matrix(c(1,0,0,1),nrow=2,ncol=2)

# 1st composite
# H x H x H
a = TensorProd(TensorProd(H,H),H)
# 1st CNOT
a1= CNOT3_12(a)

# 2nd composite
# I x I x T1Gate
b = TensorProd(TensorProd(I,I),T1Gate(I))
b1 = DotProduct(b,a1)
c = CNOT3_02(b1)

# 3rd composite
# I x I x TGate
d = TensorProd(TensorProd(I,I),TGate(I))
d1 = DotProduct(d,c)
e = CNOT3_12(d1)

# 4th composite
# I x I x T1Gate
f = TensorProd(TensorProd(I,I),T1Gate(I))
f1 = DotProduct(f,e)
g = CNOT3_02(f1)

#5th composite
# I x T x T
h = TensorProd(TensorProd(I,TGate(I)),TGate(I))
h1 = DotProduct(h,g)
i = CNOT3_12(h1)

#6th composite
# I x H x H
j1 = DotProduct(j,i)
k = CNOT3_12(j1)

# 7th composite
# I x H x H
l1 = DotProduct(l,k)
m = CNOT3_12(l1)
n = CNOT3_02(m)

#8th composite
# T x H x T1
o1 = DotProduct(o,n)
p = CNOT3_02(o1)
result = measurement(p)
plotMeasurement(result)

The measurement is identical to the that of IBM Quantum Experience

Conclusion:  This post looked at more Quantum gates. I have implemented all the gates in my QCSimulator which I hope to release in a couple of months.

Disclaimer: This article represents the author’s viewpoint only and doesn’t necessarily represent IBM’s positions, strategies or opinions

References
1. http://www1.gantep.edu.tr/~koc/qc/chapter4.pdf


text:
# If n is odd and n|(a-b) and n|(a+b) then n|a and n|b

Prove: If n is odd and n|(a-b) and n|(a+b) then n|a and n|b

since n is odd, there is some number k such that

$n=2k+1$

there are also integers p,q such that

$np=(a-b)$ and $nq = (a+b)$

(We want to show there are integers r,s such that)

$nr=a$

$ns=b$

I've only managed to make it this far into the proof, However, I've noted a few observations.

1. p and q could be either odd or even
2. a and b must be odd in order for their sum or difference to be divisible by n

Furthermore, my goal is to show that there are some integers r,s that are divisible by n.

I have attempted substitution, but it seems to be leading to dead ends.

-
n divides a-b and a+b hence n divides their sum (a-b)+(a+b)=2a. Now, you know that n divides 2a and you want to show that n divides a. Which is easy since... –  Did Nov 2 '11 at 18:36
Please do not use [homework] as the only tag in a post. Thank you. –  Arturo Magidin Nov 2 '11 at 18:37
Apologies @Arturo, I tried adding common-divisors as a tag, but deleted it after I realized it wasn't a valid tag. What would be a good alternative tag? –  user17366 Nov 3 '11 at 3:09
@arete: I added the [elementary-number-theory] tag, which covers this. –  Arturo Magidin Nov 3 '11 at 13:08

since $n|a-b$ and $n|a+b$ then $n|2a$ now since $n$ is odd then $n$ and $2$ are coprime, hence by Euclid's Lemma we have that $n|a$.
Similarly $n|-2b$ hence $n|2b$ and again by Euclid's Lemma $n|b$.
HINT. If $d$ divides $x$ and $y$, then it divides $x+y$ and $x-y$.
HINT 2. If $d$ divides $xz$, and $\gcd(d,z)=1$, then $d$ divides $x$.


text:
# Solving Linear Programming Via Big M method

$$\text{Max } x_1 +3x_2$$ $$\text{s.t } 3x_1+x_2\leq 3$$ $$x_1-x_2\geq 2$$ $$x_1,..,x_2\geq 0$$

So we first transform to the standard form:

$$\text{Min } -x_1 -3x_2$$ $$\text{s.t } 3x_1+x_2+x_3= 3$$ $$x_1-x_2-x_4= 2$$ $$x_1,..,x_4\geq 0$$

Because there is no basic feasible solution we will add and artificial variable to move to a canonical form:

$$\text{Min } -x_1 -3x_2$$ $$\text{s.t } 3x_1+x_2+x_3= 3$$ $$x_1-x_2-x_4+x_5= 2$$ $$x_1,..,x_5\geq 0$$

Using Big M we will add $$mx_5$$ to the objective function:

$$\text{Min } -x_1 -3x_2+mx_5$$ $$\text{s.t } 3x_1+x_2+x_3= 3$$ $$x_1-x_2-x_4+x_5= 2$$ $$x_1,..,x_5\geq 0$$

But now, it is is not in basic form as the objective function need to be written in the non basic variables, so we will add -M times equation 2 to the objective function:

$$\text{Min } (-m-1)x_1 +(m-3)x_2+mx_4=-m-1$$ $$\text{s.t } 3x_1+x_2+x_3= 3$$ $$x_1-x_2-x_4+x_5= 2$$ $$x_1,..,x_5\geq 0$$

Now we can start the simplex:

moving $$x_1$$ to the basis we get:

But here all the coefficients of the objective function is positive as $$m$$ is a big positive number, how can we continue? or we can not and we still have $$x_5$$ is the basis so there is no feasible solution?

• The latter: there is no feasible solution. – LinAlg Jul 8 at 12:39


text:
# Homework Help: Green's Theorem and polar coordinates

1. Dec 10, 2009

### sailsinthesun

1. The problem statement, all variables and given/known data
Using Green's Theorem, (Integral over C) -y^2 dx + x^2 dy=____________
with C: x=cos t y=sin t (t from 0-->2pi)

2. Relevant equations
(Integral over C) Pdx + Qdy=(Double integral over D) ((partial of Q w.r.t. x)-(partial of P w.r.t. y))dxdy

3. The attempt at a solution

I'm stuck from here. I remember the professor said to use polar coordinates which makes sense to get the limits on D, but how do I convert the integral (-y^2 dx + x^2 dy) to polar?

In my method I go from the original integral over C to (double integral over D) 2x+2y dxdy. I convert this to polar to get limits on D and I get (integral from 0 to 2pi)(integral from 0 to 1) 2rcos(theta)+2rsin(theta)*r*dr*d(theta). Once you calculate all of this you get 0 which I don't believe is correct. Any help?

