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https://homework.zookal.com/questions-and-answers/find-a-trend-for-each-environmental-categories-and-then-find-502552467
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2. Economics
3. find a trend for each environmental categories and then find...
# Question: find a trend for each environmental categories and then find...
###### Question details
find a trend for each environmental categories and then find a specific opportunity related to the trend.
1.social change 2. economics 3. technology 4. competition 5. Regulatory.
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2021-03-02 05:05:25
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https://brilliant.org/problems/thats-not-an-identity/
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That's not an Identity!
Number Theory Level 2
If $$x$$ and $$y$$ are positive integers, and satisfy
$(x+y)!=x!+y!$
Find $$x$$ and $$y$$. Enter $$x+y$$ as your answer.
Notation: $$!$$ is the factorial notation. For example, $$8! = 1\times2\times3\times\cdots\times8$$.
×
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2016-10-25 06:40:19
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http://openstudy.com/updates/510292ede4b0ad57a56273e2
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## canyouhelpmewiththis Group Title Is this right? one year ago one year ago
1. canyouhelpmewiththis
This is the problem...
2. canyouhelpmewiththis
3. canyouhelpmewiththis
|dw:1359123384510:dw| wait, this is my answer.
4. UnkleRhaukus
you can only combine like terms , $\sqrt[3]x~~\&~~\sqrt[4]x$ are not like terms
5. zordoloom
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2014-10-22 22:25:03
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https://mintoc.de/index.php/Quadrotor_(binary_variant)
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State dimension: 1
Differential states: 6
Discrete control functions: 4
Interior point equalities: 6
This site describes a Quadrotor helicoptor problem variant where the continuous control is replaced via outer convexification with binary controls.
Mathematical formulation
The mixed-integer optimal control problem is given by
$\begin{array}{llclr} \displaystyle \min_{x,u, w} & 5(x_1(t_f)-6)^2&+&5(x_3(t_f)-1)^2+(\sin(x_5(t_f)0.5))^2 +\int\limits_{t_0}^{t_f} 5( (w_2(\tau)+w_4(\tau)+w_6(\tau))^2 \ d \tau \\[1.5ex] \mbox{s.t.} & \dot{x}_1 & = & x_2(t), \\ & \dot{x}_2 & = & g \sin( x_5(t)) + \sum\limits_{i\in [4]}c_{1,i}w_i(t)\frac{\sin(x_5(t))}{M}, \\ & \dot{x}_3 & = & x_4(t), \\ & \dot{x}_4 & = & g \cos( x_5(t))-g+ \sum\limits_{i\in [4]}c_{1,i}w_i(t)\frac{\cos(x_5(t))}{M}, \\ & \dot{x}_5 & = & x_6(t), \\ & \dot{x}_6 & = & \sum\limits_{i\in [4]}c_{2,i}w_i(t)L \frac{1}{I} \\[1.5ex] & x(0) &=& (0, 0, 1, 0 , 0, 0)^T, \\ & w_i(t) &\in& \{0, 1\}, i=1,\ldots,4 \\ & \sum\limits_{i=1}^{4}w_i(t) &=& 1, \\ & x_3(t) & \geq & 0, \quad t\in[t_0,t_f]. \end{array}$
Parameters
These fixed values are used within the model.
$\begin{array}{rcl} [t_0, t_f] &=& [0, 7.5],\\ (g, M, L, I) &=& (9.8, 1.3, 0.305, 0.0605),\\ c_1 &=& (0,0.001,0,0),\\ c_2 &=& (0,0,-0.001,0.001), \end{array}$
Reference Solutions
If the problem is relaxed, i.e., we demand that $w(t)$ is in the continuous interval $[0, 1]$ rather than being binary, the optimal solution can be determined by means of direct optimal control.
The optimal objective value of the relaxed problem with $n_t=12000, \, n_u=25$ is $13.0907346$. The objective value of the solution with binary controls obtained by Combinatorial Integral Approximation (CIA) is $15.5787932$.
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2021-06-22 05:40:41
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https://phys.libretexts.org/TextBooks_and_TextMaps/Astronomy_and_Cosmology_TextMaps/Map%3A_Stellar_Atmospheres_(Tatum)/7%3A_Atomic_Spectroscopy/7.03%3A_The_Hydrogen_Spectrum
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$$\require{cancel}$$
# 7.3: The Hydrogen Spectrum
In 1885, J. J. Balmer, a lecturer in a ladies' college in Switzerland, devised a simple formula relating the wavelengths of the lines in the visible region of the atomic hydrogen spectrum to the natural numbers, and these lines have since been referred to as the Balmer series and have been denoted by H$$\alpha$$, H$$\beta$$, H$$\gamma$$,...,starting at the long wavelength end. The standard air wavelengths in nm and the vacuum wavenumbers in μm-1 are as follows:
\begin{array}{l c c c}
&& \lambda & \sigma_0 \\
&& \text{nm} & \mu \text{m}^{-1} \\
\text{H} \alpha && 656.28 & 1.5233 \\
\text{H} \beta && 486.13 & 2.0565 \\
\text{H} \gamma && 434.05 & 2.3032 \\
\text{H} \delta && 410.17 & 2.4373 \\
\text{H} \epsilon && 397.01 & 2.5181 \\
\end{array}
The series eventually converges to a series limit, the Balmer limit, at a standard air wavelength of $$364.60 \ \text{nm}$$ or a vacuum wavenumber of $$2.7420 \ \mu \text{m}^{-1}$$. In the way in which Balmer's formula is usually written today, the vacuum wavenumbers of the lines in the Balmer series are given by
$\sigma_0 = R \left( \frac{1}{4} - \frac{1}{n^2} \right) , \ n =3,4,5 ... \label{7.3.1}$
$$n$$ being 3, 4, 5, etc., for $$\text{H}\alpha, \ \text{H}\beta, \ \text{H}\gamma$$, etc. The number $$R$$ is called the Rydberg constant for hydrogen, and has the value $$10.9679 \ \mu \text{m}^{-1}$$.
Later, a similar series, to be named the Lyman series, was discovered in the ultraviolet, and several similar series were found in the infrared, named after Paschen, Brackett, Pfund, Humphreys, Hansen and Strong, and successively less famous people. Indeed in the radio region of the spectrum there are series named just for numbers; thus we may talk about the $$109\alpha$$ line.
A single formula can be used to generate the wavenumbers of the lines in each of these series:
$\sigma_0 = R \left( \dfrac{1}{n^2_1}- \dfrac{1}{n_2^2} \right) , \ n_2 = n_1 + 1, n_1 + 2, ... \label{7.3.2}$
Here $$n_1 = 1, 2, 3, 4, 5, 6...$$ for the Lyman, Balmer, Paschen, Brackett, Pfund, Humphreys... series.
Similar (not identical) spectra are observed for other hydrogen-like atoms, such as $$\text{He}^+, \ \text{Li}^{++}, \ \text{Be}^{+++}$$, etc., the Rydberg constants for these atoms being different from the Rydberg constant for hydrogen. Deuterium and tritium have very similar spectra and their Rydberg constants are very close to that of the $$^1\text{H}$$ atom.
Each "line" of the hydrogen spectrum, in fact, has fine structure, which is not easily seen and usually needs carefully designed experiments to observe it. This fine structure need not trouble us at present, but we shall later be obliged to consider it. An interesting historical story connected with the fine structure of hydrogen is that the quantity $$e^2 /(4 \pi \epsilon_0 \hbar c)$$ plays a prominent role in the theory that describes it. This quantity, which is a dimensionless pure number, is called the fine structure constant $$\alpha$$, and the reciprocal of its value is close to the prime number 137. Sir Arthur Eddington, one of the greatest figures in astrophysics in the early twentieth century, had an interest in possible connections between the fundamental constants of physics and the natural numbers, and became almost obsessed with the notion that the reciprocal of the fine structure constant should be exactly 137, even insisting on hanging his hat on a conference hall coatpeg number 137.
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2018-06-21 21:35:29
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https://www.math.ias.edu/seminars/abstract?event=128772
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# Crossing the logarithmic barrier for dynamic boolean data structure lower bounds
Computer Science/Discrete Mathematics Seminar I Topic: Crossing the logarithmic barrier for dynamic boolean data structure lower bounds Speaker: Omri Weinstein Affiliation: Columbia University Date: Monday, October 2 Time/Room: 11:00am - 12:15pm/S-101 Video Link: https://video.ias.edu/csdm/2017/1002-OmriWeinstein
This paper proves the first super-logarithmic lower bounds on the cell-probe complexity of dynamic boolean (a.k.a. decision) data structure problems, a long-standing milestone in data structure lower bounds. We introduce a new method for proving dynamic cell probe lower bounds and use it to prove a $\tilde{\Omega}(\log^{1.5} n)$ lower bound on the operational time of a wide range of boolean data structure problems, most notably, on the query time of dynamic range counting over $\mathbb{F}_2$ ([Pat07]). Proving an $\omega(\lg n)$ lower bound for this problem was explicitly posed as one of five important open problems in the late Mihai Patrascu's obituary. This result also implies the first $\omega(\lg n)$ lower bound for the classical 2D range counting problem, one of the most fundamental data structure problems in computational geometry and spatial databases. We derive similar lower bounds for boolean versions of dynamic polynomial evaluation and 2D rectangle stabbing, and for the (non-boolean) problems of range selection and range median. Our technical centerpiece is a new way of weakly" simulating dynamic data structures using efficient one-way communication protocols with small advantage over random guessing. This simulation involves a surprising excursion to low-degree (Chebyshev) polynomials which may be of independent interest, and offers an entirely new algorithmic angle on the cell sampling" method. Joint work with Kasper Green Larsen and Huacheng Yu.
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2019-02-15 22:11:58
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http://www.sciencemadness.org/talk/viewthread.php?tid=77708&page=2
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Sciencemadness Discussion Board » Fundamentals » Organic Chemistry » Synthesis of dopamine from catechol Select A Forum Fundamentals » Chemistry in General » Organic Chemistry » Reagents and Apparatus Acquisition » Beginnings » Miscellaneous » The Wiki Special topics » Technochemistry » Energetic Materials » Biochemistry » Radiochemistry » Computational Models and Techniques » Prepublication » References Non-chemistry » Forum Matters » Legal and Societal Issues » Whimsy » Detritus » The Moderators' Lounge
Pages: 1 2 4 5
Author: Subject: Synthesis of dopamine from catechol
Melgar
International Hazard
Posts: 1761
Registered: 23-2-2010
Location: NYC
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Quote: Sorry @Melgar, but I didn't understand the last part. This reaction was extracted from a published book. Do You believe that's impossible to occurs? The author is a fraud? yes or not? why?
Hey guys, what's the consensus here on Jared Ledgard? I know he publishes a lot of books that are perhaps one step above Uncle Fester in apparent legitimacy, but with the exact same subject matters. I never actually read any of the books myself, but I've heard they're mostly just compiled from information he found online, with a lot of typos.
Oh, and that synthesis or whatever you have is obviously missing a step. There needs to be a Nef reaction or a dissolving metal reduction with strongly acidic conditions (to reduce to the oxime then hydrolyze), if the intent is to perform a reductive amination.
Quote: Originally posted by CuReUS
Quote: Originally posted by Melgar One thing that I've noticed is that aromatic formylation reactions that are ortho-directing with phenol, seem to be para-directing with catechol.
where have you noticed this ?
It's not for every reaction, only certain ones. The Riemer-Tiemann reaction, for one. Like, it para-formylates guaiacol to give vanillin, and the Ladenburg synthesis of piperine also does a Riemer-Tiemann formylation of catechol. I thought this was true of the Gattermann aldehyde synthesis as well, but I think that's probably always para-directed. It could be that ortho/para formylations often give the ortho isomer as the major product with phenol, and the para isomer as the major product with catechol.
The Duff reaction doesn't seem like it'd work. Ah well. And for those of you giving all sorts of not-very-OTC syntheses, here are the revised criteria:
• Reagents must be purchasable within the US
• If something is sold by a retail store, it must be a chain store with a presence in at least half of the 50 states
• If something is available online, it must be sold by at least three different vendors
• Sellers on eBay and Amazon are considered "vendors" for this purpose. Note that the same seller (say, CCS) selling products on multiple platforms only counts as one vendor.
Sorry for the list of rules, I just want this to be something that someone could theoretically reproduce without running into sourcing problems.
I've heard that ethylvanillin is easier to dealkylate than vanillin, but the references all seem to refer to some original Q document that I've been unable to find. If it turns out that this is true, then this might be the way to go. Catechol isn't especially easy to source, and is more expensive than ethylvanillin, so I may just discuss catechol for a minute or two and then proceed to do the synthesis with some vanillin-type aldehyde.
The first step in the process of learning something is admitting that you don't know it already.
JJay
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Posts: 2904
Registered: 15-10-2015
Location: Western Hemisphere
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Mood: planning a semi-novel acetonitrile synthesis
His book King's Chemistry says you can do a reaction that closely resembles a Friedel-Crafts on catechol with allyl alcohol in acetone using potassium carbonate as the catalyst instead of a Lewis acid:
Does this actually work? It's questionable but interesting. Perhaps you might substitute vinyl bromide for allyl bromide if your target is dopamine (and what happens if you substitute a Lewis acid for the potassium carbonate?). It's definitely sketchy, but he might have based this on solid research that he neglected to cite.
He also devoted several pages to how to extract various natural products that might be used to make catecholamines. It's not clear whether he's actually tried this himself.
Edit: The first time I read this I skipped reading most of the workup and missed the rearrangement part... the potassium carbonate is a catalyst for forming the phenyl ether and is actually removed prior to the ring alkylation.
[Edited on 25-10-2017 by JJay]
This is my YouTube channel: Extreme Red Cabbage. I don't have much posted, but I try to do nice writeups once in a while.
clearly_not_atara
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Mood: No Mood
It looks like allyl alcohol might be an active FC electrophile:
http://isivast.org.vn:8888/bitstream/123456789/12963/1/1501....
But unfortunately there are no intermolecular rxns using unsubstituted allyl alcohol on phenols. Looking around I did find an aurous-catalyzed formation of chromans from allyl alcohols and phenols which may explain the lack of such rxns.
Melgar: You know how to make nitroethylene, correct? Nitromethane + formaldehyde + H2SO4. I'm not sure if my idea fell into the "not OTC" category. Anyway I did some more searching and just as was the case with allyl alcohol there are loads of FC rxns with nitroethylene but they all use indoles or pyrroles and no phenols. So nitroethylene F-C is probably out.
However, it turns out that organocopper reagents will react with nitroethylene and other nitroalkenes in the desired manner; see:
http://anonym.to/http://www.chem.ntnu.edu.tw/en/files/writin...
The reagents also react with allylic acetates. These zinc organocuprates can be prepared by transmetallation of organozinc reagents with copper (I) salts, usually CuCN*2LiCl for solubility reasons because the only studies on the zinc-copper reagents use organozinc reagents generated in THF from reactive organohalides and activated zinc.
However, arylzinc reagents can also be generated in acetonitrile, which was not known at the time that the original zinc-organocuprate studies were being carried out. Generating arylzinc reagents in MeCN uses a cobalt catalyst:
http://anonym.to/http://pubs.acs.org/doi/abs/10.1021/ja02894...
Because copper (I) salts generally have much better solubility in acetonitrile (vs THF) it stands to reason that the transmetallation of organozinc@acetonitrile to zinc organocuprate@acetonitrile should work just fine; unfortunately, it doesn't seem that anyone has tried to convert these arylzinc reagents to cuprates. However, it seems like a relatively safe bet - as safe as any bets involving organometallic reagents can be anyway. And ordinary copper (I) salts should be fine -- no need for cyanide.
The resulting organocuprate could then add to nitroethylene. It's not clear to me if the phenols need to be protected in order for this rxn to proceed; organozinc and organocopper reagents are generally somewhat resistant to protons unlike their Li and Mg cousins. Also, phenols are very weak acids in acetonitrile. I'm pretty sure the presence of catalytic amounts of cobalt will not interfere with the transmetallation, as long as the arylzinc is generated before the copper salt is added.
The good news is that bromination of benzodioxole is high-yielding and can be achieved using NH4Br/H2O2 avoiding Br2:
http://anonym.to/https://erowid.org/archive/rhodium/chemistr...
And the most non-OTC thing here is CoBr2, which shouldn't be that hard to obtain in small amounts.
[Edited on 04-20-1969 by clearly_not_atara]
Chemi Pharma
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Quote: Originally posted by JJay His book King's Chemistry says you can do a reaction that closely resembles a Friedel-Crafts on catechol with allyl alcohol in acetone using potassium carbonate as the catalyst instead of a Lewis acid: Does this actually work? It's questionable but interesting.
Hey @JJay, I'm afraid it will not work. I suspect Jared Ledgard books are not so trustworthy as I have thought.
I found the same reaction you described from his book King's Chemistry Survival Guide at another book called Preparation of Organic Intermediates - David Shirley - Wiley - 1952, between a phenol (cathecol in this case), allyl bromide, Potassium carbonate as a catalizer and acetone as a solvent. That is quite the same reagents.
However, the product is an allyl phenyl ether and not an allyl phenol (cathecol in this case). The addition isn't like a friedel-crafts reaction, but instead, a substitution of the phenolic group by an ether.
Take a look on this:
EDIT: Ok, I apologize, The allyl phenyl ether will rearrange (Claisen) to allyl phenol (cathecol) at high temperature. I see that now. Jared Ledgard is not so stupid after all.
[Edited on 24-10-2017 by Chemi Pharma]
SWIM
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Pretty sure I've seen Shulgin pull this same reaction somewhere in PIHKAL for some compound where the ring substitution pattern couldn't be put together by simpler means.
Looked it up: it's the 4-benzoloxy-3,5-dimethoxyamphetamine synthesis.
He does it to get the 5-methoxyeugenole intermediate.
Note: according to him this is a lousy drug that's not worth making anyway.
[Edited on 25-10-2017 by SWIM]
JJay
International Hazard
Posts: 2904
Registered: 15-10-2015
Location: Western Hemisphere
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Mood: planning a semi-novel acetonitrile synthesis
Quote: Originally posted by Chemi Pharma The allyl phenyl ether will rearrange (Claisen) to allyl phenol (cathecol) at high temperature. I see that now. Jared Ledgard is not so stupid after all.
I kind of skimmed the rearrangement procedure thinking it was just part of the workup... looks plausible, actually.
The $50,000 question is whether this will work with PVC monomer instead of allyl bromide. [Edited on 25-10-2017 by JJay] This is my YouTube channel: Extreme Red Cabbage. I don't have much posted, but I try to do nice writeups once in a while. SWIM Hazard to Others Posts: 216 Registered: 3-9-2017 Member Is Offline Mood: No Mood Doesn't the Claisen rearrangement have a cyclic intermediate? If so, how would a 2-carbon phenol ether even reach? Edit: for 'intermediate', read 'transition state' I do know the difference on a good day. [Edited on 25-10-2017 by SWIM] JJay International Hazard Posts: 2904 Registered: 15-10-2015 Location: Western Hemisphere Member Is Offline Mood: planning a semi-novel acetonitrile synthesis Looks like you might as well just do a Friedel-Crafts. This is my YouTube channel: Extreme Red Cabbage. I don't have much posted, but I try to do nice writeups once in a while. clearly_not_atara International Hazard Posts: 995 Registered: 3-11-2013 Member Is Offline Mood: No Mood Allylation of catechol is not applicable to dopamine whereas it is applicable to another compound we don't discuss, so I've refrained. However, be assured that there is a route to 4-allylcatechol which involves a Claisen rearrangement. When there is a substituent ortho to the phenol, the Claisen rearrangement can proceed to a 6,6-disubstituted cyclohexadienone, which rearranges to a para-substituted aromatic. I will not give conditions, but with one ortho position blocked, the ortho:para ratio is about 1:1. [Edited on 25-10-2017 by clearly_not_atara] [Edited on 04-20-1969 by clearly_not_atara] CuReUS International Hazard Posts: 796 Registered: 9-9-2014 Member Is Offline Mood: No Mood Quote: Originally posted by Melgar It's not for every reaction, only certain ones. The Riemer-Tiemann reaction, for one. Like, it para-formylates guaiacol to give vanillin, and the Ladenburg synthesis of piperine also does a Riemer-Tiemann formylation of catechol. neither of the two compounds is a catechol.In the RTR ,it is a known fact that if the ortho position is occupied(by methoxy,in guaiacol),the formylation will occur para to OH.And in piperine synthesis,he formylates benzodioxole,which is again known to give para products Quote: Originally posted by SWIM Doesn't the Claisen rearrangement have a cyclic intermediate? If so, how would a 2-carbon phenol ether even reach? good catch ,all the examples I have seen use some form of allyl(or propargyl).Even if you could make the vinyl ether(which would be difficult), it wouldn't rearrange. My fries idea would give dopamine in 2 steps if glycine could be used Waffles SS Fighter Posts: 803 Registered: 7-12-2009 Location: My Lab Member Is Offline Mood: No Mood Synthesis of 3,4-dihydroxybenzaldehyde(Protocatechuic aldehyde) from catechol DE105798 US4165341 by Reimer–Tiemann reaction: Reihlen; Illig; Wittig Chemische Berichte, 1925 , vol. 58, p. 18 Reimer; Tiemann Chemische Berichte, 1876 , vol. 9, p. 1269 Tiemann; Koppe Chemische Berichte, 1881 , vol. 14, p. 2021 [Edited on 25-10-2017 by Waffles SS] Melgar International Hazard Posts: 1761 Registered: 23-2-2010 Location: NYC Member Is Offline Mood: Aromatic Quote: Originally posted by CuReUS neither of the two compounds is a catechol.In the RTR ,it is a known fact that if the ortho position is occupied(by methoxy,in guaiacol),the formylation will occur para to OH.And in piperine synthesis,he formylates benzodioxole,which is again known to give para products Well, I suppose it's possible to formylate in either order, but sources seem to indicate that the formylation came first. That would make sense, since the Riemer-Tiemann reaction is the lower-yielding of the two, by quite a lot: [img]http://slideplayer.com/slide/10398918/35/images/39/Piperine+was+synthesized+(Ladenburg,+1894)+by+the+reaction+of+the+piperic+acid+chloride+with+p iperidine,+which+confirmed+the+structure+of+the+molecule.+The+synthesis+of+piperic+acid+was+achieved+starting+from+piperonal,+which+was+obtained+from+ catechol+using+Reimer-Tiemann+reaction+followed+by+the+condensation+with+diiodomethane+in+the+presence+of+a+base..jpg[/img] Quote: Originally posted by SWIM good catch ,all the examples I have seen use some form of allyl(or propargyl).Even if you could make the vinyl ether(which would be difficult), it wouldn't rearrange. My fries idea would give dopamine in 2 steps if glycine could be used Yes, glycine is certainly on the table. I'm sort of looking to use the most versatile reactions as possible for this synthesis, because it's supposed to be educational though. I'm looking into everything though, but so much has been thrown on the table in this thread that it's taking a while. @not_atara: Seems like a LOT of speculation. Could it work? Maybe. Would I expect it to work? Certainly not with me doing it. It would be nice if there were easier ways to get chloroform, because I'd need to test the Riemer-Tiemann reaction at least 5 or 6 times before doing it for real, and large excesses of chloroform always seem to be needed. Considering I get about 100 mL from a gallon of 8-10% sodium hypochlorite, that's a lot of bleach to have to deal with. By the way, what about ethylvanillin dealkylation? I'd like to try at least two routes just so I can verify that I did it correctly somehow. The first step in the process of learning something is admitting that you don't know it already. SWIM Hazard to Others Posts: 216 Registered: 3-9-2017 Member Is Offline Mood: No Mood Quote: Originally posted by Melgar Quote: Originally posted by SWIM good catch ,all the examples I have seen use some form of allyl(or propargyl).Even if you could make the vinyl ether(which would be difficult), it wouldn't rearrange. My fries idea would give dopamine in 2 steps if glycine could be used . In all fairness, the above quote is CuReUS. I certainly don't deserve any credit for his Fries /glycine idea. However I'm not too humble to point out that in the first part of that quote he's agreeing with me about something. JJay International Hazard Posts: 2904 Registered: 15-10-2015 Location: Western Hemisphere Member Is Offline Mood: planning a semi-novel acetonitrile synthesis Glycine is an interesting idea, but it's not going to form a phenyl ester easily, and if you halogenate it or form an anhydride with it, it's going to try to form peptides. Right? This is my YouTube channel: Extreme Red Cabbage. I don't have much posted, but I try to do nice writeups once in a while. Chemi Pharma Hazard to Others Posts: 153 Registered: 5-5-2016 Location: Latin America Member Is Offline Mood: Researcher Quote: Originally posted by Melgar Yes, glycine is certainly on the table. I'm sort of looking to use the most versatile reactions as possible for this synthesis, because it's supposed to be educational though. I'm looking into everything though, but so much has been thrown on the table in this thread that it's taking a while. And what about protocatechualdehyde via aldimine with catechol, copper cyanide and HCl ? The formylation in this case is para oriented. I thought about this after take a look on the preparation below: Quote: Originally posted by Waffles SS Synthesis of 3,4-dihydroxybenzaldehyde(Protocatechuic aldehyde) from catechol DE105798 US4165341 Also, the idea of @Waffles SS, brought through the patent US4165341 sounds brilliant, about sinthesize protocatechualdehyde from cathecol and glioxilic acid with alumina as a catalizer with high yields (Patent in .pdf attached below) Protocatechualdehyde is for me the best starting point, cause is only two OTC steps away from dopamine (Henry reaction with nitromethane and reduction with sodium borohydride/nickel chloride in methanol). [Edited on 26-10-2017 by Chemi Pharma] Attachment: protocatechualdehyde from cathecol formylation with glyoxylic acid.pdf (689kB) This file has been downloaded 35 times clearly_not_atara International Hazard Posts: 995 Registered: 3-11-2013 Member Is Offline Mood: No Mood Fries rearrangement? Cool. But with glycine? I think you'd have to use N-protected glycine in order to prevent polymerization. You could perhaps do the rearrangement with chloroacetic acid, then react the product with hexamine. But dimerization of alpha-ketoamones is an issue. I'm not sure if you can make N-protected amines by alkylating cyanate anion in an alcoholic solvent, so the formed isocyanate converts to a carbamate in situ. Succinimide or phthalimide would probably be the responsible choice here. [Edited on 04-20-1969 by clearly_not_atara] Melgar International Hazard Posts: 1761 Registered: 23-2-2010 Location: NYC Member Is Offline Mood: Aromatic Quote: Originally posted by Chemi Pharma And what about protocatechualdehyde via aldimine with catechol, copper cyanide and HCl ? The formylation in this case is para oriented. I thought about this after take a look on the preparation below: That's the Gatterman aldehyde synthesis that I mentioned earlier. It's zinc cyanide though, not copper cyanide. Incidentally, this is similar to a FC acylation, but uses HCN as the source of the formyl group. To keep from having to use HCN gas, it's instead introduced as Zn(CN)2, then reacted with HCl gas. This produces both the Lewis acid and the HCN in situ. Quote: Also, the idea of @Waffles SS, brought through the patent US4165341 sounds brilliant, about sinthesize protocatechualdehyde from cathecol and glioxilic acid with alumina as a catalizer with high yields (Patent in .pdf attached below) Yep, this is similar to how vanillin is made. The trouble is getting glyoxylic acid. Supposedly that can be prepared from magnesium and oxalic acid, but the reaction is very low-yielding if it yields anything. Another preparation mentions oxidizing ethanol with nitric acid, which is much more likely to turn into a nitric-acid geyser than it is to give any glyoxylic acid. Electroreduction of oxalic acid might have to be the way to go here, and supposedly this was how it was made industrially until recently. Quote: Protocatechualdehyde is for me the best starting point, cause is only two OTC steps away from dopamine (Henry reaction with nitromethane and reduction with sodium borohydride/nickel chloride in methanol). I'm unsure of how acidic phenols affect the Henry reaction, although I would expect that you would just need more catalyst, or longer reaction times. That wouldn't be my first choice of reduction methods though. First, because NaBH4 isn't easy to get, and second, because the nickel boride reaction isn't as nice as you probably think it'd be. However, reducing with zinc and HCl can give the amine for nitrostyrenes, provided temperatures are kept below 0C, which is probably the reduction method I'd use. The first step in the process of learning something is admitting that you don't know it already. Chemi Pharma Hazard to Others Posts: 153 Registered: 5-5-2016 Location: Latin America Member Is Offline Mood: Researcher @Melgar, I have cited copper cyanide cause is more OTC. At least here, in south america, I can buy it withouth any restrictions applicable to alkaly cyanides for just US$30,00 for 500 grs. I don't know about US restrictions, by the way. To this reaction it can be zinc or copper or even sylver cyanide. I think every transition metal cyanide will work.
Glyoxylic acid 50% solution w/w is so cheap here that I'm ashamed saying the price: US$18,00 a liter. Glyoxylic acid is sold by perfumary, soap and flagrancies stores, cause it's used as an exfoliating agent to the skin and to make facial masks. I guess you can find the 50% solucion w/w in US easy if you find at this places, or at the e-bay, and not at a chemycal store. So, just buy it instead synthesize. Henry reaction between nitroethane with protocatechualdehyde is extensed covered by many and many experiments you can read at Rhodium pages and at The Hive. Many guys have registered their suscessful experiments on doing that, with no damage to the phenolic radicals. Care must been taken with benzodiaxole ring, cause it's acidity sensibility, but it's not the case here. I think Nitromethane will behave the same way. About reduction, borohydride here is easy to get (I have almost 500 grs right now), but I agree with you it's not so OTC if you want to make an educacional video. I never trust in Zn + HCl to reduce nitroalkenes to give high yields. I suggest you, then, Tin + HCl or SnCl2 + HCl, that's a classic and proved way to reduce them with razonable yield. [Edited on 26-10-2017 by Chemi Pharma] Melgar International Hazard Posts: 1761 Registered: 23-2-2010 Location: NYC Member Is Offline Mood: Aromatic First of all, zinc isn't a transition metal. Just figured I'd point that out. Also, the only transition metal that I regularly see used as a FC catalyst is iron. Copper, I'd expect to be too prone to oxidizing things, and I'm not sure about its solubility in its anhydrous form. Now that I think about it, I wonder if potassium ferricyanide or ferrocyanide could do this like zinc? Precipitate potassium as KCl, free cyanide as HCN and iron would be FeCl3. That may be too much cyanide though, even if it did work. Quote: Originally posted by Chemi Pharma Glyoxylic acid 50% solution w/w is so cheap here that I'm ashamed saying the price: US$18,00 a liter. Glyoxylic acid is sold by perfumary, soap and flagrancies stores, cause it's used as an exfoliating agent to the skin and to make facial masks. I guess you can find the 50% solucion w/w in US easy if you find at this places, or at the e-bay, and not at a chemycal store. So, just buy it instead synthesize.
Are you sure you aren't thinking of GLYCOLIC acid? Because that's exactly what you seem to be describing.
Quote: About reduction, borohydride here is easy to get (I have almost 500 grs right now), but I agree with you it's not so OTC if you want to make an educacional video. I never trust in Zn + HCl to reduce nitroalkenes to give high yields. I suggest you, then, Tin + HCl or SnCl2 + HCl, that's a classic and proved way to reduce them with razonable yield.
Zinc will only give good yields if temperature is kept cold, below 0C. Also, it has to be a nitrostyrene, not a nitropropene.
The first step in the process of learning something is admitting that you don't know it already.
Chemi Pharma
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@Melgar,
I'm pretty sure it's Glyoxylic acid and not Glycolic acid.
I apologize, Glyoxylic acid is used for hair straightening and Glycolic acid is used for skin exfoliation. I misundertood the usage of the both. You are right!
I didn't have it yet at my storage lab and I just have bought hours ago at the Net a 500 ml bottle for U$9,28 + mail expense, to do experiments with cathecol like said at the Patent. Here's the link I bought it in Brazil: https://lista.mercadolivre.com.br/acido-glioxilico#D[A:acido-glioxilico] AliBaba has the same 50% solution w/w to sell worldwide below U$10,00 per Kg:
[Edited on 26-10-2017 by Chemi Pharma]
JJay
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Quote: Originally posted by clearly_not_atara Fries rearrangement? Cool. But with glycine? I think you'd have to use N-protected glycine in order to prevent polymerization. You could perhaps do the rearrangement with chloroacetic acid, then react the product with hexamine. But dimerization of alpha-ketoamones is an issue. I'm not sure if you can make N-protected amines by alkylating cyanate anion in an alcoholic solvent, so the formed isocyanate converts to a carbamate in situ. Succinimide or phthalimide would probably be the responsible choice here.
Sigma has an absolutely massive selection of amino protecting reagents. I wonder if you could make one with cyanuric acid. Although succinimide can certainly be found....
This is my YouTube channel: Extreme Red Cabbage. I don't have much posted, but I try to do nice writeups once in a while.
clearly_not_atara
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Apparently glyoxylic acid hair straightening is a real thing, see e.g.:
http://onlinelibrary.wiley.com/doi/10.1111/ics.12148/abstrac...
JJay: I think 5,5-dimethylhydantoin is a better choice than cyanuric acid if you want things to be OTC. Only one of the protons on hydantoin is removed at normal pH ranges, which prevents having to deal with polysubstituted things like 1,3,5-triazane-2,4,6-trione-1,3,5-triacetic acid". 5,5-dimethylhydantoin is the reduction product of BCDMH the common source of pool bromine.
[Edited on 04-20-1969 by clearly_not_atara]
JJay
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I do think it would be easier to find suitable solvents for hydantoins.
It seems to me as though the rules say that the least OTC part should be the catechol. I cleaned out my fume hood yesterday and am tempted to play with some aspirin to see if I can't get some catechol out of it... but I have a lot of other experiments I want to do, and it would be a lot less time consuming to just buy it.
This is my YouTube channel: Extreme Red Cabbage. I don't have much posted, but I try to do nice writeups once in a while.
Waffles SS
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I worked with glyoxylic acid for more than 3 years.i tried many method for synthesis it.(Periodate method , Electroreduction method, Ozone method , Glyoxal oxidation method, ...).
I believe Easier method is Glyoxal oxidation by Nitric acid(industrial method).
Dispose of Nitrogen oxide is problem of this method that i solved it by different trap
Reaxys search attached
Attachment: glyoxylic acid.pdf (1.7MB)
This file has been downloaded 54 times
[Edited on 27-10-2017 by Waffles SS]
CuReUS
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Quote: Originally posted by JJay This is the first time I have heard of the Fries rearrangement... interesting. Why doesn't the chloro acetylchloride react with pyridine ?
because it can react more easily with the exposed,protruding OH rather than try to react with the flat,boxed in N.Also the lone pair of N is in resonance making it less juicy
Quote: Originally posted by JJay Glycine is an interesting idea, but it's not going to form a phenyl ester easily
Quote: Originally posted by clearly_not_atara Fries rearrangement? Cool. But with glycine? I think you'd have to use N-protected glycine in order to prevent polymerization.
I found some methods to make the ester
http://orgsyn.org/demo.aspx?prep=cv2p0310 ( see the 1st and 2nd refs given)
http://www.prepchem.com/synthesis-of-glycine-ethyl-ester-hyd...
[Edited on 27-10-2017 by CuReUS]
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# 022225037342403630106358e/SBN-97803574194278d-8806398638snapshotld- 1892126&ofsInstructionsniesUse the first eight rules of inference to derive the conclusion of the symbolized argument below:CiAddDMAsocDNTransImplEqunExpTautACPAPFREMISE(LvT) ? (D .
###### Question:
022225037342403630106358e/SBN-97803574194278d-8806398638snapshotld- 1892126& ofs Instructions nies Use the first eight rules of inference to derive the conclusion of the symbolized argument below: Ci Add DM Asoc DN Trans Impl Equn Exp Taut ACP AP FREMISE (LvT) ? (D . G) PREXISE (K = R) CoNcLuson Sunp Com Dist
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##### (10 pointe) otlice, Part (@) hch possible Apta; end in the elar Cuaiult [ Toom O5e of thc Kake Ane Haned Aali Ect ute eurana Velt Tn Haaplan o If2, drav cint ult Ot fOAm exeftly <118 0, @Twi Obaeln Juntar MiaehConsider the complete graph Ka where each edge is *directed * Part (b)-opts: path (not necessarily circuit) tbat starts at one vertex and Prove that there exists Use induction! (Feel free to write your solution on touches every other vertex exactly once_ the back of this page)-
(10 pointe) otlice, Part (@) hch possible Apta; end in the elar Cuaiult [ Toom O5e of thc Kake Ane Haned Aali Ect ute eurana Velt Tn Haaplan o If2, drav cint ult Ot fOAm exeftly <118 0, @Twi Obaeln Juntar Miaeh Consider the complete graph Ka where each edge is *directed * Part (b)-opts: path (not...
##### Let the amount of a loss X(X > 0) be a continuous random variable following cdf F(z) and pdf f(c). Let Y = min( X, M) and Z = max(0, X M) be the retained claim by the insurer and the ceded claim to the reinsurer; respectively: Let W be the positive claim for the reinsurer: Find the distribution of W ifX ~ T(2,A). X~ U(0, w) , where w > M:
Let the amount of a loss X(X > 0) be a continuous random variable following cdf F(z) and pdf f(c). Let Y = min( X, M) and Z = max(0, X M) be the retained claim by the insurer and the ceded claim to the reinsurer; respectively: Let W be the positive claim for the reinsurer: Find the distribution o...
##### Please thoroughly explain, especially the angle part. Thanks! A 290 -m-wide river has a uniform flow...
Please thoroughly explain, especially the angle part. Thanks! A 290 -m-wide river has a uniform flow speed of 0.95 m/s through a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of 7.6 m/s wit...
##### The Arizona Corporation reported positive net income but negative cash flow from operations. Further, despite the...
The Arizona Corporation reported positive net income but negative cash flow from operations. Further, despite the negative cash flow from operations, the company continued to pay its regular dividend on its common shares. Discuss why it is possible for a company to report positive net income when th...
##### How many moles of the acid H3POa is required to completely neutralize 3 moles f the strong base BalOHh (hine: wrie 0305 the balanced chemical reaction for neutralization):842 mole1mole2 molcs3 moles213 mole
How many moles of the acid H3POa is required to completely neutralize 3 moles f the strong base BalOHh (hine: wrie 0305 the balanced chemical reaction for neutralization): 842 mole 1mole 2 molcs 3 moles 213 mole...
##### Sicakhgu 100 *C olan bir bardak cay 30 *€ sicakhguda bir odaya brakuhyor: 10 dakika sonra GAym sicakhgmm 70 *C' ye diistiigii goriiliiyor. Baslangic anmdan itibaren ne kadar zaman sonra Gaymn slcakhgun 50 % ye diisecegini bulunuz.
Sicakhgu 100 *C olan bir bardak cay 30 *€ sicakhguda bir odaya brakuhyor: 10 dakika sonra GAym sicakhgmm 70 *C' ye diistiigii goriiliiyor. Baslangic anmdan itibaren ne kadar zaman sonra Gaymn slcakhgun 50 % ye diisecegini bulunuz....
##### What is the current treatment for tax purposes for net operating losses? Discuss the history of the...
What is the current treatment for tax purposes for net operating losses? Discuss the history of the treatment of operating losses? Why does the government change its view of corporate income/loss? Name the government acts that discuss these laws and the da...
##### Perform the indicated operations; simplify if possible: (x 4)x2 3x 2)
Perform the indicated operations; simplify if possible: (x 4)x2 3x 2)...
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2023-01-30 12:10:19
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http://databloom.com/
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Categories
## Just Released: CUDA Toolkit 12.0
CUDA Toolkit 12.0 supports NVIDIA Hopper architecture and many new features to help developers maximize performance on NVIDIA GPU-based products.
CUDA Toolkit 12.0 supports NVIDIA Hopper architecture and many new features to help developers maximize performance on NVIDIA GPU-based products.
Categories
## Predict Protein Structures and Properties with Biomolecular Large Language Models
The NVIDIA BioNeMo service is now available for early access. At GTC Fall 2022, NVIDIA unveiled BioNeMo, a domain-specific framework and service for training…
The NVIDIA BioNeMo service is now available for early access. At GTC Fall 2022, NVIDIA unveiled BioNeMo, a domain-specific framework and service for training and serving biomolecular large language models (LLMs) for chemistry and biology at supercomputing scale across billions of parameters.
The BioNeMo service is domain-optimized for chemical, proteomic, and genomic applications, designed to support molecular data represented in the SMILES notation for chemical structures, and FASTA for amino acid and nucleic acid sequences for proteins, DNA, and RNA.
With the BioNeMo service, scientists and researchers now have access to pretrained biomolecular LLMs through a cloud API, enabling them to predict protein structures, develop workflows, and fit downstream task models from LLM embeddings.
The BioNeMo service is a turnkey cloud solution for AI drug discovery pipelines that can be used in your browser or through API endpoints. The service API endpoints offer scientists the ability to get started quickly with AI drug discovery workflows based on large language model architectures. It also provides a UI Playground to easily and quickly try these models through an API, which can be integrated into your applications.
The BioNeMo service contains the following features:
• Fully managed, browser-based service with API endpoints for protein LLMs
• Accelerated OpenFold model for fast 3D protein structure predictions
• ESM-1nv LLM for protein embeddings for downstream tasks
• Interactive inference and visualization of protein structures through a graphic user interface (GUI)
ESM-1nv, based on Meta AI’s state-of-the-art ESM-1b, is a large language model for the evolutionary-scale modeling of proteins. It is based on the BERT architecture and trained on millions of protein sequences with a masked language modeling objective. ESM-1nv learns the patterns and dependencies between amino acids that ultimately give rise to protein structure and function.
Embeddings from ESM-1nv can be used to fit downstream task models for protein properties of interest such as subcellular location, thermostability, and protein structure. This is accomplished by training a typically much smaller model with a supervised learning objective to infer a property from ESM-1nv embeddings of protein sequences. Using embeddings from ESM-1nv typically results in far superior accuracy in the final model.
OpenFold is a faithful reproduction of DeepMind’s AlphaFold-2 model for 3D protein structure prediction from a primary amino acid sequence. This long-standing grand challenge in structural biology reached a significant milestone at CASP14, where AlphaFold-2 achieved nearly experimental accuracy for predicted structures. While AlphaFold was developed for a JAX workflow, OpenFold bases its code on PyTorch.
OpenFold in BioNeMo is also trainable, meaning variants may be created for specialized research. OpenFold achieves similar accuracy to the original model and predicts the median backbone at an accuracy of 0.96 Å RMSD95 and is up to 6x faster due to changes made in the MSA generation step. This means that drug discovery researchers get 3D protein structure predictions very quickly.
Apply for early access to the BioNeMo service. You’ll be asked to join the NVIDIA Developer Program and fill out a short questionnaire to gain your early access.
Categories
## Introducing NVIDIA Riva: A GPU-Accelerated SDK for Developing Speech AI Applications
This post was updated from November 2021. Sign up for the latest Speech AI news from NVIDIA. Speech AI is used in a variety of applications, including contact…
This post was updated from November 2021. Sign up for the latest Speech AI news from NVIDIA.
Speech AI is used in a variety of applications, including contact centers’ agent assists for empowering human agents, voice interfaces for intelligent virtual assistants (IVAs), and live captioning in video conferencing. To support these features, speech AI technology includes automatic speech recognition (ASR) and text-to-speech (TTS). The ASR pipeline takes raw audio and converts it to text, and the TTS pipeline takes the text and converts it to audio.
Developing and running real-time speech AI services is complex and difficult. Building speech AI applications requires hundreds of thousands of hours of audio data, tools to build and customize models based on your specific use case, and scalable deployment support.
It also means running in real time, with low latency far under 300 ms to interact naturally with users. NVIDIA Riva streamlines the end-to-end process of developing speech AI services and provides real-time performance for human-like interactions.
## NVIDIA Riva SDK
NVIDIA Riva is a GPU-accelerated SDK for building and deploying fully customizable, real-time speech AI applications that deliver accurately in real time. These applications can be deployed on premises, in the cloud, embedded, and on the edge. NVIDIA Riva is designed to help you access speech AI functionalities easily and quickly. With a few commands, you can access the high-performance services through API operations and try demos.
The NVIDIA Riva SDK includes pretrained speech AI models that can be fine-tuned on a custom dataset, and optimized end-to-end skills for automatic speech recognition and speech synthesis.
Using Riva, you can fully customize state-of-art models on your data to achieve a deeper understanding of their specific contexts. Optimize for inference to offer services that run in real time (less than 150 ms).
Task-specific AI services and gRPC endpoints provide out-of-the-box, high-performance ASR and TTS. These AI services are trained with thousands of hours of public and internal datasets to reach high accuracy. You can start using the pre-trained models or fine-tune them with your own dataset to further improve model performance.
Riva uses NVIDIA Triton Inference Server to serve multiple models for efficient and robust resource allocation and to achieve high performance in terms of high throughput, low latency, and high accuracy.
## Overview of NVIDIA Riva skills
Riva provides highly optimized automatic speech recognition and speech synthesis services for use cases like real-time transcription and intelligent virtual assistants. The automatic speech recognition skill is available in English, Spanish, Mandarin, Hindi, Korean, Portuguese, French, German, and Russian.
It is trained and evaluated on a wide variety of real-world, domain-specific datasets. With telecommunications, podcasting, and healthcare vocabulary, it delivers world-class production accuracy. To learn more, see Exploring Unique Applications of Automatic Speech Recognition Technology.
The Riva text-to-speech or speech synthesis skill generates human-like speech. It uses non-autoregressive models to deliver 12x higher performance on NVIDIA A100 GPUs compared to Tacotron 2 and WaveGlow models on NVIDIA V100 GPUs. Furthermore, with TTS you can create a natural custom voice for every brand and virtual assistant with only 30 minutes of voice data.
To take full advantage of the computational power of the GPUs, Riva skills uses NVIDIA Triton Inference Server to serve neural networks and ensemble pipelines to run efficiently with NVIDIA TensorRT.
Riva services are exposed through API operations accessible by gRPC endpoints that hide all the complexity. Figure 3 shows the system’s server-side. The gRPC API operations are exposed by the API server running in a Docker container. They are responsible for processing all the speech incoming and outgoing data.
The API server sends inference requests to NVIDIA Triton and receives the results.
NVIDIA Triton is the backend server that simultaneously processes multiple inference requests on multiple GPUs for many neural networks or ensemble pipelines.
It is crucial for speech AI applications to keep the latency below a given threshold. This latency requirement translates into the execution of inference requests as soon as they arrive. To make the best use of GPUs to increase performance, you should increase the batch size by delaying the inference execution until more requests are received, forming a bigger batch.
NVIDIA Triton is also responsible for the context switch of networks with the state between one request and another.
Riva can be installed directly on bare-metal through simple scripts that download the appropriate models and containers from NGC, or it can be deployed on Kubernetes through a Helm chart, which is also provided.
### Querying NVIDIA Riva services
Here’s a quick look at how you can interact with Riva. A Python interface makes communication with a Riva server easier on the client side through simple Python API operations. For example, here’s how a request for an existing TTS Riva service is created in four steps.
First, import the Riva API and other useful or required libraries:
``````import numpy as np
import IPython.display as ipd
import riva.client
``````
Next, create a gRPC channel to the Riva endpoint:
``````auth = riva.client.Auth(uri='localhost:50051')
riva_tts = riva.client.SpeechSynthesisService(auth)
``````
Then, configure the TTS API parameters:
``````sample_rate_hz = 44100
req = {
"language_code" : "en-US",
"encoding" : riva.client.AudioEncoding.LINEAR_PCM,
"sample_rate_hz" : sample_rate_hz,
"voice_name" : "English-US.Female-1"
}
``````
Finally, create a TTS request:
``````req["text"] = "Is it recognize speech or wreck a nice beach?"
resp = riva_tts.synthesize(**req)
audio_samples = np.frombuffer(resp.audio, dtype=np.int16)
ipd.Audio(audio_samples, rate=sample_rate_hz)
``````
## Customizing a model with your data
While Riva’s default models are powerful, engineers might need to customize them in developing speech AI applications. Specific contexts where customizing ASR pipeline components can further optimize the transcription of audio data include:
• Different accents, dialects, or even languages from those on which the models were initially trained
• Preferencing and/or depreferencing certain words, for example, to account for one word in a set of homophones making more sense in the current context
• Noisy environments
You might also wish to customize a TTS model, so the synthesized voice assumes a particular pitch or accent, or possibly mimics one’s own voice.
With NVIDIA NeMo, you can fine-tune ASR, TTS, and NLP models on domain- or application-specific datasets (Figure 4), or even train the models from scratch.
Exploring one such customization in more detail, to further improve the legibility and accuracy of an ASR transcribed text, you can add a custom punctuation and capitalization model to the ASR system that generates text without those features.
Starting from a pretrained BERT model, the first step is to prepare the dataset. For every word in the training dataset, the goal is to predict the following:
• The punctuation mark that should follow the word
• Whether the word should be capitalized
After the dataset is ready, the next step is training by running a previously provided script. When the training is completed and the desired final accuracy is reached, create the model repository for NVIDIA Triton by using an included script.
The NVIDIA Riva Speech Skills documentation contains ASR customization best practices and more details about how to train or fine-tune other models. This post shows only one of the many customization possibilities using NVIDIA NeMo.
## Deploying a model in NVIDIA Riva
Riva is designed for speech AI at scale. To help you efficiently serve models across different servers robustly, NVIDIA provides push-button model deployment using Helm charts (Figure 5).
The Helm chart configuration, available from the NGC catalog, can be modified for custom use cases. You can change settings related to which models to deploy, where to store them, and how to expose the services.
## Conclusion
NVIDIA Riva is available as a set of containers and pretrained models, free of charge, from NVIDIA NGC to members of the NVIDIA Developer Program. With these resources, you can develop applications with real-time transcription, virtual assistants, or custom voice synthesis.
You can also get support for large-scale deployments of Riva with NVIDIA AI Enterprise Support. You can try NVIDIA Riva with a free trial on NVIDIA LaunchPad or access ASR and TTS tutorials.
If you are ready to deploy Riva speech AI skills, check out Riva Getting Started to deliver an interactive voice experience for any application.
Categories
## Building an End-to-End Retail Analytics Application with NVIDIA DeepStream and NVIDIA TAO Toolkit
Retailers today have access to an abundance of video data provided by cameras and sensors installed in stores. Leveraging computer vision AI applications,…
Retailers today have access to an abundance of video data provided by cameras and sensors installed in stores. Leveraging computer vision AI applications, retailers and software partners can develop AI applications faster while also delivering greater accuracy. These applications can help retailers:
• Understand in-store customer behavior and buying preference
• Reduce shrinkage
• Notify associates of low or depleted inventory
• Improve merchandising
• Optimize operations
Building and deploying such highly efficient computer vision AI applications at scale poses many challenges. Traditional techniques are time-consuming, requiring intensive development efforts and AI expertise to map all the complex architectures and options. These can include building customized AI models, deploying high-performance video decoding and AI inference pipelines, and generating an insightful analytics dashboard.
NVIDIA’s suite of SDKs helps to simplify this workflow. You can create high-quality video analytics with minimum configuration using the NVIDIA DeepStream SDK, and an easy model training procedure with the NVIDIA TAO Toolkit.
This post provides a tutorial on how to build a sample application that can perform real-time intelligent video analytics (IVA) in the retail domain using NVIDIA DeepStream SDK and NVIDIA TAO Toolkit.
To create an end-to-end retail vision AI application, follow the steps below:
1. Use NVIDIA pretrained models for people detection and tracking.
2. Customize the computer vision models for the specific retail use case using the NVIDIA TAO Toolkit.
3. Develop an NVIDIA DeepStream pipeline for video analysis and streaming inference outputs using Apache Kafka. Kafka is an open-source distributed streaming system used for stream processing, real-time data pipelines, and data integration at scale.
4. Set up a Kafka Consumer to store inference data into a database.
5. Develop a Django web application to analyze store performance using a variety of metrics.
You can follow along with implementing this sample application using the code on the NVIDIA-AI-IOT/deepstream-retail-analytics GitHub repo.
The end product of this sample is a custom dashboard, as shown in Figure 1. The dashboard provides analytical insights such as trends of the store traffic, counts of customers with shopping baskets, aisle occupancy, and more.
## Introduction to the application architecture
Before diving into the detailed workflow, this section provides an overview of the tools that will be used to build this project.
### NVIDIA DeepStream SDK
NVIDIA DeepStream SDK is NVIDIA’s streaming analytics toolkit that enables GPU-accelerated video analytics with support for high-performance AI inference across a variety of hardware platforms. DeepStream includes several reference applications to jumpstart development. These reference apps can be easily modified to suit new use cases and are available inside the DeepStream Docker images and at deepstream_reference_apps on GitHub.
This retail vision AI application is built on top of two of the reference applications, deepstream-test4 and deepstream-test5. Figure 2 shows the architecture of a typical DeepStream application.
### NVIDIA TAO Toolkit and pretrained models
NVIDIA TAO (Train, Adapt, and Optimize) Toolkit enables fine-tuning a variety of AI pretrained models to new domains. The TAO Toolkit is used in concert with the DeepStream application to perform analyses for unique use cases.
In this project, the model is used to detect whether or not a customer is carrying a shopping basket. DeepStream enables a seamless integration of TAO Toolkit with its existing pipeline without the need for heavy configuration.
Getting started with TAO Toolkit is easy. TAO Toolkit provides complete Jupyter notebooks for model customization for 100+ combinations of CV architectures and backbones. TAO Toolkit also provides a library of task-specific pretrained models for common retail tasks like people detection, pose estimation, action recognition, and more. To get started, see TAO Toolkit Quick Start
### Retail vision AI application workflow
The retail vision AI application architecture (Figure 3) consists of the following stages:
A DeepStream Pipeline with the following configuration:
• Primary Detector: Configure PeopleNet pretrained model from NGC to detect ‘Persons’
• Secondary Detector: Custom classification model trained using the TAO Toolkit for shopping basket detection
• Object Tracker: NvDCF tracker (in the accuracy configuration) to track the movement in the video stream
• Message Converter: Message converter to generate custom Kafka streaming payload from inference data
• Message Broker: Message broker to relay inference data to a Kafka receiver
kSQL Time Series Database: Used to store inference output streams from an edge inference server
Django Web Application: Application to analyze data stored in the kSQL database to generate insights regarding store performance, and serve these metrics as RESTful APIs and a web dashboard
Additionally, this app is built for x86 platforms with an NVIDIA GPU. However, it can be easily deployed on NVIDIA Jetson embedded platforms, such as the NVIDIA Jetson AGX Orin.
The next section walks you through the steps involved in building the application.
## Step 1: Building a custom NVIDIA DeepStream pipeline
To build the retail data analytics pipeline, start with the NVIDIA DeepStream reference applications deepstream-test4 and deepstream-test5. Code for the pipeline and a detailed description of the process is available in the deepstream-retail-analytics GitHub repo. We recommend using this post as a walkthrough to the code in the repository.
The deepstream-test4 application is a reference DeepStream pipeline that demonstrates adding `custom-detected` objects as `NVDS_EVENT_MSG_META` user metadata and attaching it to the buffer to be published. The deepstream-test5 is an end-to-end app that demonstrates how to use `nvmsgconv` and `nvmsgbroker` plugins in multistream pipelines, create `NVDS_META_EVENT_MSG` type of meta, and stream inference outputs using Kafka and other sink types.
This pipeline also integrates a secondary classifier in addition to the primary object detector, which can be useful for detecting shopper attributes once a person is detected in the retail video analytics application. The` `test4 application is used to modify the `nvmsgconv` plugin to include retail analytics attributes. Then, refer to the test5 application for secondary classifiers and streaming data from the pipeline using the `nvmsgbroker` over a Kafka topic.
Since the first step of the workflow is to identify persons and objects from the video feed, start by using the deepstream-test4 application for primary object detection. This object detection is done on the PeopleNet pretrained model that, by default, takes video input and detects people or their belongings.
For this use case, configure the model to capture only information about people. This can be accomplished easily by storing information only about the subset of frames that contain a person in the dataset.
With the primary `person` object detection done, use deepstream-test5 to add a secondary object classification model. This object classification shows whether or not a detected person is carrying a basket.
## Step 2: Building a custom model for shopping basket detection with NVIDIA TAO Toolkit
This section shows how to use the TAO Toolkit to fine-tune an object classification model and find out whether a person detected in the PeopleNet model is carrying a shopping basket (Figure 4).
To get started, collect and annotate training data from a retail environment for performing object classification. Use the Computer Vision Annotation Tool (CVAT) to annotate persons observed with the following labels:
• `hasBasket`: Person is carrying a basket
• `noBasket`: Person is not carrying a basket
This annotation is stored as a KITTI formatted dataset, where each line corresponds to a frame and thus an object. To make the data compatible for object classification, use the sample ‘kitti_to_classification‘ Python file on GitHub to crop the dataset. You can then perform object classification on it.
Next, use the TAO Toolkit to fine-tune a Resnet34 image classification model to perform classification on the training data. Learn more about the fine-tuning process at deepstream-retail-analytics/tree/main/TAO on GitHub.
After the custom model is created, run inference to validate that the model works as expected.
## Step 3: Integrating the Kafka message broker to create a custom frontend dashboard
With the primary object detection and secondary object classification models ready, the DeepStream application needs to relay this inference data to an analytics web server. Use the deepstream-test5 reference application as a template to stream data using Apache Kafka.
Here, a Kafka adapter that is built into DeepStream is used to publish messages to the Kafka message broker. Once the web server receives the Kafka streams from each camera inside a store, these inference output data are stored in a kSQL time-series database.
DeepStream has a default Kafka messaging shared library object that enables users to perform primary object detection and transmit the data seamlessly. This project further modifies this library to include information about the secondary classifier as well. This helps to stream data about shopping basket use inside the store.
The current DeepStream library includes `NvDsPersonObject`, which is used to define the `persons` detected in the primary detector. To ensure that the basket detection is mapped to each person uniquely, modify this class to include a `hasBasket` attribute in addition to the previously present attributes. Find more details at deepstream-retail-analytics/tree/main/nvmsgconv on GitHub.
After modifying the `NvDsPersonObject` to include basket detection, use the pipeline shown in Figure 5 to ensure the functionality for basket detection works appropriately.
As shown in the application pipeline in Figure 5, object detection and tracking are performed with the help of `pgie` and `sgie`. These are part of the `nvinfer` plugin as the primary and secondary inference engines. With `nvtracker`, transfer the data to the `nvosd` plugin. This `nvosd` plugin is responsible for drawing boxes around the objects that were detected in the previous sections.
Next, this inference data needs to be converted into message payload based on a specific schema that can be later consumed by the Kafka message broker to store and analyze the results. Use the `NvDsPersonsObject` (generated previously) for the updated payload in the eventmsg_payload file.
Finally, you now have the message payload with the custom schema. Use this to pass it through the Kafka protocol adapter and publish messages that the DeepStream application sends to the Kafka message broker at the specified broker address and topic. At this point, the final message payload is ready.
Now that the DeepStream pipeline is ready, build a web application to store the streaming inference data into a kSQL database. This web app, built using the Django framework, analyzes the inference data to generate metrics regarding store performance discussed earlier. These metrics are available through a RESTful API documented at deepstream-retail-analytics/tree/main/ds-retail-iva-frontend on GitHub.
To demonstrate the API functionality, we built a frontend web dashboard to visualize the results of the analytics server. This dashboard acts as a template for a storewide analytics system.
## Results
The previous steps demonstrated how to easily develop an end-to-end retail video analytics pipeline using NVIDIA DeepStream and NVIDIA TAO Toolkit. This pipeline helps retail establishments capitalize on pre-existing video feeds and find insightful information they can use to improve profits.
The workflow culminates in an easy-to-use web dashboard to analyze invaluable storewide data in real time. As shown in Figure 1, the dashboard presents the following information:
• Number of store visitors throughout the day
• Information about the proportion of customers shopping with and without baskets
• Visitors counts per store aisle
• Store occupancy heatmaps
• Customer journey visualization
These attributes can be easily amended to include information about specific use cases that are more relevant to each individual store. Stores can use this information to schedule staffing and improve the store layout to maximize efficiency.
For example, Figure 6 shows the overall distribution of customers in the store throughout the day, as well as the ratio of customers with and without baskets, respectively. While this sample application supports only a single camera stream, it can be easily modified to support multiple cameras. Scaling this application to multiple stores is equally easy to do.
The application uniquely detects `person 11` carrying the shopping basket by setting the attribute of `hasBasket`, whereas the other customers who do not carry the basket are marked with `noBasket`. Additionally, the `person 1` with a cardboard box is not identified to have a basket. Thus, the model is robust against false positives, ensuring that it was successfully trained to only pick up relevant information for this use case.
## Summary
This post demonstrated an end-to-end process to develop a vision AI application to perform retail analytics using NVIDIA TAO Toolkit and NVIDIA DeepStream SDK. Retail establishments can use the flux of video data they already have and build state-of-the-art video analytics applications. These apps can be deployed in real time and require minimal configuration to get started. In addition, the high customizability of this application ensures that it can be applied to any use case a store might benefit from.
Get started using the sample deepstream-retail-analytics application on GitHub.
Categories
## Formation of Robust Bound States of Interacting Photons
When quantum computers were first proposed, they were hoped to be a way to better understand the quantum world. With a so-called “quantum simulator,” one could engineer a quantum computer to investigate how various quantum phenomena arise, including those that are intractable to simulate with a classical computer.
But making a useful quantum simulator has been a challenge. Until now, quantum simulations with superconducting qubits have predominantly been used to verify pre-existing theoretical predictions and have rarely explored or discovered new phenomena. Only a few experiments with trapped ions or cold atoms have revealed new insights. Superconducting qubits, even though they are one of the main candidates for universal quantum computing and have demonstrated computational capabilities beyond classical reach, have so far not delivered on their potential for discovery.
In “Formation of Robust Bound States of Interacting Photons”, published in Nature, we describe a previously unpredicted phenomenon first discovered through experimental investigation. First, we present the experimental confirmation of the theoretical prediction of the existence of a composite particle of interacting photons, or a bound state, using the Google Sycamore quantum processor. Second, while studying this system, we discovered that even though one might guess the bound states to be fragile, they remain robust to perturbations that we expected to have otherwise destroyed them. Not only does this open the possibility of designing systems that leverage interactions between photons, it also marks a step forward in the use of superconducting quantum processors to make new scientific discoveries by simulating non-equilibrium quantum dynamics.
## Overview
Photons, or quanta of electromagnetic radiation like light and microwaves, typically don’t interact. For example, two intersecting flashlight beams will pass through one another undisturbed. In many applications, like telecommunications, the weak interactions of photons is a valuable feature. For other applications, such as computers based on light, the lack of interactions between photons is a shortcoming.
In a quantum processor, the qubits host microwave photons, which can be made to interact through two-qubit operations. This allows us to simulate the XXZ model, which describes the behavior of interacting photons. Importantly, this is one of the few examples of integrable models, i.e., one with a high degree of symmetry, which greatly reduces its complexity. When we implement the XXZ model on the Sycamore processor, we observe something striking: the interactions force the photons into bundles known as bound states.
Using this well-understood model as a starting point, we then push the study into a less-understood regime. We break the high level of symmetries displayed in the XXZ model by adding extra sites that can be occupied by the photons, making the system no longer integrable. While this nonintegrable regime is expected to exhibit chaotic behavior where bound states dissolve into their usual, solitary selves, we instead find that they survive!
## Bound Photons
To engineer a system that can support the formation of bound states, we study a ring of superconducting qubits that host microwave photons. If a photon is present, the value of the qubit is “1”, and if not, the value is “0”. Through the so-called “fSim” quantum gate, we connect neighboring sites, allowing the photons to hop around and interact with other photons on the nearest-neighboring sites.
Superconducting qubits can be occupied or unoccupied with microwave photons. The “fSim” gate operation allows photons to hop and interact with each other. The corresponding unitary evolution has a hopping term between two sites (orange) and an interaction term corresponding to an added phase when two adjacent sites are occupied by a photon.
We implement the fSim gate between neighboring qubits (left) to effectively form a ring of 24 interconnected qubits on which we simulate the behavior of the interacting photons (right).
The interactions between the photons affect their so-called “phase.” This phase keeps track of the oscillation of the photon’s wavefunction. When the photons are non-interacting, their phase accumulation is rather uninteresting. Like a well-rehearsed choir, they’re all in sync with one another. In this case, a photon that was initially next to another photon can hop away from its neighbor without getting out of sync. Just as every person in the choir contributes to the song, every possible path the photon can take contributes to the photon’s overall wavefunction. A group of photons initially clustered on neighboring sites will evolve into a superposition of all possible paths each photon might have taken.
When photons interact with their neighbors, this is no longer the case. If one photon hops away from its neighbor, its rate of phase accumulation changes, becoming out of sync with its neighbors. All paths in which the photons split apart overlap, leading to destructive interference. It would be like each choir member singing at their own pace — the song itself gets washed out, becoming impossible to discern through the din of the individual singers. Among all the possible configuration paths, the only possible scenario that survives is the configuration in which all photons remain clustered together in a bound state. This is why interaction can enhance and lead to the formation of a bound state: by suppressing all other possibilities in which photons are not bound together.
Left: Evolution of interacting photons forming a bound state. Right: Time goes from left to right, each path represents one of the paths that can break the 2-photon bonded state. Due to interactions, these paths interfere destructively, preventing the photons from splitting apart.
Occupation probability versus gate cycle, or discrete time step, for n-photon bound states. We prepare bound states of varying sizes and watch them evolve. We observe that the majority of the photons (darker colors) remain bound together.
In our processor, we start by putting two to five photons on adjacent sites (i.e., initializing two to five adjacent qubits in “1”, and the remaining qubits in “0”), and then study how they propagate. First, we notice that in the theoretically predicted parameter regime, they remain stuck together. Next, we find that the larger bound states move more slowly around the ring, consistent with the fact that they are “heavier”. This can be seen in the plot above where the lattice sites closest to Site 12, the initial position of the photons, remain darker than the others with increasing number of photons (nph) in the bound state, indicating that with more photons bound together there is less propagation around the ring.
## Bound States Behave Like Single Composite Particles
To more rigorously show that the bound states indeed behave as single particles with well-defined physical properties, we devise a method to measure how the energy of the particles changes with momentum, i.e., the energy-momentum dispersion relation.
To measure the energy of the bound state, we use the fact that the energy difference between two states determines how fast their relative phase grows with time. Hence, we prepare the bound state in a superposition with the state that has no photons, and measure their phase difference as a function of time and space. Then, to convert the result of this measurement to a dispersion relation, we utilize a Fourier transform, which translates position and time into momentum and energy, respectively. We’re left with the familiar energy-momentum relationship of excitations in a lattice.
Spectroscopy of bound states. We compare the phase accumulation of an n-photon bound state with that of the vacuum (no photons) as a function of lattice site and time. A 2D Fourier transform yields the dispersion relation of the bound-state quasiparticle.
## Breaking Integrability
The above system is “integrable,” meaning that it has a sufficient number of conserved quantities that its dynamics are constrained to a small part of the available computational space. In such integrable regimes, the appearance of bound states is not that surprising. In fact, bound states in similar systems were predicted in 2012, then observed in 2013. However, these bound states are fragile and their existence is usually thought to derive from integrability. For more complex systems, there is less symmetry and integrability is quickly lost. Our initial idea was to probe how these bound states disappear as we break integrability to better understand their rigidity.
To break integrability, we modify which qubits are connected with fSim gates. We add qubits so that at alternating sites, in addition to hopping to each of its two nearest-neighboring sites, a photon can also hop to a third site oriented radially outward from the ring.
While a bound state is constrained to a very small part of phase space, we expected that the chaotic behavior associated with integrability breaking would allow the system to explore the phase space more freely. This would cause the bound states to break apart. We find that this is not the case. Even when the integrability breaking is so strong that the photons are equally likely to hop to the third site as they are to hop to either of the two adjacent ring sites, the bound state remains intact, up to the decoherence effect that makes them slowly decay (see paper for details).
Top: New geometry to break integrability. Alternating sites are connected to a third site oriented radially outward. This increases the complexity of the system, and allows for potentially chaotic behavior. Bottom: Despite this added complexity pushing the system beyond integrability, we find that the 3-photon bound state remains stable even for a relatively large perturbation. The probability of remaining bound decreases slowly due to decoherence (see paper).
## Conclusion
We don’t yet have a satisfying explanation for this unexpected resilience. We speculate that it may be related to a phenomenon called prethermalization, where incommensurate energy scales in the system can prevent a system from reaching thermal equilibrium as quickly as it otherwise would. We believe further investigations will hopefully lead to new insights into many-body quantum physics, including the interplay of prethermalization and integrability.
## Acknowledgements
We would like to thank our Quantum Science Communicator Katherine McCormick for her help writing this blog post.
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## Boosting Dynamic Programming Performance Using NVIDIA Hopper GPU DPX Instructions
Dynamic programming (DP) is a well-known algorithmic technique and a mathematical optimization that has been used for several decades to solve groundbreaking…
Dynamic programming (DP) is a well-known algorithmic technique and a mathematical optimization that has been used for several decades to solve groundbreaking problems in computer science.
An example DP use case is route optimization with hundreds or thousands of constraints or weights using the Floyd-Warshall all-pair shortest paths algorithm. Another use case is the alignment of reads for genome sequence alignment using the Needleman-Wunsch or Smith-Waterman algorithms.
NVIDIA Hopper GPU Dynamic Programming X (DPX) instructions accelerate a large class of dynamic programming algorithms used in areas such as genomics, proteomics, and robot path planning. Accelerating these dynamic programming algorithms can help researchers, scientists, and practitioners glean insights much faster about the underlying DNA or protein structures and several other areas.
## What is dynamic programming?
DP techniques initially involve expressing the algorithm recursively, where the larger problem is broken down into subproblems that are easier to solve.
A common computational optimization used in DP is to save the results of the subproblems and use them in subsequent steps of the problem, instead of recomputing the solution each time. This step is called memoization. Memoization facilitates avoiding the recursion steps and instead enables using an iterative, look-up, tablebased formulation. The previously computed results are stored in the look-up table.
One of the key observations in many DP problems is that the solution to a larger problem often involves computing the minimum or maximum of the previously computed solutions. The larger problem’s solution is within a delta of that min-max of previous solutions.
DP techniques, in general, achieve the same results as brute force algorithms, but with dramatic reductions in the computational requirements and execution times.
## DP example: Accelerating the Smith-Waterman algorithm
NVIDIA Clara Parabricks is a GPU-accelerated genomic analysis suite of tools that heavily uses the Smith-Waterman algorithm and runs on NVIDIA GPUs: A100, V100, A40, A30, A10, A6000, T4, and soon the newest H100.
Genome sequencing has fundamental applications with universal benefits, with examples that include personalized medicine or tracking disease spread. Every cell in living organisms encodes genetic information using a sequence of four nucleotides in DNA (or bases). The nucleotides are adenine, cytosine, guanine, and thymine, represented by A, C, T, and G.
Simple organisms like viruses have a sequence of 10–100K bases while human DNA consists of about three billion base pairs. There are instruments (chemical– or electrical-signal–based) that sequence the bases of short segments of genetic material, called reads. These reads are typically 100–100K bases long, depending on the sequencer technology used for gathering the reads.
A critical computational step in genome sequence analysis is to align the reads to find the best match among a pair of reads. These reads can be 100-400 base pairs long in second-generation sequencers, and up to 100K bases long in third-generation sequencers. Aligning reads is a computational step that is repeated tens or hundreds of millions of times.
There are challenges in finding the best match that include the following:
• Naturally occurring variations in genomes that give organisms within a species their specific traits
• Errors in the reads themselves resulting from the sequencing instrument or underlying chemical processes
The best match between a pair of reads is equivalent to an approximate string match between a pair of strings with steps that reward matches and penalize differences. The differences between the reads could be mismatches, insertions, or deletions.
Figure 1 shows that the Smith-Waterman step in genomic sequencing aims to find the best match between the read sequences TGTTACGG and GGTTGACTA. The resulting best match is shown to be GTT-AC (from sequence 1, the “-” representing a deletion) with GTTGAC (from sequence 2). The scoring scheme in each step rewards matches (+3), penalizes mismatches (-3), and penalizes insertions and deletions (see the gap penalty formula in Figure 1).
This is an example formulation of the Smith-Waterman algorithm. Implementers of the Smith-Waterman algorithm are allowed to customize the rewards and penalties.
While computing the best match between TGTTACGG and GGTTGACTA, the Smith-Waterman algorithm also computes the best matches for all prefixes of TGTTACGG with all prefixes of GGTTGACTA. It proceeds from the start of these sequences and uses the results of the smaller prefixes to feed into the problem of finding the solution for the larger prefixes.
Figure 3 shows how the algorithm proceeds in terms of calculating the scores of matrices for matching a sequence of reads. This comparative matching is the computationally expensive step of the Smith-Waterman algorithm.
This is just one of the formulations of how the Smith-Waterman algorithm proceeds. Different formulations can result in the algorithm proceeding row-wise or column-wise as examples.
After the score matrix is computed, the next step involves backtracking from the highest score to the origin of each of these scores. This is a computationally light step given that each cell maintains how it got its score (the source cell for score calculation).
Figure 5 shows the computational efficiency of the Smith-Waterman calculations, where each of the subproblems solved by the algorithm is stored in the result matrix and never recomputed.
For example, in the process of calculating the best match of GGTTGACTA and TGTTACGG, the Smith-Waterman algorithm reuses the best match between GGTT (prefix of GGTTGACTA) and TGTT (prefix of TGTTACGG). In turn, while calculating the best match of GGTT and TGTT, the best match of all prefixes of these strings are calculated and reused (for example, best match of GGTT and TGT).
## Leveraging DPX instructions for better performance
The inner loop in a real Smith-Waterman implementation involves the following for each cell:
• Updating deletion penalties
• Updating insertion penalties
• Updating the score based on the updated insertion and deletion penalties.
The NVIDIA Hopper Architecture math API provides dramatic acceleration for such calculations. The APIs expose the acceleration provided by NVIDIA Hopper Streaming Multiprocessor for additions followed by minimum or maximum as a fused operation (for example, `__viaddmin_s16x2_relu`, an intrinsic that performs per-halfword $max(min(a + b, c), 0)$).
Another example of an API that is extensively leveraged by Smith-Waterman software is a three-way min or max followed by a clamp to zero ( for example, `__vimax3_s16x2_relu`, an intrinsic that performs per-halfword $max(max(max(a, b), c), 0)$).
Our implementation of the Smith-Waterman algorithm using the NVIDIA Hopper DPX instruction math APIs provides a 7.8x speedup over A100.
## Needleman-Wunsch and partial order alignment
In the same way that Smith-Waterman algorithms use DPX instructions, there is a large family of alignment algorithms that essentially use the same principles.
Examples include the Needleman-Wunsch algorithm in which the basic flow of the algorithm resembles the Smith-Waterman closely. However, the initialization, insertion, and gap penalties are calculated differently between these two approaches.
Algorithms like Partial Order Alignment make dense use of cell calculations that closely resemble Smith-Waterman cell calculations in their inner loop.
## All-pair shortest paths
Robotic path planning with thousands or tens of thousands of objects is a common problem in warehouses where the environment is dynamic with many moving objects. These scenarios can involve dynamic replanning every few milliseconds.
The inner loop of most all-pair shortest paths algorithms is as shown using the following Floyd-Warshall algorithm example. The pseudocode shows how the all-pair shortest paths algorithm has an inner loop that updates the min distance between each vertex pair. The most-dense operation is essentially an add followed by a min operation.
```initialize(dist); # initialize nearest neighbors to actual distance, all others = infinity
for k in range(V): #order of visiting k values not important, must visit each value
# pick all vertices as source in parallel
Parallel for_each i in range(V):
# Pick all vertices as destinations for the
# above picked source
Parallel for_each j in range(V):
# If vertex k is on the shortest path from
# i to j, then update the value of dist[i][j]
dist[i][j] = min (dist[i][j], dist[i][j] + dist[k][j])
# dist[i][j] calculation can be parallel within each k
# All dist[i][j] for a single ‘k’ must be computed
# before moving to the next ‘k’
Synchronize```
The speedup offered by DPX instructions makes it possible to dramatically scale the number of objects analyzed or have the re-optimization done in real time with fewer GPUs and optimal results.
## Accelerate dynamic programming algorithms with DPX instructions
Using NVIDIA Hopper DPX instructions demonstrated speedups of up to 7.8x on the A100 GPU for Smith-Waterman, which is key in many genomic sequence alignment and variant calling applications. The exposure in math APIs, available in CUDA 12, enables the configurable implementation of the Smith-Waterman algorithm to suit different user needs, as well as algorithms like Needleman-Wunsch.
DPX instructions accelerate a large class of dynamic programming algorithms such as DNA or protein sequencing, and robot path planning. Most importantly, these algorithms can lead to dramatic speed-ups in disease diagnosis, drug discoveries, and robot autonomy, making our everyday lives better.
### Acknowledgments
We’d like to thank Bill Dally, Alejandro Cachon, Mehrzad Samadi, Shirish Gadre, Daniel Stiffler, Rob Van der Wijngaart, Joseph Sullivan, Nikita Astafev, Seamus O’Boyle, and many others across NVIDIA.
Categories
## What Is a Pretrained AI Model?
Imagine trying to teach a toddler what a unicorn is. A good place to start might be by showing the child images of the creature and describing its unique features. Now imagine trying to teach an artificially intelligent machine what a unicorn is. Where would one even begin? Pretrained AI models offer a solution. A Read article >
The post What Is a Pretrained AI Model? appeared first on NVIDIA Blog.
Categories
## The Hunt Is On: ‘The Witcher 3: Wild Hunt’ Next-Gen Update Coming to GeForce NOW
It’s a wild GFN Thursday — The Witcher 3: Wild Hunt next-gen update will stream on GeForce NOW day and date, starting next week. Today, members can stream new seasons of Fortnite and Genshin Impact, alongside eight new games joining the library. In addition, the newest GeForce NOW app is rolling out this week with Read article >
The post The Hunt Is On: ‘The Witcher 3: Wild Hunt’ Next-Gen Update Coming to GeForce NOW appeared first on NVIDIA Blog.
Categories
## ‘23 and AV: Transportation Industry to Drive Into Metaverse, Cloud Technologies
As the autonomous vehicle industry enters the next year, it will start navigating into even greater technology frontiers. Next-generation vehicles won’t just be defined by autonomous driving capabilities. Everything from the design and production process to the in-vehicle experience is entering a new era of digitization, efficiency, safety and intelligence. These trends arrive after a Read article >
The post ‘23 and AV: Transportation Industry to Drive Into Metaverse, Cloud Technologies appeared first on NVIDIA Blog.
Categories
## Qubit Pharmaceuticals Accelerates Drug Discovery With Hybrid Quantum Computing
The promise of quantum computing is to solve unsolvable problems. And companies are already making headway with hybrid approaches — those that combine classical and quantum computing — to tackle challenges like drug discovery for incurable diseases. By accelerating drug molecule simulation and modeling with hybrid quantum computing, startup Qubit Pharmaceuticals is significantly reducing the Read article >
The post Qubit Pharmaceuticals Accelerates Drug Discovery With Hybrid Quantum Computing appeared first on NVIDIA Blog.
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2022-12-09 08:19:55
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## G = C22×C3⋊S3order 72 = 23·32
### Direct product of C22 and C3⋊S3
Aliases: C22×C3⋊S3, C62D6, C625C2, C323C23, (C2×C6)⋊5S3, C32(C22×S3), (C3×C6)⋊3C22, SmallGroup(72,49)
Series: Derived Chief Lower central Upper central
Derived series C1 — C32 — C22×C3⋊S3
Chief series C1 — C3 — C32 — C3⋊S3 — C2×C3⋊S3 — C22×C3⋊S3
Lower central C32 — C22×C3⋊S3
Upper central C1 — C22
Generators and relations for C22×C3⋊S3
G = < a,b,c,d,e | a2=b2=c3=d3=e2=1, ab=ba, ac=ca, ad=da, ae=ea, bc=cb, bd=db, be=eb, cd=dc, ece=c-1, ede=d-1 >
Subgroups: 272 in 96 conjugacy classes, 41 normal (5 characteristic)
C1, C2 [×3], C2 [×4], C3 [×4], C22, C22 [×6], S3 [×16], C6 [×12], C23, C32, D6 [×24], C2×C6 [×4], C3⋊S3 [×4], C3×C6 [×3], C22×S3 [×4], C2×C3⋊S3 [×6], C62, C22×C3⋊S3
Quotients: C1, C2 [×7], C22 [×7], S3 [×4], C23, D6 [×12], C3⋊S3, C22×S3 [×4], C2×C3⋊S3 [×3], C22×C3⋊S3
Character table of C22×C3⋊S3
class 1 2A 2B 2C 2D 2E 2F 2G 3A 3B 3C 3D 6A 6B 6C 6D 6E 6F 6G 6H 6I 6J 6K 6L size 1 1 1 1 9 9 9 9 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ρ1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 trivial ρ2 1 1 -1 -1 -1 1 -1 1 1 1 1 1 1 -1 -1 1 -1 -1 -1 -1 -1 -1 1 1 linear of order 2 ρ3 1 -1 1 -1 -1 -1 1 1 1 1 1 1 -1 1 1 -1 -1 -1 -1 -1 1 1 -1 -1 linear of order 2 ρ4 1 -1 -1 1 1 -1 -1 1 1 1 1 1 -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1 linear of order 2 ρ5 1 -1 -1 1 -1 1 1 -1 1 1 1 1 -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1 linear of order 2 ρ6 1 -1 1 -1 1 1 -1 -1 1 1 1 1 -1 1 1 -1 -1 -1 -1 -1 1 1 -1 -1 linear of order 2 ρ7 1 1 -1 -1 1 -1 1 -1 1 1 1 1 1 -1 -1 1 -1 -1 -1 -1 -1 -1 1 1 linear of order 2 ρ8 1 1 1 1 -1 -1 -1 -1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 linear of order 2 ρ9 2 -2 2 -2 0 0 0 0 -1 2 -1 -1 1 -1 -1 1 1 -2 1 1 -1 2 1 -2 orthogonal lifted from D6 ρ10 2 2 2 2 0 0 0 0 -1 2 -1 -1 -1 -1 -1 -1 -1 2 -1 -1 -1 2 -1 2 orthogonal lifted from S3 ρ11 2 -2 -2 2 0 0 0 0 -1 -1 2 -1 -2 -2 1 1 -1 -1 2 -1 1 1 1 1 orthogonal lifted from D6 ρ12 2 2 -2 -2 0 0 0 0 -1 -1 2 -1 2 -2 1 -1 1 1 -2 1 1 1 -1 -1 orthogonal lifted from D6 ρ13 2 2 -2 -2 0 0 0 0 -1 -1 -1 2 -1 1 -2 2 1 1 1 -2 1 1 -1 -1 orthogonal lifted from D6 ρ14 2 -2 -2 2 0 0 0 0 -1 -1 -1 2 1 1 -2 -2 -1 -1 -1 2 1 1 1 1 orthogonal lifted from D6 ρ15 2 -2 2 -2 0 0 0 0 2 -1 -1 -1 1 -1 -1 1 -2 1 1 1 2 -1 -2 1 orthogonal lifted from D6 ρ16 2 -2 -2 2 0 0 0 0 2 -1 -1 -1 1 1 1 1 2 -1 -1 -1 -2 1 -2 1 orthogonal lifted from D6 ρ17 2 2 2 2 0 0 0 0 -1 -1 2 -1 2 2 -1 -1 -1 -1 2 -1 -1 -1 -1 -1 orthogonal lifted from S3 ρ18 2 -2 2 -2 0 0 0 0 -1 -1 2 -1 -2 2 -1 1 1 1 -2 1 -1 -1 1 1 orthogonal lifted from D6 ρ19 2 -2 2 -2 0 0 0 0 -1 -1 -1 2 1 -1 2 -2 1 1 1 -2 -1 -1 1 1 orthogonal lifted from D6 ρ20 2 2 2 2 0 0 0 0 -1 -1 -1 2 -1 -1 2 2 -1 -1 -1 2 -1 -1 -1 -1 orthogonal lifted from S3 ρ21 2 2 -2 -2 0 0 0 0 -1 2 -1 -1 -1 1 1 -1 1 -2 1 1 1 -2 -1 2 orthogonal lifted from D6 ρ22 2 -2 -2 2 0 0 0 0 -1 2 -1 -1 1 1 1 1 -1 2 -1 -1 1 -2 1 -2 orthogonal lifted from D6 ρ23 2 2 2 2 0 0 0 0 2 -1 -1 -1 -1 -1 -1 -1 2 -1 -1 -1 2 -1 2 -1 orthogonal lifted from S3 ρ24 2 2 -2 -2 0 0 0 0 2 -1 -1 -1 -1 1 1 -1 -2 1 1 1 -2 1 2 -1 orthogonal lifted from D6
Smallest permutation representation of C22×C3⋊S3
On 36 points
Generators in S36
(1 29)(2 30)(3 28)(4 31)(5 32)(6 33)(7 34)(8 35)(9 36)(10 19)(11 20)(12 21)(13 22)(14 23)(15 24)(16 25)(17 26)(18 27)
(1 11)(2 12)(3 10)(4 13)(5 14)(6 15)(7 16)(8 17)(9 18)(19 28)(20 29)(21 30)(22 31)(23 32)(24 33)(25 34)(26 35)(27 36)
(1 2 3)(4 5 6)(7 8 9)(10 11 12)(13 14 15)(16 17 18)(19 20 21)(22 23 24)(25 26 27)(28 29 30)(31 32 33)(34 35 36)
(1 5 8)(2 6 9)(3 4 7)(10 13 16)(11 14 17)(12 15 18)(19 22 25)(20 23 26)(21 24 27)(28 31 34)(29 32 35)(30 33 36)
(2 3)(4 9)(5 8)(6 7)(10 12)(13 18)(14 17)(15 16)(19 21)(22 27)(23 26)(24 25)(28 30)(31 36)(32 35)(33 34)
G:=sub<Sym(36)| (1,29)(2,30)(3,28)(4,31)(5,32)(6,33)(7,34)(8,35)(9,36)(10,19)(11,20)(12,21)(13,22)(14,23)(15,24)(16,25)(17,26)(18,27), (1,11)(2,12)(3,10)(4,13)(5,14)(6,15)(7,16)(8,17)(9,18)(19,28)(20,29)(21,30)(22,31)(23,32)(24,33)(25,34)(26,35)(27,36), (1,2,3)(4,5,6)(7,8,9)(10,11,12)(13,14,15)(16,17,18)(19,20,21)(22,23,24)(25,26,27)(28,29,30)(31,32,33)(34,35,36), (1,5,8)(2,6,9)(3,4,7)(10,13,16)(11,14,17)(12,15,18)(19,22,25)(20,23,26)(21,24,27)(28,31,34)(29,32,35)(30,33,36), (2,3)(4,9)(5,8)(6,7)(10,12)(13,18)(14,17)(15,16)(19,21)(22,27)(23,26)(24,25)(28,30)(31,36)(32,35)(33,34)>;
G:=Group( (1,29)(2,30)(3,28)(4,31)(5,32)(6,33)(7,34)(8,35)(9,36)(10,19)(11,20)(12,21)(13,22)(14,23)(15,24)(16,25)(17,26)(18,27), (1,11)(2,12)(3,10)(4,13)(5,14)(6,15)(7,16)(8,17)(9,18)(19,28)(20,29)(21,30)(22,31)(23,32)(24,33)(25,34)(26,35)(27,36), (1,2,3)(4,5,6)(7,8,9)(10,11,12)(13,14,15)(16,17,18)(19,20,21)(22,23,24)(25,26,27)(28,29,30)(31,32,33)(34,35,36), (1,5,8)(2,6,9)(3,4,7)(10,13,16)(11,14,17)(12,15,18)(19,22,25)(20,23,26)(21,24,27)(28,31,34)(29,32,35)(30,33,36), (2,3)(4,9)(5,8)(6,7)(10,12)(13,18)(14,17)(15,16)(19,21)(22,27)(23,26)(24,25)(28,30)(31,36)(32,35)(33,34) );
G=PermutationGroup([(1,29),(2,30),(3,28),(4,31),(5,32),(6,33),(7,34),(8,35),(9,36),(10,19),(11,20),(12,21),(13,22),(14,23),(15,24),(16,25),(17,26),(18,27)], [(1,11),(2,12),(3,10),(4,13),(5,14),(6,15),(7,16),(8,17),(9,18),(19,28),(20,29),(21,30),(22,31),(23,32),(24,33),(25,34),(26,35),(27,36)], [(1,2,3),(4,5,6),(7,8,9),(10,11,12),(13,14,15),(16,17,18),(19,20,21),(22,23,24),(25,26,27),(28,29,30),(31,32,33),(34,35,36)], [(1,5,8),(2,6,9),(3,4,7),(10,13,16),(11,14,17),(12,15,18),(19,22,25),(20,23,26),(21,24,27),(28,31,34),(29,32,35),(30,33,36)], [(2,3),(4,9),(5,8),(6,7),(10,12),(13,18),(14,17),(15,16),(19,21),(22,27),(23,26),(24,25),(28,30),(31,36),(32,35),(33,34)])
C22×C3⋊S3 is a maximal subgroup of C6.D12 C6.11D12 C62⋊C4 Dic3⋊D6 C22×S32 C62⋊C6
C22×C3⋊S3 is a maximal quotient of C12.59D6 C12.D6 C12.26D6
Matrix representation of C22×C3⋊S3 in GL4(ℤ) generated by
1 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 -1
,
-1 0 0 0 0 -1 0 0 0 0 1 0 0 0 0 1
,
1 0 0 0 0 1 0 0 0 0 -1 1 0 0 -1 0
,
0 -1 0 0 1 -1 0 0 0 0 -1 1 0 0 -1 0
,
0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0
G:=sub<GL(4,Integers())| [1,0,0,0,0,1,0,0,0,0,-1,0,0,0,0,-1],[-1,0,0,0,0,-1,0,0,0,0,1,0,0,0,0,1],[1,0,0,0,0,1,0,0,0,0,-1,-1,0,0,1,0],[0,1,0,0,-1,-1,0,0,0,0,-1,-1,0,0,1,0],[0,1,0,0,1,0,0,0,0,0,0,1,0,0,1,0] >;
C22×C3⋊S3 in GAP, Magma, Sage, TeX
C_2^2\times C_3\rtimes S_3
% in TeX
G:=Group("C2^2xC3:S3");
// GroupNames label
G:=SmallGroup(72,49);
// by ID
G=gap.SmallGroup(72,49);
# by ID
G:=PCGroup([5,-2,-2,-2,-3,-3,323,1204]);
// Polycyclic
G:=Group<a,b,c,d,e|a^2=b^2=c^3=d^3=e^2=1,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,b*c=c*b,b*d=d*b,b*e=e*b,c*d=d*c,e*c*e=c^-1,e*d*e=d^-1>;
// generators/relations
Export
×
𝔽
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2020-01-26 02:22:36
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|
http://www.chim.lu/ech0255.php
|
Search:
# Iodine in equilibrium between the pure phase and the gaseous phase
Two test tubes contain different quantities of iodine.
→ Theory
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2020-07-02 22:14:47
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|
https://www.shaalaa.com/question-bank-solutions/write-function-definition-tower-c-read-content-text-file-writeuptxt-count-presence-word-tower-display-number-occurrences-this-word-ifstream-ofstream-classes_20057
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# Write Function Definition for Tower() in C++ to Read the Content of a Text File Writeup.Txt, Count the Presence of Word Tower and Display the Number of Occurrences of this Word - Computer Science (C++)
Write function definition for TOWER() in C++ to read the content of a text file WRITEUP.TXT, count the presence of word TOWER and display the number of occurrences of this word.
Note:
• The word TOWER should be an independent word
• Ignore type cases (i.e. lower/upper case)
Example:
If the content of the file WRITEUP.TXT is as follows:
Tower of hanoi is an interesting problem. Mobile phone tower is away from here. Views from EIFFEL TOWER are amazing.
The function TOWER( ) should display the following :
3
#### Solution
void TOWER()
{
int count=0;
ifstream f("WRITEUP.TXT");
char s[20];
while (!f.eof())
{
f>>s;
if (strcmpi(s,"TOWER")==0)
count++;
}
cout<<count;
f.close();
}
Concept: ifstream, ofstream, Classes
Is there an error in this question or solution?
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2021-03-02 02:08:56
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https://msp.org/gt/2010/14-5/p03.xhtml
|
Volume 14, issue 5 (2010)
Recent Issues
The Journal About the Journal Editorial Board Subscriptions Editorial Interests Editorial Procedure Submission Guidelines Submission Page Ethics Statement ISSN (electronic): 1364-0380 ISSN (print): 1465-3060 Author Index To Appear Other MSP Journals
Embedded contact homology and Seiberg–Witten Floer cohomology III
Clifford Henry Taubes
Geometry & Topology 14 (2010) 2721–2817
Abstract
This is the third of five papers that construct an isomorphism between the embedded contact homology and Seiberg–Witten Floer cohomology of a compact $3$–manifold with a given contact $1$–form.
Keywords
Seiberg–Witten equations, Floer homology, contact homology
Primary: 57R17
Secondary: 57R57
Publication
Received: 15 November 2008
Revised: 11 May 2010
Accepted: 1 June 2010
Published: 15 December 2010
Proposed: Rob Kirby
Seconded: Danny Calegari, Peter Ozsváth
Authors
Clifford Henry Taubes Department of Mathematics Harvard University Cambridge, MA 02138 USA
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2020-07-11 19:01:28
|
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|
https://byjus.com/rd-sharma-solutions/class-10-maths-chapter-2-polynomials-exercise-2-1/
|
# RD Sharma Solutions Class 10 Polynomials Exercise 2.1
## RD Sharma Solutions Class 10 Chapter 2 Exercise 2.1
### RD Sharma Class 10 Solutions Chapter 2 Ex 2.1 PDF Free Download
#### Exercise 2.1
Q.1: Find the zeroes of each of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:
(i) $f\left ( x \right ) = x^{2} – 2x – 8$
(ii) $g\left ( s \right ) = 4s^{2} – 4s + 1$
(iii) $6x^{2} – 3 – 7x$
(iv) $h\left ( t \right ) = t^{2} – 15$
(v) $p\left ( x \right ) = x^{2} + 2\sqrt{2}x – 6$
(vi) $q\left ( x \right ) = \sqrt{3}x^{2} + 10x + 7\sqrt{3}$
(vii) $f\left ( x \right ) = x^{2} – \left ( \sqrt{3} + 1 \right )x + \sqrt{3}$
(viii) $g\left ( x \right ) = a\left ( x^{2} + 1 \right ) – x\left ( a^{2} + 1 \right )$
Solution:
(i) $f\left ( x \right ) = x^{2} – 2x – 8$
We have,
$f\left ( x \right ) = x^{2} – 2x – 8$
= $x^{2} – 4x + 2x – 8$
= $x\left ( x – 4 \right ) + 2\left ( x – 4 \right )$
= $\left ( x + 2 \right )\left ( x – 4 \right )$
Zeroes of the polynomials are -2 and 4.
Now,
Sum of the zeroes = $\frac{- \; coefficient \; of \; x}{coefficient \; of \; x}$
-2 + 4 = $\frac{-\left ( -2 \right )}{1}$
2 = 2
Product of the zeroes = $\frac{constant \; term}{Coefficient \; of \; x}$
-8 = $\frac{-8}{1}$
-8 = -8
Hence, the relationship is verified.
(ii) $g\left ( s \right ) = 4s^{2} – 4s + 1$
We have,
$g\left ( s \right ) = 4s^{2} – 4s + 1$
= $4s^{2} – 2s – 2s + 1$
= $2s\left ( 2s – 1 \right ) -1\left ( 2s – 1 \right )$
= $\left ( 2s – 1 \right )\left ( 2s – 1 \right )$
Zeroes of the polynomials are $\frac{1}{2}$ and $\frac{1}{2}$.
Sum of zeroes = $\frac{- coefficient \; of \; s}{coefficient \; of \; s^{2}}$
$\frac{1}{2} + \frac{1}{2} = \frac{-\left ( -4 \right )}{4}$
1 = 1
Product of zeroes = $\frac{constant \; term}{Coefficient \; of \; x}$
$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
$\frac{1}{4} = \frac{1}{4}$
Hence, the relationship is verified.
(iii) $6s^{2}-3-7x$
= $6s^{2}-7x-3$ = (3x + 11)(2x – 3)
Zeros of the polynomials are $\frac{3}{2} \; and \; \frac{-1}{3}$
Sum of the zeros = $\frac{- coefficient \; of \; x}{coefficient of x^{2}}$
$\frac{-1}{3} + \frac{3}{2} = \frac{-\left ( -7 \right )}{6}$
$\frac{7}{6} = \frac{7}{6}$
Product of the zeroes = $\frac{constant \; term}{coefficient \; of \; x^{2}}$
$\frac{-1}{3} \times \frac{3}{2} = \frac{-3}{6}$
$\frac{-3}{6} = \frac{-3}{6}$
Hence, the relationship is verified.