Last edited: Dec 10, 2009
2. Dec 10, 2009

### LCKurtz

Why don't you think 0 is correct? It's quite a nice number, one of my favorites, actually.

3. Dec 10, 2009

### Staff: Mentor

People sometimes want answers in round numbers. You can't get any rounder than zero.


text:
Volume 301 - 35th International Cosmic Ray Conference (ICRC2017) - Session Solar & Heliospheric. SH-Terrestrial effects
Effect of field-aligned beam on upstream wave excitation and particle scattering in the earth’s foreshock: One-dimensional PIC simulation
F. Otsuka,* S. Matsukiyo, T. Hada
*corresponding author
Full text: pdf
Pre-published on: August 16, 2017
Published on: August 03, 2018
Abstract
The Earth's foreshock extends to a large domain of upstream quasi-parallel bow shock, and is characterized by a presence of field-aligned beams (FABs), diffuse ions, ultra-low frequency (ULF) waves, high frequency whistler waves, shocklets, and so on. Kinetic self-consistent numerical simulation is one of the key tools to analyze detailed physics of the foreshock which has not been clearly understood. Because of the necessity of the large simulation domain, a full particle-in-cell (PIC) simulation of quasi-parallel shock has seldom been performed.
In this paper we show preliminary results of a long-term and large-scale one-dimensional full PIC simulation of the quasi-parallel collisionless shock with the Alfven Mach number 6.6 and shock angle $20$ degrees. The FAB component is observed far upstream with the beam velocity of 10.5 times the Alfven velocity and the beam density of 0.5 % of the background plasma. This FAB generates right-handed Alfven waves in the plasma rest frame via resonant mode instability, and the excited waves are amplified as approaching the shock during the plasma convection. The number densities of energetic particles for both electrons and ions also increase as approaching the shock.
DOI: https://doi.org/10.22323/1.301.0078
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access


text:
# Sequence of polynomials that converges to $|x|$ over $[-1,1]$

Suppose I want to construct a sequence of polynomials that converges to $|x|$ pointwise.

I am pretty good on proving that sequences of functions converge to things pointwise, but I am having trouble actually coming up with a sequence of polynomials that converges to $|x|$. Someone told me to try using the Taylor series expansion of $(1−x)^{1/2}$, but that just confused me, since I am very rusty on Taylor series. Would anyone mind explaining this to me a little?

Thanks!!

$$P_{2n}(x)=\frac{2}{\pi}-\sum_{j=1}^{n}\frac{4(-1)^j}{\pi(4j^2-1)}T_{2j}(x)$$
converges uniformly to $|x|$ over the compact set $[-1,1]$, in virtue of the Fourier-Chebyshev series (the Chebyshev polynomials of the first kind are a complete base of orthonormal functions wrt the scalar product $<f,g>=\int_{-1}^{1}\frac{f(x)g(x)dx}{\sqrt{1-x^2}}$).
$$\sup_{x\in(-1,1)}||x|-P_{2n}(x)|=O\left(\frac{1}{n}\right).$$


text:
## Section Exercises

1. What situations are best modeled by a logistic equation? Give an example, and state a case for why the example is a good fit.

2. What is a carrying capacity? What kind of model has a carrying capacity built into its formula? Why does this make sense?

3. What is regression analysis? Describe the process of performing regression analysis on a graphing utility.

4. What might a scatterplot of data points look like if it were best described by a logarithmic model?

5. What does the y-intercept on the graph of a logistic equation correspond to for a population modeled by that equation?

For the following exercises, match the given function of best fit with the appropriate scatterplot. Answer using the letter beneath the matching graph.

6. $y=10.209{e}^{-0.294x}$

7. $y=5.598 - 1.912\mathrm{ln}\left(x\right)$

8. $y=2.104{\left(1.479\right)}^{x}$

9. $y=4.607+2.733\mathrm{ln}\left(x\right)$

10. $y=\frac{14.005}{1+2.79{e}^{-0.812x}}$

11. To the nearest whole number, what is the initial value of a population modeled by the logistic equation $P\left(t\right)=\frac{175}{1+6.995{e}^{-0.68t}}$? What is the carrying capacity?

12. Rewrite the exponential model $A\left(t\right)=1550{\left(1.085\right)}^{x}$ as an equivalent model with base e. Express the exponent to four significant digits.

13. A logarithmic model is given by the equation $h\left(p\right)=67.682 - 5.792\mathrm{ln}\left(p\right)$. To the nearest hundredth, for what value of p does $h\left(p\right)=62$?

14. A logistic model is given by the equation $P\left(t\right)=\frac{90}{1+5{e}^{-0.42t}}$. To the nearest hundredth, for what value of t does $P\left(t\right)=45$?

15. What is the y-intercept on the graph of the logistic model given in the previous exercise?

For the following exercises, use this scenario: The population P of a koi pond over x months is modeled by the function $P\left(x\right)=\frac{68}{1+16{e}^{-0.28x}}$.

16. Graph the population model to show the population over a span of 3 years.

17. What was the initial population of koi?

18. How many months will it take before there are 20 koi in the pond?

19. Use the intersect feature to approximate the number of months it will take before the population of the pond reaches half its carrying capacity.

For the following exercises, use this scenario: The population P of an endangered species habitat for wolves is modeled by the function $P\left(x\right)=\frac{558}{1+54.8{e}^{-0.462x}}$, where x is given in years.

20. Graph the population model to show the population over a span of 10 years.

21. What was the initial population of wolves transported to the habitat?

22. How many wolves will the habitat have after 3 years?

23. How many years will it take before there are 100 wolves in the habitat?

24. Use the intersect feature to approximate the number of years it will take before the population of the habitat reaches half its carrying capacity.

For the following exercises, refer to the table below.

 x f(x) 1 1125 2 1495 3 2310 4 3294 5 4650 6 6361

25. Use a graphing calculator to create a scatter diagram of the data.

26. Use the regression feature to find an exponential function that best fits the data in the table.

27. Write the exponential function as an exponential equation with base e.

28. Graph the exponential equation on the scatter diagram.

29. Use the intersect feature to find the value of x for which $f\left(x\right)=4000$.

For the following exercises, refer to the table below.

 x f(x) 1 555 2 383 3 307 4 210 5 158 6 122

30. Use a graphing calculator to create a scatter diagram of the data.

31. Use the regression feature to find an exponential function that best fits the data in the table.

32. Write the exponential function as an exponential equation with base e.

33. Graph the exponential equation on the scatter diagram.

34. Use the intersect feature to find the value of x for which $f\left(x\right)=250$.

For the following exercises, refer to the table below.

 x f(x) 1 5.1 2 6.3 3 7.3 4 7.7 5 8.1 6 8.6

35. Use a graphing calculator to create a scatter diagram of the data.

36. Use the LOGarithm option of the REGression feature to find a logarithmic function of the form $y=a+b\mathrm{ln}\left(x\right)$ that best fits the data in the table.

37. Use the logarithmic function to find the value of the function when = 10.

38. Graph the logarithmic equation on the scatter diagram.

39. Use the intersect feature to find the value of x for which $f\left(x\right)=7$.

For the following exercises, refer to the table below.

 x f(x) 1 7.5 2 6 3 5.2 4 4.3 5 3.9 6 3.4 7 3.1 8 2.9

40. Use a graphing calculator to create a scatter diagram of the data.

41. Use the LOGarithm option of the REGression feature to find a logarithmic function of the form $y=a+b\mathrm{ln}\left(x\right)$ that best fits the data in the table.

42. Use the logarithmic function to find the value of the function when = 10.

43. Graph the logarithmic equation on the scatter diagram.

44. Use the intersect feature to find the value of x for which $f\left(x\right)=8$.

For the following exercises, refer to the table below.

 x f(x) 1 8.7 2 12.3 3 15.4 4 18.5 5 20.7 6 22.5 7 23.3 8 24 9 24.6 10 24.8

45. Use a graphing calculator to create a scatter diagram of the data.

46. Use the LOGISTIC regression option to find a logistic growth model of the form $y=\frac{c}{1+a{e}^{-bx}}$ that best fits the data in the table.

47. Graph the logistic equation on the scatter diagram.

48. To the nearest whole number, what is the predicted carrying capacity of the model?

49. Use the intersect feature to find the value of x for which the model reaches half its carrying capacity.
For the following exercises, refer to the table below.

 x f(x) 0 12 2 28.6 4 52.8 5 70.3 7 99.9 8 112.5 10 125.8 11 127.9 15 135.1 17 135.9

50. Use a graphing calculator to create a scatter diagram of the data.

51. Use the LOGISTIC regression option to find a logistic growth model of the form $y=\frac{c}{1+a{e}^{-bx}}$ that best fits the data in the table.

52. Graph the logistic equation on the scatter diagram.

53. To the nearest whole number, what is the predicted carrying capacity of the model?

54. Use the intersect feature to find the value of x for which the model reaches half its carrying capacity.

55. Recall that the general form of a logistic equation for a population is given by $P\left(t\right)=\frac{c}{1+a{e}^{-bt}}$, such that the initial population at time t = 0 is $P\left(0\right)={P}_{0}$. Show algebraically that $\frac{c-P\left(t\right)}{P\left(t\right)}=\frac{c-{P}_{0}}{{P}_{0}}{e}^{-bt}$.

56. $\frac{c-P\left(t\right)}{P\left(t\right)}=\frac{c-\frac{c}{1+a{e}^{-bt}}}{\frac{c}{1+a{e}^{-bt}}}=\frac{\frac{c\left(1+a{e}^{-bt}\right)-c}{1+a{e}^{-bt}}}{\frac{c}{1+a{e}^{-bt}}}=\frac{\frac{c\left(1+a{e}^{-bt}-1\right)}{1+a{e}^{-bt}}}{\frac{c}{1+a{e}^{-bt}}}=1+a{e}^{-bt}-1=a{e}^{-bt}$

57. Working with the right side of the equation we show that it can also be rewritten as $a{e}^{-bt}$. But first note that when t = 0, ${P}_{0}=\frac{c}{1+a{e}^{-b\left(0\right)}}=\frac{c}{1+a}$. Therefore,

58. Use a graphing utility to find an exponential regression formula $f\left(x\right)$ and a logarithmic regression formula $g\left(x\right)$ for the points $\left(1.5,1.5\right)$ and $\left(8.5,\text{ 8}\text{.5}\right)$. Round all numbers to 6 decimal places. Graph the points and both formulas along with the line = x on the same axis. Make a conjecture about the relationship of the regression formulas.

59. Verify the conjecture made in the previous exercise. Round all numbers to six decimal places when necessary.

60. Find the inverse function ${f}^{-1}\left(x\right)$ for the logistic function $f\left(x\right)=\frac{c}{1+a{e}^{-bx}}$. Show all steps.

61. Use the result from the previous exercise to graph the logistic model $P\left(t\right)=\frac{20}{1+4{e}^{-0.5t}}$ along with its inverse on the same axis. What are the intercepts and asymptotes of each function?


text:
# Linked Questions

12 questions linked to/from Units of mass on the atomic scale
1answer
1k views

### Are atomic mass unit (amu), unified mass unit (u) and Dalton (Da) the same unit but with different names? [duplicate]

I recently read up on some chemistry and found these units of mass: atomic mass unit (amu) unified atomic mass unit (u) Dalton (Da) Are they all the same unit but just with different names or are ...
1answer
106 views

### Why is the atomic mass unit less than the mass of both a neutron and a proton? [duplicate]

The atomic mass unit is $1.6605 \times 10^{-27}$ kg. This is less than the mean of the masses of 6 protons and 6 neutrons. How do we account for the lower mass ? My understanding is that some of the ...
0answers
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### Atomic weight and its units [duplicate]

I am quite confused when it comes to this, there are quite a lot of definitions of the above terms in numerous books and websites. However the definitions do not match each other. So, here are the two ...
3answers
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### What is the difference between “molecular mass”, “average atomic mass” and “molar mass”?