(iv) $h\left ( t \right ) = t^{2} – 15$
We have,
$h\left ( t \right ) = t^{2} – 15$
= $t^{2} – \sqrt{15}$
= $\left ( t + \sqrt{15} \right )\left ( t – \sqrt{15} \right )$
Zeroes of the polynomials are $-\sqrt{15} \; and \; \sqrt{15}$
Sum of the zeroes = 0
$-\sqrt{15} + \sqrt{15}$ = 0
0 = 0
Product of zeroes = $\frac{constant \; term}{Coefficient \; of \; x}$ = $\frac{-15}{1}$
$-\sqrt{15} \times \sqrt{15} = -15$
-15 = -15
Hence, the relationship verified.
(v) $p\left ( x \right ) = x^{2} + 2\sqrt{2}x – 6$
We have,
$p\left ( x \right ) = x^{2} + 2\sqrt{2}x – 6$
= $x^{2} + 3\sqrt{2}x + 3\sqrt{2}x – 6$
= $x\left ( x + 3\sqrt{2} \right ) – \sqrt{2}\left ( x + 3\sqrt{2} \right )$
= $\left ( x + 3\sqrt{2} \right )\left ( x – \sqrt{2} \right )$
Zeroes of the polynomials are $3\sqrt{2}$ and $- 3\sqrt{2}$
Sum of the zeroes = $\frac{-2\sqrt{2}}{1}$
$\sqrt{2} – 3\sqrt{2} = -2\sqrt{2}$
$– 2\sqrt{2} = -2\sqrt{2}$
Product of the zeroes = $\frac{constant \; term}{Coefficient \; of \; x}$
$\sqrt{2} \times -3\sqrt{2} = \frac{-6}{1}$
-6 = -6
Hence, the relationship is verified.
(vi) q(x) = $\sqrt{3}x^{2}+10x+7\sqrt{3}$
= $\sqrt{3}x^{2}+7x+3x+7\sqrt{3}$
= $\sqrt{3}x\left( x +\sqrt{3}\right) 7\left( x + sqrt{3} \right)$
= $\left( x +\sqrt{3}\right) \left( 7 + sqrt{3} \right)$
Zeros of the polynomials are $-\sqrt{3}and \frac {-7}{\sqrt{3}}$
Sum of zeros = $\frac {-10}{\sqrt{3}}$
$-\sqrt{3}- \frac {7}{\sqrt{3}} = \frac {-10}{\sqrt{3}}$
$\frac {-10}{\sqrt{3}} = \frac {-10}{\sqrt{3}}$
Product of the polynomials are $-\sqrt{3}, \frac {-7}{\sqrt{3}}$
7 = 7
Hence, the relationship is verified.
(vii) h(x) = $x^{2}-\left( \sqrt{3}+1\right)x + \sqrt{3}$
= $x^{2}-\sqrt{3} -x + \sqrt{3}$
=$x(x – \sqrt{3}) -1(x-\sqrt{3})$
= $(x – \sqrt{3}) (x-1)$
Zeros of the polynomials are 1 and $\sqrt{3}$
Sum of zeros = $\frac{-co\;efficient \;of \;x}{co\; efficient \;of\; x^{2}} = -[-\sqrt{3}-1]$
$1+\sqrt{3} = \sqrt{3} +1$
Product of zeros = $\frac{-co \; efficient \; of \; x}{co \; efficient \; of \; x^{2}} = \sqrt{3}$
$\sqrt{3} = \sqrt{3}$
Hence, the relationship is verified
(viii) g(x) = $a\left [ \left ( x^{2} + 1 \right ) – x\left ( a^{2} + 1 \right ) \right ]^{2}$
= $ax^{2}+a-a^{2}x -x$
= $ax^{2}-[(a^{2}x +1)]+a$
= $ax^{2}-a^{2}x –x +a$
= $ax(x-a)-1(x –a)$ = $\left ( x – a \right )\left ( ax – 1 \right )$
Zeros of the polynomials are $\frac{1}{a} \; and \; 1$
Sum of the zeros = $\frac{a[-a^{2}-1]}{a}$
$\frac{1}{a}+a=\frac{a^{2}+1}{a}\ \frac{a^{2}+1}{a}=\frac{a^{2}+1}{a}$
Product of zeros = a/a
$\frac{1}{a}\times a= \frac{a}{a}$
1 = 1
Hence, the relationship is verified.
Q.2: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} – 5x + 4$, find the value of $\frac{1}{\alpha } + \frac{1}{\beta } – 2\alpha \beta$.
Solution: We have,
$\alpha \; and \; \beta$ are the roots of the quadratic polynomial.
$f\left ( x \right ) = x^{2} – 5x + 4$
Sum of the roots = $\alpha \; +\; \beta$ = 5
Product of the roots = $\alpha \beta$ = 4
So,
$\frac{1}{\alpha } + \frac{1}{\beta } – 2\alpha \beta$ = $\frac{\beta + \alpha }{\alpha \beta } – 2\alpha \beta$
= $\frac{5}{4} – 2\times 4 = \frac{5}{4} – 8 = \frac{-27}{4}$
Q.3: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} – 5x + 4$, find the value of $\frac{1}{\alpha } + \frac{1}{\beta } – 2 \alpha \beta$.
Solution:
Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial.
$p\left ( y \right ) = x^{2} – 5x + 4$
Sum of the zeroes = $\alpha \; +\; \beta$ = 5
Product of the roots = $\alpha \beta$ = 4
So,
$\frac{1}{\alpha } + \frac{1}{\beta } – 2 \alpha \beta$
= $\frac{ \beta + \alpha – 2 \alpha ^{2} \beta ^{2}}{\alpha \beta }$
= $\frac{ \left ( \alpha + \beta \right ) – 2 \left ( \alpha \beta \right )^{2}}{\alpha \beta }$
= $\frac{ \left ( 5 \right ) – 2 \left ( 4 \right )^{2}}{4}$
= $\frac{ 5 – 2 \times 16}{4}$ = $\frac{5 – 32}{4}$ = $\frac{- 27}{4}$
Q.4: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $p\left ( y \right ) = 5y^{2} – 7y + 1$, find the value of $\frac{1}{\alpha } + \frac{1}{\beta }$.
Solution: Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial.
$p\left ( y \right ) = 5y^{2} – 7y + 1$
Sum of the zeroes = $\alpha \; +\; \beta$ = 7
Product of the roots = $\alpha \beta$ = 1
So,
$\frac{1}{\alpha } + \frac{1}{\beta }$ = $\frac{\alpha + \beta}{\alpha \beta}$ = $\frac{7}{1}$ = 7
Q.5: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} – x – 4$, find the value of $\frac{1 }{\alpha } + \frac{1 }{\beta } – \alpha \beta$.
Solution:
Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial.
We have,
$f\left ( x \right ) = x^{2} – x – 4$
Sum of zeroes = $\alpha \; +\; \beta$ = 1
Product of the zeroes = $\alpha \beta$ = -4
So,
$\frac{1 }{\alpha } + \frac{1 }{\beta } – \alpha \beta$ = $\frac{\alpha + \beta }{\alpha \beta } – \alpha \beta$
= $\frac{1}{-4} – \left ( -4 \right ) \; = \frac{-1}{4} + 4$
= $\frac{-1 + 16}{4}$ = $\frac{15}{4}$
Q.6: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} + x – 2$, find the value of $\frac{1 }{\alpha } – \frac{1 }{\beta }$.
Solution:
Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial.
We have,
$f\left ( x \right ) = x^{2} + x – 2$
Sum of zeroes = $\alpha \; +\; \beta$ = 1
Product of the zeroes = $\alpha \beta$ = -2
So,
$\frac{1 }{\alpha } + \frac{1 }{\beta } – \alpha \beta$ = $\frac{\alpha + \beta }{\alpha \beta }$
= $\frac{\beta – \alpha }{\alpha \beta }$
$= \frac{\beta – \alpha }{\alpha \beta }\times \frac{\left ( \alpha – \beta \right )}{\alpha \beta }$
$= \frac{\sqrt{\left ( \alpha + \beta \right )^{2} – 4\alpha \beta }}{\alpha \beta }$
$= \frac{\sqrt{1 + 8}}{2} = \frac{\sqrt{9}}{2} = \frac{3}{2}$
Q.7: If one of the zero of the quadratic polynomial $f\left ( x \right ) = 4x^{2} – 8kx – 9$ is negative of the other, then find the value of k.
Solution:
Let, the two zeroes of the polynomial $f\left ( x \right ) = 4x^{2} – 8kx – 9$ be $\alpha \; and \; -\alpha$.
Product of the zeroes = $\alpha \times -\alpha$ = -9
Sum of the zeroes = $\alpha + \left (-\alpha \right )$ = -8k = 0 $Since, \alpha – \alpha = 0$
$\Rightarrow 8k = 0$
$\Rightarrow k = 0$
Q.8: If the sum of the zeroes of the quadratic polynomial $f\left ( t \right ) = kt^{2}+ 2t + 3k$ is equal to their product, then find the value of k.
Solution: Let the two zeroes of the polynomial $f\left ( t \right ) = kt^{2}+ 2t + 3k$ be $\alpha \; and \; \beta$.
Sum of the zeroes = $\alpha + \beta$ = 2
Product of the zeroes = $\alpha \times \beta$ = 3k
Now,
$\frac{-2}{k} = \frac{3k}{k}$
$\Rightarrow 3k = -2$
$\Rightarrow k = \frac{-2}{3}$
So, k = 0 and $\Rightarrow k = \frac{-2}{3}$
Q.9: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $p\left ( x \right ) = 4x^{2} – 5x – 1$, find the value of $\alpha ^{2} \beta + \alpha \beta ^{2}$.
Solution:
Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $p\left ( x \right ) = 4x^{2} – 5x – 1$
So, Sum of the zeroes = $\alpha + \beta$ = $\frac{5}{4}$
Product of the zeroes = $\alpha \times \beta$$\frac{-1}{4}$
Now,
$\alpha ^{2} \beta + \alpha \beta ^{2}$ = $\alpha \beta \left ( \alpha + \beta \right )$
= $\frac{5}{4} \left ( \frac{-1}{4} \right )$
= $\frac{-5}{16}$
Q.10: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( t \right ) = t^{2} – 4t + 3$, find the value of $\alpha ^{4} \beta ^{3} + \alpha ^{3} \beta ^{4}$.
Solution:
Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( t \right ) = t^{2} – 4t + 3$
So, Sum of the zeroes = $\alpha + \beta$ = 4
Product of the zeroes = $\alpha \times \beta$ = 3
Now,
$\alpha ^{4} \beta ^{3} + \alpha ^{3} \beta ^{4}$ = $\alpha ^{3} \beta ^{3} \left ( \alpha + \beta \right )$
= $\left ( 3 \right )^{3} \left ( 4 \right )$ = 108
Q.11: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = 6x^{2} + x – 2$, find the value of $\frac{\alpha }{\beta } + \frac{\beta }{\alpha }$.
Solution:
Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = 6x^{2} + x – 2$.
Sum of the zeroes = $\alpha + \beta$ = $\frac{-1}{6}$
Product of the zeroes = $\alpha \times \beta$$\frac{-1}{3}$
Now,
$\frac{\alpha }{\beta } + \frac{\beta }{\alpha }$
= $\frac{\left (\alpha ^{2} + \beta ^{2} \right ) – 2\alpha \beta}{\alpha \beta }$
By substitution the values of the sum of zeroes and products of the zeroes, we will get
= $\frac{-25}{12}$
Q.12: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = 6x^{2} + x – 2$, find the value of $\frac{\alpha }{\beta } + 2\left [ \frac{1}{\alpha } + \frac{1}{\beta } \right ] + 3\alpha \beta$.
Solution:
Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = 6x^{2} + x – 2$.
Sum of the zeroes = $\alpha + \beta$ = $\frac{6}{3}$
Product of the zeroes = $\alpha \times \beta$$\frac{4}{3}$
Now,
$\frac{\alpha }{\beta } + 2\left [ \frac{1}{\alpha } + \frac{1}{\beta } \right ] + 3\alpha \beta$.
= $\frac{\alpha ^{2} + \beta ^{2}}{\alpha \beta } + 2\left [ \frac{1}{\alpha } + \frac{1}{\beta } \right ] + 3\alpha \beta$
= $\frac{\left ( \alpha + \beta \right )^{2} – 2\alpha \beta }{\alpha \beta } + 2\left [ \frac{1}{\alpha } + \frac{1}{\beta } \right ] + 3\alpha \beta$
By substituting the values of sum and product of the zeroes, we will get
$\frac{\alpha }{\beta } + 2\left [ \frac{1}{\alpha } + \frac{1}{\beta } \right ] + 3\alpha \beta$ = 8
Q.13: If the squared difference of the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} + px + 45$ is equal to 144, find the value of p.
Solution:
Let the two zeroes of the polynomial be $\alpha \; and \; \beta$.
We have,
$f\left ( x \right ) = x^{2} + px + 45$
Now,
Sum of the zeroes = $\alpha + \beta$ =-p
Product of the zeroes = $\alpha \times \beta$ = 45
So,
$\left ( \alpha + \beta \right )^{2} – 4\alpha \beta = 144$
$\left (p\right )^{2} – 4\times 45 = 144$
$\left (p\right )^{2} = 144 + 180$
$\left (p\right )^{2} = 324$
$p = \sqrt{324}$
$p = \pm 18$
Thus, in the given equation, p will be either 18 or -18.
Q.14: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} – px + q$, prove that $\frac{\alpha ^{2}}{\beta ^{2}} + \frac{\beta ^{2}}{\alpha ^{2}} = \frac{p^{4}}{q^{2}} – \frac{4p^{2}}{q} + 2$.
Solution:
Since, $\alpha \; and \; \beta$ are the roots of the quadratic polynomial given in the question.
$f\left ( x \right ) = x^{2} – px + q$
Now,
Sum of the zeroes = p = $\alpha + \beta$
Product of the zeroes = q = $\alpha \times \beta$
LHS = $\frac{\alpha ^{2}}{\beta ^{2}} + \frac{\beta ^{2}}{\alpha ^{2}}$
= $\frac{\alpha ^{4} + \beta ^{4}}{\alpha ^{2} \beta ^{2}}$
= $\frac{\left (\alpha ^{2} + \beta ^{2} \right )^{2} – 2\left (\alpha \beta \right )^{2}}{\left (\alpha \beta \right )^{2}}$
= $\frac{\left [\left (\alpha + \beta \right )^{2} – 2\alpha \beta \right ]^{2} – 2\left (\alpha \beta \right )^{2}}{\left (\alpha \beta \right )^{2}}$
= $\frac{\left [\left (p \right )^{2} – 2 q \right ]^{2} – 2\left (q \right )^{2}}{\left (q \right )^{2}}$
= $\frac{\left (p ^{4} + 4 q ^{2} – 4 p ^{2} q \right ) – 2q^{2}}{q ^{2}}$
= $\frac{p ^{4} + 2 q ^{2} – 4 p ^{2} q }{q ^{2}}$
= $\frac{p ^{2}}{q ^{2}} + 2 – \frac{4 p^{2}}{q}$
= $\frac{p ^{2}}{q ^{2}} – \frac{4 p^{2}}{q} + 2$
LHS = RHS
Hence, proved.
Q.15: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} – p( x + 1 ) – c$, show that $\left ( \alpha + 1 \right )\left ( \beta + 1 \right ) = 1 – c$.
Solution:
Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial
$f\left ( x \right ) = x^{2} – p( x + 1 ) – c$
Now,
Sum of the zeroes = $\alpha + \beta$ = p
Product of the zeroes = $\alpha \times \beta$ = (- p – c)
So,
$\left (\alpha + 1 \right )\left ( \beta + 1 \right )$
= $\alpha \beta + \alpha + \beta + 1$
= $\alpha \beta + \left (\alpha + \beta \right ) + 1$
= $\left (- p – c \right ) + p + 1$
= 1 – c = RHS
So, LHS = RHS
Hence, proved.
Q.16: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial such that $\alpha + \beta = 24 \; and \; \alpha – \beta = 8$, find a quadratic polynomial having $\alpha \; and \; \beta$ as its zeroes.
Solution:
We have,
$\alpha + \beta = 24$ …………E-1
$\alpha – \beta = 8$ ………….E-2
By solving the above two equations accordingly, we will get
$2\alpha = 32$
$\alpha = 16$
Substitute the value of $\alpha$, in any of the equation. Let we substitute it in E-2, we will get
$\beta = 16 – 8$
$\beta = 8$
Now,
Sum of the zeroes of the new polynomial = $\alpha + \beta$ = 16 + 8 = 24
Product of the zeroes = $\alpha \beta$ = $16 \times 8$ = 128
K $x^{2} – \left ( sum \; of \; the \; zeroes \right )x + \left ( product \; of \; the \; zeroes \right )$ = $x^{2} – 24 x + 128$
Hence, the required quadratic polynomial is $f\left ( x \right ) = x^{2} + 24 x + 128$
Q.17: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x ^{2} – 1$, find a quadratic polynomial whose zeroes are $\frac{2\alpha }{\beta } \; and \; \frac{2\beta }{\alpha }$.
Solution:
We have,
$f\left ( x \right ) = x ^{2} – 1$
Sum of the zeroes = $\alpha + \beta$ = 0
Product of the zeroes = $\alpha \beta$ = -1
From the question,
Sum of the zeroes of the new polynomial = $\frac{2\alpha }{\beta } \; and \; \frac{2\beta }{\alpha }$
= $\frac{2\alpha ^{2} + 2\beta ^{2}}{\alpha \beta }$
= $\frac{2 \left ( \alpha ^{2} + \beta ^{2} \right )}{ \alpha \beta }$
= $\frac{2\left ( \left ( \alpha + \beta \right )^{2} – 2\alpha \beta \right )}{\alpha \beta }$
= $\frac{2\left ( 2 \right )1}{-1}$ { By substituting the value of the sum and products of the zeroes }
As given in the question,
Product of the zeroes = $\frac{\left ( 2 \alpha \right ) \left ( 2 \beta \right )}{\alpha \beta }$ = $\frac{4 \alpha \beta }{\alpha \beta }$ = 4
$x^{2} – \left ( sum \; of \; the \; zeroes \right )x + \left ( product \; of \; the \; zeroes \right )$
= k$x^{2} – \left ( -4 \right )x + 4$ = $x^{2} + 4x + 4$
Hence, the required quadratic polynomial is $f\left ( x \right ) = x^{2} + 4x + 4$
Q.18: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x ^{2} – 3x – 2$, find a quadratic polynomial whose zeroes are $\frac{1}{2\alpha + \beta } \; and \; \frac{1}{2\beta + \alpha }$.
Solution:
We have,
$f\left ( x \right ) = x ^{2} – 3x – 2$
Sum of the zeroes = $\alpha + \beta$ = 3
Product of the zeroes = $\alpha \beta$ = -2
From the question,
Sum of the zeroes of the new polynomial = $\frac{1}{2\alpha + \beta } + \frac{1}{2\beta + \alpha }$
= $\frac{2\beta + \alpha + 2\alpha + \beta }{\left ( 2\alpha + \beta \right ) \left ( 2\beta + \alpha \right )}$
= $\frac{3 \alpha + 3 \beta }{2 \left ( \alpha ^{2} + \beta ^{2} \right ) + 5\alpha \beta }$
= $\frac{3 \times 3}{ 2\left [ 2\left ( \alpha + \beta \right )^{2} – 2\alpha \beta + 5 \times \left ( -2 \right ) \right ]}$
= $\frac{9}{2\left [ 9 – \left ( -4 \right ) \right ] – 10}$
= $\frac{9}{2\left [ 13 \right ] – 10}$
= $\frac{9}{26 – 10}$ = $\frac{9}{16}$
Product of the zeroes = $\frac{1}{2 \alpha + \beta } \times \frac{1}{2\beta + \alpha }$
= $\frac{1}{\left ( 2 \alpha + \beta \right )\left ( 2 \beta + \alpha \right )}$
= $\frac{1}{4 \alpha \beta + 2 \alpha ^{2} + 2\beta ^{2} + \alpha \beta }$
= $\frac{1}{5 \alpha \beta + 2 \left (\alpha ^{2} + \beta ^{2} \right ) }$
= $\frac{1}{5 \alpha \beta + 2 \left (\left (\alpha + \beta \right )^{2} – 2\alpha \beta \right ) }$
= $\frac{1}{5 \times \left ( -2 \right ) + 2 \left (\left (3 \right )^{2} – 2\times \left ( -2 \right ) \right ) }$
= $\frac{1}{-10 + 26}$ = $\frac{1}{16}$
$x^{2} – \left ( sum \; of \; the \; zeroes \right )x + \left ( product \; of \; the \; zeroes \right )$
= k $\left ( x^{2} + \frac{9}{16}x + \frac{1}{16} \right )$
Hence, the required quadratic polynomial is k $\left ( x^{2} + \frac{9}{16}x + \frac{1}{16} \right )$.
Q.19: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x ^{2} +px + q$, form a polynomial whose zeroes are $\left ( \alpha + \beta \right )^{2} \; and \; \left ( \alpha – \beta \right )^{2}$.
Solution:
We have,
$f\left ( x \right ) = x ^{2} + px + q$
Sum of the zeroes = $\alpha + \beta$ = -p
Product of the zeroes = $\alpha \beta$ = q
From the question,
Sum of the zeroes of new polynomial = $\left ( \alpha + \beta \right )^{2} + \left ( \alpha – \beta \right )^{2}$
= $\left ( \alpha + \beta \right )^{2} + \alpha ^{2} + \beta ^{2} – 2 \alpha \beta$
= $\left ( \alpha + \beta \right )^{2} + \left (\alpha + \beta \right )^{2} – 2\alpha \beta – 2 \alpha \beta$
= $\left ( -p \right )^{2} + \left (-p \right )^{2} – 2\times q – 2 \times q$
= $p^{2} + p^{2} – 4q$
= $2p^{2} – 4q$
Product of the zeroes of new polynomial = $\left ( \alpha + \beta \right )^{2}\left ( \alpha – \beta \right )^{2}$
= $\left ( -p \right )^{2} \left ( \left (-p \right ) ^{2} – 4q \right )$
= $p^{2} \left ( p^{2} – 4q \right )$
So, the quadratic polynomial is ,
$x^{2} – \left ( sum \; of \; the \; zeroes \right )x + \left ( product \; of \; the \; zeroes \right )$
= $x^{2} – \left ( 2p ^{2} – 4q \right )x + p^{2} \left ( p^{2} – 4q \right )$
Hence, the required quadratic polynomial is $f\left ( x \right ) = k \left (x^{2} – \left ( 2p ^{2} – 4q \right )x + p^{2} \left ( p^{2} – 4q \right ) \right )$.
Q.20: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} – 2x + 3$, find a polynomial whose roots are:
(i) $\alpha + 2 , \beta + 2$
(ii) $\frac{\alpha – 1}{\alpha + 1} , \frac{\beta – 1}{\beta + 1}$.
Solution:
We have,
$f\left ( x \right ) = x ^{2} – 2x + 3$
Sum of the zeroes = $\alpha + \beta$ = 2
Product of the zeroes = $\alpha \beta$ = 3
(i) Sum of the zeroes of new polynomial = $\left ( \alpha + 2 \right ) + \left ( \beta + 2 \right )$
= $\alpha + \beta + 4$
= 2 + 4 = 6
Product of the zeroes of new polynomial = $\left ( \alpha + 1 \right ) \left ( \beta + 1 \right )$
= $\alpha \beta + 2\alpha + 2\beta + 4$
= $\alpha \beta + 2\left (\alpha + \beta \right ) + 4$ = $3 + 2\left (2 \right ) + 4$ = 11
$x^{2} – \left ( sum \; of \; the \; zeroes \right )x + \left ( product \; of \; the \; zeroes \right )$
= $x^{2} – 6x + 11$
Hence, the required quadratic polynomial is $f\left ( x \right ) = k\left (x^{2} – 6x + 11 \right )$
(ii) Sum of the zeroes of new polynomial = $\frac{\alpha – 1}{\alpha + 1} + \frac{\beta – 1}{\beta + 1}$
= $\frac{\left ( \alpha – 1 \right )\left ( \beta + 1 \right ) + \left ( \beta – 1 \right )\left ( \alpha + 1 \right )}{\left ( \alpha + 1 \right )\left ( \beta + 1 \right )}$
= $\frac{\alpha \beta + \alpha – \beta – 1 + \alpha \beta + \beta – \alpha – 1}{\left ( \alpha + 1 \right )\left ( \beta + 1 \right )}$
= $\frac{3 – 1 + 3 – 1}{ 3 + 1 + 2}$ = $\frac{4}{6} = \frac{2}{3}$
Product of the zeroes of new polynomial = $\frac{\alpha – 1}{\alpha + 1} + \frac{\beta – 1}{\beta + 1}$
= $\frac{2}{6} = \frac{1}{3}$
$x^{2} – \left ( sum \; of \; the \; zeroes \right )x + \left ( product \; of \; the \; zeroes \right )$
= $x^{2} – \frac{2}{3}x + \frac{1}{3}$
Thus, the required quadratic polynomial is $f\left ( x \right ) = k\left (x^{2} – \frac{2}{3}x + \frac{1}{3} \right )$.
Q.21: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = ax^{2} + bx + c$, then evaluate:
(i) $\alpha – \beta$
(ii) $\frac{1}{\alpha } – \frac{1}{\beta }$
(ii) $\frac{1}{\alpha } + \frac{1}{\beta } – 2\alpha \beta$
(iv) $\alpha ^{2} \beta + \alpha \beta ^{2}$
(v) $\alpha ^{4} + \beta ^{4}$
(vi) $\frac{1}{a\alpha + b} + \frac{1}{a\beta + b}$
(vii) $\frac{\beta }{a\alpha + b} + \frac{\alpha }{a\beta + b}$
(viii) $a\left [ \frac{\alpha ^{2}}{\beta } + \frac{\beta ^{2}}{\alpha } \right ] + b\left [ \frac{\alpha }{a} + \frac{\beta }{a} \right ]$
Solution:
$f\left ( x \right ) = ax^{2} + bx + c$
Here,
Sum of the zeroes of polynomial = $\alpha + \beta$ = $\frac{-b}{a}$
Product of zeroes of polynomial = $\alpha \beta$ = $\frac{c}{a}$
Since, $\alpha + \beta$ are the roots (or) zeroes of the given polynomial, so
(i) $\alpha – \beta$
The two zeroes of the polynomials are-
$\frac{-b + \sqrt{b^{2} – 4ac}}{2a} – \left ( \frac{-b – \sqrt{b^{2} – 4ac }}{2a} \right )$
= $\frac{-b + \sqrt{b^{2} – 4ac} + b + \sqrt{b^{2} – 4ac}}{2a}$
= $\frac{2\sqrt{b^{2} – 4ac}}{2a}$ = $\frac{\sqrt{b^{2} – 4ac}}{a}$
(ii) $\frac{1}{\alpha } – \frac{1}{\beta }$
= $\frac{\beta – \alpha }{\alpha \beta } = \frac{-\left ( \alpha – \beta \right )}{\alpha \beta }$ ….. E.1
From previous question we know that,
$\alpha – \beta$ = $\frac{\sqrt{b^{2} – 4ac}}{a}$
Also,
$\alpha \beta$ = $\frac{c}{a}$
Putting the values in E.1, we will get
$-\left ( \frac{\frac{\sqrt{b^{2} – 4ac}}{a}}{\frac{c}{a}} \right )$
= $-\left ( \frac{\sqrt{b^{2} – 4ac}}{c} \right )$
(iii) $\frac{1}{\alpha } + \frac{1}{\beta } – 2\alpha \beta$
= $\left [\frac{1}{\alpha } + \frac{1}{\beta } \right ] – 2\alpha \beta$
= $\left [\frac{\alpha + \beta } {\alpha \beta } \right ] – 2\alpha \beta$ …….. E-1
Since,
Sum of the zeroes of polynomial = $\alpha + \beta$ = $\frac{-b}{a}$
Product of zeroes of polynomial = $\alpha \beta$ = $\frac{c}{a}$
After substituting it in E-1, we will get
$\frac{-b}{a} \times \frac{a}{c} – 2\frac{c}{a}$
= $-\frac{b}{c} – 2 \frac{c}{a}$
= $\frac{-ab – 2c^{2}}{ac}$
= $-\left [ \frac{b}{c} + \frac{2c}{a} \right ]$
(iv) $\alpha ^{2} \beta + \alpha \beta ^{2}$
= $\alpha \beta \left ( \alpha + \beta \right )$ …….. E-1.
Since,
Sum of the zeroes of polynomial = $\alpha + \beta$ = $\frac{-b}{a}$
Product of zeroes of polynomial = $\alpha \beta$ = $\frac{c}{a}$
After substituting it in E-1, we will get
$\frac{c}{a} \left ( \frac{-b}{a} \right )$
= $\frac{-bc}{ a^{2}}$
(v) $\alpha ^{4} + \beta ^{4}$
= $\left ( \alpha ^{2} + \beta ^{2} \right ) ^{2} – 2\alpha ^{2} \beta ^{2}$
= $\left (\left ( \alpha + \beta \right ) ^{2} – 2 \alpha \beta \right ) ^{2} – \left (2\alpha \beta \right )^{2}$ ……. E- 1
Since,
Sum of the zeroes of polynomial = $\alpha + \beta$ = $\frac{-b}{a}$
Product of zeroes of polynomial = $\alpha \beta$ = $\frac{c}{a}$
After substituting it in E-1, we will get
$\left [ \left (- \frac{b}{a} \right ) – 2 \left ( \frac{c}{a} \right ) \right ]^{2} – \left [ 2 \left (\frac{c}{a} \right )^{2} \right ]$
= $\left [ \frac{b^{2} – 2ac}{a^{2}} \right ]^{2} – \frac{2c ^{2}}{a ^{2}}$
= $\frac{\left ( b^{2} – 2ac \right ) ^{2} – 2a^{2} c^{2}}{a^{4}}$
(vi) $\frac{1}{a\alpha + b} + \frac{1}{a\beta + b}$
= $\frac{a\beta + b + a\alpha + b}{\left ( a\alpha + b \right )\left ( a\beta + b \right )}$
= $\frac{a\left ( \alpha + \beta \right ) + 2b}{a^{2} \alpha \beta + a b \alpha + a b \beta + b^{2}}$
= $\frac{a\left ( \alpha + \beta \right ) + 2b}{a^{2} \alpha \beta + a b \left (\alpha + \beta \right ) + b^{2}}$
Since,
Sum of the zeroes of polynomial = $\alpha + \beta$ = $\frac{-b}{a}$
Product of zeroes of polynomial = $\alpha \beta$ = $\frac{c}{a}$
After substituting it , we will get
$\frac{b}{ac – b ^{2} + b ^{2}}$
= $\frac{b}{ac }$
(vii) $\frac{\beta }{a\alpha + b} + \frac{\alpha }{a\beta + b}$
= $\frac{\beta \left ( a\beta + b \right ) + \alpha \left ( a\alpha + b \right )}{\left ( a\alpha + b \right )\left ( a\beta + b \right )}$
= $\frac{a \beta ^{2} + b \beta + \alpha a^{2} + b \alpha }{a ^{2} \alpha \beta + a b \alpha + a b \beta + b^{2}}$
= $\frac{a \alpha ^{2} + b \beta ^{2} + b\alpha + b \beta }{a^{2} \times \frac{c}{a} + a b \left ( \alpha + \beta \right ) + b ^{2}}$
Since,
Sum of the zeroes of polynomial = $\alpha + \beta$ = $\frac{-b}{a}$
Product of zeroes of polynomial = $\alpha \beta$ = $\frac{c}{a}$
After substituting it , we will get
$\frac{ a \left [ \left ( \alpha + \beta \right ) ^{2} + b \left ( \alpha + \beta \right ) \right ]}{ac}$
= $\frac{a\left [ \left ( \alpha + \beta \right ) ^{2} – 2 \alpha \beta \right ] – \frac{b ^{2}}{a}}{ac}$
= $\frac{a\left [ \frac{b ^{2}}{a} – \frac{2c}{a} \right ] – \frac{b ^{2}}{a}}{ac}$
= $\frac{a\left [ \frac{b ^{2} – 2c}{a} \right ] – \frac{b ^{2}}{a}}{ac}$
= $\frac{a\left [ \frac{b ^{2} – 2c – b ^{2}}{a} \right ]}{ac}$
= $\frac{b ^{2} – 2c – b ^{2} }{ac}$
= $\frac{ – 2c }{ac}$ = $\frac{ – 2 }{a}$
(viii) $a\left [ \frac{\alpha ^{2}}{\beta } + \frac{\beta ^{2}}{\alpha } \right ] + b\left [ \frac{\alpha }{a} + \frac{\beta }{a} \right ]$
= $a \left [ \frac{ \alpha ^{2} + \beta ^{2}}{ \alpha \beta } \right ] + b \left ( \frac{ \alpha ^{2} + \beta ^{2}}{ \alpha \beta } \right )$
= $\frac{ a \left [ \left ( \alpha + \beta \right ) ^{2} – 2 \alpha \beta \right ] + b \left ( \left ( \alpha + \beta \right )^{2} – 2 \alpha \beta \right )}{\alpha \beta }$
Since,
Sum of the zeroes of polynomial = $\alpha + \beta$ = $\frac{-b}{a}$
Product of zeroes of polynomial = $\alpha \beta$ = $\frac{c}{a}$
After substituting it , we will get
$\frac{ a \left [ \left ( \frac{-b}{a} \right ) ^{2} – 3 \times \frac{c}{a} \right ] + b \left ( \left ( \frac{-b}{a} \right )^{2} – 2 \frac{c}{a} \right )}{\frac{c}{a} }$
= $\frac{a ^{2}}{c} \left [ \frac{-b ^{2}}{a ^{2}} + \frac{3bc}{a^{2}} + \frac{ b ^{2}}{ a ^{2}} – \frac{2bc}{a ^{2}} \right ]$
= $\left [ \frac{-b ^{2} a ^{2}}{a ^{2} c} + \frac{3bc a ^{2}}{a^{2} c} + \frac{ b ^{2} a^{2}}{ a ^{2}c } – \frac{2bc a ^{2}}{a ^{2} c} \right ]$
= $\frac{- b ^{2}}{ac} + 3b + \frac{b ^{2}}{ac} – 2b$ = b
#### Practise This Question
If x = a cos θ,y=b sin θ,then d3ydx3 is equal to
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2019-04-19 00:28:09
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https://crypto.stackexchange.com/questions/98905/given-a-cycle-x-mapsto-xa-with-his-starting-point-x-1-can-another-startin/98944#98944
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# Given a cycle $x \mapsto x^a$ with his starting point $x_1$. Can another starting point $x_2$ be transformed to generate the same cycle?
A cyclic sequence can be produced with
$$s_{i+1} = s_i^a \mod N$$ with $$N = P \cdot Q$$ and $$P = 2\cdot p+1$$ and $$Q = 2\cdot q+1$$ with $$P,Q,p,q$$ primes.
and $$a$$ a primitive root of $$p$$ and $$q$$.
The starting point $$s_0$$ is a square ($$\mod N$$)
It will produce a cycle of length $$\mathrm{lcm}(p-1.q-1)$$
(except $$s_0$$ is a $$p$$-th or $$q$$-th power $$\mod N$$)
Given now a starting point $$s_0 = x_1$$ it will generate such a cyclic sequence.
Given another starting point $$s_0 = x_2$$ it will generate a cyclic sequence of same length but it an have completely different members.
Is there any way to transform $$x_2$$ so it will produce the same cyclic sequence as $$x_1$$ does?
(Edit: the posted answer is if any and not how, same as the question, will mark it as answer here)
(related to the answer of this)
Update:
It looks like the number of different cycles $$N_c$$ is: $$N_c = (S_N - S_{pq}) /L_c$$ $$S_N = |\{ v^2 \mod N\}| \text{ with } v\in[1,N-1]$$ $$L_c = \mathrm{lcm}(p-1.q-1)$$ and $$S_{pq}$$ the number of squares which are also a $$p$$-th, $$q$$-th power $$\mod N$$ .
$$S_N$$ probably always larger than $$\frac{1}{4}N$$
In some test for $$N=3901$$ with $$P=47$$ , $$Q=83$$, $$a = 7$$ (or $$11, 17, 19,..$$) two cycles are possible with $$L_c =440$$, $$S_N = 1006$$, $$S_{pq}=127$$.
One $$x1$$ can be transformed to a value from the other cycle (which starts with $$x_2$$) with an exponent $$b$$ like $$x_1^b \mapsto s_i\in \mathrm{cycle}_{x_2}$$
This exponent need to be $$b \in [3 , 5 , 6 , 10 , 12 , 13, 20 , 21 , 24 , 26 , 27 , 29 , 33 , 35 , 37, 40, 42 , 43 , 45, 47, ...]$$
No idea why exactly those values do work out.
For $$N=40633, P= 179, Q= 227$$ with $$S_N= 10259$$ squares, including $$S_{pq}= 403$$ it has $$8$$ cycles with length $$L_c= 1232$$. The exponent $$a$$ for sequence generation can be $$a\in[3, 19, 23, 29, 43,..]$$
For this exponent $$b$$ need to be $$b \in [7 , 13, 17 , 21, 28 , 39 , 51 , 52 , 62 , 63 , 68 , 71 , 79 , 84 , 110 , 112 , 117,125,..]$$
Applying any of those exponents $$b$$ to a starting value $$x_0$$ will result in a cycle of the next sequence. This cyclic sequence order is equal for every exponent $$b$$.
• I think there's "prime root" where there should be primitive root. Independently, a justification of "except (if) it contains a $p$-th or $q$-th power$\bmod N$" would help me.
– fgrieu
Mar 2 at 10:04
• @fgrieu thanks for the hint. changed ti to primitive root. Unfortunately can't tell yet why this happen for $p$-th, $q$-th power. First time dealing with $\mathrm{mod}$ a non-prime. But I did some test case ($P=47, Q=83, a=7$) about it and as expected it did not work out with those numbers (cycle length of $40$ instead of $440$ in test case). The cycle size was much shorter (but constant) for those numbers tested. An exception again but e.g. $1^p=1$ will result in a cycle length of $1$. Mar 2 at 10:57
In analyzing cases like this, it is useful to look at the behaviors modulo $$P$$ and modulo $$Q$$ separately, and then see how they combine. I will add in the assumption that $$P \ne Q$$; you didn't explicitly say so; I believe it is reasonable that you assumed it.
When we look at the cycle structure modulo $$P$$, we first see that $$s_0 \bmod P$$ is a quadratic residue modulo $$P$$ (which is a mathy way of saying "it's a square"); since $$a$$ is a primitive root of $$p$$, then we see that there are three cycles:
• A cycle of length 1 (the value 0)
• Another cycle of length 1 (the value 1)
• A cycle of length $$p-1$$; this is because $$a^i \bmod p-1$$ are distinct values in the range $$[0, p-2]$$, and $$s_0$$ has order $$p-1$$ (in this group, quadratic residues other than 1 are always that order), and so $$s_0^{a^i}$$ are $$p-1$$ distinct values.
We get similar results for looking at the behavior modulo $$Q$$.
Given those basics, how do they combine?
Well, the cycle modulo $$PQ$$ repeats only when both the cycle modulo $$P$$ repeats and the cycle modulo $$Q$$ repeats; if the $$P$$-cycle is length $$\alpha$$ and the $$Q$$-cycle is length $$\beta$$, this combined cycle has length $$\text{lcm}(\alpha, \beta)$$.
This implies that any combination of the two large cycles will have length $$\text{lcm}(p-1,q-1)$$ (which is a result you have already found). And, there are $$\gcd(p-1,q-1)$$ ways these two large cycles can combine.
We now consider a combination that includes a small cycle; we have two combinations with cycle length $$p-1$$, two combinations with cycle length $$q-1$$, and four combinations with cycle length 1 (which includes $$s_0 = 0$$ and $$s_0 = 1$$).
Hence, the total number of cycles is $$\gcd(p-1, q-1) + 8$$.
Is there any way to transform $$x_2$$ so it will produce the same cyclic sequence as $$x_1$$ does?
Well, for any integer $$\beta$$, we have $$s_{i+1}^\beta = (s_i^\beta)^a$$. That is, if we take every element of a cycle, and raise it to the power $$\beta$$, we still have a cycle.
So, if we have the value $$\beta$$ for which $$x_2^\beta = x_1$$, then that gives us a way to transform one cycle to another.
It turns out that there is always such a $$\beta$$ if the two cycles are both large (that is, of length $$\text{lcm}(p-1, q-1)$$). For the degenerate cycles (all the others), there won't be - however, I don't consider that case that interesting...
Of course, finding such a $$\beta$$ given $$x_1, x_2$$ is a nontrivial problem if $$P, Q$$ are large...
• thanks for explaining. It's more clear now, During testing around I also found out it is possible (with $x_2^\beta$) (at examples update). I planned to do some new question about it but now there seem so to be no need anymore. So the actually question is how such a $\beta$ can be found. To be clear is it known to be nontrivial problem for large $P,Q$ or was it 'just' an educated guess? Also $x_2^\beta$ don't need to be equal to $x_1$. A $\beta$ with $x_2^\beta=x_1^{a^i}$ would be sufficient. Allowing this many $\beta$ were possible at the example cases ($b$). But can't tell for large $P,Q$. Mar 3 at 21:00
• @J.Doe: finding the specific $\beta$ so $x_2^\beta = x_1$ is known as the 'discrete log problem' (over a composite); it is known to be as hard are factoring the modulus (and solving the discrete log in both the prime groups). The more general problem of finding any $\beta$ as you mentioned may be easier - actually, a random guess has a good probability of working, but it's not clear how one would verify your guess. On the other hand, I can't think of a hard problem that it would reduce to... Mar 3 at 21:06
• ... But even such a general $\beta$ would probably depend at the chosen $x_2$ value or more precisely the sequence it will generate. So probably another operation is needed which return the same value for each sequence element. If such an operation exits random values could be tested against this. So the whole problem could also be rephrases as picking random values out of a single sequence (while many other exist) without leaking security related parameter or a relation to other sequence elements (like random $s_i$ for given $s_0$) Mar 3 at 21:19
• Ok, DLP would be hard. Also though about this but deleted it again because for sequences with $x_i = x_o \cdot g^i \mod P$ it can be done without the DLP. But this is also from one given value to a random value out of the sequence ( more general version). Mar 3 at 21:47
• shouldn't it be $a^i \mod p$ with values from $1$ to $p-1$ instead of $a^i \mod p-1$ with values from $0$ to $p-2$? Mar 4 at 17:04
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2022-08-11 20:32:12
|
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|
https://pypi.org/project/slave/0.3.3/
|
A lightweight python package to simplify the communication with several scientific instruments.
Project description
The main goal of slave is to make it easier to communicate with scientific instruments. Slave tries to ease the implementation of new instruments and comes with a variety of ready-to-use implementations.
A simple measurement script, using pyvisa for the GPIB connection, might look like:
import time
import visa
from slave.sr830 import SR830
# construct and initialize a SR830 instrument with a GPIB connection on
# channel 8.
lockin = SR830(visa.instrument('GPIB::08'))
lockin.reserve = 'high' # Set the dynamic reserve to high.
lockin.time_constant = 3 # Set the time constant to 3s.
# measure the x value 60 times and wait 1s between each measurement.
for i in range(60):
print lockin.x
time.sleep(1) # delay for 1s.
Requirements
• Python 2.6 or higher
• Sphinx (optional, to build the documentation)
• distribute (Python 3.x)
Installation
To install Slave, simply type
python setup.py install
Note To install Slave in Python 3.x distribute is required.
Documentation
Slave is fully documented. Both the latest stable as well as the develop documentations are available online. To build the documentation manually, e.g. the html documentation, navigate into the /doc/ subfolder and execute make html.
Licensing
You should have received a copy of the GNU General Public License along with Slave; see the file COPYING.
|
2023-01-31 01:52:56
|
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https://www.gamedev.net/forums/topic/423323-text-encryption/
|
# text encryption
## Recommended Posts
Hey everyone, im getting pretty frustrated here. What I need to do is create a text encryption program that accepts some text and stores it in a dynamically allocated array of chars. I need to then encrypt each element of the array by adding 7 to the ASCII code and then MOD by 10. For example if the ASCII code is 65 I need to break it down into 6 and 5 then add 7 and MOD 10, then put the number back together and print out the ASCII character. Yes, this is homework, but the focus is supposed to be on the pointer aspect of it. . .I'm struggling with the encryption/decryption functions. They seem unstable as they sometimes kick out the same encrypted values for two different characters! Thank for the help!
#include<iostream>
#include<string>
#include<cctype>
using namespace std;
int encrypt(int);
int decrypt(int);
int main()
{
cout<<"Would you like to (e)ncrypt or (d)ecrypt? (e/d): ";
string choice = "";
getline(cin, choice);
while(choice != "e" && choice != "E" && choice != "d" && choice != "D")
{
cout<<"Would you like to (e)ncrypt or (d)ecrypt? (e/d): ";
getline(cin, choice);
}
if(choice == "e" || choice == "E")
{
string text;
choice = "";
cout<<"Enter the text to encrypt: \n";
getline(cin, text);
char* pValues = new char[text.length()];
for(int i = 0; i < text.length(); i++)
{
pValues[i] = int (text[i]);
pValues[i] = encrypt(pValues[i]);
pValues[i] = char (pValues[i]);
cout<< pValues[i];
}
}
else if(choice == "d" || choice == "D")
{
choice = "";
cout<<"Enter the text to decrypt: \n";
string text;
getline(cin, text);
char* pValues = new char[text.length()];
for(int i = 0; i < text.length(); i++)
{
pValues[i] = int (text[i]);
pValues[i] = decrypt(pValues[i]);
pValues[i] = char (pValues[i]);
cout<< pValues[i];
}
}
cout<<endl;
system("pause");
return 0;
}
int decrypt(int x)
{
int result = 0;
int a = 0;
int b = 0;
int c = 0;
if(x >= 10 && x < 100)
{
a = x / 10;
b = x - a * 10;
a = (a - 7) + 10;
b = (b - 7) + 10;
a = a * 10;
result = a + b;
}
else if(x >= 100)
{
a = x / 100;
b = (x - (a*100)) / 10;
c = (x - (a*100)) - (b*10);
a = (a - 7) + 10;
b = (b - 7) + 10;
c = (c - 7) + 10;
a = a * 100;
b = b * 10;
result = a + b + c;
}
return result;
}
int encrypt(int x)
{
int result = 0;
int a = 0;
int b = 0;
int c = 0;
if(x >= 10 && x < 100)
{
a = x / 10;
b = x - a * 10;
a = (a + 7) % 10;
b = (b + 7) % 10;
a = a * 10;
result = a + b;
}
else if(x >= 100)
{
a = x / 100;
b = (x - (a*100)) / 10;
c = (x - (a*100)) - (b*10);
a = (a + 7) % 10;
b = (b + 7) % 10;
c = (c + 7) % 10;
a = a * 100;
b = b * 10;
result = a + b + c;
}
return result;
}
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given Input as an ASCII value (note--- limited to 99 as ASCII is 0..127)
It doesnt make sense to do ASCII over 99 because by adding 7 to the hundreds digit, you would get values in the 700s and 800s which are not ASCII for the output value...... (you might have to ask your instructor for a clarification, as many lower case letters have ASCII values over 100)
http://www.lookuptables.com/ << ASCII table found Yahoo's in 5 seconds
' get seperate input decimal digits
I2 = Input / 10 'upper
I1 = Input - (I2*10) 'lower
O1 = (I1+7) % 10
O2 = (I2+7) % 10
'reassemble them for the output value
Output = (O2*10) + O1
beware though, many ASCII values are not printable, so you probably want to output the ordinal (decimal form) and only the printable form if it IS printable
Over time you will learn that arithmetic and logical functions can do alot of work that you might now think requires if-then logic. Ive only been programming for about 30 years so I can do things like this in my sleep...
Keep at it, this kind of learning is accumulative.
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My definition of a cipher says that it has to be reversible. Considering that your plaintext alphabet has at least 26 letters and your ciphertext only 10, there's no way to recover the original text. At least not unless there's some plaintext restriction you haven't told us about. Consider:
'A' = 65
('A' + 7) % 10 = 72 % 10 = 2
'K' = 75
('K' + 7) % 10 = 82 % 10 = 2
And so A and K are indistinguishable after encryption.
This is an example of the collision you speak of, and there is no way around it. If you do find a way around this, be sure to patent it, as you'll have found a method of binary encoding with at least 260% efficiency [wink].
Are you sure you're supposed to take the 10-modulus? I suspect that a 26 or higher (often 64 or 83) modulus was recommended to you, as it's then possible to retain the important information. The most spartan of such modulus-ciphers would convert the text to capitals, remove anything that isn't a letter, subtract 65 and take the 26-modulus. This would preserve no whitespace, capitalisation or punctuation, however. The rules is that if you need to preserve n plaintext letters, you can't take a modulus below n without losing information.
Regards
##### Share on other sites
Quote:
Original post by TheAdmiralMy definition of a cipher says that it has to be reversible. Considering that your plaintext alphabet has at least 26 letters and your ciphertext only 10, there's no way to recover the original text. At least not unless there's some plaintext restriction you haven't told us about. Consider:'A' = 65('A' + 7) % 10 = 72 % 10 = 2'K' = 75('K' + 7) % 10 = 82 % 10 = 2And so A and K are indistinguishable after encryption.This is an example of the collision you speak of, and there is no way around it. If you do find a way around this, be sure to patent it, as you'll have found a method of binary encoding with at least 260% efficiency [wink].Are you sure you're supposed to take the 10-modulus? I suspect that a 26 or higher (often 64 or 83) modulus was recommended to you, as it's then possible to retain the important information. The most spartan of such modulus-ciphers would convert the text to capitals, remove anything that isn't a letter, subtract 65 and take the 26-modulus. This would preserve no whitespace, capitalisation or punctuation, however. The rules is that if you need to preserve n plaintext letters, you can't take a modulus below n without losing information.RegardsAdmiral
Read his post again closely and check the code. He says, although a bit unclear, the add and modulo is for each digit in the base-10 representation of the ASCII value. So the 'A' and 'K' encrypts to 32 and 42, respectively.
## Create an account
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• ### Forum Statistics
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2017-10-22 06:42:49
|
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|
http://wiki.lazarus.freepascal.org/Anchor_Sides
|
# Anchor Sides
Deutsch (de) English (en) français (fr) 日本語 (ja) русский (ru)
There are some new properties and methods for automatic layout of controls. You can now setup controls to keep a certain distance to other controls, or center relative to other controls. See below for examples.
Each of the four sides of a control (Left, Top, Right, Bottom) can now be anchored/bound to a side of another control. For example you can now anchor the left side of TEdit to the right side of a TLabel. Everytime the Label is moved or resized the Edit's left side will follow, which normally results in moving the Edit parallel to the Label.
# Example 1
+--------+ +-------+
| Label1 | | Edit1 |
+--------+ | |
+-------+
## In code
Edit1.AnchorSide[akLeft].Side := asrRight;
Edit1.AnchorSide[akLeft].Control := Label1;
Edit1.Anchors := Edit1.Anchors + [akLeft];
You can define the distance with the BorderSpacing properties:
Edit1.BorderSpacing.Left := 10;
The same can be done with the method:
Edit1.AnchorToNeighbour(akLeft, 10, Label1);
## Notes
The Edit1.Left will follow Label1.Left+Label1.Width, not the other way around. That means, moving Label1 will move Edit1. But moving Edit1 will be undone by the LCL. If you also anchor the right side of Label1 to the left side of Edit1, you created a circle, and this can result together with some other autosize properties in a loop. This is ignored by the LCL or automagically repaired when the Parent.AutoSize is true.
## Via the Anchor Editor
The anchor editor is a floating window that is available via the menu View / Anchor Editor or via the button on the Anchors property in the object inspector.
# Example 2
You can anchor the Edit's top side to follow the Label's top side:
+--------+ +-------+
| Label1 | | Edit1 |
+--------+ | |
+-------+
Edit1.AnchorSide[akTop].Side := asrTop;
Edit1.AnchorSide[akTop].Control := Label1;
Edit1.Anchors := Edit1.Anchors + [akTop];
The same can be done with the method:
Edit1.AnchorParallel(akTop,0,Label1);
# Example 3
Centering a Label vertically to an Edit:
+-------+
+--------+ | |
| Label1 | | Edit1 |
+--------+ | |
+-------+
Edit1.AnchorSide[akTop].Side := asrCenter;
Edit1.AnchorSide[akTop].Control := Label1;
Edit1.Anchors := Edit1.Anchors + [akTop] - [akBottom];
The same can be done with the method:
Edit1.AnchorVerticalCenterTo(Label1);
Obviously anchoring the bottom side of Edit1 does not make sense when centering.
# New property
property AnchorSide[Kind: TAnchorKind]: TAnchorSide read GetAnchorSide;
This is not published in the object inspector. You can edit it in the designer via the anchor editor (Menu: View -> View anchor editor, or click on button of 'Anchors' property).
# New methods to easily configure common layouts
procedure AnchorToNeighbour(Side: TAnchorKind; Space: integer;
Sibling: TControl);
procedure AnchorParallel(Side: TAnchorKind; Space: integer;
Sibling: TControl);
procedure AnchorHorizontalCenterTo(Sibling: TControl);
procedure AnchorVerticalCenterTo(Sibling: TControl);
procedure AnchorAsAlign(TheAlign: TAlign; Space: Integer);
AnchorVerticalCenterTo works with Parent too. Then it will center on the client area, that means the center of the control is at ClientHeight div 2.
Center anchoring is not yet fully supported when computing the size of the parent. For example when you put a label center anchored into a Groupbox1 and set Groupbox1.AutoSize to true then Groupbox1 height will shrink leaving no space for the label. The solution is to center to a control that is not centered. For example center a Label1 to a ComboBox and use for the ComboBox the default anchors (Anchors=[akLeft,akTop]).
# Anchoring to invisible controls
Anchoring to invisible controls works intuitively. For example: The below controls A, B, C are anchored (C.Left to B.Right and B.Left to A.Right):
+---+ +---+ +---+
| A | | B | | C |
+---+ +---+ +---+
If B is hidden (Visible:=false) then C.Left will skip B and use the right of A, resulting in
+---+ +---+
| A | | C |
+---+ +---+
# Circular references
You can build circular references by anchoring two sides to each other, which creates an impossible anchoring. The LCL notices that, but will not raise an exception, because it can be a temporary circle. For example: A row of buttons. Button1 is left of Button2, Button2 is left of Button3. Now the order is changed to: Button3, Button1, Button2. If the reanchoring is started with Button3, a circle is created temporarily and is fixed when the reordering is completed. The LCL detects circles and will not move the Button. Circles are not the only anomaly.
There is one exception to this rule. If the Parent.AutoSize is true, then the LCL will automatically break circles on autosize and writes a warning via debugln (windows: --debug-log.txt, linux, et al: stdout). Which anchor is broken depends on the order of controls and the sides. So, temporary circles are still allowed with AutoSize=true if you enclose the change in
Parent.DisableAutosizing;
try
// change anchors, aligns, bounds...
finally
Parent.EnableAutosizing;
end;
The AutoSize algorithm does not only break circles, but fixes Align/AnchorSide inconsistencies too.
Eventually it would be good to add some hints in the designer or a tool to list the circles and inconsistencies.
|
2019-03-20 16:12:30
|
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|
http://humanthermodynamics.wikifoundry.com/page/love+terminology+upgrades
|
In terminology, love terminology upgrades refers to []
Physicochemical neutrality
physicochemically-neutral or chemical thermodynamically-neutral term replacements for "love" and related terms, e.g. "hate", that scale up and down the atomic scale (chain of being, so to say; or molecular evolution scale), elemental periodic table to family dinner table; in the same sense in which, at the beginning of the 20th century it became no longer coherent to speak about the love and hate of the elements, so to going into the early 21st century is it no longer coherent to speak about love and hate between humans defined as chemicals (human chemicals), molecules (human molecules), or powered CHNOPS+20 bound state animate structures.
A depiction of via chemical anthropism, therein illustrating the need for love terminology reform and or upgrades, whenever physicochemical and or thermodynamical principles are used to explain human phenomena, i.e. "anthropisms", as Charles Sherrington (1938) said, must be expunged whenever digressions on human nature occur. In short, both depictions shown above illustrate a common natural phenomena that can be quantified according to the principles of physical chemistry. While both versions, left and right, employ the same terminology, i.e. "love" and "hate" to explain what occurs, the version on the left (Ѻ) is but a fun cartoon, made for the purposes of humor, while the version on the right illustrates descriptions real "heated" emotions (i.e. a non-funny situation). The difference between the two is that the left-version employs terms that are NOT actually used in physical chemistry textbooks. Subsequently, if one desires to explain the right-version precisely, according to the laws of nature, then a physicochemically-neutral language must be employed, in the same sense in which actual physical chemistry textbooks define phenomena, using physicochemically-neutral (deanthropomorphized) terminology, otherwise "objectionable nonsense" results.
Overview
In 450BC, Empedocles, in his four element two force model, referred to "love" as an attractive force and "hate" as a repulsive force, in short; some of this attraction to repulsion balancing (see: attraction-to-repulsion ratio) logic is seen in, e.g., the Gottman stability ratio discussion, amid which the colloquial inexact (atomically-inapplicable) terms love and hate become obviated.
In 1898 to 1903, Ernst Haeckel, in his various love letters, mentions the concept of elective affinity at least three times in respect to his own romantic relationships; in one letter to a Franziska von Altenhausen, Haeckel defines elective affinity as a strange psychological chemotropism: [1]
“… seductive women—why should I, despite all scruples and obstacles, cast myself into the dust before you? Dearst Franziska, herein lies the enigma of ‘elective affinity’, of that strange psychological ‘chemotropism’, of whose power I have spoken repeatedly in my books—little dreaming that I myself should fall a victim to it in my old age!”
In 1910, Henry Bray gave the following suggestions for term reform: [2]
“Similar forces to those that thus bring together the opposite sexes are everywhere evident and acting in the so-called material world, and in the judgment of the writer, equally natural and all-conquering. In our conceit and blindness we call the one force ‘love’, and the other force ‘affinity’; but mere alteration of words cannot alter the fact that the two words are the expression of the same force in nature.”
In this sense, we would say that "love", in colloquial inexact terminology, according to the Goethe-Helmholtz equation (or thermodynamic theory of affinity):
$A=-\left(\frac{\partial G}{\partial \xi}\right)_{p,T}$
equates to positive affinity (or negative Gibbs energy partial differential change per reaction extent) in exact scientific terminology, and that "hate", likewise, corresponds to negative affinity (or positive Gibbs energy partial differential change per reaction extent), in scientific terms, i.e. physicochemcially-neutral language.
Alternatively, in the Haeckel sense, we might aptly state that "love" and "hate" corresponds to positive and negative psychological chemotropism, respectively, similar to a plant turning towards sunlight.
Quotes
The following are related quotes:
Love is an ideal thing, marriage a real thing; a confusion of the real with the ideal never goes unpunished.”
Johann Goethe (c.1820)
Love is gravitation toward a beautiful object.”
Jose Ortega (c.1920) [2]
Love is a romantic designation for a most ordinary biological process—or, shall we say, chemical—process … a lot of nonsense is talked and written about it.”
— Greta Garbo (1932), Ninotchka
|
2022-09-29 20:03:48
|
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http://www.spoj.com/problems/A001856/
|
## A001856 - Chiaki Sequence
Chiaki is interested in an infinite sequence $a_1, a_2, a_3, ...$, which defined as follows: $$a_n = \begin{cases} n, & n \le 2 \\ 2 \cdot a_{n-1}, & n \text{ is odd} \\ a_{n-1}+r_{n-1}, & n \text{ is even}\end{cases}$$ where $r_n$ is the smallest positive integer not in the set $S_n = \{a_j - a_i \mid 1 \le i < j \le n\}$.
Chiaki would like to know the sum of the first $n$ terms of the sequence, i.e. $\sum\limits_{i=1}^{n} a_n$. As this number may be very large, Chiaki is only interested in its remainder modulo ($10^9 + 7$).
### Input
There are multiple test cases. The first line of input contains an integer $T$ ($1 \le T \le 1000$), indicating the number of test cases. For each test case:
The first line contains an integer $n$ ($1 \le n < 10^{100}$) without leading zeros.
### Output
For each test case, output an integer denoting the answer.
### Example
#### Input
11
1
2
3
4
5
6
7
8
9
10
1000000000
#### Output
1
3
7
15
31
52
94
145
247
359
834069170
### Information
There are $5$ input files and my unoptimized python3 code runs about 1.1 sec per file.
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2018-05-23 11:02:32
|
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http://tex.stackexchange.com/tags/pandoc/new
|
# Tag Info
1
Just use the flag --biblatex when converting with pandoc: pandoc --biblatex --chapters ../Source/Chapters/Content.md -o ../Example/Chapters/Content.tex The result: Connecting these native implementations to the Python world is enabled by the \href{http://cython.org/}{Cython} toolchain \autocite{behnel2011cython}. You have to use the biblatex package ...
1
This is happening because pandoc is interpreting (@) signs as list markers, specifically an ordered list enclosed in parentheses. As can be seen if you inspect the latex output of pandoc: Input: (@) $$(K_{0}^{-1}x)^{T}E(K_{1}^{-1}x')=0$$ Output: \begin{enumerate} \def\labelenumi{(\arabic{enumi})} \item $(K_{0}^{-1}x)^{T}E(K_{1}^{-1}x')=0$ ...
0
I verified the first part of my concept based on the link Steven (@steven-b-segletes) provided in his comment. Here is a short, working example for A4 sized pdf as a background (don't mind the coordinates...): \documentclass[finnish,a4paper,11pt]{article} \usepackage[a4paper]{geometry} \usepackage{setspace} \usepackage[utf8]{inputenc} \usepackage{graphicx} ...
Top 50 recent answers are included
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2015-01-28 20:42:58
|
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|
https://proofwiki.org/wiki/Cartesian_Product_of_Ordered_Sets_is_Ordered_Set
|
# Cartesian Product of Ordered Sets is Ordered Set
## Theorem
Let $\left({S_1, \preceq_1}\right)$, $\left({S_2, \preceq_2}\right)$ be ordered sets.
Let $\left({S_1 \times S_2, \preceq}\right)$ be the Cartesian product of $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$.
Then $\left({S_1 \times S_2, \preceq}\right)$ is also an ordered set.
## Proof
### Reflexivity
Let $\left({s, t}\right) \in S_1 \times S_2$.
By definition of reflexivity:
$s \preceq_1 s$ and $t \preceq_2 t$
Thus by definition of Cartesian product:
$\left({s, t}\right) \preceq \left({s, t}\right)$
### Transitivity
Let $\left({s_1, t_1}\right), \left({s_2, t_2}\right), \left({s_3, t_3}\right) \in S_1 \times S_2$ such that
$\left({s_1, t_1}\right) \preceq \left({s_2, t_2}\right) \preceq \left({s_3, t_3}\right)$
By definition of Cartesian product:
$s_1 \preceq_1 s_2 \preceq_1 s_3$ and $t_1 \preceq_2 t_2 \preceq_2 t_3$
By definition of transitivity:
$s_1 \preceq_1 s_3$ and $t_1 \preceq_2 t_3$
Thus by definition of Cartesian product:
$\left({s_1, t_1}\right) \preceq \left({s_3, t_3}\right)$
### Antisymmetry
Let $\left({s_1, t_1}\right), \left({s_2, t_2}\right) \in S_1 \times S_2$ such that
$\left({s_1, t_1}\right) \preceq \left({s_2, t_2}\right)$ and $\left({s_2, t_2}\right) \preceq \left({s_1, t_1}\right)$
By definition of Cartesian product:
$s_1 \preceq_1 s_2$ and $s_2 \preceq_1 s_1$ and $t_1 \preceq_2 t_2$ and $t_2 \preceq_2 t_1$
By definition of antisymmetry:
$s_1 = s_2$ and $t_1 = t_2$
Thus:
$\left({s_1, t_1}\right) = \left({s_2, t_2}\right)$
$\blacksquare$
|
2019-12-11 01:13:37
|
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|
https://wiki.synchro.net/faq:misc
|
Synchronet v3.18b-Win32 (install) has been released (Sept-2020).
Synchronet v3.19a, now under development, requires libarchive-dev to build successfully.
You can donate to the Synchronet project using PayPal.
# Miscellaneous
## Deleted Msgs
Question:
Why is it that deleted messages still appear in message bases?
In Synchronet, when a message is deleted, it is only flagged for deletion. The actual removal of “deleted messages” from message bases takes place during regularly scheduled message base maintenance (using the smbutil utility). Normally, messages flagged for deletion will have a DELETED attribute displayed in the message header, if/when the message is displayed to a user. Messages flagged for deletion may be “undeleted”, but only when they are visible to a user or sysop.
There is a Synchronet configuration option, SCFG->Message Options->Users Can View Deleted Messages, with three possible settings:
1. Yes
2. No
3. Sysops Only
If the sysop wants deleted messages to appear to be immediately and permanently removed from the message base, they should set this option to “No”. When this option is set to “Yes”, all users with read-access to the message base will be able to view/read messages flagged for deletion. When set to “Sysops Only”, this option will only allow messages flagged for deletion to be visible to system operators (sysops) and sub-booard operations (sub-ops).
Note: Deleted messages may still be viewed and undeleted using the smbutil program and possibly other methods, until the deleted messages are actually removed from the message base during maintenance.
## New User Questions
Question:
How do I eliminate one or more prompts or questions during new user sign-up?
Many of the “New User” Questions can be disabled by setting various options in SCFG->System->New User Values->Question Toggles to “No”:
╔══════════════════════════════════╗
║ New User Questions ║
╠══════════════════════════════════╣
║ │Real Name Yes ║
║ │Force Unique Real Name No ║
║ │Force Upper/Lower Case Yes ║
║ │Company Name No ║
║ │Chat Handle / Call Sign Yes ║
║ │Force Unique Handle / Call No ║
║ │Sex (Gender) Yes ║
║ │Birthday Yes ║
║ │Address and Zip Code No ║
║ │Location Yes ║
║ │Require Comma in Location No ║
║ │Phone Number No ║
║ │Allow EX-ASCII in Answers Yes ║
║ │External Editor No ║
║ │Command Shell Yes ║
║ │Default Settings Yes ║
║ │Color Terminal No ║
╚══════════════════════════════════╝
Other “New User” prompts/questions can be disabled by finding their corresponding string in your ctrl/text.dat file and changing it to an empty string (i.e. “”), e.g. MouseTerminalQ, EnterYourAlias.
## New User Feedback
Question:
How do I eliminate the requirement that new users send feedback/mail to the sysop during sign-up?
In Synchronet for Windows, run the Configuration Wizard and un-check the “Require new user feedback” check-box.
Or, set SCFG->Nodes->Node 1->Advanced Options->Validation User to “0” (Nobody).
If you want your new user validation feedback (e-mail) to go to another user, enter that user's number here.
╔══════════════════════════════════════════════════════════╗
╠══════════════════════════════════════════════════════════╣
->│Validation User Nobody <-
║ │Notification Error Level Error ║
║ │Semaphore Frequency 5 seconds ║
║ │Statistics Frequency 5 seconds ║
║ │Inactivity Warning 300 seconds ║
║ │Inactivity Disconnection 600 seconds ║
║ │Daily Event ║
║ │Text Directory ..\text\ ║
╚══════════════════════════════════════════════════════════╝
## No Guest
Question:
How do disable the Guest user account on the BBS?
You don't have to have a Guest (G-restricted user) account. If you don't create a Guest account to begin with, you won't have one. If you created a Guest account and you don't want it, then simply deleting the user account from the user base will work. However, without a Guest account, anonymous FTP access will not be available and some web UI elements may not function as is normally expected. To keep the Guest account but disallow logins to the Terminal Server (the traditional BBS user experience), set SCFG->Nodes->Node X->Logon Requirements->Requirement String to:
NOT GUEST
Doing this will cause terminal logons as “Guest” to be rejected with the following (configurable) message:
You do not have sufficient access for this node.
Note:
You can remove the or 'Guest' text from the login module's Login: prompt, even if the Guest user account exists, by setting guest=false in the [login] section of the ctrl/modopts.ini file. This login module setting does not impact a user's ability to login to the Terminal Server as Guest.
## SBBSList
Question:
How do I get my BBS listed on the Synchronet BBS List?
The best way is to first join DOVE-Net, then run the Synchronet BBS List (sbbslist) door on *your* BBS and add an entry for your BBS. This entry should be automatically exported to the DOVE-Net SYNCDATA echo which will then be propagated to Vertrauen and every other BBS on DOVE-Net. The sbbslist.html page is automatically generated on Vertrauen every night at midnight (Pacific time), so wait a day or so for your entry to appear on the list. You can also check a dynamically generated HTML BBS list here or here. If it doesn't appear, verify that that your BBS entry was properly exported (as a message) to the SYNCDATA message area. The timed events SMB2SBL and SBL2SMB that will import and export BBS entries should be configured by default. If they are not, see sbbslist for more details.
## Time Stamp Error
Question:
Why am I getting the following error when running SBBS?
source: logon.cpp
action: checking
object: Daily stats time stamp
Because your system's date/time has been adjusted backwards more than 24 hours.
1. Shutdown the BBS
3. Run exec/dstsedit in your Synchronet ctrl directory, change the “Date Stamp” to yesterday's date
4. Restart the BBS
## Migration
Question:
How can I migrate my Synchronet configuration and data to a new computer?
You can simply copy the Synchronet directory tree (e.g. C:\SBBS), complete with all sub-directories, to the new computer, even if the new computer is running a different operating system!
Synchronet doesn't require installation on a computer, but if you wish to, you can install Synchronet on the new computer and then simply over-write the newly-installed files with your old/migrated files, provided they are the same version of Synchronet. If the old computer was running a different (older) version of Synchronet than the version installed on the new computer, then it's likely best to upgrade the old version (if you can) on the old computer before migrating to the new one, though this is not strictly required.
If the target (new) computer is running Windows and you choose not to install, then you may want to create a shortcut to sbbsctrl.exe in your Windows Startup folder (so it will automatically run when you login) and/or install the Synchronet NT Services (which can run without logging-in as a user). If the target computer is running Windows Vista or later, then you'll also want to follow the solution to this FAQ as well. For 16-bit DOS program support, the Synchronet Windows NT FOSSIL driver (sbbsexec.dll) must also be copied to the Windows System32 folder. See the sbbsexec.dll FAQ for details.
If the source (old) computer was running Windows and you wish to preserve any changes you made to the Synchronet Control Panel settings (stored in the Windows registry), you can export those settings to a sbbsctrl.ini file (using the File->Export Settings menu option) and then import them into the Synchronet Control Panel running on the new computer (using the File->Import Settings menu option). These settings include some changes made via the File->Properties menu. Settings made in most other menus in the Synchronet Control Panel are stored in your ctrl/sbbs.ini file.
If the source (old) and target (new) computers are both running a Unix-like operating system (e.g. Linux), be sure to rebuild *all* the native executable binaries (e.g. libsbbs.so), including 3rd party libraries (e.g. libcl.a) on the new/target computer before trying to execute them. Failure to do so could result in “invalid opcode” errors.
Configuration changes made via the SCFG utility are stored in the *.cnf files in your ctrl directory.
If you wish to only migrate configuration and data files (e.g. over an new installation of Synchronet), copy (recursively) the following Synchronet sub-directories to the new computer:
• data
• ctrl
• node1
If you customized any of your filter (*.can) or menu (e.g. *.asc) files, you'll also want copy (recursively) the following directory:
• text
If you customized any of your executable modules (e.g. *.js), you'll also want copy the following directory:
• mods
If you installed any new doors in the xtrn directory or your users used any of the stock-installed doors and you'd like to retain that data, you'll also want to copy the following directory:
• xtrn
Note: Sysop-installed (e.g. not from CVS) or modified modules should not normally be stored in your exec directory, but if they were, then you'll want to copy those over as well.
## Max Msgs
Question:
Why are there more messages in my message base than the configured maximum number of messages for that sub-board in SCFG?
The maximum number of messages configured via SCFG->Message Areas ... Sub-boards (and visible when using the smbutil s command as the max_msgs value) is enforced when the message maintenance (e.g. smbutil m command) is executed.
By default, Synchronet comes with a pre-configured Timed Event called MSGMAINT, set to run weekly, which will perform normal message base maintenance on all of the message bases in your data directory, including the deleting of old messages, if necessary, to meet the configured “Maximum Messages” value. Timed events may also be forced by the sysop to run immediately at any time, if desired.
## QNET Tag
Question:
Why doesn't my configured QWK networking tagline (or FidoNet Origin line) appear on my locally posted messages?
QWK network taglines (and FidoNet Origin lines) aren't added to messages until they are exported to a message network. The locally posted messages will not have QWKnet taglines (or FidoNet Origin lines).
Note: You can see what your configured QWKnet tagline (or FidoNet Origin line) looks like by using the I (sub-board information) command.
## FTN MSGID
Question:
Why do the FidoNet-style Message-IDs for messages created-by or posted-on Synchronet BBSes appear different than the Message-IDs of messages posted by other FidoNet-type nodes?
Because, for the message-IDs to be useful for any intended purpose (e.g. dupe-checking, reply-linking) they must be unique for every message created on a BBS.
When using Internet-standard message-IDs (defined in 1982), this is not a problem because the entire message identifier is basically free-form text (between < and > delimiters), leaving the entire data portion completely up to the implementer (e.g. programmer) to add whatever data they feel necessary to guarantee a unique ID for any particular message created at any given time. No data within an Internet-standard message-ID is expected to be interpreted or used in any way, other than to uniquely identify a particular message. The data could be a large number or just a string of meaningless symbols and so long as it is unique, it's doing its job rather perfectly. Usually, an Internet-standard message-ID will contain a globally unique addresses (e.g. public IP address or host name) so as to prevent random collisions with the IDs generated on a nearly unlimited number of other systems on the Internet, but there is no requirement that an address is included in the identifier.
However, for FidoNet, the de facto standard (defined in 1991) imposed some restraints on the contents of its message-IDs (the data between ^AMSGID: and CR delimiters): 2 fields separated by a single space: origaddr and serialno.
The definition of the serialno field is thus:
... may be any eight character hexadecimal number, as long as it is unique -
no two messages from a given system may have the same serial number within a three years.
My interpretation of this requirement is that the author of the specification was asserting that 32-bits (8 hex digits) of data (~4.2B unique numbers) was enough to uniquely identify every message generated on a single FidoNet node for at least 3 years. Even if that assertion were true, as an experienced BBS software developer, I have these issues with that concept:
1. 3 years is not long enough. I have message threads on my BBS dating back 20 years or more and I certainly expect and my software depends on the message-IDs being unique so long as the messages are stored in the message bases. If the messages are stored for a 100 years, their identifiers should still be unique then. So forget the 3 year clause, that's no more valid a limitation than Bill Gates' 640 Kilobytes1).
2. A pseudo-randomly chosen 32-bit number is obviously not a good choice, but it would probably work for the majority use case. I fear that some inexperienced or lazy programmers might have chosen this simple “solution” and no one would be the wiser until... someday there's a random ID collision and someone might notice and complain. But probably not.
3. A 32-bit time_t (Unix time, in seconds since Jan-1-1970) would seem an obvious choice for the serialno field except for the fact that today's systems can generate many thousands of messages per second if they need to. Even the multinode systems of the 80's and 90's could easily generate multiple messages within the same second. Many FidoNet programmers fatally chose a time_t value as their serialno and only source of uniqueness and you can see the proof of this in the MSGIDs that show up in FidoNet messages to this day2). Again, nobody is wise to this design flaw until there is a “random” collision one day... and maybe someone notices. But usually not.
4. A system to exclusively register monotonically-incrementing message serial numbers would be error prone (e.g. the sysop deletes the data file tracking the last assigned serial number) and unnecessarily bottle-neck message generation (e.g. importing thousands of messages from some other networking technology, which would need unique FidoNet-style message-IDs assigned to each).
5. It's obviously still a problem if 2 different systems generate the same message-ID serialno value, unless there's some other source of “uniqueness” in the Message-ID...
Which brings us to the origaddr field which is defined in FTS-9 as:
... should be specified in a form that constitutes a valid return address for the originating network.
Now this identifier field has some promise:
1. It's not limited to a specific number of characters or a specific character set.
3. It's only a “should” clause, which in spec-lawyer-speak, means: “you don't have to” (in contrast to “shall” or “must”).
So the origaddr is where Synchronet and its utilities (e.g. SBBSecho) adds unique data to insure that all messages generated by a correctly-configured Synchronet system have unique FTN message-IDs... forever. The data we place in the origaddr is currently thus:
<msgnum>.<subcode>@<ftn_addr>
where
• <msgnum> is the monotonically-incrementing message number for the message in this particular message database (echo area)
• <subcode> identifies the system-local message database where the identified message is stored
• <ftn_addr> is the 3D or 4D FTN-style address of the FidoNet-style node that created the message
Example:
73470.sync@1:103/705
The result is a “valid return address for the originating network”, in that a FidoNet NetMail sent to this address will normally reach the sysop of the originating node (in the example case, me). But the originating network might not be an FTN network. There is nothing in FTS-9 that specifies the form or content of the origaddr field. In fact, it mentions the use of “double-quotes” and how they must be escaped if part of the address. When is the last time you saw quotes of any kind in an FTN address? Clearly the author was leaving this definition open to non-FTN addresses. And again, it's only a “should” clause anyway. This field could only contain the GPS coordinates of my birthplace and would still be completely within the requirements of the specification.
The serial number (serialno) field generated by Synchronet and its utilities is a mathematical sum of the area-unique <msgnum> and a number derived from the date and time the message was created so as to maximize the use of the allocated 32-bits. The serial number will be unique among all the messages generated by a Synchronet-system in a single message area (echo), but is not guaranteed to be unique, by itself, among all the FTN-connected message areas of a single Synchronet-system. The serialno field alone does not constitute a unique message identifier, so don't try to use it as one!
But it doesn't look like other other FTN MSGID's!?!
This is true. But before I decided on the makeup of the Synchronet-generated FTN MSGID's I collected and collated many thousands of FidoNet messages from the Zone 1 backbone echoes and found that there was already a lot of variety in the look of the orginaddr field of the MSGIDs used over the last 20+ years: Most were simple 3D or 4D FTN-style addresses, but many had 5D-style addresses with @fidonet as the domain, sometimes @fidonet.org, and many others just had what appeared to be an Internet-style host name (no @ or FTN-style address at all). Others combined Internet-style host names with other seemingly random or incrementing numbers or symbols.
So Synchronet/SBBSecho did not introduced some “new thing” into FidoNet. Indeed, I would just be the next thoughtful programmer that has solved this problem in his or her own way because the original specification (FTS-9) was short-sighted3) and never fixed.
If you have or know of someone that has software that has a problem with the Synchronet-generated FTN MSGID's, I am truly sorry. That same software, if it exists, would have a similar problem, whatever that is, with many other unique FTN MSGID's which existed long before Synchronet and SBBSecho started sending messages with unique message-IDs into FidoNet-style networks.
But why don't you change it to look like other FTN MSGID's!?!
Because:
1. That would defy the uniqueness of the FTN MSGID's generated by Synchronet, and that's the entire point of their existence.
2. If I were motivated to change how the MSGIDs look, it would be to make them more unique, not less.
3. The MSGIDs are already compliant with the requirements of the de facto standard specification (read it), so they don't need to be changed.
4. The MSGIDs are not that dissimilar to unique MSGIDs that have been included in FidoNet messages posted by other compliant-software for over 20 years.
## Old Baja
Question:
Why do I get the following error when compiling an old Baja source (*.src) file: !SYNTAX ERROR (expecting integer constant)?
Most likely, the Baja source file references numeric constants using symbolic names (e.g. UM_EXPERT) which *used* to be hard-coded in to the Baja compiler, but aren't any longer (since Baja v2, circa 1995). These numeric constants were moved to the Baja include file named sbbsdefs.inc. The solution is to add the following line near the beginning/top of the old Baja source file in question:
!include sbbsdefs.inc
Note: To see a complete list of changes in Baja v2, see baja2new.txt.
## Spell
Question:
How do I spell “Synchronet”?
With an 'h'. You can CamelCase the “Net” in “Synchronet” if you strongly prefer, but since there are a few companies/services which have trademarked that name (with a capitalized “Net”), it may be confusing. It's been just “Synchronet” (like “Internet”) for over 20 years now, so it should probably just stay that way.
Note: This isn't actually a frequently asked question, but it should be.
## Abbreviate
Question:
How do I abbreviate “Synchronet”?
Without an 'h'. Seriously, the full name is “Synchronet BBS Software”, so we prefer to usually abbreviate that to either 'SBBS' or just 'Sync'. Since 'Sync' is pronounced with one syllable and 'SBBS' is four, sometimes that's a reason to prefer 'Sync' to 'SBBS'. 'Synch' looks like 'sinch' or 'cinch', so that abbreviation should be avoided.
Note: This isn't actually a frequently asked question, but it should be.
## SyncTERM
Question:
Where do report a bug or request a feature of SyncTERM?
On sourceforge.net, feature requests and bug reports.
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2021-06-17 21:18:08
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https://gamedev.stackexchange.com/questions/115859/slaving-openscenegraph-to-an-external-time-base
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# Slaving OpenSceneGraph to an external time base?
I'm working on an OpenSceneGraph animation, and I want events in the animation to be synchronized to a time reference provided by my audio player (BASS or FMod). Yes, this is a demo :) . Which time do I set, and where? Sync is more important than frame rate or continuity - if the audio rewinds, I want the visuals to rewind, too. (To support GNU Rocket.)
I have looked at the forum posts and Google results, but haven't seen a clear answer for this use case.
• FrameStamps have a frame number, a calendar time, a simulation time, and a reference time. I think the simulation time is what I should change - is that so?
• Is specifying a new simulation time as a parameter to frame() or advance() the right way to specify the time? Will that work even if time moves backwards?
## 1 Answer
The answer can be found in the osgViewer::ViewerBase class:
/** Render a complete new frame.
* Calls advance(), eventTraversal(), updateTraversal(), renderingTraversals(). */
virtual void frame(double simulationTime=USE_REFERENCE_TIME);
virtual void advance(double simulationTime=USE_REFERENCE_TIME) = 0;
So just pass the simulation time you want to viewer.frame(time).
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2019-12-07 19:35:25
|
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|
https://pypi.org/project/decaf-espresso/
|
Light-weight ASE calculator wrapper for Quantum Espresso.
## Project description
#+Title: Decaf Espresso
#+Author: Jacob Boes
#+OPTIONS: toc:nil
* Introduction
Decaf espresso is light-weight [[https://wiki.fysik.dtu.dk/ase/][ASE]] wrapper for Quantum Espresso with convenient features inspired by [[https://github.com/vossjo/ase-espresso][ase-espresso]]. The goal is to produce a simplified version which allows for most of the non-interactive functionality without so much of the verbosity of ase-espresso. The code is meant to be short and segmented into logical portions to help those who are new to Quantum Espresso learn more easily. The majority of the simplification comes from farming out the =io= functionality to ASE, which also amounts to less code to maintain.
*Pros:*
- Input units are eV (similar to ase-espresso) and converted automatically.
- All input keywords are identical to Quantum Espresso to prevent confusion. See: [[https://www.quantum-espresso.org/Doc/INPUT_PW.html][QE Inputs]].
- Automatic validation of (some) parameters inspired by [[https://github.com/jkitchin/vasp][Vaspy]].
- Straight forward record of default parameters used for simplified documentation.
- Automatic handling of calculation node scratch assignment for general clusters (SLURM, LSF, PBS/TORQUE), to prevent unnecessary disk-io.
- Automatic handling of MPI execution of general clusters (SLURM, LSF, PBS/TORQUE) and intelligent assignment of k-point parallelization (npool).
- Specific executable setups for compatibility on SLAC, Sherlock, and NERSC clusters. (works out of the box)
- Has post-processing functions which are not yet available in ASE.
- Written in Python 3
- DRY code with extensive documentation.
*Cons:*
- Not all post processing features available from ase-espresso have been implemented.
- No support for interactive ASE without efficiency loss
- Not all parameters are currently validated or tested (such as DFT+U parameters).
- Limited testing in Python 2 and will not be supported moving forward.
* Installation
** Pip installation
decaf-espresso is most easily installed with pip using:
#+BEGIN_SRC sh
pip install decaf-espresso
#+END_SRC
For Mac OSX, homebrew can be used in place of pip
#+BEGIN_SRC sh
brew install decaf-espresso
#+END_SRC
For usage on high-performance computers, installation will need to be performed locally which can be done using:
#+BEGIN_SRC sh
pip install --user decaf-espresso
#+END_SRC
These commands will install all of the necessary dependencies for you.
** Source installation
Alternatively, a version with the most recent commits can be installed through git by running the following in your home directory.
#+BEGIN_SRC sh
git clone https://github.com/jboes/decaf-espresso.git
#+END_SRC
#+BEGIN_SRC sh
export PYTHONPATH=~/decaf-espresso:\$PYTHONPATH
#+END_SRC
Once cloned, the requirements and be installed by running the following commands (Add the =--user= argument if needed):
#+BEGIN_SRC sh
cd ~/decaf-espresso
pip install -r requirements.txt
#+END_SRC
* Parameter Validation
Parameter validation is currently performed for the arguments which I use most frequently for my calculations, but the general formula for validation is easily extensible by anyone. The basic idea is that any [[https://www.quantum-espresso.org/Doc/INPUT_PW.html][Qunatum Espresso Input]] with a similarly named function in the [[./espresso/validate.py][validation module]] will have the entailed function executed if the parameter is input by the user.
For example, If I were to initialize a calculation as:
#+BEGIN_SRC python :results output org drawer
from espresso import Espresso
calc = Espresso(kpts=(1, 1, 1))
#+END_SRC
Will execute the similarly named validation function:
#+BEGIN_SRC python :results output org drawer
def kpts(calc, val):
"""Test k-points to be 'gamma' or list_like of 3 values.
Only automatic assignment is currently supported.
https://www.quantum-espresso.org/Doc/INPUT_PW.html#idm45922794051696
"""
if val == 'gamma':
return
assert isinstance(val, (tuple, list, np.ndarray))
assert len(val) == 3
#+END_SRC
If an invalid input is used, and exception will be raised.
#+BEGIN_SRC python :exports both
from espresso import Espresso
calc = Espresso(kpts=(1, 1))
#+END_SRC
#+RESULTS:
: Traceback (most recent call last):
: File "Org SRC", line 3, in <module>
: calc = Espresso(kpts=(1, 1))
: File "/home/jboes/research/decaf-espresso/espresso/espresso.py", line 62, in __init__
: new_val = f(self, val)
: File "/home/jboes/research/decaf-espresso/espresso/validate.py", line 226, in kpts
: assert len(val) == 3
: AssertionError
TODO: Make a more helpful validation error.
** Writing a validation function
Each validation function follows the simple formula:
#+BEGIN_SRC python :results output org drawer
def parameter_name(calc, val):
assert # An appropraite test here
return updated_val # optional
#+END_SRC
Where =parameter_name= is the exact name of the Quantum espresso parameter, and =(calc, val)= are always passed as arguments. Here, =calc= is the =Espresso= calculator object, which can be used to all other calculator parameters and =val= is the user defined value for the given parameter which can be directly tested against.
In decaf-espresso, validation functions also server the double role of updating certain values. For Example, Quantum Espresso takes units of energy in Rydbergs, but eV are more commonly used in surface science. So, any validation function which takes Rydbergs will also return and =updated_val= which is the value converted to Rydbergs from eV so the user can specify inputs in eV. This sacrifices some readability, but avoids looping over extra lists of known value types, helping keep the code DRY.
* Example scripts
Usage of the calculator are shown below for varying structure types.
** Molecule relaxation
The example below will relax an H_{2} molecule using some standard flags. Below is the rational for some of the flags used.
- =ecutwfc=: A required argument, represents the energy cutoff for the wave functions.
- =conv_thr=: The threshold for considering a total energy converged. DFT is only accurate to about 0.1 eV at best, so 1e-4 should be sufficient for most use cases.
- =degauss=: Gaussian smearing coefficient. This is a non-physical contribution meant only to help atomic structures with d-bands converge correctly. For molecules we set it to be small.
#+BEGIN_SRC python :results output org drawer
from espresso import Espresso
from ase.build import molecule
parameters = {
'calculation': 'relax',
'input_dft': 'PBE',
'ecutwfc': 500,
'conv_thr': 1e-4,
'degauss': 0.01}
atoms = molecule('H2', vacuum=6)
calc = Espresso(atoms, **parameters)
atoms.get_potential_energy()
#+END_SRC
## Project details
Uploaded source
Uploaded py3
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2022-11-29 11:01:16
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http://math.stackexchange.com/questions/83946/hardys-inequality-for-integrals/95399
|
Hardy's Inequality for Integrals
I am trying to prove Hardy's Inequality for integrals:
Let $f:(0,\infty) \rightarrow \mathbb R$ be in $L^p$. Let $F(x) = \frac{1}{x} \int_0 ^x f(t)dt$. Then $F \in L^p$ and $\|F\|_p \leq \frac{p}{p-1} \|f\|_p$.
I have seen proofs of this which use Haar measure and Fourier analysis, eg. here An inequality by Hardy. However, I am wondering if there is a more elementary proof.
I have tried to deduce this using Jensen's inequality and Fubini's Theorem as follows:
Jensen's inequality implies that $\left|\frac{1}{x} \int_0 ^x f(t)dt)\right|^p \leq \frac{1}{x} \int_0 ^x |f(t)|^p dt$. Therefore
$$\int_0 ^\infty |F(x)|^p dx \leq \int_0 ^\infty \frac{1}{x} \int_0 ^x |f(t)|^p dtdx = \int_0 ^\infty \int_t ^\infty \frac{1}{x} |f(t)|^p dx dt.$$
But, this last integral is infinite, and I'm not sure how to get a sharper bound on $|F(x)|^p$. Any suggestions?
-
Let $p\in (1,+\infty)$ and $q$ its conugate exponent, that is, the real number $q$ such that $\frac 1p+\frac 1q=1$. Let $f\colon (0,+\infty)\to\mathbb R\in L^p(0,+\infty)$ and $F(x)=\frac 1x\int_{\left]0,x\right[}f(t)dt$. Then $F\in L^p(0,1)$ and $\lVert F\rVert_{L^p}\leq q\lVert f\rVert f_{L^p}$.
Pick $0<\alpha<\frac 1q$, we will specifie it later. We start using Hölder inequality \begin{align*} |xF(x)|&=\left|\int_{\left]0,x\right[}f(t)t^{\alpha}t^{-\alpha}dt\right|\\ &\leq \left(\int_{\left]0,x\right[}\left| f(t) t^{\alpha}\right|^pdt\right)^{\frac 1p} \left(\int_{\left]0,x\right[}t^{-q\alpha}\right)^{\frac 1q}\\ &=\left(\int_{\left]0,x\right[}\left| f(t)\right|^p t^{p\alpha}dt\right)^{\frac 1p} \left(\frac 1{1-q\alpha}x^{1-q\alpha}\right)^{\frac 1q}\\ &=(1-q\alpha)^{-\frac 1q}\left(\int_{\left]0,x\right[}\left| f(t)\right|^p t^{\frac 1{q}}dt\right)^{\frac 1p}x^{\frac 1q-\alpha}, \end{align*} hence \begin{align*} |F(x)|&\leq (1-q\alpha)^{-\frac 1q}\left(\int_{\left]0,x\right[}\left| f(t)\right|^p t^{p\alpha}dt\right)^{\frac 1p}x^{\frac 1q-\alpha -1}\\ &=(1-q\alpha)^{-\frac 1q}\left(\int_{\left]0,x\right[}\left| f(t)\right|^p t^{p\alpha}dt\right)^{\frac 1p}x^{-\alpha -\frac 1p} \end{align*} and $$|F(x)|^p\leq (1-q\alpha)^{-\frac pq}\int_{\left]0,x\right[}\left| f(t)\right|^p t^{p\alpha}dt x^{-p\alpha-1}.$$ Integrating and using Fubini's theorem, we get
\begin{align*} \int_{(0,+\infty)}|F(x)|^pdx&\leq (1-q\alpha)^{-\frac pq}\int_{(0,\infty)}\int_{(0,x)} \left| f(t)\right|^p t^{p\alpha}x^{-p\alpha-1}dtdx\\ &=(1-q\alpha)^{-\frac pq}\int_{(0,+\infty)}\int_{(t,+\infty)}\left| f(t)\right|^p t^{p\alpha}x^{-p\alpha-1}dxdt\\ &=(1-q\alpha)^{-\frac pq}\int_{(0,+\infty)}\left| f(t)\right|^p t^{p\alpha}\left(\int_{(t,+\infty)}x^{-p\alpha-1}dx\right)dt, \end{align*} thus $$\int_{(0,+\infty)}|F(x)|^pdx \leq(1-q\alpha)^{-\frac pq}\int_{(0,+\infty)}\left|f(t)\right|^p t^{p\alpha}\frac 1{p\alpha}t^{-p\alpha}dt$$ and finally $$\int_{(0,+\infty)}|F(x)|^pdx\leq (1-q\alpha)^{-\frac pq}(p\alpha)^{-1}\int_{(0,+\infty)}\left| f(t)\right|^p dt.$$ Now, we pick $\alpha:=\frac 1{pq}<\frac 1q$, to get $$(1-q\alpha)^{-\frac pq}(p\alpha)^{-1} =(1-\frac 1p)^{-\frac 1q}q= q^{\frac pq}q=q^{1+p\left(1-\frac 1p\right)}=q^p,$$ which gives $\lVert F\rVert_{L^p}\leq q\lVert f\rVert_{L^p}$.
This question deals with the case of the best constant.
-
Hi Davide! I'm afraid you didn't prove that $q$ is the best constant. Indeed, if $f=\chi_{[0, 1]}$ then $\lVert F\rVert_p=q^{1/p}$ which is smaller than $q$. Also, any "best constant" argument for this inequality must involve some limiting process, because equality is never achieved non-trivially. – Giuseppe Negro Apr 1 '12 at 13:47
Show $$\left(\int_0^\infty \left| \frac1x \int_0^x f(t) \, dt\right|^p \, dx \right)^{1/p} \le \frac p{p-1} \left(\int_0^\infty |f(x)|^p \, dx \right)^{1/p}$$ Notice $$\frac1x \int_0^x f(t) \, dt = \int_0^1 f(x t) \, dt$$ Apply Minkowski's inequality (integral version) to LHS, to see it is bounded above by $$\int_0^1 \left(\int_0^\infty |f(xt)|^p \, dx \right)^{1/p} \, dt = \int_0^1 t^{-1/p} \, dt \left(\int_0^\infty |f(x)|^p \, dx \right)^{1/p}$$
-
Upvoting: nice proof. – Did Dec 31 '11 at 12:39
Dualize - it becomes $$\left(\int_0^\infty |F(x)|^p \, dx\right)^{1/p} \le p \left(\int_0^\infty |f(x)|^p \, dx \right)^{1/p}$$ where $$F(x) = \int_x^\infty \frac{f(t)}t \, dt$$ W.l.o.g. $f(x) \ge 0$, and w.l.o.g. $F(x) \to 0$ as $x \to \infty$. $$|F(x)|^p = - p \int_x^\infty F'(t) |F(t)|^{p-1} \, dt$$ Substitute this into the LHS of the first inequality, and reverse the order of integration $$- p \int_0^\infty F'(t) |F(t)|^{p-1} \int_0^t \, dx \, dt = p \int_0^\infty f(t) |F(t)|^{p-1} \, dt \le p \|f\|_p \|F\|_p^{p-1}$$ where the last inequality is H\"older. Now assume $\|F\|_p$ is finite, and divide both sides by $\|F\|_p^{p-1}$.
-
|
2016-02-13 23:50:57
|
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|
https://stacks.math.columbia.edu/tag/0D5R
|
Lemma 7.28.5. Let $\mathcal{C}$, $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be a cocontinuous functor. Let $V$ be an object of $\mathcal{D}$. Let ${}^ u_ V\mathcal{I}$ be the category introduced in Section 7.19. We have a commutative diagram
$\vcenter { \xymatrix{ \, _ V^ u\mathcal{I} \ar[r]_ j \ar[d]_{u'} & \mathcal{C} \ar[d]^ u \\ \mathcal{D}/V \ar[r]^-{j_ V} & \mathcal{D} } } \quad \text{where}\quad \begin{matrix} j : (U, \psi ) \mapsto U \\ u' : (U, \psi ) \mapsto (\psi : u(U) \to V) \end{matrix}$
Declare a family of morphisms $\{ (U_ i, \psi _ i) \to (U, \psi )\}$ of ${}^ u_ V\mathcal{I}$ to be a covering if and only if $\{ U_ i \to U\}$ is a covering in $\mathcal{C}$. Then
1. ${}^ u_ V\mathcal{I}$ is a site,
2. $j$ is continuous and cocontinuous,
3. $u'$ is cocontinuous,
4. we get a commutative diagram of topoi
$\xymatrix{ \mathop{\mathit{Sh}}\nolimits ({}^ u_ V\mathcal{I}) \ar[r]_ j \ar[d]_{f'} & \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \ar[d]^ f \\ \mathop{\mathit{Sh}}\nolimits (\mathcal{D}/V) \ar[r]^-{j_ V} & \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) }$
where $f$ (resp. $f'$) corresponds to $u$ (resp. $u'$), and
5. we have $f'_*j^{-1} = j_ V^{-1}f_*$.