I don't understand the difference between "molecular mass" and "average atomic mass". They seem like the same thing to me. Is it that average atomic mass is just the weighted average of the "weights"/...
2answers
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### Quick and simple explanation of molar mass, molecular mass and atomic mass

As a physics student, I hardly deal with such quantities and when I do, I blunder through. I think it’s a good time to be given a good explanation for these since I’m in a class of Nuclear physics. I ...
2answers
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### Mass number, (relative) atomic mass, average mass

What is the difference between mass number, atomic mass and average atomic mass? I know the mass number is the amount of protons + the amount of neutrons. The average mass is the weighed average of ...
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### When to use Da or u for mass units

Following up on the question Units of mass on the atomic scale, Daltons (Da) and unified mass units (u) are equivalent, but which is used seems to depend on context. For example, when discussing ...
1answer
907 views

### why “amu” is used as a unit in relative atomic mass? [duplicate]

Unified atomic mass unit (amu) is defined as $1/12$ of the mass of a Carbon-12. I want to know is this amu equal to $\pu{1.99e-23 g}$ and why do we use amu which is $1/12$ the mass of carbon? Why use "...
1answer
2k views

### Difference between atomic mass, mass number, atomic weight, and relative atomic mass [duplicate]

I think this question is common, but i want a little help. I'm little confused. What I concluded. Relative atomic mass/Atomic weight (both are exactly the same) - It is the mass of atom calculated as ...
1answer
227 views

### Why the term 'unified' in “unified mass unit”? [closed]

Why different unit symbols - amu, u and dalton are used for representing atomic mass? And what does the term 'unified' mean in "unified mass unit"? Thank you.
1answer
110 views

### The unit of relative atomic mass [closed]

The mass of Carbon-12 atom is 12.0000000 a.m.u. For relative isotopic mass, e.g. Mg-24 atom is 24 compare with the c-12 as standard. However, why is it unitless? Shouldn't it be 24 a.m.u.?
1answer
104 views

### Can anyone give me detailed logical solution to my questions? [closed]

What is 1 amu? From where it is derived? Why it is compared with carbon-12? Why it was once compared with hydrogen-1? and now its not? Can anyone please explain me logically? And what would have ...


text:
# [Quick] States of interacting and non-interacting Hamiltonian

1. ### QuantumDevil

29
Eigenstates of interacting and non-interacting Hamiltonian

Have multi-particle state of full Hamiltonian and one-particle state of free Hamiltonian non-zero scalar product? Intuitively one can say that scalar product of such states should be zero because each of these states mentioned above belongs to different (orthogonal) subspaces of the Fock space.
Do you know any reference discussing this problem?

P.S. these states are eigenvectors of appropriate Hamiltonians

Last edited: Dec 25, 2007
2. ### olgranpappy

1,273
The multi-particle state of the full hamiltonian can be expanded as a sum of free single-particle and many-particle states with various coefficients. Are any of the coefficient in front of the free single-particle states non-zero?

if that were the case then the scalar product would obviously be zero. but are you sure that is the case? If they were eigenvectors of *the same* operator with different eigenvalues, then they would certainly be orthogonal, but they are eigenvectors of different operators.

3. ### QuantumDevil

29
I don't know. Therefore I asked question in the first message.

4. ### strangerep

2,246
[This is definitely not a "quick" discussion topic. :-)] If you're talking about full multiparticle
interacting QFT in 4D, the Fock space constructed by polynomials of (free) creation operators
acting on a vacuum state is disjoint from the Hilbert space corresponding to eigenstates
of the full interacting Hamiltonian. The latter Hilbert space is not merely a subspace of
the free Fock space, but a completely disjoint space (hence scalar products between states
from the two spaces are conventionally defined to be zero).

If you search for "Haag's theorem" and "unitarily inequivalent representations" that
should turn up more info on this, at least in the context of orthodox formulations of QFT.

5. ### QuantumDevil

29
I dont' actually understand the notion "full multiparticle
interacting QFT in 4D" introduced by You and the difference between it and "full interacting Hamiltonian". (Maybe the reason is that I'm not native English speaker).
Could you explain it in more detailed way or give me some references?

By "full interacting Hamiltonian" I don't mean interaction term, of course.

Last edited: Dec 28, 2007
6. ### strangerep

2,246
I'll write $H_0$ to mean the free Hamiltonian. That is, the Hamiltonian in the
absence of interactions.

I'll write $H_{int}$ to mean the interaction. (Some people write $V$ for

Then the full interacting Hamiltonian is $H = H_0 + H_{int}$

By "multiparticle Hamiltonian", I meant a Hamiltonian that describes the dynamical
evolution of many particles (rather than just a single particle). Actually, this is a poor
term, and I should have said "infinite degrees of freedom" instead of "multiparticle".

"4D" just means that it's for 3+1 spacetime.

Peskin & Schroeder is the obvious reference that immediately comes to mind
for basic QFT stuff. That should clarify the distinction between the "free" and
"interaction" parts of the Hamiltonian.

But P&S don't talk much about the disjointness between "free" and "interacting"
Hilbert spaces for infinite degrees of freedom. Try Umezawa for that.


text:
# American Institute of Mathematical Sciences

July  2016, 21(5): 1435-1444. doi: 10.3934/dcdsb.2016004

## Convergence rate of free boundary of numerical scheme for American option

 1 Department of Mathematics, University of Pittsburgh, Pittsburgh, PA 15260 2 Department of Applied and Computational Mathematics and Statistics, University of Notre Dame, Notre Dame, IN 46556 3 Department of Mathematics, Tongji University, Shanghai 200092 4 School of Mathematical Sciences, Shanxi University, Taiyuan, Shanxi 030006

Received  September 2013 Published  April 2016

Based on the optimal estimate of convergence rate $O(\Delta x)$ of the value function of an explicit finite difference scheme for the American put option problem in [6], an $O(\sqrt{\Delta x})$ rate of convergence of the free boundary resulting from a general compatible numerical scheme to the true free boundary is proven. A new criterion for the compatibility of a generic numerical scheme to the PDE problem is presented. A numerical example is also included.
Citation: Xinfu Chen, Bei Hu, Jin Liang, Yajing Zhang. Convergence rate of free boundary of numerical scheme for American option. Discrete & Continuous Dynamical Systems - B, 2016, 21 (5) : 1435-1444. doi: 10.3934/dcdsb.2016004
##### References:

show all references

##### References:
 [1] Wen Li, Song Wang. Pricing American options under proportional transaction costs using a penalty approach and a finite difference scheme. Journal of Industrial & Management Optimization, 2013, 9 (2) : 365-389. doi: 10.3934/jimo.2013.9.365 [2] Wen Chen, Song Wang. A finite difference method for pricing European and American options under a geometric Lévy process. Journal of Industrial & Management Optimization, 2015, 11 (1) : 241-264. doi: 10.3934/jimo.2015.11.241 [3] Walter Allegretto, Yanping Lin, Ningning Yan. A posteriori error analysis for FEM of American options. Discrete & Continuous Dynamical Systems - B, 2006, 6 (5) : 957-978. doi: 10.3934/dcdsb.2006.6.957 [4] Xinfu Chen, Huibin Cheng. Regularity of the free boundary for the American put option. Discrete & Continuous Dynamical Systems - B, 2012, 17 (6) : 1751-1759. doi: 10.3934/dcdsb.2012.17.1751 [5] Matúš Tibenský, Angela Handlovičová. Convergence analysis of the discrete duality finite volume scheme for the regularised Heston model. Discrete & Continuous Dynamical Systems - S, 2021, 14 (3) : 1181-1195. doi: 10.3934/dcdss.2020226 [6] Donny Citra Lesmana, Song Wang. A numerical scheme for pricing American options with transaction costs under a jump diffusion process. Journal of Industrial & Management Optimization, 2017, 13 (4) : 1793-1813. doi: 10.3934/jimo.2017019 [7] María Teresa V. Martínez-Palacios, Adrián Hernández-Del-Valle, Ambrosio Ortiz-Ramírez. On the pricing of Asian options with geometric average of American type with stochastic interest rate: A stochastic optimal control approach. Journal of Dynamics & Games, 2019, 6 (1) : 53-64. doi: 10.3934/jdg.2019004 [8] Navnit Jha. Nonpolynomial spline finite difference scheme for nonlinear singuiar boundary value problems with singular perturbation and its mechanization. Conference Publications, 2013, 2013 (special) : 355-363. doi: 10.3934/proc.2013.2013.355 [9] Xiaoyu Xing, Hailiang Yang. American type geometric step options. Journal of Industrial & Management Optimization, 2013, 9 (3) : 549-560. doi: 10.3934/jimo.2013.9.549 [10] Ian Knowles, Ajay Mahato. The inverse volatility problem for American options. Discrete & Continuous Dynamical Systems - S, 2020, 13 (12) : 3473-3489. doi: 10.3934/dcdss.2020235 [11] Ugur G. Abdulla. On the optimal control of the free boundary problems for the second order parabolic equations. II. Convergence of the method of finite differences. Inverse Problems & Imaging, 2016, 10 (4) : 869-898. doi: 10.3934/ipi.2016025 [12] Ronald E. Mickens. A nonstandard finite difference scheme for the drift-diffusion system. Conference Publications, 2009, 2009 (Special) : 558-563. doi: 10.3934/proc.2009.2009.558 [13] Yaru Xie, Genqi Xu. The exponential decay rate of generic tree of 1-d wave equations with boundary feedback controls. Networks & Heterogeneous Media, 2016, 11 (3) : 527-543. doi: 10.3934/nhm.2016008 [14] Nan Li, Song Wang, Shuhua Zhang. Pricing options on investment project contraction and ownership transfer using a finite volume scheme and an interior penalty method. Journal of Industrial & Management Optimization, 2020, 16 (3) : 1349-1368. doi: 10.3934/jimo.2019006 [15] Jian Su, Yinnian He. The almost unconditional convergence of the Euler implicit/explicit scheme for the three dimensional nonstationary Navier-Stokes equations. Discrete & Continuous Dynamical Systems - B, 2017, 22 (9) : 3421-3438. doi: 10.3934/dcdsb.2017173 [16] Shenglan Xie, Maoan Han, Peng Zhu. A posteriori error estimate of weak Galerkin fem for second order elliptic problem with mixed boundary condition. Discrete & Continuous Dynamical Systems - B, 2021, 26 (10) : 5217-5226. doi: 10.3934/dcdsb.2020340 [17] Weizhu Bao, Chunmei Su. Uniform error estimates of a finite difference method for the Klein-Gordon-Schrödinger system in the nonrelativistic and massless limit regimes. Kinetic & Related Models, 2018, 11 (4) : 1037-1062. doi: 10.3934/krm.2018040 [18] Yoshiho Akagawa, Elliott Ginder, Syota Koide, Seiro Omata, Karel Svadlenka. A Crank-Nicolson type minimization scheme for a hyperbolic free boundary problem. Discrete & Continuous Dynamical Systems - B, 2021  doi: 10.3934/dcdsb.2021153 [19] Pierangelo Ciurlia. On a general class of free boundary problems for European-style installment options with continuous payment plan. Communications on Pure & Applied Analysis, 2011, 10 (4) : 1205-1224. doi: 10.3934/cpaa.2011.10.1205 [20] Anouar El Harrak, Hatim Tayeq, Amal Bergam. A posteriori error estimates for a finite volume scheme applied to a nonlinear reaction-diffusion equation in population dynamics. Discrete & Continuous Dynamical Systems - S, 2021, 14 (7) : 2183-2197. doi: 10.3934/dcdss.2021062

2020 Impact Factor: 1.327


text:
×

# Number Theory Problem

What are the 2 numbers below 260 that would divide $$2^{32}-1$$?

Note by Swapnil Das
1 year, 11 months ago

Sort by:

This number is divided by 3, 5. ,15, 17 etc @Swapnil Das · 1 year, 11 months ago

Ok, I get it. Thanks! · 1 year, 11 months ago


text:
# All Questions

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text:
## Return to Question

2 Corrected link

It is known that there are non-Hausdorff spaces which admit unique limits for all convergent sequence (see here) and it is also known that unique limits for nets implies Hausdorff.

What I am wondering is, if there is a (somehow weak) condition which one should add to "unique limits of sequences" to obtain a Hausdorff space. Would, for example, some countability help?

Somehow in the same direction: What is the central property which is needed for a space such that it can be non-Hausdorff but has unique sequence limits? Is there a whole class of non-Hausdorff spaces which admit unique limits for convergent sequence?

1

# Unique limits of sequences plus what implies Hausdorff?

It is known that there are non-Hausdorff spaces which admit unique limits for all convergent sequence (see here) and it is also known that unique limits for nets implies Hausdorff.

What I am wondering is, if there is a (somehow weak) condition which one should add to "unique limits of sequences" to obtain a Hausdorff space. Would, for example, some countability help?

Somehow in the same direction: What is the central property which is needed for a space such that it can be non-Hausdorff but has unique sequence limits? Is there a whole class of non-Hausdorff spaces which admit unique limits for convergent sequence?


text:
# MOBILE Proof of Coverage

## Introduction

The Helium blockchain uses a novel work algorithm called "Proof of Coverage" ("PoC") to verify that Hotspots are located where they claim. Put another way, PoC tries to verify, on an ongoing basis, that Hotspots are honestly representing their location and the wireless network coverage they are creating from that location.