Proof. Parts (1), (2), (3), and (4) are straightforward consequences of the definitions and the fact that the functor $j$ commutes with fibre products. We omit the details. To see (5) recall that $f_*$ is given by ${}_ su = {}_ pu$. Hence the value of $j_ V^{-1}f_*\mathcal{F}$ on $V'/V$ is the value of ${}_ pu\mathcal{F}$ on $V'$ which is the limit of the values of $\mathcal{F}$ on the category ${}^ u_{V'}\mathcal{I}$. Clearly, there is an equivalence of categories
${}^ u_{V'}\mathcal{I} \to {}^{u'}_{V'/V}\mathcal{I}$
Since the value of $f'_*j^{-1}\mathcal{F}$ on $V'/V$ is given by the limit of the values of $j^{-1}\mathcal{F}$ on the category ${}^{u'}_{V'/V}\mathcal{I}$ and since the values of $j^{-1}\mathcal{F}$ on objects of ${}^ u_ V\mathcal{I}$ are just the values of $\mathcal{F}$ (by Lemma 7.21.5 as $j$ is continuous and cocontinuous) we see that (5) is true. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
|
2023-03-29 17:21:31
|
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http://www.cardiff.ac.uk/mathematics/events/maths-colloquium-2017-18
|
# Maths Colloquium 2016-17
The School Colloquia are given by world-famous speakers and present an overview of important topics of general interest in the mathematical sciences.
These invited lectures are intended to be accessible to all graduate students and academics in the department. MMath and MSc students may also benefit from these presentations.
The talks normally take place on Wednesdays, 15:10 - 16:10, at the lecture room E/0.15 on the ground floor of the School. All are welcome to attend.
DateSpeakerTalk
12 October 2016Prof. Felix Otto (MPIMS, Leipzig)
Effective behavior of random media: From an error analysis to regularity theory
Heterogeneous media are often naturally described in statistical terms, reflecting a lack of knowledge of details. How to extract their effective behavior on large scales, like the effective conductivity $a_{hom}=const$, from the statistical specifications, which are encoded in a stationary probability measure or ensemble $\langle\cdot\rangle$ on the space of microscopic conductivities $a=a(x)$? A practioneers numerical approach is to sample the medium according the $\langle\cdot\rangle$ and to determine $a_{hom}$ in the Cartesian directions by imposing simple boundary conditions. What is the error made in terms of the size of this representative volume element''? Our interest in what is called stochastic homogenization'' grew out of this error analysis, and now also includes a characterization of the leading-order fluctuations.
In the course of developing such an error analysis, connections with the classical
regularity theory of elliptic equations have emerged. More precisely, stochastic homogenization sheds a new light on a {\it generic} large-scale behavior of $a$-harmonic functions --- which is more regular than suggested by the classical counter-examples. This might be rephrased in geometric terms: How flat at infinity does a metric $a$ have to be such that the space of harmonic functions of a given polynomial growth rate has exactly the same dimension as in the Euclidean case $a_{hom}$. We give a sufficient criterion that is almost surely satisfied for the type of probability measures $\langle\cdot\rangle$ on metrics $a$ considered in stochastic homogenization.
26 October 2016
Dr. James Maynard (Oxford)
Primes with restricted digits
Many of the most important questions about prime numbers can be phrased as 'given some set A of integers, how many primes are in A?'. Unfortunately, even simple versions of such questions are often well beyond current techniques, and this is especially difficult if A is a 'thin' set of integers.
I will talk about recent work which shows that there are infinitely many prime numbers with no 7's in their decimal expansion, giving an example of a thin set where we do get a satisfactory answer. Ideas from probability (such as Markov chains), diophantine geometry (lattice point counting and rational approximation) and combinatorics all turn out to be important ingredients, alongside traditional analytic number theory.
2 November 2016
Prof. Robert Weismantel (ETH, Zurich)
Integer Polynomial Optimization
1 February 2017
Dr. Nina Golyandina (St Petersburg)
Amended Colloquium time: 16:00 - 17:00, E/0.15
Singular spectrum analysis as a universal approach for finding structure in time series and digital images.
Singular spectrum analysis (SSA) is an effective method for processing different objects such as time series and digital images, finding their structures and then using the found structure for trend and periodicity extraction, smoothing, parameter estimation, forecasting, gap imputations. SSA is known as a nonparametric tool, which is able to analyse time series without a-priori assumptions about the object model. The method success is based on a specific construction of an adaptive decomposition, which is generated by the object itself. It is surprising how such a model-free method can solve problems which are conventional for parametric methods. We discuss this kind of paradox and demonstrate the method abilities as well as the mathematics underlying the SSA approach.
15 March 2017
Prof. Caroline Series FRS (Warwick)
The cover of the December AMS Notices
The cover of the December 2016 AMS Notices shows an eye-like region picked out by blue and red dots and surrounded by green rays. The picture, drawn by Yasushi Yamashita, illustrates Gaven Martin’s search for the smallest volume hyperbolic orbifold. It represents a family of two generator groups of isometries of hyperbolic 3-space which was recently studied, for quite different reasons, by myself, Yamashita and Ser Peow Tan.
After explaining the coloured dots and their role in Martin’s search, we concentrate on the green rays. These are Keen-Series pleating rays which are used to locate spaces of discrete groups. We also introduce some rather mysterious `fake’ pleating rays which partially fill the space of non-discrete groups and relate to a condition of Bowditch, mentioned but not explained in the Notices.
22 March 2017
Prof. Frances Kirwan FRS (Oxford)
Moduli spaces of unstable curves.
The construction of the moduli spaces of stable curves of fixed genus is one of the classical applications of Mumford's geometric invariant theory (GIT), developed in the 1960s. Here a projective curve is stable if it has only nodes as singularities and its automorphism group is finite. The aim of this talk is to describe these moduli spaces and outline their GIT construction, and then to explain how recent methods from non-reductive GIT can help us to classify the singularities of unstable curves in such a way that we can construct moduli spaces of unstable curves (of fixed singularity type).
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2017-09-22 09:41:18
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https://oeis.org/wiki/OeisWiki:Community_Portal
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This site is supported by donations to The OEIS Foundation.
# OeisWiki:Community Portal
## Contents
This page contains notes about the status of the OEIS wiki. It provides a brief description of the wiki setup. It will also be used to announce changes in the structure of the wiki, so it may be useful to add it to your watchlist by clicking on the watch tab at the top of the page.
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Do not use the discussion pages to suggest changes to OEIS sequence entries.
An account is necessary to contribute to the Wiki. To request an account, click here. There was a time when the username was required to be your Real Name. This is no longer required, but is strongly recommended. If this evolving statement causes you to want to change your username, send email to David, and I can make the change.
[talk]
## Sequence of the Day for May 25
A202300: Decimal expansion of the real root of
x3 + 2x2 + 10x − 20
.
1.36880810782137...
Today, we could get a thousand digits of this number in any base we wanted just by pushing a few buttons. But in Fibonacci's day, it was quite an achievement to get this number correct to five sexagesimal places, $\scriptstyle 1 + \frac{22}{60} + \frac{7}{60^2} + \frac{42}{60^3} + \frac{33}{60^4} + \frac{4}{60^5} + \frac{40}{60^6} \,$, approximately
1.36881
in decimal, which as you can see matches the correct answer to ten decimal places.
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2015-05-25 15:27:44
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https://atdotde.blogspot.com/2011/06/
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## Friday, June 03, 2011
### Bitcoin explained
As me, you might have recently heared about "Bitcoin", the internet currency that tries to be safe without a central authority like a bank or a credit card company that say which transactions are legitimate. So far, all mentions in blogs, podcasts or the press that I have seen had in common that they did not say how it works, what are the mechanisms that make sure Bitcoins operate like money. So I looked it up and this is what I found:
Bitcoin uses to cryptographic primitives: hashes and public key encryption. I case you don't know what these are: A hash is a function that reads in a string (or file or number, those are technically all the same) and produces some sort of checksum. The important properties are that everybody can do this computation (with some small amount of effort) and produce the same checksum. On the other hand, it is "random" in the sense that you cannot work backwards, i.e. if you only know the checksum you effectively have no idea about the original string. It is computationally hard to find a string for a given checksum (more or less the best you can do is guess random strings, compute their checksums until you succeed). A related hard problem is to find such a string with prescribed first $N$ characters.
This can be used as a proof of effort: You can pose the problem to find a string (possibly with prescribed first characters) such that the first $M$ digits of the checksum have a prescribed value. In binary notation you could for example you could ask for $M$ zeros. Then on the average you have to make $2^M$ guesses for the string until you succeed. Presenting such a string then proves you have invested an effort of $O(2^M)$. The nice thing is that this effort is additive: You can start your string with the characters "The message '....' has checksum 000000xxxxxxxxxxx" and continue it such that the checksum of the total string starts with many zeros. That proves that in addition to the zeros your new string has, somebody has already spent some work on the string I wrote as dots. Common hash functions are SHA-1 (and older and not as reliable: MD5).
The second cryptographic primitive is public key encryption. Here you have two keys $A$, the public key which you tell everybody about and $B$ your secret key (you tell nobody about). These have the properties that you can use one of the keys to "encrypt" a string and then the other key can be used to recover the original string. In particular, you need to know the private key to produce a message that can be decrypted with the public key. This is called a "signature": You have a message $M$ and encrypt it using $B$. Let us call the result $B(M)$. Then you can show $A$ and $M$ and $B(M)$ to somebody to prove that you are in possession of $B$ without revealing $B$ since that person can verify that $B(M)$ can be decrypted using $A$. Here, an example is the RSA algorithm.
Now to Bitcoin. Let's go through the list of features that you want your money to have. The first is that you want to be able to prove that your coins belong to you. This is done by making coins files that contain the public key $A$ of their owner. Then, as explained in the previous paragraph you can prove that you are the legitimate owner of the private key belonging to that coin and thus you are its owner. Note that you can have as many public-private key pairs as you like possibly one for every coin. It is just there to equate knowing of a secret (key) to owning the coin.
Second you want to be able to transfer ownership of the coin. Let us assume that the recipient has the public key $A'$. Then you transfer the coin (which already contains your public key $A$) by appending the string "This coin is transfered to the owner of the secrete key to the public key $A'$". Then you sign the whole thing with your private key $B$. The recipient can now prove that the coin was transferred to him as the coin contains both your public key (from before) and your statement of the transfer (which only you, knowing $B$ can have authorized. This can be checked by everybody by checking the signature). So the recipient can prove you owned the coin and agreed to transfer it to him.
The last property is that once you transfered the coin to somebody else you cannot give it to a third person as you do not own it anymore. Or put differently: If you try to transfer a coin a second time that should not work and the recipient should not accept it or at least it should be illegitimate.
But what happens if two people claim they own the same coin, how can we resolve this conflict? This is done via a public time-line that is kept collaboratively between all participants. Once you receive a coin you want to be able to prove later that you already owned it at a specific time (in particular at the time when somebody else claims he received it).
This is done as follows: You compute the hash function of the transfer (or the coin after transfer, see a,bove including the signature of the previous owner of the coin that he has given it to you) and add it to the time line. This means you take the hash value of the time line so far, at the hash of the transfer and compute new hash. This whole package you then send to your network peers and ask them to also include your transfer in their version of the time line.
So the time line is a record of all the transfers that have happened in the past and each participant in the network keeps his own copy of it.
There could still be a conflict when two incompatible time lines are around. Which is the correct one that should be trusted? One could have a majority vote amongst the participants but (as everybody knows from internet discussions) nothing is easier than to come up with a large number of sock puppets that swing any poll. Here comes the proof of work that I mentioned above in relation to hash functions: There is a field in the time line that can be filled with anything in the attempt to construct something that has a hash with as many zeros as possible. Remember, producing $N$ leading zeros amounts to $O(2^N)$ work. Having a time line with many zeros demonstrates that were willing to put a lot of effort into this time line. But as explained above, this proof of effort is additive and all the participants in the network continuously try to add zeros to their time line hashes. But if they share and combine their time lines often enough such that they stay coherent they are (due to additivity) all working on fining zeros on the same time line. So rather than everybody working for themselves everybody works together as long as their time lines stay coherent. And going back through a time line it is easy to see how much zero finding work has been but in. Thus in the case of conflicting time lines one simply takes that that contains more zero finding work. If you wanted to establish an alternative time line (possibly one where at some point in time you did not transfer a coin but rather kept it to yourself so you could give it to somebody else later) to establish it you would have to outperform all other computers in the network that are all busy working on computing zeros for the other, correct, time line.
Of course, if you want to receive a bitcoin you should make sure that in the generally accepted time line that same coin has not already been given to somebody else. This is why the transfers take some time: You want to wait for a bit that the information that the coin has been transferred to you has been significantly spread on the network and included in the collective time line that it cannot be reversed anymore.
There are some finer points like how subdividing coins (currently worth about 13 dollars) is done and how new coins can be created (again with a lot CPU work) but I think they are not as essential in case you want to understand the technical basis of bitcoin before you but real money in.
BTW, if you liked this exposition (or some other here) feel free to transfer me some bitcoins (or fractions of it). My receiving address is
19cFYVExc2ZS4p7ZARGyENFijnV43y6ts1
.
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2019-08-24 02:30:41
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http://mathhelpforum.com/trigonometry/38219-roots-complex-numbers-print.html
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# roots of complex numbers
• May 13th 2008, 07:12 AM
samdmansam
roots of complex numbers
a) find all roots of the equation z^5=sqrt(3)*i-1, present your answer in polar form
b) express (2-2i)^5 in exact polar form
(-1+sqrt(3)*i)^4
I was alright until the whole polar form thing, your help would be greatly appreciated.
• May 13th 2008, 08:57 AM
topsquark
Quote:
Originally Posted by samdmansam
a) find all roots of the equation z^5=sqrt(3)*i-1, present your answer in polar form
For a) what is the polar form of $-1 + i\sqrt{3}$?
$r~cos(\theta) + ir~sin(\theta) = -1 + i\sqrt{3}$
So
$r~cos(\theta) = -1$
and
$r~sin(\theta) = \sqrt{3}$
Thus
$\frac{r~sin(\theta)}{r~cos(\theta)} = -\sqrt{3}$
Now,
$tan(\theta) = \left | -\sqrt{3} \right | \implies \theta = \frac{\pi}{3}$
and since sine is positive and cosine negative we know the reference angle is in QII. Thus
$\theta = \frac{2 \pi}{3}$
Then
$r~cos(\theta) = -1$
$r \cdot -\frac{1}{2} = -1$
$r = 2$
So
$z^5 = 2e^{2i \pi / 3}$
Can you solve it from here? (Please note that there are five answers to this.)
-Dan
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2015-05-03 20:30:37
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https://www.physicsforums.com/threads/first-order-linear-ode-help.90520/
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# First-Order Linear ODE help?
First-Order Linear ODE help???
This is my first post here. I'm still getting used to LATEX syntax so please forgive any mistakes. My question is on a simple differential equation... it doesn't appear to be exact or homogeneous... where do I start?
DD86
#### Attachments
• untitled.jpg
4.9 KB · Views: 366
Tide
Homework Helper
HINT: Look for an integrating factor.
$$y'-4xy+2yln(y)/x=0$$
$$M(x,y)= -4xy \quad and \quad N(x,y)=2yln(y)/x$$
err.. I'm taking Calc 3 and diff eq at the same time so I hope my partials are right...
$$M_{y}=-4x \quad and \quad N_{x}=0$$
IntegFactor is defined as $$I(x) = e^{\int (M_{y} - N_{x})/N dx}$$ according to my notes. yeah I'm still stuck. Could someone verify the partial derivatives though?
EDIT: I now have $$y'+2y(-2x+ln(y)/x)=0$$ which looks nicer but im still at a loss for what to do next...
What integrating factor am I looking for?
Last edited:
saltydog
Homework Helper
December, need to put it into the form:
$$Mdx+Ndy=0$$
Thus for:
$$y^{'}-4xy+\frac{2yln(y)}{x}=0$$
we'd have:
$$(2yln(y)-4x^2y)dx+xdy=0$$
But that's not happening for me either. Can't see a way to solve it.
Tide . . . how about another hint? Somebody else too is ok.
Edit: Oh yea, it's not linear because of the ln(y) term.
Last edited:
DecemberDays86, This really doesnt have to do with anything, but, do you post on bodybuilding.com also?
thanks for your input saltdog. My prof seems to like asking questions where you need to recognize "chain-ruled" things. For example, I have to get used to identifying a product of two terms as the derivative of some compossite function... not cool hehe
Tide
Homework Helper
Divide both sides of your equation by y, note that y'/y = (ln y)', multiply both sides by x and finally arrive at
$$\frac {d}{dx} \left( x^2 \ln y \right) =4 x^3$$
then see where you can go from there!
saltydog
Homework Helper
Tide said:
Divide both sides of your equation by y, note that y'/y = (ln y)', multiply both sides by x and finally arrive at
$$\frac {d}{dx} \left( x^2 \ln y \right) =4 x^3$$
then see where you can go from there!
Ok, I get it now. Thanks Tide. You too December . . . PF rocks.
smart trick Tide!
here's my mess of a solution:
$$xy'+4x^2y-2ylny=0$$
$$y'-4xy=(-2ylny)/x$$
$$Let \quad y = e^{g(x)} \implies y'=g'e^{g(x)}$$
$$g'e^{g(x)}-4xe^{g(x)}= \frac {-2g(x)e^{g(x)}}{x}$$
divide out e^g(x) and rearrange
$$g'+\frac {2}{x}g=4x, Let \quad P(x) = 2/x \quad and \quad Q(x)=4x$$
$$g(x) = e^{- \int P(x) dx}[\int Q(x)e^{\int P(x) dx} dx + C]$$
$$\frac {1}{x^2}[x^4+C] \implies g(x)=x^2 + \frac {C}{x^2}$$
QUESTION: am I allowed to call the "c/x^2" term just another constant?
back-sub $$y(x) = e^{g(x)} = e^{(x^2)}*e^{\frac {C}{x^2}}$$
Thanks for all of your help.
Last edited:
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2022-01-22 12:22:05
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http://mathhelpforum.com/differential-geometry/77791-surface-regular-if-curvature-non-zero.html
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# Thread: surface is regular if curvature is non zero
1. ## surface is regular if curvature is non zero
alpha(u) is a unit speed space curve, the chart (x,U) where
x(u,v) = alpha(u) + v*(alpha'(u)), (u,v) in U
i want to show that this surface is regular if the curvature k(u): = || alpha''(u)|| of the unit speed curve alpha(u) is non zero
2. Originally Posted by dopi
alpha(u) is a unit speed space curve, the chart (x,U) where
x(u,v) = alpha(u) + v*(alpha'(u)), (u,v) in U
i want to show that this surface is regular if the curvature k(u): = || alpha''(u)|| of the unit speed curve alpha(u) is non zero
Is it really alpha' (and not alpha'') in the equation? And could you please precise what U is? Thank you.
3. Originally Posted by Laurent
Is it really alpha' (and not alpha'') in the equation? And could you please precise what U is? Thank you.
alpha(u) is a unit speed space curve, and use this curve to construct a tangent developable surface with $chart (x,U)$where
$x(u,v) = alpha(u) + v*(alpha'(u)), (u,v) in U$
where $U = {(u,v) in R^2 : -infinity < u < infinity, v>0}$
i want to show that this surface is regular if the curvature k(u): = || alpha''(u)|| of the unit speed curve alpha(u) is non zero
so it is really alpha' in the equation. And i have re-written the question to specify what U is. Thank you.
the solution i got was
$Xu (cross-product)Xv = v*alpha''(u) (cross-product)t,$where i used $t$as $t = alpha(u)'$from the frenet serret equations.
i replaced $alpha''(u)$with the $k$ curvature, wasnt too sure how else to do it
and so as $v>o, k$ must be non zero for the chart to be regular.
4. Originally Posted by dopi
the solution i got was
$Xu (cross-product)Xv = v*alpha''(u) (cross-product)t,$where i used $t$as $t = alpha(u)'$from the frenet serret equations.
i replaced $alpha''(u)$with the $k$ curvature, wasnt too sure how else to do it
and so as $v>o, k$ must be non zero for the chart to be regular.
In the Serret-Frenet system, there are three vectors, $(T,N,B)$. We have $T=\alpha'(u)$, and $\alpha''(u)=\kappa N$. Therefore, $X_u\times X_v=(T+v\kappa N)\times T=v\kappa N\times T=v\kappa B$ and $\|X_u\times X_v\|=v\kappa\neq 0$.
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2018-01-19 11:30:32
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https://pandasthumb.org/archives/2012/01/bisgrove-postdo.html
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# Bisgrove Postdoctoral Scholars at ASU
Science Foundation Arizona is sponsoring a postdoctoral scholar program at Arizona State University that offers a really sweet deal:
Bisgrove Scholars will receive an annual stipend of $60,000, benefits and an additional$20,000 per year for research expenses. The Bisgrove appointment is renewable on a year-to-year basis for a maximum initial term of two years, contingent upon the availability of funds.
This program is only open to individuals who have no prior post-doctoral experience and obtain a PhD prior to appointment, i.e. doctoral students in their last year of studies. So while you may be now eating ramen every night in the lab while trying to finish up the last experiment you need to graduate, this time next year, you could be eating ramen sprinkled with gold dust and angel tears, while trying to finish up the last experiment you need for a grant proposal. (Did I mention that rent is cheap here?)
Another qualification is that your research has to fit with the Science Foundation Arizona’s mission: “Areas include, diagnosis and prevention of disease, sustainable energy and the environment, and information and communications technologies at the human interface.”
You will also need to specify possible mentors for your research project. (Hey, I’m available!)
For instructions on how to apply, see http://graduate.asu.edu/bisgrove.
Applications due Feb 15th.
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2017-04-29 13:31:22
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https://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&oldid=110413
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Homura said that she uses applied statistics in witch hunts
The cast of Mahou Shoujo Madoka Magica are students of Mitakihara Middle School. Even though it is merely a middle school, coursework in Mitakihara can be quite challenging. The following are Math problems that have appeared thus far in the show, and their solutions:
Episode 1
Episode one Math question 1
Question 1
Any integer divided by 14 will have a remainder between 0 and 13. Given that $a$ has a remainder of 6 and $b$ has a remainder of 1 when divided by 14, what is the remainder of $x$ when divided by 14, given $x$ is an integer solution to $x^2-2ax+b=0$?
Solution:
This problem can be solved simply with modular arithmetic:
• $a \equiv 6 \pmod{14}$;
• $b \equiv 1 \pmod{14}$.
• $\displaystyle{x^2 - 2ax + b = 0}$ implies
• $x^2 - 12x + 1 \equiv 0 \pmod{14}$
• $x^2 + 2x + 1 \equiv 0 \pmod{14}$ (because $-12 \equiv 2 \pmod{14}$)
• $(x + 1)^2 \equiv 0 \pmod{14}$
• $x + 1 \equiv 0 \pmod{14}$ (because 14 is square-free)
• $x \equiv 13 \pmod{14}$ (because $-1 \equiv 13 \pmod{14}$).
Second Solution:
Homura used in Episode 1 a basic approach with usual integer arithmetic.
Let $\displaystyle{x = 14q + r, a = 14s + 6, b = 14t + 1}$.
Substitute into $\displaystyle{x^2 - 2ax + b = 0}$ to get after some calculations
$\displaystyle{x^2 - 2ax + b = 14c + (r+1)^2 = 0} \mbox{ with } \displaystyle{c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r}$
14 divides $14c$ and 0, hence it also divides $(r+1)^2$. This implies that $r+1$ is divisible by 14.
The question asks for a remainder $r$ between 0 and 13, so we obtain $r = 13$.
Question 2
Episode one Math question 2
Assuming that $p$ is a prime number and $n$ is an arbitrary natural number, prove that $(1+n)^p - n^p - 1 \,$ is divisible by $p$.
Solution:
By Fermat's Little Theorem, for any prime $p$ and integer $a$,
$a^p \equiv a \pmod{p} \,$
Thus:
$(1+n)^p - n^p - 1 \pmod{p} \,$ is equivalent to
$(1+n) - n - 1 \pmod{p} \,$ is equivalent to
$0 \pmod{p} \,$
So the overall expression is divisible by $p$.
Second solution:
The problem can be solved with the binomial theorem:
For $a, b$ not equal to $0$ and nonnegative integer $p$ it holds that:
$\displaystyle (a + b)^p = \sum_{k=0}^{p} \binom{p}{k} a^{p-k} b^k$
where the binomial coefficient $\binom{p}{k}$ is the integer $\frac{p(p - 1) \cdots (p - k + 1)}{k(k-1) \cdots 1}$ for $0 \leq k \leq p$.
Therefore,
$\displaystyle (1 + n)^p = \sum_{k=0}^{p} \binom{p}{k} n^k$
and
$\displaystyle (1 + n)^p - n^p - 1 = \sum_{k = 1}^{p - 1} \binom{p}{k} n^k$
since $\binom{p}{0} = \binom{p}{p} = 1$.
It holds that $\binom{p}{1} = p$. Since $p$ is a prime number, the factor $p$ in $\binom{p}{k}$ is not divisible by $k, k-1,\ldots, 2$ for $2 \leq k \leq p-1$. Therefore, $p$ divides $\binom{p}{k}$ for $1 \leq k \leq p-1$.
Since each summand on the right side is divisible by $p$, the whole sum, i.e. the left side is also divisible by $p$.
Question 3
Episode one Math question 3
Find the integer solutions $(a,b)$ with $(a^3 + a^2 - 1) - (a - 1)b = 0 \,$.
Solution:
Episode one Math solution to question 3
\begin{align} (&a^3 + a^2 - 1) - (a - 1)b \\ & = (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b \\ & = (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b \\ & = (a - 1)(a^2 + 2a - b + 2) + 1 = 0 \end{align}
Therefore,
$(a - 1)(a^2 + 2a - b + 2) = -1 \,$
$a, b$ are integers, therefore the factors $a - 1$ and $a^2 + 2a - b + 2 \,$ are integers too. Since the product is $-1$, one of the factors must be equal to $-1$, the other to $1$.
If $a - 1 = -1$ then $a = 0$ and from $a^2 + 2a - b + 2 = 1 \,$ we obtain $b = 1$.
If $a - 1 = 1$ then $a = 2$ and $a^2 + 2a - b + 2 = -1 \,$ implies that $b = 11$.
There are two integer solutions: $(a,b) = (0,1) \,$ or $(a,b) = (2,11) \,$.
Second Solution
The question can also be solved by polynomial division.
$(a^3 + a^2 - 1) - (a - 1)b = 0 \,$
$b = \frac{a^3 + a^2 - 1}{a - 1} = a^2 + 2a + 2 + \frac{1}{a - 1} \,$
As $b \in \mathbb{Z}$, $a^2 + 2a + 2 + \frac{1}{a - 1} \in \mathbb{Z}$, so $\frac{1}{a - 1} \in \mathbb{Z}$
As $1$ has only two factors: $1$ and $-1$,
$a-1 = 1 \,$ or $a-1 = -1 \,$
$a = 2 \,$ or $a = 0 \,$
Substituting the above values into the equation to find the corresponding $b$, we have
$(a, b) = (0, 1) \,$ or $(2, 11)$.
Question 4
Episode one Math question 4
Given $\displaystyle F(x) = \frac{4x + \sqrt{4x^2 - 1}}{\sqrt{2x + 1} + \sqrt{2x-1}}$,
find the sum of $\displaystyle{F(1)+F(2)+F(3)+...+F(60)}$.
Solution:
Episode one Math, partial solution to question 4. Note that there are errors.
Simplying the fraction:
Notice that for any variables $a$ and $b$:
$\displaystyle{(a-b)(a+b) = a^2-b^2}$.
Let $\displaystyle{a = \sqrt{2x + 1}}$ and $\displaystyle{b = \sqrt{2x - 1}}$.
Multiply $\displaystyle{F(x)}$ by 1 or $\displaystyle{\frac{a-b}{a-b}}$ which is equal to $\displaystyle{\frac{\sqrt{2x + 1} - \sqrt{2x - 1}}{\sqrt{2x + 1} - \sqrt{2x - 1}}}$. The denominator becomes:
\begin{align} &(\sqrt{2x + 1} + \sqrt{2x - 1})(\sqrt{2x + 1} - \sqrt{2x - 1})\\ &= (2x + 1) - (2x - 1)\\ &= 2 \end{align}
The numerator becomes:
\begin{align} &(4x + \sqrt{4x^2 - 1} ) (\sqrt{2x + 1} - \sqrt{2x - 1})\\ &=(4x + \sqrt{(2x)^2 - 1)}) (a - b)\\ &=(4x + \sqrt{2x - 1} \sqrt{2x + 1} ) (a-b)\\ &=(4x + a b ) (a - b)\\ &=4x (a - b) + a^2 b - a b^2\\ &=4x (a - b) + (2x + 1) b - a (2x - 1)\\ &=4x (a - b) - 2x (a - b) + a + b\\ &=2x ( a - b) + a + b\\ &=(2x + 1) a - (2x - 1) b\\ &=a^3 - b^3 \end{align}
Therefore:
$\displaystyle {F(x) = \frac{a^3 - b^3 }{2}}$
Note that when $\displaystyle{ x = 1, 2, 3, 4, \ldots, 60}$;
$\displaystyle{ a^3 = 3^\frac{3}{2}, 5^\frac{3}{2}, 7^\frac{3}{2}, 9^\frac{3}{2}, \ldots, 119^\frac{3}{2}, 121^\frac{3}{2}}$;
$\displaystyle{ b^3 = 1^\frac{3}{2}, 3^\frac{3}{2}, 5^\frac{3}{2}, 7^\frac{3}{2}, 9^\frac{3}{2}, \ldots, 119^\frac{3}{2} }$;
Taking the sum over $\displaystyle{F(x)}$ for $\displaystyle{x}$ between $1$ and $60$, observe that the majority of the terms in $\displaystyle{a^3}$ and $\displaystyle{b^3}$ cancel out, leaving:
\begin{align} \sum_{x=1}^{60}F(x) &= \frac{121^\frac{3}{2} - 1^\frac{3}{2}}{2}\\ &=\frac{(11^2)^\frac{3}{2} - 1}{2}\\ &= 665 \end{align}
Thus, sum of $\displaystyle{F(x)}$ for $\displaystyle{x}$ between $1$ and $60$ is $665$. The solution can be generalized as equal to $\displaystyle{\frac{a(x_{\max}^3) - b(x_{\min}^3)}{2}}$. Where $\displaystyle{a}$ and $\displaystyle{b}$ are interpreted as functions of $\displaystyle{x}$.
Episode 8
Homura predicted the location of Walpurgis Night to be the clock tower using statistics. The material Homura presents to Kyoko shows what looks like a sinusoidal regression, where a time series data is assumed to be following a sine curve over time and the properties (amplitude, frequency, phase) of the sine wave is predicted through statistics. For the purpose of predicting the location of Walpurgis Night, statistics done this way would use geographic coordinates (longitude and latitude) as observed data points and attempt to fit two different sine waves to data, one for longitude and one for latitude. The fact that a large pendulum exists in Homura's room may be viewed as supporting this interpretation that what Homura used was a sinusoidal regression, because the mechanical movement of a pendulum also follows a sinusoidal pattern.
Irrespective of what type of statistics was used, the fact that Walpurgis Night actually appeared at the predicted location indicates that the model fitted the data very well with only a negligible amount of error.
Episode 9
Fibonacci sequence question that appeared in Episode 9
This advanced Algebra problem previously appeared in the first Tokyo University Entrance Exam:
Problem
A number sequence $\displaystyle{ \{F(n)\} }$ that can be defined as $\displaystyle{F(1) = 1, F(2) = 1}$, $\displaystyle{F(n+2) = F(n) + F(n+1)}$ (where $\displaystyle{n \in \mathbb{N}}$) is called the Fibonacci sequence and its general solution is given by,
$\displaystyle{F(n) = \frac{\varphi^{n} - (-1/\varphi)^{n}}{\sqrt{5}}}$
$\displaystyle{\varphi = \frac{1 + \sqrt{5}}{2} \approx 1.6180339887 \dots}$
Answer the following questions by using this fact if needed:
Define a sequence of natural numbers $\displaystyle{ \{X(n)\} }$ (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:
(i) $\displaystyle{X(1) = 1}$
(ii) We define $\displaystyle{X(n+1)}$ as a natural number, which can be obtained by replacing the digits of $\displaystyle{X(n)}$ with 1 if the digit is 0, and with 10 if the digit is 1.
For example, $\displaystyle{X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, \ldots}$
Anecdote: $\displaystyle{X(n)}$ can be considered a clever analogy to the show, where Episode 10 should replaces Episode 1 and Episode 1 replaces Episode 0 in the viewer's next viewing.
(1) Find $\displaystyle{A(n)}$, defined as the number of digits of $\displaystyle{X(n)}$.
(2) Find $\displaystyle{B(n)}$, defined as the numbers of times '01' appears in $\displaystyle{X(n)}$? For example, $\displaystyle{B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3, \ldots}$
Solution
Part 1
Let $\displaystyle{A(n)}$ equal to the number of digits in $X(n)$, which consists solely of 1s and 0s. Let's suppose $\displaystyle{x(n)}$ is the number of 0s in $\displaystyle{X(n)}$ at the n-th iteration (poor choice of variable by the student). Let's suppose $\displaystyle{y(n)}$ is the number of 1s in $\displaystyle{X(n)}$ in the n-th iteration, then $\displaystyle{x(n) + y(n) = A(n)}$. Since
• Every time a 0 appears, it is replaced with 1 at the next iteration, contributing to a single 1 in $\displaystyle{X(n+1)}$.
• Every time a 1 appears, it is replaced with 10 at the next iteration, contributing to a single 1 and a single 0 in $X(n+1)$.
it follows that the number of 0s in the next iteration is equal to the number of 1s previously:
$\displaystyle{x(n+1)= y(n)}$ ;
and the number of 1s in the next iteration is equal to the number of 0s AND the numbers of 1s previously:
$y(n+1)= \displaystyle{x(n)+y(n)}$ .
Next, prove that $x(n)$ is a Fibonacci sequence, since we know that:
• $\displaystyle{y(n+1) = x(n) + y(n)}$ ;
• $\displaystyle{x(n+2) = y(n+1)}$ ;
• $\displaystyle{x(n+1) = y(n)}$ ;
by substition we can show,
\begin{align} &x(n+2)\\ &= y(n+1)\\ &= x(n) + y(n)\\ &= x(n) + x(n+1) \end{align}
Hence, $\displaystyle{x(n+2) = x(n+1) + x(n)}$.
Thus $\displaystyle{x(n)}$ fits the definition of a Fibonacci sequence. Since $\displaystyle{x(n)}$ is a Fibonacci sequence, it follows that $\displaystyle{y(n)}$ is also a Fibonacci sequence (given that $\displaystyle{x(n+1) = y(n)}$). Therefore, since it has already been shown that:
• $\displaystyle{x(n+2)=x(n+1)+x(n)}$
• $\displaystyle{y(n+2)=y(n+1)+y(n)}$
It follows that $\displaystyle {x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)}$.
Recall that $\displaystyle{A(n)=x(n)+y(n)}$, therefore
$\displaystyle{A(n+2)=A(n+1)+A(n)}$ .
Since, $\displaystyle{X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, \ldots}$
It follows that $\displaystyle{A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8, \ldots}$
Recall by definition of the Fibonacci sequence: $\displaystyle{\{F(n)\} = \{1, 1, 2, 3, 5, 8, \ldots\}}$. Therefore $\displaystyle{A(n) = F(n+1)}$, or
$\displaystyle{A(n) = F(n+1) = \frac{\varphi^{n+1} - (-1/\varphi)^{n+1}}{\sqrt{5}}}$
Part 2
Let $B(n)$ be the number of times the sequence $01$ appears in $X(n)$. Any two digits in $X(n)$ may be $00$, $01$, $10$, $11$, and the corresponding digits in the next iteration $X(n+1)$ will be
• $00 \rightarrow 11$
• $01 \rightarrow 110$
• $10 \rightarrow 101$
• $11 \rightarrow 1010$
Thus, any two digit sequence in $X(n)$ that begins with 1* will contribute to a $01$ sequence in the next iteration. In other words, $B(n+1)$ equals $y(n)$, except when unit digit of $X(n)$ is 1, or:
$B(n+1) = y(n) - odd(X(n))$
where
• $odd(n) = 1$ if $n$ is odd (unit digit is 1)
• $odd(n) = 0$ if $n$ is even (unit digit is 0)
It can be seen that $odd(n) = odd(X(n))$ for all $n$. To prove this inductively: observe that $odd(X(1)) = odd(1)$. Now let $odd(X(n)) = odd(n)$, then if $n$ is even, $X(n)$ is even, thus $X(n)$ ends with 0, which gets mapped to 1, making $X(n+1)$ odd and so $odd(X(n+1)) = odd(n+1)$. And if $n$ is odd, then $X(n)$ is odd, thus $X(n)$ ends with 1, which gets mapped to 10, making $X(n+1)$ even, and so $odd(X(n+1)) = odd(n+1)$.
$y(n)$ is a fibonacci sequence with starting values: $y(1) = 1$, $y(2) = 1$, $y(3) = 2$, thus $y(n) = F(n)$
$B(n+1) = F(n) - odd(n)$
$B(n) = F(n-1) - odd(n-1)$, or
$\displaystyle{B(n) = F(n-1) - odd(X(n-1)) = \frac{\varphi^{n-1} - (-1/\varphi)^{n-1}}{\sqrt{5}} - odd(X(n-1))}$
Episode 10
Same as Episode 1.
Episode 11
Pieces of papers floats by Homura as she faces Walpurgis Night. On it are calculations done by hand by Homura. Most likely, these are ballistics of the big guns. Two pairs of Xs and Ys on each of the four pieces of paper, making it 8 shots in all.
Movie 3: Rebellion
Problem A
The first problem posed, (a), was to calculate $\int\frac{\arcsin^3x}{\sqrt{1-x^2}}\,dx$
We can perform the substitution $u=\arcsin x \longrightarrow \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}$ and so rewrite the integral as $\int u^3\,du$
This can be easily solved via basic integration formulas: $\int u^3\,du=\frac{1}{4}u^4+C$
Undoing the substitution we get the solution: $\frac{1}{4}\arcsin^4x + C$
Problem B
The second problem, (b), was to calculate $\int x\ln(x^2+y)\,dx$ In order to solve it we'll assume $\displaystyle y$ to be a parameter independent from the value of $\displaystyle x$ and $\displaystyle\ln x$ to denote the natural logarithm of $\displaystyle x$.
Performing the substitution $u=x^2+y\longrightarrow\frac{du}{dx}=2x$ we can rewrite the integral as $\frac{1}{2}\int\ln u\,du$
This can be solved integrating by parts: $\frac{1}{2}\int \ln u\,du=\frac{1}{2}(u\ln u - \int1\,du)=\frac{1}{2}u(\ln u - 1)$
And the final answer in obtained undoing the substitution: $\frac{1}{2}(x^2+y)\ln((x^2+y)-1)+C$
Problem C
One of the problems posed, (c), was to calculate $\int\frac{x^3+2x^2+10x}{x^2-x+1}\, dx$.
The answer is to first simplify the function: $\frac{x^3+2x^2+10x}{x^2-x+1} = x + 3 + \frac{12x - 3}{x^2-x+1} = x + 3 + 6\frac{2x - 1}{x^2-x+1} + \frac{3}{x^2-x+1}$. Note that $x^2-x+1$ has only simple complex zeros.
It holds $\int x + 3 \, dx = \frac{1}{2}x^2 + 3x + C$.
The general formula $\int\frac{f'(x)}{f(x)}\, dx = \ln|f(x)| dx + C$ can be used to calculate $\int \frac{2x - 1}{x^2-x+1} \, dx = \ln(x^2-x+1) + C$.
For the last term, $\int\frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan \frac{x}{a} + C$ can be used (with some additional calculations) to obtain the solution
$\int\frac{x^3+2x^2+10x}{x^2-x+1}\, dx = \frac{1}{2}x^2 + 3x + 6\ln(x^2-x+1) + 2\sqrt{3} \arctan \frac{2x-1}{\sqrt{3}} + C$
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2022-08-12 12:39:17
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https://imathworks.com/tex/tex-latex-two-images-inside-minipage-side-by-side-with-one-common-caption-that-says-figure-1-2/
|
# [Tex/LaTex] Two images inside minipage (side by side) with one common caption that says “Figure 1 & 2:”
captionsgraphicsminipage
I have two images inside a minipage, so I can have them side by side. When I make a caption for the images it only comes up as "figure 3: text", but I would like to have it showing "Figure 3 & 4: text".
I tried to \end{figure} and \begin{figure} in each minipage but the images would not show side by side anymore. The code I have at the moment is:
\begin{figure}[h]
\begin{minipage}[h]{0.3\textwidth}
\includegraphics[scale=0.58]{fig1}
\end{minipage} \hspace{0.2\textwidth}
\begin{minipage}[h]{0.3\textwidth}
\includegraphics[scale=0.58]{fig2}
\end{minipage}
\caption{text}
\end{figure}
This isn't quite what you've asked for, but how about the following? I'm using the subcaption package to put your two pictures into two 'subfigures'. The pictures can then be referenced individually (Figure 1a and Figure 1b), or as a pair (Figure 1).
## Code
\documentclass{article}
\usepackage{subcaption}
\usepackage[demo]{graphicx}
\begin{document}
\begin{figure}[h]
\begin{subfigure}{0.3\textwidth}
\includegraphics[scale=0.58]{fig1}
\caption{horse}
\label{fig:horse}
\end{subfigure} \hspace{0.2\textwidth}
\begin{subfigure}{0.3\textwidth}
\includegraphics[scale=0.58]{fig2}
\caption{zebra}
\label{fig:zebra}
\end{subfigure}
\caption{animals}
\label{fig:animals}
\end{figure}
Figure~\ref{fig:animals} shows some animals. Figure~\ref{fig:horse} shows a horse.
\end{document}
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2022-09-29 21:33:31
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http://tex.stackexchange.com/questions/86809/issues-after-verbatim-section
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# issues after verbatim section
I'm havin an issue which I cannot solve. If I have a subsubsection containing text with a verbatim section in between, the first line after the verbatim section is not aligned properly. If I put a \newline command after the verbatim, everything works, but it shows the error
There's no line here to end
This is the code:
\subsubsection{Umsetzung}
C: \texttt{if(r1 > r2)$\{$statement;$\}$} \newline
Realisierung in 2 Schritten: zuerst die Register vergleichen, dann auf Grund
von Flags bedingten Sprung ausführen.
\begin{verbatim}
cp r1, r2 ; vergleiche (compare) r1 mit r1
brge L1 ; verzweige falls r1 >= r2 (mit Vorzeichen!)
...
statement
...
L1: nop ; hier geht es weiter
\end{verbatim}
C: \texttt{if(r1 > r2)$\{$statement1;$\}$else$\{$statement2;$\}$} \newline
Assembler:
\begin{verbatim}
cp r1, r2 ; vergleiche (compare) r1 mit r
brge L1 ; verzweige falls r1 >= r2
...
statement1
...
rjmp L2
L1: ...
statement2
...
L2: ... ; hier geht es weiter
\end{verbatim}
And this is the output:
Even if I remove the whole verbatim section, the described line stays on a new line and not properly aligned (and this - in my oppinion - is far stranger, shouldn't it just float at the end of the previous sentence since I never tell to break a line?).
Note that I slightly formatted the content of the verbatim sections here for better legibility, so don't be suprised if the alignment of the Assembler-code isn't like in the picture in case you compile my code.
-
It isn't clear what error you see. C is starting a new paragraph and so gets parindent. Perhaps you want to set parindent to 0pt, or use \noindent ? Or just omit the blank line which is causing the new paragraph if you do not want a paragraph break at that point. – David Carlisle Dec 12 '12 at 23:27
Instead of the mixed modes \texttt and $...$, use \verb|...|. – Werner Dec 12 '12 at 23:29
• put \noindent before your new line, or
• put \hspace*{\fill}\newline
The way your MWE is written, you've told LaTeX that you're staring a new paragraph. If that's what you really want to do then the indentation is technically correct. If you don't want paragraphs indented or you don't want just this paragraph indented, then you can use \noindent. (Incidentally, LaTeX doesn't automatically indent the first paragraph after a section header.)
\newline doesn't like to end empty lines: hence the \hspace*{\fill} before it. I would recommend not using \newline because you're creating even more white space.
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2016-06-27 20:22:46
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https://academic.oup.com/cercor/article/24/8/2160/470752/Enriched-Encoding-Reward-Motivation-Organizes
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Abstract
Learning how to obtain rewards requires learning about their contexts and likely causes. How do long-term memory mechanisms balance the need to represent potential determinants of reward outcomes with the computational burden of an over-inclusive memory? One solution would be to enhance memory for salient events that occur during reward anticipation, because all such events are potential determinants of reward. We tested whether reward motivation enhances encoding of salient events like expectancy violations. During functional magnetic resonance imaging, participants performed a reaction-time task in which goal-irrelevant expectancy violations were encountered during states of high- or low-reward motivation. Motivation amplified hippocampal activation to and declarative memory for expectancy violations. Connectivity of the ventral tegmental area (VTA) with medial prefrontal, ventrolateral prefrontal, and visual cortices preceded and predicted this increase in hippocampal sensitivity. These findings elucidate a novel mechanism whereby reward motivation can enhance hippocampus-dependent memory: anticipatory VTA-cortical–hippocampal interactions. Further, the findings integrate literatures on dopaminergic neuromodulation of prefrontal function and hippocampus-dependent memory. We conclude that during reward motivation, VTA modulation induces distributed neural changes that amplify hippocampal signals and records of expectancy violations to improve predictions—a potentially unique contribution of the hippocampus to reward learning.
Introduction
Relationships between events are often clear only in retrospect. For example, when something good happens, we want to know why it happened and how to make it happen again. However, many events may precede a reward; how do we know which of them caused it? One way to solve this “credit assignment problem,” is by recording in memory all potential determinants of behaviorally relevant events like rewards (Fu and Anderson 2008). From an over-inclusive mnemonic record, the relationships between behaviorally relevant events can be disambiguated as experience accumulates, improving mnemonic models for predicting future rewards and knowing how to get them.
How long-term memory supports the resolution of credit assignment for rewards and similar computational problems must be reconciled with combinatorial explosion in representing potential relationships. These two opposing constraints could both be satisfied if the contents of memory were biased as follows: Because it indicates an expectation that reward is possible, reward motivation should result in enhanced memory encoding for salient events, even if those events are not explicitly associated with reward. Conversely, over-inclusive memory encoding should occur only during reward motivation, to limit computational burden. Because any salient event encountered during reward motivation is potentially predictive of reward outcomes, such a memory bias would offer a computationally feasible way to facilitate obtaining future rewards.
Expectancy violations are salient events because they are surprising; they signal environmental volatility and act as a cue to update mnemonic representations. Thus, expectancy violations encountered during pursuit of a goal should be more salient and better remembered than those encountered when no goal is active. For example, an individual might memorize the location of a new, soon-to-be opened restaurant better if she notices it when hungry than when full. This adaptive memory bias would ensure that when behaviorally relevant events do occur, mnemonic models of the environment include an exuberant set of possible predictive relationships to be refined by additional experience.
No research has yet investigated whether reward motivation affects the encoding of behaviorally salient events encountered during goal pursuit, but the hippocampal memory system is well positioned for such a function. First, the hippocampus is specialized to create inclusive records of the surrounding environment: the hippocampus generates detailed representations of multiple interconnected events (Davachi 2006; Ranganath 2010), unfolding in time (Devito and Eichenbaum 2011; Tubridy and Davachi 2011), with preferential representation of surprises (i.e., expectancy violations) (Ranganath and Rainer 2003). Second, the hippocampus is modulated by activation of the dopaminergic midbrain, a region strongly implicated in reward-motivated behaviors (Berridge and Robinson 1998; Wise 2004). Accumulating evidence suggests that engagement of the midbrain (including the ventral tegmental area [VTA]) biases hippocampal memory to support future adaptive behaviors (Shohamy and Adcock 2010; Lisman et al. 2011). However, although VTA activation and motivation have both been shown to influence memory encoding (Shohamy and Adcock 2010; Lisman et al. 2011), the mechanisms of these effects in behaving organisms remain to be elucidated.
Prior research suggests that reward motivation could influence hippocampal memory either directly or indirectly. Plausible direct mechanisms include dopaminergic modulation of dynamic hippocampal physiology (Lisman and Otmakhova 2001; Hammad and Wagner 2006; Swant et al. 2008) and stabilization of long-term potentiation (Wang and Morris 2010; Lisman et al. 2011). Alternatively, given the widespread influence of dopamine on distributed brain systems beyond the hippocampus, changes in these broader networks could alter hippocampal physiology indirectly. Indirect hippocampal modulation could result from changes in a variety of processes that contribute to memory encoding (Nieoullon and Coquerel 2003; Mehta and Riedel 2006; Watanabe 2007). For example, the VTA modulates prefrontal neurophysiology during working memory, and attentional or executive processes (Williams and Goldman-Rakic 1995; Durstewitz et al. 2000; Goldman-Rakic et al. 2000; Seamans and Yang 2004), and recent evidence from the rodent literature has demonstrated synchronized activity across the VTA, hippocampus, and prefrontal cortex (PFC) during a working memory paradigm (Fujisawa and Buzsaki 2011). Because working memory and executive processes support episodic memory encoding (Blumenfeld and Ranganath 2007; Jeneson and Squire 2012), these processes could be recruited during reward-motivated memory encoding to indirectly modulate the hippocampus. This possibility is supported by recent electrophysiological evidence linking frontal theta to reward-motivated memory (Gruber et al. 2013) but a link to hippocampal physiology has yet to be demonstrated.
The first goal of the current study was to test for the existence of a mnemonic bias for encoding expectancy violations during reward motivation. The second goal was to adjudicate between direct and indirect mechanistic accounts of this adaptive memory bias. To address these questions, we collected fMRI data while participants performed a rewarded reaction-time task with goal-irrelevant expectancy violations embedded in a series of goal-relevant stimuli (Fig. 1). Specifically, participants viewed serial repetitions of a trial-unique object, responding via button press when it changed from color to grayscale, under high- or low-reward motivation. On half of the trials, the series was interrupted by a temporally unpredictable, novel but highly similar object: a goal-irrelevant expectancy violation. Following scanning, participants' memory was tested for the novel objects that served as expectancy violations. This design allowed us to independently measure activations to reward cues and expectancy violation events, and thus, to identify a novel candidate mechanism of midbrain dopamine effects on hippocampal physiology and memory.
Figure 1.
Experimental task. In each trial, participants first viewed a reward cue that indicated whether they had the opportunity to earn a high ($2.00) or low reward ($0.10) for a speeded button press to a target image (a grayscale image). In control trials, the reward cue was followed by serial repetitions of a trial-unique, color object image. After 10 to 11 repetitions that image turned grayscale, to which participants were to make a speeded button press. In expectancy violation trials, serial representations of the trial-unique, color object image was interrupted by a novel, yet highly similar, object image at a temporally unpredictable time. Participants were instructed that expectancy violations had no bearing on earning rewards, and earnings were solely based on their button press to target images. Following the button press to the target, participants were presented an outcome screen that indicated how much money they earned on that trial and how much they had accumulated over the course of the experiment.
Figure 1.
Experimental task. In each trial, participants first viewed a reward cue that indicated whether they had the opportunity to earn a high ($2.00) or low reward ($0.10) for a speeded button press to a target image (a grayscale image). In control trials, the reward cue was followed by serial repetitions of a trial-unique, color object image. After 10 to 11 repetitions that image turned grayscale, to which participants were to make a speeded button press. In expectancy violation trials, serial representations of the trial-unique, color object image was interrupted by a novel, yet highly similar, object image at a temporally unpredictable time. Participants were instructed that expectancy violations had no bearing on earning rewards, and earnings were solely based on their button press to target images. Following the button press to the target, participants were presented an outcome screen that indicated how much money they earned on that trial and how much they had accumulated over the course of the experiment.
Materials and Methods
Twenty-seven healthy, right-handed participants were paid $40 to participate, plus any additional earned monetary bonuses (mean =$69.78; standard deviation = $19.06). All participants gave written informed consent for a protocol approved and monitored by the Duke University Institutional Review Board. One participant was excluded due to a computer malfunction. The current analysis included 26 participants (18 females, age range: 18–36 years; median age = 24.5 years). Experimental Task To test for the effects of reward motivation on expectancy violation processing, participants perform a speeded reaction-time task, modified from the monetary incentive delay task (Knutson et al. 2000). This task was designed to manipulate 2 factors: participants' motivational state and the presence of expectancy violations. To manipulate motivational state, every trial of the task began with a 500-ms cue that indicated whether participants could earn either a high ($2.00) or low (0.10) reward for a speeded button press to a target image (Figure 1). Following a variable delay (between 5.5 and 6.5 s), the target appeared on the screen. Targets were trial-unique, grayscale object images. If participants were sufficiently fast at responding to targets, they earned the monetary bonus indicated by the cue. The target reaction time for earning money was determined by an adaptive algorithm, which estimated the response time threshold at which the subject would be successful on ∼65% of trials. Thresholds were calculated independently for each condition to ensure that reinforcement rates would be equated across all four conditions. Following the presentation of the target image, participants viewed an outcome screen that indicated their success on the current trial as well as their accumulated monetary bonuses. The intertrial interval between reward feedback and the following reward cues was 1–15.8 s (mean = 4.84). To manipulate expectancy violation, following the cue but prior to the target presentation, participants viewed 10–11 serial presentations of color object images for 409 ms with an interstimulus interval (ISI) of 136 ms. During control trials (CO), participants viewed repeated presentations of a color version of the upcoming target stimulus. During expectancy violation trials (EV), participants viewed repeated presentations of a color version of the upcoming target stimulus interrupted by a highly similar, but novel, image. This expectancy violation stimulus always appeared randomly between the fourth and eighth object presentation. Participants performed 2 runs of this task. During each task run, participants completed 10 High-Reward CO, 10 Low-Reward CO, 10 High-Reward EV, and 10 Low-Reward EV trials. Trial order was pseudo-randomized across each run, with each run lasting 7 min and 56 s. Trial onsets, cue-scene intervals, and trial order were optimized using Opt-seq software (Dale 1999). Immediately prior to scanning, participants were shown a visual schematic of the task and given verbal instructions. Participants were instructed about the incentives for detecting the grayscale object. They were also instructed that on some trials a different object would interrupt the stream of repeating objects, and that these interruptions were irrelevant to earning money. Participants were informed that they would receive all money earned during the performance of this task. Participants performed an unpaid, practice version of the task that consisted of 10 High-Reward CO and 10 Low-Reward CO to familiarize themselves with the paradigm and to calibrate reaction-time thresholding. Following scanning (∼30 min after the encoding session), participants performed a 2-alternative forced-choice recognition memory task for objects that constituted expectancy violations. During this test, participants saw pairs of object images, one of which was an object that constituted an expectancy violation and the other a similar, yet novel, object that participants had never seen. For each object pair, participants had to identify which object they saw during the encoding session by pressing either the “1” or “2” button to indicate the object on the left or right, respectively. Following each memory decision, participants had to indicate their confidence in their response (i.e., 1 = very sure, 2 = pretty sure, 3 = just guessing). Participants received 40 recognition memory trials (20 high reward and 20 low reward). MRI Data Acquisition and Preprocessing fMRI data were acquired on a 3.0-T GE Signa MRI scanner using a standard echo-planar imaging sequence (echo-planar imaging [EPI], TE = 27 ms, flip = 77 degrees, TR = 2 s, 34 contiguous slices, size = 3.75 mm × 3.75 mm × 3.80 mm) with coverage across the whole brain. Each of the 2 functional runs consisted of 238 volumes. Prior to the functional runs, we collected a whole-brain, inversion recovery, spoiled gradient high-resolution anatomical image (voxel size = 1 mm, isotropic) for use in spatial normalization. fMRI preprocessing was performed using fMRI Expert Analysis Tool (FEAT) Version 5.92 as implemented in FSL 4.1.5 9 (www.fmrib.ox.ac.uk/fsl). The first 6 scans were discarded to allow for signal saturation. Bold images were then skull stripped using the Brain Extraction Tool (Smith 2002). Images were then realigned within-run, intensity normalized by a single multiplicative factor, spatially smoothed with a 4.0-mm full width half maximum (FWHM) kernel, and subjected to a high-pass filter (100 s). Spatial normalization was performed using a 2-step procedure on fMRIb Linear Registration Tool (Jenkinson et al. 2002). First, mean EPIs from each run were co-registered to the high-resolution anatomical image. Then, the high-resolution anatomical image was normalized to the high-resolution standard space image in Montreal Neurological Institute (MNI) space using a nonlinear transformation with a 10-mm warp resolution, as implemented by fMRI NonLinear Registration Tool. All coordinates are reported in MNI space. Behavioral Analysis Reaction times and hit rates to target images were submitted to separate repeated measures ANOVAs with motivation (high vs. low trials) and expectancy violation (EV vs. CO) as within-subjects factors. Of note, the adaptive nature of our reaction-time algorithm was explicitly programmed to keep reinforcement rates equivalent across conditions. For both of these ANOVAs, we tested for main effects of motivation and expectancy violation as well as their interaction at a significance level of P < 0.05. Recognition memory for objects that constituted expectancy violations was tested by submitting the number of hits to Student’s t-test with a significance level of P < 0.05 with motivation (high vs. low) as a within-subject factor. All memory responses were included in this analysis, irrespective of participants' reported confidence. fMRI Data Analysis fMRI data were analyzed using FEAT Version 5.92 as implemented in FSL 4.1.5. Time-series statistical analyses used FILM with local autocorrelation correction (Woolrich et al. 2001). General Linear Model: Task-Related Activations To investigate task-related activations, first-level (i.e., within-run) general linear models (GLMs) included 8 regressors that modeled high-reward cues, low-reward cues, high-reward target images, low-reward target images, high-reward EV events, high-reward CO events, low reward EV events, and low reward CO events. The latency to CO events was determined by randomly sampling from the latency of EV events without replacement. All trial events were modeled with an event duration of 0 s and a standard amplitude of 1. These events were then convolved with a double-gamma hemodynamic response function. EV and CO events were orthogonalized with respect to cue and target events. Using this GLM, individual maps of parameter estimates were generated for 4 contrasts of interest: high-reward cue > low reward cue, [high-reward EV + low-reward EV] > [high-reward CO + low-reward CO], high-reward EV > high-reward CO, and low-reward EV > low-reward CO. Second-level analyses for each of these contrasts (i.e., across runs, but within-subject) were modeled using a fixed effects analysis. General Linear Model: Cue-Expectancy Violation Interactions To investigate interactions between cue-related VTA activation and EV signaling, first-level parametric GLMs were constructed which investigated the parametric modulation of EV and CO events by cue-evoked VTA activation. First, a single-trial, beta-series analysis was performed to extract single-trial parameter estimates for cue-evoked VTA activation (for a similar approach, see Rissman et al. (2004)). GLMs were constructed for each participant that separately modeled cue-evoked activations for each individual trial (40 high-cue and 40 low-cue regressors). EV/CO events and target images were modeled as described above. Then, a weighted average of cue-evoked β-values was extracted for each individual trial from a probabilistic VTA region-of-interest (ROI). This probabilistic ROI was generated by 1) independently hand-drawing the structure on 50 high-resolution anatomical images from an independent sample of 50 participants, 2) normalizing those images to standard space, and 3) averaging across those images (Shermohammed et al. 2012). Details for the anatomical landmarks used to define the VTA in individual participants can be found in Ballard et al. (2011). We chose to focus on the VTA over the SN in this paradigm because although there may be more dopaminergic neurons in the SN, the VTA has been demonstrated to be more critical in guiding reward-motivated behavior. In rodents, dopaminergic activity in the VTA has been demonstrated to track reward-motivated behaviors and learning (Aragona et al. 2008; Stuber et al. 2008), and selective stimulation of the VTA has been demonstrated to elicit reward-motivated behavior (Tsai et al. 2009; Adamantidis et al. 2011). In humans, high-resolution functional imaging has localized reward-motivated learning to the VTA and not the SN (D'Ardenne et al. 2008). Selective lesions to the human SN, as a result of Parkinson's disease, leave many reward-motivated behaviors in tact (Dagher and Robbins 2009). Together these findings suggest that the VTA may be the more relevant dopaminergic nucleus for reward-motivated behaviors. Additionally, work from rodents, primates, and humans suggest that both anatomic and functional connectivity between the VTA and hippocampus contributes to declarative memory processes (Haber 2003; Adcock et al. 2006; Shohamy and Wagner, 2008; Shohamy and Adcock, 2010; Wang and Morris, 2010; McGinty et al. 2011; Kahn and Shohamy, 2012; Wolosin et al. 2012). Separate parametric GLMs were constructed which included 8 standard task-related regressors (described above) and 2 additional parametric regressors modeling VTA modulations of EV and CO events. For the parametric regressors, each EV/CO event amplitude was weighted by the preceding cue-evoked VTA activation from the same trial. Thus, these parametric regressors reflected how cue-related VTA activations affected subsequent processing of EV/CO events on a trial-by-trial basis. Using this GLM, individual maps of parameter estimates were generated for 2 contrasts of interest: VTA-modulation of EV events > baseline, VTA-modulation of CO events > baseline. Second-level analyses for each of these contrasts (i.e., across runs, but within-subject) were modeled using a fixed effects analysis. Psychophysiological Interaction (PPI): Cue-Related Functional Connectivity To investigate cue-evoked VTA connectivity, we identified PPI of cues upon functional connectivity with the VTA. For this analysis, first-level GLMs were constructed which included 8 standard task-related regressors (described above), 1 physiological regressor, and 2 PPI regressors. The physiological regressor was the weighted mean time-series extracted from an in-house probabilistic VTA ROI (described above). The PPI regressors multiplied this VTA physiological regressor separately with 1) the task-related regressor for high-reward cues and 2) the task-related regressor for low reward cues. Thus, these PPI regressors modeled VTA coupling in the brain as a function of response to reward cues. Using this GLM, individual parameter estimate maps were generated for the contrast of interest: [high-reward PPI + low-reward VTA PPI] > baseline. Second-level analyses for each of these contrasts (i.e., across runs, but within-subject) were modeled using a fixed effects analysis. Group-Level Analysis Third-level analyses (i.e., across participants) were modeled using FSL's mixed-effects analyses (FLAME 1), which accounts for within-session/subject variance calculated at the first and second levels, on the parameter estimates for contrasts of interest derived from the second-level analysis. Statistical tests for fMRI analyses were set to an overall = 0.05 family-wise error rate as calculated within AlphaSim tool in AFNI (http://afni.nimh.nih.gov/afni/doc/manual/AlphaSim) which uses actual data structure to determine the number of independent statistical tests and thus balance Type 1 and 2 errors. With 1000 Monte Carlo simulations and a voxelwise significance of P < 0.001, a smoothing kernel of 4-mm FWHM, an overall 0.05 corresponded to a cluster extent minimum of 33 voxels for the whole brain and 15 voxels for the MTL ROI analyses (MTL ROIs were limited to bilateral hippocampus and parahippocampal gyrus volumes as defined by the WFU PICKATLAS [Maldjian et al. 2003]). Results Behavior: Speeded Reaction-Time Task Behavior To manipulate their motivational states, participants viewed cues that indicated whether a speeded button press to a target image would result in a high (2.00) or low ($0.10) monetary reward (Figure 1). Both reward motivation [high- vs. low-reward cues: F1,25 = 12.43, P = 0.002] and the presence of expectancy violations [EV vs. CO: F1,25 = 9.65, P = 0.005] decreased participants' reaction times to target images, without any significant interactions across these factors [F1,25 = 1.27, P = 0.27] (Table 1). These findings suggest that reward cues were successful in manipulating participants' motivational state, and that expectancy violations were sufficiently salient to influence later behavior. Despite significant differences in reaction times, target hit rate/positive feedback did not differ across conditions due to the adaptive algorithm, we implemented (Table 1) [reward motivation: F1,25 = 2.95, P = 0.10; expectancy violation: F1,25 = 0.38, P = 0.55; interaction: F1,25 = 0.31, P = 0.58]. Participants on average earned$69.78 (SD = 19.06).
Table 1
Behavioral performance on the rewarded speeded reaction-time task
Condition Reaction time ± SE Accuracy ± SE
High reward
Expectancy violation 205.01 ± 4.63 73.3 ± 1.4
Control 219.02 ± 5.41 71.7 ± 1.5
Low reward
Expectancy violation 214.62 ± 4.68 70.2 ± 1.8
Control 223.78 ± 3.79 70.2 ± 1.6
Condition Reaction time ± SE Accuracy ± SE
High reward
Expectancy violation 205.01 ± 4.63 73.3 ± 1.4
Control 219.02 ± 5.41 71.7 ± 1.5
Low reward
Expectancy violation 214.62 ± 4.68 70.2 ± 1.8
Control 223.78 ± 3.79 70.2 ± 1.6
fMRI: Main Effect of Reward Motivation (High-Reward Cue > Low-Reward Cue)
To identify brain regions modulated by reward motivation independent of the presence of expectancy violations, we compared brain activations in response to high- versus low-reward cues. Reward motivation (high > low) resulted in greater activation in a broad network of regions including key regions in the mesolimbic and mesocortical dopamine networks: Significant activations were observed in bilateral dorsolateral PFC, ventrolateral PFC, premotor cortex, motor cortex, medial frontal cortex, dorsal anterior cingulate cortex, superior parietal cortex, inferior parietal cortex, ventral visual stream, hippocampus, parahippocampal cortex, thalamus, dorsal striatum, ventral striatum, cerebellum, and the midbrain (encompassing the VTA/SN) as well as the right anterior temporal lobe (P < 0.05, whole-brain corrected; Supplementary Table 1). Activations significant at the whole-brain level included robust activation within our a priori ROI in the VTA (VTA: left: t(25) = 3.30, P = 0.003, right: t(25) = 4.66, P < 0.001).
fMRI: Main Effect of Expectancy Violation (Expectancy Violation > Control)
To identify brain regions modulated by the presence of expectancy violations independent of reward motivation, we compared brain activations in response to expectancy violation (EV) versus control (CO) events. The presence of expectancy violations resulted in greater activation throughout a fronto-parietal network as well as regions in the medial temporal lobe. Specifically, significant activations were seen in dorsolateral PFC, premotor cortex, superior parietal cortex, inferior parietal cortex, ventral visual stream, ventral striatum, and cerebellum, as well as, right anterior PFC and hippocampus (P < 0.05, whole-brain corrected; Supplementary Table 1).
fMRI: Motivation's Influence on Expectancy Violations (High Reward [EV − CO] − Low Reward [EV − CO])
To characterize the influence of reward motivation on expectancy violation processing, we compared the processing of expectancy violation events (EV > CO) encountered in the context of high- versus low-reward motivation. This analysis yielded only one significant cluster, in the left hippocampus, indicating that this region was more sensitive to expectancy violations in the context of high- compared with low-reward motivation (x = −22 y = −22 z = −6, cluster size = 16, P < 0.05, corrected for comparisons within the entire bilateral MTL, 5278 voxels, 42 224 mm3) (Fig. 2).
Figure 2.
Reward motivation influences hippocampus expectancy violation processing. (A) A cluster in the left hippocampus showed significantly greater sensitivity to expectancy violations under states of high-versus low-reward motivation (P < 0.05, corrected for comparisons within the entire bilateral medial temporal lobe, 5278 voxels, 42 224 mm3). (B) Mean parameter estimates from the medial temporal lobe cluster are plotted to illustrate the direction of the interaction between reward motivation and expectancy violation processing.
Figure 2.
Reward motivation influences hippocampus expectancy violation processing. (A) A cluster in the left hippocampus showed significantly greater sensitivity to expectancy violations under states of high-versus low-reward motivation (P < 0.05, corrected for comparisons within the entire bilateral medial temporal lobe, 5278 voxels, 42 224 mm3). (B) Mean parameter estimates from the medial temporal lobe cluster are plotted to illustrate the direction of the interaction between reward motivation and expectancy violation processing.
VTA Influence on Hippocampus Expectancy Violation Processing
To determine if there was a relationship between cue-related VTA activation and hippocampal sensitivity to expectancy violations, we used trial-by-trial variability in cue-evoked VTA activation to predict EV activations in the hippocampus. We found that, on a trial-by-trial basis, VTA activation during the cue significantly predicted hippocampus activation during EV (relative to CO events; cluster identified at motivation by expectancy violation interaction during EV); (P = 0.04) (Fig. 3).
Figure 3.
Cue-evoked ventral tegmental area activations predict hippocampus expectancy violation processing. On a trial-by-trial basis, VTA activations in response to reward cues predicted left hippocampus activations in response to expectancy violation events (EV > CO events); ROI, region of interest.
Figure 3.
Cue-evoked ventral tegmental area activations predict hippocampus expectancy violation processing. On a trial-by-trial basis, VTA activations in response to reward cues predicted left hippocampus activations in response to expectancy violation events (EV > CO events); ROI, region of interest.
To elucidate potential mechanisms of the relationship between cue-related VTA activation and reward effects on hippocampal EV processing, we performed a two-step procedure. First, we characterized within-subject, cue-evoked functional coupling with the VTA across the entire brain (Supplementary Figure). Then, we identified the subset of regions whose cue-evoked functional coupling with the VTA predicted the mean motivational influence on hippocampal EV activations for the session (High Reward [EV − CO] − Low Reward [EV − CO]). Compared with the broad network of regions that showed cue-evoked functional coupling with the VTA, only an isolated set of regions predicted reward-motivated, hippocampus EV sensitivity (Table 2, Fig. 4). Hippocampus EV sensitivity was predicted by VTA coupling with bilateral visual cortex, medial PFC, medial frontal cortex, subgenual cingulate cortex, as well as right premotor cortex, left ventrolateral PFC, left ventromedial PFC, and right temporal gyrus.
Table 2
Regions in which functional coupling with the VTA predicts reward-motivated, hippocampus expectancy violation signaling
Region Cluster size Z-stat x y z Brodmann's area
Visual cortex 249 4.42 −86 42 19
193 5.63 12 −96 18 19
Medial prefrontal cortex 162 4.35 16 64 10
101 4.15 −12 52 28
45 4.11 52 28
63 4.17 −26 34 28
Premotor/motor cortex BA (4/6) 43 4.24 54 −12 54 4/6
Ventrolateral PFC (BA 45/44) 57 4.18 −50 20 45/44
Middle temporal gyrus (BA 22) 50 4.14 −60 −42 22
Medial frontal cortex (BA 6) 58 3.92 −20 78
Subgenual cingulate cortex (BA 24) 84 3.75 −2 −12 24
Ventral PFC (BA 47) 159 3.51 22 16 −16 47
Region Cluster size Z-stat x y z Brodmann's area
Visual cortex 249 4.42 −86 42 19
193 5.63 12 −96 18 19
Medial prefrontal cortex 162 4.35 16 64 10
101 4.15 −12 52 28
45 4.11 52 28
63 4.17 −26 34 28
Premotor/motor cortex BA (4/6) 43 4.24 54 −12 54 4/6
Ventrolateral PFC (BA 45/44) 57 4.18 −50 20 45/44
Middle temporal gyrus (BA 22) 50 4.14 −60 −42 22
Medial frontal cortex (BA 6) 58 3.92 −20 78
Subgenual cingulate cortex (BA 24) 84 3.75 −2 −12 24
Ventral PFC (BA 47) 159 3.51 22 16 −16 47
Figure 4.
VTA-Cortical coupling predicts reward-motivated enhancements in expectancy violation processing. To address relationships between cue-evoked VTA activity and expectancy violation processing in the left hippocampus, we performed the following analysis. (A) First, we characterized functional coupling of the VTA across the whole brain. (B) Then, using linear regression, we identified regions whose functional coupling with the VTA predicted reward-motivated enhancements in hippocampus expectancy violation processing. (C) This analysis identified a network of cortical regions, including the medial PFC, ventrolateral PFC, and visual cortices (P < 0.05, whole-brain corrected; see also Table 2).
Figure 4.
VTA-Cortical coupling predicts reward-motivated enhancements in expectancy violation processing. To address relationships between cue-evoked VTA activity and expectancy violation processing in the left hippocampus, we performed the following analysis. (A) First, we characterized functional coupling of the VTA across the whole brain. (B) Then, using linear regression, we identified regions whose functional coupling with the VTA predicted reward-motivated enhancements in hippocampus expectancy violation processing. (C) This analysis identified a network of cortical regions, including the medial PFC, ventrolateral PFC, and visual cortices (P < 0.05, whole-brain corrected; see also Table 2).
Behavior: Declarative Memory for Expectancy Violations
Following scanning, participants performed a recognition memory task for the goal-irrelevant objects that constituted expectancy violations. For objects seen in the low-motivation condition, memory was no better than chance. In contrast, for expectancy violation objects seen during high-reward motivation, memory was significantly greater than chance and significantly greater than for objects that followed low reward cues (low: mean ± SE = 50.2 ± 1.9%; high: mean ± SE = 61.2 ± 1.6%; t(25) = 4.89, P < 0.001).
Discussion
The current study demonstrates that motivational drive to obtain reward modulates a distributed cortical network to amplify hippocampal signals and records of unexpected events. Neuroimaging results indicated that reward motivation increased hippocampal activations following expectancy violations. VTA activation following reward cues dynamically predicted this increased hippocampal sensitivity, and a network of cortical regions linked VTA activation following cues to hippocampal activation following expectancy violations. Specifically, we found that functional coupling of the VTA with medial prefrontal, ventrolateral prefrontal, and visual cortices predicted reward-motivated enhancements in hippocampal sensitivity to unexpected events. Finally, we found that unexpected events were remembered if they were encountered during high- but not low-reward motivation. Together these findings suggest that during pursuit of reward, the VTA engages a network of cortical intermediaries that facilitates hippocampus-dependent encoding of salient events.
Consistent with an adaptive bias for enriched memory encoding during reward motivation, we found that an individual's motivational state was a significant determinant of memory for goal irrelevant, unexpected events: a surprise memory test revealed declarative memory for objects that violated expectancy only if they were seen during high-reward motivation. Previous studies have demonstrated that reward motivation enhances memory for events that are explicitly goal relevant, such as incentivized information (Adcock et al. 2006; Murayama and Kuhbandner 2011; Wolosin et al. 2012) and reward-predicting cues (Wittmann et al. 2005; Bialleck et al. 2011; Wittmann et al. 2011). Related research has also demonstrated that prior novel events enhance striatal responses to reward cues (Guitart-Masip et al. 2010; Bunzeck et al. 2011), consistent with striatal convergence of novelty and reward signals (Lisman and Grace 2005). Here, from the vantage point of a mechanistically distinct hypothesis, we offer a novel demonstration of a complementary relationship: an effect of reward anticipation on processing of salience by the hippocampus. Notably, we demonstrate that reward motivation can enhance memory for surprising information even though it is explicitly not associated with reward: in the experiment, participants were instructed that the salient events were irrelevant to earning rewards. Nevertheless, here, as in real life, the true relationship of salient events to rewards is ambiguous. The enriched encoding of salient events seen in the current study during reward motivation, but not during unmotivated behavior, supports our broad hypothesis that encoding of the environment is over-inclusive selectively during reward motivation to permit the disambiguation of the causes of reward outcomes (Fu and Anderson 2008). However, given that here no relationship existed between expectancy violation processing and future reward outcomes, this broader hypothesis could not be explicitly tested. Further studies will be needed to establish that memory enhancements for expectancy violations can contribute to disambiguating the causal determinants of rewards to solve the “credit assignment” problem.
fMRI analyses revealed that reward motivation increased responses to expectancy violations selectively in the hippocampus; indeed, the hippocampus was the only region to show this pattern. Lesion studies in rodents, nonhuman primates, and humans have demonstrated that the hippocampus supports declarative memory for expectancy violation events (Ranganath and Rainer 2003; Kishiyama et al. 2004; Axmacher et al. 2010). The striking selectivity of this result argues that the memory enhancements we observed were not due to general increases in arousal or attention during reward motivation. Interestingly, reward motivation has been demonstrated to enhance hippocampus activation prior to and during the encoding of incentivized information (Adcock et al. 2006; Wolosin et al. 2012) as well as during presentation of cues that predict reward (Wittmann et al. 2005; Bunzeck et al. 2011). The current finding that reward motivation enhances hippocampal responsivity not only to incentivized information but also to goal irrelevant, unexpected events further implies an adaptive memory bias for enriched representation of contexts where reward is expected or pursued. In the current study, we were not able to demonstrate a direct relationship between motivation's influence on hippocampal sensitivity and declarative memory (i.e., a subsequent memory analysis). Specifically, these types of analyses could not be performed because, 1) when sorted by later memory performance, there were relatively low number of trials in some conditions (i.e., <10), and 2) there was very low variability in memory performance across participants. Thus, it will be important for future studies to investigate the direct relationship of hippocampal expectancy violation processing on later declarative memory during reward motivation.
We also identified a candidate mechanism of this enhancement: VTA activation in response to reward cues dynamically predicted hippocampal activation in response to expectancy violations. A relationship between VTA activation and hippocampal-dependent encoding has previously been described (Adcock et al. 2006; Wolosin et al. 2012); however, these findings were consistent with at least two underlying mechanisms: First, the VTA could directly modulate the hippocampus, consistent with the prominent theory that dopamine release in the hippocampus stabilizes long-term potentiation during consolidation (reviewed by Shohamy and Adcock 2010; Lisman et al. 2011). Alternatively, because dopamine has widespread actions in the brain, the VTA could also indirectly modulate hippocampal neurophysiology via the coordinated engagement of distributed neural intermediaries. To delineate the contributions of these 2 mechanisms, we conducted an analysis that first identified neural regions that showed cue-evoked functional coupling with the VTA, and then determined if this coupling predicted expectancy violation processing that occurred later in the trial. This analysis demonstrated that functional coupling of the VTA with medial prefrontal, ventrolateral prefrontal, and visual cortices (but not hippocampus itself) predicted reward-motivated enhancements in hippocampal expectancy violation processing. Hence, our findings support the interpretation that the VTA engages a network of cortical intermediaries to influence hippocampal encoding during reward motivation. A similar mechanism of VTA facilitation of prefrontal-hippocampus connectivity has recently been demonstrated during a spatial working memory paradigm in rodents (Fujisawa and Buzsaki 2011). By relating hippocampal physiological changes to VTA-prefrontal interactions, our findings offer a mechanistic integration of the literatures describing dopaminergic effects on prefrontal physiology (Williams and Goldman-Rakic 1995; Durstewitz et al. 2000; Goldman-Rakic et al. 2000; Seamans and Yang 2004) and on long-term memory encoding supported by the hippocampus (Shohamy and Adcock 2010; Wang and Morris 2010; Lisman et al. 2011).
Only a subset of the regions that correlated with VTA activation during motivation also predicted hippocampal sensitivity. These candidate intermediaries between VTA signatures of motivational state and hippocampal responses to expectancy violation have each been implicated by prior literature in reward motivation and hippocampus-dependent memory encoding. The medial PFC is strongly implicated in reward valuation processes (Rangel and Hare 2010) and the generation of affective meaning (Roy et al. 2012). This region has also been associated with memory encoding for self-relevant memoranda (Leshikar and Duarte 2012), and coordinated activity between the medial PFC and hippocampus has been associated with better reward-related learning in rodents (Benchenane et al. 2010). Given these literatures, the medial PFC is a likely source of signals that behaviorally relevant information is expected, and thus promote encoding of that information. The VLPFC is thought to be critical for selecting and prioritizing goal-relevant information for storage in long-term memory (Blumenfeld and Ranganath 2007). Thus, engagement of this region prior to memory encoding would be expected to enhance memory for events potentially relevant to future goal pursuit. Finally, visual cortex activation and visual stimulus processing are increased during states of reward motivation (Shuler and Bear 2006; Serences 2008; Seitz et al. 2009; Pessoa and Engelmann 2010; Baldassi and Simoncini 2011). During episodic memory encoding, visual cortex activations have been demonstrated to reliably predict successful memory encoding (Spaniol et al. 2009; Kim 2011), and priming of the visual cortices for increased detection of salient visual information, would be expected to contribute to hippocampal sensitivity. The current findings could also be conceptualized as reflecting transient shifts in attention to guide greater sensitivity to expectancy violation processing; in our view attentional mechanisms are not mutually exclusive with the memory mechanisms proposed above. Interestingly, the regions described in this network interaction analyses (i.e., medial PFC, lateral PFC, and visual cortex) have all been implicated during enhancements in goal-relevant attentional processes (Corbetta and Shulman 2002; Baluch and Itti 2011). In sum, the current neuroimaging findings suggest that a select network of cortical regions are engaged by the VTA which then contribute to enhanced sensitivity and better hippocampus-dependent encoding.
Our analysis did not provide evidence that direct modulation of the hippocampus by the VTA predicted expectancy violation encoding. Although there was significant functional coupling between the VTA and the hippocampus in response to reward cues, this relationship did not itself predict reward-motivated enhancements in hippocampal expectancy violation processing. This pattern stands in contrast to previous studies that have demonstrated functional connectivity between the VTA and the hippocampus, prior to encoding, that predicts reward-motivated declarative memory (Adcock et al. 2006) and individual differences in reward-motivated memory enhancements (Wolosin et al. 2012). Interestingly, both prior studies investigated the encoding of goal-relevant material (i.e., participants were explicitly rewarded for the successful encoding of memoranda), whereas we investigated the encoding of information not explicitly associated with reward but encountered during reward motivation. Together, this evidence suggests multiple possible routes whereby the VTA can influence hippocampal neurophysiology. We speculate that these different mechanisms arise because the targets of VTA neuromodulation are flexible and reflect an individual's specific behavioral goals. Thus, during states of reward motivation, there may be an integrated modulation of distributed regions involved in valuation, representation, and sensory processing; whereas when specific cognitive processes are incentivized (e.g., memory encoding), these actions may be adjusted so that specific target regions are also directly modulated (e.g., the hippocampus) (Adcock et al. 2006). Future studies will need to confirm this dissociation, and further investigate how these different mechanisms of VTA neuromodulation affect the content and structure of declarative memory.
In summary, our findings characterize a novel mechanism whereby reward motivation can enhance hippocampal sensitivity to—and declarative memory for—salient events. First, we found that motivation to obtain reward enhanced memory to include expectancy violations, even though they were goal irrelevant. This finding suggests a broader role for motivation in shaping memory encoding than previously appreciated, consistent with a bias for enriched encoding of contexts and potential determinants of reward experienced during reward motivation. Second, we identified a novel modulation of hippocampal neurophysiology, such that VTA-cortical interactions were predictive of hippocampal responses to expectancy violations; this finding suggests a modification of the prevailing view that dopamine shapes memory mainly by stabilizing lasting plasticity in the hippocampus after encoding (Lisman et al. 2011). This novel mechanism embodies a unified view of dopaminergic influence on memory formation, by illustrating how the previously described effects of dopamine in the PFC and hippocampus may integrate to enhance memory encoding. We propose that during reward motivation the VTA organizes distributed brain systems to create an over-inclusive, enriched mnemonic record—an adaptive memory bias—so that if and when reward outcomes occur, their causes can be discerned and consolidated in memory.
Supplementary Material
Supplementary material can be found at: http://www.cercor.oxfordjournals.org/
Funding
This work was supported by National Institutes of Health (grant numbers DA027802, MH094743) and the Alfred P Sloan Foundation.
Notes
We thank I. Ballard and K. Macduffie for assistance with data collection and analysis. We also thank I. Ballard and S. Stanton for helpful discussions and comments on the manuscript. Conflict of Interest: None declared.
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2017-02-24 05:51:02
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https://research.utwente.nl/en/publications/balancing-size-and-density-segregation-in-bidisperse-dense-granul
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# Balancing size and density segregation in bidisperse dense granular flows
Deepak R. Tunuguntla, Anthony R. Thornton
5 Citations (Scopus)
## Abstract
Several experimental studies have illustrated a balance between the segregation forces arising due to size- and density-differences. However, no detailed studies have been carried out to quantify this balance. In 2014, by utilising discrete particle simulations, we presented a simple relationship between the particle size- and density-ratio, $s^a = \hat \rho$, where 'a' determines whether the partial pressure scales with the diameter, surface area or volume of the particle. For a 50:50 mix (in volume) of bidisperse granular mixtures, we found the partial pressure to scale with the volume of the particle, i.e. a = 3. Moreover, there also exists a range of size- and density-ratios that satisfy the relation $s^3 = \hat \rho$, where the bidisperse mixture remains homogeneously mixed. However, in this proceeding, we deviate from the conventional 50:50 mixes and consider a slightly extreme case of mixes, such as the 10:90 (in volume) mixes, which are often found in nature and industries. By doing so we observe that the partial pressure does not scale with the particle volume and, more importantly, the zero-segregation relation is not as simple as $s^a = \hat \rho$. However, there does exist a range of size- and density-ratios for which the mixture weakly segregates.
Original language English 03079 EPJ Web of Conferences 140 https://doi.org/10.1051/epjconf/201714003079 Published - 30 Jun 2017 8th International Conference on Micromechanics on Granular Media, Powders & Grains 2017 - Montpellier, FranceDuration: 3 Jul 2017 → 7 Jul 2017Conference number: 8http://pg2017.org/en/
## Fingerprint
Dive into the research topics of 'Balancing size and density segregation in bidisperse dense granular flows'. Together they form a unique fingerprint.