While current 5G Hotspots are also able to participate in the IOT Network and PoC, the CBRS Small Cell and 5G aspects of a 5G Hotspot require a new approach to validating PoC.

Modeled Coverage introduces a fundamental difference in using hexes for the MOBILE Network compared to the IOT Network where hexes determine the locations and number of Hotspots within the given hex. In the MOBILE Network, hexes determine where the coverage exists. Modeled Coverage data allows us to un-tie a Radio's location from its signal strength and propagation.

One Way Traffic

A CBRS Small Cell Radio ("Radio") can only operate as a transmitter; they cannot receive incoming data and thus can not "hear" one another directly.

LoRaWAN Hotspots utilize a PoC method that benefits from Hotspots being used as senders ("Beacon") and receivers ("Witness") of PoC activity; read more here: Proof of Coverage

## Proof of Coverage Milestones

Determining useful coverage and the impact on MOBILE PoC rewards will be based on a number of factors:

These factors will be rolled out in stages over the next twelve months and explained in detail to ensure Network Builders can take concrete steps to plan or improve their setups. Check out the MOBILE PoC Roadmap here.

## Reward Algorithm

The proposed new algorithm for MOBILE Reward calculation in the MOBILE Rewards Oracle is as follows:

1. Supply the declared transmitter power of each Radio and its location to the Obstruction Data Oracle.
2. Get all hexes that have coverage from Outdoor Radios based on the information from the Obstruction Data Oracle.
3. Based on the location of Indoor Radios, get all hexes with Indoor coverage and all adjacent hexes.
4. Use projected signal loss information from the Obstruction Data Oracle to determine the potential signal strength of each Outdoor Radio in each hex.
5. For each hex, get at most 5 Outdoor Radios with the top signal strength of the same level.
• If there are more than 5 Radios with the same signal strength level, use the coverage_claim_time value to determine the top 5 oldest installations where coverage_claim_time is the timestamp when the Radio received the spectrum access grant for the first time.
• To prevent rewarding "dead" Radios, we propose to reset coverage_claim_time if the Radio was not generating a Heartbeat for more than 72 hours and use the time of the last Heartbeat as the new coverage_claim_time.
1. Get max 5 Indoor Radios using the same approach as above for Outdoor Radios.
2. Based on Modeled Coverage Reward Tiers, sum up all estimated coverage points earned by each Radio in all hexes and multiply that by speedtest_multiplier for each Radio.
3. Divide the total number of MOBILE Rewards emitted during the Rewards Period by the sum of multiples of heartbeat_multiplier, estimated_reward_points, and speedtest_multiplier for each Radio to determine MOBILE Rewards per estimated_reward_point for Radio with a passing Speed Test and valid Heartbeats.
4. Multiply the MOBILE per estimated_reward_point by the sum of each Radio's reward points to determine the MOBILE Rewards for each Radio.

The new formula for Reward calculation per Radio:

$W = k_H \times k_S \times C_E \times W_p$
VariableDescription
$W$Total rewards
$k_H$Heartbeat multiplier
$k_S$Speedtest multiplier
$C_E$Estimated coverage points
$W_p$Epoch rewards per coverage point

### Calculation Example

For simplicity, assume that the total MOBILE Rewards per period is 10,000.

RadioHeartbeatHeartbeat multiplier $k_H$Speed TestSpeedtest Multiplier $k_S$Hex 1 - Hex 10Hex 11 - Hex 20Hex 21 - Hex 220Total Coverage PointsTotal Reward Points
1 (Outdoor)Ok1Acceptable1160808001,0401,040
2 (Outdoor)Ok1Poor0.25804012030
Points1,070

Table 3. Simplified data for two Outdoor Radios with Heartbeat, Speed Test, and Estimated Coverage Points for one Reward Period.

RadioHeartbeatHeartbeat multiplier $k_H$Speed TestSpeedtest Multiplier $k_S$Hex 1Hex 2Hex 3Hex 4Total Coverage PointsTotal Reward Points
Points350

Table 4. Simplified data for one Indoor Radio with Heartbeat, Speed Test, and Estimated Coverage Points for one Reward Period.

$R \times (1 \times 1 \times 1040 + 1 \times 0.25 \times 120 + 1 \times 0.5 \times 700) = 10,000$

Where R is the reward per one estimated coverage point.

$R = 6.54$
CalcuationTotal MOBILE Rewards
Radio 1 (Outdoor)6.54 x 1 x 1 x 10406,806.30
Radio 2 (Outdoor)6.54 x 1 x 0.25 x 120903.14
Radio 3 (Indoor)6.54 x 1 x 0.5 x 7002,290.58

## Modeled Coverage

With the passing of HIP-74 Modeled Coverage aims to allow for more appropriate Rewards and incentivizes the deployment of Radios at optimal locations.

This new algorithm uses the location of the Radio to calculate MOBILE Rewards based on the number of res12 hexes receiving coverage from a given Radio.

The MOBILE Rewards Oracle will calculate a coverage model for each Radio by considering the physical location, antenna direction, radio max transmit power, and Obstruction Data Oracle inputs to calculate the number of res12 hexes receiving coverage from each Radio.

Depending on the number of res12 hexes and the resulting signal power in each, the MOBILE Rewards Oracle will allot Coverage Points to a given Radio for calculating its MOBILE Rewards instead of assuming a fixed Radio Type multiplier Reward.

### Reward Hexagon Sizing

Analysis of the coverage by a single Outdoor Radio with Modeled Coverage data shows that the res8 hex used in the IOT Network is too large to represent a Radio's coverage. Using res12 hexes better aligns with what an Outdoor Radio can reasonably cover, with an average area of 0.0003071 km² and an edge length of 9.4 m, slightly bigger than an average single-family home.

To summarize:

• Hexes in the MOBILE Network identify the level of coverage from all the different Radios providing coverage to each res12 hex.
• A Radio can provide coverage for a hex it is not located within.
• A Radio can earn Rewards by providing coverage in multiple hexes.

To help visualize the significant difference between these two resolutions, below is a map with a green res8 hex and a purple res12 hex inside.

### Modeled Coverage Reward Tiers

info

Each Radio is given a coverage points value depending on the number and signal strength of the res12 hexes receiving coverage from the Radio.

Modeled Coverage replaces Radio Type multipliers with coverage points represented by four tiers of potential signal power as modeled by the Obstruction Data Oracle - High, Medium, Low, and None.