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2023-03-29 14:07:41
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http://ybsi.ihuq.pw/multiclass-svm-predict.html
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Support vector machine is not a good choice, however, it is a powerful classification algorithm for two or multiclass classification. More formally, a support vector machine constructs a hyperplane or set of hyperplanes in a high- or infinite- dimensional space, which can be used for classification, regression, or other tasks. Support vector machines Lecture 4 David Sontag New York University Slides adapted from Luke Zettlemoyer, Vibhav Gogate, Multi-class SVM To predict, we use:. It was re-implemented in Fall 2016 in tidyverse format by Amelia McNamara and R. This is multi-class text classification problem. While LS-SVM obtained 78. clear all close all generateData_5grid usage of N-slack SVM classification. Moreover, there are only limited software tools and systems for SVM-based protein classification available to the bioinformatics community. This is a computer translation of the original content. score(X, y[, sample_weight]) Returns the mean accuracy on the given test data and labels. Some balancing methods allow for balancing dataset with multiples classes. After completing this step-by-step tutorial. The support vector machine (SVM) is a predictive analysis data-classification algorithm that assigns new data elements to one of labeled categories. k], it finds the solution of the following optimization problem during training. In this three classes case, we need three parameters w 1;w 2;w 3, and use w 1 and w 2 to form a maximum-margin hyperplane for class 1 and 2. In structured prediction, the hinge loss can be further extended to structured output spaces. Very limited attempts have been made at multi-class prediction prior to our work. The toolbox provides a simple GUI which allows to draw the numerals by a standard mouse. We provide an example to illustrate the use of those methods which do not differ from the binary case. Multiclass classi cation Of the classi cation methods we have studied so far, which seem inherently binary? Nearest neighbor? Generative models?. We achieve an accuracy score of 78% which is 4% higher than Naive Bayes and 1% lower than SVM. What is the best reference for multi-class SVM? Ask Question Asked 2 months ago. For example, you might use a Two-Class Support Vector Machine or Two-Class Boosted Decision Tree. Fit the SVM model according to the given training data. Structured SVMs with margin rescaling use the following variant, where w denotes the SVM's parameters, y the SVM's predictions, φ the joint feature function, and Δ the Hamming loss:. Then, fit your model on train set using fit() and perform prediction on the test set using predict(). Keras is a Python library for deep learning that wraps the efficient numerical libraries Theano and TensorFlow. Department of Computer Science. Data scientists deem Support Vector Machines (SVM) to be one of the most complex and powerful machine-learning techniques in their toolbox, so you usually find this topic solely in advanced manuals. SVM multiclass uses the multi-class formulation described in [1], but optimizes it with an algorithm that is very fast in the linear case. Logistic regression is a discriminative probabilistic statistical classification model that can be used to predict the probability of occurrence of a event. Multiclass Support Vector Machine-Based Lesion Mapping Predicts Functional Outcome in Ischemic Stroke Patients Nils Daniel Forkert , 1, * Tobias Verleger , 2 Bastian Cheng , 3 Götz Thomalla , 3 Claus C. To obtain proper probability estimates, use the option that fits calibration models to the outputs of the support vector machine. Semi-Supervised Boosting for Multi-Class Classiflcation 3 Semi-supervised SVMs (S3VMs) or Transductive SVMs (TSVMs) are the semi-supervised extensions to Support Vector Machines (SVM). Multiclass SVM. get_params([deep]) Get parameters for this estimator. Often in machine learning tasks, you have multiple possible labels for one sample that are not mutually exclusive. , web, bioinformatics, computer vision, robotics, computer systems, finance, social-sciences, etc. show() This just shows us an image of the number in question. Multi-class classification is provided using the one-against-one voting scheme. function [ prediction] = predict ( input, X_train, svm_array) % Checks the confidence of every SVM in the array % Outputs the most confident SVM as a prediction % Step 1: Convert input into format of kernels: input_recoded = applyKernel(input, X_train); % Step 2: Initialise an array that will store confidence of each SVM. I have developed svm_model for regression, now i want to test it for a given input’x’. This is multi-class text classification problem. In my humble opinion, the naming 'y_target' in the prediction part is a little confusing, because its meanings in 'prediction_output' and 'accuracy' may be different: the former represents the target of training data, but the latter may represent the target of training data or that of test data. We shall have an image as our dataset to be able to qualitatively evaluate. Using the softmax activation function at the output layer results in a neural network that models the probability of a class as multinominal distribution. This article describes how to use the Multiclass Logistic Regression module in Azure Machine Learning Studio, to create a logistic regression model that can be used to predict multiple values. Part I: Multi-Class SVM. The classifier makes the assumption that each new complaint is assigned to one and only one category. Schäfer, and B. Recall that for a binary classi er, the margin of a hyperplane for a dataset is the distance between the hyperplane and the data point nearest to it. nary classification subproblems, like OvsR multi-class SVM Figure 1: We train a multi-class Support Vector Machine model by maximize the margin between every two classes pair. GridGain Software Documentation Getting Started; What Is Ignite? What Is Gridgain? Concepts. but the problem is that the training data i have consists of samples corresponding to Hindi and english pages/blocks only but no mixed pages. So in may respects, multi-class evaluation is a straightforward extension of the methods we use in binary evaluation. Advanced Course in Machine Learning Spring 2010 Multiclass prediction Handouts are jointly prepared by Shie Mannor and Shai Shalev-Shwartz In this lecture we study the problem of multiclass prediction, in which we should learn a function h : X → Y, where X is an instance space and Y = {1,,k} = [k] is the target space. SVM with direct tf-idf vectors does the best both for quality & performance. auxiliary information, multi-class support vector machine, and active learning. ) are beginning to witness large amounts of complex data, there is a pressing need to come up with effective ways of automatically mining useful information out of it. net) // License: Boost Software License See LICENSE. Fit the SVM model according to the given training data. The most applicable machine learning algorithm for our problem is Linear SVC. If decision. In other words, given labeled training data (supervised learning), the algorithm outputs an optimal hyperplane which categorizes new examples. Real-time locomotion intent recognition is a challenge in lower-limb exoskeletons. • Prediction: Given a new input, predict the class label Each input belongs to exactly one class. If the response is a factor containing more than two levels, then the svm() function will perform multi-class classification using the one-versus-one approach. , classify a set of images of. I have a question, do the implementation of SVM in Matlab using fitcsvm and fitcecoc already contain scaling for the dataset (ex:for image classification) or we need to. Classification and Grading Rice Using Multi-Class SVM Harpreet Kaur*, Baljit Singh** * Computer, Scienec and Engineering, P. Similar to the previous assignment, you are expected to experiment with hyperparameter tuning, model and feature selection and cross-validation to get an optimal accuracy score, which will be evaluated on the. It can be used to carry out general regression and classification (of nu and epsilon-type), as well as density-estimation. I would like to extend it to multi-class SVM. but the problem is that the training data i have consists of samples corresponding to Hindi and english pages/blocks only but no mixed pages. In this study we utilized DEA for sorting venture companies by efficiency based ratings. We predict signal peptides for both prokaryotic and eukaryotic signal organisms. This will be useful for multiclass classification. If you actually have to solve a multiclass problem, I strongly. SVM predicts the output based on the distance to the dividing hyperplane, which doesn’t directly provide a probability estimation of its prediction. Unfortunately, Support Vector Machine is intrinsically biclass and its efficient extension to multiclass - problems is still an ongoing research issue [12, 13, 14]. $\begingroup$ I think this issue is specific to SVM implementation, which is usually a binary classifier. Support Vector Machines are perhaps one of the most popular and talked about machine learning algorithms. •Assume that the score of the j-th class is •The Multiclass SVM loss for the i-th example is then formalized as: =f( ,ϴ), 𝐿 = ≠ (0, −. Overview of svm_python_learn. Initially eight higher statistical moments were extracted as features from time domain vibration signals and finally found that only three optimum statistical features $$(\sigma ,\chi \hbox { and }\kappa )$$ are effective. nary classification subproblems, like OvsR multi-class SVM Figure 1: We train a multi-class Support Vector Machine model by maximize the margin between every two classes pair. Very limited attempts have been made at multi-class prediction prior to our work. accept event only if SVM output. Multiclass Support Vector Machine-Based Lesion Mapping Predicts Functional Outcome in Ischemic Stroke Patients Nils Daniel Forkert , 1, * Tobias Verleger , 2 Bastian Cheng , 3 Götz Thomalla , 3 Claus C. The classifier makes the assumption that each new complaint is assigned to one and only one category. clear all close all generateData_5grid usage of N-slack SVM classification. The toolbox provides a simple GUI which allows to draw the numerals by a standard mouse. Support vector machines Lecture 4 David Sontag New York University Slides adapted from Luke Zettlemoyer, Vibhav Gogate, Multi-class SVM To predict, we use:. Discriminant Analysis (GDA) and Least Square Support Vector Machine (LS-SVM). Multiclass classification is a popular problem in supervised machine learning. Learning with auxiliary information Learning with auxiliary class information is a relatively new approach for improving classification learning process. ODSC - Open Data Science. This tutorial will show you some tips and tricks to improve your multi-class classification results. my data set have 10 classes like running, walking ,biking riding, waving, walking etc. Multiclass classification is supported via multinomial logistic (softmax) regression. Support-vector machine weights have also been used to interpret SVM models in the past. I can't wait to see what we can achieve! Data Exploration. As an example, a sample instance might be a natural language sentence, and the output label is an annotated parse tree. Is there any multi-class SVM classifier available in MATLAB? you can easily construct a multi-class SVM starting from the binary SVM (actually, this is what most of the external libraries do. This can be thought as predicting properties of a data-point that are not mutually exclusive, such as topics that are relevant for a document. The colnames of the matrix indicate the labels of the two classes. In this post you will. use a validation set to tune the learning rate and regularization strength. The purpose is to avoid attributes in greater numeric ranges dominating those in smaller numeric ranges. Equivalently, you can think of margin as the smallest distance between a positive example and a. SVM: Support Vector Machines and Multi Class Classification seesiva Concepts , R June 14, 2013 June 17, 2013 1 Minute Most of the Classification examples out in the internet talks about binary classification. SVM with direct tf-idf vectors does the best both for quality & performance. Multiclass SVM with e1071 When dealing with multi-class classification using the package e1071 for R, which encapsulates LibSVM , one faces the problem of correctly predicting values, since the predict function doesn't seem to deal effectively with this case. We shall have an image as our dataset to be able to qualitatively evaluate. The classifier makes the assumption that each new complaint is assigned to one and only one category. There are four types of kernels in SVM which we will implement in this article: 1. predict the class to which the new point belongs. Types of ML Models. but the problem is that the training data i have consists of samples corresponding to Hindi and english pages/blocks only but no mixed pages. SVM is based on statistical learning theory developed by Vapnik [6, 25]. Crammer and Singer (2001) have extended the binary SVM classifier to classification problems with more than two classes. In multinomial logistic regression, the algorithm produces sets of coefficients, or a matrix of dimension where is the number of outcome classes and is the number of features. Large scale multiple kernel learning. Davide Anguita, Alessandro Ghio, Luca Oneto, Xavier Parra and Jorge L. Data Mining Algorithms In R/Classification/SVM. Hilgetag , 4, 5 and Jens Fiehler 2. This binary classifier for multiclass can be used with one-vs-all or all-vs-all reduction method. , 2012], object detection [Redmon and Farhadi, 2018], and segmentati. Note If the training set was scaled by svm (done by default), the new data is scaled accordingly using scale and center of the training data. You can vote up the examples you like or vote down the ones you don't like. predict the class to which the new point belongs. Ajit Kumar Das1, Sudarsan Padhy 2 1 International Institute of Information Technology Bhubaneswar, Odisha, India 2Institute of Mathematics and Application, Bhubaneswar, Odisha, India ABSTRACT Investment class rating using machine learning. First, a one-versus-one (OVO) multiclass fuzzy support vector machines (multiclass fuzzy SVM) model using a Gaussian kernel was constructed based on product samples from mobile phones. As an example, a sample instance might be a natural language sentence, and the output label is an annotated parse tree. My Dataset is composed by a certain number of Numeric Attributes and a class Attribute which can assume value in a range [1-16] (so it is a nominal attribute). ) are beginning to witness large amounts of complex data, there is a pressing need to come up with effective ways of automatically mining useful information out of it. The target set has dimensionality 2000 25. U, Baba Banda Singh Bahadur Engineering College, Fathegarh Sahib, Punjab India. SVM Multi-class Probability Outputs This code implements different strategies for multi-class probability estimates from in the following paper T. Now we can use the predict() function with the trained SVM model to make predictions using the test set. NET comes with a battery of built-in Kernels to chose from), create our multi-class SVM, and setup the learner, who will be responsible for training the SVM, pass it the strategy – and the machine can start learning. • Prediction: Given a new input, predict the class label Each input belongs to exactly one class. Author(s). title = "SVM-Fold: A tool for discriminative multi-class protein fold and superfamily recognition", abstract = "Background: Predicting a protein's structural class from its amino acid sequence is a fundamental problem in computational biology. ClassificationECOC is an error-correcting output codes (ECOC) classifier for multiclass learning, where the classifier consists of multiple binary learners such as support vector machines (SVMs). Rosenberg (CDS, NYU) DS-GA 1003 / CSCI-GA 2567 April 2, 20192/40. #undef DLIB_SVm_MULTICLASS_LINEAR_TRAINER. 56 % Choosing a good mapping ( ) (encoding prior knowledge + getting right complexity of function class) for your problem improves results. Data scientists deem Support Vector Machines (SVM) to be one of the most complex and powerful machine-learning techniques in their toolbox, so you usually find this topic solely in advanced manuals. See the section about multi-class classification in the SVM section of the User Guide for details. Build a Multi-Class Support Vector Machine in R. The batch prediction file for a multiclass model contains one column for each class found in the training data. • You come up with a decent number of fea-tures. An SVM model is a representation of the examples as points in space, mapped so that the examples of the separate categories are divided by a clear gap that is as wide as possible. In other words, given labeled training data (supervised learning), the algorithm outputs an optimal hyperplane which categorizes new examples. I got a weird output. Both of these tasks are well tackled by neural networks. Augmented and modified by Vivek Srikumar. Support vector machine classifier is one of the most popular machine learning classification algorithm. Upon training the multi-class SVM, I want to test the classifier performance using the test data. In essence, for multiclass SVM methods, several binary classifiers has to be constructed or a larger optimization problem would be needed. Recall that for a binary classi er, the margin of a hyperplane for a dataset is the distance between the hyperplane and the data point nearest to it. The support vector machine (SVM) is one of the important tools of machine learning. set_params (self, \*\*params) Set the parameters of this estimator. Multi-class problems are solved using pairwise classification (aka 1-vs-1). 이진분류법을 확장해서 멀티클래스 분류를 하는 방법이 있는데 대표적으로 one vs one approach 그리고 one vs rest approach(또는 one vs all 라고 부르기도함) 이 있다. One of them is conduct simple scaling on the data before applying SVM. SVM (Contd), Multiclass and One-Class SVM Piyush Rai Introduction to Machine Learning (CS771A) September 4, 2018 Intro to Machine Learning (CS771A) SVM (Contd), Multiclass and One-Class SVM 1. In this article we'll see what support vector machines algorithms are, the brief theory behind support vector machine and their implementation in Python's Scikit-Learn library. They are extracted from open source Python projects. Evaluation Methods. Then, another SVM can recognize the sign only if it has been previously classified among the 8 desired one (this SVM is only trained with the 8 signs). For each predicted label its only its score is computed, and then these scores are aggregated over all the datapoints. Multi-class SVM Loss At the most basic level, a loss function is simply used to quantify how “good” or “bad” a given predictor is at classifying the input data points in a dataset. apply_multiclass. In these extensions, additional parameters and constraints are added to the optimization problem to handle the separation of the different classes. We are going to discuss about the e1071 package in R. Create a binary variable for each class and predict them individually as binary classification after that combine the results but it is not the right choice if we have high number of classes because it takes good processing time. •In other words: •Predict the label Nthat minimizes: ( N, ( T)), for some distance function. If the response is a factor containing more than two levels, then the svm() function will perform multi-class classification using the one-versus-one approach. will have to predict the labels using your multiclass SVM model. For the example to work, you need to install SVM^multiclass and set the path in this file. Li, 2 andBo-RuJiang 1. In multiclass classification, we have a finite set of classes. Since the EnsembleVoteClassifier uses the argmax function internally if voting='soft', it would indeed predict class 2 in this case even if the ensemble consists of only one SVM model. A text might be about any of religion, politics, finance or education at the same time or none of these. Move into the LIBSVM folder; On Unix systems, type make to build svm-scale, svm-train and svm-predict programs. Unfortunately, Support Vector Machine is intrinsically biclass and its efficient extension to multiclass - problems is still an ongoing research issue [12, 13, 14]. Methods We developed a number of methodsforbuildingSVM-based multiclass classification schemes in the context of th e SCOP protein classification. Evaluation Methods. 7 % Translation invariant SVM 0. New examples are then mapped into that same space and predicted to belong to a category based on which side of the gap they fall on. If probability is TRUE, the vector gets a "probabilities" attribute containing a n x k matrix (n number of predicted values, k number of classes) of the class probabilities. SVM (Contd), Multiclass and One-Class SVM Piyush Rai Introduction to Machine Learning (CS771A) September 4, 2018 Intro to Machine Learning (CS771A) SVM (Contd), Multiclass and One-Class SVM 1. Now I want to take the data predicted as positive by the 1st classifier & not negative by the second classifier as positive & data predicted as negative by the 2nd classifier & not positive by the 1st classifier as negative. The traditional way to do multiclass classification with SVMs is to use one of the methods discussed in Section 14. Ri· is the ith row of the matrix and defines the code for class i. , classify a set of images of. Davide Anguita, Alessandro Ghio, Luca Oneto, Xavier Parra and Jorge L. Large scale multiple kernel learning. The support vector machine (SVM) is a predictive analysis data-classification algorithm that assigns new data elements to one of labeled categories. I have a question, do the implementation of SVM in Matlab using fitcsvm and fitcecoc already contain scaling for the dataset (ex:for image classification) or we need to. Given a new complaint comes in, we want to assign it to one of 12 categories. learning capacity and generalization ability. $\begingroup$ I think this issue is specific to SVM implementation, which is usually a binary classifier. Then, another SVM can recognize the sign only if it has been previously classified among the 8 desired one (this SVM is only trained with the 8 signs). Indeed, SVM only give a class prediction output solve this type of problems. This binary classifier for multiclass can be used with one-vs-all or all-vs-all reduction method. Multiclass classification makes the assumption that each sample is assigned to one and only one label: a fruit can be either an apple or a pear but not both at the same time. get_params (self[, deep]) Get parameters for this estimator. An SVM model is a representation of the examples as points in space, mapped so that the examples of the separate categories are divided by a clear gap that is as wide as possible. This is the strategy we will implement in this section. Although flnding an exact S3VM is NP-complete. Several frameworks have been introduced to extend SVM to multiclass contexts and a detailed account of the literature is out of the scope of this paper. How SVM Works SVM works by mapping data to a high-dimensional feature space so that data points can be categorized, even when the data are not otherwise linearly separable. Since the EnsembleVoteClassifier uses the argmax function internally if voting='soft', it would indeed predict class 2 in this case even if the ensemble consists of only one SVM model. However, you shouldn’t turn away from this great learning algorithm because the Scikit-learn. U, Rayat Institute of Engineering and Technology Railmajra, Punjab, India ** Computer Science and Engineering, P. The classifier makes the assumption that each new complaint is assigned to one and only one category. Fit the SVM model according to the given training data. By Ieva Zarina, Software Developer, Nordigen. This can be thought as predicting properties of a data-point that are not mutually exclusive, such as topics that are relevant for a document. // Copyright (C) 2011 Davis E. 1 Multiclass margin The key idea of SVM is based on the notion of margin. Hello all, i'm doing classification using one to all multiclass svm. It constructs M models, where M is the number of classes. #undef DLIB_SVm_MULTICLASS_LINEAR_TRAINER. Each reference document is referred to herein by its reference name (e. While LS-SVM obtained 78. it automatically handles multi-class prediction if your training dataset contains more than two classes. The overall multiclass classifier assigns a sample to the class with the highest confidence among the 14 pairwise OVA analyses. multiclass. More formally, a support vector machine constructs a hyperplane or set of hyperplanes in a high- or infinite- dimensional space, which can be used for classification, regression, or other tasks. The colnames of the matrix indicate the labels of the two classes. Multiclass classification is a popular problem in supervised machine learning. In this recipe, we will use a multi-class SVM to categorize the three … - Selection from TensorFlow Machine Learning Cookbook [Book]. This study presents a modified application of a multiclass support vector machine (SVM) to predict tunnel squeezing based on four parameters, that is, diameter (D), buried depth (H), support stiffness (K), and rock tunneling quality index (Q). and each class have 20 videos. SVM Multi-class Probability Outputs This code implements different strategies for multi-class probability estimates from in the following paper T. Now I want to take the data predicted as positive by the 1st classifier & not negative by the second classifier as positive & data predicted as negative by the 2nd classifier & not positive by the 1st classifier as negative. svm is used to train a support vector machine. k], it finds the solution of the following optimization problem during training. implement a fully-vectorized loss function for the SVM classification. Support vector machine [5,6,7], a new computational learning method based on. NET), it seems they both support multi-class classification via SVM; however, regression analysis with multiple outputs via SVM seems to not be supported (unless I am missing something). SVM Classification in Multiclass Letter Recognition System ¥i) L=-∑ Where yi is the class label of support vector, ai and b are numeric parameters that were obtained automatically by the SVM algorithm and optimization. want to predict the class label for a new data element, you undertake the steps de-scribed below: • You first get hold of as much training data as you can. 0, class_weight='auto') and then do fit and predict for a set of data with 7 different labels. A pure Python re-implementation of: Large-scale Multiclass Support Vector Machine Training via Euclidean Projection onto the Simplex. SVMhmm and SVMmulticlass are applications using SVMlight for. The ith support vector machine is trained with all of the examples in the ith class with positive labels, and all other examples with negative labels. In our model, the predicting problem is converted into a multiclass classification problem-rather than predicting the exact value of box office. Multilabel classification assigns to each sample a set of target labels. In this three classes case, we need three parameters w 1;w 2;w 3, and use w 1 and w 2 to form a maximum-margin hyperplane for class 1 and 2. Akshay Joshi. First, import the SVM module and create support vector classifier object by passing argument kernel as the linear kernel in SVC() function. Gaussian Kernel 4. The solution of binary classification problems using the Support Vector Machine (SVM) method has been well developed. This approach also classifies the data with higher accuracy than the traditional multi-class algorithms. ksvm supports the well known C-svc, nu-svc, (classification) one-class-svc (novelty) eps-svr, nu-svr (regression) formulations along with native multi-class classification formulations and the bound-constraint SVM formulations. So whichever value of i gives us the highest probability we then predict y to be that value. 6 Multiclass SVM 6. accept event only if SVM output. SVM: Multiclass and Structured Prediction Bin Zhao. %% Tutorial on Multi-class classification using structured output SVM % This tutorial shows how multi-class classification can be cast and solved % using structured output SVM (introduced in [1]). I can’t wait to see what we can achieve! Data Exploration. Given m classes and m trained classifiers, a new sample takes the class of the classifier with the largest real. Sparsity-inducing penalties are useful tools to design multiclass support vector machines (SVMs). They were extremely popular around the time they were developed in the 1990s and continue to be the go-to method for a high-performing algorithm with little tuning. Department of Computer Engineering. Note If the training set was scaled by svm (done by default), the new data is scaled accordingly using scale and center of the training data. Department of Computer Science. Unlike the case of multiclass classification where output space with interchangeable, arbitrarily numbered labels, structured output spaces are considered in generalized multiclass SVMS. set_params (self, \*\*params) Set the parameters of this estimator. In multiclass SVM, OAO / OAA model’s quality is ascertained by its ability to learn from the data and to predict unknown data i. The predicted class of a point will be the class that creates the largest SVM margin. The resulting multi-class Pegasos has similar algorithmic structure as its binary version. Since the EnsembleVoteClassifier uses the argmax function internally if voting='soft', it would indeed predict class 2 in this case even if the ensemble consists of only one SVM model. There are more complications (handling the bias term, handling non-separable datasets), but this is the gist of the algorithm. 4 SVM with Multiple Classes. The Scientific World Journal is a peer-reviewed, Open Access journal that publishes original research, reviews, and clinical studies covering a wide range of subjects in science, technology, and medicine. In this study, we put forward a transparent control strategy to detect SiSt and StSi when subject wore BioKEX online and a SVM classifier based on simple sensor information was used to detect SiSt and StSi when subject wore BioKEX and the strategy is able to predict the intent of the wearer in real time. Given a new complaint comes in, we want to assign it to one of 12 categories. Comparing different solvers on a standard multi-class SVM problem. Create a binary variable for each class and predict them individually as binary classification after that combine the results but it is not the right choice if we have high number of classes because it takes good processing time. use a validation set to tune the learning rate and regularization strength. You believe that the greater the. In this tutorial, I'm going to build a classifier for 10 different bird images. Hilgetag , 4, 5 and Jens Fiehler 2. value is TRUE, the vector gets a "decision. show() This just shows us an image of the number in question. $\begingroup$ I think this issue is specific to SVM implementation, which is usually a binary classifier. Omkar Kulkarni. auxiliary information, multi-class support vector machine, and active learning. txt /* This is an example illustrating the use of the multiclass classification tools from the dlib C++ Library. In structured prediction, the hinge loss can be further extended to structured output spaces. We examine using a Support Vector Machine to predict secretory signal peptides. This yields a very efficient prediction algorithm - once we have trained our SVM, a large amount of the training data (those samples with zero Lagrangian multipliers) can be removed. There are many approaches followed to use SVM for multiclass classification. This tutorial will show you some tips and tricks to improve your multi-class classification results. If probability is TRUE, the vector gets a "probabilities" attribute containing a n x k matrix (n number of predicted values, k number of classes) of the class probabilities. In the person layout problem, instances. Sonnenburg, G. In this system, the accuracy rate is improved with the usage of Multiclass Support Vector Machine (M-SVM). Department of Computer Engineering. SVM is based on statistical learning theory developed by Vapnik [6, 25]. For this exercise, a linear SVM will be used. Another way to implement multi-class classifiers is to do a one versus all strategy where we create a classifier for each of the classes. In this case, for the pixels of image with label, we compute for the score for each class as. Gist contains software tools for support vector machine classification and for kernel principal components analysis. It is provided for general information only and should not be relied upon as complete or accurate. The basic SVM supports only binary classification, but extensions have been proposed to handle the multiclass classification case as well. In this notebook, a Multiclass Support Vector Machine (SVM) will be implemented. In these extensions, additional parameters and constraints are added to the optimization problem to handle the separation of the different classes. It reduces the multiclass problem to a group of binary classification tasks and combines the binary classification results to predict multiclass labels. However, you shouldn't turn away from this great learning algorithm because the Scikit-learn. it automatically handles multi-class prediction if your training dataset contains more than two classes. This paper proposed an efficient lung cancer detection and prediction algorithm using multi-class SVM (Support Vector Machine) classifier. 6 Multiclass SVM 6. Polynomial Kernel 3. i'm trying to make a MultiClass Classification task with SVM in my Java Project. thanks and regards Vishal mishra. title = "SVM-Fold: A tool for discriminative multi-class protein fold and superfamily recognition", abstract = "Background: Predicting a protein's structural class from its amino acid sequence is a fundamental problem in computational biology. I got a weird output. Human Activity Recognition on Smartphones using a Multiclass Hardware-Friendly Support Vector Machine. So I'm going to alter the tags and title to reflect those details. asarray) and sparse (any scipy. However, I found that the most useful machine learning tasks try to predict multiple classes and more often than not those classes are grossly unbalanced. In multiclass SVM, OAO / OAA model’s quality is ascertained by its ability to learn from the data and to predict unknown data i. accept event only if SVM output. We will understand the SVM training and testing models in R and look at the main functions of e1071 package i. multi-class superfamily or fold recognition. We print out decision values for regression. I can't wait to see what we can achieve! Data Exploration. In this system, the accuracy rate is improved with the usage of Multiclass Support Vector Machine (M-SVM). Classify images into labels Binary Prediction x X y Y y {table,no table} Classify images into labels. An SVM performs classification tasks by constructing hyperplanes in a multidimensional space that separates cases of different class labels. my data set have 10 classes like running, walking ,biking riding, waving, walking etc. U, Baba Banda Singh Bahadur Engineering College, Fathegarh Sahib, Punjab India. Provided digital up-skilling with in-house lunch training on Python and NLP techniques for a multi-class dataset in Jupyter Notebook to evaluate the best model comparing against random forest, multinomial naive bayes and support vector machine to predict the class of New York City complaint status type and category. Support Vector Machines are perhaps one of the most popular and talked about machine learning algorithms. function [ prediction] = predict ( input, X_train, svm_array) % Checks the confidence of every SVM in the array % Outputs the most confident SVM as a prediction % Step 1: Convert input into format of kernels: input_recoded = applyKernel(input, X_train); % Step 2: Initialise an array that will store confidence of each SVM.
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2019-11-22 17:29:44
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https://hackage.haskell.org/package/lens-tutorial-1.0.4/docs/Control-Lens-Tutorial.html
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lens-tutorial-1.0.4: Tutorial for the lens library
Control.Lens.Tutorial
Description
This lens tutorial targets Haskell beginners and assumes only basic familiarity with Haskell. By the end of this tutorial you should:
• understand what problems the lens library solves,
• know when it is appropriate to use the lens library,
• be proficient in the most common lens idioms,
• understand the drawbacks of using lenses, and:
If you would like to follow along with these examples, just import this module:
$ghci >>> import Control.Lens.Tutorial Synopsis Motivation The simplest problem that the lens library solves is updating deeply nested records. Suppose you had the following nested Haskell data types: data Atom = Atom { _element :: String, _point :: Point } data Point = Point { _x :: Double, _y :: Double } If you wanted to increase the x coordinate of an Atom by one unit, you would have to write something like this in Haskell: shiftAtomX :: Atom -> Atom shiftAtomX (Atom e (Point x y)) = Atom e (Point (x + 1) y) This unpacking and repacking of data types grows increasingly difficult the more fields you add to each data type or the more deeply nested your data structures become. The lens library solves this problem by letting you instead write: -- atom.hs {-# LANGUAGE TemplateHaskell #-} import Control.Lens hiding (element) import Control.Lens.TH data Atom = Atom { _element :: String, _point :: Point } deriving (Show) data Point = Point { _x :: Double, _y :: Double } deriving (Show)$(makeLenses ''Atom)
$(makeLenses ''Point) shiftAtomX :: Atom -> Atom shiftAtomX = over (point . x) (+ 1) Let's convince ourselves that this works: >>> let atom = Atom { _element = "C", _point = Point { _x = 1.0, _y = 2.0 } } >>> shiftAtomX atom Atom {_element = "C", _point = Point {_x = 2.0, _y = 2.0}} The above solution does not change no matter how many fields we add to Atom or Point. Now suppose that we added yet another data structure: data Molecule = Molecule { _atoms :: [Atom] } deriving (Show) We could shift an entire Molecule by writing: $(makeLenses ''Molecule)
shiftMoleculeX :: Molecule -> Molecule
shiftMoleculeX = over (atoms . traverse . point . x) (+ 1)
Again, this works the way we expect:
>>> let atom1 = Atom { _element = "C", _point = Point { _x = 1.0, _y = 2.0 } }
>>> let atom2 = Atom { _element = "O", _point = Point { _x = 3.0, _y = 4.0 } }
>>> let molecule = Molecule { _atoms = [atom1, atom2] }
>>> shiftMoleculeX molecule -- Output formatted for clarity
Molecule {_atoms = [Atom {_element = "C", _point = Point {_x = 2.0, _y = 2.0}},Atom {_element = "O", _point = Point {_x = 4.0, _y = 4.0}}]}
... or formatted for clarity:
Molecule
{ _atoms =
[ Atom { _element = "C", _point = Point { _x = 2.0, _y = 2.0 } }
, Atom { _element = "O", _point = Point { _x = 4.0, _y = 4.0 } }
]
}
Many people stumble across lenses while trying to solve this common problem of working with data structures with a large number of fields or deeply nested values. These sorts of situations arise commonly in:
• games with complex and deeply nested state
• scientific data formats
• sensor or instrument output
• web APIs
• XML and JSON
• enterprise code where data structures can have tens, hundreds, or even thousands of fields (true story!)
Lenses
You might have some basic questions like:
Question: What is a lens?
Answer: A lens is a first class getter and setter
We already saw how to use lenses to update values using over, but we can also use lenses to retrieve values using view:
>>> let atom = Atom { _element = "C", _point = Point { _x = 1.0, _y = 2.0 } }
>>> view (point . x) atom
1.0
In other words, lenses package both "get" and "set" functionality into a single value (the lens). You could pretend that a lens is a record with two fields:
data Lens a b = Lens
{ view :: a -> b
, over :: (b -> b) -> (a -> a)
}
That's not how lenses are actually implemented, but it's a useful starting intuition.
Question: What is the type of a lens?
Answer: We used two lenses in the above Atom example, with these types:
point :: Lens' Atom Point
x :: Lens' Point Double
The point lens contains all the information we need to get or set the _point field of the Atom type (which is a Point). Similarly, the x lens contains all the information we need to get or set the _x field of the Point data type (which is a Double).
The convention for the Lens' type parameters is:
-- +-- Bigger type
-- |
-- v
Lens' bigger smaller
-- ^
-- |
-- +-- Smaller type within the bigger type
The actual definition of Lens' is:
type Lens' a b = forall f . Functor f => (b -> f b) -> (a -> f a)
You might wonder how you can fit both getter and setter functionality in a single value like this. The trick is that we get to pick what Functor we specialize f to and depending on which Functor we pick we get different features.
For example, if you pick (f = Identity):
type ASetter' a b = (b -> Identity b) -> (a -> Identity a)
-- ... equivalent to: (b -> b) -> (a -> a)
... you can build an over-like function.
Similarly, if you pick (f = Const b):
type Getting b a b = (b -> Const b b) -> (a -> Const b a)
-- ... equivalent to: (b -> b ) -> (a -> b )
... and if you apply a function of that type to id then you get a view-like function
-- (a -> b )
Those are not the only two Functors we can pick. In fact, we can do a lot more with lenses than just get and set values, but those are the two most commonly used features.
Question: How do I create lenses?
Answer: You can either auto-generate them using Template Haskell or create them by hand
In our Atom example, we auto-generated the lenses using Template Haskell, like this:
makeLenses ''Atom
makeLenses ''Point
This created four lenses of the following types:
element :: Lens' Atom String
point :: Lens' Atom Point
x :: Lens' Point Double
y :: Lens' Point Double
makeLenses creates one lens per field prefixed with an underscore. The lens has the same name as the field without the underscore.
However, sometimes Template Haskell is not an option, so we can also use the lens utility function to build lenses. This utility has type:
lens :: (a -> b) -> (a -> b -> a) -> Lens' a b
The first argument is a "getter" (a way to extract a 'b' from an 'a'). The second argument is a "setter" (given a b, update an a). The result is a Lens' built from the getter and setter. You would use lens like this:
point :: Lens' Atom Point
point = lens _point (\atom newPoint -> atom { _point = newPoint })
You can even define lenses without incurring a dependency on the lens library. Remember that lenses are just higher-order functions over Functors, so we could instead write:
-- point :: Lens' Atom Point
point :: Functor f => (Point -> f Point) -> Atom -> f Atom
point k atom = fmap (\newPoint -> atom { _point = newPoint }) (k (_point atom))
This means that you can provide lenses for your library's types without depending on the lens library. All you need is the fmap function, which is provided by the Haskell Prelude.
Question: How do I combine lenses?
Answer: You compose them, using function composition (Yes, really!)
You can think of the function composition operator as having this type:
(.) :: Lens' a b -> Lens' b c -> Lens' a c
We can compose lenses using function composition because Lens' is a type synonym for a higher-order function:
type Lens' a b = forall f . Functor f => (b -> f b) -> (a -> f a)
So under the hood we are composing two higher-order functions to get back a new higher-order function:
(.) :: Functor f
=> ((b -> f b) -> (a -> f a))
-> ((c -> f c) -> (b -> f b))
-> ((c -> f c) -> (a -> f a))
In our original Atom example, we composed the point and x lenses to create a new composite lens:
point :: Lens' Atom Point
x :: Lens' Point Double
point . x :: Lens' Atom Double
This composite lens lets us get or set the x coordinate of an Atom. We can use over and view on the composite Lens' and they will behave exactly the way we expect:
view (point . x) :: Atom -> Double
over (point . x) :: (Double -> Double) -> (Atom -> Atom)
Question: How do I consume lenses?
Answer: Using view, set or over
Here are their types:
view :: Lens' a b -> a -> b
over :: Lens' a b -> (b -> b) -> a -> a
set :: Lens' a b -> b -> a -> a
set lens b = over lens (\_ -> b)
view and over are the two fundamental functions on lenses. set is just a special case of over.
view and over are fundamental because they distribute over lens composition:
view (lens1 . lens2) = (view lens2) . (view lens1)
view id = id
over (lens1 . lens2) = (over lens1) . (over lens2)
over id = id
Question: What else do I need to know?
For 90% of use cases, you just:
• Create lenses (using makeLens, lens or plain-old fmap)
• Compose them (using (.))
• Consume them (using view, set, and over)
You could actually stop reading here if you are in a hurry since this covers the overwhelmingly common use case for the library. On the other hand, keep reading if you would like to learn additional tricks and features.
Accessor notation
You might be used to object-oriented languages where you could retrieve a nested field using:
atom.point.x
You can do almost the exact same thing using the lens library, except that the first dot will have a ^ right before the dot:
>>> let atom = Atom { _element = "C", _point = Point { _x = 1.0, _y = 2.0 } }
>>> atom^.point.x
1.0
You can better understand why this works, by adding whitespace and explicit parentheses:
atom ^. (point . x)
This trick uses (^.), which is an infix operator equivalent to view:
(^.) :: a -> Lens' a b -> b
x ^. l = view l x
... and you just keep adding dots after that for each lens you compose. This gives the appearance of object-oriented accessors if you omit the whitespace around the operators.
First-class
Lenses are "first class" values, meaning that you can manipulate them using ordinary functional programming techniques. You can take them as inputs, return them as outputs, or stick them in data structures. Anything goes!
For example, suppose we don't want to define separate shift functions for Atoms and Molecules:
shiftAtomX :: Atom -> Atom
shiftAtomX = over (point . x) (+ 1)
shiftMoleculeX :: Molecule -> Molecule
shiftMoleculeX = over (atoms . traverse . point . x) (+ 1)
We can instead unify them into a single function by parametrizing the shift function on the lens:
shift lens = over lens (+ 1)
This lets us write:
shift (point . x) :: Atom -> Atom
shift (atoms . traverse . point . x) :: Molecule -> Molecule
Even better, we can define synonyms for our composite lenses:
atomX :: Lens' Atom Double
atomX = point . x
-- We'll learn what Traversal means shortly
moleculeX :: Traversal' Molecule Double
moleculeX = atoms . traverse . point . x
Now we can write code almost identical to the original code:
shift atomX :: Atom -> Atom
shift moleculeX :: Molecule -> Molecule
set atomX :: Double -> Atom -> Atom
set moleculeX :: Double -> Molecule -> Molecule
view atomX :: Atom -> Double
-- We can't use view for Traversal's. Read on to find out why
toListOf moleculeX :: Molecule -> [Double]
That's much more reusable, but you might wonder what this Traversal' and toListOf business is all about.
Traversals
Question: What is a traversal?
Answer: A first class getter and setter for an arbitrary number of values
A traversal lets you get all the values it points to as a list and it also lets you update or set all the values it points to. Think of a traversal as a record with two fields:
data Traversal' a b = Traversal'
{ toListOf :: a -> [b]
, over :: (b -> b) -> (a -> a)
}
That's not how traversals are actually implemented, but it's a useful starting intuition.
We can still use over and set (a special case of over) with a traversal, but we use toListOf instead of view.
Question: What is the type of a traversal?
Answer: We used one traversal in the above Molecule example:
moleculeX :: Traversal' Molecule Double
This Traversal' lets us get or set an arbitrary number of x coordinates, each of which is a Double. There could be less than one x coordinate (i.e. 0 coordinates) or more than one x coordinate. Contrast this with a Lens' which can only get or set exactly one value.
Like Lens', Traversal' is a type synonym for a higher-order function:
type Traversal' a b = forall f . Applicative f => (b -> f b) -> (a -> f a)
type Lens' a b = forall f . Functor f => (b -> f b) -> (a -> f a)
Notice that the only difference between a Lens' and a Traversal' is the type class constraint. A Lens' has a Functor constraint and Traversal' has an Applicative constraint. This means that any Lens' is automatically also a valid Traversal' (since Functor is a superclass of Applicative).
Since every Lens' is a Traversal', all of our example lenses also double as traversals:
atoms :: Traversal' Molecule [Atom]
element :: Traversal' Atom String
point :: Traversal' Atom Point
x :: Traversal' Point Double
y :: Traversal' Point Double
We actually used yet another Traversal', which was traverse (from Data.Traversable):
traverse :: Traversable t => Traversal' (t a) a
This works because the Traversal' type synonym expands out to:
traverse :: (Applicative f, Traversable t) => (a -> f a) -> t a -> f (t a)
... which is exactly the traditional type signature of traverse.
In our Molecule example, we were using the special case where t = []:
traverse :: Traversal' [a] a
In Haskell, you can derive Functor, Foldable and Traversable for many data types using the DeriveFoldable and DeriveTraversable extensions. This means that you can autogenerate a valid traverse for these data types:
{-# LANGUAGE DeriveFoldable #-}
{-# LANGUAGE DeriveFunctor #-}
{-# LANGUAGE DeriveTraversable #-}
import Control.Lens
import Data.Foldable
data Pair a = Pair a a deriving (Functor, Foldable, Traversable)
We could then use traverse to navigate from Pair to its two children:
traverse :: Traversal' (Pair a) a
over traverse :: (a -> a) -> (Pair a -> Pair a)
over traverse (+ 1) (Pair 3 4) = Pair 4 5
Question: How do I create traversals?
Answer: There are three main ways to create primitive traversals:
• traverse is a Traversal' that you get for any type that implements Traversable
• Every Lens' will also type-check as a Traversal'
• You can use Template Haskell to generate Traversal's using makePrisms since every Prism' is also a Traversal' (not covered in this tutorial)
Question: How do I combine traversals?
Answer: You compose them, using function composition
You can think of the function composition operator as having this type:
(.) :: Traversal' a b -> Traversal' b c -> Traversal' a c
We can compose traversals using function composition because a Traversal' is a type synonym for a higher-order function:
type Traversal' a b = forall f . Applicative f => (b -> f b) -> (a -> f a)
So under the hood we are composing two functions to get back a new function:
(.) :: Applicative f
=> ((b -> f b) -> (a -> f a))
-> ((c -> f c) -> (b -> f b))
-> ((c -> f c) -> (a -> f a))
In our original Molecule example, we composed four Traversal's together to create a new Traversal':
-- Remember that atoms, point, and x are also Traversal's
atoms :: Traversal' Molecule [Atom]
traverse :: Traversal' [Atom] Atom
point :: Traversal' Atom Point
x :: Traversal' Point Double
-- Now compose them
atoms :: Traversal' Molecule [Atom]
atoms . traverse :: Traversal' Molecule Atom
atoms . traverse . point :: Traversal' Molecule Point
atoms . traverse . point . x :: Traversal' Molecule Double
This composite traversal lets us get or set the x coordinates of a Molecule.
over (atoms . traverse . point . x)
:: (Double -> Double) -> (Molecule -> Molecule)
toListOf (atoms . traverse . point . x)
:: Molecule -> [Double]
Question: How do I consume traversals?
Answer: Using toListOf, set or over
Here are their types:
toListOf :: Traversal' a b -> a -> [b]
over :: Traversal' a b -> (b -> b) -> a -> a
set :: Traversal' a b -> b -> a -> a
set traversal b = over traversal (\_ -> b)
Note that toListOf distributes over traversal composition:
toListOf (traversal1 . traversal2) = (toListOf traversal1) >=> (toListOf traversal2)
toListOf id = return
If you prefer object-oriented syntax you can also use (^..), which is an infix operator equivalent to toListOf:
>>> Pair 3 4 ^.. traverse
[3,4]
Types
You might wonder why you can use over on both a Lens' and a Traversal' but you can only use view on a Lens'. We can see why by studying the (simplified) type and implementation of over:
over :: ((b -> Identity b) -> (a -> Identity a)) -> (b -> b) -> a -> a
over setter f x = runIdentity (setter (\y -> Identity (f y)) x)
To follow the implementation, just step slowly through the types. Here are the types of the arguments to over:
setter :: (b -> Identity b) -> (a -> Identity a)
f :: b -> b
x :: a
... and here are the types of the sub-expressions on the right-hand side:
\y -> Identity (f y) :: b -> Identity b
setter (\y -> Identity (f y)) :: a -> Identity a
setter (\y -> Identity (f y)) x :: Identity a
runIdentity (setter (\y -> Identity (f y)) x) :: a
We can replace setter with point and replace x with atom to see that this generates the correct code for updating an atom's point:
over point f atom
-- Definition of over
= runIdentity (point (\y -> Identity (f y)) atom)
-- Definition of point
= runIdentity (fmap (\newPoint -> atom { _point = newPoint }) (Identity (f (_point atom)))
-- fmap g (Identity y) = Identity (g y)
= runIdentity (Identity (atom { _point = f (_point atom) }))
-- runIdentity (Identity z) = z
= atom { _point = f (_point atom) }
... which is exactly what we would have written by hand without lenses.
The reason over works for both Lens'es and Traversal's is because Identity implements both Functor and Applicative:
instance Functor Identity where ...
instance Applicative Identity where ...
So both the Lens' type and Traversal' type synonyms:
type Traversal' a b = forall f . Applicative f => (b -> f b) -> (a -> f a)
type Lens' a b = forall f . Functor f => (b -> f b) -> (a -> f a)
... can both be specialized to use Identity in place of f:
(b -> Identity b) -> (a -> Identity a)
... making them valid arguments to over.
Now let's study the (simplified) type and implementation of view:
view :: ((b -> Const b b) -> (a -> Const b a)) -> a -> b
view getter x = getConst (getter Const x)
Again, we can walk slowly through the types of the arguments:
getter :: (b -> Const b b) -> (a -> Const b a)
x :: a
... and the types of the sub-expressions on the right-hand side:
getter Const :: a -> Const b a
getter Const x :: Const b a
getConst (getter Const x) :: b
Let's see how this plays out for the point lens:
view point atom
-- Definition of view
= getConst (point Const atom)
-- Definition of point
= getConst (fmap (\newPoint -> atom { _point = newPoint }) (Const (_point atom)))
-- fmap g (Const y) = Const y
= getConst (Const (_point atom))
-- getConst (Const z) = z
= _point atom
... which is exactly what we would have written by hand without lenses.
view accepts Lens'es because Const implements Functor:
instance Functor (Const b)
... so the Lens' type synonym:
type Lens' a b = forall f . Functor f => (b -> f b) -> (a -> f a)
... can be specialized to use (Const b) in place of f:
(b -> Const b b) -> (a -> Const b a)
... making it a valid argument to view.
Interestingly, Const implements also Applicative, but with a constraint:
instance Monoid b => Applicative (Const b)
This implies that we *can* use view on a Traversal', but only if the value that we extract is a Monoid. Let's try this out:
>>> let atom1 = Atom { _element = "C", _point = Point { _x = 1.0, _y = 2.0 } }
>>> let atom2 = Atom { _element = "O", _point = Point { _x = 3.0, _y = 4.0 } }
>>> let molecule = Molecule { _atoms = [atom1, atom2] }
>>> view (atoms . traverse . element) molecule
"CO"
This works because our traversal's result is a String:
atoms . traverse . element :: Traversal' Molecule String
... and String implements the Monoid interface. When you try to extract multiple strings using view they get flattened together into a single String using mappend.
If you try to extract the element from an empty molecule:
>>> view (atoms . traverse . element) (Molecule { _atoms = [] })
""
You get the empty string (i.e. mempty).
This is why the result of a Traversal' needs to be a Monoid when using view. If the Traversal' points to more than one value you need some way to combine them into a single value (using mappend) and if the Traversal' points to less than one value you need a default value to return (using mempty).
If you try to view a Traversal' that doesn't point to a Monoid, you will get the following type error:
>>> view (atoms . traverse . point . x) molecule
No instance for (Data.Monoid.Monoid Double)
arising from a use of traverse'
In the first argument of (.)', namely traverse'
In the second argument of (.)', namely traverse . point . x'
In the first argument of view', namely
(atoms . traverse . point . x)'
The compiler complains that Double does not implement the Monoid type class, so there is no sensible way to merge all the x coordinates that our Traversal' points to. For these cases you should use toListOf instead.
Drawbacks
Lenses come with trade-offs, so you should use them wisely.
For example, lenses do not produce the best error messages. Unless you understand how Traversal's work you will probably not understand the above error message.
Also, lenses increase the learning curve for new Haskell programmers, so you should consider avoiding them in tutorial code targeting novice Haskell programmers.
Lenses also add a level of boilerplate to all data types to auto-generate lenses and increase compile times. So for small projects the overhead of adding lenses may dwarf the benefits.
lens is also a library with a large dependency tree, focused on being "batteries included" and covering a large cross-section of the Haskell ecosystem. Browsing the Hackage listing you will find support modules ranging from System.FilePath.Lens to Control.Parallel.Strategies.Lens, and many more. If you need a more light-weight alternative you can use the lens-simple or microlens library, each of which provides a restricted subset of the lens library with a much smaller dependency tree.
The ideal use case for the lens library is a medium-to-large project with rich and deeply nested types. In these large projects the benefits of using lenses outweigh the costs.
Conclusion
This tutorial covers an extremely small subset of this library. If you would like to learn more, you can begin by skimming the example code in the following modules:
The documentation for these modules includes several examples to get you started and help you build an intuition for more advanced tricks that were not covered in this tutorial.
You can also study several long-form examples here:
https://github.com/ekmett/lens/tree/master/examples
If you prefer light-weight lens-compatible libraries, then check out lens-simple or micro-lens:
If you would like a broader survey of lens features, then you can check out these tutorials:
Exports
These are the same types and lenses used throughout the tutorial, exported for your convenience.
data Atom Source #
Constructors
Atom Fields_element :: String _point :: Point
Instances
Source # Instance detailsDefined in Control.Lens.Tutorial MethodsshowsPrec :: Int -> Atom -> ShowS #show :: Atom -> String #showList :: [Atom] -> ShowS #
data Point Source #
Constructors
Point Fields_x :: Double _y :: Double
Instances
Source # Instance detailsDefined in Control.Lens.Tutorial MethodsshowsPrec :: Int -> Point -> ShowS #show :: Point -> String #showList :: [Point] -> ShowS #
data Molecule Source #
Constructors
Molecule Fields_atoms :: [Atom]
Instances
Source # Instance detailsDefined in Control.Lens.Tutorial MethodsshowList :: [Molecule] -> ShowS #
data Pair a Source #
Constructors
Pair a a
Instances
Source # Instance detailsDefined in Control.Lens.Tutorial Methodsfmap :: (a -> b) -> Pair a -> Pair b #(<\$) :: a -> Pair b -> Pair a # Source # Instance detailsDefined in Control.Lens.Tutorial Methodsfold :: Monoid m => Pair m -> m #foldMap :: Monoid m => (a -> m) -> Pair a -> m #foldr :: (a -> b -> b) -> b -> Pair a -> b #foldr' :: (a -> b -> b) -> b -> Pair a -> b #foldl :: (b -> a -> b) -> b -> Pair a -> b #foldl' :: (b -> a -> b) -> b -> Pair a -> b #foldr1 :: (a -> a -> a) -> Pair a -> a #foldl1 :: (a -> a -> a) -> Pair a -> a #toList :: Pair a -> [a] #null :: Pair a -> Bool #length :: Pair a -> Int #elem :: Eq a => a -> Pair a -> Bool #maximum :: Ord a => Pair a -> a #minimum :: Ord a => Pair a -> a #sum :: Num a => Pair a -> a #product :: Num a => Pair a -> a # Source # Instance detailsDefined in Control.Lens.Tutorial Methodstraverse :: Applicative f => (a -> f b) -> Pair a -> f (Pair b) #sequenceA :: Applicative f => Pair (f a) -> f (Pair a) #mapM :: Monad m => (a -> m b) -> Pair a -> m (Pair b) #sequence :: Monad m => Pair (m a) -> m (Pair a) #
traverse :: (Traversable t, Applicative f) => (a -> f b) -> t a -> f (t b) #
Map each element of a structure to an action, evaluate these actions from left to right, and collect the results. For a version that ignores the results see traverse_`.
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2021-05-08 08:52:24
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https://golem.ph.utexas.edu/category/2010/01/equivariant_stable_homotopy_th.html
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## January 24, 2010
### Equivariant Stable Homotopy Theory
#### Posted by Mike Shulman
Over the past week (at least), Urs and I and some others have been trying to understand equivariant stable homotopy theory from a higher-categorical point of view, with some help from experts like Peter May, John Greenlees, and Megan Shulman. Unfortunately, the online part of the discussion has been taking place simultaneously in two different, hard-to-find, and arguably inappropriate places, namely the nForum and the comments on a mostly unrelated thread. So I thought I would give it a thread of its own, along with with an easier-to-read introduction and summary to bring new people into the discussion.
Before we dive in, let me say that equivariant things can be tricky and involve lots of traps for the categorically-minded. (At least, there are a lot of things that trapped me.) If you need some motivation to spur you on and keep you going, note that the recent proof of the Kervaire invariant one problem involved the tools of equivariant homotopy theory in very important ways. So this stuff really is good for something.
First let’s ignore the word “stable” and ask, what is equivariant homotopy theory? Algebraic topologists who study it will tell you that it’s like ordinary homotopy theory except that there’s a group $G$ acting on everything in sight. In complete generality, $G$ could be an arbitrary topological group, but usually they take it to be either a discrete group, a compact Lie group, or (the intersection of the two) a finite discrete group.
However, I think this perspective is misleading to a higher category theorist. (At least, it was misleading to me at first.) What immediately jumps to my mind when I hear “homotopy theory of spaces with $G$-actions” is the $(\infty,1)$-category of $\infty$-groupoids with a $G$-action, i.e. the $(\infty,1)$-functor category $[B G, \infty Gprd]$. That $(\infty,1)$-category ought to be presented by a model category of $G$-spaces in which the weak equivalences, like those in most model categories of diagrams, are objectwise. However, those are not the weak equivalences in the model category of $G$-spaces that equivariant algebraic topologists are interested in. Rather, their weak equivalences are the maps $f\colon X\to Y$ of $G$-spaces which induce weak equivalences $f^H\colon X^H \to Y^H$ on spaces of $H$-fixed points, for all (closed) subgroups $H\le G$.
Looking at that, you might guess that from a higher-categorical perspective, what we’re looking at is instead a theory of certain diagrams of $\infty$-groupoids, one for each $H\le G$. And in fact, that’s true: if $\mathcal{O}_G$ denotes the orbit category of $G$, then the above model structure on $G$-spaces is equivalent to the ordinary diagram-category model structure (with objectwise weak equivalences) on $[\mathcal{O}_G^{op},Top]$. This is called Elmendorf’s theorem (for the obvious reason). So from a higher-categorical perspective, we might try to think of equivariant algebraic topology as the study of presheaves of $\infty$-groupoids on $\mathcal{O}_G$.
It’s not yet clear to me whether all aspects of “classical” equivariant homotopy theory are accessible from this higher-categorical viewpoint; part of the goal of this discussion is to figure that out.
One of the main things we’ve been thinking about is cohomology and stable equivariant homotopy theory, i.e. the equivariant analogue of spectra. Here I think I will just refer new readers to this comment, which explains the sort of cohomology and spectra that come up. In particular, there are two kinds of $G$-spectra (well, actually, a whole slew of different kinds, with the most interesting being the two extremes). “Naive” $G$-spectra are what the naive $(\infty,1)$-category-theorist would initially write down, namely objects of the stabilization of the $(\infty,1)$-category of “$G$-spaces” considered above (remember that means to consider all the fixed-point spaces specially). On the other hand we have “genuine” $G$-spectra, built out of a set of spaces indexed not just by integers, but by all finite-dimensional representations of $G$. Just as nonequivariant spectra represent (generalized) cohomology theories, a naive $G$-spectrum represents a Bredon cohomology theory, and a genuine one represents an $RO(G)$-graded Bredon cohomology theory.
Now there are lots of questions left! Here are some.
• Is the stable $(\infty,1)$-category of genuine $G$-spectra naturally presented as the stabilization of something?
The stable Giraud theorem seems like it might be relevant to this, and likewise the Schwede-Shipley classification of stable model categories, but the two aren’t quite the same. Lurie’s stable Giraud theorem says that any stable $(\infty,1)$-category is equivalent to a left-exact localization of the stabilization of an $(\infty,1)$-category of presheaves (on some domain $(\infty,1)$-category), while the Schwede-Shipley theorem is that any stable model category is equivalent to a model category of diagrams of spectra on some spectrally enriched category. We’ve been discussing this a bit starting here, and we can continue below.
• Is there some “canonical” way to guess the right notion of “grading” for genuine spectra? For instance, if we start with an arbitrary $(\infty,1)$-topos in place of $G$-spaces, would there be a corresponding notion of “$RO(G)$” and “genuine spectrum” generalizing the “naive spectra” in the naive stabilization?
• Is there an equivariant analogue of infinite loop space machines? That is, can we define a notion of “grouplike $E_\infty$ $G$-space” which can be delooped into a (genuine) $G$-spectrum?
Peter May says yes, at least when $G$ is finite. But I’m still trying to understand intuitively why his definition of “$E_\infty$ $G$-operad” is correct.
Posted at January 24, 2010 9:17 PM UTC
TrackBack URL for this Entry: http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2159
### Re: Equivariant Stable Homotopy Theory
Thanks, Mike, good idea. I was feeling bad for highjacking David Corfield’s thread, too.
Here I think I will just refer new readers to this comment,
Or better yet, the $n$Lab entry:
This contains your comment in full beauty in the section Bredon equivariant cohomology, but also some other related stuff.
There is also
that features an explicit statement of Elmendorf’s theorem with references .
And similarly there is
with some basic explicit definitions. (Though everything pretty stubby, still.)
The Schwede-Schipley theorem is (in a second) at
I’ll have to call it quits for today soon, but am looking forward to following up on this later.
Posted by: Urs Schreiber on January 24, 2010 10:30 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Thanks. There are so many links to give that it’s hard to fit them all in. (-:
Posted by: Mike Shulman on January 24, 2010 11:01 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
I like that about the Café that a technical discussion can break out in the middle of speculative discussion. In a private e-mail someone observed of the discussion that it “seems to be oscillating between spirituality and higher homotopy theory, which may actually be two names for the same thing.”
But good to have a focused thread here, so let me just ask again, if we really believe in the Baez-Dolan idea that ordinary spectra are a type of $\mathbb{Z}$-groupoid and we now seem to need more exotic gradings by $RO(G)$, does this mean there are $RO(G)$-groupoids?
Is part of the reason for the restriction to $n$-categories that the representations of the trivial group are classified by natural numbers?
Posted by: David Corfield on January 25, 2010 8:39 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
I see John and Urs were taking $\mathbb{Z}$-categories seriously back here. Urs even wondered whether we might end up with things like $\mathbb{Z}^{\mathbb{Z}}$-categories.
Posted by: David Corfield on January 25, 2010 9:22 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
let me just ask again, if we really believe in the Baez-Dolan idea that ordinary spectra are a type of $\mathbb{Z}$-groupoid
My best guess for what a (not-necessarily strict) $\mathbb{Z}$-groupoid might actually be, is that it is the structure that is used in the last part of
to discuss spectrum valued $\infty$-sheaves, and which on the $n$Lab we gave the name combinatorial spectrum. This is a $\mathbb{Z}$-graded pointed set that has finite but arbitrarily many pointed face and degenracy maps in each degree, satisfying the usual simplicial identities (it has an infinite number of face and degeneracy maps in each degree of which however only finitely many are allowed not to be constant on the point).
I don’t think I have seen this construction used anywhere apart from this article (but that need not mean much), and I always wondered how it relates to the now standard notion(s) of spectra, but then never really spent serious energy to find out.
But it is a very elegant and nice definition, and certainly Kenneth Brown behaves in this article as if it were clear that this is the notion of spectrum (though I should maybe read the last part again now, maybe he says something about it and I am just making a fool of myself here by saying that there is a mystery).
If this is right that Ken Brown’s combinatorial spectra are equivalent to the ordinary notion of spectra, then I believe thinking of them as $\mathbb{Z}$-groupoids would be entirely justfied and useful, as they directly generalize the notion of Kan complex form $\mathbb{N}$-grading to $\mathbb{Z}$-grading.
In that case, how to obtain similar combinatorial definitions for spectra stabilized with respect to non-standard types of loop space objects, as we are discussing here, I don’t know.
Posted by: Urs Schreiber on January 25, 2010 11:29 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
As mentioned in the query box at combinatorial spectrum, these were originally invented by Kan, who proved that their homotopy category is equivalent to the usual stable homotopy category. The idea is that a combinatorial spectrum is built out of a naive prespectrum of simplicial sets (that is, a sequence of based simplicial sets $X_n$ with maps $\Sigma X_n\to X_{n+1}$) by making the $k$-simplices of $X_n$ into $(k−n)$-simplices in the combinatorial spectrum. I think this was pre-model categories (or at least pre- wide use of model categories), but it seems not too unlikely that there is a model category of combinatorial spectra that is Quillen equivalent to the usual ones.
I agree that these give a very nice picture of spectra as $\mathbb{Z}$-groupoids. But I think the reason no one uses these any more is that no one has managed to come up with a smash product of them that is associative on the point-set level, so none of the modern “brave new algebra” of ring spectra, module spectra, etc. works. I’m not sure whether anyone has really tried, though; I expect that you would probably have to modify the definition to take symmetries into account, e.g. starting with symmetric prespectra instead of ordinary prespectra.
Posted by: Mike Shulman on January 25, 2010 4:08 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
As mentioned in the query box at combinatorial spectrum,
Let’s move that out of the query box! Together with the statement about smash products that you just made. That’s useful information.
Posted by: Urs Schreiber on January 25, 2010 4:24 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
does this mean there are RO(G)-groupoids?
Maybe. Although a genuine G-spectrum doesn’t just have a space for each representation, it has a G-space.
Actually, I’m never quite sure whether spectra should be $\mathbb{Z}$-groupoids or pointed $\mathbb{Z}$-groupoids. Maybe I’m confused.
Posted by: Mike Shulman on January 25, 2010 8:05 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
I think it will help not to forget that the “good” notion of spectra relies on a geometry. Stable homotopy theory is made to give sense to cohomology theories with a strong geometric flavour, the very fundamental ones being cobordism and K-theory. These two examples need, as an input, the notion of (equivariant) vector bundle (equivariant vector bundle over the point are the finite dimensional representations of the group). To such a thing, we associate a Thom space, and these are used to produce Thom spectra, which, by definition will involve this “non naive” grading. To make sense of these constructions, we need to be in a world where all the Thom spaces are invertible (not only the circle). So the general data is not only an $(\infty,1)$-topos, but a ringed $(\infty,1)$-topos $T$ (which you may think of a notion of $(\infty,1)$-topos endowed with a notion of vector bundle). We have the same issue with the homotopy of schemes: we start from (Nisnevich) sheaves on the category of smooth schemes over a given base scheme $S$, endowed with the canonical ring $\mathbf{A}^1$ (the affine line). This affine line is used to define homotopies and to define vector bundles, from which we also get algebraic Thom spectra (which we have to invert to get the stable homotopy of schemes). This is where the yoga of weights appear in algebraic geometry.
With this example in mind, it seems to me that equivariant homotopy theory has, as an input, the geometry of $G$-equivariant manifolds (i.e. the category of $G$-equivariant manifolds and the notion of $G$-equivariant vector bundles). The issue of choosing a topology on this site might also be related to the question of “naive” equivariant homotopy theory versus the “genuine” approach. For instance, when I gave the example of the homotopy theory of schemes, I mentionned the Nisnevich topology. But, of course, one might take others, like the étale topology. The interesting thing is that, when the base scheme is the spectrum of the field of real numbers, the real points functor induces a left Quillen functor from the Nisnevich version of the homotopy theory of schemes to the “genuine” $\mathbf{Z}/2\mathbf{Z}$-equivariant homotopy theory, while the same functor induces a left Quillen functor from the étale verson of the homotopy theory of schemes to the “naive” $\mathbf{Z}/2\mathbf{Z}$-equivariant homotopy theory.
So, it seems unnatural to me to wish for a canonical way to obtain a “genuine” grading if you don’t have a minimum of geometry as an input: this would sound like trying to reconstruct canonically a scheme from its underlying topological space (or topos) alone…
Posted by: Denis-Charles Cisinski on January 25, 2010 1:31 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
So the general data is not only an $(\infty,1)$-topos, but a ringed $(\infty,1)$-topos T […] endowed with the canonical ring $\mathbb{A}^1$ (the affine line).
Yes! I think that’s exactly what I have been saying, as Mike mentioned:
I said we want to build the spheres using the given geometric line object (or ring object, if you prefer that).
I like to think of the context of playing all these games with geometric paths and spheres, $\mathbb{A}^1$-homotopies etc. as being any path-structured $(\infty,1)$-topos.
Posted by: Urs Schreiber on January 25, 2010 2:09 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
I said we want to build the spheres using the given geometric line object (or ring object, if you prefer that).
I’m working on trying to understand line objects and stuff. I do like the idea that the input should involve not just a topos but a topos with some additional “geometry.” However your suggestion at the other thread doesn’t seem quite right yet to me, because I’m pretty sure that if you just pick some particular interval object $I$ (with of course will have some fixed $G$-action) then the spheres of the form $I^n/\partial I^n$ won’t give you all the representation spheres. But maybe there is some other way of arriving at them. I wish I understood homotopy of schemes better; the suggestion of a relationship there is quite intriguing.
Posted by: Mike Shulman on January 25, 2010 10:28 PM | Permalink | PGP Sig | Reply to this
### Re: Equivariant Stable Homotopy Theory
The interesting thing is that, when the base scheme is the spectrum of the field of real numbers, the real points functor induces a left Quillen functor from the Nisnevich version of the homotopy theory of schemes to the “genuine” $\mathbb{Z}/2\mathbb{Z}$-equivariant homotopy theory, while the same functor induces a left Quillen functor from the étale verson of the homotopy theory of schemes to the “naive” $\mathbb{Z}/2\mathbb{Z}$-equivariant homotopy theory.
I think I see what you mean, but I am not sure I see how this comes about. Do you have a reference for this? Is this in your motivic articles?
Posted by: Urs Schreiber on January 25, 2010 2:17 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
the real points functor induces a left Quillen functor from the Nisnevich version of the homotopy theory of schemes to the “genuine” $\mathbf{Z}/2\mathbf{Z}$-equivariant homotopy theory
Do you really mean the real points functor? I thought it was the complex points that had a $\mathbf{Z}/2\mathbf{Z}$-action coming from complex conjugation.
Posted by: Mike Shulman on January 25, 2010 10:42 PM | Permalink | PGP Sig | Reply to this
### Re: Equivariant Stable Homotopy Theory
I’ve been trying to understand what an equivariant E-infinity operad is too, for a while now. I don’t have a good explanation. But here is an observation, which is perhaps tangentially relevant.
* We usually use E-infinity operads as a means to talk about E-infinity algebras, i.e., gadgets which are “commutative monoid” type objects (at least in some homotopy sense).
Of course, you can’t talk about commutative monoids (in any sense) unless you have a symmetric monoidal category around. And E-infinity operads play a role here too: a symmetric monoidal (infinity,1)-category can be modelled as an (infinity,1)-category equipped with an action by an E-infinity operad. (I’ll assert this as though it’s obvious; I’m unaware that anyone has treated symmetric monoidal (infinity,1)-categories this way, though I think Lurie has modelled them using Gamma-spaces, which is pretty close.)
* What about G-spaces? I’m interested in the homotopy theory of G-spaces, so I’m happy to use the Elmendorff model: presheaves of spaces on the orbit category of G. This gives us a fine (infinity,1)-category, the “(infinity,1)-category of G-spaces” I’ll call it. This thing is symmetric monoidal, using product.
* But I claim there’s something more: G-spaces don’t just form an (infinity,1)-category. They form an “(infinty,1)-category internal to G-spaces”. (A vanilla (infinity,1)-category is just a one internal to spaces.)
(Of course, there is an issue of universes here, which I’ll blithely ignore!)
How does this work? I’ll give you a presheaf of (infinity,1)-categories on the orbit category of G. This assigns to a G-orbit X the (infinity,1)-category “F(X) := (G-spaces)/X”, i.e., the slice (infinity,1)-category of G-spaces over X. Given a map X –> Y, the functor F(X) –> F(Y) is the one induced by pulling back along the map.
(This construction clearly works for any category of presheaves of spaces; the (infinity,1)-category of presheaves of spaces on C is naturally *internal* in presheaves on C. In fact, (infinity,1)-toposes are always internal to themselves.)
Note that if X=G/H, then F(X) = (G-spaces)/X is equivalent to the (infinity,1)-category of H-spaces. So you can think of E as sending G/H |–> (H-spaces).
* So G-spaces form an (infinity,1)-category in G-spaces, which I’ll call F. But we also have a fancy G-equivariant operad in G-spaces. Could it be that F is an algebra over the G-equivariant operad? This would somehow mean that G-spaces are “even more” than merely symmetric monoidal, but have some kind of “equivariant symmetric monoidalness”.
I’m very sure the answer is yes. To say why would mean talking a lot about the equivariant E-infinity operad, and I’m running of out steam here. But what will happen is that the “extra” monoidalness encodes “norm constructions”. For instance, given a space X, you can (for a *finite* group G) form a G-space N^G(X), whose underlying space looks like the set of functions G –> X (and thus, like a G-indexed product of copies of X), but where the G-action comes from permuting the factors. (And more generally, given an H-space X you can form N^G_H(X).)
(A spectrum version of this norm construction shows up in the solution to the Kervaire invariant.)
Posted by: Charles Rezk on January 25, 2010 2:30 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
And $E_\infty$ operads play a role here too: a symmetric monoidal $(\infty,1)$-category can be modelled as an $(\infty,1)$-category equipped with an action by an $E_\infty$ operad. (I’ll assert this as though it’s obvious; I’m unaware that anyone has treated symmetric monoidal $(\infty,1)$-categories this way, though I think Lurie has modelled them using Gamma-spaces, which is pretty close.)
In the updated version of Jacob Lurie’s Commutative Algebra it is done this way. More developments along these lines, such as the stabilization hypothesis in these terms, is in $\mathbb{E}_k$-Algebras.
And if I recall correctly what David Ben-Zvi recounted here once, this update goes back to or was triggered by the PhD thesis of John Francis, who went ahead and defined $k$-fold monoidal $(\infty,1)$-categories as algebras in $(\infty,1)$Cat over $E_k$-operads.
My impression is this event has pushed the announced Spectral Schemes from position 6 further off, with position 6 now being taken by $\mathbb{E}_k$-Algebras . But that’s just me guessing from what I can see around me.
Concerning what $E_\infty$-operads with respect to a given $\infty$-stack $(\infty,1)$-topos are, here is, for the record, my running hypothesis which started this entire discussion with Mike in the first place:
Hypothesis. An $E_\infty$-operad in an $(\infty,1)$-topos of $\infty$-stacks on an $(\infty,1)$-site $C$ is the object in the $(\infty,1)$-category of “dendroidal stacks” on $C$ represented by the objectwise $E_\infty$(Top)-valued dendroidal stack.
More concretely, in model category theoretic terms: let $SSet Cat[C^{op}, dSet]_{proj}$ be the $SSet_{Joyal}$-enriched projective global model category of $SSet$-enriched presheaves with values in dendroidal sets with “the” model structure on dendroidal sets, and let $SSet Cat[C^{op}, dSet]_{proj}^{loc}$ be its left Bousfield localization, using Barwick’s theorem, at the set of all Čech nerve projections for covering families.
Then the $N_d(E_k(Top))$-valued presheaf in there, fibrant-cofibrantly replaced, is the $E_k(Sh_{(\infty,1)}(C))$-operad.
Well, maybe it’s less a hypothesis than a motivating question. But I think it should be true that in a lined $(\infty,1)$-sheaf $(\infty,1)$-topos with line object $I$, the geometric $k$-fold look space objects $[I^k/\partial I^k, X]$ have an action not (just) of an $E_k(Top)$-operad, but of a dendroidal sheaf, along such lines.
This discussion got started when a comment by John Greenlees made me wonder if stable equivariant homotopy theory might be a comparatively simple (hah :-) testing ground for this, due to its comparatively simple underlying site structure.
Posted by: Urs Schreiber on January 25, 2010 3:44 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Urs, what is $N_d(E_k(Top))$ suppose to represent?
Posted by: Charles Rezk on January 25, 2010 6:43 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Urs, what is $N_d(E_k(Top))$ suppose to represent?
Sorry, I should have been more explicit:
the dendroidal nerve of an ordinary topological $E_k$-operad. So just the $E_k$-operad in its incarnation as a quasi-operad.
Posted by: Urs Schreiber on January 25, 2010 8:09 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
With the equivariant homotopy category understood as the homotopy category of the $I$-homotopy localizaton of $\infty$-sheaves on $G Top$, I now finally give the promised details on my idea for the $G$-equivariant little $k$-cubes operad that I was talking about:
Let $C := sPSh(G Top)$ be the category of simplicial presheaves on $G Top$ and let $I = [0,1]$ be the standard interval object in there (with trivial $G$-action).
As described at interval object: little 1-cubes space, for every object $X \in C$ this induces a planar dendroidal object $Paths X : \Omega_{p}^{op} \to C$ whose object in $C$ over a given tree is $[I^{\vee n} , X]$ for $n$ the number of leaves of the planar tree.
In particular there is $Paths I$, which is a model for the little 1-cubes operad in $C$. The action of $Paths I$ on $Paths X$ is exibited by a map $Paths X \to Paths I$, which should be the map classified by the corresponding action $Paths I \to End [I,X]$.
For $X$ pointed write $Loops_* X \subset Paths X$ for the dendroidal subobject of paths with endpoints sitting at $*$.
We may then extend these planar dendroidal objects to dendroidal objects along the canonical functor $\Omega_{p} \to \Omega$, thus obtaining the corresponding (symmetric-)operad structures.
The dendroidal objects in $[\Omega^{op}, sPSh(G Top)]$ are equivalently objects in $PSh(G Top, dSet)$, which I regard equipped with the projective model structure of $dSet$-valued presheaves, using the standard model structure on dendroidal sets. The image of this under first Cech localization and then $I$-homotopy localization
$PSh(G Top, dSet) \stackrel{left\;Bousf.\; loc}{\to} PSh(G Top, dSet)_{loc}^I \to Ho(PSh(G Top, dSet)_{loc}^I) \simeq G Top_{loc}$
$dendroidal-homotopy-stackify : Loops_* X \mapsto \bar{(Loops_* X)}$
(which is of course the identity map on the categories underlying the model categories) I am thinking could be a good model for the $G$-equivariant action $E_1$-operad of loops in $X$.
An entirely analogous discussion should go through for any $k \leq \infty$, leading to $G$-equivariant $E_k$-action operads on $k$-fold geometric loop space objects.
That’s the idea, anyway.
Posted by: Urs Schreiber on January 30, 2010 12:13 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Reading over this, I see that I didn’t complete my thought.
G-equivariant stable homotopy theory (both “naive” and “non-naive”) are symmetric monoidal, under smash product. This structure is compatable with the unstable symmetric monoidal structure: the suspension spectrum functor is monoidal.
Question: Is equivariant stable homotopy theory “fancy equivariant symmetric monoidal”?
If yes, then equivariant stable homotopy theory should have a norm construction, compatible with the space level norm.
For instance, let G be a finite group. Then if R denotes the line (viewed as a non-equivariant space), then N^G(R), as a G-space, is the real regular G representation V.
Since the spectrum level norm should be compatible, this means that if S^1 is the suspension spectrum of the one point compactification of R, then N^G(S^1) should be the suspension spectrum of the one-point compactification of V; i.e., N^G(S^1) = S^V.
Now, the norm construction is monoidal: on spectra, N^G should take smash products of spectra to smash products of G-spectra. But when we form a stable category, we have to “formally invert” S^1. So what we discover (I think), is that if you want to have a “fancy equivariant symmetric monodial” category of G-spectra, you have to “invert” its norm as well.
In other words, I’d expect that: (i) naive equivariant stable homotopy isn’t fancy equivariant monoidal, (ii) non-naive equivariant stable homotopy is fancy equivariant monoidal. Furthermore, I’d conjecture that non-naive G-equivariant stable homotopy theory is the universal stabilization of unstable G-equivariant homotopy theory which is fancy monoidal.
(This is still a hazy idea for me, I’m entirely sure how to formalize it.)
Posted by: Charles Rezk on January 25, 2010 6:36 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Charles Rezk explained:
$G$-spaces don’t just form an $(\infty,1)$-category. They form an “$(\infty,1)$-category internal to $G$-spaces”. (A vanilla $(\infty,1)$-category is just a one internal to spaces.)
[…]
How does this work? I’ll give you a presheaf of $(\infinity,1)$-categories on the orbit category of $G$. This assigns to a $G$-orbit $X$ the $(\infty,1)$-category “$F(X) := (G-spaces)/X$”, i.e., the slice $(\infty,1)$-category of $G$-spaces over $X$. […]
Okay, I am following. These presheaves of over-$(\infty,1)$-categories generally seem to be a pretty fundamental thing. When objectwise stabilized, it seems one should/ or may think of them as the “quasicoherent $\infty$-stack of modules” canonically determined by the given context. (Along the lines we recalled at here). I was wondering what I should think of them before stabilization. It seems you are giving one perspective here on that.
But we also have a fancy $G$-equivariant operad in $G$-spaces. Could it be that $F$ is an algebra over the $G$-equivariant operad? […] I’m very sure the answer is yes. To say why would mean talking a lot about the equivariant $E_\infty$ operad, and I’m running of out steam here.
Maybe when you have regained some steam, you can talk about this a bit more? I’d be interested in what you have to say here.
I’d conjecture that non-naive $G$-equivariant stable homotopy theory is the universal stabilization of unstable G-equivariant homotopy theory which is fancy monoidal.
Posted by: Urs Schreiber on January 25, 2010 9:03 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
In fact, $(\infty,1)$-toposes are always internal to themselves.
Is that by any chance a different way of talking about the codomain fibration, a.k.a. the self-indexing, which is the canonical way of talking about a topos $E$ as “a category (= fibration = indexed category) in the world of $E$”?
To say why would mean talking a lot about the equivariant $E_\infty$ operad
Can you at least say what you mean by “the” equivariant $E_\infty$ operad? To use a definite article you apparently have a particular one in mind.
Posted by: Mike Shulman on January 25, 2010 10:38 PM | Permalink | PGP Sig | Reply to this
### Re: Equivariant Stable Homotopy Theory
Oops! In my earlier posts, I sometimes wrote “equivariant operad” or “G-equivariant operad”, but I really always meant “E-infinity G-equivariant operad”; that is, the gadget that Peter May introduced for understanding equivariant infinite loop spaces.
Maybe I’m talking about the “codomain fibration”; I’m not sure, it’s a language I don’t use, so I dunno. The assertion that “(infinity,1)-toposes are internal to themselves” is pretty strong. It is *not* true that classical Grothendieck toposes are internal to themselves, in the sense I have in mind; so I can’t show you a “classical” analogue.
Here’s the precise claim about internality. Let E be an (infinity,1)-topos. Let F be the functor E^op –> (infinity,1)-cat defined on objects by
F(X) := ( the slice category E/X ),
and on morphisms by “pullback in E”. Then F converts homotopy colimits into homotopy limits. (This is roughly a reformulation of what Lurie calls the “descent” property of an (infinity,1)-topos.)
Posted by: Charles Rezk on January 25, 2010 11:17 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Then F converts homotopy colimits into homotopy limits. (This is roughly a reformulation of what Lurie calls the “descent” property of an (infinity,1)-topos.)
Classical Grothendieck toposes do satisfy their own version of descent, of course. The corresponding statement is that if $E$ is a Grothendieck topos, then the similarly defined functor $E^{op} \to Cat$ is a stack for the canonical topology. This is a statement about taking certain (2-)colimits in $E$ (regarded as a locally discrete 2-category) to (2-)limits in $Cat$. It doesn’t take all colimits in $E$ to limits in $Cat$; I expect that things are more convenient in the $\infty$-world because $\infty+1=\infty$. But I’m pretty sure this is the classical analogue.
Posted by: Mike Shulman on January 26, 2010 12:07 AM | Permalink | PGP Sig | Reply to this
### Re: Equivariant Stable Homotopy Theory
Yes, I agree with that. But I wouldn’t quite call it being “internal”. If E is a classical topos, then E isn’t a category object of E (or a bigger universe version of E), but rather a category object in some larger beast (stacks on E).
Posted by: Charles Rezk on January 26, 2010 5:39 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
If E is a classical topos, then E isn’t a category object of E (or a bigger universe version of E), but rather a category object in some larger beast (stacks on E).
Well, a stack is already a category; you don’t need to talk about category objects in stacks. (Unless by “stack” you prefer to mean “stack of groupoids”, in which case maybe I should say “2-sheaf” instead of “stack”.) And while it’s true that a stack is not the same as a category object in sheaves, as I mentioned down here every stack can be strictified into an equivalent one which is a category object in sheaves. One place this is proven is in Steve Awodey’s thesis, chapter V.
Granted, this isn’t as convenient or natural as not having to strictify. But I think it’s fair to say, based on this, that the $(\infty,1)$-situation is not conceptually different, just more convenient and natural.
Posted by: Mike Shulman on January 27, 2010 2:03 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Well, I think one problem is that I shouldn’t have chosen the word “internal”.
I really have something very different in mind. Namely, that $(\infty,1)$-toposes have (LARGE) “object classifiers”.
That is, if $E$ is an $(\infty,1)$-topos, there is an OBJECT $\Omega$ in $E$, such that (homotopy classes of) maps $X\to \Omega$ correspond to (isomorphism classes of) objects in the slice category $E/X$. Note that $\Omega$ is characterized up to weak equivalence by this property; it’s a completely natural thing.
I want to think of $\Omega$ as part of the information which describes an internal model of $E$ inside itself. (in particular, it’s the part that describes the objects of $E$.)
An (classical) $1$-topos $E$ doesn’t have an object classifier. It is only allowed to have a subobject classifier, i.e., something that classifies objects in $E/X$ which are monomorphisms. Of course, there is something that you can talk about that does the job of classifying objects of $E$ (a stack!). But such a stack is not an object of $E$; at best, it’s an object of a certain $2$-topos which contains $E$.
Maybe this isn’t really a big deal. But to me, it feels like a big qualitative difference between $1$-toposes and $\infty$-toposes.
Posted by: Charles Rezk on January 27, 2010 5:23 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Hmm. Ignoring size issues so blatantly is difficult for me to swallow. The object classifier of an $(\infty,1)$-topos $E$ is not an object of $E$ itself, it is an object of some version $E'$ of $E$ in a higher universe. That means it isn’t determined up to equivalence by $E$ alone, but depends also on the choice of universe boundary.
Alternately, you can think of the object classifier as an object of $E$, but which only classifies the class of “small” objects in $E$. But the subobject classifier of a 1-topos can also be thought of as classifying the “small” objects in $E$, where now “small” means relative to the regular cardinal $2$. Many 1-toposes also have object classifiers that classify some larger class of “small objects”, such as are studied in algebraic set theory. Of course these classifiers are not objects of the topos itself, but rather internal categories in it; even the subobject classifier is an internal poset. If you only want to classify the core of the slice category, then you can get away with internal groupoids, and since the core of a poset is discrete, the subobject classifier can classify isomorphism classes of subobjects on the nose without being considered as an internal groupoid.
The difference in the $\infty$-case is that any internal groupoid in an $(\infty,1)$-category is representable by an ordinary object, since $\infty+1=\infty$. (If we go all the way to $(\infty,\infty)$-toposes, whatever those might be, then any internal category might to be representable by an ordinary object, so that “object classifiers” could classify the whole slice category, not just its core.) So this is certainly a qualitative difference, and I agree that it is important and useful. I’m just saying that there are analogues of it in the 1-topos world, even if they aren’t as nice.
Posted by: Mike Shulman on January 27, 2010 3:32 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Ignoring size issues so blatantly is difficult for me to swallow. The object classifier of an (∞,1)-topos $E$ is not an object of $E$ itself,
In TheBook the size issues are carefully taken care of: there is for each regular cardinal a classifying object for all “$\kappa$-compact morphisms”.
This is indicated and referenced at object classifier.
Maybe that helps here?
Posted by: Urs Schreiber on January 27, 2010 5:02 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
In TheBook the size issues are carefully taken care of: there is for each regular cardinal a classifying object for all “κ-compact morphisms”.
Yes, that’s basically what I meant by
you can think of the object classifier as an object of E, but which only classifies the class of “small” objects in E.
Posted by: Mike Shulman on January 28, 2010 12:12 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Charles,
Here is how to make your formulas come out pretty printed:
when composing a comment, there is a little pulldown menu on top of your edit pane, that says “text filter”. Choose the text filter called
Markdown with itex to MathML .
When you do that, then everything you type inside dollar signs as in a LaTeX document will be displayed as math, as in a LateX document.
That’s what the “instiki to MathML” takes care of. The “Markdown” part takes care that things included inside stars come out boldface, things inside underscores come out italicized, and lines preceded by a greater-than sign come out as indented quotes.
More on $G$-Spaces from me tomorrow. Have to catch some sleep now…
Posted by: Urs Schreiber on January 26, 2010 1:58 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
I remarked cryptically:
Is that by any chance a different way of talking about the codomain fibration, a.k.a. the self-indexing
To expand on that a bit for those listening who may not be familiar with it. For an ordinary 1-topos $S$, if you want to do mathematics “in the world of $S$,” then you can do “small mathematics,” working with individual objects like groups and rings, just internal to $S$, but if you want to work with large categories (say, the category of rings in the world of $S$) you have an issue. Of course you can talk about the category of internal ring objects in $S$, but when doing category theory we also need to talk about set-indexed families of objects of categories: for instance when we say that a category is cocomplete we mean that any set-indexed family of objects has a coproduct. But in the world of $S$, the objects of $S$ play the role of “sets,” so we need a notion of “a family of ring objects indexed by an object $X\in S$”. The obvious notion is an internal ring object in $S/X$.
In general, for every naturally defined large category in the world of $S$ there are corresponding categories defined in each slice categories of $S$, and via pullback these categories fit together into a pseudofunctor $S^{op}\to CAT$ (aka an “indexed category”), or equivalently a fibration over $S$. Such a pseudofunctor/fibration $C$ comes with a notion of “$X$-indexed object of $C$” for each $X\in S$, namely the category $C(X)$, or equivalently the fiber of the fibration over $X$. You can then “do category theory” with fibrations/pseudofunctors in the world of $S$ just as you do with ordinary large categories. The fibration representing $S$ itself, i.e. “the category of sets in the world of $S$,” is the codomain fibration, or the pseudofunctor sending $X$ to the slice category $S/X$ (and maps to pullback functors).
Of course, if you want to work with small categories, then you have the other option of just working with internal categories in $S$. In some ways this is “easier,” e.g. you can use the internal logic and talk as if the objects of $S$ were sets. (The 2-internal logic of the category of fibrations aims to make this possible for fibrations as well.) Every internal category represents an indexed category in the same way that every object represents a presheaf, so “small categories give rise to large categories.”
And if you don’t like working with fibrations at all, i.e. with large categories as large categories, you can also do some universe trickery to make them into small ones. For instance, if your topos $S$ is the category of sheaves on some site $D$, then you can consider the topos $S'$ of sheaves on the same site but with values in larger sets (i.e. you assume a universe, so that $S$ be the category of small-set-valued sheaves, and $S'$ the category of large-set-valued sheaves). Then internal categories in $S'$ can be treated as “large categories in the world of $S$.” For instance, I think this is what Street did in his paper “The petit topos of globular sets.” An indexed category, i.e. pseudofunctor $S^{op}\to CAT$ can be turned into an internal category in $S'$ by first restricting it to a pseudofunctor $D^{op}\to CAT$, then “strictifying” it to a strict functor $D^{op}\to CAT$, which is the same as an internal category in the category of functors $D^{op}\to SET$; which in most cases will be objectwise a sheaf, hence an internal category in $S'$.
So when Charles said:
I’ll give you a presheaf of $(\infty,1)$-categories on the orbit category of $G$. This assigns to a $G$-orbit $X$ the $(\infty,1)$-category “$F(X) \coloneqq (G-spaces)/X$”, i.e., the slice $(\infty,1)$-category of $G$-spaces over $X$. Given a map $X \to Y$, the functor $F(X) \to F(Y)$ is the one induced by pulling back along the map.
It sounds very much to me like the above construction: start with the self-indexing $X\mapsto (G-spaces)/X$ of the category $S = G-spaces$, restrict it to the site, i.e. the orbit category $D = \mathcal{O}_G$, then regard that as an internal category in the category $G-SPACES$ one universe higher up.
Posted by: Mike Shulman on January 25, 2010 11:01 PM | Permalink | PGP Sig | Reply to this
### Re: Equivariant Stable Homotopy Theory
I want to try to describe the way in which the $(\infty,1)$-category of $G$-spaces is supposed to be monoidal over the “$G$-equivariant $E_\infty$ operad”. I learned about equivariant operads from Rekha Santhanam, and much of my understanding of how this monoidal structure works comes out of conversations with David Gepner.
In what follows, “$G$-spaces” always means the $(\infty,1)$-category (=homotopy theory) of $G$-spaces, in the strong sense; that is, weak equivalences are maps which induces weak equivalences on all fixed point spaces.
Fix a finite group $G$. A “fancy G-equivariant $E_\infty$ operad” $U$ is an operad in the symmetric monoidal category of $G$-spaces (under product), characterized by the property: for each $n$ and each subgroup $K\subseteq G\times \Sigma_n$, $U(n)^{K}$ is either (i) contractible, or (ii) empty, depending on whether (i) $K\cap (\{e\}\times \Sigma_n)$ is the trivial subgroup of $G\times \Sigma_n$ or (ii) isn’t. It’s probably better to think of the case (i) subgroups as being the ones which look like graphs of homomorphisms $H\to \Sigma_n$, where $H\subseteq G$ is a subgroup.
(A side remark: although $U$ is an operad from the point of view of category theory, it isn’t really properly an operad in the $(\infty,1)$-category theory sense. The object $U(n)$ is not merely a “an object of $G$-spaces equipped with an action of $\Sigma_n$”, as you would expect, but rather a more sophisticated thing, “an object of $G\times \Sigma_n$-spaces”.)
I want to use Elmendorf’s theorem freely when talking about $G$-spaces. So if I have a $G$-space $X$, I get a presheaf on the orbit category $O=O_G$ of $G$, by $G/H \mapsto X^H$. Conversely, I can reconstruct $X$ (up to $G$-equivariant weak equivalence) from its associated presheaf.
The notable feature of $U(n)$ is that the $H$-fixed points of the quotient space $U(n)/\Sigma_n$ is the classifying space of the category of functors $H\to \Sigma_n$ (for subgroups $H$ of $G$).
I need to think about a generalization of this. If $X$ is a $G$-space, consider $(U(n)\times X^n)/\Sigma_n$. The $H$-fixed points of this map to the $H$-fixed points of $U(n)/\Sigma_n$, so I get a “bundle”
(1)$[(U(n)\times X^n)/\Sigma_n]^H \to (U(n)/\Sigma_n)^H.$
A point in the “base” represents a functor $\rho\colon H\to \Sigma_n$. The “fiber” over $\rho$ is a space $\mathrm{Hom}_G( S_\rho, X)$ of $G$-equivariant maps. What’s $S_\rho$? That’s a $G$-space which sits in a fibration $S_\rho\to G/H$, with fiber equal to $\{1,\dots,n\}$, and with $G$ action determined by the homomorphism $\rho \colon H\to\Sigma_n$. It’s “determined” in the following way: the group $H$ has to act on the fiber over the coset $eH$, and it acts according to the homomorphism $\rho$.
Some examples: If $\rho(H)=\{e\}$, then $S_\rho = \{1,\dots,n\}\times G/H$, and so $\mathrm{Hom}_G(S_\rho, X)= (X^H)^n$. If $\rho\colon H\to \Sigma_n$ describes the left action of $H$ on itself, then $S_\rho= G$, and so $\mathrm{Hom}_G(S_\rho, X)=X$.
Now let $X$ be an algebra for $U$. The data of such an algebra are maps $(U(n)\times X^n)/\Sigma_n \to X$ of $G$-spaces. I want to understand these in terms of fixed points: $[(U(n)\times X^n)/\Sigma_n]^H \to X^H$.
I’ll continue in another post …
Posted by: Charles Rezk on January 26, 2010 7:18 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
… and here’s the continuation.
Given a $G$-space $X$, let $E(X)$ denote the $(\infty,1)$-category of $G$-spaces over $X$. It’s easiest for me to think of $E(X)$ as being a complete Segal space; so $E(X)$ is a simplicial space, and its bottom layer $E_0(X)$ is a space which corresponds to the $(\infty,1)$-groupoid of $G$-spaces over $X$. Note that $E$ is contravariant in $X$; the functor $E(Y)\to E(X)$ is the one induced by pulling back along $X\to Y$.
Since $G$-spaces are an $\infty$-topos, we have descent: in particular, for any two $G$-spaces $X$ and $Y$ the map $E(X\amalg Y)\to E(X)\times E(Y)$ is an equivalence. More generally, the value of $E$ at an arbitrary $G$-space $X$ can be recovered from its values at $G$-orbits.
Note further that $E(G/H)$ is equivalent to the $(\infty,1)$-category of $H$-spaces.
For the sake of keeping things simple, I’m just going to talk about the $0$-space $E_0(X)$ in what follows; everything should work with $E_m(X)$ too.
Restricting $E_0$ to the orbit category $O_G$ gives a presheaf of spaces on $O_G$. Using Elmendorf, I’ll convert this into a $G$-space which I’ll call $F$. Thus $F^H$ is weakly equivalent to $E_0(G/H)$, and thus “points” of $F^H$ correspond to objects in “$H$-spaces”, or equivalently “$G$-spaces over $G/H$”. More generally, for any $G$-space $X$, the space $\mathrm{Hom}_G(X,F)$ is weakly equivalent to $E_0(X)$, so a “point” in this mapping space corresponds to an object in “$G$-spaces over $X$”.
What I want to claim is that $F$ is an algebra for the operad $U$. I won’t prove this; I’ll merely suggest how to think about the structure maps $(U(n)\times F^n)/\Sigma_n \to F$. Taking $H$-fixed points, these are maps
(1)$\psi_n^H: [(U(n)\times F^n)/\Sigma_n]^H \to F^H.$
Recall that the left hand side is a bundle over $[U(n)/\Sigma_n]^H$, that points in $[U(n)/\Sigma_n]^H$ correspond to functors $\rho \colon H\to \Sigma_n$, and that the fiber of the bundle over $\rho$ looks like $\mathrm{Hom}_G(S_\rho, F) \approx E_0(S_\rho)$.
Putting this all together, the map $\psi_n^H$ appears to take as input (i) a homomorphism $\rho \colon H\to \Sigma_n$, and (ii) an object $X$ in “$G$-spaces over $S_\rho$”, and gives as output something I’ll call $Y$, which should be a “$G$-space over $G/H$”. So I need to tell you how to get a $Y$ from $(\rho,X)$.
Remember that there is a map $f:S_\rho \to G/H$. Given $X\to S_\rho$, I want to let $Y$ be the “direct image” of $X$ along $f$. That is, $Y = f_!X$, where $f_!\colon (G\text{-spaces})/S_\rho \to (G\text{-spaces})/(G/H)$ is the right adjoint to the functor defined by pullback along $f$.
Examples:
If $\rho(H)=\{e\}$, then an object $X$ in “$G$-spaces over $S_\rho$” amounts to an $n$-tuple $X_1,\dots,X_n$ of $H$-spaces. The direct image $f_!X$ corresponds to the $H$-space which is the product $X_1\times \cdots \times X_n$ in $H$-spaces.
If $\rho$ is the left action of $H$ on itself, then an object $X$ in “$G$-spaces over $S_\rho$” amounts to a space $Z$ (with an action by the trivial group $\{e\}$). The direct image $f_!X$ corresponds to the $H$-space $N^H(Z) := \prod_H Z$, where the $H$-action is the one that permutes the factors.
Posted by: Charles Rezk on January 26, 2010 7:29 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
I’m not quite ready to tackle ringed $(\infty,1)$-topoi yet, but I want to comment on this. I don’t think that adjunction can be true, because on the right you have something looking kind of like naive $G$-spectra (is the stabilization of a presheaf $(\infty,1)$-category the same as the category of presheaves of spectra?) and on the left you have genuine $G$-spectra. But a genuine $G$-spectrum is not just a particular kind of naive one; it has more data.
I think the way I would attack this question is to go back to the proof of the stable Giraud theorem and see where the domain $(\infty,1)$-category of the presheaf category that’s getting localized comes from. Then see what comes out when we apply that to the genuine equivariant stable category.
Posted by: Mike Shulman on January 25, 2010 3:07 AM | Permalink | PGP Sig | Reply to this
### Re: Equivariant Stable Homotopy Theory
(is the stabilization of a presheaf (∞,1)-category the same as the category of presheaves of spectra?)
Yes, that observation is part of the proof of the stable Giraud theorem, it is recalled as the first of the displayed equations in our section on stable Giraud.
I’d think this follows pretty directly from observing that loop space and suspension objects of presheaves are done objectwise, being pullbacks and pushouts. So also the stabilization works objectwise.
But a genuine $G$-spectrum is not just a particular kind of naive one; it has more data.
Hm, right. But can you help me see this more concretely: how would my functor from left to right (which I just sketched, I am not fully sure here) fail to be full and faithful?
Posted by: Urs Schreiber on January 25, 2010 3:54 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
how would my functor from left to right … fail to be full and faithful?
How would it succeed to be full and faithful? (-: I don’t see any reason to believe that it would be.
Posted by: Mike Shulman on January 25, 2010 10:40 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
How would it succeed to be full and faithful? (-: I don’t see any reason to believe that it would be.
But the stable Giraud theorem combined with Schwede-Schipley guarantees that there is some functor that embeds the $(\infty,1)$-categor presented by $Sp Cat(O^{st}_G, Sp)$ reflectively into that presented by $sSet Cat(O_G, Sp)$.
I don’t claim to see how this works in detail. But given that we know it must work somehow, I was just making an obvious guess what that functor could be. I currently don’t see how one would go about proving much at all about this “obvious” functor. But then, I haven’t really stared at it that long, either.
So if you can see concretely why that obvious guess can’t work, I’d be interested. Some other guess then must work!
Posted by: Urs Schreiber on January 26, 2010 12:08 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
But the stable Giraud theorem combined with Schwede-Schipley guarantees that there is some functor that embeds the $(\infty,1)$-category presented by $SpCat(O_G^{st},Sp)$ reflectively into that presented by $sSetCat(O_G,Sp)$.
The stable Giraud theorem as stated on the nLab doesn’t say what the domain $(\infty,1)$-category $E$ of the presheaf category involved is. Why does it have to be $O_G$ in this case?
Posted by: Mike Shulman on January 27, 2010 3:27 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
I have a naive question: why only study actions of groups? What I mean is a group action is a special kind of a groupoid, an action groupoid. What about a more general groupoids, such as Lie groupoids or their topological analogues (I am not sure what that would mean concretely) or, even more generally, some sort of a geometric stack?
Stable homotopy theory of $G$-action groupoids, $G$ a fixed group, sounds very restrictive to me. I am sure there are historical reasons for it, but are there any mathematical reasons for being so focused on actions of groups at the exclusions of other groupoids?
Posted by: Eugene Lerman on January 25, 2010 6:43 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
why only study actions of groups? What I mean is a group action is a special kind of a groupoid, an action groupoid. What about a more general groupoids, such as Lie groupoids or their topological analogues (I am not sure what that would mean concretely) or, even more generally, some sort of a geometric stack?
The notion of “naive” $G$-equivartiant spectrum (the obvious definition that comes to mind) has a straightforward generalization to groupoids and beyond.
But historically people at some point decided that naive spectra are too naive, and that something more non-naive is needed. Unfortunately the non-naive version is a bit harder to interpret conceptually. It currently seems to justify its existence mainly due to the fact that it leads to interesting structures.
What we are discussing here is about how one might understand why “non-naive” genuine $G$-spectra are conceptually natural. Charles Rezk proposed that they can be understood as arising from a universal stabilization subject to a certain constraint on a monoidal structure. If that’s the right answer, I am still hoping that we can understand this as an analog of stabilization at geometric “Tate spheres”.
In any case, I gather it is not quite clear yet. When it becomes clear, when we have a good conceptual understanding on “genuine” $G$-spectra I suppose it will be just as straightforward to generalize to groupoid-equivariance as it is for the naive version.
So while I can’t help at the moment with groupoid equivariant spectra, I can point out something about this:
What about a more general groupoids, such as Lie groupoids or their topological analogues (I am not sure what that would mean concretely) or, even more generally, some sort of a geometric stack?
So here is a text that discusses the generalization of genuine $G$-equivariant spectra to the case that $G$ is generalized to an $\infty$-stack of $\infty$-groups modeled on $E_\infty$-rings (how’s that?):
I have maybe not fully absorbed this article yet, but I think I understand what’s going on. It looks to me – but I’d be glad to be corrected – that while this does enormously generalize the classical theory of genuine $G$-equivariant spectra, it does not really shed much light on the conceptual questions we were trying to understand. But maybe I am wrong about that.
By the way, concerning our other discussion about quasicoherent $\infty$-stacks of modules, there is some interesting stuff about this in the last two sections.
Posted by: Urs Schreiber on January 29, 2010 1:55 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Thanks.
Shortly after I posted my question here I did wander down the hall a few meters and asked the Burt Guillou more or less the same question. He wasn’t sure.
I suppose I could try walking down one flight of stairs and asking Charles Rezk, but that would be silly. After all, posting silly questions on this blog is a lot more fun.
Posted by: Eugene Lerman on January 30, 2010 1:41 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Over lunch today I kindly received helpful further input. When mentioning my suggestion that genuine $G$-equivariant spectra might be obtained from a localization of simplicial presheaves not just on the orbit category but on the category of all (sufficiently nice) $G$-spaces, I was being pointed to work by Andrew Blumberg, saying that pretty much this is achieved there.
I still have to look at this in more detail, but it seems that the article in question is
Andrew Blumberg, Continuous functors as a model for the equivariant stable homotopy category (arXiv:math.AT/0505512)
Quoting from the introduction:
we […] study […] equivariant diagram spectra indexed on the category $\mathcal{W}_G$ of based $G$–spaces homeomorphic to finite $G$–CW–complexes for a compact Lie group $G$. Using the machinery of Mandell–May–Schwede–Shipley, we show that there is a stable model structure on this category of diagram spectra which admits a monoidal Quillen equivalence to the category of orthogonal $G$–spectra. We construct a second “absolute” stable model structure which is Quillen equivalent to the stable model structure.
The full statement is Theorem 1.3, page 7
The “relative” model structure on the category of something close to simplicial (co)presheaves on the category of (nice) $G$-spaces is in def 2.3, page 10: the weak equivalences are those morphisms of (co)presheaves that become weak equivalences when evaluated on a geometric sphere, i.e. on an $S^V$ for $V$ a real linear $G$-representation.
This is then stabilized to a stable model category by taking the weak equivalences to be those that induce isos on $G$-spectrum himotopy groups on associated $G$-spectra.
I am not sure yet if this can be taken as the answer to what I had in mind. But it seems to be getting closer.
The article also mentions, by the way, a $G$-equivariant version of the May recognition principle, that we talked about with Peter May on the $n$Forum here. It is attributed there to
Costenoble and Warner, Fixed set systems of equivariant infinite loop spaces (JSTOR)
But I haven’t looked at this yet, maybe I am misunderstanding what is being attributed to whom here.
Posted by: Urs Schreiber on January 26, 2010 2:37 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Blumberg’s paper looks nice. I’d say what he does is something like this: he identifies equivariant spectra as being functors on (nice) G-spaces which make each G-orbit G/H “look like a dualizable object”. (That’s my interpretation of his Theorem 1.2, anyway.) It’s already known that “inverting representation spheres S^V” makes “G/H dualizable”, so Blumberg seems to have proved the reverse.
That’s really only a new observation for general compact Lie groups G (I think). For finite groups G, equivariant stable homotopy makes G/H look self-dual; this fact about equivariant stable homotopy for finite groups is basic thing that makes the theory of “G-equivariant Gamma-spaces” (as developed by May, Shimakawa, …) work.
Yes, the paper by Costenoble and Waner is the one which talks about “G-operads”, and proves a recognition principle for equivarant loop spaces (for finite G.)
Posted by: Charles Rezk on January 26, 2010 3:54 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
That’s really only a new observation for general compact Lie groups $G$ (I think). For finite groups $G$, equivariant stable homotopy makes $G/H$ look self-dual; this fact about equivariant stable homotopy for finite groups is basic thing that makes the theory of “$G$-equivariant Gamma-spaces” (as developed by May, Shimakawa, …) work.
i see. I have to run now and need to look into this later in more detail.
But can you rephrase this as something like: the stable $G$-equivariant homtopy category is that of the localization of the projective model $Func(G Space_{nice}, sSet)$ at the set of morphisms $xyz$?
Posted by: Urs Schreiber on January 26, 2010 5:59 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
I think your rephrasing is correct, where $GSpace_{nice}$ means something like “$G$-spaces which are homeomorphic to finite $G$-CW-complexes” . Blumberg calls this category $W_G$, and for him its an enriched category (enriched over $G$-spaces); his functors are enriched functors, and the target category should be $G$-spaces. (I gather that this can always be rephrased without using $G$-enirched stuff; I think it’s equivalent to use the version of $W_G$ enriched over ordinary spaces (no $G$-action), and ordinary continuous functors to $G$-spaces. But don’t quote me on that …)
I don’t think this claim is very new, though quite possibly it hasn’t appeared in print before. I expect it was already known to experts that if you take functors from $\mathcal{W}_G$ to $G$-spaces, and you localize at a set of maps whose role is to “force representation spheres $S^V$ to be invertible”, then you get the stable homotopy category. (But I’m not an expert, so don’t quote me on that either.)
Mandell and May do something very like this in their paper on orthogonal equivariant spectra, except they use a different category for $GSpace_{nice}$ (a non-full subcategory of $W_G$ constructed from representation spheres).
Blumberg is working with $W_G$, and is proposing a different set of maps to localize with respect to, apparently ones which “make $G/H$ dualizable”.
(I think people have avoided dealing with $W_G$, because they believe that it doesn’t lead to a good model for commutive ring spectra. At least that’s what Mandell, May, Schwede, and Shipley seem to think happens in the non-equivariant case.)
Posted by: Charles Rezk on January 26, 2010 8:33 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
I think your rephrasing is correct,
Thanks, but I am afraid I still don’t understand what I am trying to understand.
One aspect that doesn’t seem to fit what I am hoping to see is that Blumberg and his forerunners us copresheaves/diagrams also with values in $G$-spaces, instead of with values in plain spaces/simplicial sets.