The reward tiers for Outdoor Radios will be as follows:

Tier 1Tier 2Tier 3Tier 4
Potential Signal Power$P > -95 dBm$$-95 dBm \ge P > -105 dBm$$-105 dBm \ge P > -115 dBm$$P \le -115 dBm$
Potential Signal LevelHighMediumLowNone
Estimated coverage points16840
Modeled Coverage Distance Limit for Outdoor Radios

The Modeled Coverage algorithm limits the radius of coverage for a given Radio to 1000 meters. This value results from evaluating a selection of "ideal case" Radios, which provided usable coverage to roughly 500 meters and doubled that value. A larger radius would also be too heavy for the computation and is a current trade-off.

Like Outdoor, Indoor Radios can collect MOBILE Rewards by providing coverage in multiple hexes.

The Obstruction Data Oracle cannot evaluate coverage by Indoor Radios as there are no good and reliable data sources about obstacles in indoor spaces. Instead, the Modeled Coverage Oracle uses an approximation based on the data gathered while testing certified Indoor Radios in various indoor settings.

From this testing, the hex in which an Indoor Radio is physically located receives maximum signal strength coverage, and all adjacent hexes receive lower signal strength. Therefore, the algorithm errs on generosity to ensure equitable potential MOBILE Rewards for Indoor Radio coverage.

Estimated per-hex coverage points for Indoor Radios are intentionally significantly higher than those awarded to Outdoor Radios to balance the importance of Indoor Radios in the MOBILE Network with their relatively smaller coverage footprints.

The reward tiers for Indoor Radios will be as follows:

Tier 1Tier 2
Potential Signal LevelHighLow
Estimated coverage points400100

These values were chosen by taking all the Outdoor Radios in an example area and analyzing the resulting distribution of estimated coverage points per Radio . Estimated coverage point values were chosen such that Indoor Radios receive roughly ¼th the estimated coverage points as the 95th percentile Outdoor Radios and about ½ the points of the average Outdoor Radio.

These results align with the current Radio Type Multiplieres approach 1:2.5:4 for Indoor Radios, Outdoor Radios, and High-Power Outdoor Radios. Figure 2 below is the cumulative distribution function plot of the sample market Outdoor Radio estimated coverage points.

Cumulative distribution function of Outdoor Radio estimated coverage points for 904 Radios in the L.A. market area.

### Modeled Coverage Explorer

check coverage

Modeled Coverage Explorer is available here.

Modeled Coverage data will play a key role in providing information to visualize it, with more data added as additional external sources (e.g., Mobile Mappers) come into play.

HIP-74 proposed the creation of a new Mobile Explorer dedicated exclusively to the MOBILE Network as adding more features and data for MOBILE Rewards to the IOT Network Explorer becomes practically impossible as the MOBILE Network grows. The first iteration of the MOBILE Explorer proposes a basic map overlay of 5G data coverage with signal strength.

## Uptime Heartbeats

A "heartbeat" is data sent by the 5G Hotspot indicating that a connected Radio is authorized to transmit 5G coverage.

Heartbeats occur every 60 seconds and transmit authorizations are valid for 240 seconds. This rolling overlap ensures that a Small Cell Radio can broadcast on an ongoing basis while allowing for brief interruptions.

To encourage reliable signal availability, 5G Hotspots and connected CBRS Small Cell Radios must generate at least one heartbeat per hour in 12 unique hours of the 24-hour Reward Period.

### Heartbeat Reward Tiers

Each Radio is given a heartbeat multiplier of either 0 or 1.

Each Radio is awarded 1 point for a successful heartbeat in each hour, with a maximum of 24 points in the Reward Period. All Radios with at least 12 points in the 24-hour Reward Period are given a heartbeat multiplier of 1 and will be eligible to earn MOBILE Rewards.

### Heartbeat Examples

Choose from the options below to see some heartbeat example scenarios and the impact on MOBILE rewards.

The Radio generated a heartbeat every hour of the day, earning 24 points.

This Radio will be eligible for rewards during the Reward Period.

## Backhaul Speed Tests

A 5G Hotspot automatically performs Speed Tests, with the results averaged by the MOBILE PoC Oracle. The MOBILE PoC Oracle includes these Speed Test metrics in the Rewards calculations for each 5G Hotspot.

The 5G Hotspot will randomly measure download, upload, and latency to the MOBILE verifier twice in a given 24-hour period. If a test can not complete successfully, the 5G Hotspot automatically retries every 30 minutes until a Speed Test is successful. The first Speed Test for a newly installed 5G Hotspot is performed within the first 12 hours after powering it on.

A minimum of two and maximum of six Speed Tests are required in the previous 48 hours to calculate a Moving average results for each download, upload, and latency, after which a new series of Speed Test results must be collected before a Moving Average can be calculated.

### Speed Test Reward Tiers

Speed test impact on Radio rewards

Each 5G Hotspot is given a speedtest multiplier of either 0.0, 0.25, 0.5 or 1.0.

This multiplier is applied to all connected Radios in rewards calculations.

Many locations where CBRS connectivity is being deployed, including some rural areas, do not always have the high-speed Internet connectivity needed to meet the acceptable Internet requirements for Genesis rewards consistently.

Often these areas do not have good cellular coverage either. That's why we believe it is essential to incentivize Helium deployments in less well-connected areas.

In response to community feedback and further exploration by Chris from DeWiPulse for his analysis of and commentary on public Internet data in the United States, Speed Test results are categorized into one of four Tiers - Acceptable, Degraded, Poor, and Fail.

Speed Test TierSpeedtest MultiplierRequirement (speeds in Mbps, latency in ms)
Acceptable1.00x100+ Download, AND 10+ Upload, AND <50 Latency
Degraded0.50x50+ Download, AND 5+ Upload, AND <75 Latency
Poor0.25x30+ Download, AND 2+ Upload, AND <100 Latency
Fail0.00x<30 Download, OR <2 Upload, OR >100 Latency

Tiered Speed Test values are used as a multiplier in Rewards calculations as follows:

• Speed Test Results are put into Tiers based on the minimum value of each Download, Upload, and Latency (logical AND).
• Speed Test results that do not meet the minimum requirements for any Download, Upload, or Latency are considered to have Failed and are not eligible for Rewards until the Speed Test Average is improved (logical OR).
Requirements Will Evolve

As the needs and realities of the Network change, backhaul requirements will evolve to better suit the Network and its users.

### Speedtest Examples

Choose from the options below to see some speedtest example scenarios and the impact on MOBILE rewards.