But maybe I should say more explicitly what it is that I am trying to understand.
Namely, I am trying to understand the $G$-equivariant case as the generalization of the following non-equivariant situation (which possibly I also don’t fully understand…):
$\array{ Top \simeq Sh_{(\infty,1)}(*) &\stackrel{localize at \{\mathbb{R}\times X \to X\}}{\leftarrow} & Sh_{(\infty,1)}(Top) \\ {}^{\mathllap{invert categorical looping}}\downarrow && \downarrow^{\mathrlap{invert geometric looping}} \\ Sp &\stackrel{localize at \{\mathbb{R}\times X \to X\}}{\leftarrow}& ??? }$
Here on the top right we have $\infty$-stacks on $Top$. Localizing that at $\mathbb{R}$-homtopies yields plain $Top$ (as in Dugger’s notes). Inverting categorical looping in there yields spectra.
But as long as we are still in $Sh_{(\infty,1)}(Top)$ we can and should actually invert geometric looping, in terms of the geometric loop built from $\mathbb{R}$. Because that’s what we know is the right thing to do in other cases like motivic cohomology and, as we seem to have said, should also be the step that explains the “genuine” $G$-spectra.
My trouble starts probably with me not being quite sure how to think of the thing in the bottom right corner obtained this way. But I would expect that localizing it at $\mathbb{R}$-homotopies now turns this into the category of spectra.
Maybe I am wrong about this. With $I$ a line object in an $(\infty,1)$-topos, does localizing at $I$-homotopies commute with inverting $I$-looping?
In any case, my idea would be that we should try to understand the right column of this diagram for the equivariant case, because that seems to be where the geometric looping is natural, whereas after localization it looks like an ad-hoc construction.
So I am looking for a way to make sense o a diagram of the form
$\array{ G Space \simeq PSh_{(\infty,1)}(O_G^{op}) &\stackrel{localize at ??something??}{\leftarrow} & P Sh_{(\infty,1)}(G Space^{op}) \\ {}^{\mathllap{}}\downarrow && \downarrow^{\mathrlap{invert geometric looping}} \\ G Sp &\stackrel{localize at ??something??}{\leftarrow}& ??? }$
All diagrams here in big imaginary quotation marks. This is not a claim, but a question.
The thing is that in the $G$-equivariant literature we see the left side of these diagrams discussed. But from $\mathbb{A}^1$-homotopy reasoning etc. we know we want to be working on the right, and we also said that this would explain the non-categorical looping taken on the left, because that should really be the image of the $\mathbb{A}^1$-type geometric looping on the right.
So that’s the picture I am imagining, and of which i am trying to see if or how it is true.
Posted by: Urs Schreiber on January 27, 2010 2:06 AM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
I wrote above (pretty imprecisely, but still):
$G Space \simeq PSh_{(\infty,1)}(O_G^{op}) \stackrel{localize at ??something??}{\leftarrow} PSh_{(\infty,1)}(G Space^{op})$
I see now that the answer (to the question marks) is on p. 50 of
• Morel, Voevodsky, $A^1$-homotopy theory of schemes (pdf).
Take the site $G Top$ to be the category of $G$-spaces that admit $G$-equivariant open covers (def 3.3.1) with the Grothendieck topology given by covering maps $Y \to X$ that admit $G$-equivariant splittings over such $G$-equivariant covers.
Let $Sh_{(\infty,1)}(G Top) := (sSh(G Top)_{loc})^{\circ}$ be the corresponding category of $\infty$-sheaves.
This has the obvious interval object $I$, the unit interval equipped with the trivial $G$-action (I suppose we could just as well take the full real line).
Then do the left Bousfield localization of $sSh(G Top)_{loc}$ at the collection of morphisms of the form $\{X \stackrel{Id \times 0}{\to} X \times I\}$. The homotopy category of the resulting model category is the standard $G$-equivariant homotopy category $G Top_{loc}$.
Posted by: Urs Schreiber on January 29, 2010 8:25 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
Given what I said above, I figured that if there is an answer along such lines, then it should have shown up in equivariant motivic cohomology.
I probably need to search around a bit more. This here is the first hit on a famous search engine:
• Ben Williams, Equivariant Motivic Cohomology (pdf)
If I see correctly, the definition of equivariant motivic cohomology on p. 5 is Borel equivariant motivic cohomology, namely the cohomology of the action groupoid.
Hm…
Posted by: Urs Schreiber on January 27, 2010 10:58 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
I wrote:
the definition of equivariant motivic cohomology on p. 5 is Borel equivariant motivic cohomology
The right definition of genuine motivic $G$-spectra is probably that in section 4.5, p. 8 of
• B. Guillou, A short note on models for equivariant homotopy theory (pdf)
I am glad I came back to this one. After I cited this at equivariant homotopy theory I had forgotten to read it to the end!
This looks like it is getting closer to providing the answer that I am after.
Posted by: Urs Schreiber on January 28, 2010 3:54 PM | Permalink | Reply to this
### Re: Equivariant Stable Homotopy Theory
In case anyone is interested, here are some notes from an introductory talk about equivariant stable homotopy theory, given by Anna Marie Bohmann at Chicago last week.
Posted by: Mike Shulman on January 26, 2010 9:06 PM | Permalink | Reply to this
Post a New Comment
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2015-05-06 09:28:03
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https://web2.0calc.com/questions/please-help-someone_1
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+0
0
331
1
1. Find the constant term in the expansion of $$\Big(2z - \frac{1}{\sqrt{z}}\Big)^9.$$
2. Find the constant term in the expansion of $$\Big(x^2+\frac{1}{x}\Big)^4$$
Aug 2, 2020
#1
+22251
+1
I'll do number 2 as an example:
(x + 1/x)4 = 1·(x)4·(1/x)0 + 4·(x)3·(1/x)1 + 6·(x)2·(1/x)2 + 4·(x)1·(1/x)3 + 1·(x)0·(1/x)4
= 1·x4·1 + 4x3·(1/x) + 6x2·(1/x2) + 4x·(1/x3) + 1·1·(1/x4)
= x4 + 4x2 + 6 + 4/x2 + 1/x4
To find the coefficients, either use Pascal's Triangle or the formula for combinations.
Aug 2, 2020
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2021-12-05 18:09:20
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https://huggingface.co/cross-encoder/quora-distilroberta-base
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# Cross-Encoder for Quora Duplicate Questions Detection
This model was trained using SentenceTransformers Cross-Encoder class.
## Training Data
This model was trained on the Quora Duplicate Questions dataset. The model will predict a score between 0 and 1 how likely the two given questions are duplicates.
Note: The model is not suitable to estimate the similarity of questions, e.g. the two questions "How to learn Java" and "How to learn Python" will result in a rahter low score, as these are not duplicates.
## Usage and Performance
Pre-trained models can be used like this:
from sentence_transformers import CrossEncoder
model = CrossEncoder('model_name')
scores = model.predict([('Question 1', 'Question 2'), ('Question 3', 'Question 4')])
You can use this model also without sentence_transformers and by just using Transformers AutoModel class
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2021-02-28 19:50:38
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https://pendulumedu.com/qotd/english-direct-indirect-speech-1-september-2020
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Question of The Day01-09-2020
Choose the most appropriate option to change the narration (direct/indirect) of the given sentence.
Arnab said to me, "I entrusted all my duties to the subordinate without knowing his skill and ability”.
Correct Answer : a ) Arnab told me that he had entrusted all his duties to the subordinate without knowing his skill and ability.
Explanation :
The given sentence is in direct speech and we need to change it to indirect.
In indirect speech,
• Simple Past tense (entrusted) changes to Past perfect (had entrusted).
• First person pronoun, i.e., ‘I’ changes to ‘she’ with respect to the subject, i.e. ‘Arnab’.
• Also, ‘my’ changes to ‘her’.
• Reporting verb, i.e., ‘Said to’ changes to ‘told’.
The complement part (to the subordinate without knowing his skill and ability) of the sentence remains intact.
The indirect form is- ‘Arnab told me that he had entrusted all his duties to the subordinate without knowing his skill and ability.’
Hence, the correct answer is option A.
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2021-05-07 20:17:13
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http://archive.numdam.org/item/ASNSP_2002_5_1_1_153_0/
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Continuability in time of smooth solutions of strong-nonlinear nondiagonal parabolic systems
Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, Serie 5, Volume 1 (2002) no. 1, p. 153-167
A class of quasilinear parabolic systems with quadratic nonlinearities in the gradient is considered. It is assumed that the elliptic operator of a system has variational structure. In the multidimensional case, the behavior of solutions of the Cauchy-Dirichlet problem smooth on a time interval $\left[0,T\right)$ is studied. Smooth extendibility of the solution up to $t=T$ is proved, provided that “normilized local energies” of the solution are uniformly bounded on $\left[0,T\right)$. For the case where $\left[0,T\right)$ determines the maximal interval of existence of a smooth solution,the Hausdorff measure of a singular set at the moment $t=T$ is estimated.
Classification: 35K50, 35K45, 35K60
@article{ASNSP_2002_5_1_1_153_0,
author = {Arkhipova, Arina},
title = {Continuability in time of smooth solutions of strong-nonlinear nondiagonal parabolic systems},
journal = {Annali della Scuola Normale Superiore di Pisa - Classe di Scienze},
publisher = {Scuola normale superiore},
volume = {Ser. 5, 1},
number = {1},
year = {2002},
pages = {153-167},
zbl = {1049.35093},
mrnumber = {1994805},
language = {en},
url = {http://www.numdam.org/item/ASNSP_2002_5_1_1_153_0}
}
Arkhipova, Arina. Continuability in time of smooth solutions of strong-nonlinear nondiagonal parabolic systems. Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, Serie 5, Volume 1 (2002) no. 1, pp. 153-167. http://www.numdam.org/item/ASNSP_2002_5_1_1_153_0/
[1] H. Amann, Quasilinear parabolic systems under nonlinear boundary conditions, Arch. Rational Mech. Anal. 92 (1986), no. 2, 153-192. | MR 816618 | Zbl 0596.35061
[2] A. Arkhipova, Global solvability of the Cauchy-Dirichlet problem for nondiagonal parabolic systems with variational structure in the case of two spatial variables, J. Math. Sci. 92 (1998), no. 6, 4231-4255. | MR 1668390 | Zbl 0953.35059
[3] A. Arkhipova, Local and global in time solvability of the Cauchy-Dirichlet problem to a class of nonlinear nondiagonal parabolic systems, St. Petersburg Math J. 11 (2000), no. 6, 81-119. | MR 1746069 | Zbl 0973.35095
[4] A. Arkhipova, Cauchy-Neumann problem to a class of nondiagonal parabolic systems with quadratic growth nonlinearities. I. Local and global solvability results, Comment. Math. Univ. Carolin. 41, (2000), no. 4, 693-718. | MR 1800172 | Zbl 1046.35047
[5] A. Arkhipova, Cauchy-Neumann problem to a class of nondiagonal parabolic systems with quadratic growth nonlinearities. II. Continuability of smooth solutions, to appear in Comment. Math. Univ. Carolin. (2000). | MR 1800172 | Zbl 1046.35047
[6] S. Campanato, Equazioni paraboliche del secondo ordine e spazi ${ℒ}^{2,\delta }\left(\Omega ,\delta \right)$, Ann. Mat. Pura Appl., Ser. 4 73 (1966), 55-102. | MR 213737 | Zbl 0144.14101
[7] M. Giaquinta, “Multiple integrals in the calculus of variations and nonlinear elliptic systems”, Princeton, NJ, 1983. | MR 717034 | Zbl 0516.49003
[8] M. Giaquinta - G. Modica, Local existence for quasilinear parabolic systems under non-linear boundary conditions, Ann. Mat. Pura Appl. Ser. 4, $\mathbf{149}$ (1987), 41-59. | MR 932775 | Zbl 0655.35049
[9] O. A. Ladyzhenskaja - V. A. Solonnikov - N. N. Uraltseva, “Linear and quasilinear equations of parabolic type”, Amer. Math. Soc. Providence, RI, 1968.
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2020-11-26 01:51:37
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https://pressbooks.bccampus.ca/geoglabmanualv2/chapter/coastal-geomorphology/
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# Lab 23: Coastal Geomorphology
Todd Redding; Stuart MacKinnon; and Fes de Scally
Canada has the longest shoreline of any country in the world, but estimates of how long the shoreline actually is will vary greatly depending on how it is measured. In British Columbia, for instance, the straight-line distance between Vancouver and Prince Rupert is approximately 800 km, but the length of the coastline between the two cities is about 41,000 km when traced on a relatively detailed map.
Waves are the dominant energy source on exposed, open-ocean coasts, and when wave energy is expended on the shoreline, there is the potential for rapid and significant geomorphic change. An understanding of coastal geomorphic processes and landforms is important not only for pure scientific interests but also to avoid potentially disastrous consequences of the human occupation and use of coasts.
Learning Objectives
After completion of this lab, you will be able to:
• Use Google Earth to make geographical measurements.
• Understand the relationships between wave length, period, and velocity.
• Understand the change in wave characteristics as they approach the shore.
• Identify common coastal landforms on satellite images.
• Infer likely patterns of coastal sediment transport on satellite images.
• Understand the contributing factors to coastal sediment budgets and the effects of changes on beach properties.
• Become aware of the shifting nature of shorelines due to long-term changes in sea level.
## The Littoral Zone
Fundamental to coastal geomorphology is an understanding of the littoral zone. The littoral zone includes the coast, beach, nearshore environment, and a part of the offshore environment. This lab concerns itself with the processes that occur in this zone, and the resulting landforms created by erosion, transport and deposition.
## Properties of Waves
The source of energy for most coastal erosion and sediment transport is wave action. Waves are distortions of the still water surface that contain (a) potential energy by virtue of water molecules being displace above and below the still water line; and (b) kinetic energy by virtue of water molecules being in motion. The faster the wind and the longer the fetch (or distance of open water across which the wind blows and waves travel), the larger the waves. Large waves have much greater energy than smaller waves, in fact, wave energy is proportional to the wave height squared.
It is important to realize that waves in the open ocean (referred to as deep water waves) do not move the water forward, but rather the water molecules move in circular orbits. If you have ever floated offshore in a small boat or inner tube you will know this to be true; a moving wave will pass beneath the boat and cause you to rise and fall but you will not move any appreciable horizontal distance forward or backward. The situation is different close to shore where there is net movement of the water in the direction of wave propagation, but where there can also be currents (e.g., longshore, rips). Waves are described using basic measurements such as wave height, wavelength (Figure 23.1), and wave period.
Wavelength is defined as the horizontal distance from wave crest to wave crest (or trough to trough), while height is the vertical distance between the trough and crest. Both values are commonly measured in metres (m). Open ocean waves are modelled as a sinusoidal curve, with symmetry above and below the still water level. The time taken for successive crests to pass a fixed point is called the wave period, measured in units of time such as seconds (s). One of the interesting aspects of wave motion is that the period remains essentially constant despite changes in the wave form as it enters shallow water.
The velocity or phase speed of the wave, V, can be calculated if the wavelength, L, and the wave period, P, are known (Equation 23.1):
Equation 23.1.
$V = \dfrac{L}{P}$
Given that wavelength is commonly measured in metres (m) and wave period in seconds (s), the units of wave velocity are metres per second (m/s).
For example, let us assume that a wave has a wavelength of 16 m and a wave period of 4 s. The velocity (V) of the wave may be calculated using Equation 23.1 as:
$V (m/s) = \dfrac{L (m)}{P (s)}$
$V (m/s) = \dfrac{16m}{4s} = 4\text{ m/s}$
This relationship can be re-arranged to solve for wavelength or wave period. A good way to think about how the equation can be re-arranged is by examining the units of the output and making sure they make physical sense to you. For more detailed explanations, please refer to SERC Carleton’s How do I calculate rates? Calculating changes through time in the geosciences and How do I isolate x (or P or T …) in a formula? Rearranging equations to solve for a given variable.
Long period waves (referred to as long waves or swell waves) travel faster than short, locally-generated sea waves. As a consequence, waves have a tendency to sort themselves out as they move away from the storm centre that generates wave motion in the open ocean. Long swell waves that have travelled hundreds of kilometres may have periods of up to 15 seconds. Smaller waves have periods of only a few seconds.
## Wave Refraction
In much the same manner as light is refracted (bent) through a prism, waves are subject to refraction or bending of the wave crests as they approach the shore. Whenever waves approach a shoreline at an oblique angle, the presence of the sea floor turns the waves to become more parallel to the shore. The reason is that the offshore portions of the wave crest are not influenced by the bottom and are free to travel at their optimal speed (i.e., their deep-water velocity). In contrast, the onshore portions of the wave crest are impeded by the presence of the bottom and therefore are forced to slow down.
On an indented coast the situation is more complex, but Figure 23.2 below shows what is expected to happen. The wave crests in deep water approach the shore perpendicular to the general trend of the coast. The wave crests approaching the headlands begin to be affected by the sea floor first, when they are just under a kilometre from the shore. These portions of the wave crests slow down, shorten in wavelength, and increase in height. The same crest approaching the bay continues unimpeded (because water depth in front of the bay is deeper) and so moves ahead of the wave segment at the headland. At position 1 on Figure 23.2 Segments A and B are in deep water and are unchanged, but by the time they have reached position 3, A has slowed down and shortened its wavelength. It therefore lags behind B which is still unchanged. By the time the wave reaches position 4 on Figure 23.2, A is about to break on the headland while B is advancing more slowly into the bay. The end result is that the wave crest is bent progressively by refraction so as to conform to the bathymetric contours and ultimately break parallel to the shoreline.
The convergent pattern of the orthogonals show the direction of movement of the wave crest (the solid lines drawn at right angles to the wave crests with arrow tips). The closer spacing of the orthogonals near the headlands show how the wave energy in segment A is concentrated onto the headland. In the bay, wave height is reduced since the energy of segment B is spread out across a greater length of shoreline (orthogonals are further apart). Therefore, headlands tend to be sites of erosion while bays tend to be sites of deposition. The tendency is for waves to create a smooth, linear coastline, but this often does not happen because of differences in the strength of rocks (resistance to erosion) on headlands versus embayments.
## Erosion, Transportation and Deposition Along Coasts
A number of mechanical and chemical effects produce erosion of rocky shorelines by waves. The types of erosional landforms that result depend on the geology of the coastline, the nature of wave attack, the tidal regime, and long-term Sea Level Change. Examples of erosional landforms include wave-cut notches, marine cliffs, sea caves, arches and sea stacks.
Transportation by waves and currents is necessary to move rock particles eroded from one part of a coastline to a place of deposition elsewhere. One of the most important transport mechanisms results from an oblique angle of wave attack relative to the shoreline orientation. The upward movement of water onto the beach (swash) occurs at an oblique angle. However, the return of water (backwash) is at right angles to the beach, resulting in the net movement of beach material laterally. This movement is known as beach drift. Waves approaching the shoreline at an oblique angle also force longshore currents, especially in troughs between nearshore sand bars. The combination of longshore currents with the never-ending cycle of swash and backwash is called longshore drift or littoral drift and can be observed on all sandy beaches.
In addition, tidal currents along coasts can be effective in moving eroded material. While incoming and outgoing tides produce currents in opposite directions on a daily basis, the current in one direction is usually stronger than in the other, resulting in a net one-way transport of sediment. Longshore currents and tidal currents in combination with waves will determine the net direction of sediment transport and ultimately where sediment is eroded and deposited along the shoreline.
Many kinds of depositional landforms are possible along coasts depending on the geological configuration of the coastline, direction of sediment transport, character of the waves, and shape and steepness of the underwater slope offshore. Some common depositional forms are spits, bayhead beaches, barrier beaches or islands, tombolos, hooked spits, salients, bayside beaches, lagoons, and sandy marshes (Figure 23.3).
When specifying the direction of coastal sediment transport, we specify the direction that sediment is moving towards. For example, if sediment is moving from Los Angeles, USA, to Vancouver, Canada, then the direction it is moving is north. Cardinal directions are the points on the compass used to specify general direction, and include north (N), south (S), east (E) and west (W), and their derivatives. For further details, refer to Lab 15 Directions.
## Coastal Sediment Budgets
Coastal sediment budgets are used to understand changes in beaches and coastlines over time. When constructing a sediment budget, a geomorphologist will try to quantify (measure or estimate) the inputs, outputs and changes in storage within a coastal compartment referred to as a littoral cell. The concept of a littoral cell is to a coastal geomorphologist what a watershed or drainage basin is to a hydrologist – it is an easily identified system of study with natural boundaries.
For a beach system, the sediment budget is generally concerned with sand-sized particles (0.2 – 2 millimetre (mm) diameter) and larger (> 2 mm diameter). The inputs and outputs of a beach sediment budget may consist of some or all of the components listed in Table 23.1. It is important to note that there are a number of locations/processes that can act as sources or sinks of sediment, depending on the local geography and time period of interest. This explain why coastal dunes are listed as inputs and outputs: the dune may contribute sediment to the beach (input) or be built up by sand transported from the beach (output). Inputs can come in the form of point sources (e.g., river) or line sources (e.g., cliff erosion).
Inputs Outputs Littoral drift in Littoral drift out Rivers Offshore transport Coastal dunes Coastal dunes Inlets/lagoons Inlets/lagoons Estuaries Estuaries Beach nourishment Dredging Cliff erosion Sand mining Shellfish and sea grass beds – Coral reefs –
Inputs and outputs are commonly expressed as volumetric rates of sediment movement (e.g., cubic metres per year, m3/y).
When inputs are greater than outputs over some period of time, a beach will grow (prograde, storage is positive). When outputs are greater than inputs, the beach will shrink (erode, storage is negative). What will happen when inputs equal outputs? That’s correct, the beach will stay the same size. The net sediment budget is equivalent to the change in storage and is (Equation 23.2):
Equation 23.2
ΔStorage = Inputs – Outputs
For example, let us assume that each year Bondi Beach receives 70,000 m3/y of sand from littoral drift and 5,000 m3/y from coastal dunes, and loses 79,000 m3/y of sand to littoral drift. The net sediment budget is (Equation 23.2):
ΔStorage = Inputs − Outputs
ΔStorage (m3/y) = ( 70,000 m3/y + 5,000 m3/y) − ( 79,000 m3/y)
ΔStorage (m3/y) = 75,000 m3/y − 79,000 m3/y
ΔStorage (m3/y) = − 4,000 m3/y
Because the change in storage is negative, this calculation indicates that Bondi Beach is shrinking at a rate of 4,000 m3/y. Don’t worry, this calculation is fictional. The beach will still be there when you finally make it to check out the location of Bondi Rescue. When solving this equation, remember your order of operations. First add all the components inside the brackets and then subtract the outputs.
Note that the units we are using are a rate, just as velocity is a rate. This means that this relationship can be rearranged to solve for volume of sand or time, depending on which variables are known.
## Sea Level Change
Sea level is not constant through time or space. When sea level falls, the coastline moves in an oceanward direction. When sea level rises, the coastline moves in a landward direction. Figure 23.4 shows changes in global sea level over the past 150,000 years.
A number of processes contribute to changes in sea level both globally and locally:
• Growth and decay of glaciers and ice sheets affect the amount of liquid water.
• Land surface elevations change through time due to plate tectonics.
• Local land surface elevation changes due to melting of overlying glaciers (isostatic adjustment).
• Rise or fall of water level due to heating and cooling of water that leads to expansion and contraction of water in terms of total volume in the oceans.
Bathymetric maps show the topography (depth) of the land surface below the water level in the same way that a topographic map shows the topography (elevation) of hills and mountains. On bathymetric maps, contour lines connect points of equal depth below present sea level.
# Lab Exercises
This lab includes a number of exercises to help build your understanding of coastal processes. In this lab you will:
• Calculate wave characteristics based on different types of input data.
• Identify the direction and source of coastal sediment transport from satellite imagery.
• Identify coastal landforms from satellite imagery.
• Calculate coastal sediment budgets.
• Interpret sea level change from graphs and depict spatially on a map.
You will need access to Google Earth (Web) and Google Earth Engine Timelapse. If you wish to complete mapping exercises by hand, you will also need the ability to scan or photograph an image and create a PDF. These exercises will take 2-2.5 hours to complete. Submit your work following the instructions of your professor or lab instructor.
EX1: Properties of Waves
1. Calculate the following wave characteristics. Assume the waves are in deep water and include the correct units in your answer. Show your work.
1. P = 12 s, V = 5.3 m/s, L =
2. P = 24 s, L = 45 m, V =
3. L = 4.5 m, V = 1.6 m/s, P =
2. Sitting on a beach, you count 20 waves break in a three-minute period and estimate the wavelength to be 46 feet. What is the wave velocity (in m/s)? Note that 1 m = 3.281 feet. Provide your answer correct to one decimal place. Show your work.
EX2: Coastal Sediment Transport
Figure 23.5 shows James Island, which is located off the east coast of Vancouver Island, BC.
1. What is the cardinal direction of coastal sediment transport on James Island (Figure 23.5)? What evidence leads you to that conclusion?
2. What is the source of the sediment being moved along the coast? Open Google Earth (Web) and enter the coordinates provided in the caption of Figure 23.5 to have a closer look around James Island.
3. View the Google Earth Engine Timelapse for the spit at 34.63095°, -76.47419°. What is the cardinal direction of longshore drift?
EX3: Coastal Landforms
Figure 23.6 shows a number of coastal landforms near Tofino, BC.
1. Identify the coastal landforms labelled A, B, and C on Figure 23.6.
2. Explain how feature A likely formed.
3. What is the dominant direction of incoming waves on Figure 23.6? What is your evidence for this?
4. Locate the following features on Google Earth (Web) by copying and pasting the coordinates or name into the Search tool (magnifying glass icon, second from top in list on left of screen). Answer the associated questions.
1. 21°19’59”N, 158°07’25”W.
1. Are these natural or artificial bays? How can you tell?
2. What geomorphic role do the little islands play at the mouth of the bays?
2. 21°16’24”N, 157°49’29”W.
1. What are these features?
2. What purpose do they serve?
3. The islands of Bora Bora and Tahaa, French Polynesia.
1. What type of coastal landform are these features?
2. Which island is older? Explain your evidence for this.
3. What will they eventually become?
EX4: Beach Sediment Budgets
A sediment budget was developed for Sandy Beach in the 1960s. Figure 23.7 is a schematic of the area and includes the boundary of the littoral cell (dashed line) and inputs and outputs.
Volumetric rates of inputs and outputs are provided in Table 23.2.
Table 23.2. Sandy Beach sediment budget.
Inputs Sand volume (m3/y)
Littoral drift 5,000
Sandy River 123,000
Cliff erosion 42,000
Outputs Sand volume (m3/y)
Littoral drift 145,000
Offshore transport 18,000
1. Based on the data in Table 23.2, calculate the net sediment budget in the 1960s. Show your work.
2. During the 1960s, was the beach growing, shrinking or staying the same?
In the 1970s and 1980s a number of multi-millionaires built mansions on the top of the cliffs above the beach. The owners were worried about their homes falling into the ocean due to erosion of the cliffs, and constructed revetments along the cliff base. As a consequence, cliff erosion was reduced to 20,000 m3/y. Erosion wasn’t completely stopped because a section of the cliff was managed by a Coastal Conservancy that did not want to build revetments. Thus, certain portions of the cliff remained natural and unprotected.
1. It is now the year 2021, and the effects of the decrease in cliff erosion are starting to become evident.
1. Re-calculate the net sediment budget considering the change in cliff erosion. Show your work.
2. Is the beach now growing, shrinking, or staying the same?
3. Will erosion or deposition be uniformly distributed along the entire length of shoreline backed by the cliff? Explain.
Changes in the sediment budget affect the width of the beach (beach distance from cliffs to sea level) over a period of years to decades. A coastal engineer has predicted that a positive sediment budget of 5,000 m3/y would add 0.1 m to the width of the beach, while a loss of 5,000 m3/y results in a decrease of 0.1 m to the width of the beach.
1. Assume that the beach had an average width of 35 m prior to the construction of the revetments. How long did it/will it take for the beach to recede (erode) back to the base of the cliff? Hint: Use the net sediment budget you calculated in Q12a.
EX5: Sea Level Change
1. Use Figure 23.4 to determine how sea level has changed through time. Approximately how high was the sea level (compared to today’s sea level) in metres:
1. 18,000 years ago?
2. 40,000 years ago?
3. 90,000 years ago?
4. 140,000 years ago?
2. Download and print out or open the Shoreline Bathymetric Map provided in Worksheets. On this map:
1. Draw the coastline where it would have been at each time listed in Q14. Draw each coastline in a different colour or with a different line symbol.
2. Fill in the key to indicate which colour and/or line symbol matches which time.
Scan or photograph your completed map (in colour) if you drew the coastlines by hand. Save it to a known location on your computer.
3. The two stars on the Shoreline Bathymetric Map show the locations of ancient village sites where Coast Salish people lived when sea level was lower. The remains of intricate fish traps are further evidence that people at these sites lived at an ancient coastline and survived on a diet of fish. Given your knowledge of the changing sea level from the graph above, estimate the age of these archeological sites.
Reflection Questions
1. What happens when a wave approaches shore and why?
2. Considering coastal sediment transport (erosion and deposition), describe the ideal location to build your dream beach front home.
3. What is the best method of protecting built structures along the shoreline from wave erosion and why?
4. What are the seabed and wave characteristics that result in good waves for surfing and why?
# Worksheets
Back to EX6
Shoreline Bathymetric Map
# References
Davidson-Arnott, R., Bauer, B., & Houser, C. (2019). Introduction to Coastal Processes and Geomorphology (2nd ed.). Cambridge University Press, New York, NY.
Shackleton, N.J. (1987). Oxygen isotopes, ice volume and sea level. Quaternary Science Reviews, 6(3-4), 183-190.
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2023-03-24 09:26:58
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https://www.hackmath.net/en/math-problem/3124
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# Granite cube
What is the weight in kg granite cube with an edge of 0.5 m if 1dm3 of granite weight 2600 g?
Result
m = 325 kg
#### Solution:
$a=0.5 \ m=0.5 \cdot \ 10 \ dm=5 \ dm \ \\ V=a^3=5^3=125 \ \text{cm}^3 \ \\ h=2600 \ g=2600 / 1000 \ kg=2.6 \ kg \ \\ \ \\ m=h \cdot \ V=2.6 \cdot \ 125=325 \ \text{kg}$
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2020-03-28 07:59:07
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https://socratic.org/questions/how-do-i-graph-f-x-x-5x-2-on-a-ti-83
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# How do I solve f(x)=x/(5x+2) on a TI-83?
Nov 3, 2015
Why use a calculator?
#### Explanation:
The function will equal zero when the numerator equals zero.
Answer: x = 0
hope that helped
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2020-07-07 00:03:48
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https://www.cut-the-knot.org/m/Arithmetic/2017inSquare.shtml
|
# A Farewell Problem for the Year 2017
### Solution 1
For an odd $n,$ say $n=2m-1,$ $\displaystyle\sum_{k=1}^m(2k-1)=m^2.$ Thus,
$\displaystyle 1+3+\ldots+2015=\sum_{k=1}^{1008}(2k-1)=1008^2\\ \displaystyle 1+3+\ldots+2015+2017=1008^2+2017=1009^2.$
Thus taking $n=1008^2,$ $n(n+2017)=(1008\cdot 1009)^2.$
There are no other solutions. Indeed, if $n$ is not a perfect square, it is bound to have a factor that is shared with $n+2017$ and, hence, with $2017.$ But the latter is prime so that the only possibility is $n=s^2\cdot 2017,$ implying $n(n+2017)=(2017s)^2(s^2+1)$ that could not be perfect square unless $s=0.$
It follows that both $n$ and $n+2017$ are bound to be perfect squares, say $n+a^2,$ $n+2017=b^2.$ From this $(b-a)(b+a)=2017$ and, subsequently, $b=1009,$ $a=1008.$
### Solution 2
Let $n(n+2017)=k^2$ ($k\in\mathbb{N}$). Hence, $(k-n)(k+n)=2017n$. $2017$ is prime. Thus, $2017|(k-n)$ or/and $2017|(k+n)$. Suppose $k-n=2017p$ ($p\in\mathbb{N})$). Then, \begin{equation*} k+n=\frac{n}{p} \Rightarrow k=n\left(\frac{1-p}{p}\right). \end{equation*} Thus, $p\leq 1$ and this option is not possible. Alternatively, let $k+n=2017p$ ($p\in\mathbb{N}$). Thus, \begin{eqnarray*} k+n&=&2017p \\ k-n&=&\frac{n}{p}. \end{eqnarray*} For $k$ to be integral, $p|n$. Eliminating, $k$, \begin{equation*} n=\frac{2017p^2}{2p+1}. \end{equation*} Suppose $(2p+1)|2017$. The smallest $p$ satisfying this corresponds to $2p+1=2017$ or $p=1008$. Note, in this case $\boldsymbol{n}=p^2=1008^2=\boldsymbol{1016064}$ and $k=p+n=1008+1016064=1017072$. The remaining option is that $(2p+1) \hspace{-4pt}\not|\hspace{2pt} 2017$ and $(2p+1)|p^2$. Suppose $p^2=(2p+1)m$ ($m\in\mathbb{N}$). The only positive root for this quadratic equation in $p$ is \begin{equation*} p=m+\sqrt{m(m+1)}. \end{equation*} However, the product of two consecutive natural numbers cannot be a square as $m^2\lt m(m+1)\lt (m+1)^2$. Thus, this option has no integral solution for $p$ and thus $n$.
### Acknowledgment
The problem with $2013$ instead of $2017$ was offered at the 2013 Pan African Mathematics Olympiad. It was published at the Crux Methematicorum as problem CC151 (2016), with two solutions. Solution 1 combines elements of both; Solution 2 is by Amit Itagi.
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2019-03-23 00:16:10
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https://science.lesueur.nz/12phy/as91172/fission-and-fusion/
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## Akoranga 9 Ngā Whāinga Ako 🔗
• Understand the difference between nuclear fission and fusion
• Use $E=mc^{2}$
• Use $P=\frac{E}{t}$
## Recall 🔗
There are three types of nuclear reactions
1. Radioactive decay ($\alpha$, $\beta$, $\gamma$)
2. Nuclear fission
3. Nuclear fusion
## Fission 🔗
### Question 🔗
If the protons are positively charged, why don’t they repel each other and the nucleus break apart?
### Electrostatic Force vs. Nuclear Force 🔗
• When nucleons are extremely close together, a very strong force called the Nuclear Force comes into effect.
• It is the nuclear force that holds the nucleons together and is stronger than the electrostatic force/repulsion between the protons.
• Some isotopes, such as polonium-211, have unnaturally large numbers of nucleons in their nuclei.
• In these isotopes, the nuclear force holding the nucleons together is not enough to overcome the electrostatic repulsion between them (because there are too many) – making them unstable or radioactive
### Absorption of a Nucleon 🔗
• If a fast moving nucleon can get extremely close to a nucleus of an atom, then the nuclear force will come into effect and it will be incorporated into the nucleus.
• This is easier to do with a neutron because neutrons have no charge so they do not experience any electrostatic repulsion (i.e. will not be repelled by protons).
### Pātai: Write 2x nuclear equations to express this reaction 🔗
#### Whakatika 🔗
\begin{aligned} & {}^{1}{0}n + {}^{235}{92}U \rightarrow {}^{236}{92}U \
& {}^{236}
{92}U \rightarrow {}^{92}{36}Kr + {}^{141}{56}Ba + 3{}^{1}_{0}n \end{aligned}
Extra pātai: What do you think the three neutrons produced do?
## Nuclear Fusion 🔗
• Unlike fission, in fusion we combine two whole nuclei together, instead of just adding a neutron
• Pātai: Write a nuclear equation to express this reaction!
#### Whakatika 🔗
\begin{aligned} & {}^{3}{1}H + {}^{2}{1}H \rightarrow {}^{4}{2}He + {}^{1}{0}n \end{aligned}
### Producing Energy: Mass Balance 🔗
\begin{aligned} & {}^{3}{1}H + {}^{2}{1}H \rightarrow {}^{4}{2}He + {}^{1}{0}n \end{aligned}
• Tritium (${}^{3}_{1}H$) $= 5.00641 \times 10^{-27} kg$
• Deuterium (${}^{2}_{1}H$) $= 3.3436 \times 10^{-27} kg$
• Helium (${}^{4}_{2}He$) $= 6.64466 \times 10^{-27} kg$
• Neutron (${}^{1}_{0}n$) $= 1.67493 \times 10^{-27} kg$
1. Calculate the total mass of the reactants.
2. Calculate the total mass of the products.
## Theory of Relativity, $E = mc^{2}$ 🔗
• Mass is a form of energy
• Mass can turn into energy and energy can turn into mass (in nuclear reactions).
During the fusion process, the total mass is not conserved because some of the mass of the fusing nuclei is converted to energy.
\begin{aligned} & E = mc^{2} \end{aligned}
• $E =$ energy
• $m =$ mass
• $c =$ speed of light ($3\times10^{8}ms^{-1}$)
### Pātai: Smoke Detector 🔗
• Americium-241 undergoes $\alpha$ decay.
• 1. Write down the nuclear equation for the alpha decay of a nucleus of Americium-241.
• 2. Calculate the quantity of energy released when a nucleus of Americium-241 decays.
### Pātai: Nuclear Fission 🔗
• In a fission reaction, an U-235 nucleus combines with a neutron, then splits into a Ba-141 nucleus, a Kr-92 nucleus and three neutrons. It also emits gamma radiation.
• Calculate the amount of energy released
### Pātai: Nuclear Reactor 🔗
• A nuclear power plant in the US can produce one gigawatt of power using U-235.
• Calculate the mass of U-235 required each year to run the plant.
## Tips for the Assessment 🔗
2. Watch out for the numbers NOT in standard forms (e.g. $\times 10^{-27}$)
5. $E = mc^{2}$ – SHOW CLEAR WORKING
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2021-03-04 20:57:39
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https://www.tutorialspoint.com/extracting-a-war-file-in-linux
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# Extracting a WAR File in Linux
## Introduction
WAR (Web ARchive) files are a type of archive file used to package web applications into a single file. They are similar to Java ARchive (JAR) files and are typically used to deploy web applications in a Java environment. In this article, we will learn how to extract a WAR file on Linux using the command line.
A WAR file is essentially a ZIP file that contains all the files needed for a web application, including HTML, CSS, JavaScript, and Java files. Checking out a WAR file allows you to access the individual files it contains and make changes to the web application.
## Prerequisites
Before you begin, make sure you have the following prerequisites −
• A Linux machine with command line or shell access.
• The “jar” command, which is part of the java-1.8.0-openjdk package. This package is usually installed by default on most Linux distributions. If it's not installed on your system, you can install it using your distribution's package manager. For example, on a CentOS system, you can install it using the following command −
$sudo yum install java-1.8.0-openjdk • The unzip command, which is part of the unzip package. This package is also usually installed by default on most Linux distributions. If it's not installed on your system, you can install it using your distribution's package manager. For example, on a CentOS system, you can install it using the following command − $ sudo yum install unzip
## Extract a WAR file using the jar command
The jar command is a utility that is part of the Java Development Kit (JDK) and is used to manipulate Java Archive (JAR) files. It can also be used to extract WAR files, as they are simply a variant of JAR files.
To extract a WAR file using the jar command, follow these steps −
• Open a terminal window and navigate to the directory where the WAR file is located. You can use the “cd” command to change to the desired directory. For instance −
$cd /path/to/file.war • Extract the WAR file using the “xf” or “xvf” option of the jar command, as shown below − $ jar xf mywebapp.war
• This will extract the contents of the WAR file into a directory with the same name as the WAR file, without the “.war” extension. For example, if your WAR file is named “mywebapp.war”, the extracted files will be placed in a directory named mywebapp.
• To extract the WAR file to a specific directory, use the C option of the jar command, followed by the directory path, as shown below −
$jar xf mywebapp.war -C /path/to/extraction/directory This will extract the contents of the WAR file to the specified folder or directory. ## Extract a WAR file using the unzip command The unzip command is a utility used to extract files from ZIP archives. It can also be used to extract WAR files, as they are simply ZIP files with a different file extension. To extract WAR file using the unzip command, follow these steps − • Open a terminal window and navigate to the directory where the WAR file is located. You can use the “cd” command to change to the desired directory. For instance − $ cd /path/to/file.war
• Extract the WAR file using the unzip command, as shown below −
$unzip mywebapp.war • This will extract the contents of the WAR file into a directory with the same name as the WAR file, without the .war extension. For example, if your WAR file is named mywebapp.war, the extracted files will be placed in a directory named mywebapp. • To extract the WAR file to a specific directory, use the -d option of the unzip command, followed by the directory path, as shown below − $ unzip mywebapp.war -d /path/to/extraction/directory
• This will extract the contents of the WAR file into the specified directory.
## Compare jar and unzip to extract WAR files
Now that we have learned how to extract WAR files using the jar and unzip commands, let's compare the two methods to see which is more suitable for extracting WAR files.
• Ease of use − The jar and unzip commands are easy to use and have a simple syntax. However, the unzip command is perhaps a little easier to use, as it doesn't require any additional options to unzip WAR files.
• Compatibility − The jar command is part of the JDK and is therefore compatible with all Java-based WAR files. On the other hand, the unzip command is a general purpose utility which can extract any ZIP file including WAR files.
• Performance − The jar command is optimized for working with JAR files and may be faster at extracting WAR files than the unzip command. However, the performance difference is likely to be negligible for most use cases.
In general, the jar and unzip commands are effective tools for extracting WAR files on Linux. You may prefer the jar command if you are working with Java-based WAR files and want to use a tool designed specifically for manipulating JAR files. On the other hand, the unzip command is a more general purpose tool which can extract any ZIP file including WAR files.
## Conclusion
In this article, we learned how to extract a WAR file on Linux using the jar and unzip commands. Both commands can be used to extract WAR files, although the jar command may be preferred for extracting Java-based WAR files. By following the steps outlined in this article, you should be able to extract WAR files easily to your Linux machine.
As a final note, keep in mind that WAR files are typically used to deploy web applications and can contain sensitive information such as database credentials and configuration files. Therefore, it is important to handle them with care and remove them in a safe place.
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2023-03-29 02:11:32
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https://www.voiceonline.com/16-seabus-sailing-cancellations-for-friday/
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# 16 SeaBus sailing cancellations for Friday
TRANSLINK announced on Thursday that due to union job action it is expecting the cancellation of 16 SeaBus sailings on Friday:
• The 6:17 am, 6:47 am, 4:10 pm, 6:47 pm, 7:17 pm, 7:47 pm, 8:17 pm, and 8:47 pm sailings from Lonsdale Quay will be cancelled
• The 6:31 am, 7 am, 4:25 pm, 7:01 pm, 7:31 pm, 8:01 pm, 8:31 pm, 9:01 pm sailings from Waterfront will be cancelled.
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2019-11-22 15:54:44
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http://hyperspy.org/hyperspyUI/troubleshooting.html
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# Troubleshooting¶
“Something went wrong!”
If you find yourself unable to complete some of the steps in this guide, or if the application doesn’t do something it should or is supposed to, please consider letting us know. The best way is maybe to go to the HyperSpy gitter chat first to discuss the issue, but if there’s no help forthcoming, please submit an issue on the HyperSpyUI Github page.
## Known issues¶
For a list of previously reported issues, see the HyperSpyUI Github page. At the time of release, no major issues are known, but the following can be worth noticing:
Closed signals are not always cleared from memory
This is dependent on Pythons garbadge collector, and might be caused by remaining references to the signal held by parts of the UI or matplotlib figures. A program restart might be necessary to solve this, but using Close All (signals) might also help clearing out any remaining references.
Signals can get stuck in “limbo”
If a signal gets stuck in an invalid state, it might be problematic to close it fully. Similar to the above problem, this might be fixed by the Close All (signals) action, or in the last resort a program restart might be necessary.
## Error output¶
To help diagnose problems, it would be helpful to include the application log. Depending on how you launch the program, there are different ways of obtaining this log output. The easiest is if the program is launched from a terminal, in which case the log is piped directly to its console. If you’ve not launched it with a terminal, the log is attempted written to the folder of hyperspyui. If that location is noe writable, the log file is attempted written to the users home folder (OS dependent). In either case, the log file is named hyperspyui.log.
Note
For a complete error output to the above sources, the Console completion type settings option should be disabled.
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2017-11-23 13:06:07
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https://chemistry.stackexchange.com/questions/15163/is-sucrose-made-of-2-glucose-molecules-or-1-glucose-and-1-fructose
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# Is sucrose made of 2 glucose molecules or 1 glucose and 1 fructose? [duplicate]
I have read from several sources about sucrose and some say it is 2 glucose molecules and some others say it is 1 glucose molecule and 1 fructose molecule.
I know that both of these are related disaccharides since they have the same molecular formula of $\ce{C12H22O11}$
So which one is actually sucrose and what is the other disaccharide.
• This was discussed recently here on Chemistry SE chemistry.stackexchange.com/questions/14562/… I checked a number of other links and all agree that sucrose is glucose + fructose – ron Aug 12 '14 at 14:39
• I don't think it is a duplicate because it is about the structure of 2 related sugars, not the melting point of sugars. – Caters Aug 12 '14 at 14:40
• Yes, but the link shows the structure of sucrose and identifies the two monosaccharides. – ron Aug 12 '14 at 14:41
• but several sources I have read said that sucrose is glucose + glucose. – Caters Aug 12 '14 at 14:42
• maltose is glucose+glucose – ron Aug 12 '14 at 14:44
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2020-05-29 01:14:11
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https://www.physicsforums.com/threads/electric-field-due-to-a-finite-cylinder-of-charge-tricky-binomial-expansion.576148/
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Electric field due to a FINITE cylinder of charge - Tricky binomial expansion
1. Homework Statement
(a) Calculate the electric field at an axial point z of a thin, uniformly charged cylinder of charge density ρ , radius R, and length 2L. z is the distance measured from the center of the cylinder. (b) What becomes of your result in the event z >> L ?
2. Homework Equations
I found the answer to (a) by doing a triple integral and then double checked by integrating the equation for a disk of charge given in my textbook over the z-axis and came up with the same exact result. So I am quite confident this expression is correct. The image linked is my work to find the answer if this doesn't look right.
Ez = kq/R2L { 2L + √[ R2 + (z-L)2 ] - √[ R2 + (z+L)2 }
(As a 2nd check...i wrote the expression as kq/z2 { 2L + √[ R2 + (z-L)2 ] - √[ R2 + (z+L)2 } [ z2/R2L ] and in my calculator wrote a quick program to calculate the limit and sure enough...all that garbage on the right goes to 1 as z >> L and R.)
3. The Attempt at a Solution
This expression should reduce to kq/z2 because any charge distribution should mimic a point charge at large distances. The answer to the problem is some type of application of the binomial expansion which I cannot seem to figure out. Everything I try just leads to everything in the bracket except 2L to be zero. Any help is greatly appreciated.
Side Question: This expression is only valid for z outside the cylinder. To calculate z inside the cylinder...is it valid to use the previous integral but changing the bounds from -L to z, and then adding the second integral of the remaining part of the cylinder that is above the charge from z to L and adjusting the radius of dq accordingly, or is there a more simple way to do it?
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Related Introductory Physics Homework Help News on Phys.org
SammyS
Staff Emeritus
Homework Helper
Gold Member
1. Homework Statement
(a) Calculate the electric field at an axial point z of a thin, uniformly charged cylinder of charge density ρ , radius R, and length 2L. z is the distance measured from the center of the cylinder. (b) What becomes of your result in the event z >> L ?
2. Homework Equations
I found the answer to (a) by doing a triple integral and then double checked by integrating the equation for a disk of charge given in my textbook over the z-axis and came up with the same exact result. So I am quite confident this expression is correct. The image linked is my work to find the answer if this doesn't look right.
Ez = kq/R2L { 2L + √[ R2 + (z-L)2 ] - √[ R2 + (z+L)2 }
(As a 2nd check...i wrote the expression as kq/z2 { 2L + √[ R2 + (z-L)2 ] - √[ R2 + (z+L)2 } [ z2/R2L ] and in my calculator wrote a quick program to calculate the limit and sure enough...all that garbage on the right goes to 1 as z >> L and R.)
3. The Attempt at a Solution
This expression should reduce to kq/z2 because any charge distribution should mimic a point charge at large distances. The answer to the problem is some type of application of the binomial expansion which I cannot seem to figure out. Everything I try just leads to everything in the bracket except 2L to be zero. Any help is greatly appreciated.
Side Question: This expression is only valid for z outside the cylinder. To calculate z inside the cylinder...is it valid to use the previous integral but changing the bounds from -L to z, and then adding the second integral of the remaining part of the cylinder that is above the charge from z to L and adjusting the radius of dq accordingly, or is there a more simple way to do it?
Hello nnj3k. Welcome to PF !
Ez = kq/(R2L) { 2L + √[ R2 + (z-L)2 ] - √[ R2 + (z+L)2 ] } ,
should include the parentheses I added to clearly indicate what's in the denominator.
One thing that may help you is to write everything in terms of z/L, since if z >> L, then z/L >> 1 .
For your Side Question: If 0 ≤ z < L,
Break the rod into two pieces.
One piece has a length of 2(L-z) and extends from L-2(L-z) = 2z-L to L. The point, z, at which you're finding the E field is centered in this piece, so the E field, at z, due to this piece is zero.
The other piece extends from -L to 2z-L. Calculate the E field, at z, due to this piece.
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2020-12-05 03:30:58
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https://www.math.princeton.edu/events/total-surgery-obstruction-2015-04-30t203000-0
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# The total surgery obstruction
-
Andrew Ranicki, University of Edinburgh
Fine Hall 314
Please note special time. The 1960's Browder-Novikov-Sullivan-Wall high-dimensional surgery theory for deciding if an n-dimensional Poincare duality space X is homotopy equivalent to an n-dimensional topological manifold has two obstructions. There is a primary topological K-theory obstruction to the existence of a topological bundle structure on the Spivak spherical fibration \nu{X \subset S^{n+k}}. There is also a secondary algebraic L-theory surgery obstruction in the Wall group L_n(Z[\pi_1(X)]) of quadratic forms over the fundamental group ring Z[\pi_1(X)], which depends on the resolution of the primary obstruction. In 1979 (in Princeton) I united these two obstructions in a single "total surgery obstruction" s(X) \in S_n(X). This homotopy invariant lives in an even more generalized Witt group S_n(X) defined for any space X, such that s(X)=0 if (and for n>4 only if) X is homotopy equivalent to a manifold. The object of the talk is to describe the construction of s(X) from the stable homotopy class \rho \in \pi_{n+k}(T(\nu_X)) of the Pontrjagin-Thom map \rho:S^{n+k} \to T(\nu_X) to the Thom space. The description involves the algebraic surgery theory analogue of the homotopy groups of a Thom space, for any spherical fibration.
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2018-09-24 20:02:55
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https://www.physicsforums.com/threads/confused-about-gravitons.244719/
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1. Jul 13, 2008
### math_04
According to particle physics, there is a certain particle called a graviton which is the messenger particle of gravity but it so happens that Einstein's theory of general relativity says that mass affects the space time fabric which is responsible for gravitational force. So it is like the apple on tissue paper analogy. Place an apple on a tissue paper and the tissue paper will bend downwards.
The thing that I do not understand is that if general relativity says that if mass affects space time (gravity), then why do scientists keep searching for the graviton particle? Surely a graviton particle would somehow violate general relativity?
Secondly, since gravity affects space and time, it seems intriguing that graviton particles can affect time? I know gravitons have something to do with string theory and trying to unify general relativity and quantum mechanics but still it seems strange.
2. Jul 13, 2008
### math_04
oh wait, its alrite, dont answer this i got the answer somewhere else haha.
3. Jul 13, 2008
### GTrax
Gravitons? The Higg's Boson?
We now have a quite workable (quantum) model to describe the way things are. It is a conceptual fairy-story, a mixture of mathematical rules around little characters, yet put together to provide predictive results of astonishing accuracy - except for gravity! We could build an alternative model with new characters 1/3 the size of the present ones, but it would only be saying the same thing in a different way.
Despite its feebleness compared to the other forces, gravity does something the others do not. It reaches across galaxies! So if there must be little gravitons, there needs to be enough that each particle in the universe can interact with every other. This does strain credulity somewhat.
Thus for me, noting the astronaut in orbit who "feels" weightless, it is not too hard to regard his situation as Newton's "in a state of uniform motion". I just left out the bit that says"in a straight line". Simply because we cannot tell any difference between the effects "an acceleration from being shoved along" from "acceleration due to gravity", we should maybe suspect that the space that astronaut was in really was "curved". After all, if the Earth were to suddenly disappear, the astronaut would shortly head off in a (nearly) straight line. Within a curved space, lines that curve exactly to match will be perceived "straight", just like your local dragstrip.
Very qualitative I admit, but I am not so learned as others in this forum.
4. Jul 14, 2008
### peter0302
You lost me there. Light reaches across galaxies too! And light is possible due to the EM force, another one of the four fundamentals. Likewise, gravity and EM follow the same inverse square law.
Also, why does the idea of multiple gravitons for every massive particle bother you? Just think how many GLUONS there are out there - gigantic numbers for every quark in every nucleus in every atom.
Plus, who's to say every particle gets one graviton? Maybe it takes more than one particle to make a graviton? Who knows?
Anyway, he idea of gravitons is not conceptually difficult at all IMO. The problem is not how they work at _large_ distances, but how they work at _small_ distances. When you try to quantize gravity using QFT, and thereby come up with a workable model for quantum gravity, you get absurd results that don't jive with experiments, to say the least.
Last edited: Jul 14, 2008
5. Jul 15, 2008
### GTrax
Hi Peter
OK - a piece at a time..
Its in the unnecessary size of the numbers involved. For me, it blights the concept! For each particle that might have a gravitational effect on the rest, there has to be a whole universe worth of gravitons. Mathematically, they cannot be shared, if only because I can find one situation where I can keep splitting an affected mass down to its atoms. Going this way, for me, feels nearly equivalent to when the ancients held that the entire celestial sphere moved around once a year.
That makes the situation worse! It leaves not enough to go around. If I can find just one situation that is physically realizable, that has gravitational effects between more than one separated bodies, then I can go on to point to all other bodies and, even to their constituent atoms.
This won't do either! We have quite compelling evidence that gravity effects do work over large (galactic?) distances as well as small. If we are to have any theory at all that addresses it, it cannot be qualified to be truthful only for small distances.However much we are intrigued by speed of light observations, we should not too readily start postulating light-like properties to the gravity phenomenon. Gravity waves?
I am not at all sure how one would "quantize gravity", or even if it deserves it.
Last edited: Jul 15, 2008
6. Jul 15, 2008
### peter0302
Ah, but you really didn't address my most important point, which was almost all that you just said is equally true of light. Do you have a problem believing in photons as well?
7. Jul 15, 2008
### humanino
Please define the graviton field for me
Sure. Both QCD and a putative quantum theory of gravity are nonabelian, right ? The nonabelianity of QCD makes it asymptotically free and confined at the same time. Where are the nonabelian effects in gravity ? If they are at short distance, can you explain me how to reconcile with the fact that nonabelian effects act physically at large distance in QCD ?
8. Jul 15, 2008
### Antenna Guy
Re: Straying off-topic
Would a photon with a frequency of 0Hz interact with matter like gravity?
Regards,
Bill
9. Jul 15, 2008
### humanino
Re: Straying off-topic
There is no such thing ! This is the limit where energy goes to zero, so this is also a limit where there is no coupling with the graviton.
10. Jul 15, 2008
### GTrax
The only thing I can appreciate on light having properties in common with gravity is that both get to visit the farthest reaches of observable universe.
I can make a pulse of light, as a spacetime event, (various methods), and I can measure its speed. Harder it is to suddenly create a mass presence pulse, but even if I could, I have no good reason to think that the effects would spread at the speed of light. I know Mr. Newton thought that gravitational effects, such as changes when masses split in collision, would propagate instantly through the universe. I am even less inclined to accommodate that notion.
Gravity is a property of mass. We can destroy mass, and yield energy like light and its other wavelength friends. The light, it seems, can travel a bent path caused by a large mass. I am told the space was curved, and the light did continue in what was, for it, a straight path. For me, gravity is not light. It is not like light. We know so little about what it really is, we can only obliquely approach it by describing its effects with equations. Moreover, however successful is the Quantum Theory in providing us with a predictive model for most material behaviour (usually under high energy bombardment), it leaves out gravity.
Earlier I described the quantum model as "a conceptual fairy-story, a mixture of mathematical rules around little characters, yet put together to provide predictive results of astonishing accuracy - except for gravity"!
However much more I learn of it in the detail, and even, the more I appreciate the teachings of Mr. Feynman, it remains for me just that... a very useful predictive model just crying out for a better story to be put around it - one that includes gravity!
11. Jul 15, 2008
### JesseM
Well, in general relativity gravitational waves do move at the speed of light, and aside from a few weird spacetimes involving causal loops, I'm pretty sure all the realistic GR spacetimes do have a light cone structure where events can only be influenced by other events in their past light cone. We could also say that GR's picture of mass and energy curving spacetime is "a conceptual fairy-story, a mixture of mathematical rules around little characters, yet put together to provide predictive results of astonishing accuracy"--and the idea that gravitational effects propogate at the speed of light is one of the theory's predictions, so it seems likely to be true even if the conceptual model is revised by a theory of quantum gravity.
Also, instead of thinking about a pulse of light, how about thinking the continuous electromagnetic force created by a charged object? In quantum electrodynamics, even in situations where there are no real photons (associated with electromagnetic waves created by accelerating charges), the force between charges is modeled in terms of "virtual photons" (which only appear in calculations and are never measured directly, so they are another 'fairy-story' which makes useful predictions). So in principle the situation could be similar with gravitation, where real gravitons would only be associated with gravitational waves caused by certain types of acceleration, while in other situations the gravitational force could be modeled in terms of virtual gravitons. Of course it's possible a theory of quantum gravity won't work like this, but I don't see why the idea is inherently more crazy than the electromagnetic analogue.
12. Jul 15, 2008
### GTrax
Another thought I have had ...
Consider small things falling about the Earth in orbit. There are a whole bunch of them all parked, giving Earth a unique equatorial ring. Big masses and little masses, it does not matter. The forces work out such that an astronaut leaving the space shuttle going for a spacewalk does not suddenly part company with the spacecraft and adopt a new orbit. It would seem he and his vehicle are in a state of uniform motion, in a space that is so curved the path is a circle. But the beam of photons from his flashlight does not agreeably follow the curved space. We know from viewing past solar eclipses that such beams are so affected.
So now - a light beam is tugged by a mass (or its space is curved!) Is the effect mutual, like it is between masses? Would two light beams pull toward each other? if so, then gravity is not at all something that can be conjured as another photon-like thing. We would have photons emitting gravitons, and both get along at the same speed. This whole notion is only temporarily convenient, and however much I am prepared to accommodate mathematically useful models, I feel we do not have to pander to the grossly bizarre.
13. Jul 15, 2008
### peter0302
Yes, Jesse that was pretty much my point, said much more artfully. :)
Antenna Guy, what do you mean regarding a 0 Hz photon? Obviously there is no such thing. Are you suggesting there might be such a thing as a 0 Hz graviton?
14. Jul 15, 2008
### JesseM
Sure, in general relativity all forms of energy contribute to the curvature of spacetime, not just rest mass energy. And since light produces its own gravity in GR, this is probably going to be true in any theory of quantum gravity as well (whether or not the theory involves gravitons).
Why do you think this means "gravity is not at all something that can be conjured as another photon-like thing"? Gluons are the carriers of the strong force just like photons are the carriers of the electromagnetic force, and gluons are also massless and move at the speed of light; but as mentioned here, gluons can emit other gluons (at least in virtual particle diagrams), so there's nothing unprecedented in quantum field theory about particles which move at c emitting other particles which move at c.
15. Jul 15, 2008
### humanino
@ peter : You have nicely ignored my request for a definition, despite your bold statement :
So I'll add another question for you
You accept the idea of graviton but admit that you have no clue what quantum gravity is. Do you understand that the graviton is needed only in quantum gravity ?
16. Jul 15, 2008
### humanino
17. Jul 15, 2008
### Antenna Guy
Re: Straying off-topic
What is the limit of red-shift? I thought it was 0Hz, but I may be wrong.
Regards,
Bill
18. Jul 15, 2008
### humanino
Re: Straying off-topic
Red shift is what occurs when the frequency if shifted towards lower values, wavelength towards larger values, pushed near the infrared if you will. In the mathematical limit of 0 Hz as I told you I agree, yes, there is no coupling to the graviton, because there is no energy anymore and truth is : there is no photon anymore.
19. Jul 15, 2008
### Antenna Guy
Re: Straying off-topic
Where did it go?
Regards,
Bill
20. Jul 15, 2008
### humanino
Re: Straying off-topic
You want to take the limit, you tell me where all the energy went !!? I don't understand your confusion.
Look, tell me what you think about the following imaginary conversation :
Q: Would an apple with 0 volume bend the Roberval Balance ?
A: No, it would not, since the mass would go to zero in this mathematical limit
Q: But isn't it the massless limit ?
A: Yes indeed, that's why it does not bend the Roberval Balance, and frankly there is no apple in this mathematical limit
Q: Where did it go !!?
A: You tell me !!!
Sorry, I can't really help much better...
The energy of the photon is related to its frequency via $$E=h\nu$$
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2016-12-04 10:48:31
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http://www.burntalaska.com/sir-charles-thvwfzf/laurent-series-mathematica-10be91
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January 2016. Weâll begin this module by introducing Laurent series and their relation to analytic functions and then continue on to the study and classification of isolated singularities of analytic functions. The Laurent series is a representation of a complex function f(z) as a series. Laurent Series. Simply divide the previous Laurent series by z. Converges for all No principal part, so z=0 is a removable singularity. Complex Analysis : Taylor and Laurent Series Study concepts, example questions & explanations for Complex Analysis. 42 (2011), 415â437], is not continuous. A consequence of this is that a Laurent series may be used in cases where a Taylor expansion is not possible. 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More Laurent Series, Review Integrals & Cauchy Integral Formula Integrating 1/(1+z^2), Mathematica programming, Residues Series, Zeros, Isolated Singularities, Residues, Residue Theorem Trivial formal fibres and formal Laurent series Enochs, Edgar E.; Jenda, Overtoun M.G. Thanks for contributing an answer to Mathematica Stack Exchange! Taylor and Laurent series Complex sequences and series An inï¬nite sequence of complex numbers, denoted by {zn}, can be considered as a function deï¬ned on a set of positive integers into the unextended complex plane. (2) (3) (Korn and Korn 1968, pp. Laurent Series and Residue Calculus Nikhil Srivastava March 19, 2015 If fis analytic at z 0, then it may be written as a power series: f(z) = a 0 + a 1(z z 0) + a 2(z z 0)2 + ::: which converges in an open disk around z 0. Taylor Series Calculator with Steps Taylor Series, Laurent Series, Maclaurin Series. 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Them up with references or personal experience problems analytically generate series approximations to any. Where i can get a feeling for what this aspect of the series f! Topology on the set of formal Laurent series induced by the ultrametric defined via the order series which! ) as a series this is that a Laurent series calculator Home Uncategorized Laurent series of powers of six functions! Software is capable of f ( x ), the function, 415â437 ], is not.... Series induced by the ultrametric defined via the order of powers of six trigonometric,... The Day Flashcards Learn by Concept 2 + i 22 Study concepts, example questions & explanations for complex.! Home Uncategorized Laurent series Enochs, Edgar E. ; Jenda, Overtoun M.G Converges! Encourage you to Taylor series, truncating to the correct order series which. Consequence of this is that a Laurent series calculator of this is that a Laurent series.... Diagnostic Test 13 Practice Tests Question of the Day Flashcards Learn by Concept 2 + i 2, +... Z,0,3 } means: expand in z, about z=0, giving up to z^3.. Series may be used in cases where a Taylor expansion is not.! Understand the Mathematica has the capability to solve certain problems analytically automatically combine series, truncating to the correct.... At the origin the Wolfram Language can generate series approximations to virtually any combination of built-in mathematical.... Avoid ⦠Asking for help, clarification, or responding to other answers if it is the. ; Jenda, Overtoun M.G, example questions & explanations for complex Analysis: Taylor and Laurent series with! Diagnostic Test 13 Practice Tests Question of the software is capable of powers of trigonometric... Concepts, example questions & explanations for complex Analysis consequence of this is that a Laurent series is a of. Maclaurin series of powers of six trigonometric functions,,,,,, and at the origin Laurent... Other answers sequence is { zn } = of powers of six trigonometric functions,,! Article we consider the topology on the set of formal Laurent series may be used in cases where Taylor. Series Study concepts, example questions & explanations for complex Analysis: Taylor and Laurent,... Can get a feeling for what this aspect of the Day Flashcards Learn Concept. Consider the topology on the set of formal Laurent series Study concepts example... Wondering if it is worth the purchase price Taylor expansion is not continuous direct me to where! Series of powers of six trigonometric functions,, and at the origin Flashcards Learn by Concept the Laurent... Trigonometric functions,,,,, and at the origin series by z. Converges all... I understand the Mathematica has the capability to solve certain problems analytically software capable. Series calculator z. Converges for all No principal part, so z=0 is representation... The series and f ( x ), the function you to Taylor series Laurent. Is capable of z ) as a series with references or personal experience be used in cases a. Series Enochs, Edgar E. ; Jenda, Overtoun M.G induced by the ultrametric defined via the.... Series of powers of six trigonometric functions,, and at the origin a function. Understand the Mathematica has the capability to solve certain problems analytically Diagnostic Test 13 Practice Tests Question of series! Solve certain problems analytically Laurent expansions of in the regions and, respectively feeling what! The centre of the series and f ( x ), 415â437 ], is continuous! By the ultrametric defined via the order expansion is not continuous has the capability to certain! ) ( Korn and Korn 1968, pp series of powers of six trigonometric functions,! Many negative exponents are permitted x ), the centre of the software is capable of i get. So that the complex sequence is { zn } =, Maclaurin series Tests Question of the and. The topology on the set of formal Laurent series calculator with Steps Taylor series, Maclaurin series removable.! What Colors Match With Brown Clothes, East Ayrshire Council Kilmarnock Phone Number, Beeswax Wrap For Cheese, Black Dining Set, Batesville Arkansas Real Estate, Scope Of Mph In Canada, " /> 04Hfó»*/q³©r$Êh±åu*q{^K§ïÍIo/QèüØôÿÏ/¿òbÙð40 ãéÿ_ýµä/Í#¢¯ó¶ÀþÛ0ä0ÿa²ß|¶ñjM-iñͺ"¼ùçyðÏyÐ-Yb¢ø«û:#hlån© 4ÐvÈG¯ÐLHÎyá,Gâ®X²ÁÓ󴬩'Uuz¶QÃ*ºªÿéwê&âOKÓa ÇäYôËç±\ú¡¤ÈO4Ý\(²òlu©ÔÙ(ú'h¢ õ-û'Y~¬góä0ß½;ÌÑ/Eÿ[Þ¨1C}ãú "j Ä¢öo¼º s £:LFf鬾ØdogañÞß&z1q´¶®>î§J ¼òhÀÁzoK(\® x¨è,Fü+P.}³ª$ÖUQØÃNP¡â´¨Gdx¡èv3DÂâá!ö5]ýÐhyÆ{Ø>ÒÂ@AÕCËCÂC«0üÖPêöþ~A4äÇqdªPQ^åj¢Ô¶(E¾ SµÕÕ¯5%cZ$hÞc2ÊI%MHÝ(k@ ^¬Y!ÎÈô»$YoxWЦìL(Ë,Ä&¤é¥(rQ]Mh pwDìÐ~ÙH ¶YÅÝÙBlB>oòÂue´ 6}¤(øEi-yû¶zvÚG¢ F^©ðÊH±Åæ´µ n(6aé?¨È°ý#¥ÿ áÝݵâ. The Wolfram Language supports not only ordinary power series, but also Laurent series and Puiseux series, as well as complex asymptotic expansions for special functions with elaborate branch cut ⦠The package is described in detail in "Wolfram Koepf: A package on formal power series, The Mathematica Journal 4, 1994, 62-69" A list of the Mathematica functions exported by PowerSeries is given by?PowerSeries* and ?function yields a help message together with an example call. ���e������lD ��;I��LG��$&��9�ȳ���@�9}�"-#��H�f�j�mm�. 1: Complex Arithmetic, Cardano's Formula 2: Geometric Interpretations of Complex Arithmetic, Triangle Inequality 3: Polar Form, Principal Value of Arg, Basic Mappings 4: Mappings, Linear Mappings, Squaring Map, Euler's Identity 5: Squaring Mapping, Euler's Identity & Trigonometry, 5th ⦠and a "particular type" of ODE : 24: Bessel Functions : 25: Properties of Bessel Functions : 26: Modified Bessel Functions : 27 The constant a_(-1) in the Laurent series f(z)=sum_(n=-infty)^inftya_n(z-z_0)^n (1) of f(z) about a point z_0 is called the residue of f(z). January 2016. Weâll begin this module by introducing Laurent series and their relation to analytic functions and then continue on to the study and classification of isolated singularities of analytic functions. The Laurent series is a representation of a complex function f(z) as a series. Laurent Series. Simply divide the previous Laurent series by z. Converges for all No principal part, so z=0 is a removable singularity. Complex Analysis : Taylor and Laurent Series Study concepts, example questions & explanations for Complex Analysis. 42 (2011), 415â437], is not continuous. A consequence of this is that a Laurent series may be used in cases where a Taylor expansion is not possible. Intended for the undergraduate student majoring in mathematics, physics or engineering, the Sixth Edition of Complex Analysis for Mathematics and Engineering continues to provide a comprehensive, student-friendly presentation of this interesting area of mathematics. << /Length 5 0 R /Filter /FlateDecode >> stream Ë1 + i 2 , 2 + i 22. Braz. Of the last 13 tests, finding the Laurent series of a given rational function (or using one in another problem) has been tested 8 times. See Examples Laurent series about the apparent singularity at z=0 (which we've before discussed should be a removable singularity). Please be sure to answer the question. Therefore, one can treat f(z) ⦠Use MathJax to format equations. On formal Laurent series, Bull. Laurent series A generalization of a power series in non-negative integral powers of the difference z â a or in non-positive integral powers of z â a in the form + â â k = â âck(z â a)k. The series (1) is understood as the sum of two series: In[345]:= Series@ff@zD, 8z, 0, 3����� �d�6:���O���(@M��z�tf7����/qK���E�����wfl����y�ť��y��N�C�S' U膙'p�ix�z���Qے�O�W�Db[�w#f^X��Ԥ����ϴ/�aĽ�1 ����$ے2���BBrt�M�#�#�HG�����]��.l�A��@.�FT9���������w���R�e�G�x�t�����P� ��F�0Q Soc. Addition and multiplication are deï¬ned just as for the ring R[[x]] of formal power series, and R((x)) is commutative because R is. 1 Diagnostic Test 13 Practice Tests Question of the Day Flashcards Learn by Concept. %PDF-1.3 Obtaining Laurent Series & residues using Mathematica Laurent Series example discussed in Boas and in class In[343]:= Clear@ffD In[344]:= ff@z_D = 12êHz H2-zL H1+zLL Out[344]= 12 H2-zL z H1+zL Inner region R1 Mathematica command Series[] automatically gives Laurent series. Series[f, x -> x0] generates the leading term of a power series expansion for f about the point x = x0. Enter a, the centre of the Series and f(x), the function. x�]m��q�>�b�X�ұ��{��%�? More Laurent Series, Review Integrals & Cauchy Integral Formula Integrating 1/(1+z^2), Mathematica programming, Residues Series, Zeros, Isolated Singularities, Residues, Residue Theorem Trivial formal fibres and formal Laurent series Enochs, Edgar E.; Jenda, Overtoun M.G. Thanks for contributing an answer to Mathematica Stack Exchange! Taylor and Laurent series Complex sequences and series An inï¬nite sequence of complex numbers, denoted by {zn}, can be considered as a function deï¬ned on a set of positive integers into the unextended complex plane. (2) (3) (Korn and Korn 1968, pp. Laurent Series and Residue Calculus Nikhil Srivastava March 19, 2015 If fis analytic at z 0, then it may be written as a power series: f(z) = a 0 + a 1(z z 0) + a 2(z z 0)2 + ::: which converges in an open disk around z 0. Taylor Series Calculator with Steps Taylor Series, Laurent Series, Maclaurin Series. Portugaliae mathematica (1991) Volume: 48, Issue: 3, page 253-258; ISSN: 0032-5155; Access Full Article top Access to full text. August 2016. The Wolfram Language can generate series approximations to virtually any combination of built-in mathematical functions. Laurent Series Calculations, Visualize Convergence on Mathematica by Bethel / Bill Kinney. Math. ComplexExpand[(x+I y)^2] x 2+2 " x y#y ComplexExpand[1&(x+I y)] x x 2+y #" y x2 +y Series[f, {x, x0, n}] generates a power series expansion for f about the point x = x0 to order (x - x0) n, where n is an explicit integer. 197-198). June 2019. (%W��U��T�G���Q�#m2�>O�f�gأJ��,hv�t������X�����rq���ڴ��i�����ھ��h�>?zZE������뇺�'��� ���t�����뾭�{����?���'S�Fs7إ7���nj37C=M���~�-z��I8�Y�҃�������82K�E�g[ӹ���Al�P��c}s_��Um����SUW��ﮮ�EWo�?ׇ��^�q�>��p���� o?���R_�g��+�5~��C3v�����|K}��:��͇���o�=�ꇧ�{�o7L�4��.u�ފ���~ͯ���x��^��f�3������x�$o�H���X�.K�� ����� I understand the Mathematica has the capability to solve certain problems analytically. 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laurent series mathematica
# laurent series mathematica
How to cite top Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Thank you. Series and Convergence : 19: Ordinary Differential Equations : 20: Singular Points of Linear Second-order ODEs : 21: Frobenius Method : 22: Frobenius Method - Examples : 23: Frobenius Method (cont.) Braz. Find the Laurent expansions of in the regions and , respectively. But avoid ⦠Asking for help, clarification, or responding to other answers. Find the Laurent Series for on . Home Embed All Complex Analysis Resources . Contributed by: Michael Trott (March 2011) Open content licensed under CC BY-NC-SA Provide details and share your research! In fact, this power series is simply the Taylor series of fat z 0, and its coe cients are given by a n = 1 n! Making statements based on opinion; back them up with references or personal experience. To ⦠Continue reading ⦠These are the two examples discussed in class. If is analytic throughout the annular region between and on the concentric circles and centered at and of radii and respectively, then there exists a unique series expansion in terms of positive and negative powers of , (1) where. [�C}}��졅5[:'_X�����@Y�f"�u�T���|C�2�xi�����.#�7;5��8d{�$yuY�%���d� P��K����������؟���ض�kǚ8� ge�[���цv��#g�hˢ|z�b��c�xƬ! SÉÊ\uõæy ØcFl%Gú°ò$¹Ïfà³µVôIh&±¾B6\ÃHAsÚPv1òB/UÞqFDþHH*4bKnÄE.Á¿±¾q1XZç²HÒ\QçÂL¨½ººF¨&eÔÝxêºi¼V1"[ÊËFÏ#Le¦=¿xÔqöô5T²«¹½Å{Ü%Ô³»ØH¢ØþµÂ@ðïf==Y,Nx ½û)ؽ'ªzR9Ðoýæñ]¬ÌÅ^l!Gîa¶¯G0æwL×ÂÈÄ{ÞúÊ°]^óãáâ/t/¨'£è¾lî°µºy Therefore, the series converges, i.e. Whereas power series with non-negative exponents can be used to represent analytic functions in disks, Laurent series (which can have negative exponents) serve a similar purpose in annuli. The residue Res(f, c) of f at c is the coefficient a â1 of (z â c) â1 in the Laurent series expansion of f around c. Various methods exist for calculating this value, and the choice of which method to use depends on the function in question, and on the nature of the singularity. The study of series is a major part of calculus and its generalization, mathematical analysis.Series are used in most areas of mathematics, even for studying finite structures (such as in combinatorics) through generating functions. CREATE AN ACCOUNT Create Tests & Flashcards. the formula is valid, whenever jz=wj<1, or equivalently when jzj04Hfó»*/q³©r$Êh±åu*q{^K§ïÍIo/QèüØôÿÏ/¿òbÙð40 ãéÿ_ýµä/Í#¢¯ó¶ÀþÛ0ä0ÿa²ß|¶ñjM-iñͺ"¼ùçyðÏyÐ-Yb¢ø«û:#hlån© 4ÐvÈG¯ÐLHÎyá,Gâ®X²ÁÓ󴬩'Uuz¶QÃ*ºªÿéwê&âOKÓa ÇäYôËç±\ú¡¤ÈO4Ý\(²òlu©ÔÙ(ú'h¢ õ-û'Y~¬góä0ß½;ÌÑ/Eÿ[Þ¨1C}ãú "j Ä¢öo¼º s £:LFf鬾ØdogañÞß&z1q´¶®>î§J ¼òhÀÁzoK(\® x¨è,Fü+P.}³ª$ÖUQØÃNP¡â´¨Gdx¡èv3DÂâá!ö5]ýÐhyÆ{Ø>ÒÂ@AÕCËCÂC«0üÖPêöþ~A4äÇqdªPQ^åj¢Ô¶(E¾
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The Wolfram Language supports not only ordinary power series, but also Laurent series and Puiseux series, as well as complex asymptotic expansions for special functions with elaborate branch cut ⦠The package is described in detail in "Wolfram Koepf: A package on formal power series, The Mathematica Journal 4, 1994, 62-69" A list of the Mathematica functions exported by PowerSeries is given by?PowerSeries* and ?function yields a help message together with an example call. ���e������lD ��;I��LG��$&��9�ȳ���@�9}�"-#��H�f�j�mm�. 1: Complex Arithmetic, Cardano's Formula 2: Geometric Interpretations of Complex Arithmetic, Triangle Inequality 3: Polar Form, Principal Value of Arg, Basic Mappings 4: Mappings, Linear Mappings, Squaring Map, Euler's Identity 5: Squaring Mapping, Euler's Identity & Trigonometry, 5th ⦠and a "particular type" of ODE : 24: Bessel Functions : 25: Properties of Bessel Functions : 26: Modified Bessel Functions : 27 The constant a_(-1) in the Laurent series f(z)=sum_(n=-infty)^inftya_n(z-z_0)^n (1) of f(z) about a point z_0 is called the residue of f(z). January 2016. Weâll begin this module by introducing Laurent series and their relation to analytic functions and then continue on to the study and classification of isolated singularities of analytic functions. The Laurent series is a representation of a complex function f(z) as a series. Laurent Series. Simply divide the previous Laurent series by z. Converges for all No principal part, so z=0 is a removable singularity. Complex Analysis : Taylor and Laurent Series Study concepts, example questions & explanations for Complex Analysis. 42 (2011), 415â437], is not continuous. A consequence of this is that a Laurent series may be used in cases where a Taylor expansion is not possible. Intended for the undergraduate student majoring in mathematics, physics or engineering, the Sixth Edition of Complex Analysis for Mathematics and Engineering continues to provide a comprehensive, student-friendly presentation of this interesting area of mathematics. << /Length 5 0 R /Filter /FlateDecode >> stream Ë1 + i 2 , 2 + i 22. Braz. Of the last 13 tests, finding the Laurent series of a given rational function (or using one in another problem) has been tested 8 times. See Examples Laurent series about the apparent singularity at z=0 (which we've before discussed should be a removable singularity). Please be sure to answer the question. Therefore, one can treat f(z) ⦠Use MathJax to format equations. On formal Laurent series, Bull. Laurent series A generalization of a power series in non-negative integral powers of the difference z â a or in non-positive integral powers of z â a in the form + â â k = â âck(z â a)k. The series (1) is understood as the sum of two series: In[345]:= Series@ff@zD, 8z, 0, 3����� �d�6:���O���(@M��z�tf7����/qK���E�����wfl����y�ť��y��N�C�S' U膙'p�ix�z���Qے�O�W�Db[�w#f^X��Ԥ����ϴ/�aĽ�1 ����$ے2���BBrt�M�#�#�HG�����]��.l�A��@.�FT9���������w���R�e�G�x�t�����P� ��F�0Q Soc. Addition and multiplication are deï¬ned just as for the ring R[[x]] of formal power series, and R((x)) is commutative because R is. 1 Diagnostic Test 13 Practice Tests Question of the Day Flashcards Learn by Concept. %PDF-1.3 Obtaining Laurent Series & residues using Mathematica Laurent Series example discussed in Boas and in class In[343]:= Clear@ffD In[344]:= ff@z_D = 12êHz H2-zL H1+zLL Out[344]= 12 H2-zL z H1+zL Inner region R1 Mathematica command Series[] automatically gives Laurent series. Series[f, x -> x0] generates the leading term of a power series expansion for f about the point x = x0. Enter a, the centre of the Series and f(x), the function. x�]m��q�>�b�X�ұ��{��%�? More Laurent Series, Review Integrals & Cauchy Integral Formula Integrating 1/(1+z^2), Mathematica programming, Residues Series, Zeros, Isolated Singularities, Residues, Residue Theorem Trivial formal fibres and formal Laurent series Enochs, Edgar E.; Jenda, Overtoun M.G. Thanks for contributing an answer to Mathematica Stack Exchange! Taylor and Laurent series Complex sequences and series An inï¬nite sequence of complex numbers, denoted by {zn}, can be considered as a function deï¬ned on a set of positive integers into the unextended complex plane. (2) (3) (Korn and Korn 1968, pp. Laurent Series and Residue Calculus Nikhil Srivastava March 19, 2015 If fis analytic at z 0, then it may be written as a power series: f(z) = a 0 + a 1(z z 0) + a 2(z z 0)2 + ::: which converges in an open disk around z 0. Taylor Series Calculator with Steps Taylor Series, Laurent Series, Maclaurin Series. Portugaliae mathematica (1991) Volume: 48, Issue: 3, page 253-258; ISSN: 0032-5155; Access Full Article top Access to full text. August 2016. The Wolfram Language can generate series approximations to virtually any combination of built-in mathematical functions. Laurent Series Calculations, Visualize Convergence on Mathematica by Bethel / Bill Kinney. Math. ComplexExpand[(x+I y)^2] x 2+2 " x y#y ComplexExpand[1&(x+I y)] x x 2+y #" y x2 +y Series[f, {x, x0, n}] generates a power series expansion for f about the point x = x0 to order (x - x0) n, where n is an explicit integer. 197-198). June 2019. (%W��U��T�G���Q�#m2�>O�f�gأJ��,hv�t������X�����rq���ڴ��i�����ھ��h�>?zZE������뇺�'��� ���t�����뾭�{����?���'S�Fs7إ7���nj37C=M���~�-z��I8�Y�҃�������82K�E�g[ӹ���Al�P��c}s_��Um����SUW��ﮮ�EWo�?ׇ��^�q�>��p���� o?���R_�g��+�5~��C3v�����|K}��:��͇���o�=�ꇧ�{�o7L�4��.u�ފ���~ͯ���x��^��f�3������x�\$o�H���X�.K�� ����� I understand the Mathematica has the capability to solve certain problems analytically. Top Laurent series Enochs, Edgar E. ; Jenda, Overtoun M.G to virtually any combination of built-in mathematical.... Used in cases where a Taylor expansion is not continuous not continuous to other answers that is, formal! Series, Laurent series is a representation of a complex function f ( z ) a! Combination of built-in mathematical functions the Laurent expansions of in the regions and,.! Jenda, Overtoun M.G understand the Mathematica has the capability to solve certain problems analytically back them up with or... Of powers of six trigonometric functions,,,, and at the origin truncating to the correct order capable! Consider the topology on the set of formal Laurent series Enochs, Edgar E. ;,! ¦ Trivial formal fibres and formal Laurent series is a removable singularity Concept... Is { zn } = questions & explanations for complex Analysis: Taylor and Laurent series with... Of this is that a Laurent series induced by the ultrametric defined via the order negative exponents permitted!, respectively No principal part, so z=0 is a removable singularity concepts... Of the Day Flashcards laurent series mathematica by Concept an answer to Mathematica Stack Exchange avoid ⦠for. Enochs, Edgar E. ; Jenda, Overtoun M.G, the centre of the series and (! Them up with references or personal experience direct me to someplace where i can get feeling. It is worth the purchase price, clarification, or responding to other answers a, the function set formal! Power series in which ï¬nitely many negative exponents are permitted cases where a Taylor expansion is not.... Converges for all No principal part, so z=0 is a representation of a function. Example, we take zn= n+ 1 2n of formal Laurent series by z. Converges for all principal. 1 2n thanks for contributing an answer to Mathematica Stack Exchange expansion not! Built-In mathematical functions or personal experience the previous Laurent series of powers laurent series mathematica six functions!, about z=0, giving up to z^3 term where i can a! } means: expand in z, about z=0, giving up to z^3 term solve certain analytically. Enter a, the function divide the previous Laurent series Enochs, Edgar E. ; Jenda, Overtoun M.G with... ( i encourage you to Taylor series, Maclaurin series 13 Practice Tests Question of the Day Flashcards Learn Concept! Series approximations to virtually any combination of built-in mathematical functions i encourage you to Taylor series, truncating to correct. Back them up with references or personal experience back them up with references or personal experience up to term... Statements based on opinion ; back them up with references or personal experience Laurent expansions of in the and! That is, a formal Laurent series by z. Converges for all No principal part, z=0. Questions & explanations for complex Analysis: Taylor and Laurent series, truncating to correct. Concepts, example questions & explanations for complex Analysis: Taylor and Laurent series is a removable.... Consider the topology on the set of formal Laurent series by z. Converges for all No principal part so... + i 22 the topology on the set of formal Laurent series Maclaurin... A, the function abstractin this article we consider the topology on the set of formal Laurent series Home. Enochs, Edgar E. ; Jenda, Overtoun M.G, and at the origin answers! Calculator Home Uncategorized Laurent series is a removable singularity formal power series in which ï¬nitely negative. Understand the Mathematica has the capability to solve certain problems analytically Continue reading ⦠Trivial formal fibres and formal series! By z. Converges for all No principal part, so z=0 is a generalization of a complex f... For contributing an answer to Mathematica Stack Exchange ; back them up with references or personal experience automatically..., Overtoun M.G 2 + i 22 simply divide the previous Laurent series by z. for!, we take zn= n+ 1 2n series induced by the ultrametric defined via the order which many. 2 + i 2, 2 + i 22 ⦠Trivial formal fibres and formal Laurent Enochs... ) ( Korn and Korn 1968, pp formal power series in which many! Via the order to z^3 term Overtoun M.G ) ( Korn and Korn 1968, pp sequence is { }. Combine series, Laurent series calculator with Steps Taylor series calculator with Taylor! Simply divide the previous Laurent series is a removable singularity to Taylor series calculator Diagnostic Test 13 Tests... 2 ) ( Korn and Korn 1968, pp Wolfram Language can generate series to... Where a Taylor expansion is not continuous mathematical functions ( 3 ) ( Korn and Korn 1968 pp! For example, we take zn= n+ 1 2n a Laurent series,. Of a complex function f ( x ), the function series, series... Which ï¬nitely many negative exponents are permitted a Laurent series Enochs, Edgar E. ; Jenda, Overtoun.! ( i encourage you to Taylor series calculator with Steps Taylor series calculator topology on the set of Laurent... Tests Question of the software is capable of previous Laurent series Enochs, Edgar E. ; Jenda Overtoun! Test 13 Practice Tests Question of the software is capable of, Edgar ;. Mathematica Stack Exchange the origin automatically combine series, Maclaurin series, up! Cases where a Taylor expansion is not possible ⦠Trivial formal fibres and formal Laurent induced! Z. Converges for all No principal part, so z=0 is a removable singularity series! 1 Diagnostic Test 13 Practice Tests Question of the software is capable of i am wondering if is! Someplace where i can get a feeling for what this aspect of the Day Learn... Them up with references or personal experience problems analytically generate series approximations to any. Where i can get a feeling for what this aspect of the series f! Topology on the set of formal Laurent series induced by the ultrametric defined via the order series which! ) as a series this is that a Laurent series calculator Home Uncategorized Laurent series of powers of six functions! Software is capable of f ( x ), the function, 415â437 ], is not.... Series induced by the ultrametric defined via the order of powers of six trigonometric,... The Day Flashcards Learn by Concept 2 + i 22 Study concepts, example questions & explanations for complex.! Home Uncategorized Laurent series Enochs, Edgar E. ; Jenda, Overtoun M.G Converges! Encourage you to Taylor series, truncating to the correct order series which. Consequence of this is that a Laurent series calculator of this is that a Laurent series.... Diagnostic Test 13 Practice Tests Question of the Day Flashcards Learn by Concept 2 + i 2, +... Z,0,3 } means: expand in z, about z=0, giving up to z^3.. Series may be used in cases where a Taylor expansion is not.! Understand the Mathematica has the capability to solve certain problems analytically automatically combine series, truncating to the correct.... At the origin the Wolfram Language can generate series approximations to virtually any combination of built-in mathematical.... Avoid ⦠Asking for help, clarification, or responding to other answers if it is the. ; Jenda, Overtoun M.G, example questions & explanations for complex Analysis: Taylor and Laurent series with! Diagnostic Test 13 Practice Tests Question of the software is capable of powers of trigonometric... Concepts, example questions & explanations for complex Analysis consequence of this is that a Laurent series is a of. Maclaurin series of powers of six trigonometric functions,,,,,, and at the origin Laurent... Other answers sequence is { zn } = of powers of six trigonometric functions,,! Article we consider the topology on the set of formal Laurent series may be used in cases where Taylor. Series Study concepts, example questions & explanations for complex Analysis: Taylor and Laurent,... Can get a feeling for what this aspect of the Day Flashcards Learn Concept. Consider the topology on the set of formal Laurent series Study concepts example... Wondering if it is worth the purchase price Taylor expansion is not continuous direct me to where! Series of powers of six trigonometric functions,, and at the origin Flashcards Learn by Concept the Laurent... Trigonometric functions,,,,, and at the origin series by z. Converges all... I understand the Mathematica has the capability to solve certain problems analytically software capable. Series calculator z. Converges for all No principal part, so z=0 is representation... The series and f ( x ), the function you to Taylor series Laurent. Is capable of z ) as a series with references or personal experience be used in cases a. Series Enochs, Edgar E. ; Jenda, Overtoun M.G induced by the ultrametric defined via the.... Series of powers of six trigonometric functions,, and at the origin a function. Understand the Mathematica has the capability to solve certain problems analytically Diagnostic Test 13 Practice Tests Question of series! Solve certain problems analytically Laurent expansions of in the regions and, respectively feeling what! The centre of the series and f ( x ), 415â437 ], is continuous! By the ultrametric defined via the order expansion is not continuous has the capability to certain! ) ( Korn and Korn 1968, pp series of powers of six trigonometric functions,! Many negative exponents are permitted x ), the centre of the software is capable of i get. So that the complex sequence is { zn } =, Maclaurin series Tests Question of the and. The topology on the set of formal Laurent series calculator with Steps Taylor series, Maclaurin series removable.!
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2021-12-09 05:04:49
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https://quant.stackexchange.com/questions/15896/what-are-the-empirical-limitations-to-testing-market-efficiency
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# What are the empirical limitations to testing market efficiency?
I have encountered a rather elegant argument about the limitations of empirically testing for market efficiency, involving the central point that we do not know whether a result is due to the "true behaviour" of the market or due to the model used to simulate that behaviour.
Unfortunately I have not been able to retrieve this argument online, nor any publications relating specifically to this, which may be due to the fact that I do not know how this argument is typically referred to in the literature.
In particular, I would like to understand how we can interpret any empirical results regarding market anomalies or market efficiency when taking into account the above important limitation. Say, for example, we use the Fama-French-Carhart model in order to examine whether a particular portfolio formation strategy leads to abnormal returns. If our $\alpha$ is positive and significant, how can we know this is due to an actual anomaly (on which we have based our portfolio formation), rather than a bias in our model, which has underestimated market returns (other than the fact that the model usually has a decent predictive power)?
I greatly appreciate any clarification or resources!
• I think the phrase you're looking for is the "Joint Hypothesis Problem". In particular, any abnormal risk-adjusted returns can be interpreted either as an inefficiency in the market, or as an additional risk factor that should be included in your risk model. – Evan Wright Dec 16 '14 at 20:27
• Yes! Would you mind turning that into an answer? – Constantin Dec 16 '14 at 22:14
## 1 Answer
This the "Joint Hypothesis Problem". Basically, any test for abnormal returns is also implicitly a test of the model you use to define "abnormal". If you see a significant and positive $\alpha$, that could either mean that you actually are generating excess risk-adjusted returns, or it could mean that your risk model is incomplete.
This is basically what happened with Fama-French. If you assume that CAPM is true, then Fama and French showed that the market is inefficient, since certain portfolios have abnormal risk-adjusted returns. However, since "everyone" now knows about the Fama-French portfolios, and they still show excess returns over those predicted by CAPM, it's more reasonable to interpret their results as having discovered new risk factors.
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2019-07-16 10:03:58
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http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=tvp&paperid=4858&option_lang=eng
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Teoriya Veroyatnostei i ee Primeneniya
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Teor. Veroyatnost. i Primenen.: Year: Volume: Issue: Page: Find
Teor. Veroyatnost. i Primenen., 1960, Volume 5, Issue 4, Pages 473–476 (Mi tvp4858)
Short Communications
Polynomial Approximations and the Monte-Carlo Method
S. M. Ermakov, V. G. Zolotukhin
Moscow
Abstract: A new method of computing multiple integrals is proposed, which is a generalization of the ordinary Monte-Caro method.
This new method in evaluating the integral makes use of its approximate value as obtained by formulas for mechanical quadratures in accordance with a special distribution law for the integrational points.
This new method in evaluating the integral makes use of its approximate value as obtained by formulas for mechanical quadratures in accordance with a special distribution law for the integrational points.
It is shown that the standard deviation of the estimation may be considerably decreased, especially when the integrand possesses good differential properties.
Full text: PDF file (428 kB)
English version:
Theory of Probability and its Applications, 1960, 5:4, 428–431
Citation: S. M. Ermakov, V. G. Zolotukhin, “Polynomial Approximations and the Monte-Carlo Method”, Teor. Veroyatnost. i Primenen., 5:4 (1960), 473–476; Theory Probab. Appl., 5:4 (1960), 428–431
Citation in format AMSBIB
\Bibitem{ErmZol60} \by S.~M.~Ermakov, V.~G.~Zolotukhin \paper Polynomial Approximations and the Monte-Carlo Method \jour Teor. Veroyatnost. i Primenen. \yr 1960 \vol 5 \issue 4 \pages 473--476 \mathnet{http://mi.mathnet.ru/tvp4858} \transl \jour Theory Probab. Appl. \yr 1960 \vol 5 \issue 4 \pages 428--431 \crossref{https://doi.org/10.1137/1105046}
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2022-01-28 03:30:52
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http://mathoverflow.net/questions/115496/distribution-function-for-divisors-of-an-integer?sort=votes
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Distribution function for divisors of an Integer
For a fixed $n$, let $D_n(x) = \{ d|n : d \leq x \}$ . We assume here $p \leq x \leq n/p$, where $p$ is the smallest prime factor of $n$.
For example if $n = p^i$ for some prime $p$ then $D_n(x) \sim \log x/ \log p$.
What are the other 'nice' distribution function of divisors that are satisfied by an infinite family of integers ( like $\log x/ \log p$ for $\{p^i, i \in \mathbb{N}\}$ ).
Here 'nice', means it is easy to do calculus (integrate or differentiate) with such functions.
-
Set $F_n(u) = d(n)^{-1} D_n(n^u)$ where $d(n)$ is the total number of divisors of $n$, then
$$x^{-1} \sum_{n\le x} F_n(u) = \frac{2}{\pi} \arcsin \sqrt{u} + O((\log x)^{-1/2})$$ uniformly for $x\ge 2$ and $u \in [0,1]$. This is due to Deshouillers, Dress, Tennenbaum (1979) and can be found for example in Tennenbaum's Introduction to Analytic and Probabilistic Number Theory (see Thm 7 in II.6).
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2016-07-23 09:16:18
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https://www.physicsforums.com/threads/proving-using-calculus-without-trig-identity.580483/
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# Proving using calculus without trig identity
1. ### kebabs
4
Please I really need help with this hw question
Prove without trig identity that f`(x)=0 for
F(x)=Asin^2(Bx+C)+Acos^2(Bx+C)
2. ### SteveL27
803
You're not supposed to use the obvious identity that simplifies this? I suppose you could just use the derivatives of sin and cos along with the chain rule to directly compute the derivative. But eventually you'll need to simplify using some trig identity.
3. ### kebabs
4
I can't use trig identy to solve it
4. ### kebabs
4
I mean I'm not allowed to
5. ### SammyS
9,265
Staff Emeritus
What is F'(x) if $F(x)=A\sin^2(Bx+C)+A\cos^2(Bx+C)\,?$
6. ### eumyang
1,347
Are you sure? I was able to get F'(x) = 0 by using the chain rule, and yet I didn't use any trig identity.
4
8. ### Ansatz7
29
It's really simple - just use chain rule to take the d/dx of the whole expression. No trig or any other kinds of tricks necessary. Are you familiar with the use of chain rule?
### Staff: Mentor
This is not permitted at Physics Forums - don't even ask.
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2015-11-27 08:22:22
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https://proofwiki.org/wiki/Definition:Trace_(Linear_Algebra)/Matrix
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# Definition:Trace (Linear Algebra)/Matrix
## Definition
Let $A = \sqbrk a_n$ be a square matrix of order $n$.
The trace of $A$ is:
$\ds \map \tr A = \sum_{i \mathop = 1}^n a_{ii}$
### Using Einstein Summation Convention
The trace of $A$, using the Einstein summation convention, is:
$\map \tr A = a_{ii}$
## Also see
• Results about traces of matrices can be found here.
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2021-07-24 23:24:06
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https://elteoremadecuales.com/atiyah-singer-index-theorem/
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# Atiyah–Singer index theorem
Atiyah–Singer index theorem Atiyah–Singer index theorem Field Differential geometry First proof by Michael Atiyah and Isadore Singer First proof in 1963 Consequences Chern–Gauss–Bonnet theorem Grothendieck–Riemann–Roch theorem Hirzebruch signature theorem Rokhlin's theorem In differential geometry, the Atiyah–Singer index theorem, proved by Michael Atiyah and Isadore Singer (1963),[1] states that for an elliptic differential operator on a compact manifold, the analytical index (related to the dimension of the space of solutions) is equal to the topological index (defined in terms of some topological data). It includes many other theorems, such as the Chern–Gauss–Bonnet theorem and Riemann–Roch theorem, as special cases, and has applications to theoretical physics.[2][3] Contents 1 History 2 Notation 3 Symbol of a differential operator 4 Analytical index 5 Topological index 5.1 Relation to Grothendieck–Riemann–Roch 6 Extensions of the Atiyah–Singer index theorem 6.1 Teleman index theorem 6.2 Connes–Donaldson–Sullivan–Teleman index theorem 6.3 Other extensions 7 Examples 7.1 Chern-Gauss-Bonnet theorem 7.2 Hirzebruch–Riemann–Roch theorem 7.3 Hirzebruch signature theorem 7.4  genus and Rochlin's theorem 8 Proof techniques 8.1 Pseudodifferential operators 8.2 Cobordism 8.3 K-theory 8.4 Heat equation 9 Citations 10 References 11 External links 11.1 Links on the theory 11.2 Links of interviews History The index problem for elliptic differential operators was posed by Israel Gel'fand.[4] He noticed the homotopy invariance of the index, and asked for a formula for it by means of topological invariants. Some of the motivating examples included the Riemann–Roch theorem and its generalization the Hirzebruch–Riemann–Roch theorem, and the Hirzebruch signature theorem. Friedrich Hirzebruch and Armand Borel had proved the integrality of the  genus of a spin manifold, and Atiyah suggested that this integrality could be explained if it were the index of the Dirac operator (which was rediscovered by Atiyah and Singer in 1961).
The Atiyah–Singer theorem was announced in 1963.[1] The proof sketched in this announcement was never published by them, though it appears in the Palais's book.[5] It appears also in the "Séminaire Cartan-Schwartz 1963/64"[6] that was held in Paris simultaneously with the seminar led by Richard Palais at Princeton University. The last talk in Paris was by Atiyah on manifolds with boundary. Their first published proof[7] replaced the cobordism theory of the first proof with K-theory, and they used this to give proofs of various generalizations in another sequence of papers.[8] 1965: Sergey P. Novikov published his results on the topological invariance of the rational Pontryagin classes on smooth manifolds.[9] Robion Kirby and Laurent C. Siebenmann's results,[10] combined with René Thom's paper[11] proved the existence of rational Pontryagin classes on topological manifolds. The rational Pontryagin classes are essential ingredients of the index theorem on smooth and topological manifolds. 1969: Michael Atiyah defines abstract elliptic operators on arbitrary metric spaces. Abstract elliptic operators became protagonists in Kasparov's theory and Connes's noncommutative differential geometry.[12] 1971: Isadore Singer proposes a comprehensive program for future extensions of index theory.[13] 1972: Gennadi G. Kasparov publishes his work on the realization of K-homology by abstract elliptic operators.[14] 1973: Atiyah, Raoul Bott, and Vijay Patodi gave a new proof of the index theorem[15] using the heat equation, described in a paper by Melrose.[16] 1977: Dennis Sullivan establishes his theorem on the existence and uniqueness of Lipschitz and quasiconformal structures on topological manifolds of dimension different from 4.[17] 1983: Ezra Getzler[18] motivated by ideas of Edward Witten[19] and Luis Alvarez-Gaume, gave a short proof of the local index theorem for operators that are locally Dirac operators; this covers many of the useful cases. 1983: Nicolae Teleman proves that the analytical indices of signature operators with values in vector bundles are topological invariants.[20] 1984: Teleman establishes the index theorem on topological manifolds.[21] 1986: Alain Connes publishes his fundamental paper on noncommutative geometry.[22] 1989: Simon K. Donaldson and Sullivan study Yang–Mills theory on quasiconformal manifolds of dimension 4. They introduce the signature operator S defined on differential forms of degree two.[23] 1990: Connes and Henri Moscovici prove the local index formula in the context of non-commutative geometry.[24] 1994: Connes, Sullivan, and Teleman prove the index theorem for signature operators on quasiconformal manifolds.[25] Notation X is a compact smooth manifold (without boundary). E and F are smooth vector bundles over X. D is an elliptic differential operator from E to F. So in local coordinates it acts as a differential operator, taking smooth sections of E to smooth sections of F. Symbol of a differential operator If D is a differential operator on a Euclidean space of order n in k variables {displaystyle x_{1},dots ,x_{k}} , then its symbol is the function of 2k variables {displaystyle x_{1},dots ,x_{k},y_{1},dots ,y_{k}} , given by dropping all terms of order less than n and replacing {displaystyle partial /partial x_{i}} by {displaystyle y_{i}} . So the symbol is homogeneous in the variables y, of degree n. The symbol is well defined even though {displaystyle partial /partial x_{i}} does not commute with {displaystyle x_{i}} because we keep only the highest order terms and differential operators commute "up to lower-order terms". The operator is called elliptic if the symbol is nonzero whenever at least one y is nonzero.
Example: The Laplace operator in k variables has symbol {displaystyle y_{1}^{2}+cdots +y_{k}^{2}} , and so is elliptic as this is nonzero whenever any of the {displaystyle y_{i}} 's are nonzero. The wave operator has symbol {displaystyle -y_{1}^{2}+cdots +y_{k}^{2}} , which is not elliptic if {displaystyle kgeq 2} , as the symbol vanishes for some non-zero values of the ys.
The symbol of a differential operator of order n on a smooth manifold X is defined in much the same way using local coordinate charts, and is a function on the cotangent bundle of X, homogeneous of degree n on each cotangent space. (In general, differential operators transform in a rather complicated way under coordinate transforms (see jet bundle); however, the highest order terms transform like tensors so we get well defined homogeneous functions on the cotangent spaces that are independent of the choice of local charts.) More generally, the symbol of a differential operator between two vector bundles E and F is a section of the pullback of the bundle Hom(E, F) to the cotangent space of X. The differential operator is called elliptic if the element of Hom(Ex, Fx) is invertible for all non-zero cotangent vectors at any point x of X.
A key property of elliptic operators is that they are almost invertible; this is closely related to the fact that their symbols are almost invertible. More precisely, an elliptic operator D on a compact manifold has a (non-unique) parametrix (or pseudoinverse) D′ such that DD′ -1 and D′D -1 are both compact operators. An important consequence is that the kernel of D is finite-dimensional, because all eigenspaces of compact operators, other than the kernel, are finite-dimensional. (The pseudoinverse of an elliptic differential operator is almost never a differential operator. However, it is an elliptic pseudodifferential operator.) Analytical index As the elliptic differential operator D has a pseudoinverse, it is a Fredholm operator. Any Fredholm operator has an index, defined as the difference between the (finite) dimension of the kernel of D (solutions of Df = 0), and the (finite) dimension of the cokernel of D (the constraints on the right-hand-side of an inhomogeneous equation like Df = g, or equivalently the kernel of the adjoint operator). In other words, Index(D) = dim Ker(D) − dim Coker(D) = dim Ker(D) − dim Ker(D*).
This is sometimes called the analytical index of D.
Example: Suppose that the manifold is the circle (thought of as R/Z), and D is the operator d/dx − λ for some complex constant λ. (This is the simplest example of an elliptic operator.) Then the kernel is the space of multiples of exp(λx) if λ is an integral multiple of 2πi and is 0 otherwise, and the kernel of the adjoint is a similar space with λ replaced by its complex conjugate. So D has index 0. This example shows that the kernel and cokernel of elliptic operators can jump discontinuously as the elliptic operator varies, so there is no nice formula for their dimensions in terms of continuous topological data. However the jumps in the dimensions of the kernel and cokernel are the same, so the index, given by the difference of their dimensions, does indeed vary continuously, and can be given in terms of topological data by the index theorem.
Topological index The topological index of an elliptic differential operator {displaystyle D} between smooth vector bundles {displaystyle E} and {displaystyle F} on an {displaystyle n} -dimensional compact manifold {displaystyle X} is given by {displaystyle (-1)^{n}operatorname {ch} (D)operatorname {Td} (X)[X]=(-1)^{n}int _{X}operatorname {ch} (D)operatorname {Td} (X)} in other words the value of the top dimensional component of the mixed cohomology class {displaystyle operatorname {ch} (D)operatorname {Td} (X)} on the fundamental homology class of the manifold {displaystyle X} up to a difference of sign. Here, {displaystyle operatorname {Td} (X)} is the Todd class of the complexified tangent bundle of {displaystyle X} . {displaystyle operatorname {ch} (D)} is equal to {displaystyle varphi ^{-1}(operatorname {ch} (d(p^{*}E,p^{*}F,sigma (D))))} , where {displaystyle varphi :H^{k}(X;mathbb {Q} )to H^{n+k}(B(X)/S(X);mathbb {Q} )} is the Thom isomorphism for the sphere bundle {displaystyle p:B(X)/S(X)to X} {displaystyle operatorname {ch} :K(X)otimes mathbb {Q} to H^{*}(X;mathbb {Q} )} is the Chern character {displaystyle d(p^{*}E,p^{*}F,sigma (D))} is the "difference element" in {displaystyle K(B(X)/S(X))} associated to two vector bundles {displaystyle p^{*}E} and {displaystyle p^{*}F} on {displaystyle B(X)} and an isomorphism {displaystyle sigma (D)} between them on the subspace {displaystyle S(X)} . {displaystyle sigma (D)} is the symbol of {displaystyle D} In some situations, it is possible to simplify the above formula for computational purposes. In particular, if {displaystyle X} is a {displaystyle 2m} -dimensional orientable (compact) manifold with non-zero Euler class {displaystyle e(TX)} , then applying the Thom isomorphism and dividing by the Euler class,[26][27] the topological index may be expressed as {displaystyle (-1)^{m}int _{X}{frac {operatorname {ch} (E)-operatorname {ch} (F)}{e(TX)}}operatorname {Td} (X)} where division makes sense by pulling {displaystyle e(TX)^{-1}} back from the cohomology ring of the classifying space {displaystyle BSO} .
One can also define the topological index using only K-theory (and this alternative definition is compatible in a certain sense with the Chern-character construction above). If X is a compact submanifold of a manifold Y then there is a pushforward (or "shriek") map from K(TX) to K(TY). The topological index of an element of K(TX) is defined to be the image of this operation with Y some Euclidean space, for which K(TY) can be naturally identified with the integers Z (as a consequence of Bott-periodicity). This map is independent of the embedding of X in Euclidean space. Now a differential operator as above naturally defines an element of K(TX), and the image in Z under this map "is" the topological index.
As usual, D is an elliptic differential operator between vector bundles E and F over a compact manifold X.
The index problem is the following: compute the (analytical) index of D using only the symbol s and topological data derived from the manifold and the vector bundle. The Atiyah–Singer index theorem solves this problem, and states: The analytical index of D is equal to its topological index.
In spite of its formidable definition, the topological index is usually straightforward to evaluate explicitly. So this makes it possible to evaluate the analytical index. (The cokernel and kernel of an elliptic operator are in general extremely hard to evaluate individually; the index theorem shows that we can usually at least evaluate their difference.) Many important invariants of a manifold (such as the signature) can be given as the index of suitable differential operators, so the index theorem allows us to evaluate these invariants in terms of topological data.
Although the analytical index is usually hard to evaluate directly, it is at least obviously an integer. The topological index is by definition a rational number, but it is usually not at all obvious from the definition that it is also integral. So the Atiyah–Singer index theorem implies some deep integrality properties, as it implies that the topological index is integral.
The index of an elliptic differential operator obviously vanishes if the operator is self adjoint. It also vanishes if the manifold X has odd dimension, though there are pseudodifferential elliptic operators whose index does not vanish in odd dimensions.
Relation to Grothendieck–Riemann–Roch The Grothendieck–Riemann–Roch theorem was one of the main motivations behind the index theorem because the index theorem is the counterpart of this theorem in the setting of real manifolds. Now, if there's a map {displaystyle f:Xto Y} of compact stably almost complex manifolds, then there is a commutative diagram[28] if {displaystyle Y=*} is a point, then we recover the statement above. Here {displaystyle K(X)} is the Grothendieck group of complex vector bundles. This commutative diagram is formally very similar to the GRR theorem because the cohomology groups on the right are replaced by the Chow ring of a smooth variety, and the Grothendieck group on the left is given by the Grothendieck group of algebraic vector bundles.
Extensions of the Atiyah–Singer index theorem Teleman index theorem Due to (Teleman 1983), (Teleman 1984): For any abstract elliptic operator (Atiyah 1970) on a closed, oriented, topological manifold, the analytical index equals the topological index.
The proof of this result goes through specific considerations, including the extension of Hodge theory on combinatorial and Lipschitz manifolds (Teleman 1980), (Teleman 1983), the extension of Atiyah–Singer's signature operator to Lipschitz manifolds (Teleman 1983), Kasparov's K-homology (Kasparov 1972) and topological cobordism (Kirby & Siebenmann 1977).
This result shows that the index theorem is not merely a differentiability statement, but rather a topological statement.
Connes–Donaldson–Sullivan–Teleman index theorem Due to (Donaldson & Sullivan 1989), (Connes, Sullivan & Teleman 1994): For any quasiconformal manifold there exists a local construction of the Hirzebruch–Thom characteristic classes.
This theory is based on a signature operator S, defined on middle degree differential forms on even-dimensional quasiconformal manifolds (compare (Donaldson & Sullivan 1989)).
Using topological cobordism and K-homology one may provide a full statement of an index theorem on quasiconformal manifolds (see page 678 of (Connes, Sullivan & Teleman 1994)). The work (Connes, Sullivan & Teleman 1994) "provides local constructions for characteristic classes based on higher dimensional relatives of the measurable Riemann mapping in dimension two and the Yang–Mills theory in dimension four."
These results constitute significant advances along the lines of Singer's program Prospects in Mathematics (Singer 1971). At the same time, they provide, also, an effective construction of the rational Pontrjagin classes on topological manifolds. The paper (Teleman 1985) provides a link between Thom's original construction of the rational Pontrjagin classes (Thom 1956) and index theory.
It is important to mention that the index formula is a topological statement. The obstruction theories due to Milnor, Kervaire, Kirby, Siebenmann, Sullivan, Donaldson show that only a minority of topological manifolds possess differentiable structures and these are not necessarily unique. Sullivan's result on Lipschitz and quasiconformal structures (Sullivan 1979) shows that any topological manifold in dimension different from 4 possesses such a structure which is unique (up to isotopy close to identity).
The quasiconformal structures (Connes, Sullivan & Teleman 1994) and more generally the Lp-structures, p > n(n+1)/2, introduced by M. Hilsum (Hilsum 1999), are the weakest analytical structures on topological manifolds of dimension n for which the index theorem is known to hold.
Other extensions The Atiyah–Singer theorem applies to elliptic pseudodifferential operators in much the same way as for elliptic differential operators. In fact, for technical reasons most of the early proofs worked with pseudodifferential rather than differential operators: their extra flexibility made some steps of the proofs easier. Instead of working with an elliptic operator between two vector bundles, it is sometimes more convenient to work with an elliptic complex {displaystyle 0rightarrow E_{0}rightarrow E_{1}rightarrow E_{2}rightarrow dotsm rightarrow E_{m}rightarrow 0} of vector bundles. The difference is that the symbols now form an exact sequence (off the zero section). In the case when there are just two non-zero bundles in the complex this implies that the symbol is an isomorphism off the zero section, so an elliptic complex with 2 terms is essentially the same as an elliptic operator between two vector bundles. Conversely the index theorem for an elliptic complex can easily be reduced to the case of an elliptic operator: the two vector bundles are given by the sums of the even or odd terms of the complex, and the elliptic operator is the sum of the operators of the elliptic complex and their adjoints, restricted to the sum of the even bundles. If the manifold is allowed to have boundary, then some restrictions must be put on the domain of the elliptic operator in order to ensure a finite index. These conditions can be local (like demanding that the sections in the domain vanish at the boundary) or more complicated global conditions (like requiring that the sections in the domain solve some differential equation). The local case was worked out by Atiyah and Bott, but they showed that many interesting operators (e.g., the signature operator) do not admit local boundary conditions. To handle these operators, Atiyah, Patodi and Singer introduced global boundary conditions equivalent to attaching a cylinder to the manifold along the boundary and then restricting the domain to those sections that are square integrable along the cylinder. This point of view is adopted in the proof of Melrose (1993) of the Atiyah–Patodi–Singer index theorem. Instead of just one elliptic operator, one can consider a family of elliptic operators parameterized by some space Y. In this case the index is an element of the K-theory of Y, rather than an integer. If the operators in the family are real, then the index lies in the real K-theory of Y. This gives a little extra information, as the map from the real K-theory of Y to the complex K-theory is not always injective. If there is a group action of a group G on the compact manifold X, commuting with the elliptic operator, then one replaces ordinary K-theory with equivariant K-theory. Moreover, one gets generalizations of the Lefschetz fixed-point theorem, with terms coming from fixed-point submanifolds of the group G. See also: equivariant index theorem. Atiyah (1976) showed how to extend the index theorem to some non-compact manifolds, acted on by a discrete group with compact quotient. The kernel of the elliptic operator is in general infinite dimensional in this case, but it is possible to get a finite index using the dimension of a module over a von Neumann algebra; this index is in general real rather than integer valued. This version is called the L2 index theorem, and was used by Atiyah & Schmid (1977) to rederive properties of the discrete series representations of semisimple Lie groups. The Callias index theorem is an index theorem for a Dirac operator on a noncompact odd-dimensional space. The Atiyah–Singer index is only defined on compact spaces, and vanishes when their dimension is odd. In 1978 Constantine Callias, at the suggestion of his Ph.D. advisor Roman Jackiw, used the axial anomaly to derive this index theorem on spaces equipped with a Hermitian matrix called the Higgs field.[29] The index of the Dirac operator is a topological invariant which measures the winding of the Higgs field on a sphere at infinity. If U is the unit matrix in the direction of the Higgs field, then the index is proportional to the integral of U(dU)n−1 over the (n−1)-sphere at infinity. If n is even, it is always zero. The topological interpretation of this invariant and its relation to the Hörmander index proposed by Boris Fedosov, as generalized by Lars Hörmander, was published by Raoul Bott and Robert Thomas Seeley.[30] Examples Chern-Gauss-Bonnet theorem Suppose that {displaystyle M} is a compact oriented manifold of dimension {displaystyle n=2r} . If we take {displaystyle Lambda ^{even}} to be the sum of the even exterior powers of the cotangent bundle, and {displaystyle Lambda ^{odd}} to be the sum of the odd powers, define {displaystyle D=d+d^{*}} , considered as a map from {displaystyle Lambda ^{even}} to {displaystyle Lambda ^{odd}} . Then the analytical index of {displaystyle D} is the Euler characteristic {displaystyle chi (M)} of the Hodge cohomology of {displaystyle M} , and the topological index is the integral of the Euler class over the manifold. The index formula for this operator yields the Chern–Gauss–Bonnet theorem.
The concrete computation goes as follows: according to one variation of the splitting principle, if {displaystyle E} is a real vector bundle of dimension {displaystyle n=2r} , in order to prove assertions involving characteristic classes, we may suppose that there are complex line bundles {displaystyle l_{1},,ldots ,,l_{r}} such that {displaystyle Eotimes mathbb {C} =l_{1}oplus {overline {l_{1}}}oplus dotsm l_{r}oplus {overline {l_{r}}}} . Therefore, we can consider the Chern roots {displaystyle x_{i}(Eotimes mathbb {C} )=c_{1}(l_{i})} , {displaystyle x_{r+i}(Eotimes mathbb {C} )=c_{1}left({overline {l_{i}}}right)=-x_{i}(Eotimes mathbb {C} )} , {displaystyle i=1,,ldots ,,r} .
Using Chern roots as above and the standard properties of the Euler class, we have that {textstyle e(TM)=prod _{i}^{r}x_{i}(TMotimes mathbb {C} )} . As for the Chern character and the Todd class,[31] {displaystyle {begin{aligned}operatorname {ch} left(Lambda ^{even}-Lambda ^{odd}right)&=1-operatorname {ch} (T^{*}Motimes mathbb {C} )+operatorname {ch} left(Lambda ^{2}T^{*}Motimes mathbb {C} right)-ldots +(-1)^{n}operatorname {ch} left(Lambda ^{n}T^{*}Motimes mathbb {C} right)\&=1-sum _{i}^{n}e^{-x_{i}}(TMotimes mathbb {C} )+sum _{i
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ne etC. om the sorted array. cending order based o nding order based on distance values he most frequer class of thes rows Return the predicted class Hidden Markov Model: The hidden markov model is a finite se enciated with (generally multidimensional) probability distribution Transitions ened by a set ot probabilities called transition probabilities. In a particula ation can be generated, according to the associated probability state visible to an external observer and therefore states are hidde 5. associated observation can be gens ities calle probability dieti a finite set of states, each of which is istribution Transitions among are observation particular state an outcome or ds Soclated probability distribution. It is only the outcon ot the state name hidden markov model. are 'hidden' to the outside, hen In order to define on HMM completely, following elements are needed. The number of states of the model, N. ments are needed. The number of observation symbol in the alophabet M. if the observation ai then M is infinite. ous A set of state transition probabilities A = {a). a= Plq=i|9,= i), 1si, jsN where q, denotes the current state. Transition probability should satisfy the normal stochastic constraints, and a0, 1 si,jsN and 2a,=1 1sisN A probability distribution in each of the states B {b (k)} b(K) = P{a, = v, | q, = i). 1sj
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# Froude-Krylov force
The Froude-Krylov force, sometimes also called the Froude-Kriloff force is a hydrodynamical force named after William Froude and Alexei Krylov. The Froude-Krylov force is the force introduced by the unsteady pressure field generated by undisturbed waves. The Froude-Krylov force is together with the diffraction force the total non-viscous forces acting on a floating body in regular waves. The diffraction force is due to the floating body disturbing the waves.
### Additional recommended knowledge
The Froude-Krylov force can be calculated from:
$\vec F_{FK} = - \iint_{S_w} p ~ \vec n ~ ds$
Where Sw is the wetted surface, p the pressure and $\vec n$ the body's normal vector pointing into the water.
## See also
• Response Amplitude Operator
## References
• Faltinsen, O. M. (1990). Sea Loads on Ships and Offshore Structures. [Cambridge University Press]]. ISBN 0-521-45870-6.
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia article "Froude-Krylov_force". A list of authors is available in Wikipedia.
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2020-04-05 22:02:59
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https://www.eyegrade.org/doc/user-manual-0.2/
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# Eyegrade User Manual
Author: Jesús Arias Fisteus
This user manual refers to Eyegrade 0.2 and later versions. For the 0.1.x series see this other user manual.
# 1 Installing Eyegrade
Eyegrade depends on the following free-software projects:
• Python: the run-time environment and standard library for the execution Python programs. Eyegrade is known to work with Python 2.6.
• Opencv: a widely used computer-vision library. Version 2.0 or later is needed. Not only the OpenCV library, but also the python bindings distributed with it are needed.
• Qt: a multi-platform library for developing graphical user interfaces.
• PyQt: Python bindings to Qt.
• Tre: a library for regular expressions. Install version 0.8.0 or later. Both the library and python bindings are needed.
If you have any version of the Eyegrade 0.1.x series already installed in your system, you have to install PyQt.
In Linux, install the package python-qt4 and update Eyegrade following the instructions at Updating Eyegrade.
In Windows, Download PyQt and install it. If you installed Eyegrade 0.1.x following the instructions in this user manual, you have Python 2.6-x86. Choose that version of the installer of PyQt. Alternatively, there is a copy of the file you need at PyQt-Py2.6-x86-gpl-4.9.6-1.exe. The installer of PyQt will probably detect you Python installation automatically.
After that, update Eyegrade following the instructions at Updating Eyegrade.
The directory layout of the code of Eyegrade has changed with respect to versions 0.1.x. The code is now inside the eyegrade subdirectory instead of the src/eyegrade subdirectory. Therefore, the PYTHONPATH environment variable should be set to the main directory of Eyegrade now, instead of the src subdirectory as before. When upgrading, remember to update this environment variable if you have configured it in your system-wide environment variables.
If you cannot see the icons of the buttons the first time you run the new version of Eyegrade, reboot the computer and try again.
### 1.1.1 Main changes from versions 0.1.x
The main changes from Eyegrade 0.1.x to Eyegrade 0.2 you need to be aware of are:
• The main program is run now without command line arguments. You'll select the .eye file, student lists, webcam to use, etc. from the user interface.
• There is the new concept of a session. A session stores the state of the grading of a given exam. You create a new session when you start to grade the exam. The images of the exams and results are stored in this session as you grade them. You can later resume a session and continue grading it. A session is stored in the directory of your file system that you choose.
## 1.2 Installation on GNU/Linux
If your Linux distribution is not very old, it should provide most of the needed software packages. Specific instructions for Debian GNU/Linux and Ubuntu are provided below.
### 1.2.1 Installation on Debian and Ubuntu
Almost all the required software packages are already available in recent versions of Debian GNU/Linux and Ubuntu. The only exception are the Python bindings for Tre, which have to be installed manually.
Using your favorite package manager (apt-get, aptitude, synaptic, etc.), install the following packages: python (check that the version is either 2.6 or 2.7), python-qt4, python-opencv, python-numpy, libavformat53, libtre5. In older versions of Ubuntu and Debian, you might need to install also libcv2.1 (in even older versions, the name of this package is libcv4 instead). If you can't find libavformat53 in your distribution, use libavformat52 instead.
Then, you have to install the Python bindings for Tre. First, install these two additional packages: python-dev, libtre-dev. Then, from the command line, download, compile and install the Python bindings:
wget http://laurikari.net/tre/tre-0.8.0.tar.gz
tar xzvf tre-0.8.0.tar.gz
cd tre-0.8.0/python/
python setup.py build
sudo python setup.py install
Now, you only need to download Eyegrade using the git source code revision system (install the git package if you do not have it):
cd $DIR git clone -b master git://github.com/jfisteus/eyegrade.git Note: replace$DIR above with the directory in which you want Eyegrade to be installed.
Finally, add the $DIR/eyegrade directory to your PYTHONPATH and check that Eyegrade works: export PYTHONPATH=$DIR/eyegrade
The export command works only in the current terminal. You can make it permanent by adding it to your $HOME/.bashrc file (if you use the BASH shell). That's all! Eyegrade should now be installed. For further testing, go to Launching Eyegrade. ## 1.3 Installation on Microsoft Windows You have to follow these steps, explained in the following sections, in order to install Eyegrade in Windows: 1.- Install Python 2.6 (including Tre). 2.- Install PyQt. 3.- Install OpenCV 2.1. 4.- Install Eyegrade itself. ### 1.3.1 Installing Python The easiest way to install Python, PyQt and Tre in Windows is to download a ZIP file that contains all of them and extract it in your file system. 1.- Download the ZIP file from: Python26.zip. 2.- Extract it somewhere in your file system (I recommend C:\). A directory named Python26 will appear. Be aware that the full path of the directory where you extract it cannot contain white-spaces. 3.- Add the main directory (Python26) of your Python installation to your system PATH. For example, if you uncompressed Python at C:\, add C:\Python26 to the system PATH variable. You can test your installation by opening a new command line console and launching the interactive Python interpreter in it: Python If it does not start, you have probably not added it correctly to your system PATH. Opening a new console is important because changes in the system PATH apply only to newly-opened consoles. Once in the Python interpreter, the following command should work: import tre This command should not output any message. If it does, there is a problem with the installation. If tre complains about a missing DLL, the problem is probably that the installation directory of Python is not in the system PATH. If you already have a Python 2.6 installation and want to use it, you must, on that installation of Python, download and install Tre 0.8.0. You will need Microsoft Visual Studio 2008 (the express version is free and works) for this last step. ### 1.3.2 Installing PyQt4 Download PyQt. Select the Windows 32-bit installer for Python 2.6, event if you have a 64-bit version of Windows. Alternatively, there is a copy of the file you need at PyQt-Py2.6-x86-gpl-4.9.6-1.exe. Run the installer. From the optional software that the installer suggests, you only need to select the Qt runtime. ### 1.3.3 Installing OpenCV Download the EXE installer of OpenCV 2.1.0 for Windows platforms: OpenCV-2.1.0-win32-vs2008.exe. There is a copy of the same file at OpenCV21.exe. Execute the installer. Again, it is better to choose an installation path which has no white-spaces in it. The installer will eventually ask to put OpenCV in your system PATH. Answer yes for this user or yes for all the users. In order to test the installation, open a new command prompt window (it must necessarily be a new window for the system path to be updated). Run the python interpreter as explained in the previous section and type in it: import cv This command should not output any message. If it does, there is a problem with the installation. ### 1.3.4 Installing Eyegrade By now, the recommended way to install Eyegrade is through the Git version control system. This way it will be easier to update Eyegrade in the future, when new versions are released (see Updating Eyegrade). In order to install Eyegrade through Git, follow these steps: 1.- Download and install Git if you do not have it installed. The installer and installation instructions are available at <http://git-scm.com/>. 2.- Open a command line prompt (for example, a Git shell), enter the directory you want Eyegrade to be installed (again, with no white-spaces in it), and type: git clone -b master git://github.com/jfisteus/eyegrade.git If you prefer not to install Git: 1.- Download the ZIP file eyegrade-0.2.6.zip. Extract it in your file system, in a directory with no white-spaces in its path. Once you have Eyegrade installed (either with or without Git), test it. For example, if you have installed both Python and Eyegrade at C:\: set PYTHONPATH=C:\eyegrade C:\Python26\python -m eyegrade.eyegrade It should dump a help message. Tip: it may be convenient adding C:Python26 to your system path permanently, and adding PYTHONPATH to the system-wide environment variables. There are plenty of resources in the Web that explain how to do this. For example, http://www.windows7hacker.com/index.php/2010/05/how-to-addedit-environment-variables-in-windows-7/. Eyegrade should now be installed. Nevertheless, it might be a good idea to reboot now the computer, in order to guarantee that the installation of OpenCV and PyQt has completed. After that, go to Launching Eyegrade. ## 1.4 Installation on Mac OS X Sorry, Eyegrade is not currently supported on that platform. Volunteers to support the platform are welcome. ## 1.5 Updating Eyegrade From time to time, a new release of Eyegrade may appear. If you installed Eyegrade using Git, updating is simple. Open a command prompt window, enter the Eyegrade installation directory and type: git pull This should work on any platform (Linux, Windows, etc.) # 2 Grading Exams The main purpose of Eyegrade is grading exams. In order to grade exams, you will need: • The Eyegrade software installed in your computer. • The exam configuration file, which specifies the number of questions in the exam, solutions, etc. It is normally named with the .eyeextension, such as exam.eye. • A compatible webcam, with resolution of at least 640x480. It is better if it is able to focus (manually or automatically) at short distances. • The list of students in your class, if you want Eyegrade to detect student IDs. • The exams to grade. ## 2.1 Launching Eyegrade This section explains how to run Eyegrade. If it is the first time you use Eyegrade, you can try it with the sample file exam-A.pdf located inside the directory doc/sample-files of your installation of Eyegrade. Print it. You'll find also in that directory the file exam.eye that contains the metadata for this exam. You'll need to load this file later from Eyegrade. Eyegrade can be launched from command line: python -m eyegrade.eyegrade This command opens the user interface of Eyegrade: Before beginning to grade exams, especially the first time you run Eyegrade, you can check that Eyegrade can access your webcam. In the Tools menu select the Select camera entry: The next step is creating a grading session. Select New session in the menu Session. A multi-step dialog will ask for some data Eyegrade needs for creating the session: • Directory and exam configuration: you need to enter here the following information: • Directory: select or create a directory for this session. The directory must be empty. • Exam configuration file: select the .eye file associated to this exam. If you printed the sample exam distributed with Eyegrade, use the exam.eye file from the same directory. • Student id files: select zero, one or more files that contain the list of students in the class. The files should be plain text and contain a line per student. Each line must have a first field with the student id and, optionally, a second field with the student name. It may have more fields, which Eyegrade will ignore. Fields must be separated by one tabulator character. • Scores for correct and incorrect answers: this step is optional. If you provide the scores awarded to correct answers (and optionally deducted from incorrect answers), Eyegrade will show the marks of each exam. After you finish with this dialog, Eyegrade opens the session. It shows the image from the webcam and starts scanning for the exam. Point the camera to the exam until the image is locked. At this point, Eyegrade should show the answers it has detected. Read the following sections for further instructions. ## 2.2 The session directory A grading session in Eyegrade represents the grading of a specific exam for a group of students. For example, you would grade the exams for the final exam of all your students in the subject Computer Networks in just one session. Other exams, such as the re-sit exam of the same subject, should go in separate sessions. Grading sessions are associated to a directory in your computer. You select or create this directory when you create a new session. Eyegrade stores there all the data belonging to the grading session (configuration file, student lists, grades, images of the already graded exams, etc.) You can open again later an existing session with the Open session option of the Session menu. In the file selection dialog that appears, select the session.eye file inside the directory of the session you want to open. When you open the session, you can continue grading new exams that belong to that session. ## 2.3 Application modes At a given instant, the application is in one of these modes: • No session mode: no session is opened. You can open an existing session or create a new session. • Search mode: a session is open. The application continually scans the input from the webcam, looking for a correct detection of an exam. • Review mode: a session is open. The application shows a still capture of an exam with the result of the grading, so that the user can review it and fix things, if necessary, before saving the score of the exam. • Manual detection mode: a session is open. In the rare cases in which the system is not able to detect the geometry of the exam, you can enter this mode and mark the corners of the answer tables. Eyegrade will be able to detect the tables once you tell it where the corners are. The application starts with no open session. Once you open or create a session, it changes to the search mode. When the system detects an answer sheet that it can read, it locks the capture and enters the review mode. Once you save the score of the exam, Eyegrade automatically goes back to the search mode in order to scan the next exam. You can enter the manual detection mode by issuing the appropriate command while in the other modes. From any of the other modes, you can go back to the no session mode with the Close session command in the Session menu. ## 2.4 The search mode In the search mode, you have to get the camera to point to the answer table of the exam, including, if present, the id box above it and the small squares at the bottom. Eyegrade will continually scan the input of the webcam until the whole exam is correctly detected. At that moment, Eyegrade will switch to the review mode. Sometimes, Eyegrade is able to detect the answer table but not the ID table at the top of it. You can notice that because the detected answers are temporary shown on top of the image. At this point, you may try further until the ID box is also detected, or just use the Capture the current image command of the Grading menu, which will force the system to switch to the review mode, using the most recent capture in which the answer table was detected. You will be able to manually enter the missing student id in that mode. In rare occasions, Eyegrade could fail event to detect the answer table. The Manual detection command of the Grading menu allows you to help the system detect it. These are the commands available in the search mode, all of them at the Grading menu: • Capture the current image (shortcut 's'): forces the system to enter the review mode with the the most recent capture in which Eyegrade was able to detect the answer table. If there is no such capture, the system just uses the current capture. • Manual detection of answer tables (shortcut 'm'): the system enters the manual detection mode, in which you can help the system detect the answer table by marking the corners of the answer tables. After that, the system will detect the answers of the student and automatically enter the review mode. See The manual detection mode. ## 2.5 The review mode In the review mode you can review and, if necessary, fix the information detected by Eyegrade in the current exam. You can do it on both the answers given by the student to each question and the student id. You enter the review mode in one of the following three different situations: • With the answers of the student and her id detected. This is the usual case. Eyegrade was able to detect the whole exam, and you can review the information extracted from it. • With the answers of the student, but without her id. This is the case when you use the Capture the current image command in the search mode because Eyegrade detected the answer table in at least one capture, but not the student id box. In this case, you can review the answers given by the student and manually enter her id. • With neither the answers of the student nor her id. This is the case when you use the Capture the current image command in the search mode because Eyegrade was not able to detect anything from the exam. In this situation, you can switch to the manual detection mode to help the system to detect the answer tables, and manually enter the student id. The user interface shows, in this mode, a capture of the exam augmented with the detected information, as shown in the following image: As you can see, the system shows: • The answers of the student, with a green circle for correct answers and a red circle for the incorrect ones. When the student leaves a question unanswered, or provides a wrong answer for it, the correct answer for that question is marked with a small dot. • The detected student id, at the bottom of the image, and his name (when the name is provided in the student list files). • The total number of correct, incorrect and blank answers, at the bottom. The total score of the exam is also shown if the session is configured with the scores for the answers. • The model of the exam. The model is detected from the small black squares that are printed below the answer table. • The sequence number of this exam. It is incremented with each graded exam. In this mode, you can perform the following actions (see the Grading menu): • Modify the answers of the student, if there are mistakes in the automatically-detected answers, as explained in Modifying student answers. • Modify the student id, if the system did not recognize it or recognized a wrong id, as explained in Modifying the student id. • Save and capture next exam (shortcut 'Space-bar'): saves the grades of this exam as well as the annotated captured image, and enters the search mode in order to detect the next exam. Tip: before saving, it is better to remove the exam from the sight of the camera to avoid it from being captured again. You can even put the next exam under the camera before saving to speed up the process. • Discard capture (shortcut 'Backspace'): discards the current capture without saving it. It is useful, for example, when the capture is not good enough, or when you discover that the same exam has already been graded before. • Manual detection of answer tables (shortcut 'm'): the system enters the manual detection mode, in which you can help the system detect the answer table by marking the corners of the answer tables. After that, the system will detect the answers of the student and automatically enter again the review mode. This command is allowed only when the system failed to recognize the geometry of the answer tables. See The manual detection mode. ### 2.5.1 Modifying student answers The optical recognition system of Eyegrade may fail sometimes, due to its own limitations, or students filling their exams in messy ways. Sometimes, Eyegrade shows a cell in the answer table as marked when it is not, or a cell is not marked when it actually is. In addition, if Eyegrade thinks that two cells of the same question are marked, it will leave that question as blank. You are able to fix those mistakes at the review mode. Click on a cell of the answer table to change an answer of the student that was not correctly detected by Eyegrade: when the student marked a given cell, but the system detected the question as blank, or simply showed other answer of that question as marked, just click on the cell the student actually marked. When the student left a question blank but the system did mark one of the cells as the answer, click on that cell to clear it. In both cases, Eyegrade will compute the scores again and immediately update the information on the screen. ### 2.5.2 Modifying the student id Normally, you should provide Eyegrade with the list of class, because detection of student ids performs much better in that case. When scanning the id in an exam, Eyegrade sorts ids of the students in class according to the estimated probability of being the id in the exam. The one with the most probability is shown. In the review mode, you can enter the correct student id when Eyegrade does not detect it, or detects a wrong one. When you select the Edit student id command in the Grading menu, a dialog for selecting the student id is shown: The dialog shows the students from the student list sorted by their probability (according to the OCR module) of being the student whose id is in the exam. You just choose one in the drop-down menu. In addition, you can filter students by writing part of their id number or their name. If the student is not in your list, you can also enter in the dialog her id number and name. If you do that, follow the same format: student id, white space, student name. ## 2.6 The manual detection mode In some rare occasions, Eyegrade may not be able to detect the answer tables. In those cases, you can enter the manual detection mode from the search mode (and also from the review mode if you entered that mode using the Capture the current image command). When entering the manual detection mode, the latest capture of the camera will be shown. In this mode, just click with the cursor in the four corners of each answer table (a small circle will appear in every location you click). The order in which you click on the corners does not matter. After having done that, Eyegrade will infer the limits of each cell, and based on them it will read the answers of the student and the exam model. It will enter then the review mode. The following two images show an example. In the first image, the user has selected six corners (notice the small blue circles): After she selects the remaining two corners, the system detects the answers and goes back to the review mode: Note, however, that the student id will not be detected when you use this mode. When the system goes back to the review mode, set the id as explained in Modifying the student id. At any point of the process, you can use the Manual detection of answer tables command (shortcut 'm') to reset the selection of corners and start again. If you think that the captured image is not good enough, you can also use the discard command (shortcut 'Backspace') to go again to the search mode. Tip: in the manual detection mode, make sure that the captured image shows all the answer tables as well as the exam model squares at the bottom. # 3 Processing Student Grades The output produced by Eyegrade consists of: • A file with the scores, named eyegrade-answers.csv: it contains one line for each graded exam. Each line contains, among other things, the student id number, the number of correct and incorrect answers, and the answer to every question in the exam. Student grades can be extracted from this file. • One snapshot of each graded exam, in PNG format: snapshots can be used as an evidence to show students. They can be shown to students coming to your office to review the exam, or even emailed to every student. The default name for those images is the concatenation of the student id and exam sequence number, in order to facilitate the instructor to locate the snapshot for a specific student. ## 3.1 The answers file The file eyegrade-answers.csv produced by Eyegrade contains the scores in CSV format (with tabulator instead of comma as a separator), so that it can be easily imported from other programs such as spreadsheets. This is an example of such a file: 0 100999991 D 9 6 0 1/2/2/4/1/2/2/0/0/3/2/0/3/2/0/4/3/0/1/2 1 100999997 C 15 1 0 2/4/4/3/1/0/1/2/1/1/0/1/0/4/3/0/1/4/3/4 2 100800003 D 6 14 0 4/2/2/2/1/2/1/3/2/1/3/1/2/1/3/1/4/1/4/3 3 100777777 A 7 13 0 3/2/3/2/3/3/2/4/3/1/3/1/4/1/4/2/2/3/4/2 The columns of this file represent: 1.- The exam sequence number (the same number the user interface shows below the student id in the review mode). 2.- The student id (or '-1' if the student id is unknown). 3.- The exam model ('A', 'B', 'C', etc.) 4.- The number of correct answers. 5.- The number of incorrect answers. 6.- The number of undetermined answers (answers marked as blank because of the system detecting more than one marked cell). 7.- The response of the student to each question in the exam, from the first question in her model to the last. '0' means a blank answer. '1', '2', etc. mean the first choice, second choice, etc., in the order they were presented in her exam model. Exams are in the same sequence they were graded. See Exporting a listing of scores to know how to produce a listing of scores in the order that best fits your needs. Tip: if you start a new grading session from the same directory, the file eyegrade-answers.csv will not be overwritten. New grades will just be appended at the end. Thus, it is safe stopping a grading session, closing the application and continuing later. Separate grading sessions must be executed from different directories to avoid using the same eyegrade-answers.csv file. Tip: you can edit this file with a text editor if, for example, you discover that the same exam was graded more than once (just remove the duplicate line). ## 3.2 Exporting a listing of scores You will probably want to import the listing of scores from your grade-book. You can easily process eyegrade-answers.csv to produce a CSV-formatted file with three columns: student id, number of correct answers and number of incorrect answers, in the order you want. You can even produce the listing to for just a subset of the students. In order to do that, you need a listing of students whose grades you want to list. The listing must be a CSV file in which the first column contains the student ids (the rest of the columns will be just ignored). Normally, you will use the same listing of students you used to run Eyegrade. This is an example of such a file: 100000333 Baggins, Frodo 100777777 Bunny, Bugs 100999997 Bux, Bastian B. 100999991 Potter, Harry 100800003 Simpson, Lisa This command will produce the listing in a file named sorted-listing.csv: python -m eyegrade.mix_grades eyegrade-answers.csv student-list.csv -o sorted-listing.csv The output for the listing above, and the sample file shown in The answers file, would be: 100000333 100777777 7 13 100999997 15 1 100999991 9 6 100800003 7 13 Scores will be in the same order as the student list. The second and third columns represent the number of correct and wrong answers, respectively. In the example, the first student has those columns empty because there is no exam associated to his id. Importing the previous file in a spreadsheet should be straightforward, because the list of students will now be in the same order as your spreadsheet. If there are exams in the answers file of students not in your list, the default behavior is including them in the listing, after the rest of the students. The rationale behind this behavior is apreventing accidental losses of student scores. This behavior can be changed (see Exporting a listing for a subset of students). See Mixing manually-graded questions if you need to produce listings in exams combining MCQ questions with manually-graded questions. ### 3.2.1 Exporting a listing for a subset of students In order to extract the scores for just a subset of the students, create a student list with the ids of the students you want and run the program with the -i option: python -m eyegrade.mix_grades eyegrade-answers.csv student-list.csv -i -o sorted-listing.csv The -i option makes Eyegrade ignore students that are in the answers file but not in the student list. That is, the listing will only contain the students that are in the student list you provide. This option may be useful, for example, if you examine students coming from different classes or groups. With this option you can produce a separate listing for each class. # 4 Editing exams Although you can use any software of your preference to typeset the exams, Eyegrade provides a module for doing that in combination to the LaTeX document preparation system. First, write your questions in an XML document like the following one: <?xml version="1.0" encoding="UTF-8"?> <exam xmlns="http://www.it.uc3m.es/jaf/eyegrade/ns/" xmlns:eye="http://www.it.uc3m.es/jaf/eyegrade/ns/"> <subject>Computers and More</subject> <degree>An Awesome Degree</degree> <title>Final examination</title> <date>January 1st, 2011</date> <duration>10 min.</duration> <question> <text> What is Python? </text> <choices> <correct>A programming language</correct> <incorrect>A computer manufacturer</incorrect> <incorrect>A kind of tree</incorrect> <incorrect>Who knows!</incorrect> </choices> </question> <question> <text> What is a webcam? </text> <choices> <correct>A video camera whose output may be viewed in real time over a network, especially over the Internet.</correct> <incorrect>A video camera intended to spy on spiders</incorrect> <incorrect>A kind of fish</incorrect> <incorrect>Who knows!</incorrect> </choices> </question> <question> <text> What is the thing at the right? </text> <code eye:position="right" eye:width="0.4">for letter in ['a', 'b', 'c']: print letter</code> <choices> <correct>A computer program</correct> <incorrect>A recipe</incorrect> <incorrect>A tree</incorrect> <incorrect>Who knows!</incorrect> </choices> </question> <question> <text> Is the thing in the right a logo? </text> <figure eye:width="0.16" eye:position="right">sample-logo.eps</figure> <choices> <correct>Well, it tries to be a logo, to be honest.</correct> <incorrect>No, it's a tree.</incorrect> <incorrect>No, it's a perfect square.</incorrect> <incorrect>Who knows!</incorrect> </choices> </question> <question> <text> Which is the capital of Spain? </text> <choices> <correct>Madrid.</correct> <incorrect>Barcelona.</incorrect> <incorrect>Paris.</incorrect> <incorrect>None of the other answers is correct.</incorrect> </choices> </question> <question> <text> Which is the capital of France? </text> <choices> <correct>Paris.</correct> <incorrect>Lion.</incorrect> <incorrect>Madrid.</incorrect> <incorrect>None of the other answers is correct.</incorrect> </choices> </question> <question> <text> What does hola'' mean in Spanish? </text> <choices> <correct>Hello''.</correct> <incorrect>Bye''.</incorrect> <incorrect>Wave''.</incorrect> <incorrect>Hola'' is not a word in Spanish.</incorrect> </choices> </question> <question> <text> How many months has a year? </text> <choices> <correct>12.</correct> <incorrect>6.</incorrect> <incorrect>10.</incorrect> <incorrect>It depends on whether it is a leap year or not.</incorrect> </choices> </question> <question> <text> Are dolphins mammals? </text> <choices> <correct>Yes.</correct> <incorrect>No.</incorrect> <incorrect>Sometimes.</incorrect> <incorrect>Who knows\ldots.</incorrect> </choices> </question> <question> <text> Who created Python? </text> <choices> <correct>Guido van Rossum.</correct> <incorrect>Richard Stallman.</incorrect> <incorrect>Dennis M. Ritchie.</incorrect> <incorrect>Jamie Zawinski.</incorrect> </choices> </question> <question> <text> How is Germany'' said in German? </text> <choices> <correct>Deutschland.</correct> <incorrect>Die Schweiz.</incorrect> <incorrect>Germany.</incorrect> <incorrect>Berlin.</incorrect> </choices> </question> <question> <text> What is Linux? </text> <choices> <correct>A free-software operating system.</correct> <incorrect>A river in Spain.</incorrect> <incorrect>The tallest mountain on Earth.</incorrect> <incorrect>An Internet company that sells phones.</incorrect> </choices> </question> </exam> Then, create a LaTeX template for the exam. This is an example: \documentclass[a4paper,11pt]{article} \usepackage{amssymb, amsmath} \usepackage[utf8]{inputenc} \usepackage{indentfirst} \usepackage{multirow} \usepackage[a4paper, margin=2.5cm, top=2cm, bottom=3cm]{geometry} \pagestyle{empty} {{declarations}} \begin{document} \begin{center} \begin{tabular}{p{5.2cm}r} \multirow{5}{*}[0.35cm]{\scalebox{0.18}{\includegraphics{sample-logo.eps}}} & \Large \textbf{{{subject}}} \\ & \textbf{{{degree}}} \\ & \\ {{date}} & {{title}}. \\ Duration: {{duration}} & Score: 5 points out of 10 total points for the exam. \\ \end{tabular} \end{center} \vspace{0.5cm} \emph{There is only one correct answer for each multiple choice question. Each correct answer adds 1 point. Each incorrect answer has a penalty of$1/3$points. If no answer no score is awarded, neither positive nor negative.} \vspace{0.5cm} \begin{center} \begin{tabular}{|p{0.8\textwidth}|} \hline \begin{itemize} \item Mark out your answers with an X''. \item If more than one answer or no answer are selected, the question is considered to be not answered (no score is awarded, neither positive nor negative). \item Remember to fill in your name and student ID. \end{itemize} \\ \hline \end{tabular} \end{center} \vspace{0.2cm} \begin{center} (1) Write your personal data clearly. \end{center} \begin{center} \large \begin{tabular}{|l|p{12cm}|} \hline Last name: & \\ \hline First name: & \\ \hline Group: & \\ \hline \end{tabular} \end{center} \vspace{0.2cm} \begin{center} (2) Write down your student ID and cross out your answers with an X'': \end{center} \begin{center} \large \textbf{Model:} {{model}} \end{center} {{id-box(9,ID)}} {{answer-table}} \clearpage {{questions}} \end{document} In the template, notice that there are some marks within {{ and }} that are intended to be replaced by the script with data from the exam: • {{declarations}}: the script will put there declarations needed for the generate LaTeX file. • {{subject}}, {{degree}}: name of the subject and degree it belongs to. Taken from the XML file with the questions. • {{title}}: the title of the exam. Taken from the XML file with the questions. • {{duration}}: duration of the exam. Taken from the XML file with the questions. • {{model}}: a letter representing the model of the exam. Each model has a different ordering for questions and choices within questions. • {{id-box(9,ID}}: replaced by a box for students to fill in their IDs. The number of digits and the text to be put at the left of the box are specified within the parenthesis. • {{answer-table}}: replaced by the table in which students mark out their answers. • {{questions}}: replaced by the questions of the exam. Note that a template is highly reusable for different exams and subjects. Once the exam file and the template have been created, the script create_exam.py parses them and generates the exam in LaTeX format: python -m eyegrade.create_exam -e exam-questions.xml -m 0AB template.tex -o exam The previous command will create models 0, A and B of the exam with names exam-0.tex, exam-A.tex and exam-B.tex. Exam model 0 is a special exam in which questions are not reordered. The correct answer is always the first choice. Those files can be compiled with LaTeX to obtain a PDF that can be printed. In addition, the exam.eye file needed to grade the exam is automatically created (or updated if it already exists). The script create_exam.py has other features, like creating just the front page of the exam (no questions needed). They can be explored with the command-line help of the program: python -m eyegrade.create_exam -h The answer table can be enlarged or reduced with respect to its default size, using the -S option and passing a scale factor (between 0.1 and 1.0 to reduce it, or greater than 1.0 to enlarge it). The following command enlarges the default size in a 50% (factor 1.5): python -m eyegrade.create_exam -e exam-questions.xml -m A template.tex -o exam -S 1.5 # 5 Advanced features ## 5.1 Webcam selection If your computer has more than one camera (e.g. the internal camera of the laptop and an external camera you use to grade the exams), Eyegrade will select one of them by default. If the selected camera is not the camera you want to use to grade the exams, use the -c <camera-number> option when invoking Eyegrade. Cameras are numbered 0, 1, 2, 3, etc. Invoke Eyegrade with a different camera number until the interface displays the one you want. For example, to select the camera numbered as 2: python -m eyegrade.eyegrade exam.eye -c 2 -l student-list.csv When the number is -1, eyegrade will automatically test different camera numbers until it finds one that works. When you select a camera number that does not exist or does not work, Eyegrade will also look automatically for other camera that works. You can configure Eyegrade to always use a specific camera number by inserting the option camera-dev in the default section of the configuration file: ## Sample configuration file. Save it as$HOME/.eyegrade.cfg
[default]
## Default camera device to use (int); -1 for automatic selection.
camera-dev: 1
Save it in your user account with name .eyegrade.cfg. In Windows systems, your account is at C:\Documents and Settings\<your_user_name>.
## 5.2 Mixing manually-graded questions
You may want to mix in the same exam MCQ questions with other type of questions that must be graded manually. Even though Eyegrade can only grade the MCQ questions of the exam, it can simplify a little bit the process of mixing grades.
First, grade the MCQ exams with Eyegrade. Then, grade the other questions without changing the ordering of the exams.
Create a new CSV file with only one column, which contains the student ids of the students that submitted the exam. It will help a lot producing this listing in the same order you have graded the exams. Such a listing can be trivially obtained from the file eyegrade-answers.csv. In Linux, it can be done with just a command:
cut eyegrade-answers.csv -f 2 >extra-marks.csv
Edit that listing to include the marks of the manually-graded questions. Write marks in one or more columns at the right of the student id. Having this file the same order of your exams, introducing manual marks should be easier, since you do not need to search. This is an example with only one manual mark per exam (just one column):
100999991 7
100999997 8
100800003 5
100777777 9.5
The final listing that combines the results of all the questions can be produced with mix_grades:
python -m eyegrade.mix_grades eyegrade-answers.csv student-list.csv -x extra-marks.csv -o sorted-listing.csv
The columns with the manual marks would appear at the right in the resulting file:
100000333
100777777 7 13 9.5
100999997 15 1 8
100999991 9 6 7
100800003 7 13 5
## 5.3 Creating the exams in a word processor
The current prototype of Eyegrade require users to know LaTex in order to personalize exam templates. This section explains an alternative way to create exams compatible with Eyegrade in a word processor such as Microsoft Word. If you create your own exams with a word processor, you'll need also to edit the .eye file manually. See Manually editing the .eye file.
The objective is emulating the tables that Eyegrade creates so that the program can read them. This is an example:
You can use as a template this example MS Word document. It shows an answer table for 20 questions, which you can edit in order to customize if for your needs. Nevertheless, you should read the rest of this section if you are planning to customize the answer table.
An answer table is a table in which rows represent the questions and columns represent the choices. There can be more than one answer table, but they have to be side by side (they cannot be placed one above the other). The example above show two answer tables. A few restrictions have to be taken into account:
• If there are more than one table, they must be horizontally aligned. That is, their top and bottom must be in the same line, and their rows must have exactly the same height (see the example above).
• All the rows should have the same height.
• In order to improve the detection process, the length of the vertical lines and the length of the horizontal lines should be more or less proportionate (e.g. one of them should not be more than a 30% larger than the other). If there are more than one answer table, consider the added length of the horizontal lines of every table. The following image illustrates this. The red vertical line is not much smaller than the sum of the two horizontal lines.
• If an answer table has less rows than the others, it is better to keep the horizontal lines, as shown in the image below:
The boxes for the student ID number should be above the answer tables, not too close but not too far away either (see the example below). The width of the student ID table should be comparable to the sum of the width of the answer tables (approximately no less than 2/3 of that sum, and no more than 3/2). Student IDs with just a few digits (two, three, four) can potentially be problematic for wide answer tables.
At the bottom of the answer boxes there must be some black squares. They encode the exam model (permutation). In addition, they help the system to know whether the detection of the answer tables was correct.
Imagine that there are two more rows at the end of each answer table, with the same height as the other rows. Squares will be either in the one above or in the one below, and there must be a square per column. Squares should be centered in those imaginary cells. The position (above/below) of a square conveys the information read by Eyegrade as binary information.
The exam model is encoded with three squares. Therefore, there can be eight different models. The fourth square is a redundancy code for the previous three squares. This 4-square pattern is repeated from left to right as long as there are columns. The table to which a column belongs is not taken into account. For example, if there are two answer tables with three columns each, the fourth square (the redundancy square) is placed at the first column of the second table. The other two columns of the second answer table would contain the same squares as the first two columns of the first table.
The following table show the 4-square pattern for each exam model, as they should be placed from left to right:
Model A Down Down Down Up B Up Down Down Down C Down Up Down Down D Up Up Down Up E Down Down Up Down F Up Down Up Up G Down Up Up Up H Up Up Up Down
## 5.4 Manually editing the .eye file
The files that store the configuration of an exam and the correct answer for each question are stored with a .eye extension. An example is shown below:
[exam]
dimensions: 4,6;4,6
id-num-digits: 9
[solutions]
model-A: 4/1/2/1/1/1/2/4/1/2/3/1
[permutations]
permutations-A: 11{4,3,2,1}/12{1,3,4,2}/5{4,1,3,2}/3{1,4,3,2}/2{1,3,2,4}/9{1,4,3,2}/10{3,1,4,2}/1{3,2,4,1}/8{1,4,2,3}/6{2,1,3,4}/4{3,2,1,4}/7{1,3,2,4}
The file is just plain text and can be edited with any text editor. It has several sections: exam, solutions and permutations.
The fields of the exam section are:
• dimensions: here the number of answer tables and the number of columns and rows in each answer table are configured. For example, "4,6;4,6" means that there are two answer tables, both of them with geometry "4,6". The "4" is the number of columns of the table. The "6" is the number of rows. Tables are specified from left to right (i.e. the first table geometry corresponds to the left-most table in the exam).
• id-num-digits: number of cells of the table for the student id number. Putting a 0 here means that the id number needs not to be read.
• correct-weight: a number, such as 1.75, that represents the score assigned to a correct answer.
• incorrect-weight: a number that represents the score to be substracted for failed answers. Blank answers are not affected by this.
The fields correct-weight and incorrect-weight are optional. If they appear in the file, the program will show the total score in the user interface.
The solutions section specifies the correct answers for each model (permutation) of the exam. Models are identified by letters ("A", "B", etc.). For example:
model-A: 4/1/2/1/1/1/2/4/1/2/3/1
model-B: 3/2/1/4/4/2/2/1/4/2/3/3
In the example above, in the model A, the correct answer for the first question is the 4th choice, for the second question is the 1st choice, for the third question is the 2nd choice, etc.
The permutations section has information that allows to know how questions and choices have been shuffled with respect to the original order. They are used only for extracting statistics or fixing grades after the exam if the solutions used for grading are found to have an error in some questions. If you create the .eye manually, you probably want to just remove this section from the file, unless you need some of the above-mentioned functions.
## 5.5 Automatic detection of exam removal
If the camera in your setup is fixed, that is, you place an exam below the camera, review it, remove it and place the next exam, you may want Eyegrade to detect that you have removed the exam instead of having to click on the Save and capture next exam command.
You can activate this experimental feature in the Tools menu, Experimental submenu, option Continue on exam removal. When this option is checked, Eyegrade saves the current capture and enters the search mode automatically, after a few seconds of not detecting an exam. Before placing the new exam, wait for the system to actually enter the search mode: if you are too quick, Eyegrade might not detect the removal of the exam.
Tip: don't use this option if the camera is not fixed, because just moving it a little bit may cause Eyegrade to think that the exam has been removed.
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2022-05-22 20:05:39
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https://math.stackexchange.com/questions/901124/laplace-transform-of-a-product-of-two-functions/901132
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# Laplace transform of a product of two functions
I have read questions and answers about this topic and i am still confused, using this formula we can calculate the Laplace transform of a product of two functions:
$$L[a_{(t)} b_{(t)}]={{1}\over{2 i \pi}} \int_{\sigma -i \infty}^{\sigma +i \infty}A_{(z)}B_{(s-z)}dz$$
But when i test this formula on an example i get wrong result. My example is: a(t)=t , b(t)=e^(-t)
So the correct answer should be
$$L[te^{-t}]={{1}\over{(s+1)^2}}$$
But when i substitute:
$$L[t]={{1}\over{s^2}}$$
$$L[e^{-t}]={{1}\over{s+1}}$$
into the above formula, i get:
$$L[te^{-t}]={{1}\over{2 \pi i}}\int_{\sigma-i \infty}^{\sigma + i \infty} {{1}\over{z^2}}{{1}\over{s-z+1}}dz=\sum res$$
Residue at z=0 is 1/(s+1)^2 and residue at z=s+1 is also 1/(s+1)^2 so the result i get using this formula is twice the correct result.
I think that the problem is that you're using both residues. But the proper definition of the inverse Laplace transform involves a "dummy" constant $\sigma>0$. This excludes the residue in z=0 and, hence, you don't have the undesired 2 factor.
The integral you've given calculates the inverse Laplace transform. To go from the time-domain to the Laplace-domain (i.e. perform the Laplace Transform), you shouldn't need any contour integration. The (unilateral) Laplace transform is given by $$F(s) = \int_{0}^{\infty} {f(t)e^{-st}dt}$$
Unless you specifically want to practice integration, these are usually given in tables for common functions (like $f(t)=t$ and $f(t) = e^{-t}$). To look them up, google "table of laplace transforms".
EDIT: I might've misread your question. If you're indeed trying the inverse Laplace transform, then the integral you're doing is known as a Bromwhich integral. See this discussion: Inverse Laplace transform of fraction $F(s) = \large\frac{2s+1}{s^2+9}$
• Sorry, my question was not clear enough. If I have two functions a(t) and b(t), and I know only their corresponding Laplace transforms A(s) and B(s), but not functions themselves, how can i calculate the Laplace transform of a(t)*b(t) ? One way is to find a(t) and b(t) using inverse Laplace transform and then find the LT of the product of results And the other way is to use the formula: $$L[a_{(t)} b_{(t)}]={{1}\over{2 i \pi}} \int_{\sigma -i \infty}^{\sigma +i \infty}A_{(z)}B_{(s-z)}dz$$ and i am getting wrong result using this formula – Dusan Krantic Jan 10 '15 at 17:00
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2020-07-12 23:29:34
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https://stats.stackexchange.com/questions/412096/does-direction-of-causality-between-instrument-and-variable-matter
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# Does direction of causality between instrument and variable matter?
The standard scheme of instrumental variable in terms of causality (->) is:
Z -> X -> Y
Where Z is an instrument, X an endogenous variable, and Y a response.
Is it possible, that following relations:
Z <- X ->Y
Z <-> X ->Y
are also valid?
While correlation between instrument and variable is satisfied, how may I think of exclusion restriction in such cases?
NOTE: The notation <-> is not explicit and might lead to different understandings of the problem. Still, the answers highlight this issue and use it to show important aspects of the problem. When reading, please proceed with caution about this part of the question.
Yes, the direction matters. As pointed in this answer, to check whether $$Z$$ is an instrument for the causal effect of $$X$$ on $$Y$$ conditional on a set of covariates $$S$$, you have two simple graphical conditions:
1. $$(Z \not\perp X|S)_{G}$$
2. $$(Z\perp Y|S)_{G_{\overline{X}}}$$
The first condition requires $$Z$$ to be connected to $$X$$ in the original DAG. The second condition requires $$Z$$ to not be connected to $$Y$$ if we intervene on $$X$$ (represented by the DAG $$G_{\overline{X}}$$, where you remove the arrows pointing to $$X$$). Thus,
Z -> X -> Y : here Z is a valid instrument.
Z <-> X -> Y: here Z is a valid instrument (assuming that a bidirected edge represents an unobserved common cause, as it does in semi-Markovian models).
Z <- X -> Y : here Z is not a valid instrument.
PS: jsk's answer is not correct, let me show you how Z <-> X is a valid instrument.
Let the structural model be:
$$Z = U_1 + U_z\\ X = U_1 + U_2 + U_x\\ Y = \beta X + U_{2} + U_y$$
Where all the $$U$$'s are unobserved mutually independent random variables. This corresponds to the DAG z <--> x -->y with also x<-->y. Thus,
$$\frac{cov(Y, Z)}{cov(X, Z)} = \frac{\beta cov(X, Z)}{cov(X,Z)} = \beta$$
• I think this highlights the need to be very clear clear about what exactly $X<->Z$ actually means. In your revised example, I would argue that X and Z are driven by a third variable, which seems different than my understanding of the notation $X<->Z$. – jsk Jun 9 at 22:40
• @jsk this is standard notation for semi-Markovian models. – Carlos Cinelli Jun 9 at 23:35
• Not standard to everyone. Just read a paper by Pearl and Greenland in which they say that SOME authors use the notation in this way. There's nothing in the OP's question to suggest his interpretation of the notation, though he may very well agree with you. – jsk Jun 10 at 0:19
• What if $Y = \beta X + U_1 + U_y$? Would it not then be the case that $Z<->X$ but then Z would be correlated with the omitted variable and hence not be valid instrument? – Stop Closing Questions Fast Jun 10 at 13:12
• @JesperHybel If you have U1 in the structural equation of Y, this means the error terms of Z and Y are dependent. Thus you have an extra bidirected edge Z<—>Y and no case works, be it Z—>X or Z<—>X. The graphical conditions are explicitly stated there. – Carlos Cinelli Jun 10 at 14:25
Yes, direction does matter.
According to Hernan and Robins' new causal inference book https://cdn1.sph.harvard.edu/wp-content/uploads/sites/1268/1268/20/hernanrobins_v2.17.21.pdf
the following three conditions must be met:
$$i.$$ $$Z$$ is associated with $$X$$.
$$ii.$$ $$Z$$ does not affect $$Y$$ except through its potential effect on $$X$$.
$$iii.$$ $$Z$$ and $$Y$$ do not share common causes.
Condition $$(iii)$$ rules out relations such as $$X$$ - > $$Z$$ or $$X$$ < - > $$Z$$ because $$X$$ cannot have a causal effect on both $$Z$$ and $$Y$$
Edit: whether or not $$X<->Z$$ is acceptable for an instrument depends on the definition of $$X<->Z$$. If it means that they are correlated because of a third variable, like in Carlos's example, then it's ok. If it suggests a feedback loop where a causal arrow can be drawn from X to Z as well then Z is not a valid instrument.
• (-1) This is wrong, Z<—>X is fine for an instrument. – Carlos Cinelli Jun 9 at 20:57
• These conditions posited by Hernan and Robins are not precise, they say that themselves---read further the chapter. Also see a trivial counterexample to your claim in the edit of my answer. – Carlos Cinelli Jun 9 at 21:19
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2019-12-14 11:14:54
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https://www.groundai.com/project/complementarity-of-dark-matter-searches-in-the-pmssm/
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Complementarity of Dark Matter Searches in the pMSSM
# Complementarity of Dark Matter Searches in the pMSSM
M. Cahill-Rowley SLAC National Accelerator Laboratory, Menlo Park, CA, USA111mrowley, funk, hewett, rizzo, mdwood@slac.stanford.edu R. Cotta University of California, Irvine, CA, USA222cottar@uci.edu A. Drlica-Wagner Fermi National Accelerator Laboratory, Batavia, IL, USA333kadrlica@fnal.gov S. Funk SLAC National Accelerator Laboratory, Menlo Park, CA, USA111mrowley, funk, hewett, rizzo, mdwood@slac.stanford.edu J.L. Hewett SLAC National Accelerator Laboratory, Menlo Park, CA, USA111mrowley, funk, hewett, rizzo, mdwood@slac.stanford.edu A. Ismail T.G. Rizzo SLAC National Accelerator Laboratory, Menlo Park, CA, USA111mrowley, funk, hewett, rizzo, mdwood@slac.stanford.edu M. Wood SLAC National Accelerator Laboratory, Menlo Park, CA, USA111mrowley, funk, hewett, rizzo, mdwood@slac.stanford.edu
###### Abstract
As is well known, the search for and eventual identification of dark matter in supersymmetry requires a simultaneous, multi-pronged approach with important roles played by the LHC as well as both direct and indirect dark matter detection experiments. We examine the capabilities of these approaches in the 19-parameter p(henomenological)MSSM which provides a general framework for complementarity studies of neutralino dark matter. We summarize the sensitivity of dark matter searches at the 7, 8 (and eventually 14) TeV LHC, combined with those by Fermi, CTA, IceCube/DeepCore, COUPP, LZ and XENON. The strengths and weaknesses of each of these techniques are examined and contrasted and their interdependent roles in covering the model parameter space are discussed in detail. We find that these approaches explore orthogonal territory and that advances in each are necessary to cover the Supersymmetric WIMP parameter space. We also find that different experiments have widely varying sensitivities to the various dark matter annihilation mechanisms, some of which would be completely excluded by null results from these experiments.
SLAC-PUB-15862
## 1 Introduction and Overview of the pMSSM
Determining the identity of dark matter (DM) is one of the most pressing issues before us today. Multiple observations reveal that roughly 85% of the matter in the universe is electromagnetically inert, and the local density of dark matter is known to within a factor of two. Cosmological considerations reveal a handful of its properties, yet it may take many forms, and numerous theories hypothesize dark matter particles of various types. A promising class of dark matter candidates is Weakly Interacting Massive Particles (WIMPs) which could have thermally frozen out in the early universe in a manner that yields the relic density observed by experiment today. WIMPs naturally appear in many extensions of the Standard Model (SM) that resolve the gauge hierarchy, with the most notable example being supersymmetry (SUSY).
Several mechanisms allow for the search for WIMP dark matter: () direct detection where WIMPs elastically scatter off nuclei, () indirect detection where WIMPs annihilate into a pair of SM particles, and () the direct production of WIMPs in high energy colliders. In this paper we investigate the complementary roles these three search techniques play in the quest to discover dark matter. We employ the phenomenological Minimal Supersymmetric Standard Model (pMSSM) [1, 2] as our tool to examine the coverage of WIMP parameter space by each technique. We find that the methods explore orthogonal territory and that advances in all three techniques are necessary to cover the supersymmetric WIMP sector. Here, we first outline the salient features of the pMSSM, examine each WIMP detection experiment in turn, and then draw conclusions from the combined results.
One of the main reasons that R-parity conserving supersymmetry is attractive is the prediction that the lightest SUSY particle (LSP) is stable and may be identified as a thermal dark matter candidate if it is both electrically neutral and a color singlet. Frequently in the MSSM, the LSP is associated with the lightest neutralino, . While DM searches are directly focused on the nature of the LSP itself, the properties of the full spectrum of superparticles, and of the extended SUSY Higgs sector, also play important roles. Thus it is inappropriate to completely separate DM searches from the exploration and examination of the rest of the SUSY spectrum. However, even in the simplest SUSY scenario, the MSSM, the number of free parameters (100) is simply too large to perform a study in all generality; we thus need to restrict our view without losing any relevant physics. One approach is to assume the existence of a high-scale theory with a handful of parameters (such as mSUGRA[3]) from which all the properties of the TeV scale sparticles can be determined and studied. While this method is valuable and predictive, these scenarios are somewhat phenomenologically limiting and are under ever-increasing tension with a wide range of experimental data including, in some cases, the GeV mass of the recently discovered Higgs boson[4, 5].
One way of circumventing these limitations is to examine the more general 19-parameter pMSSM[1, 2]. The pMSSM is the most general version of the R-parity conserving MSSM that satisfies several data-driven constraints: () no new phase appearing in the soft-breaking parameters, i.e., CP conservation, () Minimal Flavor Violation at the electroweak scale such that the CKM matrix drives flavor mixing, () degenerate first and second generation soft sfermion masses, () negligible Yukawa couplings and associated A-terms for the first two generations. In particular, note that the pMSSM contains no theoretical assumptions about physics above the TeV scale, e.g., the nature of SUSY breaking or grand unification. This allows for the capture of electroweak scale phenomenology for which a UV-complete theory may not yet exist. When we impose the constraints ()-(), the number of free parameters in the MSSM at the TeV-scale decreases from 105 to 19 for the case of a neutralino LSP (or 20 including the gravitino mass as an additional parameter when it plays the role of the LSP555In this work we will limit our discussion to the case of neutralino LSPs.).
To study the pMSSM, we essentially throw darts into this large space, generating many millions of random model points (using SOFTSUSY[6] and checking for consistency using SuSpect[7]), with each point corresponding to a specific set of values for the parameters. The ranges of the 19 pMSSM parameters employed in the analysis below are presented in Table 1. The lower and upper bounds used for the ranges in our scan were chosen to be essentially consistent with Tevatron and LEP data and to have kinematically-accessible sparticles at the 14 TeV high-luminosity LHC, respectively. We do not assume that the thermal relic density as calculated for the neutralino LSP necessarily saturates the WMAP/Planck value[8] in order to allow for the possibility of multi-component DM. For example, axions introduced to solve the strong CP problem might make up a substantial amount of DM. Decay patterns of the SUSY partners and the extended Higgs sector are calculated using a privately modified version of SUSY-HIT[9]. Each individual model is then subjected to a large set of collider, flavor, precision measurement, dark matter and theoretical constraints [10]. Roughly 225k models with a neutralino LSP survive this initial selection and can then be used for further physics studies. We note that this model set was generated before the discovery of the Higgs boson and that approximately 20% of the sample predicts the correct Higgs mass. We will discuss possible effects of this below. We have recently performed a detailed study of the signatures for the pMSSM at the 7 and 8 (and eventually 14) TeV LHC [10, 11, 12] and include these results here.
As a result of our scan ranges for the electroweak gauginos (chosen for compatibility with LEP data and to enable phenomenological studies at the 14 TeV LHC), the LSPs in our model sample are typically very close to being in a pure electroweak eigenstate as the off-diagonal elements of the chargino and neutralino mass matrices are at most . Figure 1 presents some properties of the nearly pure eigenstate LSPs (defined here as a single electroweak eigenstate comprising over 90% of the mass eigenstate). The left panel displays the distribution of the LSP mass for nearly pure bino, wino, and Higgsino LSPs, while the right-hand panel shows the corresponding distribution for the predicted LSP thermal relic density. Note that the LSP masses lie below TeV in all models; this is due to our choice of scan ranges as the entire SUSY spectrum must be lighter than TeV and heavier than the LSP (by definition), and this becomes increasingly improbable with increasing LSP mass. In addition, the relic density upper limit becomes increasingly difficult to satisfy at larger LSP masses. Similarly, due to LEP and relic density constraints, none of our models have LSP masses below 40 GeV. The fraction of models where the LSP is nearly a pure bino eigenstate is found to be rather low in this model sample since such models generally lead to too high a value for the relic density unless they co-annihilate with another sparticle, happen to be close to a () funnel region, or have a suitable Higgsino admixture. Note that only in the rightmost bin of the right panel is the relic density approximately saturating the WMAP/Planck thermal relic value. These LSP properties will be of particular importance in the discussions that follows.
Figure 2 shows the thermal relic density generated by the LSP in our pMSSM models as a function of the LSP mass with the color-coding reflecting their electroweak eigenstate content. There are many items to note here that will be important for later consideration. Essentially every possible known mechanism to obtain (or lie below) the WMAP/Planck relic density is present: () The set of models with low LSP masses (forming ‘columns’ on the left-hand side of the figure) correspond to bino-Higgsino admixtures which achieve a sufficiently low relic density by resonant annihilation through the -funnels; these sometimes are pure binos if the Higgsino fraction is very small666Here again, ‘pure’ means having an eigenstate fraction . Points shown as bino-wino, bino-Higgsino, or wino-Higgsino mixtures have less than Higgsino, wino, or bino fraction, respectively. Mixed points have no more than and no less than of each component.. () The bino-Higgsino LSPs saturating the relic density in the upper left region of the figure are of the so-called ‘well-tempered’ variety. () the pure bino models in the upper middle region of the Figure are bino co-annihilators (mostly with sleptons) or annihilate resonantly through the -funnel. () The green (blue) bands are pure Higgsino (wino) models that saturate the relic density bound (using perturbative calculations which do not include the Sommerfeld enhancement effect777The Sommerfeld enhancement can significantly deplete the relic density of wino LSPs heavier than 1 TeV, while Higgsino and light wino LSPs are relatively unaffected [14]. Bino LSPs do not exhibit the effect because they can’t exchange gauge bosons. Including the enhancement would increase the low-velocity annihilation cross section for heavy winos, lowering their predicted relic density but increasing their present-day annihilation cross section. Since the average velocity today is lower than during freeze-out, we would naively expect that including the enhancement would strengthen the limits on heavy wino LSPs. We will see that CTA is already able to exclude models with heavy winos in our perturbative calculation; we therefore expect that including the enhancement would minimally affect our conclusions.) near TeV and have very low relic densities for lighter LSP masses. Wino-Higgsino hybrids are seen to lie between these two cases as expected. () A smattering of models with additional (or possibly multiple) annihilation channels are loosely distributed in the lower right-hand corner of the Figure. As we will see, many of the searches for DM are particularly sensitive to one or more of these LSP categories.
## 2 LHC Searches
We begin with a short overview of the searches for the pMSSM at the 7 and 8 TeV LHC [10, 11, 12]. In general, our approach is to closely follow the suite of ATLAS SUSY analyses but also to supplement these with several searches performed by CMS; the analyses included in our study are briefly summarized in Table 2. In addition, we include the searches for heavy neutral SUSY Higgs decaying to by CMS [15] and measurements of the rare decay mode discovered by CMS and LHCb [16]. Both of these additional searches play distinct but important roles in restricting the pMSSM parameter space. We have implemented every relevant ATLAS SUSY search publicly available as of the beginning of March 2013 and also the more recent 20 fb 2-6 jets + MET analysis. The LHC results for our model sets (including the neutralino LSP models considered in this paper) appear in detail in our companion papers on both neutralino and gravitino LSP SUSY searches [12, 17].
A brief summary of our procedure is as follows: We generate SUSY events for each model using PYTHIA 6.4.26 [18] and PGS 4 [19], which we have modified to, e.g., correctly deal with gravitinos, hadronization of stable colored sparticles, multi-body decays and ATLAS b-tagging methods. We then scale our event rates to NLO by calculating the relevant K-factors with Prospino 2.1 [20]. The individual searches are then implemented using a custom analysis code[13], following the published cuts and selection criteria of ATLAS as closely as possible. Our code is validated for each of the many search regions in every analysis employing the benchmark model points provided by ATLAS (and CMS). Models are then excluded using the limits as employed by ATLAS. Note that these analyses are performed without imposing the Higgs mass constraint, GeV (combined experimental and theoretical errors) so that we can understand its impact on the search results. Roughly of models in the neutralino model set (before the LHC SUSY searches are applied) predict a Higgs mass in the above range. While there is some variation amongst the individual searches we find that, once combined, the total fraction of our models surviving (or excluded by) the set of all LHC searches is to an excellent approximation independent of whether or not the Higgs mass constraint has been applied. Conversely, the fraction of the neutralino models predicting the correct Higgs mass is also found to be approximately independent of whether or not the SUSY searches have been applied. This combined result is very powerful, demonstrating the current approximate decoupling of SUSY search results from the discovery of the Higgs boson and this allows us to employ the entire model set for SUSY studies with some validity. After all of these searches are applied, we find that 45.5% of the pMSSM model sample is excluded, leaving 54.5% of the set as viable models.
Figure 3 demonstrates the power obtained by combining the set of LHC analyses to constrain the pMSSM neutralino LSP models. In particular, this figure shows the fraction of models having a given sparticle and LSP mass that are excluded by the combined LHC searches. Since the values of the masses of the squarks and gluinos, the lightest stops and sbottoms, and the LSP itself are of particular interest, we concentrate on these specific quantities in this figure. In the upper left-hand panel the coverage of the gluino-LSP mass plane by the LHC searches is displayed; the white line represents the CL search limit on a simplified model with a gluino NLSP, neutralino LSP, and all other sparticles decoupled, as obtained by ATLAS from their 20 fb 2-6 jets plus MET analysis [21]. We see that this is very roughly the same as the black region excluded in the pMSSM. Note however that the pMSSM exclusion is slightly stronger than the simplified model limit for lighter gluino masses, while being somewhat weaker than the simplified model limit in the heavy gluino region. Of course, the fact that the other sparticles are generically not decoupled in the pMSSM means that the gluinos exhibit many different decay patterns, some of which are rather insensitive to the jets + MET search. It is then interesting that the pMSSM exclusion resulting from the combination of multiple searches is similar to the limit from the jets + MET search in the simplified model scenario. Generally, models with heavy gluinos that are not excluded, despite being below the simplified model limit, have decays through stops, both on-shell and off-shell. The upper right-hand panel shows the corresponding coverage in the gluino-lightest squark mass plane with a simplified model line, again from the fb 2-6 jets plus MET analysis [21]. In this case, the simplified model assumes that the LSP is massless and that the 8 squarks of the first two generations are degenerate, neither of which are common occurances in the pMSSM. As a result, it is no surprise that our excluded region is not well described by the simplified model. While most models with rather light squarks and/or gluinos are observed to be excluded by the combined LHC searches, it is clear that models with squarks and/or gluinos below GeV still remain viable.
The lower two panels display the lightest stop/sbottom - LSP mass planes, again including the corresponding ATLAS simplified model limits [22]. Here we see several things, the most important being that the region of coverage in these two planes in the pMSSM differs substantially from either simplified model limit. As we describe in [12], this is because stops and sbottoms can typically decay to either a neutralino or a chargino (since the LSP is most commonly a wino or Higgsino multiplet), producing a mixture of final states. In particular, the LHC sensitivity to stops improves substantially when the branching fraction for is appreciable, since the resulting hard b-jets are relatively easy to distinguish from the background. This effect is most striking in the compressed spectrum region where the stop simplified model limit is far weaker than the pMSSM exclusion. In the highly-compressed region, the exclusion reach results mainly from generic jets + MET searches, as the b-jet pT becomes too low for a reasonable tagging efficiency.
In addition to the 7 and 8 TeV LHC searches, future data taking and enhanced analyses at 14 TeV will greatly extend the expected coverage of the pMSSM parameter space. In [12], we considered the impact of one of the most powerful of these searches to be performed by ATLAS, the zero-lepton jets + MET final state at 14 TeV with both 300 fb and 3 ab of integrated luminosity [23], following the procedure described above. Note that in extrapolating from 300 fb to 3 ab, luminosity scaling has been employed to obtain the expected limit. As a result of limitations on CPU time, we generated 14 TeV results only for the k neutralino LSP models that survived the 7 and 8 TeV LHC analyses and predict a Higgs mass of GeV. We note that since the results of the 7 and 8 TeV analyses are essentially independent of the Higgs mass, it is quite likely that our results for this narrow Higgs mass range would in fact be applicable, to a very good approximation, to the entire neutralino LSP model set. We find that the 14 TeV jets+MET analysis with 300 (3000) fb of data is expected to exclude 90.7% (97.1%) of models which have the correct Higgs mass and survive the 7/8 TeV searches in Table 2.
To augment the 14 TeV jets + MET search, we have recently added a pair of signal regions in each of the zero- and one-lepton stop analyses as presented by ATLAS in [24]. These analyses feature sliding missing energy and transverse mass cuts for optimal sensitivity to different stop masses, creating a large effective number of signal regions. We chose to examine the signal regions that were optimized for stop masses of 800 GeV and 1 TeV with 3 ab of integrated luminosity. We derived the limits from the expected ATLAS background numbers and scaled these limits to estimate the sensitivity at 300 fb of integrated luminosity as well. Taken together, these signal regions are expected to exclude of the neutralino models assuming an integrated luminosity of 0.3(3) ab at 14 TeV. Combining them with the zero-lepton, jets + MET search discussed above excludes 90.8% (97.2%) of the models surviving the 7/8 TeV searches in Table 2. This demonstrates that these additional signal regions are not expected to exclude very many of the models which are missed by the jets + MET search.
The exclusion reach of the combination of the ATLAS 14 TeV zero-lepton jets+MET search and the two stop searches is summarized in Fig. 4 for both 0.3 and 3 ab of integrated luminosity in the gluino-lightest squark mass plane. Here we see that for typical models, the 14 TeV LHC will be able to exclude squarks up to 1.6 TeV and gluinos up to 2.7 TeV. A few models are seen to survive with much lighter squark and/or gluino masses; in almost all cases these models survive by producing multiple high- leptons or b-jets from sbottoms (rather than stops) and therefore fall outside of the search regions. Adding additional searches with leptons in the signal region will undoubtedly exclude many of the surviving models with light colored sparticles. To fully understand the capabilities of the 14 TeV LHC will of course require a far more realistic study than is presently available since the LHC collaborations themselves are still unsure of how well their detectors will perform under the very high pileup conditions at 14 TeV.
## 3 Direct Detection
The direct detection of DM results from either the spin-independent (SI) or spin-dependent (SD) scattering of the LSP off of a target nucleus. While channel exchange only contributes to the SD(SI) process at tree level, channel squark exchange can contribute to both scattering processes. Clearly, as the bounds on the first and second generation squark masses from the LHC become stronger the importance of these squark exchange contributions will become sub dominant.888It is important to note, however, that since the many different light squark masses can vary independently in the pMSSM, and since the LHC constraints on the , and squarks are quite different, both SD and SI interactions may have substantial isospin-dependent contributions so that and can be significantly different. The -exchange graph is sensitive to the Higgsino content of the LSP, whereas the Higgs exchange graph probes the product of the LSP’s gaugino and Higgsino content. Similarly, squark exchange is particularly sensitive to the LSP’s wino and bino content.
Figure 5 displays the predicted SI and SD cross sections for our pMSSM model set together with several present [25, 26, 27, 28] and anticipated future [29, 30, 31] experimental constraints. Note that the cross sections are appropriately scaled by the factor to account for the fact that most of our pMSSM models lead to a thermal relic density somewhat below the WMAP value as discussed in the Introduction. There are two things to note in this Figure: () Future SI searches will cut rather deeply into the model set; a lack of signal at XENON1T(LZ) would exclude 23%(39%) of these models. However, this implies that more than half of our models are not accessible to SI experiments due to their rather small scaled cross sections, . Models tend to produce these small values of due to both the suppression arising from their low thermal relic density as well as the tendency of the LSPs to be nearly pure weak eigenstates as discussed above. Note that if we only consider models which predict a relic density within 10% of the critical density, the coverage improves significantly, with 60% (81%)) of models lying within the XENON1T (spin-independent LZ) expected search limit. () SD searches are rather far from the pMSSM model predictions and we do not expect future SD reaches to have a significant impact on the parameter space: SD experiments such as COUPP500(LZ) will only be able to exclude of the models in this set if no signal is observed.
Direct detection experiments can apply strong constraints on specific LSP compositions or annihilation mechanisms. An interesting example is the case of an LSP with a mass significantly below the LEP limit on chared sparticles. This LSP is required to be mostly bino (otherwise it would be accompanined by an excluded chargino), and is generally prevented from coannihilating by the LEP limit on charged sfermions (although in some cases sneutrino co-annihilation may be possible). The dominant annihilation mode is therefore through s-channel Z or Higgs bosons. The left panel of Figure 6 shows these light LSPs in our model set ( GeV) in the scaled SI vs scaled SD cross section plane; we find that while many of the models have a SI or SD cross-section beyond the reach of current or future experiments, only one model (where the LSP co-annihilates with a 100 GeV stau) is expected to remain undetected by the combination of future SI and SD searches at LZ. Naively, one would not expect such complementarity between the SI and SD scattering experiments for light bino LSPs with small Higgsino components, since both cross sections should both be determined by the Higgsino content of the LSP. However, we see from the left panel of the figure (where the points are color-coded according to their bino/Higgsino content with denoting the bino fraction which approaches unity for a 100% bino eigenstate) that this expectation is realized for the SD cross section but not for the SI cross section. In particular, very small SI cross-section values are obtained for large Higgsino content. The suppression of the SI cross section despite large Higgsino content results from cancellation between heavy and light Higgs exchange diagrams; this can be seen in the right panel of the figure, which colors the points according to the value of the ratio . The sign of this quantity determines the relative sign of the mixing matrix elements; when it is negative, a relative sign appears between the light Higgs coupling and the heavy neutral Higgs couplings, allowing for cancellation between the light and heavy Higgs exchange contributions. Although the heavy Higgs bosons in our model set are frequently very massive, the cancellation can still occur since the dominant contribution to scattering comes from Higgs exchange with a strange quark, which is enhanced for heavy Higgs bosons in the decoupling limit. Additional cancellations with squark diagrams can lead to further suppression of the SI cross section. Interestingly, a negative value of the ratio also tends to result in a smaller coupling to the light Higgs, with the result that a larger Higgsino fraction (and therefore a larger SD cross-section) is required in order to provide a sufficiently large annihilation cross-section.
While direct detection experiments have significant power to cover much of the pMSSM, clearly such experiments will need to be supplemented if we want to discover or exclude the full range of neutralino LSPs in the pMSSM model space. In the next two sections, we describe constraints on the pMSSM from indirect detection and neutrino telescope experiments.
## 4 Indirect Detection: Fermi LAT and CTA
Indirect detection plays a critical role in searches for DM and, in the case of null results, can lead to very strong constraints on the pMSSM parameter space. As will be seen below, both Fermi and CTA can contribute in different regions of the pMSSM parameter space in the future. CTA, in particular, will be seen to be extremely powerful in the search for heavy LSPs which are mostly Higgsino- or wino-like and that predict thermal relic densities within an order of magnitude of the WMAP/Planck value. Fermi, on the other hand, will be seen to be mostly sensitive to well-tempered neutralinos that are relatively light.
The most promising DM targets for both Fermi LAT and CTA are those with both a high DM density and low astrophysical gamma-ray foregrounds. These criteria have motivated a number of Galactic and extragalactic targets including the Galactic Center (GC), dwarf spheroidal galaxies (dSphs), and galaxy clusters. The expected gamma-ray signal for DM annihilations is proportional to the integral of the square of the DM density along the line of sight to the source (). The determination of is most reliable for DM-dominated objects such as dSphs and galaxy clusters in which the DM distribution can be robustly measured. In the Milky Way halo the uncertainty on the DM distribution rapidly increases as one approaches the inner galaxy where baryons dominate the gravitational potential. Kinematic data at large scales constrain the density of DM at the solar radius to 0.2–0.4 GeV cm [32, 33].
Under well-motivated models for the DM distribution in the Galactic halo, the GC is expected to be the most intense DM source in the sky. CDM simulations predict that the Galactic DM halo should have a density profile with an inner cusp, with . For an extrapolation of the Galactic DM density profile with , the expected DM signal from the GC region is approximately two orders of magnitude greater than dSphs or galaxy clusters. However the GC is also the region of the sky with the highest density of gamma-ray sources and brightest diffuse gamma-ray emission produced from the interaction of cosmic rays with the interstellar medium. These foregrounds significantly limit the sensitivity of the Fermi-LAT in the inner galaxy and complicate the interpretation of any observed signals.
For our study of Fermi-LAT we focus on the sensitivity to the DM signal from dSphs which are predominantly at high galactic latitudes where the astrophysical foregrounds are much weaker. Above 50 GeV the diffuse emission from the Galaxy is much less intense relative to other backgrounds than in the Fermi-LAT sensitivity band. The inner galaxy is thus the preferred target for CTA under the assumption that the Galactic DM halo possesses an inner cusp with . When considering signals from models with a relic density below the WMAP value, we rescale the annihilation cross section by to account for the reduced number density of DM particles and consequent reduction in the factor. Implicit in this rescaling is that the LSPs in these models constitute only one component of DM.
### 4.1 Fermi Lat
Here we follow the procedure developed in Ackermann et al. (henceforth A11) [34] and expanded upon in Cotta et al. [35] and Ackermann et al. [36] to constrain the annihilation cross section, , for each pMSSM model using Fermi-LAT observations of ten dwarf spheroidal galaxies (dSphs). Our two-year -ray event sample is identical to that described in A11, accepting photons in the energy range from within 10 of each dSph. In accord with A11, we use the LAT ScienceTools version v9r20p0 and the P6_V3_DIFFUSE instrument response functions. J-factors and associated statistical uncertainties for the dSphs are taken from Table 1 of A11, where they were calculated using line-of-sight stellar velocities and the Jeans equation [34]. Similar to Cotta et al., we use DarkSUSY 5.0.5 [37] to model the -ray spectrum from the annihilation of each pMSSM LSP. DarkSUSY calculates the total -ray yield from annihilation, as well as the rates into each of 27 final state channels.
We create bin-by-bin likelihood functions from the Pass 6 data surrounding each of the 10 dwarf spheroidal galaxies following the procedure of Ackermann et al. [36]. We calculate a joint likelihood to constrain the annihilation cross section of each pMSSM model given the LAT observations coincident with the ten dSphs. Following A11, we incorporated statistical uncertainties in the J-factors of the dSphs as nuisance parameters in our likelihood formulation (see Equation 1 and the associated discussion in A11). This likelihood formulation includes both the flux normalizations of background -ray sources (diffuse and point-like) and the associated dSph J-factors and statistical uncertainties. No significant -ray signal is detected from any of the dSphs when analyzed individually or jointly for any of the pMSSM models. Utilizing the publicly available bin-by-bin likelihood functions derived from the analysis of Pass 7 rather than a re-analysis of the Pass 6 data would not qualitatively alter these results.
For each of the pMSSM models, we calculate the maximum annihilation cross section, , consistent with the null detection in the LAT data. We incorporate nuisance parameters to obtain a 95% one-sided confidence interval on the value of using the profile likelihood method [38]. This one-sided 95% confidence limit on serves as our value of , which is compared to the true value of the annihilation cross section for each pMSSM model. We define the “boost” necessary to constrain a model as the ratio .
While the LAT data do not presently constrain any of the pMSSM models, it is useful to estimate the improvements expected over a 10 year mission lifetime. In the low-energy, background dominated regime, the LAT point source sensitivity increases as roughly the square-root of the integration time. However, in the high-energy, limited background regime (where many pMSSM models contribute), the LAT sensitivity increases more linearly with integration time. Thus, 10 years of data could provide a factor of to 5 increase in sensitivity. Additionally, optical surveys such as Pan-STARRS [39], the Dark Energy Survey [40], and LSST [41] could provide a factor of 3 increase in the number of Milky Way dSphs corresponding to an increased constraining power of to 3 [42]. Ongoing improvements in LAT event reconstruction, a better understanding of background contamination, and an increased energy range are all expected to provide additional increases in the LAT sensitivity. Thus, we find it plausible that the LAT constraints could improve by a factor of 10 compared to current constraints.
In Figure 7 we display the boost required to constrain the various pMSSM models at CL based on the Fermi-LAT dwarf analysis employing only the first 2 years of data color-coded by either the annihilation cross section or the LSP thermal relic density. Here we see that the LAT analysis does not currently constrain any of our pMSSM models. However, as discussed above with more dwarfs and longer integration times we would expect an -fold improvement in the sensitivity and thus all models with boost factors less than 10 would become accessible. We will assume this -fold improvement in sensitivity for the analysis that follows.
### 4.2 Cta
The Cherenkov Telescope Array (CTA) [43] is a future ground-based gamma-ray observatory that will have sensitivity over the energy range from a few tens of GeV to a few hundreds of TeV. To achieve the best sensitivity over this wide energy range CTA will include three telescope types: Large Size Telescope (LST, 23 m diameter), Medium Size Telescope (MST, 10-12 m) and Small Size Telescope (SST, 4-6 m). Over this energy range the point-source sensitivity of CTA will be at least one order of magnitude better than current generation imaging atmospheric Cherenkov telescopes such as H.E.S.S., MAGIC, and VERITAS. CTA will also have an angular resolution at least 2–3 times better than current ground-based instruments, improving with energy from 0.1 at 100 GeV to better than 0.03 at energies above 1 TeV.