Download    200 ->  AcceptableUpload       40 ->  AcceptableLatency      15 ->  Acceptable---------------------------------Speed Test Tier ->  Acceptable  -> 1.0x Multiplier

### Speed Test Example Calculations

Here are a set of tables to help visualize the components that make up the Speed Test result.

info

It may take up to 12 hours for newly connected 5G Hotspots to perform a Speed Test, and a minimum of 2 Speed Tests are required for the MOBILE PoC Oracle to calculate a moving average.

This 5G Hotspot is passing all Speed Test metrics and is providing a high-quality connection.

Test 1

This 5G Hotspot will not be eligible for rewards during the Reward Periods for which the results of Test 1 are the most recent as there is not enough data to calculate a Moving Average.

Test 2 and Later

This 5G Hotspot will be eligible for full rewards during the Reward Periods for which the results of these are the most recent.

## Statistics on Helium Explorer

Speed Test metrics are viewable on the 5G Hotspot's page on Explorer under the 5G Statistics tab. Here are examples of each status and a brief description of each metric and the resulting impact on Rewards calculations.

### Helium Explorer Examples

Choose from the options below to see some example scenarios and the impact on MOBILE rewards.

Speed Test Passing - Acceptable

The Moving Average for all Speed Test metrics is in the Acceptable range.

A 1.0x multiplier is applied for the current Reward Period.

The Moving Average for Latency metrics is in the Acceptable range, however the 10 Mbps Download and 1 Mbps Upload metrics are below the Poor Tier.
A 0.0x multiplier is applied for the current Reward Period.


text:
## LaTeX forum ⇒ Text Formatting ⇒ Refer to Theorem that contains several Parts

Information and discussion about LaTeX's general text formatting features (e.g. bold, italic, enumerations, ...)
Mohsen Avaji
Posts: 14
Joined: Mon Aug 27, 2012 12:06 am

### Refer to Theorem that contains several Parts

Hi,

in books and papers often a theorem contains several parts, labeled (i), (ii), ...
\begin{theorem}   Let $A$ be a finite set and $f$ a map from $A$ to itself. The following statements are equivalent:   \begin{itemize}     \item[(i)] $f$ is injective.     \item[(ii)] $f$ is surjective.     \item[(iii)] $f$ is a bijection.   \end{itemize}\end{theorem}

I want refer to part(i) of this theorem but I do not know how to do it. How can I do that? I'd be very grateful for any help.
Last edited by localghost on Mon Jun 24, 2013 9:53 am, edited 1 time in total.
Reason: Source code has to be tagged as such to keep a post clear and legible (see Board Rules).
Yours faithfully,
Mohsen Avaji.

Tags:

localghost
Site Moderator
Posts: 9204
Joined: Fri Feb 02, 2007 12:06 pm
Please get used to always preparing a self-contained and minimal example. People are rarely motivated to write one on their own just to test solutions for you.

In your case you are better with using an {enumerate} environment instead of {itemize}. It can be customized easily by enumitem. The single list items can be referenced by the common mechanism of the \label and \ref commands.
\documentclass[11pt]{article}\usepackage[T1]{fontenc}\usepackage{enumitem} \newtheorem{theorem}{Theorem} \begin{document}   \begin{theorem}     Let $A$ be a finite set and $f$ a map from $A$ to itself. The following statements are equivalent:     \begin{enumerate}[label={(\roman*)},itemindent=1em]       \item $f$ is injective. \label{thm:injective}       \item $f$ is surjective. \label{thm:surjective}       \item $f$ is a bijection. \label{thm:bijection}     \end{enumerate}   \end{theorem}    Reference to item \ref{thm:surjective}.\end{document}

For a deeper understanding of the involved packages take a look at their manuals.

Thorsten
LaTeX Community Moderator

¹ System: openSUSE 42.2 (Linux 4.4.52), TeX Live 2016 (vanilla), TeXworks 0.6.1

Mohsen Avaji
Posts: 14
Joined: Mon Aug 27, 2012 12:06 am
localghost wrote:In your case you are better with using an {enumerate} environment instead of {itemize}. It can be customized easily by enumitem. The single list items can be referenced by the common mechanism of the \label and \ref commands.
\documentclass[11pt]{article}\usepackage[T1]{fontenc}\usepackage{enumitem} \newtheorem{theorem}{Theorem} \begin{document}   \begin{theorem}     Let $A$ be a finite set and $f$ a map from $A$ to itself. The following statements are equivalent:     \begin{enumerate}[label={(\roman*)},itemindent=1em]       \item $f$ is injective. \label{thm:injective}       \item $f$ is surjective. \label{thm:surjective}       \item $f$ is a bijection. \label{thm:bijection}     \end{enumerate}   \end{theorem}    Reference to item \ref{thm:surjective}.\end{document}

For a deeper understanding of the involved packages take a look at their manuals.
Thorsten

Hi
In the following I have used \usepackage{theoremref}, this package refer to theorem with number (in the following example \thref refer to theorem "Theorem 1")
but I want refer to part(i) of theorem by the common mechanism of the \label and \ref and after run have ("Theorem 1(i)").
Any suggestion is appreciated.

Best,
Mohsen
     \documentclass[11pt]{article}     \usepackage[T1]{fontenc}     \usepackage{enumitem}     \usepackage{theoremref}     \newtheorem{theorem}{Theorem}      \begin{document}       \begin{theorem}\thlabel{theorem}         Let $A$ be a finite set and $f$ a map from $A$ to itself. The following statements are equivalent:         \begin{enumerate}[label={(\roman*)},itemindent=1em]           \item $f$ is injective. \label{thm:injective}           \item $f$ is surjective. \label{thm:surjective}           \item $f$ is a bijection. \label{thm:bijection}         \end{enumerate}       \end{theorem}      Reference to theorem \thref{theorem}.     \end{document}
Yours faithfully,
Mohsen Avaji.

localghost
Site Moderator
Posts: 9204
Joined: Fri Feb 02, 2007 12:06 pm
Mohsen Avaji wrote:[…] but I want refer to part(i) of theorem by the common mechanism of the \label and \ref and after run have ("Theorem 1(i)"). […]

Any reason against writing it this way?
Reference to \thref{thm:theorem}\ref{thm:surjective}.
LaTeX Community Moderator

¹ System: openSUSE 42.2 (Linux 4.4.52), TeX Live 2016 (vanilla), TeXworks 0.6.1