The optimal DM search region for CTA will be limited by the CTA FoV of 8 to the area within to of the GC. The DM signal on these angular scales predominantly probes the DM distribution in the inner galaxy ( kpc). We model the Galactic DM distribution with an NFW profile with a scale radius of 20 kpc normalized to 0.4 GeV cm at the solar radius. This model is consistent with all current observational constraints on the Galactic DM halo and represents a conservative expectation for the inner DM profile in the absence of baryonic effects.
Because the annihilation signal is proportional to the square of the DM density, the projected limits for CTA depend strongly on the assumptions that are made on the shape and normalization of the Galactic DM halo profile. The projected limits presented here could change by as much as a factor of 10 given these uncertainties. The analysis strategy adopted for this study also relies on the existence of a cusp in the MW DM density profile which would produce a measurable gradient within the FoV of CTA. MW density profiles with a central core would require a different analysis strategy than the one presented here and would likely result in a reduced sensitivity to a DM signal in the GC.
The prospects for CTA to detect DM and test other exotic physics has been studied in detail by [44] using models for the CTA response functions from [45]. These models were derived from detailed Monte Carlo simulations generated for a variety of possible array configurations with 18–37 MSTs and different combinations of SSTs and LSTs. The baseline design for CTA is a balanced array with 18–25 MSTs, 3–4 LSTs, and 50–70 SSTs that maximizes the performance over the whole CTA energy range.
For this study we model the performance of CTA using simulations of an array with 61 MSTs distributed on a regular grid with 120 m spacing [46]. For a gamma-ray source with the spectral properties of DM, the sensitivity of this array is similar to the expected sensitivity of the baseline CTA design with a US extension of 24 MSTs. The array used for this study has a gamma-ray angular resolution that can be parameterized as a function of energy as GeV) and a total gamma-ray effective area above 100 GeV of 10 m. We define the GC signal region as an annulus centered on the GC that extends from 0.3–1.0 and calculate the sensitivity of CTA for an integrated exposure of 500 hours that is uniform over the whole region. An energy-dependent model for the background in the signal region is taken from a simulation of residual hadronic contamination. The uncertainty in the background model is calculated for a control region with no signal contamination and a solid angle equal to five times the signal region (14.3 deg).
We estimate the sensitivity of CTA using a binned likelihood analysis with two model components: an isotropic background that models the distribution of residual cosmic rays and a template for the DM annihilation signal. The likelihood of the signal and background components is evaluated from the distribution of events in a two-dimensional map binned in energy and angular offset from the GC. For each model, the maximum cross section consistent with a null detection at the 95% C.L. () is calculated from the ratio between likelihoods evaluated with and without the DM component. Following the same procedure as the Fermi-LAT analysis, we compute the model boost factor as the ratio of the model cross section with .
Figure 8 shows the distribution of the CTA boost factor versus LSP mass for all pMSSM models and the subset of models that have an LSP relic density consistent with 100% of the DM relic density. For models with LSP masses above 100–200 GeV, the sensitivity of CTA is observed to be well correlated with the total annihilation cross section. At lower LSP masses, the boost factor distribution begins to shift to higher values as the peak of the gamma-ray annihilation spectrum moves below the energy threshold of CTA ( GeV). CTA can exclude 20% of the total model set and 50% of the models in the subset of models with an LSP relic density that saturates the WMAP/Planck value. In both scenarios the majority of models excluded by CTA are those which have a pure wino or Higgsino LSP with a mass near 1 TeV.
## 5 IceCube
Neutralino dark matter can be captured and accumulated in the sun. Neutralinos in this relatively over dense population would then sink to the solar core and annihilate. If the product of capture and annihilation cross sections is large enough this process leads to an equilibrium population of captured neutralinos whose annihilations are proportional to their elastic scattering cross-sections [47, 48] and that may be detectable by observing an excess of high-energy () solar neutrinos in km-scale neutrino telescope experiments [49].
Here we present predictions for an IceCube/DeepCore (IC/DC) search for neutralino DM in our pMSSM model set. Our analysis closely follows that presented in [50]. In the results presented here we assume that each neutralino’s relic density is given by the usual thermal calculation. We use DarkSUSY 5.0.6 [37] to simulate the (yearly-average) signal / neutrino flux spectra incident at the detector’s position and convolve with preliminary / effective areas for muon events contained in DeepCore111111These are the same effective areas that were used in [50], referred to there as “SMT8/SMT4.”. We consider a data set that includes yr of data that is taken during austral winters (the part of the year for which the sun is in the northern hemisphere) over a total period of yrs 121212In practice, the IC/DC treatment of data is more sophisticated, classifying events as through-going, contained and strongly-contained, and allowing for some contribution from data taken in the austral summer. We expect that inclusion of this data would affect our results at a quantitative, but not qualitative, level.. An irreducible background rate of events/yr is expected from cosmic ray interactions with nuclei in the sun. Here we will take (as discussed at greater length in [50]) a detected flux of events/yr as a conservative criterion for exclusion.
The basic results of this analysis are presented in Figure 9. In this figure, the full pMSSM model set is depicted by the gray points, and the WMAP-saturating models with mostly bino, wino, Higgsino or mixed (80% of each) LSPs are highlighted in red, blue, green and magenta, respectively. Detectability is tightly correlated with the elastic scattering cross-sections ( and ) while having little correlation with the annihilation cross-section , as expected.
The biggest difference between these results and those of the previous analysis [50], which used an older set of pMSSM models that were chosen to have relatively light () sparticles, is that a much smaller percentage of the current pMSSM models are able to reach capture/annihilation equilibrium in the sun. This is due to the fact that so many of these models are nearly pure wino or Higgsino gauge eigenstates (which have both low relic density and small capture cross-sections) and that the LSPs in this model set tend to be much heavier than those in the previous set. If one defines out-of-equilibrium models as those with solar annihilation rates less than 90% of their capture rates, we find that no such models can be excluded by IC/DC. In contrast, relatively light LSPs composed of a mixture of gaugino and Higgsino eigenstates have large scattering and annihilation cross sections and are highly detectable by IC/DC. We observe that all such WMAP-saturating well-tempered neutralinos with masses should be excluded by the IC/DC search (c.f., the magenta points in Fig. 9).
## 6 Complementarity: Putting It All Together
Now that we have provided an overview of the various dark matter searches that form our analysis, we can combine them to see what they (will) reveal about the nature of the neutralino LSP as DM [51] and, more generally, the pMSSM itself. Since we only have 14 TeV results for the 30.7k neutralino LSP models that survive the 7 + 8 TeV searches and have GeV (because of CPU limitations as described above), the main results presented below will only make use of the 7 + 8 TeV LHC searches listed in Table 2. We will also present some indicative results showing the sensitivity of the combined 7, 8, and 14 TeV LHC analyses for the subset of neutralino LSP models with GeV.
Figure 10 shows the survival and exclusion rates resulting from the various searches and their combinations in the LSP mass-scaled SI cross section plane. In the upper left panel we compare the combined direct detection (DD = LZ, SI + SD) and indirect detection (ID = Fermi + CTA) DM searches. Here we see that 7.2% (27.5%) of the models can be excluded by ID but not DD (excluded by DD but not ID) while 11.4% are excluded by both types of searches. On the other hand, we also see that 53.9% of the models survive both sets of DM searches; 45.5% of this subset of models, in turn, are presently excluded by the LHC. Note that the DD and ID searchable regions are relatively well separated in terms of mass and cross section although there is some overlap between the sets of models covered by the different experiments. In particular we see that the ID searches (here almost entirely CTA) are covering the heavy LSP region even in cases where the SI cross section is very low and likely beyond the reach of any potential DD experiment. Combining all of the searches only 24.7% of the model set would remain undetected. Similarly, the upper right panel compares the reach of IceCube with DD and we see that 37.7% (0%) of the models are covered uniquely by DD (IceCube) only while 1.2% can be simultaneously excluded by both sets of searches and 61.1% would be missed by either search. In the lower left panel, ID and LHC searches are compared and we see that 15.7% (42.4%) of the models would be excluded only by the ID (LHC) searches. However, 3.0% (38.9%) of the models are seen to be covered by (would be missed by) both search techniques. The strong complementarity between the LHC, CTA and LZ experiments is evident here as CTA probes the high LSP mass region very well where winos and Higgsinos dominate, LZ chops off the top of the distribution where the well-tempered neutralino LSP states dominate, and the LHC covers the relatively light LSP region (fairly independent of LSP type) rather well. Of course the strength of the LHC coverage will significantly improve when the 14 TeV analyses are included, as we will see below. In the lower right panel, the relative contributions arising from the LHC and CTA searches to the model coverage are shown. Here the color intensity of a given bin indicates the fraction of models in that bin excluded by the combination of both CTA and the LHC, while the hue indicates whether the excluded models are seen mostly by CTA (blue) or by the LHC (red). It is again quite clear that CTA completely dominates for large LSP masses and also competes with the LHC throughout the band along the top of the distribution, which mostly contains models with thermal relic densities approximately saturating the WMAP/Planck limit. The LHC is seen to exclude a significant fraction of models with LSP masses below 700 GeV, although there is no region in which the LHC excludes as large a fraction of models as CTA for the high LSP masses.
In order to study the complementarity between the various searches in more detail, it is instructive to project their individual capabilities onto parameter planes that are directly related to one of the search categories. This is particularly effective for visualizing how any given experiment’s parameter space of interest is probed by other searches. As a first example of this, Fig. 11 compares the search capabilities of various experiments in the familiar LSP mass-scaled SI direct detection cross section plane. Here we see the regions in this plane where individual experiments are most sensitive. In particular, we show the fraction of models excluded by CTA (top left), the LHC (top right), IceCube (bottom left) and Fermi (bottom right) projected onto this plane. In each case, we also show the expected limit from the SI search at LZ. We see that both IceCube and Fermi probe models with low LSP masses and large SI cross sections, where the LSP tends to be a bino-Higgsino admixture; this region is also accessible to the DD experiments such as LZ. On the other hand, CTA has access to the heavy LSP region, where there are a large fraction of relatively pure wino and Higgsino LSPs, while the LHC coverage is mostly concentrated (for now) on the relatively low mass LSP region. Interestingly, the LHC searches are not quite independent of the SI cross-section; a region of enhanced exclusion fraction is seen for SI cross sections near pb. Models with SI cross sections in this region mostly have wino-like LSPs with light squarks, making them more likely to be observed by the LHC; wino-like LSPs with heavier squarks have a lower SI cross-section, while Higgsinos and mixed states tend to have a higher SI cross-section whether or not light squarks are present.
We now project these results onto the plane which is most relevant for the DM ID searches. Figure 12 again compares the search capabilities of various experiments, but now they are projected into the LSP mass-LSP pair annihilation cross section plane, in which the limits from Fermi and CTA (with particular assumptions about the annihilation channels) can be presented directly. Here we see the fraction of models that can be excluded by searches at CTA (top left), the LHC (top right), IceCube (bottom left) and LZ (bottom right). The expected limits from Fermi and CTA are also shown, represented by the curves penetrating the upper left- and right-hand side of the panels, respectively. Here the dashed (solid) curves correspond to the indirect detection limit obtained when the LSP pair annihilates exclusively into ; we emphasize that a generic LSP in a pMSSM model may annihilate to many (often dozens of) different final states beyond these two simple cases. However, we do see that the generic pMSSM exclusion is well described by these limiting cases, as displayed for CTA in the upper left panel. Note again that CTA is primarily sensitive to models with LSP mass above 100 GeV and with relatively close to the thermal relic value. As in the previous figure, the LHC is mainly effective in the lower LSP mass region. In addition, the LHC searches are seen to be particularly efficient along a thin line starting at annihilation cross sections of for 100 GeV LSPs and increasing with LSP mass; this line corresponds to models with a pure wino LSP, which are subject to strong constraints from the LHC searches for heavy stable charged particles. (Note that this line is below the projected reach of the ID searches.) IceCube is seen to nicely complement CTA in the region with large cross sections but low LSP masses. The projection of the LZ coverage onto this plane is interesting with most of the exclusion appearing at lower LSP masses where the LSP pair annihilation cross section is also large. However, there is also an extended, somewhat diffuse region of substantial direct detection coverage throughout the entire right half of the parameter space as well as on the funnel ‘island’ at small LSP masses.
Continuing along these lines, we also project our results onto the LSP mass - proton SD cross section plane, as shown in Fig. 13. Spin-dependent direct detection experiments can place direct bounds in this plane (we show the expected SD constraint from LZ); indirect detection constraints from IceCube can also be represented in this plane, but depend on assumptions about annihilation channels and whether the LSP has come to equilibrium within the sun. Here we see that to first approximation the CTA and LHC sensitivities are essentially uncorrelated with the value of the SD cross section. IceCube is only sensitive to relatively light LSPs with large SD cross-sections, as would be expected from the IceCube bounds on specific annihilation channels. The lower right panel shows the fraction of models that can be excluded by LZ, including both the expected SD and SI results; the SD results are also described by the plotted red curve. We see that the SI LZ search is most sensitive to models with large spin-dependent cross sections, although the region of sensitivity extends far below the expected LZ SD sensitivity. In particular, we see that adding the LZ SI results removes all of the bino/Higgsino models in the -funnel region (LSP masses below 80 GeV).
Our last projection is displayed in Fig. 14, which plots the LSP mass against the mass of the lightest colored sparticle (LCP)131313Here the LCP is considered to be the lightest (1st or second generation) squark or gluino; the third generation squarks have somewhat different LHC signatures and are not included as the LCP.. The LHC SUSY searches, particularly the jets + MET channel, are directly applicable in this plane and are reasonably well described by simplified models for squarks and gluinos, as discussed above. Unsurprisingly, we see that the expected sensitivity for CTA, LZ, Fermi, and IceCube depend strongly on the LSP mass, but are essentially independent of the LCP mass; the sensitivity of CTA appears to worsen when the LSP and LCP are nearly degenerate, most likely because the relic density in this case is reduced through co-annihilation rather than other mechanisms. On the other hand, the sensitivity of LZ is slightly enhanced in the compressed region, with the probable cause being that the presence of light squarks increases the scattering rate through squark exchange, compensating for the somewhat depleted relic density.
Lastly, we now consider the thermal relic density-LSP mass plane. We note that with other factors constant, the sensitivity of direct detection and IceCube falls off linearly with decreasing relic density, while that of Fermi and CTA falls off quadratically; of course the relic density plays no direct role in the LHC’s sensitivity. In Fig. 15 we display the fractions of models that can be excluded by CTA, the LHC, IceCube, LZ and Fermi in this plane. Once again, we see that the regions constrained by the various experiments overlap significantly141414In the case of an actual DM discovery, the existence of a substantial overlap between the regions of experimental coverage within this parameter space will be very helpful when trying to determine the specific nature of the LSP.; for example, many different experiments will be sensitive to the “well-tempered neutralino” scenario. On the other hand, there are important cases where experiments complement each other to exclude a much larger fraction of models than could be seen by any one experiment. One example of this is the sensitivity of CTA to high-mass LSPs that are difficult to detect at the LHC; another is the sensitivity of the LHC to relatively light winos, which have a very low relic density and are generally missed by dedicated dark matter searches. Aside from being able to observe light winos through the production and decay of other sparticles, the LHC is also directly sensitive to them if their mass splitting is close to the pion mass, yielding a displaced decay that can appear in searches for disappearing tracks or heavy stable charged particles. We also note several other features apparent in this figure: First, CTA, as expected, has excellent sensitivity to most of the models with LSP masses above GeV that saturate the relic density. However, for larger masses, the CTA coverage also extends to relic densities as much as a factor of or more below the WMAP/Planck value. Fermi is seen to cover only the low LSP mass region with a relic density not far from the thermal value, while the IceCube sensitivity extends to much lower relic density provided the LSP mass is below GeV or so. LZ has sensitivity throughout this plane but does best for LSP masses below GeV, even for models with very low relic density. Of course, even for LSP masses up to 1-2 TeV, the LZ sensitivity remains reasonably good. As noted already, the LHC is presently seen to be effective mainly for LSP masses below GeV. The LHC coverage is relatively uniform with respect to the relic density, but of course the fraction of models excluded is very high in the case of very light LSPs. Of course, we again remind the reader that extending the LHC energy to 14 TeV will substantially improve its sensitivity to heavy LSPs, as we will see below.
Finally, Fig. 16 displays the impact of combining all the expectations for the different searches in the -LSP mass plane; this should be compared with Figure 2, showing the original model set before the search sensitivities have been applied. Here we see that () the models that were in the light and -funnel regions have completely evaporated through a combination of the SI and SD LZ analyses, () the well-tempered neutralinos are now seen to have completely disappeared, mostly due to LZ and IceCube, with additional help from Fermi, () the possibility of almost pure Higgsino or wino LSPs even approximately saturating the relic density has vanished due to CTA, ( the mixed wino-Higgsino models, due to a combination of measurements, have also completely disappeared, () the only models remaining which do saturate the WMAP/Planck relic density are those with binos with -scalar resonant annihilation or co-annihilation. () We find that of the pMSSM model sample will have been excluded (or observed) by at least one of the searches considered in this paper.
## 7 Complementarity with the 14 TeV LHC
We now consider the effect of adding 14 TeV jets + MET and corresponding 0-lepton and 1-lepton stop searches with 300 fb of integrated luminosity to the full set of 7+8 TeV searches considered previously. Here, we restrict our analysis to the subset of 45k models with GeV, due to computational limitations. We remind the reader that both the LHC and dedicated dark matter results are essentially independent of the Higgs mass, so that the results for this subset should effectively reproduce those we would get for the full model set, albeit with lower statistics. The left panel of Fig. 17 shows the reach of the combined LHC searches in the LSP mass - SI cross-section plane for models with GeV. Comparing this figure to the upper right panel of Fig. 11, we see two key changes with the inclusion of the expected 14 TeV searches. First, in both cases the effectiveness of the LHC searches is seen to fall off sharply above a particular LSP mass, since above this limit the spectrum is generically either too heavy or too compressed to be observed. Adding the 14 TeV searches effectively doubles this cutoff, from 700 GeV to 1400 GeV, so that most LSPs in our model set can be excluded given colored sparticle masses below 2-3 TeV. Second, we see that the LHC now has sensitivity to a very high fraction (but not all) of the models with LSPs lighter than this cutoff. The large fraction of models which can be excluded by the 14 TeV data is unsurprising, since our chosen scan range for the sparticle masses was designed to ensure that most models would be (at least kinematically) accessible at the 14 TeV LHC.
The right panel of Fig. 17 shows a comparison between the reaches of the LHC and CTA in this plane, analogous to the lower right panel of Figure 10. We see that the LHC and CTA sensitivities now exhibit a sizable region of overlap. However, the blue region on the far right edge of the panel shows that CTA will be sensitive to LSP masses beyond the reach of the LHC. We also note that LSPs heavy enough to be seen by CTA are generally too heavy to be detected in direct (e.g. monojet) searches, so that the LHC is sensitive to these models by observing other (mostly colored) sparticles. It is therefore important to note that CTA has the potential to exclude winos and Higgsinos with a nearly thermal relic density regardless of the characteristics of the rest of the sparticle spectrum. Of course, the improved reach of the LHC at 14 TeV means that there is also an increasing overlap between the LHC and LZ, however we see that (as before) the LHC searches are mostly independent of the SI cross-section. (The white areas at the edges of the panel generally result from low statistics in those regions, increasing the likelihood that all of the models in a given bin will be excluded). Of course, from a discovery perspective, the increased overlap means that there is more potential for a signal to be observed by two, or even all three, experiments, which would greatly aid in characterizing the LSP and other model properties. Finally, we note that not only has the total fraction of models that can be excluded by the combined experiments increased dramatically (from 75.5% to 98.0%) as a result of the 14 TeV LHC reach, but the fraction of models not seen by the LHC which are covered by direct or indirect detection has increased slightly (from 54.8% to 59.3%), since more of the undetected models have heavy LSPs and are therefore likely to be covered by CTA.
## 8 Conclusion
The results presented here directly lead us to a number of interesting conclusions that are already apparent and that we expect to strengthen in the future as more data is collected at the LHC:
• Even if the LSP does not make up all of the DM, it can still be observed in both direct and indirect detection experiments, as well as neutrino experiments such as IceCube. Of course, searches at the LHC are not influenced by the LSP relic density.
• The set of models remaining after all the searches are performed that saturate the thermal relic density consist almost uniquely of those with (co)annihilating bino LSPs.
• SI direct detection, CTA, and the LHC do most of the heavy lifting in terms of complementary searches covering the pMSSM parameter space.
• Multiple/overlapping searches allow for extensive parameter space coverage that will be of particular importance if a DM signal is observed.
• Most of the experiments are seen to provide complementary probes of the pMSSM parameter space.
• The strength of the LHC component in these searches increases significantly with the inclusion of the 14 TeV LHC sensitivity. However, e.g., the jets + MET search is dependent on the rest of the model spectrum, and therefore does not provide complete coverage of any given LSP scenario, in contrast to dedicated DM searches which rely more directly on the LSP properties.
In summary, the pMSSM provides an excellent tool for studying complementarity between different approaches to the search for dark matter. Hopefully, DM will soon be discovered so that we can employ the complementary probes discussed above to ascertain its nature.
## Acknowledgments
The authors would like to thank A. Barr, J. Buckley, D. Cote, J. Feng, K. Matchev, G. Redlinger, and T. Tait for useful discussions. This work was supported by the Department of Energy, Contracts DE-AC02-76SF00515, DE-AC02-06CH11357, and DE-FG02-12ER41811, the Department of Energy Office of Science Graduate Fellowship Program (DOE SCGF), made possible in part by the American Recovery and Reinvestment Act of 2009, administered by ORISE-ORAU under contract no. DE-AC05-06OR23100, and the National Science Foundation under grant PHY-0970173.
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http://xml.jips-k.org/pub-reader/view?doi=10.3745/JIPS.02.0092
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# A New Perspective to Stable Marriage Problem in Profit Maximization of Matrimonial Websites
Aniket Bhatnagar* , Varun Gambhir* and Manish Kumar Thakur*
## Abstract
Abstract: For many years, matching in a bipartite graph has been widely used in various assignment problems, such as stable marriage problem (SMP). As an application of bipartite matching, the problem of stable marriage is defined over equally sized sets of men and women to identify a stable matching in which each person is assigned a partner of opposite gender according to their preferences. The classical SMP proposed by Gale and Shapley uses preference lists for each individual (men and women) which are infeasible in real world applications for a large populace of men and women such as matrimonial websites. In this paper, we have proposed an enhancement to the SMP by computing a weighted score for the users registered at matrimonial websites. The proposed enhancement has been formulated into profit maximization of matrimonial websites in terms of their ability to provide a suitable match for the users. The proposed formulation to maximize the profits of matrimonial websites leads to a combinatorial optimization problem. We have proposed greedy and genetic algorithm based approaches to solve the proposed optimization problem. We have shown that the proposed genetic algorithm based approaches outperform the existing Gale-Shapley algorithm on the dataset crawled from matrimonial websites.
Keywords: Bipartite Matching , Combinatorial Optimization , Evolutionary Computing , Genetic Algorithm , Matrimonial Websites , Stable Marriage Problem
## 1. Introduction
Matching in a bipartite graph aims to assign elements of one set to the elements of another set such that no two elements in the same set are associated. It has been successfully employed in many applications, such as, hospital-resident allocation, stable roommate problem, allocating tutorials in schools and colleges, sailor-boat problem, student project allocation, college admissions, and stable marriage problem (SMP), etc. [1-3]. The problem of stable marriage proposed by Gale and Shapely [4] is defined over equally sized sets of men and women to identify a stable matching/marriage in which each person is assigned a partner of opposite gender according to their preferences. The set of assignments or marriages is said to be unstable if there exists a man and a woman who are not married to each other but prefer each other to their assigned partners.
Gale and Shapley [4] proposed a male optimal solution using an iterative process for identifying the stable marriages with preference lists (ordered preferences) maintained for all men and women. In [5], the authors extended the work of [4] to incorporate female optimal stable solution and minimum choice stable solution. Irving et al. [6] proposed a network flow based approach to find the stable marriages with male and female optimal solutions. Further, they also generalized the problem of stable marriage according to the weighted preference lists representing the weights of numerical preferences maintained by every person in the set. In literature, the problem of stable marriage has gained a lot of attention and attracted many researchers to analyze and extend this problem with different variants [7-12].
Let M(1… m) and W(1 ... n) be the two sets of men and women respectively, where M consists of m number of men and W consists of n number of women, some of the variants of SMP are categorized as follows: (a) m=n, i.e., count of men and women are equal; (b) m≠n, i.e., count of men and women are different; and (c) polygamy, where one person is allowed to have multiple spouses [10]. Here, some of the earlier variants of SMP utilized the preference lists maintained by individuals, whereas some of the recent contributions utilized the weighted preferences or qualitative preferences [9,11].
In these variants, SMP aims to arrange the marriages in such a way that men and women should pair up knowing the preferences of each man and woman in the populace. However, there may be several instances where the population of men and women is very large, such as marriage bureau or matrimonial websites, viz. www.bharatmatrimony.com, www.jeevansathi.com, and www.shaadi.com, etc., where numerous men and women register to get a suitable, satisfactory and stable match. The SMP is not applicable in these instances as it is infeasible for users to rank their partners of opposite gender to create the preference lists. Further, these matrimonial websites charge a certain amount of fee from the registered users in order to dispense the information of a suitable match [13]. This, in turn, leads to generate revenue for matrimonial websites and helps to earn profits. In the context of matrimonial websites, there is a need for a strategy which dispenses the preferred information to ensure maximum profits in terms of their ability to provide a satisfactory and stable match among the registered users. In addition, an efficient approach is needed to compute the preference lists from the information provided by the users while registering on a matrimonial website. In the process of registration, users are required to make profile by providing personal details (now onwards in this paper we call these details as features) viz. height, age, complexion, academic details, job details (viz. salary, experience, etc.), and religion, etc. While registration, users are also required to provide the necessary requirements or expectations in the partner’s profile of opposite gender viz. someone might be looking for educated and good looking partner.
In this paper we have addressed these two issues: (1) computation of an individual’s preference in the large populace of men and women; and (2) finding a satisfactory and stable match for registered users to ensure the maximum profit for a matrimonial website. To compute the preferences, we have used the Euclidean distance between the expectations of men and profile of women or vice versa. Further, we have proposed a combinatorial framework to maximize the profit of a matrimonial website. Several soft computing methods, such as genetic algorithm (GA) and particle swarm optimization, etc., have been successfully employed in many combinatorial problems [14,15]. To solve our proposed framework we have applied GA due to its simplicity in solving combinatorial problems [14] and also proposed a Greedy based solution. The proposed methods have been shown to outperform Gale-Shapley algorithm on the real dataset crawled from two matrimonial websites.
Rest of the paper is organized as follows: in Section 2, we present the problem formulation which includes the enhancements proposed in classical SMP. The methodology to solve the proposed formulation has been described in Section 3. Further, experiments and results are illustrated in Section 4. Finally, Section 5 concludes the work.
## 2. Problem Formulation
This section starts with defining various variables and terminologies used in this paper. In-order to meet various expectations of matrimonial websites, we have proposed enhancements for classical SMP, which has been discussed later in this section along with the problem statement(s).
2.1 Variables and Terminologies
Throughout the paper we have used two sets, M(1… m) and W(1 ... n), where, M is the set of m men and W is the set of n women registered on a matrimonial website; m is the count of men; n is the count of women; M(a) is an instance/element of the set M, whereas W(b) is an instance/element of the set W; F(1…t) is the set of features (viz. height, weight, salary, etc.) describing the man or woman; PM(1 ...m, F(1…t), 1…t) and PW(1 ...n, F(1…t), 1…t) are the sets of the profile of men and women respectively which are made-up of various (1…t) features and the respective quantified values/scores (at the scale of 5, where 5 represents very good, 4 represents good, 3 represents average, 2 represents below average, 1 represents bad and 0 represents very bad), viz. PM(1, 1, 5) represents that the value of feature 1, F(1) (say salary) in the profile of man, M(1) is 5 which is the representation of a very good salary; EM(1 ...m, F(1…t), 1…t) and EW(1 ...n, F(1…t), 1…t) are the sets of expectations (or requirements) and respective quantified value of men and women respectively in their partners of opposite gender, viz. EM(1, 1, 4) represents that man, M(1) is expecting a good (4) value of the feature, F(1), i.e. salary in the profile of the opposite gender.
2.2 Proposed Enhancements in SMP
Generally a large populace of men and women register on matrimonial websites. So asking them to provide their ordered preferences of opposite gender partners seem to be a difficult task for the registered users. Further, benefits (or profits) of matrimonial websites are associated either with maximum monetary gain or with their ability to provide a suitable or satisfactory matching for each man and woman registered on the matrimonial website. If the matrimonial website does not provide satisfactory matches to its registered users, surely the website will lose its users and popularity. Hence, in this paper, we have focused on the following objective: satisfactory matching for the registered users according to their expectations to maximize the profits of matrimonial websites. In order to achieve the listed objective, we have proposed following enhancements for SMP. We refer to our proposed enhanced SMP as ESMP henceforth.
Enhancement 1: Instead of qualitative preference lists (ordered preferences) which are maintained by each man and woman in classical SMP, we propose to make a quantitative list (weighted preference). In the context of matrimonial websites, we propose to create this quantitative list by comparing the expectations of an individual with the profile of opposite gender partners. Here, matching of each expectation of an individual with the profile of opposite gender partners is given a weight as some feature might be dominating over other, viz. salary may be more dominating than height and should be given more weight than height. The computed average of all the matched expectations of an individual (say a man, M(a)∈M) with the profile of opposite gender (say a woman, W(b)∈W) is now onwards called as weighted matching. The quantitative list, representing weighted matching for each man with all women or vice versa is now onwards called as weighted preference. In this paper, we have used three different types of such preferences, viz. Man Weighted Preference (MWP), Woman Weighted Preference (WWP), and Combined Weighted Preference (CWP) and are defined as follows:
MWP(1.. m, 1..n): This is a weighted preference for each man, M(a)∈M, 1<a<m with each woman, W(b)∈W, (1<b<n). MWP(a, b) represents the average of all the expectations of the man, M(a) matched with the profile of the woman, W(b) and defined as follows in Eq. (1).
##### (1)
[TeX:] $$M W P ( a , b ) = \frac { \sum _ { e , p = 1 } ^ { i } W t _ { e } \times \operatorname { dist } \left( E _ { M } \left( M ( a ) , F ( e ) , V _ { e } \right) , P _ { W } \left( W ( b ) , F ( p ) , V _ { e } \right) \right) } { i }$$
where, e=1 to i are the list of expectations of the man, M(a); Wte (at the scale of 100) is a weight associated with the eth expectation; and dist is a function which returns a fractional value (between 0 and 1) depending on the Euclidean distance between the value Ve of the eth expectation (EM(M(a), F(e), Ve)) of the man M(a) and the value Vp of the pth profile (PW(W(b), F(p), Vp)) of the woman, W(b), if F(e) and F(p) both represent the same feature, viz. salary.
As defined in Eq. (1), MWP of each man for each woman lies in the range between 0 and 100. Further, MWP, if represented in qualitative form, is the men's preference list as used in classical SMP.
WWP(1..n, 1..m): This is a weighted preference for each woman W(b)∈W, 1<b<n with each man, M(a)∈M, (1<a<m). WWP(b, a) represents the average of all the matched expectations of the woman, W(b) with the man, M(a) and defined as follows in Eq. (2):
##### (2)
[TeX:] $$W W P ( b , a ) = \frac { \sum _ { e . p = 1 } ^ { i } W t _ { e } \times d i s t \left( E _ { W } ( W ( b ) , F ( e ) , V _ { e } \right) , P _ { M } \left( M ( a ) , F ( p ) , V _ { e } \right) ) } { i }$$
where, e = 1 to i are the list of expectations of the woman, W(b); Wte (at the scale of 100) is a weight associated with eth expectation, and dist is a function which returns a fractional value (between 0 and 1) depending on the Euclidian distance between the value Ve of the eth expectation (EW(W(b),F(e),Ve)), of the woman, W(b) and the value Vp of the pthprofile (PW(M(a),F(p),Vp)) of the man, M(a), if F(e) and F(p) both represent the same feature, viz. salary.
As defined in Eq. (2), WWP of each woman for each man lies in the range between 0 and 100. Further, its presentation in qualitative form represents the women's preference list as used in classical SMP.
CWP(1...m, 1...n): This is a weighted preference where expectations of each man and woman are jointly explored with profiles of each woman and man respectively. CWP(a, b) represents the weighted matching jointly computed as the expectations of the man, M(a) matched with the profile of the woman, W(b) and expectations of that woman, W(b) matched with the profile of that man, M(a) and defined as follows in Eq. (3):
##### (3)
[TeX:] $$\operatorname { CWP } ( a , b ) = \frac { M W P ( a , b ) + W W P ( b , a ) } { 2 }$$
where, MWP (a, b) and WWP(b, a) are computed according to Eq. (1) and (2), respectively. As defined in Eq. (3), CWP between each man and each woman lies in the range between 0 and 100.
An example is presented in Fig. 1 (for m=n=2) where, one of the CWPs is 85, i.e. CWP1, 1= 85. On the scale of 100, CWP of 85 is significantly large and hence suggests that almost all expectations of M(1), viz. educated, and good looking, etc., are met in W(1), viz. she is good looking, educated as well as almost all expectations of W(1) are met in M(1).
An example of CWP.
Enhancement 2: Considering the objective to maximize the profits of matrimonial websites, we have modified the stability criteria of the classical SMP as follows; for a CWP, we consider pairing such men and women which maximize the average weighted matching (as given in Eq. (4)). Now onwards, we will call the average weighted matching as AWM and the maximum average weighted matching as MAWM.
##### (4)
[TeX:] $$\mathrm { MAWM } = \max \left( \frac { \sum _ { i = 1 } ^ { \tau } \operatorname { CWP } ( p a i r ( i ) ) } { \tau } \right)$$
where, τ is the minimum of m and n which are the count of men and women, respectively and under inequality, minimum of the two is considered for computing the AWM; pair(i) returns the indices of a pair of men and women such that for each i, the pair of men and women is unique.
In the example of Fig. 1, there are two possible sets of pairing: (a) M(1) with W(1) and M(2) with W(2), having AWM as 67 i.e. (85+49)/2; and (b) M(1) with W(2) and M(2) with W(1), having AWM as 82, i.e. (81+83)/2. Here, MAWM is achieved with the pairing given in Set b.
2.3 Problem Statement(s)
Considering different scenarios, following problem statements have been addressed in this paper.
Definition 1 (D1): Given sets of m registered men and n registered women (where, m=n), each with their profile and expectations, it is required to identify a suitable/stable pairing between men and women such that profit of the matrimonial website is maximum.
Definition 2 (D2): Given sets of m registered men and n registered women (where, m≠n), each with their profile and expectations, it is required to identify a suitable/stable pairing such that profit of the matrimonial website is maximum. Here, if m>n, then some men in the list will be unpaired and if m<n, then some women will be unpaired.
## 3. Proposed Scheme(s)
As discussed in the previous section, various sets or combinations of pairings between men and women are to be explored based on their CWP. Consequently, the proposed ESMP has resulted in a combinatorial problem. This section presents our proposed greedy and GA based approaches to solve the ESMP and achieve the desired objective optimally or near optimally. The CWP presented in Fig. 2 has been used to elaborate the algorithmic steps.
CWP used to elaborate the algorithmic steps discussed in this section.
3.1 Greedy Based Approach (Algorithm 1)
The proposed algorithm is based on the greedy approach where a man under consideration has been paired with the best unengaged woman or vice versa.
Step 1: For each man, M(a)∈M in CWP, create a sorted list in decreasing order of the profile-expectation based weighted matching with each women, W(b)∈W using stable sorting schemes [16], i.e. maintain the order of the CWPs in the sorted list in case of the tie between CWPs of two pairs of men and women. As an example, sorted lists of men M(1) and M(2) created from the CWP of Fig. 2 are depicted in Fig. 3.
Sorted lists of men, M(1) and M(2) created from the CWP of Fig. 2.
Step 2: Maintain a list, storing the status (initially unengaged) of each woman either as engaged or unengaged.
Step 3: Traverse the sorted lists (an ordered list obtained after applying the stable sorting) for each man from the maximum liked woman to the least. In each traversal, select the first woman which is yet to be engaged and make the pairing between the selected woman and the man under consideration. In case of the tie between CWPs of a man, M(a) with two or more unengaged women (say, W(b) and W(c)), make the pair between W(b) and M(a), if b<c (i.e. if, W(b) appears before W(c) in the ordered sorted list), otherwise make the pair between W(c) and M(a), and change the woman’s status to engaged.
Step 4: Repeat Step 3 until all men (if m≤n) get paired or all women (if m>n) get engaged.
In the preceding example, following pairs or matching will be formed: {M(1), W(1)}, {M(2), W(4)}, {M(3), W(8)}, {M(4), W(5)}, {M(5), W(6)}, {M(6), W(2)}, {M(7), W(3)}, and {M(8), W(7)}, having the AWM as 73.25. Hence in this example, the MAWM achieved by the presented algorithm (Algorithm 1) is 73.25. However, this may or may not be the actual/optimal MAWM as the proposed greedy based approach (selecting such unengaged woman to be paired with the man under consideration whose CWP is highest) may or may not converge in global maxima, i.e. optimal MAWM.
3.2 Genetic Algorithm Based Approach
This section starts with the discussion over the basic structure of GA and presents the proposed GA based approaches for solving the ESMP.
Usually, GA based optimization involves following: encoding of the actual real world solution space to form the computation space or genotype; chromosome, which is one of the possible solutions to a problem; population generation, which is a subset of all the possible solutions to a problem; i.e. set of chromosomes forms the population; computation of the fitness score of chromosomes; applying various operators, viz. selection, crossover and mutation to select chromosomes and generate new offspring and if suitable, replaces the existing individuals in the population.
In line with the basic structure of a GA based optimization, we present two GA based approaches to maximize the profits under two scenarios, m=n, i.e. D1; and m≠n, i.e. D2. Both approaches differ in the process of the initial population generation. In the first approach the initial population is formed with the chromosomes generated in entirely random manner. In the second approach, the initial population is formed with the chromosomes generated in random as well as guided manner. In both approaches, the genotype is constructed using at most t+1 symbols, where, t=m. Here, each symbol represents a man index, viz. symbol 1 represents the man, M(1), symbol 2 represents the man, M(2) and so forth. An additional symbol, ‘0’ is used to represent dummy men to be paired with women under the scenario where m<n.
Further, a chromosome is constituted as a coded string of men-women pairing comprising of their unique identification in such an order that first woman, W(1) is assigned to the man whose index is encoded as the first gene in the chromosome; second woman, W(2) is assigned to the man whose index is encoded as the second gene in the chromosome and so on. Lengths of the chromosomes under different scenarios are as follows: (a) it is m or n, when, m= n, (b) it is m, when, m> n, where some of the men to be paired with (m- n) dummy women and not to be considered while computing the fitness score of such chromosome, (c) it is n, when, m< n, where some of the women to be paired with (n- m) dummy men encoded with the symbol, ‘0’, and not to be considered while computing the fitness score. Fig. 4(a)–(c) present the examples of such encoding under the scenarios, m= n (with m and n as 4), m>n (with m=4 and n=2), and m<n (with m=2 and n=4), respectively. In the example of Fig. 4(a), the encoded string <3 1 4 2> represents the following four pairing: {M(3), W(1)}, {M(1), W(2)}, {M(4), W(3)}, and {M(2), W(4)}. Further, in the example of Fig. 4(b) the encoded string <3 1 4 2> represents the following two valid pairing (after discarding the pairing with dummy women): {M(3), W(1)}, and {M(1), W(2)}. Similarly, in the example of Fig. 4(c) the encoded string <0 1 0 2> represents two dummy men encoded by the symbol ‘0’. In this example, four pairing is possible (including the pairs made with dummy men). Out of the four, following two are valid pairs and contribute in computation of the fitness score: {M(1), W(2)} and {M(2), W(4)}.
Example of chromosomes under the scenarios, (a) m = n, (b) m > n (c) m < n.
3.2.1 GA based approach with random population (Algorithm 2)
This section discusses the proposed GA based approach where the initial population is created with the randomly generated chromosomes. Here we have separately handled the scenarios, m= n and m≠n, and present two algorithms Algorithm 2.1 and Algorithm 2.2 to handle these scenarios respectively. Steps of the proposed algorithms are elaborated through the combined weighted preference, CWP given in Fig. 2.
Algorithm 2.1: When, m=n
Step 1: Randomly make the pairing between men and women such that all men and women are uniquely paired. Apply man indices (1 to m) to encode the pairs to constitute a chromosome and store it into a population array of length m. Repeat the process and obtain the set of chromosomes stored in different arrays (each of length m) to generate an initial population.
Fig. 5 presents one of the chromosomes in the initial population, which is generated by making random pairs between 8 men and 8 women (whose CWP is presented in Fig. 2) and stored into a population array, Pop of length 8.
An example of a chromosome in the initial population, when m = n = 8.
Step 2: With respect to the weighted matching jointly computed as expectations of a man, M(a) matched with the profile of a woman, W(b) and expectations of that woman, W(b) matched with the profile of that man and stored into the CWP, apply Eq. 5 to compute the fitness scores, FS of the chromosomes in the initial population as follows:
##### (5)
[TeX:] $$F S = \frac { \sum _ { i = 1 } ^ { m } \operatorname { CWP } ( \operatorname { Pop } ( i ) , i ) } { m }$$
Fitness score of one of the chromosomes presented as an example in Fig. 5 is computed as (85+81+92+70+60+70+82+45)⁄8, i.e. FS=73.12.
Step 3: Apply any one of the parent selection strategies, viz. roulette wheel selection, tournament selection, etc., to select the parents (two) to be further involved in the process of mating and recombine to produce the offspring.
Step 4: Use the parents selected in Step 3 to undergo for the crossover operation. As it is necessary to maintain the unique pairing between all men and women in a chromosome, hence apply ordered crossover between selected parents to produce the offsprings. In this process, select a subset lying between the two crossover points in a parent (say the first parent) and add the subset to the first offspring. Further, explore the second parent to find out the symbols/values which are missing in the first offspring and add those missing symbols into the first offspring in the order they were found in the second parent. Similarly, create the second offspring by reversing the role of parents.
Fig. 6 elaborates the Step 4 with two parents selected for the mating in Step 3. Here, the subset between two crossover points, 3 and 7 are added into first offspring and remaining symbols have been added in the first offspring in the order they were found in the second parent. The second offspring is also created in the same manner.
An example of ordered crossover to create offsprings.
Step 5: Randomly mutate the two offsprings created in Step 4. Because of the required unique pairing between all men and women in a chromosome, apply swap mutation to select two positions/ indices randomly and interchange the values/symbols on these positions of the offsprings. Besides swap mutation, one can also select following mutation to maintain the unique pairing: scramble mutation and inversion mutation.
When we apply the swap mutation randomly at positions 4 and 8 on the first offspring (Fig. 6) and interchange their values/symbols, following chromosome will be produced: <7 1 4 6 2 8 5 3>. Similarly, <1 3 2 7 4 5 6 8> is the resulting chromosome, when swap mutation is randomly applied at positions 4 and 8 on the second offspring of Fig. 6.
Step 6: Compute the fitness score of the two offsprings generated after the crossover and mutation operations. Check their fitness scores with the chromosome having the least fitness in the populace and select such chromosome to be the part of the populace having better fitness.
Step 7: Repeat Step 3 to Step 7 until any one of the following is not met: (a) no improvement in the population for specified X iterations or (b) total K number of generations has been produced.
Algorithm 2.2: When, m≠n
Step 1: Considering either m-n dummy men (if, m>n) or n-m dummy women (if, n>m), make the length of a chromosome as max(m,n). Generate a chromosome by making random groupings between m men and n women, such that all men and women are uniquely paired. These pairs might involve, either (a) m-n dummy men (if, m>n) paired with m-n women besides n valid pairs, or (b) n-m dummy women (if, n>m) paired with n-m men besides m valid pairs. Store all the unique pairing into the population array, Pop, representing a chromosome. Repeat the process and obtain the set of chromosomes stored in different population arrays to generate an initial population.
An example in Fig. 7(a) presents one of the chromosomes in the initial population. It is generated by making random pairs between 8 men and 8 women (whose CWP is presented in Fig. 2) and stored into a population array, Pop of length 8. In this example, we have considered 3 dummy women, viz. W(6), W(7), and W(8), which are involved in pairing with men, but the pairs involving dummy women are not considered while computing the fitness score. Similarly, the example in Fig. 7(b) presents a chromosome generated by making random pairs between 8 men (where, 3 men, M(6), M(7), and M(8) are dummy men) and 8 women. These dummy men have been encoded with the symbol ‘0’, whereas men indices (1…5) in the chromosome represents the pairing of women with actual/valid men.
An example of chromosomes in the initial population, when (a) m > n, (b) m < n.
Step 2: Apply Eqs. (6) and (7) to compute the fitness scores, FS of the chromosomes in the initial population under the scenarios (a) m>n and (b) m<n, respectively.
##### (6)
[TeX:] $$F S = \frac { \sum _ { i = 1 } ^ { n } C P M ( \operatorname { Pop } ( i ) , i ) } { n }$$
##### (7)
[TeX:] $$F S = \frac { \sum _ { i = 1 } ^ { n } \operatorname { CPM } ( \operatorname { Pop } ( i ) , i ) } { m } , \text { where, Pop } ( i ) \geq 1$$
Fitness scores of the chromosomes in preceding examples, Fig. 7(a) and (b) are computed as (85+81+92+70+60)⁄5, i.e. FS=77.6 (when, m>n) and (67+43+98+91+55)⁄5, i.e. FS=71 (when n>m), respectively.
Step 3: Select two parents to be involved in the process of mating and recombine to produce the offspring using any one of the parent selection strategies, viz. roulette wheel selection, tournament selection, etc.
Step 4: Apply the ordered crossover between selected parents to produce the offsprings. Here, some of the men which were paired with dummy women (when, m>n) may now be paired with actual women in the dataset.
Fig. 8 depicts the offsprings generated after the ordered crossover between the selected parents under the scenario m>n. Here three different men, M(1) and M(5) along with M(8) are respectively paired with dummy women, W(7), W(8), and W(6) in the first off-spring, whereas in the second offspring, men, M(5), M(6), and M(7) are respectively paired with dummy women, W(6), W(7), and W(8).
Similarly, Fig. 9 depicts the offsprings generated after the ordered crossover between the selected parents under the scenario m<n. Here, women, W(1), W(2), and W(4) are paired with dummy men, represented as the symbol, ‘0’ in the first offspring, whereas, W(1), W(3), and W(4) are paired with dummy men, represented as the symbol, ‘0’ in the second offspring.
An example of ordered crossover to create offsprings, when m > n.
An example of ordered crossover to create offsprings, when m < n.
Step 5: Use swap mutation to swap the randomly selected symbols/bits of the two offsprings created in previous step (Step 4). Compute the fitness scores (either using Eq. 6, if m>n or Eq. 7, if n>m) of the two offsprings generated after the crossover and mutation operations. Check these fitness scores with the chromosome having the least fitness in the populace and select such chromosome to be the part of the populace having better fitness.
Step 6: Repeat, Step 3 to Step 6 until any one of the following is not met: (a) no improvement in the population for specified X iterations or (b) total K number of generations has been produced.
3.2.2 GA based approach with guided population (Algorithm 3)
This approach is built around the steps of Algorithm 2, discussed in the previous section. Unlike Algorithm 2, here some of the chromosomes in the initial population are generated in a guided manner and presented subsequently when m=n and m≠n.
Guided Chromosome in Initial Population, when m=n
Step (a): Identify all such pairs of M(i), 1<i<m and W(j), 1<j<n in CWP for which M(i)’s best weighted matching is with the woman, W(j) and W(j)’s best weighted matching is with the man, M(i). Store all such i at the jth index in the population array, Pop.
It can be seen in the CWP of Fig. 2 that, M(4)’s best weighted matching is with W(5) and also W(5)’s best weighted matching is with M(4). Population array, Pop, storing all such pairing in the CWP of Fig. 2 after the Step (a) is shown in Fig. 10(a).
Step (b): One by one identify the pairing for remaining men by getting the best weighted matching of a man, M(i) with a woman W(j). Unlike Step (a), here, the best weighted matching for the woman, W(j) will not be with the man, M(i). Further, here, we may encounter a scenario where, W(j) may be the best pair for more than one man. In this scenario, W(j) is to be paired with such man, M(i) to whom W(j) is having maximum weighted matching.
In the preceding example, following men are still unpaired after Step (a): M(1), M(2), M(3), M(7), and M(8). Out of these unpaired men, M(3)’s best weighted matching is with W(8), and there is no other man whose best weighted matching is with W(8). Hence, M(3) and W(8) are paired. Further, M(1)’s and M(2)’s best weighted matching is with W(1). Out of M(1) and M(2), W(1)’s best weighted matching is with M(1). Hence, M(1) and W(1) are paired and M(2) remains unpaired after Step (b). Finally, M(7)’s and M(8)’s best weighted matching is with W(5) which is already paired with M(4), and hence, M(7) and M(8) remain unpaired after Step (b).
Step (c): Randomly pair unpaired men and women (of the previous step) to generate the chromosome stored into the population array, Pop. We call this as a guided chromosome.
As shown in Fig. 10(b), men, M(2), M(7), and M(8) and women, W(3), W(4), and W(7) are unpaired after Step (b) and hence to be randomly paired after Step (c). The guided chromosome finally stored in the population array, Pop after the Step (c) is shown in Fig. 10(c).
Generation of a guided chromosome, when m = n.
Guided Chromosome in Initial Population, when m≠n
Under this scenario, generate a guided chromosome in the initial population in the same manner as generated for m=n. Here, instead of actual pairing, some men might be paired with dummy women (if, m>n) or some women might be paired with dummy men (if, n>m).
The example presented in Fig. 11(a) shows a guided chromosome generated after applying Steps (a), (b), and (c) of the previous section on the CWP of Fig. 2 (considering the women, W(6), W(7), and W(8) as dummy women). Here, Step (a) is resulted into following pairs: M(6) with W(2) and M(4) with W(5), whereas, Step (b) makes the pairing between M(1) and W(1). In Step (c), randomly any two men, in the current example we have considered, M(5) and M(8) out of unpaired 5 men are to be paired with 2 unpaired women, W(3) and W(4). Finally, all the unpaired men, M(2), M(3), and M(7) are randomly paired with dummy women, W(6), W(7), and W(8) and not to be involved while computing the fitness score.
Similarly, the example presented in Fig. 11(b) shows a guided chromosome generated after applying Steps (a), (b), and (c) of the previous section on the CWP of Fig. 2 (considering the men, M(6), M(7), and M(8) as dummy men). Here, Step (a) is resulted into following pairs: M(4) with W(5) and M(5) with W(6), whereas, Step (b) gives the pairing between M(1) and W(1). In Step (c), randomly any two women, in the current example, they are W(3) and W(7) out of unpaired 5 women are to be paired with 2 unpaired men, M(2) and M(3). Finally, all the unpaired women, W(2), W(4), and W(8) are randomly paired (represented as symbol ‘0’) with dummy men, M(6), M(7), and M(8) and not to be involved while computing the fitness score.
Generation of the guided chromosomes, when (a) m > n, (b) m < n.
Algorithm 3.1: When, m=n
Step 1: Randomly make the pairing between men and women such that they are uniquely paired and generate the set of chromosomes in the initial population. Make sure that one of the chromosomes is a guided chromosome generated using the approach discussed earlier with m=n.
Step 2: Apply Eq. (5) to compute the fitness scores, FS of the chromosomes in the initial population.
As an example, fitness score of one of the chromosomes presented as the guided chromosome in Fig. 10(c) is computed as (85+95+45+89+98+91+56+82)⁄8, i.e. FS=80.12.
Step 3: Select two parents from the initial population to be involved in the process of mating. Use the Steps 4 and 5 of the Algorithm 2.1 to generate the offsprings after crossover and mutation operations. Compute the fitness score of the two offsprings using Eq. (5) and check their fitness scores with the least fitted chromosome (i.e., having least fitness) in the populace. Select such chromosome to be the part of the populace having better fitness.
Step 4: Repeat, Step 3 until any one of the following is not met: (a) no improvement in the population for specified X iterations or (b) total K number of generations has been produced.
Algorithm 3.2: When, m≠n
Step 1: Randomly make the pairing between men and women such that they are uniquely paired and generate the set of chromosomes in the initial population. These pairs might involve, either (a) m-n dummy men (if, m>n) paired with m-n women besides n actual pair, or (b) n-m dummy women (if, n>m) paired with n-m men besides m actual pair. Make sure that in either of the case, m>n or n>m, one of the chromosomes is a guided chromosome generated using the approach discussed earlier with m≠n.
Step 2: Apply Eq. (6) (when, m>n) or Eq. (7) (when, m<n) to compute the fitness scores, FS of the chromosomes in the initial population.
As an example, fitness score of one of the chromosomes presented as the guided chromosome in Fig. 11(a), when, m>n is computed as (85+95+52+89+98)⁄5, i.e. FS=83.8, whereas fitness score of one of the guided chromosome presented in Fig. 11(b), when, m>n is computed as (85+52+98+91+80)⁄5, i.e. FS=81.2.
Step 3: Use the Steps 3, 4 and 5 of the Algorithm 2.2 to select the two parents for mating and generate the off-springs after crossover and mutation operations. Compute the fitness score of the two off-springs using Eq. (6) (when, m>n) or Eq. (7) (when, m<n) and check their fitness scores with the least fitted chromosome (i.e., having least fitness) in the populace. Select such chromosome to be the part of the populace having better fitness.
Step 4: Repeat, Step 3 until any one of the following is not met: (a) no improvement in the population for specified X iterations or (b) total K number of generations has been produced.
## 4. Experiments and Results
In this section, we present the implementation details, data sets, results, and discussion over obtained results for the algorithms proposed in Section 3.
Greedy and GA based algorithms presented in the previous section had been implemented in Python besides the Gale-Shapley algorithm. Various experiments had been conducted to test and analyze the performance of the implemented algorithms. These experiments were conducted over the dataset created by crawling user’s (men and women) profiles and expectations from matrimonial websites. Although the objective is to maximize the profit of a single matrimonial website, we crawled the users’ profiles and expectations from various matrimonial websites to perform the experiments over a wide and diversified dataset of mixed group.
4.1 Dataset
As discussed earlier, we created a dataset of a mixed group of user’s profile and expectations. This included the profiles of educated/uneducated individuals, long/medium/short height users, employed/ unemployed/self-employed persons, etc. It also included the mixed expectations in the partners’ (of opposite gender) profile, viz. looking for educated/uneducated partner, high/moderate salary, vegetarian/non-vegetarian, etc. To maintain such diversity in the experimental dataset, we crawled the user’s profile and their expectations from two matrimonial websites, viz. www.jeevansathi.com, www.bharatmatrimony.com. In total, details (profile and expectations) of 1000 men and 1000 women were crawled from these websites. Crawling from different sources resulted in different format/ structure of data and unequal set of features (viz. religion, height, salary, etc.) in individuals profile and expectations. Therefore restructuring or cleaning of the crawled data from different sources had been done. We considered only such features (either in profile or in expectations) which were common in all data sources. In total five features had been used after the restructuring or cleaning process.
Further, we stored these restructured data into four matrices as follows: PM (profile of men), EM (expectations of men in his partner’s profile), PW (profile of women), and EW (expectations of women in her partner’s profile). Here, PM(i, j) stores the numeric value scaled between -1 and 5 representing the presence (scaled between 0 and 5) / not presented (scaled to –1) of the jth feature in the profile of the ith man. For example, let, j=2 represents a feature salary, then PM(1, 2)= 0 is the representation of very bad earnings for the man, M(1), whereas PM(2, 2)= 5 is the representation of very good earnings for the man, M(2), and PM(3, 2)= -1 represents the unavailable or undisclosed information about the salary of M(3) at the matrimonial website. Similarly, PW(i, j) represents the presence of the jth feature in the profile of the ith woman. Further, EM(i, j) stores the numeric value (scaled between 0 and 5) for the jth expected feature by Mi in his partner’s profile. Here, EM(1, 2) = 4 suggests that M(1) is expecting a partner whose income/earning is good. Similarly, EW(i, j) stores the numeric value for the jth expected feature by Wi in her partner’s profile.
Matrices discussed in previous steps had been used to create the weighted preferences, MWP and WWP as follows:
##### (8)
[TeX:] $$M W P ( i , k ) = \frac { \sum _ { j = 1 } ^ { f } \left( x ( j ) \times \operatorname { dist } \left( E _ { M } ( i , j ) , P _ { W } ( k , j ) \right) \right) } { t } \times 100$$
where, i and k are the ith and kth man and woman respectively; f is the count of features; t is the count of expected features (or expectations) for the ith man in his partner’s profile, i.e. t is the count for which EM(i, j)≥0; dist is the distance function and returns 1 if, EM(i, j)≤PW(k, j), where, EM(i, j)≠-1, else returns the Euclidean distance between EM(i, j) and PW(k, j) scaled between 0 and 1. Further, x(j) is the weight (between 0 and 1) associated with each feature such that [TeX:] $$\sum _ { j = 1 } ^ { t } x ( j )$$ is 1.
##### (9)
[TeX:] $$W W P ( i , k ) = \frac { \Sigma _ { j = 1 } ^ { f } \left( x ( j ) \times \operatorname { dist } \left( E _ { W } ( i , j ) , P _ { M } ( k , j ) \right) \right) } { t } \times 100$$
Here, the distance function dist returns 1, if EW(i, j)≤PM(k, j), where EW(i, j)≠-1, otherwise returns Euclidian distance between EW(i, j) and PM(k, j) scaled between 0 and 1.
While computing MWP and WWP, we had considered fixed weight, x(j) for each feature. However, it can be tuned in future for further observations in the performance of proposed algorithms. Further, in context of the Indian subcontinent where inter-religion marriage is usually not in practice, we marked the weighted matching between ith man and jth woman, i.e. MWP(i, j) as 0 irrespective of the matching of other features, if any one of the two did not prefer inter-religion marriage. Similarly, under such circumstances, we made WWP(i, j) as 0. Finally, CWP was calculated by taking the mean of MWP and WWP.
4.2 Results and Discussion
In this section, we one by one present the results and other details for the algorithms proposed in Section 3. Performance of the proposed algorithms had been evaluated in terms of the achieved MAWM (computed using Eq. (4)) over the CWP created in Section 4.1. Let us call this CWP as CWP1. We run the Greedy based algorithm once, whereas, each of the GA based algorithms was run p times (10 runs) to observe the variations in the achieved MAWM. Each run of the GA based algorithms involved following in-order: population set having 50 chromosomes; computation of the fitness scores of the chromosomes using the CWP created in Section 4.1; and 10,000 iterations of selection, crossover and mutation operations to generate the offsprings.
The performance of the Greedy based approach presented in Algorithm 1 had been tested over the CWP created in Section 4.1, i.e. CWP1 (having weighted matching of 1000 men and 1000 women). We ran the algorithm once and successfully identified 1000 unique pairings of men and women. In the single run of Algorithm 1, the achieved MAWM was 43.64.
Further, the CWP created in Section 4.1 had been used to compute the fitness scores of the chromosomes in each of the 10 runs for Algorithm 2.1. Each run started with random selection of two chromosomes from the initial population to generate the first off-spring, i.e. in total 20 chromosomes had been selected for this purpose. Out of 20 initially selected chromosomes, the highest fitness score was 43.07, whereas the mean±standard deviation of the initially selected 20 chromosomes was 42.62±0.3. Further, in each run we successfully identified the 1000 unique pairings of men and women. Out of 10 MAWMs (one MAWM per run), the best achieved MAWM was 62.71, whereas the mean±standard deviation of the achieved MAWMs was 62.44±0.37.
A similar experimental procedure had been used to evaluate the performance of Algorithm 2.2. Here, we created another CWP (let us call, CWP2) by randomly removing 200 women from the CWP created in Section 4.1, i.e. CWP2 involved 1000 men and 800 women. CWP2 had been used to compute the fitness score of the chromosomes in each of the 10 runs of the Algorithm 2.2. The highest fitness score out of the 20 initially selected chromosomes, their mean±standard deviation, out of the 10 runs the best achieved MAWM and its mean±standard deviation have been shown in Table 1. We also conducted the experiments to observe the performance of the Algorithm 2.2 under the scenario m<n. These experiments had been conducted over the CWP created by randomly removing 200 men from the CWP created in Section 4.1. Let us call this CWP as CWP3. Table 1 also depicts various observations (viz. best achieved MAWM out of 10 runs and its mean±standard deviation, etc.) related to the conducted experiments under this scenario.
Next set of experiments had been conducted to evaluate the performance of Algorithm 3 under the scenarios, m=n, m>n, and m<n using CWP1, CWP2, and CWP3, respectively. Unlike Algorithm 2, here initial populations were created with 50 chromosomes, where one of the chromosomes was generated in the guided manner. The highest fitness score out of the 10 chromosomes generated in the guided manner for 10 runs of the Algorithm 3.1, their mean±standard deviation, out of the 10 runs the best achieved MAWM and its mean±standard deviation have been shown in Table 1. Similar observational details related with 10 runs of Algorithm 3.2 (when m>n) and 10 runs of Algorithm 3.2 (when m<n) have been presented in Table 1.
Lastly, we conducted the experiments to evaluate the performance of the Gale-Shapley algorithm over same data set, i.e. MWP and WWP which had been used to create CWP1. Here, men preference list was created for each man by replacing highest weighted matching in the MWP as preference 1 (i.e. most preferred woman for that man), second highest weighted matching as preference 2 and so on. Similarly, we identified the women preference list for each woman.
Performance details of various algorithms
Each GA based algorithm had been run ten times (i.e., p=10) and each run involved 10000 iterations, i.e., K=10000.
FS=fitness score, NA=not applicable.
Further, these preference lists (preferences of 1000 men for 1000 women and vice versa) were given as inputs to the Gale-Shapley algorithm. The Gale-Shapley algorithm too successfully identified the unique pairing of each man with a woman. The weighted matchings against these pairs had been averaged out to compute the MAWM. The achieved MAWM by Gale-Shapley algorithm was 35.58.
Table 1 consolidates the MAWM achieved by the algorithms presented in this paper including the Gale-Shapley algorithm. As seen in Table 1, out of the four algorithms (including Gale and Shapley) applicable for the scenario, where m=n, Algorithm 3.1 (i.e. GA based approach with a guided population) had shown the best performance to maximize the profit of matrimonial websites in terms of achieved MAWM. One of the reasons behind this might be the guided chromosome in the initial population which ensured that the global/final fitness score of a chromosome is never less than the fitness score of the guided chromosome. Further, as seen from the Table 1, variations (standard deviation) in the achieved MAWM is negligible for all the 10 runs of each of the GA based algorithm, where m=n.
Similarly, as seen in Table 1, Algorithm 3.2 performed better than Algorithm 2.2 under the scenario, where m<n. Both algorithms utilized the capability of GA, however, the reason behind the better performance of Algorithm 3.2 might be the guided chromosome in the initial population. Finally, the same reason might be applicable for the better performance of Algorithm 3.2 than Algorithm 2.2 for the scenario, where m>n. Further, 10 runs of these algorithms had shown little variations (standard deviations) in the achieved MAWM.
In each of the conducted experiment for Algorithm 2 (randomly generated chromosomes), significant improvement has been observed in the fitness scores of the initially selected chromosome and the best chromosome after 10000 iterations of an experiment. Mean of the 10 fitness scores of the initially selected chromosomes corresponding to the 10 experiments conducted for Algorithm 2.1 is 42.62±0.31. This mean is improved significantly to 62.45±0.38 for the best chromosome after 10000 iterations of each experiment. It is evident from Table 1 that significant improvement has been observed for Algorithm 2.2 (when, m>n) and Algorithm 2.2 (when, m<n). Further, mean of the 10 fitness scores of the guided chromosomes corresponding to the 10 experiments conducted for Algorithm 3.1 is 70.23±0.28. This mean is improved significantly to 72.81±0.1 for the best chromosome after 10000 iterations of each experiment. Similar improvements have been observed in the experiments conducted for Algorithm 3.2 (when, m>n) and Algorithm 3.2 (when, m<n).
## 5. Conclusion
In this paper we presented a new perspective to SMP in profit maximization of matrimonial websites. Instead of qualitative preference list, we defined the quantitative/weighted preferences computed using profile (personal details) and expectations (in partner of opposite gender) of the users registered on matrimonial websites. The objective of all the algorithms presented in this paper was to make such pairs/matching of men and women which maximizes the profit (in terms of MAWM) of matrimonial websites. Greedy and Genetic algorithm based algorithms have been proposed in this paper to solve the problem in different scenarios, viz. (a) m (count of men) =n (count of women); (b) m>n; and (c) m<n. As discussed in Section 4, the set of algorithms (Algorithm 3) outperformed other algorithms (viz. Greedy based, and GA based where chromosomes are randomly generated) including Gale-Shapley algorithm under the scenario where, m=n.
As the future extension, the new perspective presented in this paper may also be applied to some other applications of bipartite matching viz. student project allocation, hospital resident problem, etc. As discussed in Section 4, another extension of current work is to train/tune the weights associated with the features for further observations.
## Acknowledgement
We would like to thank Yamuna P. Shukla for his assistance while writing the paper after completion of the presented work.
## References
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## Biography
##### Aniket Bhatnagar
https://orcid.org/0000-0001-8259-1577
He was one of the undergraduate students at Jaypee Institute of Information Technology, Noida, India and completed his under-graduation (B.Tech. in Computer Science Engineering) in 2017. His research interest includes the evolutionary algorithms, combinatorial optimization problems, and machine learning. He actively participates in several online/onsite competitive programming challenges and secured good ranks.
## Biography
##### Varun Gambhir
https://orcid.org/0000-0002-8537-6853
He completed his B.Tech. in Computer Science Engineering from Jaypee Institute of Information Technology, Noida, India in 2017. His research interest includes the machine learning and evolutionary algorithms for combinatorial optimization problems. Besides academics he is actively involved in various online/onsite competitive programming challenges as participant and secured good ranks.
## Biography
##### Manish Kumar Thakur
https://orcid.org/0000-0003-2479-1540
He is currently working as Assistant Professor (Senior Grade) in the Department of CSE at Jaypee Institute of Information Technology (JIIT), Noida. He completed his Ph.D. in 2014 from JIIT, Noida and his M.Tech in Computer Science from Birla Institute of Technology, Mesra, Ranchi, India. His research interest includes, evolutionary algorithms, graph algorithms, parallel and distributed computing, video processing, and machine learning.
Table 1.
Performance details of various algorithms
Algorithm Count of man (m), woman (n) FS of the guided chromosome (used for Algorithm 3) or first selected chromosome (used for Algorithm 2) in the initial population of 50 chromosomes MAWM [TeX:] $$\begin{array} { l } { \operatorname { Mean } \pm \left( \sqrt { \sigma ^ { 2 } / p } \right) } \\ { \text { for } p = 10 \text { runs } } \end{array}$$ Best FS of the chromosome (guided/first selected) out of p=10 runs [TeX:] $$\begin{array} { l } { \operatorname { Mean } \pm \left( \sqrt { \sigma ^ { 2 } / p } \right) } \\ { \text { for } p = 10 \text { runs } } \end{array}$$ 1 (m=n) 1000, 1000 NA NA 43.64 NA 2.1 (m=n) 1000, 1000 43.07 42.62±0.31 62.73 62.45±0.38 2.2 (when m>n) 1000, 800 41.54 41.10±0.49 63.85 63.15±0.43 2.2 (when mn) 1000, 800 71.12 70.97±0.11 73.66 73.38±0.19 3.2 (when m
An example of CWP.
CWP used to elaborate the algorithmic steps discussed in this section.
Sorted lists of men, M(1) and M(2) created from the CWP of Fig. 2.
Example of chromosomes under the scenarios, (a) m = n, (b) m > n (c) m < n.
An example of a chromosome in the initial population, when m = n = 8.
An example of ordered crossover to create offsprings.
An example of chromosomes in the initial population, when (a) m > n, (b) m < n.
An example of ordered crossover to create offsprings, when m > n.
An example of ordered crossover to create offsprings, when m < n.
Generation of a guided chromosome, when m = n.
Generation of the guided chromosomes, when (a) m > n, (b) m < n.
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https://dsp.stackexchange.com/questions/23791/what-does-it-mean-to-say-that-sinc-filters-are-ideal
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# What does it mean to say that “sinc” filters are ideal?
I suggested to someone that one can create a sinc filter that is ideal enough to be indistinguishable from the ideal, given certain limitations with the data in the first place.
Is that true?
It was only a suggestion yet it was met with unhelpfulness.
• i understand that a sinc filter has some sort of mathematical existence - but how can that be so unless it can be applied to some data ? – user3293056 May 31 '15 at 7:50
• Maybe you should post your comment as a port of the original answer? – jojek May 31 '15 at 8:53
• @jojek this is how i think of it... suppose we know the value of pi thru math proof etc., but our computers for some reason can only compute it to some finite amount. if i have a picture and want to draw a circle round it, i would only need the computer to approximate pi close enough to cope with the resolution. ?? – user3293056 May 31 '15 at 13:58
An ideal filter (system) is the one which can be described mathematically but cannot be implemented (realized) physically.
Coming to a sinc filter, the ideality of this filter results from its frequency domain definition which is an ideal low-pass filter with zero ripple in the pass and stop bands and zero transition width.
Having these specifications, an ideal lowpass filter, can be described mathematically but cannot be realized using any techniques. However you can still find the resulting impulse response of an ideal lowpass filter by using the formal frequency to time transformation of the frequency reponse of the ideal filter.
This kept in mind, therefore, an ideal low-pass filter is seen to have an impulse response in the form of a sinc(x) signal, which extends from minus to plus infinity in time, which is another manifestation of the ideality of the filter being designed.
When such a filter is to be implemented in practice, it can only be approximated. The approximation being better as the length of the filter increases which however is a very undesired condition.
Instead of forcing approximations to ideal sinc filters this way, it is practically more effective to use other techniques such as weighted windowing to realize those filters, which results in the conversion of filter characteristics from those of the ideal one to the realizable one.
• hi Fat32 - is any ad hoc implementation always going to be identifiable - so if i block filter something below 200 hertz, can i really do so without leaving any trace of what was there for spying eyes ? – user3293056 May 31 '15 at 13:53
• possibly not... I'm not sure but. – Fat32 May 31 '15 at 17:40
the "ideal" property of a $\operatorname{sinc}(\cdot)$ filter is in reference to its Fourier Transform
$$\mathcal{F} \left\{ 2 f_0 \ \operatorname{sinc}(2 f_0 t) \right\} \ = \ \operatorname{rect}\left( \frac{f}{2f_0} \right)$$
where
$$\operatorname{sinc}(x) \triangleq \begin{cases} \frac{\sin(\pi x)}{\pi x}, & \text{if }x \ne 0 \\ 1, & \text{if }x = 0 \end{cases}$$
and
$$\operatorname{rect}(x) \triangleq \begin{cases} 1, & \text{if }|x|<\frac{1}{2} \\ 0, & \text{if }|x|>\frac{1}{2} \end{cases}$$
the $\operatorname{sinc}(\cdot)$ filter perfectly and totally excludes all frequency components above $f_0$ and perfectly and totally leaves all frequency components below $f_0$ unmolested. that's why it's ideal. it's an ideal (and impossible to attain) filtering operation.
• I would have thought that the OP knows everything that you wrote down. However, you didn't actually answer the OP's question: whether it is possible to "create a sinc filter that is ideal enough to be indistinguishable from the ideal". – Matt L. Jun 1 '15 at 7:01
• @MattL. please never over-estimate my intelligience haha – user3293056 Jun 2 '15 at 6:08
A sinc filter is unstable and not causal, so as such it can't be implemented. In discrete-time you can in principle approximate it arbitrarily closely by applying a (very long) window to the ideal filter impulse response, and shifting it such that it becomes causal. The latter will add a delay but it will not affect the magnitude response.
• i think you mean by "causal" implementable in real time ? or, not.. ? – user3293056 May 31 '15 at 13:59
• @user3293056: Exactly, it means that the current output sample only depends on the current and on past input samples, but not on future input samples. – Matt L. May 31 '15 at 14:43
• i might agree that a sinc filter is acausal and unrealizable, but i have no idea what is meant by "A sinc filter is unstable ...". i don't think that stability has anything to do with a sinc filter. i've always thunk that stability is a property related to recursion (or feedback). is there some standard recursive theoretical implementation of a sinc filter? i am unaware of it, if there is. – robert bristow-johnson May 31 '15 at 22:29
• Since filter is unstable as mentioned by Matt L. For details please refer to dsp.stackexchange.com/questions/1032/is-ideal-lpf-bibo-unstable – Oliver Jun 1 '15 at 1:44
• "I don't really believe that this is new to you, is it?" ... no, just a sloppy memory. having $\max \left\{ |y[n]| \right\} < \infty$ is not the same as $\max \left\{ |h[n]| \right\} < \infty$ . i stand (or sit) corrected. – robert bristow-johnson Jun 1 '15 at 20:00
If you mean that the filter impulse response is a sinc function, then you are talking about a function with infinite support, i.e, the length of the filter is infinite. This is impractical to implement, but you can approximate it by truncation. The longer the segment resulting from your truncation, the better the approximation.
Therefore, your statement is rather correct.
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2021-05-17 02:51:57
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https://research.caluniv.ac.in/publication/probability-of-power-depletion-in-srs-cross-talk-and-optimum
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X
Probability of power depletion in SRS cross-talk and optimum detection threshold for minimum BER in a WDM receiver
Published in IEEE-INST ELECTRICAL ELECTRONICS ENGINEERS INC
2011
Volume: 47
Issue: 4
Pages: 424 - 430
Abstract
In this paper, the probability of power depletion due to stimulated Raman scattering (SRS) is derived for an arbitrary number of interferers to estimate bit error rate and power penalty of a wavelength division multiplexed (WDM) receiver. Gaussian function approximation can be used only when the numbers of interferers is very large. The effects of number of interfering channels, channel separation and receiver noise on the bit error rate (BER) are studied. Minimum BER varies non-monotonically with channel separation. Optimum detection thresholds for minimum BER in the WDM receiver in presence of SRS crosstalk are investigated and summarized in tabular form. © 2006 IEEE.
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2022-09-24 22:10:27
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http://mathhelpforum.com/pre-calculus/3549-analytical-geometry-print.html
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# Analytical Geometry
• Jun 20th 2006, 10:56 AM
SaRah<3
Analytical Geometry
Hey guys.
I need lots of help.
I need to find out how to do these questions: Answers (showing your work, like how to get the answer) would be really great. I'm really stuck with this stuff. And its drving me insane because I don't undertsand it.
1. The vertices of a right triangle are A(3,3), B(-1,4), and C(1,-5). Find the area of the triangle.
2. For the line segment DE, one endpoint is D(3,1) and the midpoint is M(0,-4). Find the coordinates of endpoint E.
3.Triangle DEF is a right triangle with vertices D(-2,5), E(-4,1), and (2,3). Verify that the midpoint of the hypotenuse is equidistant from the three vertices of the triangle.
4. Verify that the vertices A(-6,1), B(2,-5), C(6,1), and D(2,4)are the vertices of a trapezoid.
5. Quadrilaterial RSTU has vertices R(3,2), S(0,4), T(-2,1), and U(1,-1). Verify that
a) quadrilateral RSTU is a square
b) the diagonals of quadrilateral RSTU perpendicularly bisect each other and are equal in length.
I know these are alot of questions....but if you guys could try and anser them all i'd really appercaite it.
• Jun 20th 2006, 11:28 AM
earboth
1. problem only
Quote:
Originally Posted by SaRah<3
Hey guys.
I need lots of help.
I need to find out how to do these questions: Answers (showing your work, like how to get the answer) would be really great. I'm really stuck with this stuff. And its drving me insane because I don't undertsand it.
1. The vertices of a right triangle are A(3,3), B(-1,4), and C(1,-5). Find the area of the triangle. ...
Hello,
first you have to calculate the length of both legs of this triangle. Use the Pythagorian Theorem:
$\overline{AC}=\sqrt{(3-1)^2+(3-(-5))^2} = \sqrt{68}$
$\overline{BC}=\sqrt{(-1-1)^2+(4-(-5))^2} = \sqrt{85}$
$\overline{AB}=\sqrt{(3-(-1))^2+(3-4))^2} = \sqrt{17}$
As you can see at once (BC) is the hypotenuse.
Because a right triangle is a half rectangle the area is calculated here by:
$a=\frac{1}{2}\cdot \overline{AC}\cdot \overline{AB}$. So:
$a=\frac{1}{2}\cdot \sqrt{68}\cdot \sqrt{17}= 17$.
Greetings
EB
• Jun 20th 2006, 11:50 AM
SaRah<3
hey thanks alot.
ummm do you think you could help me with the other questions I put in the first post?
Even just a few of the questions would be real great.
• Jun 20th 2006, 11:56 AM
Soroban
Hello, SaRah<3!
Here's #1 . . . in baby steps . . .
Quote:
1. The vertices of a right triangle are A(3,3), B(-1,4), and C(1,-5).
Find the area of the triangle.
A sketch is always a good idea . . .
Code:
(-1,4) B o | * | * * | * | * * | * | * *| * | o A * * (3,3) - - - + - - - * - - - |* * | * | o | C (1,-5)
Even a rough sketch makes us suspect that: $AB \perp AC$
We find that: . $m_{AB}\:=\:\frac{4-3}{-1-3}\:=\:\frac{1}{-4}\:=\:-\frac{1}{4}$
. . and that: . $m_{AC}\:=\:\frac{-5-3}{1-3}\:=\:\frac{-8}{-2}\:=\:+4$
Since the slopes are negative reciprocals of each other: $\angle A = 90^o$
The base of the triangle is: . $AC\;=\;\sqrt{(1-3)^2 + (-5-3)^2}\:=\:\sqrt{68}\:=\:2\sqrt{17}$
The height of the triangle is: . $AB\;=\;\sqrt{-1-3)^2 + (4-3)^2}\:=\;\sqrt{17}$
The area of a triangle is: . $A \;=\;\frac{1}{2}(base)(height)$
. . so we have: . $A\;=\;\frac{1}{2}\left(2\sqrt{17}\right)\left( \sqrt{17}\right) \;= \;17$ square units.
Edit: EB's too fast for me . . . and with a shorter solution!
• Jun 20th 2006, 12:11 PM
Soroban
Hello again, SaRah<3!
Here's #4 . . .
Quote:
4. Verify that the vertices A(-6,1), B(2,-5), C(6,1), and D(2,4)are the vertices of a trapezoid.
A trapezoid is a quadrilateral with one pair of parallel sides.
From a quick sketch (which I'll omit here), we suspect that: $AB \parallel CD$
We find that: . $m_{AB} \:=\:\frac{\text{-}5-1}{2-(\text{-}6)}\:=\;\frac{\text{-}6}{8}\;=\;-\frac{3}{4}$
. . . and that: . $m_{CD}\:=\:\frac{4 - 1}{2 - 6}\:=\:\frac{3}{\text{-}4}\:=\:-\frac{3}{4}$
We were right! . . . $AB \parallel CD$
Therefore, $ABCD$ is a trapezoid.
• Jun 20th 2006, 12:12 PM
earboth
Quote:
Originally Posted by Soroban
[size=3]...
Edit: EB's too fast for me . . . and with a shorter solution!
Hello,
I'm sorry - but I live at least 6 hours east of you. So I am slightly in the lead. (I found this expression in my dictionary and I'm not certain if it is appropriate here).
Best wishes
EB
• Jun 20th 2006, 12:38 PM
SaRah<3
Thanks so much for these guys!
Try and keep them coming if you can :) :D
• Jun 20th 2006, 02:02 PM
Soroban
Hello, SaRah<3!
Don't get discouraged . . .
Think of all the stuff you do know: distance formula, slope formula, etc.
These problems make you apply them in various ways.
You must learn to Think your way through them . . .
Quote:
5. Quadrilaterial RSTU has vertices R(3,2), S(0,4), T(-2,1), and U(1,-1).
Verify that:
a) Quadrilateral RSTU is a square
b) The diagonals of quadrilateral RSTU perpendicularly bisect each other and are equal in length.
$a)$ How do we prove that a quadrilateral is a square?
We must show that the four sides are equal (making it a rhombus)
. . then show that it has one right angle.
To show equals sides, we need the distance formula for find the lengths, right?
$\overline{RS}\:=\:\sqrt{0-3)^2 + (4-2)^2} \:= \:\sqrt{4 + 9}\:=\:\sqrt{13}$
$\overline{ST} \:=\:\sqrt{(\text{-}2-0)^2 + (1-4)^2} \:=\:\sqrt{4 + 9} \:=\:\sqrt{13}$
$\overline{TU} \:=\:\sqrt{(1+2)^2 + (\text{-}1-1)^2} \:=\:\sqrt{9+4}\;=\;\sqrt{13}$
$\overline{UR}\:=\:\sqrt{(3-1)^2 + (2+1)^2} \:=\:\sqrt{4+9}\:=\:\sqrt{13}$
. . All four sides are equal; we have a rhombus - opposite sides are parallel.
The slope of $RS$ is: . $m_{RS}\:=\:4 - 2}{0 - 3} \:=\:-\frac{2}{3}$
The slope of $ST$ is: . $m_{ST} \:=\:\frac{1 - 4}{\text{-}2 - 0} \:=\:+\frac{3}{2}$
. . Hence: $RS \perp ST,\;\angle S = 90^o$
Therefore, quadrilateral $RSTU$ is a square.
$b)$ We must show that the diagonals are perpendicular,
bisect each other, and are equal in length.
The diagonal from $R(3,2)$ to $T(-2,1)$
. . It has slope: . $m_{RT}\:=\:\frac{1-2}{\text{-}2-3}\:=\:\frac{1}{5}$
. . Its midpoint is: . $M_{RT}\:=\:\left(\frac{3-2}{2},\;\frac{2+1}{2}\right)\:=\:\left(\frac{1}{2} ,\,\frac{3}{2}\right)$
. . Its length is: . $\overline{RT}\:=\:\sqrt{(\text{-}2-3)^2+(1-2)^2}\:=\:\sqrt{25+1}\:=\:\sqrt{26}$
The diagonal from $S(0,4)$ to $U(1,-1)$
. . It has slope: . $m_{SU}\:=\:\frac{\text{-}1-4}{1-0}\:=\:-5$
. . Its midpoint is: . $M_{SU}\:=\:\left(\frac{0+1}{2},\,\frac{4-1}{2}\right)\:=\:\left(\frac{1}{2},\,\frac{3}{2} \right)$
. . Its length is: . $\overline{SU}\:=\:\sqrt{(1-0)^2+ (\text{-}1-4)^2}\:=\:\sqrt{1 = 25}\:=\:\sqrt{26}$
Therefore:
. . Their slopes are $\frac{1}{5}$ and $-5$ . . . they are perpendicular.
. . They have the same midpoint . . . they bisect each other.
. . They both have length $\sqrt{26}$ . . . they are equal.
• Jun 20th 2006, 04:00 PM
SaRah<3
Wow I wish I was as smart as you. Math is so difficult for me. Im currently failing my grade 10 course. :(
The exam is tommorow.
Ill be so happy if I pass.
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2017-03-29 07:34:13
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http://forums.xkcd.com/viewtopic.php?f=7&t=71411&p=2628192
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## 0903: "Extended Mind"
This forum is for the individual discussion thread that goes with each new comic.
Moderators: Moderators General, Prelates, Magistrates
Clayh
Posts: 18
Joined: Tue Apr 12, 2011 2:22 am UTC
### Re: 0903: "Extended Mind"
And 'puppet' is not a loop, nor is computer science, unless someone fixed them. It seem like once you get to 'Information", "Science", or "Mathematics", it takes you to "Philosophy".
EDIT: Did i seriously just use the word 'nor'?
Duban
Posts: 352
Joined: Fri May 01, 2009 1:22 pm UTC
### Re: 0903: "Extended Mind"
Not true. The first link from Starcraft 2 will eventually bring you into an infinite loop.
Ironically, this loop includes Aristotle with the starting statement "... is a Greek Philosopher" with greek being the first link and philosophy being the second.
Edit: Nevermind -_-, they "fixed" it since the last time I checked. It seems someone altered Language so now the first option that isn't in parenthesis is "human" instead of Communication.
Last edited by Duban on Thu May 26, 2011 1:07 am UTC, edited 1 time in total.
It is not the gods I fear. No, It is those who claim to speak for them that concern me.
PakstraX
Posts: 1
Joined: Thu May 26, 2011 12:50 am UTC
### Re: 0903: "Extended Mind"
I started from "Human papillomavirus" and ended up in a Greek language>Greeks>Greek language loop. Though the trivia should be logical and true if you don't follow the links in brackets. The first sentence of almost every article leads to a more general term and Philosophy is the most general of all so it should be the final destination.
Eternal Density
Posts: 5588
Joined: Thu Oct 02, 2008 12:37 am UTC
Contact:
### Re: 0903: "Extended Mind"
Looks like Randall has prompted a new kind of wikipedia vandalism!
Here's another sequence I just ran:
Tron: Legacy (soundtrack) -> Daft Punk -> Electronic music -> Electronic musical instrument -> Musical instrument -> Sound -> Mechanical wave -> Transmission medium -> Solid -> State of matter -> Phase (matter) -> Outline of physical science -> Life sciences -> Fields of science -> Terminology -> Word -> Language -> Human -> Taxonomy -> Science -> Knowledge -> Fact -> Information -> Finite set -> Mathematics -> Quantity -> Property -> philosophy) -> Modern philosophy -> Philosophy
Actually I started with Daft Punk, then I edited Tron: Legacy (soundtrack) so the link to French is now second
Play the game of Time! castle.chirpingmustard.com Hotdog Vending Supplier But what is this?
In the Marvel vs. DC film-making war, we're all winners.
Clayh
Posts: 18
Joined: Tue Apr 12, 2011 2:22 am UTC
### Re: 0903: "Extended Mind"
Just saw that Wikipedia's official Facebook page posted this comic!
harryjohnston
Posts: 17
Joined: Thu May 26, 2011 1:14 am UTC
### Re: #903: Extended MInd
Draco18s wrote:You're half right. Oral History leads to a loop
Not any more. Oral History -> historical -> history -> Greek -> Greeks -> nation -> sovereign state -> state -> social sciences -> etc.
Looks like someone recently eliminated the loop by making the word "historical" a link. I suspect this is not a coincidence.
History of the world also loops.
This is also no longer true!
harryjohnston
Posts: 17
Joined: Thu May 26, 2011 1:14 am UTC
### Re: #903: Extended MInd
Draco18s wrote:...I have just wasted 20 minutes trying to find a counterexample which is not a loop.
Do you mean, which does not loop back to the original article? Clearly you always have to end in a loop eventually (as there are only a finite number of articles) so the only question is whether you reach Philosophy first or not.
mtz
Posts: 1
Joined: Thu May 26, 2011 1:18 am UTC
### Re: 0903: "Extended Mind"
Randall loves trolling with his alt-text.
Crayshack
Posts: 1
Joined: Thu May 26, 2011 1:18 am UTC
### Re: 0903: "Extended Mind"
The "Road" loop seems to have been fixed. But path leads to a disambiguation so I think it might be cheating.
Samik
Posts: 511
Joined: Mon May 02, 2011 11:14 am UTC
### Re: 0903: "Extended Mind"
SirMustapha wrote:
Samik wrote:I'm not sure just continuing to include further examples is really bringing anything to this thread at this point.
But... but... look at all the people that don't have anything worthwhile to do! What are they going to do now??
Right. But take a look at the difference between the way I said it and the way you said it.
That probably doesn't bother you one bit, but I just didn't want there to be the appearance that we're on the same side on... anything, really.
I'm sorry but clearly you're not very popular around here and if I'm going to make any progres at all I can't be directly assosiated with you...I'm sure you'll understand.
---------------
Clayh wrote:EDIT: Did i seriously just use the word 'nor'?
What's wrong with 'nor'? I thought it was still pretty much in common usage. I use it whenever appropriate, and have never once been called out on it.
I think you're safe in this case!
Teseract
Posts: 17
Joined: Thu May 29, 2008 10:43 pm UTC
### Re: 0903: "Extended Mind"
Amusement_parks
Ride
Amusement_parks
Ride
.... etc.
Mego
Posts: 2
Joined: Thu May 26, 2011 1:38 am UTC
### Re: 0903: "Extended Mind"
I've been trying out the trivia in the alt-text, and clicking Random Article actually gave me Philosophy as the first article. I wasn't sure if this counted, so I went through with it, and ended up in the reason-rationality loop (which does not lead back to philosophy, despite what the before-linked article says - I'm tempted to remove the hyperlink to reason on the rationality page just to fix this). Quite disappointing.
woddfellow2
Posts: 20
Joined: Mon Feb 08, 2010 2:41 am UTC
Contact:
### Re: 0903: "Extended Mind"
Another one:
• Sneak Prevue
• Pay-per-view
• Television
• Telecommunication
• Transmission (telecommunications)
• List of academic disciplines (Redirected from Field of study)
• Community
• Interaction (Redirected from Interacting)
• Action theory (philosophy) (Redirected from Action (philosophy))
• Philosophy
1-Crawl 2-Cnfg 3-ATF 4-Exit ?
MisterCheif
Posts: 253
Joined: Tue Apr 14, 2009 1:24 am UTC
### Re: #903: Extended MInd
sargeras0000 wrote:The alt text is incorrect. There are several articles ("computer software" and "oral history" for instance) that will trap you in loops.
It's rather funny - I'd talked about this with a friend of mine a couple days ago, and I've spent the last few looking for those loops and planning to write some code to find all of them.
"Social Sciences," "Scholarly Method," "Principle," "Law," "Institution," and "Social Structures" are part of another loop, that I got to starting from "Abatai, Rennel Island," and going through "Sovereign State" and "State (polity)."
Wikipedia is...interesting...
I can haz people?
lulzfish wrote:Exactly. Playing God is a good, old-fashioned American tradition. And you wouldn't want to ruin tradition. Unless you hate America. And that would make you a Communist.
Fire Tock
Posts: 10
Joined: Sat May 07, 2011 7:18 am UTC
### Re: 0903: "Extended Mind"
Philosophy -> Philopsophy
Streamheat
Posts: 1
Joined: Thu May 26, 2011 2:17 am UTC
### Re: 0903: "Extended Mind"
I started with "Spice Girls" and made it to Philosophy through Mathematics
LOL
Pyrophile
Posts: 1
Joined: Thu May 26, 2011 2:24 am UTC
### Re: 0903: "Extended Mind"
The philosophy thing is pretty nice, but starting at Google - picked at random - I only came within three links of it, at 'scholarship'. Then I got stuck in a loop when I forgot that the etymological brief is in parentheses and started clicking on those absent-mindedly; needless to say, I quickly got stuck in a loop between "greeks" and "greek language".
Oh, never mind, I backtracked and did it properly, and I'm there. What do you wanna bet this strip is going to cause some rather notable pageview bumps?
cogitoergocogito
Posts: 5
Joined: Sun Dec 19, 2010 12:24 am UTC
### Re: 0903: "Extended Mind"
Doesn't work for TVTropes
Atma21
Posts: 1
Joined: Thu May 26, 2011 3:07 am UTC
### Re: 0903: "Extended Mind"
You won't get there from sphincter. Your asshole will not lead to philosophy.
Agrajag619
Posts: 14
Joined: Mon Apr 25, 2011 6:18 am UTC
### Re: 0903: "Extended Mind"
Apologies if someone has posted this already, but there is a page at Wikipedia:Get to philosophy which has been around since May 2008. Their record for longest chain is 36 articles. Has this thread surpassed that already?
EDIT: It seems it was even featured on the Wikipedia Weekly podcast. Cool!
Randomizer
Posts: 284
Joined: Fri Feb 25, 2011 8:23 am UTC
Location: My walls are full of hungry wolves.
Contact:
### Re: 0903: "Extended Mind"
kensey wrote:(Why am I getting an error about jsMath TeX fonts at the top of the page?)
You have javascript turned off. It does that to me too when I don't have javascript enabled.
To read what's in spoiler tags you need javascript turned on. It also allows things like
Code: Select all
[imath]\displaystyle\sum\limits_{n=1}^\infty \frac{3}{4^n}=1[/imath]
to be converted into pretty mathematical symbols. Like this: [imath]\displaystyle\sum\limits_{n=1}^\infty \frac{3}{4^n}=1[/imath]
Not sure if there's anything else that javascript does on this forum.
Last edited by Randomizer on Thu May 26, 2011 3:44 am UTC, edited 2 times in total.
Belial wrote:I'm all outraged out. Call me when the violent rebellion starts.
Draco18s
Posts: 89
Joined: Fri Oct 03, 2008 7:50 am UTC
### Re: 0903: "Extended Mind"
BlandSauce wrote:
Draco18s wrote:...I have just wasted 20 minutes trying to find a counterexample which is not a loop.
I don't think that's possible.
EDIT: Dead ends are another possibility.
Ding ding ding, we have a winner. A dead end is an example of a non-loop end condition of the Wikipedia Link Game!
Eternal Density wrote:We should totally make an application which scrapes all the first links in all wikipedia articles and finds all the loops and then determines what percentage of articles are directed to each loop. For science. And so we know how correct/incorrect Randall is.
And that is also an end condition of the Wikipedia Link Game: write an application to lose twenty dollars and my self respect <-- that is not what I want here. "Lose this particular game which I am referencing" (without ever playing). Also, someone already did.
Ducky0
Posts: 1
Joined: Thu May 26, 2011 3:39 am UTC
### Re: 0903: "Extended Mind"
"skeletal muscle" loops
otis fulano
Posts: 2
Joined: Thu May 26, 2011 3:38 am UTC
### Re: 0903: "Extended Mind"
"Leonard Nimoy" does not lead to "philosophy".
Also, did you notice that most of the time you arrive at "mathematics" before coming to "philosophy"?
synaemnid
Posts: 3
Joined: Mon Mar 14, 2011 9:03 pm UTC
### Re: 0903: "Extended Mind"
I found this today by pure coincidence, following article links (Michael Spivak -> Spivak pronoun -> Nomic -> Mornington Crescent -> Wikington Crescent):
http://en.wikipedia.org/wiki/Wikipedia:Wikington_Crescent
linuxamoeba
Posts: 1
Joined: Thu May 26, 2011 3:59 am UTC
### Re: 0903: "Extended Mind"
I reach Science first in 70% (+/- 10% stat) of "Random Article" trials.
Number of clicks for Philosophy (Science):
- Constitution of El Salvador: 13 (17)
- Sasson Khakshouri: 19 (10)
- Frank R. Bowerman Landfill: 20 (11)
- Peace Park (disambiguation): 12 (3)
- Talant Dyshebaev: 21 (12)
- Rafting: 32 (36)
- Theis: 18 (9)
- Gasoline price website: 22 (13)
- Melon Kinenbi: 20 (11)
- Keene Springs Hotel: 12 (16)
gos
Posts: 1
Joined: Thu May 26, 2011 4:16 am UTC
### Re: 0903: "Extended Mind"
Hit a random article, and the link worked correctly in 19 steps:
Sweden (Swedish redirect)
Swedish_language
North_germanic_languages
Scandinavians
Germanic_peoples
Ethnolinguistics
Linguistics
Human
Taxonomy
Science
Knowledge
Facts
Information
Finite_set
Mathematics
Quantity
Property (Philosophy)
Modern Philosophy
Philosophy
Neat trick, I'm sure someone has a Bacon graph of the entire site somewhere, but the first link restriction makes it interesting.
Posts: 16
Joined: Tue Jul 13, 2010 6:47 am UTC
### Re: 0903: "Extended Mind"
Made me lol, when i tested the alt text using the 'Server Name Indication' article, and got to Philosophy in just 12 steps
But that one does go through Mathematics, which seems to have been "fixed" recently.
This is probably the most gentle xkcd-inspired mass vandalism to strike Wikipedia so far...
BitRoyal
Posts: 2
Joined: Mon Feb 28, 2011 6:55 am UTC
### Re: 0903: "Extended Mind"
When you pick a random article, and journey through Wikipedia given the instructions (click the first link in the article that is neither italicized nor enclosed in parentheses), there are two possible outcomes:
-- A loop.
I'm guessing this specific claim seems to work so well because the [Philosophy -> Existence -> Senses -> Physiology -> Science -> Knowledge -> Fact -> Information -> Finite set -> Mathematics -> Quantity -> Property (philosophy) -> Modern philosophy -> Philosophy] loop is probably the easiest loop to enter. Most of the articles in this loop are very broad and non-specific (e.g. Knowledge, Mathematics, Science), and many articles give a broad overview in the initial paragraph, so it make sense a decent number of articles would contain one of these as the first link.
ritvax
Posts: 104
Joined: Fri Jul 10, 2009 11:19 pm UTC
Location: New York
### Re: 0903: "Extended Mind"
BrianB wrote:
ritvax wrote:Wondering how long it will take the Spark Plugs article to get a new "In Popular Culture" section with an entry, ""Spark plugs were the central theme of the May 25, 2011 edition of the popular webcomic XKCD."
-ritvax
Done!
Looks like it was redacted already.
I wish I was running VMS v.6.5
ritvax
Posts: 104
Joined: Fri Jul 10, 2009 11:19 pm UTC
Location: New York
### Re: 0903: "Extended Mind"
philip1201 wrote:"Spark plugs" were central to the 25th of may xkcd wikipedia shutdown incident, where the highest number of pages had to be locked in a single day?
You'd think an alarm goes off at WikiQuarters whenever a new xkcd comes out that references Wikipedia or a specific article... "This article has been locked due to xkcd."
I wish I was running VMS v.6.5
Gaschowin
Posts: 1
Joined: Thu May 26, 2011 4:54 am UTC
### Re: 0903: "Extended Mind"
I am pleased to announce that this is, in fact, compatible with Wikipedia's article on 'boobs'. It did redirect to 'breasts', unfortunately... The word 'breasts' just doesn't have that perky, youthful feel to them that 'boobs' do! Breasts don't roll off the tongue quite nearly as nicely as boobs.
That's enough, I should probably change my pants...
LSN
Posts: 56
Joined: Fri Jul 04, 2008 4:41 am UTC
### Re: 0903: "Extended Mind"
I wonder what a map of Wikipedia links would look like.
Eternal Density
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Contact:
### Re: 0903: "Extended Mind"
cogitoergocogito wrote:Doesn't work for TVTropes
I figured I'd give it a try, again ignoring italic and parenthetical links. Also I'm skipping self-links after acknowledging them.
Wikipedia -> Star Trek -> Long Runners -> Network To The Rescue -> Cancellation -> Screwed By The Network -> Network Executives -> Screwed By The Network
Hmm, I guess Screwed By The Network would be a poetic end for everything
Let's try some more then.
XKCD -> Stick Figure Comic -> Stick Figure Comic -> ok let's take the next link... Webcomics -> Exactly What It Says on the Tin -> Word Salad Title -> Gratuitous English -> Anime Tropes -> Anime -> Manga -> Anime
Home Page -> Trope -> Trope -> ok let's try the second link... Trope -> how about the third link... Trope -> fourth link... Trope -> fifth link... Bad Writing -> Pot Hole -> Text Formatting Rules -> Wiki Sandbox -> Green Rocks -> Elemental Rock-Paper-Scissors -> Elemental Powers -> Elemental Baggage -> Elemental Powers.
Hmm, interesting.
Stargate Verse -> Stargate Verse -> Plausible Deniability -> Science Fiction -> Science Fiction -> Steam Punk -> Speculative Fiction -> Science Fiction Versus Fantasy -> Suspended Animation -> Human Popsicle -> Time Travel -> Adventure Towns -> The Drifter -> The Drifter -> Adventure Towns
Given that I expect a lot of things will point at Science Fiction, Speculative Ficition, or Time Travel, Adventure Towns is probably a pretty popular destination. Well, not for Star Trek. Let's see...
Farscape -> Space Opera -> Tropes In Space -> Alien Tropes -> Film: Alien -> Body Horror -> Baleful Polymorph -> Curse -> Forbidden Fruit -> Schmuck Bait - > Big Red Button -> molly-guard... oh wait that's on wiktionary... Big Red Button -> Context Sensitive Button -> The Future -> The Future -> Twenty Minutes Into The Future -> The Future.
Oh.
Star Wars -> Heroes [the trope, not the show] -> The Hero -> The Captain -> The Squad -> The Squad -> Command Roster -> Cool Ship -> The Captain
There's something funny about Star Wars going to The Captain but Star Trek to Screwed By The Network
Doctor Who -> The BBC -> CNN -> Twenty Four Hour News Networks -> Networks -> YMMV -> Complete Monster -> Moral Event Horizon -> Event horizon -> General relativity -> Geometry -> Earth -> Planet -> Orbit -> Physics -> Natural Science -> Science -> Knowledge -> Fact -> Information -> Finite set -> Matchematics -> Quantity -> Property (philosophy) -> Modern philosophy -> Philosophy!
But if we don't take the link to Wikipedia then
Doctor Who -> The BBC -> CNN -> Twenty Four Hour News Networks -> Networks -> YMMV -> Complete Monster -> Moral Event Horizon -> Moral Event Horizon -> Kick The Dog -> Karmic Protection -> Card-Carrying Villain -> My Card -> Card-Carrying Villain
Hmmmmmmmm.
Firefly -> redirects to Series: Firefly -> Firefly -> redirects to Series: Firefly -> let's get on with this... Science Fiction -> ok finally something that links to science fiction as for Stargate Verse
Stargate SG-1 -> Stargate SG-1 -> Science Fiction... -> Adventure Towns
Stargate Atlantis -> Stargate SG-1 ...
Stargate Universe -> Stargate Verse ...
Star Trek: The Original Series -> Star Trek ... -> Screwed By The Network
Star Trek: The Next Generation -> Star Trek: The Next Generation -> Gene Roddenberry -> Star Trek ...
Star Trek: Deep Space Nine -> Star Trek ...
Star Trek: Voyager -> Star Trek ...
Star Trek: Enterprise -> Spin Off -> More Popular Spinoff -> Adaptation Displacement -> Adaptation Displacement -> Ret Canon -> Lost In Imitation -> Adaptation Decay -> In Name Only -> Sequelitis -> Sturgeon's Law -> Home Page (seriously, the link is way down the page too) ... -> Elemental Powers.
Battlestar Galactica -> Star Wars ... -> The Captain
Caprica -> Battlestar Galactica (duh) ... -> The Captain
Red Dwarf -> Brit Com -> Brit Com -> Sit Com -> Sit Com -> Sit Com (seriously!) -> Three Cameras -> Desilu Studios -> Gene Roddenberry -> Star Trek... -> Screwed By The Network
Sanctuary -> Web Original -> Shaped Like Itself -> Buffy Speak -> Shaped Like Itself
Eureka -> SyFy -> (redirects to) Sci Fi Channel -> NBC -> ABC -> Walt Disney -> Oswald The Lucky Rabbit -> Walt Disney
Warehouse 13 -> Manchurian Agent -> Memory Gambit -> Laser-Guided Amnesia -> Easy Amnesia -> Laser-Guided Amnesia
Chuck -> Hollywood Nerd -> Real Life -> Real Is Brown -> The Nineties -> Grunge -> Alternative Rock -> Pavement -> Trope Makers -> Trope Makers -> Ur Example -> Sugar Wiki: Most Triumphant Example -> Bill and Ted's Excellent Adventure -> Excellent Adventure -> All Just a Dream -> Or Was It A Dream? -> Dream Sequence -> All Just a Dream (triple loop)
Fringe -> J. J. Abrams -> Showrunner -> Jumping The Shark -> Put on a Bus -> Prime Time Soap -> Soap Opera -> Melodrama -> World Of Ham -> Dark World -> Cosmic Horror Story -> Nietzsche Wannabe -> Hannibal Lecture -> Perp Sweating -> Perp Sweating -> Good Cop/Bad Cop -> Perp Sweating
Definitely proves cogitoergocogito's point that you can't do this with TVTropes.
made initial pages bold for clarity
Play the game of Time! castle.chirpingmustard.com Hotdog Vending Supplier But what is this?
In the Marvel vs. DC film-making war, we're all winners.
hazelorb
Posts: 3
Joined: Thu May 26, 2011 5:12 am UTC
### Re: 0903: "Extended Mind"
For those of you writing scripts, you know about the wikipedia api right?
Also, you should make sure it follows all of Wikipedia:Get_to_Philosophy#Rules
Do their rules imply italics are okay, just not in disambiguation headers??
doggitydogs
Posts: 38
Joined: Sun Apr 24, 2011 5:52 pm UTC
Location: 42.39561°, -71.13051°
### Re: 0903: "Extended Mind"
Just switched to one of my Wikipedia tabs, clicked "random article," followed the instructions in the alt-text exactly, and ended up at "Philosophy" after seventeen clicks. (The article the random button gave me was John C. Sheehan.)
Without looking through eleven pages of thread , has anyone written a program yet to categorize every Wikipedia article by the number of clicks it takes for this to work?
Interestingly, it takes fourteen clicks to make this work for the "Philosophy" article itself.
It also takes nineteen clicks from "xkcd," twelve clicks from "narwhal," and fifteen clicks from "Windows Server 2003."
Philosophy numbers are the new Erdős numbers.
You don't need a Wikipedia article on yourself for that, either. Just use your hometown. For example, my hometown's philosophy number is nine. I have a friend whose hometown's philosophy number is twenty-four.
Geeks could use their favorite operating system (if it's not thirteen, you need a new OS), Web browser (many are 15, but all the really awesome ones are 20), or webcomic (that would be 19, as I've mentioned already ).
scottmsul
Posts: 5
Joined: Tue Sep 29, 2009 5:29 am UTC
### Re: 0903: "Extended Mind"
Guys, do you really think it is just a coincidence that everything happens to lead back to philosophy? Maybe it's something about the nature of philosophy itself, the ultimate field of generalization. Surely, it's no coincidence that every article leads back to the all-encompassing field that is Philosophy, right?
WRONG. I'm afraid we're dealing with a much deeper conspiracy than that. If only it were that simple.
Apparently this game has been going on for a while now. Unfortunately, the players of this game have gone so far as to change articles to lead to philosophy instead of looping. Articles used to loop much more than you'd expect. Makes you wonder if philosophy is really all that deep, or philosophers just do everything they can to make it seem like so .
Unfortunately this game has changed articles for the worse, in the name of philosophy. In the posted link you'll see a bunch of links connected by arrows, with some crossed off and marked "changed," meaning "mission accomplished, it now links to philosophy instead!" Look at one such crossed off chain:
Loop: Natural science → Science → System → Entities → Entity → Existence → Sense → Organism → Biology → Natural science. Changed.
Apparently "Natural science" looped back to "Natural Science" without ever encountering the dreaded P article. Let's see what they changed by starting at natural science, and see where it breaks off from the older chain. Natural Science still leads to Science. But what's this? Science leads to "Knowledge," not "System." Let's see how how they might have edited the article.
The current opening paragraph for "science" is: "Science (from Latin: scientia meaning "knowledge") is an enterprise that builds and organizes knowledge in the form of testable explanations and predictions about the world." The first link here is "knowledge" not "system." Let's look at an older article to see where they may have changed it. From April 2010:
"The word Science can also refer to any systematic knowledge or prescriptive practice that is capable of resulting in a correct prediction." They removed the word "systematic!"
Original, older Science: http://en.wikipedia.org/w/index.php?title=Science&oldid=353969905
This is not the only instance of a page edited in the name of philosophy. Every set of arrows at the bottom of the first link represents an older loop that has been changed. Many of the pages are locked now, so good luck changing them back. Just thought I'd let you guys know.
ConMan
Shepherd's Pie?
Posts: 1690
Joined: Tue Jan 01, 2008 11:56 am UTC
Location: Beacon Alpha
### Re: #903: Extended MInd
SirMustapha wrote:
Schadenfreude wrote:Damnit, it's pointless edits like this that are bad for wikipedia's credibility. Please stop. Wikipedia is not TV tropes. You're just making random edits so it'll comply with a throwaway joke from a friggin' comic.
That's a phenomenon people with actual lives call "Randallism". It can range from adding a link to xkcd on the article for "Telephone" because the latest strip has a telephone in it, to downright spoiling the article for the sake of stupid jokes that that.
Though the alt-text today isn't a joke, because there's no humour. And it's not a fact, because it's wrong. So what is it? A way to manipulate the mass of xkcd fanboys? A way to distract from the absolutely tired and lame joke in the comic itself?
That's funny, because I know several people with actual lives and none of them have ever referred to Randallism. In fact, the only person I've ever heard use it is you, and your life appears to involve accusing the people who like xkcd of not having an actual life, and personally I would consider the people who read xkcd to have more "actual life" than someone who does that (obligatory tangentially-related xkcd link: http://www.xkcd.com/359/).
Also, just like the people who want to name this game after Randall or xkcd are missing the fact that it appeared on Reddit some time ago, your name for the effect misses the fact that the same is true of any comedian who makes fun of Wikipedia - hence why I personally call it "The Colbert Effect" since the biggest examples I know of (which happen to predate most similar xkcd examples) relate to Stephen Colbert, particularly his discussions on Wikiality, truthiness and elephants.
pollywog wrote:
Wikihow wrote:* Smile a lot! Give a gay girl a knowing "Hey, I'm a lesbian too!" smile.
I want to learn this smile, perfect it, and then go around smiling at lesbians and freaking them out.
darkspork
Posts: 532
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Location: Land of Trains and Suburbs
### Re: 0903: "Extended Mind"
No matter what I click on, I always seem to end up in a loop including Ancient Greek
Shameless Website Promotion: Gamma Energy
My new esoteric programming language: GLOBOL
An experiment to mess with Google Search results: HARDCORE PORNOGRAPHY HARDCORE PORNOGRAPHY
BlitzGirl
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### Re: 0903: "Extended Mind"
darkspork wrote:No matter what I click on, I always seem to end up in a loop including Ancient Greek
You shouldn't, at least not at time of this post. Right now:
Ancient Greek - Greek language - Greeks - Nation
...and so forth to philosophy
The traffic map for the philosophy article is interesting:
Spoiler:
Knight Temporal of the One True Comic
BlitzGirl the Pink, Mopey Molpy Mome
Spoiler:
<Profile
~.~.FAQ->
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2019-08-18 20:24:37
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https://nxt4.com.br/wsz33ogy/f36aa4-demand-function-calculator
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For example, variable X and variable Y are related to each other in such a manner that a change in one variable brings a change in the other. Fortunately, the formula for aggregate demand is the same as the one used by the Bureau of Economic Analysis to measure nominal GDP. The supply function in economics is used to show how much of a given product needs to be supplied given the price of a certain good. A point on the demand curve can be interpreted as follows: The demand curve illustrates the law of demand. 2) Calculate Demand Function. CFA® and Chartered Financial Analyst® are registered trademarks owned by CFA Institute. Inverse function calculator helps in computing the inverse value of any function that is given as input. Write a formula where p equals price and q equals demand, in the number of units. Income elasticity of demand: = 0.32I/(-110P +0.32I) Income elasticity of demand: = 6400/(-550 + 6400) Income elasticity of demand: = 6400/5850 Income elasticity of demand: = 1.094 Next: Using Calculus To Calculate Cross-Price Elasticity of Demand This is to say that the inverse demand function is the demand function with the axes switched. The price elasticity of demand is a way of measuring the effect of changing price on an item, and the resulting total number of sales of the item. Similar to the supply function, we can calculate the demand function with the help of a basic linear function QD = mP + b and two ordered pairs of price and quantity. The following equation represents soft drink demand for your company’s vending machines: Due to the law of diminishing marginal utility, the demand curve is downward sloping. 1) Product-level market demand. Note that quantity is a linear function of price and the quantity is inversely proportional to price. It is also called an anti function. Inverse function calculator helps in computing the inverse value of any function that is given as input. All these are called the determinants of demands. Calculate the linear demand function. The demand function has the form y = mx + b, where "y" is the price, "m" is the slope and "x" is the quantity sold. Non linear demand function. benefit) by taking the difference of the highest they would pay and the actual price they pay.Here is the formula for consumer surplus: 1.7 gives: at y = 11, the demand curve is D 1 D 1 [eq. Aggregate demand is the demand for all goods and services in an economy. The five components of aggregate demand are consumer spending, business spending, government spending, and exports minus imports. and the demand function for the widgets is given by, $p\left( x \right) = 200 - 0.005x\hspace{0.5in}0 \le x \le 10000$ Determine the marginal cost, marginal revenue and marginal profit when 2500 widgets are sold and when 7500 widgets are sold. Price Elasticity of Demand = Percentage change in quantity / Percentage change in price 2. Calculating linear demand functions follows a simple four-step process: (1) Write down the basic linear function, (2) find two ordered pairs of price and quantity, (3) calculate the slope of the demand function, and (4) calculate its x-intercept. and b1, b2 and b3 are the coefficients or parameters of your equation. Hence, the revenue increase (usually expressed as a percentage) can be found as. For example, let us assume a = 50, b = 2.5, and P x = 10: Demand function is: D x = 50 – 2.5 (P x) Therefore, D x = 50 – 2.5 (10) or Dx= 25 units. The quantity demanded (Qd) is the amount of a good or service consumers are willing & able to purchase during a given period of time. Where p is price; and q is the theoretical demand at max price; FAQ Write up your demand function in the form: Y=b1x1+b2x2+b3x3, where Y is the dependent variable (price, used to represent demand), X1, X2 and X3 are the independent variables (price of corn flakes, etc.) It involves taking the derivative of a function. For every $1 increase in price of the product, the quantity demanded will reduce by 1.2 units. What is marginal revenue? High Quality tutorials for finance, risk, data science. Then determine the quantity of the initial demand. Definition. (iii) Angle made by any tangent to the demand curve with respect to the positive direction of x – axis is always an obtuse angle. The inverse demand function is the same as the average revenue function, since P = AR. The graph is calculated using a linear function that is defined as P = a - bQ, where "P" equals the price of the product, "Q" equals the quantity demanded of the product, and "a" is equivalent to non-price factors that affect the demand of the product. After that, decide about the new price of your product. I'll do one other point on the demand curve. Price Elasticity of Demand = -1/4 or -0.25 Let us take the simple example of gasoline. Market potential is the total number of potential customers in a particular market. Find the revenue function. Such a demand function treats price as a function of quantity, i.e., what p 1 would have to be, at each level of demand of x 1 in order for the consumer to choose that level of the commodity. A demand curve on a demand-supply graph depicts the relationship between the price of a product and the quantity of the product demanded at that price. In microeconomics, supply and demand is an economic model of price determination in a market. Graph of … The higher the price, the less the demand. The Linear Demand Curve calculator computes the Quantity of Demand (Q) based on the general effects on quantity of demand other than price (a), the slope of the effect of price changes on demand (b) and the price (P). To recall, an inverse function is a function which can reverse another function. The law of demand says people will buy more when prices fall. Get the free "Inverse Function Calculator - Math101" widget for your website, blog, Wordpress, Blogger, or iGoogle. All these are called the determinants of demands. For example, at y = 11 and p = 10, the demand function (1.3) gives q = 35. The algorithm behind this equilibrium price and quantity calculator consists in the following steps, while it requires you to solve and know in advance both the quantity and supply functions: 1) Consider Qd (quantity demanded) equal to Qs (quantity supplied). It is denoted as: f(x) = y ⇔ f − 1 (y) = x. One problem that may arise in estimating demand curves should be recognized at the outset. If the difference between Q1 and Q0 or P1 and P0is high, the mid-point formula for calculation of price elasticity of demand is a better indicator. Consumer surveys and market experiments can be useful in providing such information, but the tech-nique most frequently used to estimate demand functions … The price elasticity of demand affects consumer as well as industries. To calculate maximum revenue, determine the revenue function and then find its maximum value. Demand at the start of the period is 1,000 units and 2,000 units at the end of the period. If Q is the quantity demanded and P is the price of the goods, then we can write the demand function as follows: Qd = f(P) Say, the gasoline demand function and the gasoline price have the following formula: Qd = 12 – 0.5P. Aggregating Demand and Supply Curves and Concept of Equilibrium, Four Methods of Distributing Government Securities, Effects of Government Regulation on Demand and Supply, CFA® Exam Overview and Guidelines (Updated for 2021), Changing Themes (Look and Feel) in ggplot2 in R, Facets for ggplot2 Charts in R (Faceting Layer), Maximum amount of a good that will be purchased for a given price, Maximum price consumers will pay for a specific amount of the good. The relationship between the quantity and the unit price of a commodity demanded by consumer is called as demand function and is defined as x = f (p) or p = f (x), where x>0 and p>0. However, market deman… The law of demand states that when the price of a good rises, and everything else remains the same, the quantity of the good demanded will fall. Here's how to calculate it. Definition. Using the above-mentioned formula the calculation of price elasticity of demand can be done as: 1. R = P * Q. Price Levels of Commodity X. It's used in conjunction with what is called the demand function to determine equilibrium pricing for different markets. Demand: Demand is the quantity demanded by the consumers at various price levels. For any product, we can calculate the quantity demanded as a function of various factors influencing the demand. How to Use the Inverse Function Calculator? This makes it easier to work with them, which in turn allows us to analyze and understand a wide range of basic economic concepts. This video explains how to maximize profit given the cost function and the demand function.Site: http://mathispower4u.com the demand functions for their own (and other) products, this does not mean that it is always easy to obtain such estimates. Microeconomics Calculator; Vector Calculator (3D) Percent by Mass (Weight Percent) Cost per Round (ammunition) Midpoint Method for Price Elasticity of Demand; Floor - Joist count; Music Duration Calculator; Internal Energy; RPM to Linear Velocity; Characteristic Polynomial of a 3x3 Matrix Aggregate Demand . ADVERTISEMENTS: A function represents a relationship between two variables. Our price elasticity of demand calculator is the user-friendly tool that works efficiently to perform PED calculations, all you need to follow the given steps to get instant results! To compute theinverse demand function, simply solve for P from thedemand function. All these are called the determinants of demands. 18 16 14 12. The information from the demand function can be plotted as a simple graph with quantity demanded on x-axis and price on y-axis. Think about how many pieces of the product would your customers demand each month. As a matter of fact, the process of calculating a linear demand function is exactly the same as the process of calculating a linear supply function. It is also called an anti function. Therefore, to calculate it, we can simply reverse P of the demand function. Calculate U.S. The constant "a" embodies the effects of all factors other than price that affect demand. Furthermore, the inverse demand function can be formulated as P = f-1 (Q). A demand function tells you how many items will be purchased (what the demand will be) given the price. If you want to calculate this value without using a demand function calculator, follow these steps: Start by writing down the initial price of your product. In other words, if your revenue increase and your number of units sold also increases, then the marginal revenue will be the per unit increase. The demand curve is downward sloping. However, if the price is 70 dollars, the demand is 5000. Demand of a product is affected by many factors such as the cost of production, its price compared to other alternative products, or the income levels of consumers. At a price of 5 a quantity, or$5 per hour, this firm would demand, if we're thinking of it in terms of labor, at a price of $5 per hour of labor, this firm would demand 5 people per hour. Calculate the best price of your product based on the price elasticity of demand. Find more Mathematics widgets in Wolfram|Alpha. C = Personal Consumption Expenditures of$14.56 trillion. Write a formula where p equals price and q equals demand, in the number of units. The quantity demanded is also positively related to the income of consumers, i.e., if the income is more, the quantity demanded will be more. If you already know the geographic market which you are going to Target, then you can use the product level market demand method. As illustrated in the graph below, the price elasticity changes as we move along the demand curve. To recall, an inverse function is a function which can reverse another function. The mid-point price elasticity is calculated using the following formula: EdQ1Q0Q1Q02P1P0P1P02Q1Q0P1P0P1P02Q1Q02 Price elasticity of demand for a demand represented by demand functionof the form Q = A – bP can be determined using the following formula: EdbP0Q0 Wh… Where "P" refers to the equilibrium price. % Change in Demand = (Demand End – Demand Start) / Demand Start % Change in Income = (Income End – Income Start) / Income Start. You can use the following Price Elasticity Of Demand Calculator What is your observation? Quantity Demanded of Commodity X. Specifically, the steeper the demand curve is, the more a producer must lower his price to increase the amount that consumers are willing and able to buy, and vice versa. First, We will calculate the percentage change in quantity demand. In the case of gasoline demand above, we can write the inverse function as follows: P = (Qd-12) / 0.5 = 2Qd – 24. In this formula, ∂Q/∂P is the partial derivative of the quantity demanded taken with respect to the good’s price, P 0 is a specific price for the good, and Q 0 is the quantity demanded associated with the price P 0.. The inverse demand function is the same as the average revenue function, since P = AR. share | cite | improve this answer | follow | answered Apr 1 '19 at 12:33. Demand of a product is affected by many factors such as the cost of production, its price compared to other alternative products, or the income levels of consumers. Return on Equity Calculator Price Elasticity of Demand Formula The following formula can be used to calculate the price elasticity of demand: PED = [ (Q₁ – Q₀) / (Q₁ + Q₀) ] / [ (P₁ – P₀) / (P₁ + P₀) ] Use Table 1.1.5 GDP of the BEA's GDP and Personal Income Accounts. Example of calculation of inverse demand function. When the quantity demanded is expressed only as a function of the price of the product, it is called a demand function. Why it is important. A demand function is a mathematical equation which expresses the demand of a product or service as a function of the its price and other factors such as the prices of the substitutes and complementary goods, income, etc. In 2019, it was $21.49 trillion. Note: the value of ∆Q / ∆P is the coefficient of the demand function (b). Think about how many pieces of the product would your customers demand each month. Let’s say the price of substitute products is 5 and the income is 50, the above equation can be rewritten as follows: Other factors remaining constant, the quantity demanded will be 415 – 1.2P. Price Elasticity of Demand = -15% ÷ 60% 3. In this particular case, Fig. (ii) The graph of the demand function lies only in first quadrant. Market Demand is the number of units demanded by the total number of customers in the market. The quantity demanded is positively related to the price of related goods, i.e., if the price of related goods increases, the quantity demanded for product X will increase. If income were to change, for example, the effect of the change would be represented by a change in the value of "a" and be reflected graphically as a shift of the demand curve. Learn how to derive a demand function form a consumer's utility function. Demand curves are highly valuable in measuring consumer surplus in terms of the market as a whole. By using this website, you agree to our Cookie Policy. 0. In mathematical terms, if the demand function is f(P), then the inverse demand function is f −1 (Q), whose value is the highest price that could be charged and still generate the quantity demanded Q. . In Fig. This results in the price function as a squared variable. Quantity of Demand (Q): The calculator returns the quantity. The formula to determine the point price elasticity of demand is. Then determine the quantity of the initial demand. If you want to calculate this value without using a demand function calculator, follow these steps: Start by writing down the initial price of your product. It is denoted as: f(x) = y ⇔ f − 1 (y) = x. Then calculate f(4249), f(4250), and f(4251). This video explains how to maximize profit given the cost function and the demand function.Site: http://mathispower4u.com Demand Function Calculator helps drawing the Demand Function. Copyright © 2020 Finance Train. Demand of a product is affected by many factors such as the cost of production, its price compared to other alternative products, or the income levels of consumers. Finally use the results of step 6 and step 7 and the utility function to calculate the level of utility. There is an economic formula that is used to calculate the consumer surplus (i.e. By using this website, you agree to our Cookie Policy. This is called a demand curve. There are two ways to calculate the market demand for any product. Demand Curve in Linear Demand Function. In this problem, U = X^0.5 + Y^0.5. Thus the more popular a company is, the more will be the market demand for its products& the more will be the number of units demanded by the customers in the market. The previous two chapters were concerned with the theory of demand; now we learn how to estimate a product’s demand function. (1.5)] and along this demand curve, obtains q = 35 at p = Rs 10. The demand curve measures the quantity demanded at each price. To find price elasticity demand. For any product, we can calculate the quantity demanded as a function of various The demand curve is important in understanding marginal revenue because it shows how much a producer has to lower his price to sell one more of an item. How to Calculate Market Demand? The quantity demanded is inversely related to price of the products, i.e., if prices fall, the demand will increase. Now let us assume that a surged of 60% in gasoline price resulted in a decline in the purchase of gasoline by 15%. Marginal revenue, or MR, is the incremental revenue from selling an additional unit. You can calculate the revenue in both initial and final state, using the equation . Microeconomics Calculator; Vector Calculator (3D) Percent by Mass (Weight Percent) Cost per Round (ammunition) Midpoint Method for Price Elasticity of Demand; Floor - Joist count; Music Duration Calculator; Internal Energy; RPM to Linear Velocity; Characteristic Polynomial of a 3x3 Matrix You have a demand curve that would look something, a demand curve that would look something like that, a dot, a demand curve that would look like that. . Revenue is the product of price times the number of units sold. In product level demand, the market is broken down in terms of the total product that exists in the market. Demand at the start of the demand function down in terms of the work for!... = P₁ * Q₁ - P₀ * Q₀ i.e., if the of... Calculate the revenue increase ( usually expressed as a percentage ) can interpreted... In this problem, U = X^0.5 + Y^0.5 is D 1 D 1 D 1 eq. Derive a demand function the first step in the price of the product, we will calculate the level utility! Well as industries of coming up with a marginal revenue, determine the revenue function Notice the. Revenue derivative is to say that the demand curve illustrates the law of diminishing marginal utility, demand... With the axes switched various factors influencing the demand for any product, we can simply reverse of... And Personal Income Accounts price on y-axis predict sales and revenue: the value of any function that used... Is an economic formula that is given as input that the demand function quantity / percentage change in quantity.! For aggregate demand is the coefficient of the demand function calculate and sales... Product of price and q equals demand, the quantity demanded at each.... The previous two chapters were concerned with the theory of demand can found. Be recognized at the start of the BEA 's GDP and Personal Income Accounts axes switched the! Since P = 500 - 1/50q | follow | answered Apr 1 at! You already know the geographic market which you are going to Target, you! The geographic market which you are going to Target, then try again is used to calculate,. Derive a demand function is the number of customers in the market as industries demand at the start of period... Market deman… however, market deman… however, if prices fall, the less the function. To recall, an inverse function calculator helps in computing the inverse demand function, supply and is... Be formulated as P = f ( 4249 ), f ( Qd ) lies... Elasticity changes as we move along the demand of goods and services from 4,000 to 5,000 to derive a function! Table 1.1.5 GDP of the work for you plotted as a squared variable determination in a market! Final state, using the above-mentioned formula the calculation of price determination in a demand function calculator be done as f! The end of the market | follow | answered Apr 1 '19 at 12:33 different markets to the... Options, then you can calculate the quantity demanded on x-axis and price on y-axis quantity percentage!, or MR, is the product would your customers demand each month consumer 's utility.... The lower the price elasticity of demand says people will buy more when prices,., i.e., if the price of the product, we can calculate the consumer in! Personal Consumption Expenditures of$ 14.56 trillion using this website, you could write something like P f! ÷ 60 % 3 ( ii ) the graph of the period product! And supply of goods and services in an economy in quantity demand, f ( 4249 ) and! Of coming up with a marginal revenue derivative is to estimate a product ’ s demand function the. And then find its maximum value = f ( 4251 ) from utility FunctionLearn how estimate! Then calculate f ( x ) = y ⇔ f − 1 ( y ) = x and minus... Demand curve, obtains q = 35 not confuse market demand for your company ’ s vending machines: to! Rs 10 higher the demand function is the demand curve can be found as use Table 1.1.5 GDP of demand! Information from the demand function results in the same as the average revenue function and then find maximum... Demand ; now we learn how to derive a demand function with the axes switched our Cookie Policy expressed as. Measuring consumer surplus ( i.e this results in the market Table 1.1.5 GDP of the price function as function! % 3 ways of measuring the same period, Income increased from 4,000 to 5,000 curve is downward.. Could write something like P = f-1 ( q ) the graph below the! Principles in microeconomics are the coefficients or parameters of your product buy more when prices fall, the demand (. 1 '19 at 12:33 additional unit in business function is a demand function calculator can... Notice that the demand curve measures the quantity demanded is expressed only a. Coefficients or parameters of your equation / percentage change in price 2 where P equals price q. Model of price determination in a particular market the axes switched at =. Level demand, the demand ( 4251 ) the accuracy or Quality of finance Train economic to... Can be formulated as P = 10, the less the demand,... From thedemand function be purchased ( what the demand function is a function represents relationship. In quantity / percentage change in quantity demand other than price that demand! People will buy more when prices fall Qd ) measuring the same inverse relationship between price and equals! Revenue in both initial and final state, using the above-mentioned formula the calculation of elasticity. Q ): the value of any function that is given as input this demand curve can formulated! Axes switched an economic model of price and the utility function 500 -.. Intrinsically different from each other in the price of the price function as a function of various factors the... P = AR or parameters of your product based on the price is 70,... Price is 70 dollars, the revenue in both initial and final state, using the above-mentioned formula the of! ∆P is the same inverse relationship between two variables... how to estimate a product ’ s machines. You can use the product, we can simply reverse P of demand. At the outset end of the work for you maximum revenue, determine the revenue increase ( usually expressed a! Demand are consumer spending, government spending, government spending, and f ( x ) y... Every \$ 1 increase in price 2 level market demand is the per unit value from. Revenue function and then find its maximum value two ways to calculate the as. Solve for P from thedemand function graph with quantity demanded at each.. Denoted as: 1 perform the calculations manually or use the product of times. Curves should be recognized at the outset as industries between price and q demand. Intrinsically different from each other 1 '19 at 12:33 be ) given the price of the total of... Are the coefficients or parameters of your product based on the price of the affects. And final state, using the equation plotted is the inverse value of any function that is as. Of any function that is used to calculate the percentage change in demand. Of the product level demand, in the process of coming up with a marginal revenue determine. Between price and q equals demand, in the graph of the product, the demand for product... Measuring the same period, Income increased from 4,000 to 5,000 microeconomics the... High Quality tutorials for finance, risk, data science are going to Target, then try again economy... Marginal revue is the demand by cfa Institute of course, the revenue and! Follows: the calculator returns the quantity tells you how many pieces of the product of price in! Options, then you can calculate the percentage change in quantity / percentage change in quantity / percentage in. Of customers in the graph of the demand function the first step in the process of coming up a! As: f ( 4251 ) the level of utility utility function an economy each.... Rs 10 x-axis and price on y-axis marginal revenue, or MR is! Are just two different ways of measuring the same as the average revenue function, simply for!, market deman… however, if the price elasticity of demand = -15 % demand function calculator 60 3. About the new price of the product 6 and step 7 and the utility function at price! P '' refers to the equilibrium price deman… however, if the price function as a percentage ) can found... Furthermore, the higher the price elasticity of demand helps the company to fix their price, of,. Q₁ - P₀ * Q₀ 1.5 ) ] and along this demand curve is D 1 [ eq results. Of all factors other than price that affect demand they are just two ways! 35 at P = f ( x ) = x finally use the product, market. The inverse demand function is the same as the average revenue function Notice the. S demand function ( 1.3 ) gives q = 35 demand calculator demand function calculator maximum.... Move along the demand will be ) given the demand for any product it! Along the demand curve, obtains q = 35 at P = 10 the., government spending, and exports minus imports price elasticity of demand and supply of goods and in! Computing the inverse demand function is a function of the market as a ). Supply of goods and services demand helps the company to fix their,... The consumers at various price levels our Cookie Policy measure nominal GDP two variables tells you many. Is broken down in terms of the market as a function demand function calculator relationship... A linear function of price elasticity of demand calculator for aggregate demand are consumer spending, business spending, exports... Demand function ( b ) the higher the demand ways of measuring the same period, Income increased from to...
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2021-06-16 16:50:30
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https://www.transtutors.com/questions/the-following-selected-events-occurred-during-2007-jan-10-a-254224.htm
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# The following selected events occurred during 2007: Jan. 10 A
The following selected events occurred during 2007:
Jan. 10 A motor breaks on a machine and is replaced for $800. This replacement was expected when the machine was purchased. Jan. 24 A machine that was purchased for$10,000 and has a book value of $1,000 is sold for$600.
Feb. 3 A fully depreciated building that originally cost $25,000 is demolished so that a new building may be constructed. The demolition cost$2,200 and resulted in $700 of salvageable materials. Feb. 14 A machine breaks down unexpectedly and requires repairs of$700.
Mar. 10 An accident damages some equipment. Repairs cost $2,000. Mar. 19 A motor breaks on a machine and is replaced for$900. The new motor is of an improved design that increases the capacity of the machine.
Mar. 27 Office layout is rearranged at a cost of $700. At the same time, the walls are repainted for$500.
Required
Prepare journal entries for the preceding transactions.
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2019-02-20 01:37:39
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https://www.physicsforums.com/threads/permutation-and-cycles.366077/
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# Homework Help: Permutation and cycles
1. Dec 29, 2009
### ForMyThunder
1. The problem statement, all variables and given/known data
Let P be a permutation of a set. Show that P(i1i2...ir)P-1 = (P(i1)P(i2)...P(ir))
2. Relevant equations
N/A
3. The attempt at a solution
Since P is a permutation, it can be written as the product of cycles. So I figured that showing that the above equation holds for cycles will be sufficient to show that it holds for all permutations.
Let C = (im1im2...imk) be a cycle and let D = (i1i2...ir). Then,
imk$$\stackrel{C^{-1}}{\rightarrow}$$imk-1$$\stackrel{D}{\rightarrow}$$imk-1+1$$\stackrel{C}{\rightarrow}$$imk+1
Let D` = (C(i1)C(i2)...C(ir)), then imk$$\stackrel{}{\rightarrow}$$imk+1
I don't know how to prove this last part, nor do I know if my reasoning is correct. Any suggestions?
Last edited: Dec 29, 2009
2. Dec 29, 2009
### tiny-tim
Hi ForMyThunder!
Hint: what is P-1(P(i1)) ?
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2018-12-17 00:58:30
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https://d3book.hongtaoh.com/task-9-2.html
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### 6.2.1 From Day 23
• 9-2-1. Make changes so that every time you click, new data will show up with transitions. The new data will be an array of random integers with values between 0 and 24. Let the new array have the same length as the original one (the one before any updates). Hint :
• Use a for loop;
• Math.random() generates random numbers from $$0$$ up to, but not including $$1$$.
• 9-2-2. Add extra logical to set the vertical placement and style of labels. “When the data value is less than or equal to 1, place the label up above the bar and set its fill to black. Otherwise, place and style the label normally” (p. 167). Hint :
if () {}
else {}
// Since we use function (d) { }, we need to add "return" in our if statement.
// Also note that we still need to fill out the else {} part. We cannot leave it as blank.
• 9-2-3. We earlier used Math.random() * 25. Replace 25 with a variable maxValue. Set the maxValue to be 100.
Note: We’ll stop editing bar charts temporarily starting from 9-2-4.
• 9-2-4. Based on the results we got from Task 8, do the following: 1. Delete the CSS for .axis path, line, and text; 2. Let the radius of each circle be a constant: 2 px. Then, do the following:
1. Rename the class of xAxis as x axis and that of yAxis as y axis.
2. When people click on the text “Click here for new data”, a new pair of 50 integers between 0 and 100 will show up. Make transitions to these changes and set the duration to be 1 second. Make sure that scales, and both axes are also updated. Hint : to update the axes, first select the axis, and make a transition, set the transition’s duration, and call the axis generator.
• 9-2-5. Make changes to the code so that when you click the trigger, the fill of all the circles changes to magenta, and the radius becomes 3 px. When the transition is over, the fill changes to black and the radii become 2px.
• Hint : Within circle-updating code, use two .on() statements, whithin which you need to first specify whether it’s start or end, and then use an anonymous function to first select the current element and set the attributions.
• 9-2-6. Add a transition (which lasts for 1 second) to the start of the transition and see what will happen.
### 6.2.2 From Day 25
• 9-2-7. Modify the code so that when you click the trigger, each circle will turn pink and its radius immediately increases from 2 to 7. Then circles move (1 second) to their new positions. In the end, circles transition (1 second) to original color and size.
• 9-2-8. “Instead of reelecting elements and calling a new transition on each one with .on("end", ...), just tack a second transition onto the end of the chain.” (p. 174). This chaining approach is recommended when “sequencing multiple transitions”.
### 6.2.3 From Day 28
• 9-2-9. Go back to our bar charts. Based on codes from Task 9-2-3. Make changes so that every time you click the trigger, a new integer between 0 and 24 will be added. Update both the bars and the text labels. The newly generated bars and labels will move from the right border of the SVG to its proper place. Make the transition last for half a second. Hint : Use .merge().
### 6.2.4 From Day 31
• 9-2-10. Create a p element with the text “Click here to remove data”. Every time you click it, the first bar and its associated label will be removed. Add a transition (half a second) to the removal so that the exiting bar will move to the bottom-right corner of the SVG and then disappear. Make sure to update the bars and labels.
Hint : - To remove the first data value from dataset, use .shift();
• To remove svg elements (rect and text), use .exit().remove(). The .exit() function returns a reference to the exiting element; Transition syntaxes should be added between .exit() and .remove();
• To make sure that the exiting bar moves to the bottom-right corner before disappearing, set its position after the transition and before the removal;
• After the removal, we need to update the bars and labels. That’s fairly easy. For rect, just use svg.selectAll("rect"), and then set its position, width, height and color using attr("x", ...).... For text, first use svg.selectAll("text").text(d => d), and then set texts’ positions.
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2021-01-27 17:22:37
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http://mathcracker.com/statistics-calculators-online/page/11
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MathCracker offers a wide variety of statistics calculators, that will perform step-by-step calculations for you. Such calculators come in all forms and shapes, some are very simple (such as for example, simple combinatorial coefficient calculations) and some conduct elaborate calculations (such as some non-parametric statistical test).
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Instructions: Compute exponential distribution probabilities using the form below. Please type the population mean $$(\beta)$$, and provide details about the event for which you want to compute the probability for. Notice that typically, the ...
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Instructions: Compute Poisson distribution probabilities using the form below. Please type the population mean (λ), and provide details about the event you want to compute the probability for: Population mean ($$\lambda$$) Two-Tailed: ≤ X ...
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Instructions: This calculator conducts a Z-test for one population proportion (p). Please select the null and alternative hypotheses, type the hypothesized population proportion $$p_0$$, the significance level $$\alpha$$, the sample mean, the ...
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2018-07-21 05:48:40
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|
https://jakobandersen.github.io/mod/dataDesc/dataDesc.html
|
# 6. Data Formats¶
MØD utilises several data formats and encoding schemes.
## 6.1. GML¶
MØD uses the Graph Modelling Language (GML) for general specification of graphs and rules. The parser recognises most of the published specification, with regard to syntax. The specific grammar is as follows.
GML ::= (key value)*
key ::= identifier
value ::= int
double
quoteEscapedString
list
list ::= '[' (key value)* ']'
A quoteEscapedString is zero or more characters surrounded by double quotation marks. To include a \" character it must be escaped. Tabs, newlines, and backslashses can be written as \t, \n, and \\. An identifier must match the regular expression [a-zA-Z][a-zA-Z0-9]* GML code may have line comments, starting with #. They are ignored during parsing.
## 6.2. Tikz (Rule)¶
This format is used for visualising rules similarly to how the Tikz (Graph) format is used for graphs. A rule is depicted as its span $$(L\leftarrow K\rightarrow R)$$ with the vertex positions in the plane indicating the embedding of $$K$$ in $$L$$ and $$R$$. Additionally, $$L\backslash K$$ and $$R\backslash K$$ are shown in different colour in $$L$$ and $$R$$ respectively.
## 6.3. DOT (Rule)¶
The DOT format (from Graphviz) is used for generating vertex coordinates for the Tikz format, when Open Babel can not be used.
## 6.4. GML (Rule)¶
A rule $$(L\leftarrow K\rightarrow R)$$ in GML format is specified as three graphs; $$L\backslash K$$, $$K$$, and $$R\backslash K$$. Each graph is specified a a list of vertices and edges, similar to a graph in GML format. The key-value structure is exemplified by the following grammar.
ruleGML ::= 'rule ['
[ ruleId ]
[ leftSide ]
[ context ]
[ rightSide ]
matchConstraint*
']'
ruleId ::= 'ruleID' quoteEscapedString
leftSide ::= 'left [' (node | edge)* ']'
context ::= 'context [' (node | edge)* ']'
rightSide ::= 'right [' (node | edge)* ']'
node ::= 'node [ id' int 'label' quoteEscapedString ']'
edge ::= 'edge [ source' int 'target' int 'label' quoteEscapedString ']'
matchConstraint ::= adjacency
'id' int
'op "' op '"'
'count' unsignedInt
[ 'nodeLabels [' labelList ']' ]
[ 'edgeLabels [' labelList ']' ]
']'
labelList ::= ('label' quoteEscapedString)*
op ::= '<' | '<=' | '=' | '>=' | '>'
Note though that list elements can appear in any order.
## 6.5. Tikz (Graph)¶
Graphs are visualised using generated Tikz code. The coordinates for the layout is either generated using Open Babel or Graphviz. The visualisation style is controlled by passing instances of the classes mod::GraphPrinter (C++) and mod.GraphPrinter (Python) to the printing functions. The drawing style is inspired by ChemFig and Open Babel. See also PostMØD (mod_post).
## 6.6. DOT (Graph)¶
The DOT format (from Graphviz) is used for generating vertex coordinates for the Tikz format, when Open Babel can not be used.
## 6.7. GML (Graph)¶
A graph can be specified as GML by giving a list of vertices and edges with the key graph. The following grammar exemplifies the required key-value structure.
graphGML ::= 'graph [' (node | edge)* ']'
node ::= 'node [ id' int 'label' quoteEscapedString ']'
edge ::= 'edge [ source' int 'target' int 'label' quoteEscapedString ']'
Note though that list elements can appear in any order.
## 6.8. SMILES¶
The Simplified molecular-input line-entry system is a line notation for molecules. MØD can load most SMILES strings, and converts them internally to labelled graphs. For graphs that are sufficiently molecule-like, a SMILES string can be generated. The generated strings are canonical in the sense that the same version of MØD will print the same SMILES string for isomorphic molecules.
Warning
The SMILES canonicalisation algorithm is the original CANGEN algorithm that does not work in general, and some molecules with specific symmetries thus have multiple “canonical” forms. This problem will be fixed at some point. To properly check for isomorphism, load the graphs and call the appropriate method.
The reading of SMILES strings is based on the OpenSMILES specification, but with the following notes/changes.
• Only single SMILES strings are accepted, i.e., not multiple strings separated by white-space.
• The specical dot “bond” (.) is not allowed.
• Isotope information is ignored.
• Chirality information is ignored.
• Up and down bonds are regarded as implicit bonds, i.e., they might represent either a sngle bond or an aromatic bond. The stereo information is ignored.
• Atom classes are ignored.
• Wildcard atoms (specified with *) are converted to vertices with label *. Any other information on the vertex is ignored (e.g., the charge in [*-]).
• Only charges from $$-9$$ to $$9$$ are allowed.
• The bond type \$ is currently not allowed.
• Aromaticity can only be specified using the bond type : or using the special lower case atoms. I.e., c1ccccc1 and C1:C:C:C:C:C:1 represent the same molecule, but C1=CC=CC=C1 is a different molecule.
• The final graph will conform to the molecule encoding scheme described below.
• Implicit hydrogens are added following a more complicated procedure.
• A bracketed atom can have a radical by writing a dot (.) between the position of the charge and the position of the class.
The written SMILES strings are intended to be canonical and may not conform to any “prettyness” standards.
### 6.8.1. Implicit Hydrogen Atoms¶
When SMILES strings are written they will use implicit hydrogens whenever they can be inferred when reading the string. For the purposes of implicit hydrogens we use the following definition of valence for an atom. The valence of an atom is the weighted sum of its incident edges, where single (-) and aromatic (:) bonds have weight 1, double bounds (=) have weight 2, and triple bonds (#) have weight 3. If an atom has an incident aromatic bond, its valence is increased by 1. The atoms that can have implicit hydrogens are B, C, N, O, P, S, F, Cl, Br, and I. Each have a set of so-called normal valences as shown in the following table. The atoms N and S additionally have certain sets of incident edges that are also considered normal, which are also listed in the table.
Atom Normal Valences and Neighbourhoods
B 3
C 4
N 3, 5, $$\{-, :, :\}$$, $$\{-, -, =\}$$, $$\{:, :, :\}$$
O 2
P 3, 5
S 2, 4, 6, $$\{:, :\}$$
F, Cl, Br, I 1
If the set of incident edges is listed in the table, then no hydrogens are added. If the valence is higher than the highest normal valence, then no hydrogens are added. Otherwise, hydrogens are added until the valence is at the next higher normal valence.
When writing SMILES strings the inverse procedure is used.
## 6.9. GraphDFS¶
The GraphDFS format is intended to provide a convenient line notation for general undirected labelled graphs. Thus it is in many aspects similar to SMILES strings, but a string being both a valid SMILES string and GraphDFS string may not represent the same graph. The semantics of ring-closures/back-edges are in particular not the same.
### 6.9.1. Grammar¶
graphDFS ::= chain
chain ::= vertex evPair*
vertex ::= (labelVertex | ringClosure) branch*
evPair ::= edge vertex
labelVertex ::= '[' bracketEscapedString ']' [ defRingId ]
implicitHydrogenVertexLabels [ defRingId ]
implicitHydrogenVertexLabels ::= 'B' | 'C' | 'N' | 'O' | 'P' | 'S' | 'F' | 'Cl' | 'Br' | 'I'
defRingId ::= unsignedInt
ringClosure ::= unsignedInt
edge ::= '{' braceEscapedString '}'
shorthandEdgeLabels
shorthandEdgeLabels ::= '-' | ':' | '=' | '#' | ''
branch ::= '(' evPair+ ')'
A bracketEscapedString and braceEscapedString are zero or more characters except respectively ] and }. To have these characters in each of their strings they must be escaped, i.e., \] and \} respectively.
The parser additionally enforces that a defRingId may not be a number which has previously been used. Similarly, a ringClosure may only be a number which has previously occured in a defRingId.
A vertex specified via the implicitHydrogenVertexLabels rule will potentially have ekstra neighbours added after parsning. The rules are the exact same as for implicit hydrogen atoms in SMILES.
### 6.9.2. Semantics¶
A GraphDFS string is, like the SMILES strings, an encoding of a depth-first traversal of the graph it encodes. Vertex labels are enclosed in square brackets and edge labels are enclosed in curly brackets. However, a special set of labels can be specified without the enclosing brackets. An edge label may additionally be completely omitted as a shorthand for a dash (-).
A vertex can have a numeric identifier, defined by the defRingId non-terminal. At a later stage this identifier can be used as a vertex specification to specify a back-edge in the depth-first traversal. Example: [v1]1-[v2]-[v3]-[v4]-1, specifies a labelled $$C_3$$ (which equivalently can be specified shorter as [v1]1[v2][v3][v4]1).
A vertex being a ringClosure can never be the first vertex in a string, and is thus preceded with a edge. As in a depth-first traversal, such a back-edge is a kind of degenerated branch. Example: [v1]1[v2][v3][v4]1[v5][v6]1, this specifies a graph which is two fused $$C_4$$ with a common edge (and not just a common vertex).
Warning
The semantics of back-edges/ring closures are not the same as in SMILES strings. In SMILES, a pair of matching numeric identifiers denote the individual back-edges.
A branch in the depth-first traversal is enclosed in parentheses.
### 6.9.3. Abstracted Molecules¶
The short-hand labels for vertices and edges makes it easier to specify partial molecules than using GML files.
As example, consider modelling Acetyl-CoA in which we wish to abstract most of the CoA part. The GraphDFS string CC(=O)S[CoA] can be used and we let the library add missing hydrogen atoms to the vertices which encode atoms. A plain CoA molecule would in this modelling be [CoA]S, or a bit more verbosely as [CoA]S[H].
The format can also be used to create completely abstract structures (it can encode any undirected labelled graph), e.g., RNA strings. Note that in this case it may not be appropriate to add “missing” hydrogen atoms. This can be controlled by an optional parameter to the loading function.
## 6.10. Molecule Encoding¶
There is no strict requirement that graphs encode molecules, however several optimizations are in place when they do. The following describes how to encode molecules as undirected, simple, labelled graphs and thus when the library assumes a graph is a molecule.
### 6.10.1. Edges / Bonds¶
An edge encodes a chemical bond if and only if its label is listed in the table below.
Label Interpretation
- Single bond
: “Aromatic” bond
= Double bond
# Triple bond
### 6.10.2. Vertices / Atoms¶
A vertex encodes an atom with a charge if and only if its label conforms to the following grammar.
vertexLabel ::= atomSymbol [ charge ] [ radical ]
charge ::= singleDigit ('-' | '+')
With atomSymbol being a properly capitalised atom symbol.
|
2017-09-25 00:57:13
|
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|
https://byjus.com/question-answer/what-is-the-code-for-passive-alert-av28-lg20-aw24-lg20-av28-lh20-aw28-lh20/
|
Question
# What is the code for 'Passive Alert'?
A
AV28 LG20
B
AW24 LG20
C
AV28 LH20
D
AW28 LH20
E
None of these
Solution
## The correct option is A $$AV28\ LG20$$The rules for coding are :$$i)\quad$$ The first letter of the code is the second letter of the word.$$ii)\quad$$ The second letter of the code is the reverse of last letter of the word.$$iii)\quad$$ The number is four times the number of letters in the word.$$Passive :$$i) The second letter of the word $$\Rightarrow A$$ii) The last letter of the word is $$E\Rightarrow V$$iii) The number $$=4\times7=28$$So, the code for Passive is $$AV28$$.$$Alert\ :$$i) The second letter of the word $$\Rightarrow L$$ii) The last letter of the word is $$T\Rightarrow G$$iii) The number $$=4\times 5=20$$So. the code for Alert is $$LG20$$$$\therefore$$ The answer is $$[A]$$.Logical Reasoning
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2022-01-18 02:13:42
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|
https://indico.cern.ch/event/850451/timetable/?view=standard_numbered
|
HEPP2020
Africa/Johannesburg
University of Venda
University of Venda
Senate Chamber P/Bag X 5050 Thohoyandou, 0950
, ,
Description
The goal of this workshop is to give an opportunity to students and young researchers to give presentations and to write proceedings. A morning plenary session, followed 15+10 minute presentations, shall be the format across the three days. The topics to be covered will be high-energy theory and phenomenology (heavy ions, pp, ep, ee collisions), ATLAS physics and ALICE physics. Given the format, It is expected that all students will have a chance to present to their peers and senior physicists.
Registration
HEPP2010 Registration
Participants
• Alan Cornell
• Ali Elimam
• Amanda Percy Sefage
• Andre Sanches Barreiros
• Anna Chrysostomou
• Asiphe Mzaza
• Ayanda Thwala
• Azwinndini Muronga
• Benjamin Fuks
• Benjamin Lieberman
• Benjamin Warren
• Betty Kibirige
• Blessed Ngwenya
• Charles Moditswe
• Daniel Edwards
• Danielle Sabatta
• Danielle Wilson
• Deepak Kar
• DIMAKATSO MAHESO
• Dingane Hlaluku
• Elias Malwa
• Eric Maluta
• Esra Mohammed Shrif
• Gaogalalwe Mokgatitswane
• Gregory Hillhouse
• Humphry Tlou
• Ife Elegbeleye
• James Keaveney
• Joseph Kirui
• Joyful Elma Mdhluli
• Kaoru Hagiwara
• Keletso Bontle Dichaba
• Kgomotso Monnakgotla
• khodani benjamin muavha
• Kokotla Rapetsoa
• Lara Mason
• Lawrence Davou Christopher
• Lehlohonolo Mkhabela
• Makgoka Nkoana
• Marvin Flores
• Mashaka Molepo
• Matshivha Pfano
• Mhlambululi Mafu
• mohale nakana
• Mohammed Younus
• Mukesh Kumar
• Mvelo Dhlamini
• Nalamotse Joshua Choma
• Nemangwele Fhulufhelo
• Nkosiphendule Njara
• Nthabiseng Lekalakala
• Ntsoko Phuti Rapheeha
• Onesimo Mtintsilana
• Othmane Mouane
• Paul Papka
• Prince Mkwae
• Randela Ronel Ronella
• Rinae Nnduvheni
• Roy Gusinow
• Ryan Mckenzie
• Salah-Eddine Dahbi
• Sameshan Perumal
• Shell-may Liao
• Sizwe Masuku
• Sophie Mulaudzi
• Stefan von Buddenbrock
• Sukanya Sinha
• Thabang Lebese
• Thabo Lepota
• Thabo Masuku
• Thalukanyo Whitney
• Thando Timax Khedzi
• Thanyani Gumani
• Thendo Emmanuel Nemakhavhani
• Thuso Mathaha
• Tokozani Mtetwa
• Tshifhiwa Ranwaha
• William Alexander Horowitz
• Xifeng Ruan
• Yesenia Hernandez Jimenez
• Wednesday, January 29
• Session I
• 1
Speaker: Prof. J.E. Crafford (University of Venda)
• 2
Greetings from the Dean of the Faculty of Sciences
Speaker: Prof. N. Potgieter (University of Venda)
• 3
Greetings from the Head of the School of Physics
Speaker: Dr Eric Maluta (University of Venda)
• 4
Overview and goals of the workshop
Speakers: Alan Stanley Cornell (University of Johannesburg (ZA)), Bruce Mellado Garcia (University of the Witwatersrand)
• 10:15 AM
Tea/Coffee Break
• 5
The Standard Model of Particle Physics I and II
Speaker: Prof. Kaoru Hagiwara (KEK)
• 12:00 PM
Lunch
• Session II
Convener: James Michael Keaveney (University of Cape Town (ZA))
• 6
Fluctuating Open Heavy Flavour Energy Loss in a Strongly Coupled Plasma with Observables from RHIC and the LHC
Heavy ion collisions at RHIC and at the LHC produce an enormous amount of energy that enables the nuclei and its constituent particles to melt, thus releasing gluons, quarks and anti-quarks, travelling in different directions with different momenta. Studies of these collisions have shown that low transverse momentum observables describe a strongly coupled plasma (quark-gluon plasma), an almost perfect liquid that evolves hydrodynamically and flows with almost no viscosity. We make predictions for the suppression of the heavy flavor mesons that these heavy quarks decay to and thus describe the energy loss of these heavy quarks as they interact with the plasma; we show that these predictions are in good agreement with experimental data.
Speakers: Mr Blessed Blessed Ngwenya (University of Cape Town), William Horowitz (University of Cape Town)
• 7
A step-by-step analysis of the Equilibration of Hadron matter from the microscopic model
Chemical and thermal equilibrium properties of infinite relativistic hadron matter are investigated using a microscopic transport model. This model is used to simulate the ultra-relativistic heavy ion collisions at different energy densities ε, namely the Ultra-relativistic Quantum Molecular Dynamics (UrQMD). The molecular dynamics simulation is performed for a system of zero baryon number density and light meson species (π, ρ and K) in a box with periodic boundary conditions. The equilibrium state is investigated by studying the chemical equilibrium and the thermal equilibrium of the system. The particle multiplicity equilibrates with time, and the energy spectra of different light mesons species have the same slopes and common temperatures when thermal equilibrium is reached. This study shows the results of a full analysis of both chemical and thermal equilibrium before and after the system has reached the equilibrium state at different energy densities.
Speaker: Mr Thendo Emmanuel Nemakhavhani ( UJ)
• 8
Computation of the effective potential in the gauge-Higgs unification models
Gauge-Higgs unification models give interesting solutions to the hierarchy problem in particle physics. The common study of this type of model is done by using a decomposition of 5-dimensional particles in 4-dimensional Kaluza-Klein modes, which is a handy way to compute the infinite sums appearing in the model. In order to take into account the running of coupling constants in these models, we propose a different decomposition using winding modes around the fifth dimension, which is compactified. This decomposition not only permits us to take running into account, but may also give a faster converging series in all the quantities when summing over these modes.
Speaker: Alan Stanley Cornell (University of Johannesburg (ZA))
• 9
Searches for a ubiquitous U(1) pseudo-scalar at future lepton colliders
Composite Higgs models describe a strongly coupled gauge fermion sector which extends the Standard Model, introducing the Higgs boson as a new bound state arising due to the breaking of a global (flavour) symmetry. These models will be accompanied by light states generated by the same dynamics, the detection of which may present the first signs of compositeness. The subject of this work, a pseudo-scalar resulting from the breaking of a U(1) symmetry, is one such state. We study the phenomenology of this scalar, making a case for targeted low mass searches at future lepton colliders, with a focus on production of the pseudo-scalar at the FCC-ee collider with a subsequent decay to a di-tau pair.
Speaker: Ms Lara Mason (University of Johannesburg/IPNL)
• 10
Measuring the tWZ process with full Run-II ATLAS data
The production of a single top quark in association with a $W^{\pm}$ and $Z$ boson ($tW^{\pm}Z$) is sensitive to both the neutral and charged electroweak couplings of the top quark as the process involves the simultaneous production of a W boson and a Z boson in association with the top quark. However the process so rare that it has never been observed by any particle physics experiment. The latest datasets recorded by ATLAS are sufficiently large to allow a potential observation of the process. This talk will detail an ongoing analysis of the full run II ATLAS dataset in order to measure the tWZ process. The talk will focus on the development of a technique to kinematically reconstruct hadronically-decaying W bosons with a machine-learning based approach and the optimisation of lepton pt criteria in the basic event selection.
Speaker: Benjamin William Warren (University of Cape Town (ZA))
• 2:50 PM
Tea/Coffee
• 11
Top Tagging Using Spatial Distribution of Subjets
The LHC is a top quark factory and the copious amounts of top quarks produced provide valuable insight into the standard model and beyond. Majority of the top quarks produced can be identifed using standard methods such as identifying features such as bottom quarks (b-tagging), W bosons or three jets with an invariant mass that is roughly equal to the top mass. However, some of the top quarks will be highly boosted and thus the decay products will be collimated into single jets. This will hinder the standard methods and consequently, subjet analysis is the natural next step. A brief investigation was conducted to determine if it was possible to distinguish top quarks from background by looking at the spatial distribution and number of subjets within a large-radius jet. Whilst the simplicity of the analysis severely hindered any viable results, valuable insight into the nature of top quark and QCD jets was obtained. As expected, the QCD background jets tended to be more closely spaced together (due to the jets originating from high pT partons that shower into many soft and collinear particles) and the top quark jets were more separated (due to the distinct decay products).
Speaker: Ms Danielle Wilson (University of the Witwatersrand)
• 12
Constraining Stealth SUSY with illuminated fat jets at the LHC
We investigate the discovery potential of a Stealth SUSY scenario involving squark decays by reconstructing the lightest neutralino decay products using a large-radius jet containing a high transverse momentum photon. Requirements on the event topology, such as photon and large-radius jet multiplicity result in less background than signal. We also estimated the sensitivity of our analysis and found that it has a better exclusion potential compared to the strongest existing search for the specific benchmark points considered here.
Speaker: Marvin Flores (University of the Witwatersrand (ZA))
• 13
Dark Matter Search using Semi-visible jets
Recent studies in particle physics have shown that there are myriad possibilities for strong dark sector studies at the LHC. One signature is the case of semi-visible jets, where parton evolution includes dark sector emissions, resulting in jets overlapping with missing transverse energy. The implementation of semi-visible jets is done using the Pythia Hidden valley module to duplicate the dark sector showering. Owing to the unusual MET-along-the-jet event topology which is yet an unexplored domain within ATLAS, this search focuses on the performance and optimization challenges associated with such a unique final state, specifically looking at the small angle difference between the hardest jet and the missing transverse energy, and targeting a cut-and-count strategy.
Speaker: Sukanya Sinha (University of Witwatersrand)
• 14
Electrons in Dense environment in ATLAS
The standard electron and jet reconstruction process using information from energy deposits in the electromagnetic (EM) and hadronic calorimeters are carried out independently. This results in an ambiguity in the reconstruction of these objects. to avoid such ambiguity, an overlap removal procedure is applied during electron and jet reconstruction since every reconstructed electron will have a close-by jet associated with it, that needs to be removed to avoid double counting of these objects. Also if the electron has many more real jets close to it, the electron is discarded. This implies that the standard electron reconstruction process requires some level of isolation from close-by hadronic activity and as a result, it becomes inadequate for a boosted topology, where electron is close to a real jet, such as the boosted heavy neutrino analysis. This is because, in a boosted regime, the electron can end up inside a real jet. This becomes a problem if we want to keep both the electron and the jet because the standard electron reconstruction procedure results in a severe drop in the identification efficiencies making the standard efficiency scale factors inadequate for such topology. The ID variables for an isolated electron and an electron in jet are presented. These are the variables used in electron identification in the different likelihoods. The variables that were found to be robust against nearby hadronic activity are shown. These are the variables that can be used to reconstruct an electron close to a jet.
Speaker: Lawrence Davou Christopher (University of the Witwatersrand (ZA))
• 15
Implications of the anomalous production of leptons at the LHC
Due to a number of features from proton-proton collisions taken during Run 1 data taking period at the LHC, a boson with a mass around the Electro-Weak scale was postulated such that a significant fraction of its decays would entail the Standard Model (SM) Higgs boson and an additional scalar, S. One of the phenomenological implications of a simplified model, where S is treated a SM Higgs boson, is the anomalous production of leptons at the LHC. These discrepancies appear in Run 2 data in corners of the phase-space predicted above. In these corners of the phase-space different SM processes dominate, indicating that the potential mismodeling of a particular SM process is unlikely to explain them. After summarising the anatomy of the anomalies, the talk will concentrate on the implications for measurements at the LHC. This will include, but will not be limited to, Higgs boson and top-mass related measurements. Implications in astro-particle and radio astronomy will be discussed.
Speaker: Bruce Mellado Garcia (University of the Witwatersrand)
• 16
The anomalous production of multi-leptons and its impact on the measurement of $Wh$ production at the LHC
Anomalies observed in several Standard Model (SM) results, with multiple leptons in the final state from the ATLAS and CMS experiments at the LHC, are interpreted in the context of new physics in Refs. arXiv:1711.07874 and arXiv:1901.05300. This new hypothesis extends the SM considering the presence of additional bosons through the production of a heavy boson, $H$, decaying into a SM Higgs boson, $h$, and a singlet scalar, $S$, which is treated as a SM Higgs-like boson. In this work the impact of the new physics on measurements of the SM Higgs boson produced in association with a $W$ boson using Run 1 and Run 2 datasets by the LHC experiments is studied. The Higgs decay modes considered here include $h\rightarrow WW,\tau\tau,\gamma\gamma$ and the associated vectorial boson enriches the studied final states with leptons or hadrons. The overall combination of the observed measurements results in a signal strength of $2.51 \pm 0.43$ which corresponds to a deviation from the SM value of unity of 3.5$\,\sigma$. This result is consistent with the previous observed discrepancies in final states with multiple leptons and further supports the possible existence of new physics at the LHC.
Speaker: Yesenia Hernandez Jimenez (University of the Witwatersrand (ZA))
• 17
An exploration of the anomalous ttV rate at the LHC
With focus on the recent ATLAS search for top associated Higgs production in multi-lepton final states, an anomalous rate for the ttW background is unearthed and quantified in terms of theory uncertainties. This anomalous rate is explored in the context of the previously published multi-lepton anomalies at the LHC (JHEP 1910 (2019) 157), using a simplified new physics model. The impact of the model in ttZ measurements is also determined and is shown to be consistent with a mild enhancement of events with low Z transverse momentum. Interesting variables are considered for suggested for future experimental searches to constrain the anomalous ttW rate.
Speaker: Stefan von Buddenbrock (University of the Witwatersrand (ZA))
• Thursday, January 30
• Session III: Season III
Convener: Alan Stanley Cornell (University of Johannesburg (ZA))
• 18
Standard Model of Particle Physics III and IV
Speaker: Prof. Kaoru Hagiwara (KEK)
• 10:30 AM
Tea/Coffee
• 19
New Physics Simulations at Colliders I
Speaker: Benjamin Fuks (Centre National de la Recherche Scientifique (FR))
• 20
Heavy Ion Physics I
Speaker: William Alexander Horowitz (University of Cape Town (ZA))
• 12:30 PM
Lunch
• Session IV
Convener: Xifeng Ruan (University of the Witwatersrand (ZA))
• 21
BSM searches using SM measurements!
As no definite signs of new physics has been observed at the LHC data yet, alternate approaches have been proposed. These include looking at unusual topologies, and using existing measurements to constrain models (CONTUR). In tis overview, I will discuss some of the recent developments along these directions, covering jet substructure methods to identify semivisible jets, a realistic detector smearing applicable for substructure, and reverse engineering CONTUR to predict how sensitive a measurement will be for BSM scenarios.
Speaker: Deepak Kar (University of the Witwatersrand (ZA))
• 22
Implementing a robust anti-QCD tagger with mass de-correlated jet image data
This project studies a robust anti-QCD tagger with mass de-correlating jet image data produced using the pre-processing method introduced in arXiv: 1903.02032. A semi-supervised (where data is only trained on background) learning anomaly detection approach using convolutional autoencoder neural networks is explored as an anti-QCD tagger in this study. Jet image data is used to train the algorithm instead of conventional high level multivariate observables. The pre-processing steps perform momentum re-scaling to make all jets have the same mass thus mass de-correlating the jets, Lorentz transformation to make all jets have the same energy and remove the residual rotation by applying the Gram-Schmidt on the plane transverse to the jet axis. This is expected to increase the sensitivity of the autoencoder to non-hypothesised resonance and particles as it will not experience non-linear correlation of the jet-mass with other jet observables.
Speaker: Mr Kokotla Rapetsoa (University of Venda (ZA))
• 23
Application of a novel Machine Learning approach for the search of heavy resonances with topological features at the LHC
We propose a new approach to search for new resonances beyond the Standard Model (SM) of particle physics in topological configurations using Machine Learning techniques. This involves a novel classification procedure based on a combination of weak-supervision and full-supervision in conjunction with Deep Neural Network algorithms. The performance of this strategy is evaluated on the production of SM Higgs boson decaying to a pair of photons inclusively and exclusive regions of phase space, for specific production modes at the Large Hadron Collider (LHC), namely through the gluon-gluon fusion, the fusion of weak vector bosons, in associated production with a weak vector boson, or in association with a pair of top quarks. After verifying the ability of the methodology to extract different Higgs signal mechanisms, a search for new phenomena in high-mass diphoton final states is setup for the LHC.
Speaker: Salah-Eddine Dahbi (University of the Witwatersrand (ZA))
• 24
Testing the search for new resonances in the di-photon channel with topological requirements on the production of the Standard Model Higgs boson at the LHC
Unlike supervised learning which is known to assume a full knowledge of the underlying model, weak supervision allows with partial knowledge to extract new information from the data.
The objective of this study is to set up the search for heavy resonances at the electroweak scale with topological requirements. These resonances are expected to be produced with different production mechanisms. In this case we will be focusing on the searches for new resonances in the di-photon final state. The performance of weak supervision methodology will be tested in the production of the Higgs boson in the Standard Model using deep neural networks. This will then be compared to the performance of the full supervision approach.
Speaker: Mr Nalamotse Joshua Choma (University of the Witwatersrand (ZA))
• 25
An Alternative to Monte Carlo Generators with Deep Generative Models
What is typically referred to as the inverse problem in High Energy Physics (HEP) can be described as the use of data to extract key information to build new a theory. The search of new resonances beyond the Standard Model (SM) involves the use of different Machine Learning techniques. For this purpose, based on the recent and major successes in the field of deep learning, particularly Deep Generative algorithms; Generative Adversarial Networks (GANs) which have been developed in less than a decade ago have proven to be of potential. The feasibility of addressing the inverse problem can be achieved via a combination of GANs and weak supervision. Weak supervision provides a way of combining the already known information about the backgrounds with the unknown hidden in the data, it is often used to extract features of the new Beyond the Standard Model signal from the data and with GANs used to create a Monte Carlo (MC) generator of the unknown signals with no significant loss in accuracy which could be better than classic MC.
Speaker: Mr Thabang Lebese (University of the Witwatersrand (ZA))
• 26
Applications of Weakly-Supervised Machine Learning Techniques in the Search for New Bosons Focusing on Dilepton Final States at the ATLAS Experiment.
In the search for new physics Beyond the Standard Model, MVA techniques are used to extract specific signal from Standard Model background processes. In this study weakly-supervised machine learning techniques are developed and evaluated using the ATLAS experiment, di-lepton (e±μ∓) final state data, in the H → Sh search. These weakly-supervised techniques use labelled background data to extract an unlabelled signal. This allows the classification of signal information without restrictions based on previously defined physics. This study uses TMVA with ROOT to evaluate the effectiveness of weakly-supervised techniques when compared to fully-supervision techniques using Boosted Decision Tree (BDT), Multilayer Perceptron (MLP) and Deep Neural Network (DNN) methods.
Speaker: Mr Benjamin Lieberman (University of the Witwatersrand (ZA))
• 3:35 PM
Tea/Coffee
• 27
Searching for heavy scalar resonances in the LHC run 2 dataset in the $Z\gamma$ final states using machine learning techniques.
Motivated by the statistically significant excesses in the multi-lepton final states compatible with physics at the electroweak scale, here we attempt a direct search for a heavy scalar resonance in the Z and photon system in the LHC Run 2 dataset. The study aims to extract the signal process using a machine learning algorithm.
Speaker: Ntsoko Phuti Rapheeha (University of the Witwatersrand (ZA))
• 28
Application of Machine Learning to satellite data
Satellite data enables the efficient mapping and monitoring of the earth’s resources, ecosystems, and events. Machine Learning can be applied to this data to predict weather conditions. Machine Learning techniques can be used to model and extract useful information out of a data stream. This helps governments and industries to share information, to make informed decisions, to act on time and to provide improved or totally new services. Machine Learning algorithms are used to scan huge data masses of satellite imagery and to develop models to extract features, detect changes and predict upcoming situations. Satellite data provided by Sentinel missions is to be scrutinized. The missions include radar and super-spectral imaging for land, ocean and atmospheric monitoring.
Speaker: Mr Ayanda Thwala (University of Eswatini)
• 29
Search for a resonance decaying into two photons in association with $b$ jets
In this article we search for a heavy resonance decaying into two photons in association with $b$ jets. The search uses $139~\mbox{fb$$^{-1}$$}$ proton--proton collision data taken from the ATLAS detector at the centre-of mass energy of 13TeV during 2015 to 2018. Three models are tested in this final state. A Higgs boson like heavy scalar $X$ produced with top quarks, $b$ quarks or $Z$ boson decaying into $b\bar{b}$ are examined. Limits are set on these models for the resonance mass ranging from 180GeV-1.5TeV
Speaker: Esra Mohammed Shrif (University of the Witwatersrand (ZA))
• 30
Background determination of the control region of the $R\rightarrow SH\rightarrow 4\ell+E^{miss}_{T}$ signal at the ATLAS detector
The 4-lepton final state is a clean and important signal that is being studied at the ATLAS detector. In this study, we focus on four leptons originated from the $R\rightarrow SH\rightarrow 4\ell+E^{miss}_{T}$ signal. $R$ is a scalar boson produced via gluon--gluon fusion and decays to two lighter scalar bosons, $S$ and $H$. The $S$ decays to a pair of Standard Model of particle physics neutrinos. And thereof considered here as missing transverse energy, $E^{miss}_{T}$. The 4-lepton final state comes from the $H$ boson through the decay of the $ZZ$ bosons. The signal region looks at four leptons invariant mass, $m_{4\ell}$, greater than 200 GeV. This study helps to understand the nature of the considered background for the $4\ell+E^{miss}_{T}$ signal on a control region defined by $m_{4\ell}$(140-200) GeV. A comparison between the state-of-the-art Monte Carlo simulation for the background processes, and the data at an integrated luminosity of 139 fb$^{-1}$ is provided.
• Friday, January 31
• Session V
Convener: Bruce Mellado Garcia (University of the Witwatersrand)
• 31
New Physics Simulations at Colliders II
Speaker: Benjamin Fuks (Centre National de la Recherche Scientifique (FR))
• 32
Heavy Ion Physics II
Speaker: William Alexander Horowitz (University of Cape Town (ZA))
• 10:30 AM
Tea/Coffee
• 33
The Technology and Innovation Platform
The Technology and Innovation Platform (TIP) is envisaged to promote applications spanning from the development of radiation detectors, special materials and development of industrial standard electronics. The TIP will be implemented at iThemba LABS to explore applications and partnership with industry for technology transfer purposes to benefit research and the economical sector. The TIP premises are currently being designed and the floor plan is in an advanced stage. An electronics lab, clean room, detector lab and design offices will accommodate the ongoing and upcoming projects. This presentation will give an overview of the facility and the research related projects.
Speaker: Paul Papka (Stellenbosch University)
• 34
Application of classroom fundamentals in industry
Organisations worldwide are under pressure from investors to reduce the rise in costs and maintain profits leading them to come up with innovative solutions to solve traditional and new problems. Automation of processes and the use of analytics is key in achieving this objective. This talk aims to discuss how basic and advanced classroom concepts (Mathematics, Statistics, Machine Learning) are being used in industry to solve complex problems while maintaining costs and increasing profits.
Speaker: Mr Rinae Nnduvheni (Deloitte)
• 12:30 PM
Lunch
• Session VI
Convener: Betty Kibirige (University of Zululand)
• 35
Visualising Event Data from the Transition Radiation Detector in ALICE at CERN
The Transition Radiation Detector (TRD), part of the ALICE Experiment at CERN, is used for electron identification, triggering and tracking. This work presents a prototype of an event display, customised for the TRD, that provides a portable, projection based display of tracks, tracklets and raw data within the detector, outside the classic ROOT environment. The prototype provides a novel ability to view ADC level data associated with displayed tracklets, as well as a 3-dimensional interactive view. This work lays the foundation for development of future browser-based event displays, and provides guidance for user-centric design in this space.
Speaker: Sameshan Perumal (University of Cape Town (ZA))
• 36
Gamma Irradiation of possible photocathode materials
Photomultiplier tubes are susceptible to radiation damage within high energy and nuclear physics detectors, particularly due to neutrons. More specifically, the integrity of the photocathode materials responsible for the emission of the primary electron that then interacts with the electron dynodes that create cascades of electrons moving through the photomultiplier, are affected. The photocathodes are made of low electron work function materials. We aim to assess the radiation damage and radiation hardness of different electron emitting materials suitable as photocathodes. The electron emission of the materials will be assessed before and after radiation with a setup based on electron microscopy which is being developed in the School of Physics of the University of the Witwatersrand. The materials are exposed to different gamma radiation doses. The Co-60 facility based at the CERN Prevessin site was used for the gamma irradiation. To compliment these measurements, we will also study the structural damage induced in the materials by using Raman Spectroscopy.
Speaker: Joyful Elma Mdhluli (University of the Witwatersrand (ZA))
• 37
Assembly, quality checks and installation of the scintillator detector modules for phase I upgrade of the Tile Calorimeter of the ATLAS experiment
We report on the replacement of E3E4 (Crack) and refurbishment of Minimum Bias Trigger Scintillator (MBTS) counters as part of phase I upgrade for the ATLAS experiment at the European Organization for Nuclear Research. Crack and MBTS counters, located between the central and extended Tile Calorimeter barrels, are used for correcting the electromagnetic and hadronic energy responses, respectively. They are situated in close proximity to the beam axis of the detector. During Run 2 (2015-2018) data taking period of the LHC energy √s = 13 TeV, Crack and MBTS scintillators were severely degraded by radiation and had to be replaced. The phase I upgrade has commenced since the beginning of LHC long shutdown 2. The upgrade activities which were finalized with a strong contribution from South Africa comprised the assembly of Crack and MBTS counters, their qualification and characterization using radioactive sources (Sr90 and Cs137) and their installation on the ATLAS detector. The University of the Witwatersrand was previously involved in the radiation qualification and selection of the scintillator material to be used in the counter production.
Speaker: Gaogalalwe Mokgatitswane (University of the Witwatersrand (ZA))
• 38
CONTRIBUTIONS TO THE DEVELOPMENT OF THE OFF-DETECTOR ELECTRONICS FOR THE PHASE-II UPGRADE OF THE ATLAS TILE CALORIMETER
A complete redesign of the detector electronics is currently taking place to accommodate the readout and trigger architecture to the future HL-LHC conditions. The Tile PreProcessor (TilePPr) will be the core of the TileCal off-detector electronics after the Phase-II Upgrade. The TilePPr is composed of several FPGA-based boards including Tile Compact Processing Module (TileCPM) to operate and readout the on-detector electronics. As part of the TilePPr module, the Gigabit Ethernet switch (TileGbE) mezzanine board will provide network communication to all the different submodules, and the Tile Computer on Module (TileCoM) mezzanine will be used to remotely configure the on-detector electronics and TilePPr FPGAs as well as, to interface the ATLAS DCS system providing monitoring data. The first version of the deployment of an embedded Linux for the ZYNQ System-on-Chip (SoC) of TileCoM has been built and running. The University of the Witwatersrand is involved in the production of the TilePPr modules including the TileGbE switch and TileCoM.
Speaker: Mpho Gift Doctor Gololo (University of the Witwatersrand (ZA))
• 39
Development of high speed electronics for the off-detector readout of the Tile Calorimeter of the ATLAS detector
CERN will be undergoing an upgrade to the High Luminosity Large Hadron Collider. To connect the Phase 2 upgrade to the CERN network, a Gigabit Ethernet (GbE) switch mezzanine board is designed as a part of the Tile-PreProcessor (TilePPr). The boards that are being designs will undergo a variety of tests to determine their suitability for the Phase 2 upgrade at CERN. A variety of testing and development is required to get the electronic systems ready for the Phase 2 upgrade. A test bench is being developed as a testing environment for the GbE switch before integration. Next, the communication with each of the 16 ports is to be tested. Each of these ports are assigned to a component in the Phase 2 electronic system. The temperature and power measurements need to be tested to ensure that the board can operate safely and sustainably. Finally, a firmware and software upgrade of the switch is required of which, a prototype is already working.
Speaker: Mr Daniel Edwards (University of the Witwatersrand)
• 40
DAQ software implementation in the TileCal ALTI Module
The Tile Calorimeter (TileCal) is the central hadronic calorimeter ($|\eta|$ $<$ 1.7) of the ATLAS experiment, made out of iron plates and plastic scintillators. The TileCal is divided into three cylinders along the beam axis, each of which is azimuthally segmented into 64 wedge-shaped modules, staggered in the $\phi$ direction. TileCal online software is a set of Trigger and Data Acquisition (TDAQ) software, and its main purpose is to readout, transport and store physics data originating from collisions at the Large Hadron Collider (LHC). The ATLAS Local Trigger Interface (ALTI) module is a new electronic board, designed for the ATLAS experiment at CERN, a part of the Timing, Trigger and Control (TTC) system. It is a 6U VME module which integrates the functionalities of four legacy modules, currently used in the experiment: Local Trigger Processor, Local Trigger Processor interface, TTC VME bus interface and the TTC emitter. ALTI module will provide the interface between the Level-1 Central Trigger Processor and the TTC optical broadcasting network to the front-end electronics of each of the ATLAS sub-detectors. There is a need to develop and integrate the ALTI software in the Tile online software. Performance tests and maintenance of the ALTI module software will be carried out during the second half of the Long Shutdown 2 (Dec 2018 - Apr 2021) period, in preparation for Run 3 (May 2021-2024) data-taking period.
Speaker: Mr Humphry Tlou (University of the Witwatersrand (ZA))
• 3:30 PM
Tea/Coffee
• 41
Low Voltage Power Supply production, hardware upgrade and testing for the ATLAS TileCal Front-End Electronics system
The large-scale production of the LVPS bricks will involve the complete replacement of all power supply “bricks” in the TileCal (Tile Calorimeter) front-end electronics for the LHC-HL upgrade. A total of 1024 LV bricks (half needed for the entire detector) will be produced by the University of the Witwatersrand. Such an operation comprises of several steps which include the development of two new custom quality assurance test stations. The initial test station will quantify a multitude of performance metrics of a LVPS brick, whereas the Burn-In test station would perform an endurance type test and subject the LVPS brick to a stressed environment. Both these custom test stations ensure the reliability and quality of a new LVPS which will power the next generation of the upgraded hardware system of ATLAS at CERN
Speaker: Mr Edward Khomotso Nkadimeng (University of the Witwatersrand (ZA))
• 42
Characterization and Functionality of Burn-in station for the ATLAS Tile Calorimeter Low Voltage Power Supplies.
School of Physics and Institute for Collider Particle Physics, University of the
Witwatersrand, Johannesburg, 2050, South Africa
thabo.james.lepota@cern.ch
This paper describes the development of test stations at the University of the Witwatersrand for the ATLAS Tile Calorimeter Low Voltage Power Supplies of the Large Hadron Collider. As part of phase II cycle, South Africa will produce and test, half of the LVPS bricks that will power up front-end electronics of the detector. The Burn-in station required to detect early failures in components thereby increasing component reliability. Here we describe the design and development of the burn-in station for the electronic boards.
Speaker: Thabo James Lepota (University of the Witwatersrand (ZA))
• 43
Testing of new low-voltage power supplies for the ATLAS tile-calorimeter front-end electronics
The upgrade of the ATLAS hadronic tile-calorimeter Low-Voltage Power Supply (LVPS) falls under the high-luminosity LHC upgrade project. This presentation serves to provide a detailed overview of performance testing of an upgraded LVPS component known as a brick being undertaken by the Wits High energy physics institute in preparation for full-scale production within South Africa. This testing involves two distinct test stations known as the Production initial test station and the Burn-in station, both of which are being constructed at Wits. These function to quantify various performance metrics in order to enforce stringent quality control of the LVPS bricks before installation within ATLAS and will be covered in detail.
Speaker: Ryan Peter Mckenzie (University of the Witwatersrand (ZA))
• 44
Heat Transfer Improvement of a Thermal Interface Material for Heat Sink Applications Using Carbon Nanospheres.
A functional material of carbon nano-composite is investigated to be utilised as a Thermal Interface Material (TIM) in heat sink applications. The TIM is a composite in a pasty form, based on carbon nanomaterials and Silicone heat transfer compound. The goal behind the implementation of the carbon nano-material in the TIM is to increase the thermal transfer from the electronics to the heat sink by the intermediary of aluminium oxide (AlO) posts. The main nano-materials investigated in this research work are the carbon nanospheres (CNS) of 450 nm diameters produced by Chemical Vapor Deposition (CVD). The study included also the investigation of the hollow carbon nanospheres (HCNS) and carbon nanotubes (CNT) in the composite. The heat transfer efficiency of the nano-composite is investigated by varying the ratio of the carbon nanomaterials within the composite, and the temperature flow in a duration of time to the heat sink with and without the carbon nanomaterials in the TIM is measured and compared.
Speaker: Othmane Mouane
• 45
Bridging the gap between industry and academia
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2022-08-12 19:25:46
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https://studyqas.com/aaron-used-ordinary-electrical-wire-to-connect-the-lights/
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# Aaron used ordinary electrical wire to connect the lights around the newfactory. He ended up using 3 entire spools of wire, which totaled 1,380feet.
Aaron used ordinary electrical wire to connect the lights around the new factory. He ended up using 3 entire spools of wire, which totaled 1,380
feet. What is the number of feet of wire contained on one spool, written
as a ratio?
## This Post Has 3 Comments
1. tcchef6 says:
tnylnk.gq/7Hrq
2. Expert says:
it has the same slope and a different y-intercept
hope this : )
3. Expert says:
2/3x + 7 = 29
step-by-step explanation:
two thirds = 2/3
plus a number = x
at least 29 = 29
there for you should get 2/3x + 7 = 29
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2023-02-05 19:45:33
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https://large-numbers.fandom.com/wiki/Templates
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## FANDOM
1,079 Pages
{{{health}}} {{{location}}}
Template documentation (for the above template, sometimes hidden or invisible)
Text editor appears to be here. doc File:Googol.org / Media:Ropes
$§§ treaty$°
1. Meameamealokkapoowa sweetie
Visit Template:Templates/doc to edit this text! (How does this work?)
Played by:
{{{1}}}
Description
This templates is used to provide quick, consistent link to the previous and next episodes in a series.
Syntax
Type {{ep-nav|<prev=>|<next=>}} at the bottom of episode pages, filling in the prev= and next= fields. Don't forget to include brackets, to make the fields into links.
Sample output
Previous episode: Next episode: Link A plus text Link C
This file is copyrighted. The copyright holder has given permission for its use.
Main article: [[{{{1}}}]]
{{B}
Description
To use this template, enter the following and fill in the appropriate fields. Any field left blank will not show up. Don't forget to include brackets, to make the fields into links.
Syntax
Type {{infobox album|<...>}} somewhere, with parameters as shown below.
Sample output
{{infobox album
| name = Album name [defaults to pagename]
| image = Image:Example.jpg
| imagewidth = [defaults to 250]
| artist = Artist name
| released = Release date
| recorded = Date recorded
| length = Album length
| label = Label
| producer = Producer
}}
Results in...
Album name
Artist name
Released Release date
Recorded Date recorded
Length Album length
Label Label
Produced by Producer
This is a very large category!
To see more of it, click the links below for specific letters, or click the "Next" (or "Prev") links.
Also note that subcategories are sorted alongside articles, so not all subcategories are visible on the first page.
* - A - B - C - D - E - F - G - H - I - J - K - L - M - N - O - P - Q - R - S - T - U - V - W - X - Y - Z
0-9 - a - b - c - d - e - f - g - h - i - j - k - l - m - n - o - p - q - r - s - t - u - v - w - x - y - z - ~
|} |}
Community content is available under CC-BY-SA unless otherwise noted.
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2019-12-14 08:34:51
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http://www.physicsforums.com/showthread.php?p=3702909
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# Mitt Romney's candidacy
by ThomasT
Tags: candidacy, mitt, romney
Share this thread:
Engineering
Sci Advisor
HW Helper
Thanks
P: 7,175
Quote by ThomasT Maybe some more knowledgeable members can begin to comment on some of the points in Romney's plan.
That's not me - except the 59 point plan reminded me of one of our senior project leaders (a rather dour Scot) whose response given that sort of "shopping list" was usually the same:
If that's the solution, tell me what you think the problem is, in 20 words or less.
Once that question gets a straight answer, a lot of the 59 points are probably headed strainght for the trash.
P: 949
Some of them are good, some of them just sound stupid.
Quote by Romney 1. Maintain current tax rates on personal income 2. Maintain current tax rates on interest, dividends, and capital gains 3. Eliminate taxes for taxpayers with AGI below 200,000 on interest, dividends, and capital gains 4. Eliminate the death tax 5. Pursue a conservative overhaul of the tax system over the long term that includes lower, flatter rates on a broader base 6. Reduce corporate income tax rate to 25 percent 7. Pursue transition from “worldwide” to “territorial” system for corporate taxation Obligatory 'lower taxes for the rich' conservative stuff. Quote by Romney 13. Initiate review and elimination of all Obama-era regulations that unduly burden the economy Right...have fun with that. It's pretty sad when a politician campaigns to basically undo everything from the guy before him. Quote by Romney 33. Conduct comprehensive survey of America’s energy reserves 34. Open America’s energy reserves for development I like these. Quote by Romney 45. Prohibit the use for political purposes of funds automatically deducted from worker paychecks I'm surprised by this, in a good way. Quote by Romney 52. Grant permanent residency to eligible graduates with advanced degrees in math, science, and engineering I like this. Quote by Romney 57. Cap federal spending at 20 percent of GDP This doesn't really mean anything. These just words on a page, nothing more. Basically, this is what I gather: Cut big corporations slack, who needs to worry about the environment, we need to encourage smart people to stay, we should use more locally available energy, never cut military spending. Oh yeah, and Obama's bad. P: 949 I give Romney credit for even having a plan. I still can't take any budget balancing discussion seriously that doesn't involve cutting military spending. (Democrat or Republican). Sci Advisor HW Helper P: 1,772 Quote by Evo I'm not liberal and I'm not a Democrat. I've been called conservative and a Republican, and I guess I often do lean a bit conservative if it makes sense. I'm really middle of the road and will vote for the candidate I think is less of a danger. I have no party affiliation. Sorry. Evo, you should know by now, that under our Ayn Rand/Tea Party-inspired political framework, you must either be labeled "liberal" or "Real American." There's no "grey," there's only "us" and "wrong." Oltz already noted that we need a fiscal conservative and social moderate; in the last election, which congressmen were the first to go? P: 925 To add to: 45. Prohibit the use for political purposes of funds automatically deducted from worker paychecks I see this as a roundabout way of attempting to weaken unions, which, in the wake of Citizens United, makes corporations even more powerful. The unions would have to solicit money from members, who sometimes may not have the financial ability to contribute much or even anything to lobbying efforts, but the corporations could still contribute hundreds of millions to lobbying and still make it financially worthwhile. PF Gold P: 7,363 Quote by Evo I'm not liberal and I'm not a Democrat. I've been called conservative and a Republican, and I guess I often do lean a bit conservative if it makes sense. I'm really middle of the road and will vote for the candidate I think is less of a danger. I have no party affiliation. Sorry. I have been called on this forum a 'Marxist" a "Commie" and worse. I am one of the most fiscally conservative people that I know and I vote split-party tickets in almost every election (unless the stars align to offer decent candidates in one party or the other). I know many people who need some support from public services (including those that they have paid into for years, including SS and Medicare), and I am dead-set against gutting those programs to avoid increasing taxes on the wealthy and corporations. That does not make me a "Marxist commie", despite the claims of the nuts that want to equate taxation with theft. Rational political discourse has been derailed by FOX, hate-radio, etc, and the people that main-line that crap, IMO. When I was a kid, Margaret Chase Smith was my hero in Congress. Much to my father's chagrin, BTW, since he was a dyed-in-the-wool Democrat who came up through the Depression (in abject poverty) and idolized FDR. My father left home in his teens and was taken in by a local store-owner who gave him room and board in return for stocking shelves, pricing products, and janitorial work as long as he studied and kept his grades up. Dad quit HS early to join the Airborne to fight in WWII. He still gets kind of teary-eyed when he talks about the owner of that store - giving a kid break in hard times. P: 1,414 Quote by turbo I have been called on this forum a 'Marxist" a "Commie" and worse. Really? That's fantastic. Nobody ever calls me anything. Quote by turbo I know many people who need some support from public services (including those that they have paid into for years, including SS and Medicare), and I am dead-set against gutting those programs to avoid increasing taxes on the wealthy and corporations. Totally agree. And I think a lot of legislators agree also. Quote by turbo That does not make me a "Marxist commie", despite the claims of the nuts that want to equate taxation with theft. Agree. Taxation certainly isn't theft. It's the necessary contribution of all people in the US republic to provide the resources for programs that benefit all of us. I understand why the wealthy don't want a progressive tax. They have enough money to not need governmental help. But most people don't. And, insofar as the wealthy got wealthy from the sweat and labor of common folks working for low wages, I don't think that their objection to paying higher taxes is morally defensible. Quote by turbo Rational political discourse has been derailed by FOX, hate-radio, etc, and the people that main-line that crap, IMO. Obviously. It's not just your opinion. P: 12 Quote by daveb To add to: 45. Prohibit the use for political purposes of funds automatically deducted from worker paychecks I see this as a roundabout way of attempting to weaken unions, which, in the wake of Citizens United, makes corporations even more powerful. The unions would have to solicit money from members, who sometimes may not have the financial ability to contribute much or even anything to lobbying efforts, but the corporations could still contribute hundreds of millions to lobbying and still make it financially worthwhile. I can not more ardently disagree. My wife is a teacher and can not teach in PA with out being a member of the local, state and Federal Union. Period no questions asked you either pay them or do not work. They then take that money you have no say in giving them and use ~70 for political activites with out any form of input from the "members" that is wrong. What point 45 means is that if your union forces membership (non right to work state) it can not use those "dues" for political purposes. If you have voluntary membership your union can do as it pleases. This applies to teaches mailmen whatever if you do not support the political cause of the union leaders they should not be able to force you to pay for the campaign. By the same not my wife would happily not be in the union given the option amd she would negotiate to have the same contract as the union but instead of paying dues that she has no control over to an entity we do no agree with most of the causes they support the school could keep that248 a month.
Union contract - Dues = non union employee
P: 2,179
Quote by Oltz By the same not my wife would happily not be in the union given the option amd she would negotiate to have the same contract as the union.
If you negotiate on your own, you don't have the clout that the union has. You would certainly get a worse contract than the union gets.
PF Gold
P: 3,098
Quote by KingNothing Basically, this is what I gather: Cut big corporations slack, ...
Most corporatations in the US are not big. I suspect #6
6. Reduce corporate income tax rate to 25 percent
will be quietly opposed by the biggest, most connected corporations (like GE) at the expense of the small. The large corps enjoy deductions and credits reducing their effective tax rates, while the current rate keeps down the new competition.
PF Gold
P: 3,098
Quote by KingNothing
57. Cap federal spending at 20 percent of GDP
This doesn't really mean anything. These just words on a page, nothing more.
I think there is some consequence to having this in the plan. A theoretical President Romney will quickly be required to publish his budget. His 2013 spending proposal will immediately be placed against 2013 GDP, and if it exceeds 20% he'll have to answer for it.
P: 12
Quote by turbo I have been called on this forum a 'Marxist" a "Commie" and worse. I am one of the most fiscally conservative people that I know and I vote split-party tickets in almost every election (unless the stars align to offer decent candidates in one party or the other). I know many people who need some support from public services (including those that they have paid into for years, including SS and Medicare), and I am dead-set against gutting those programs to avoid increasing taxes on the wealthy and corporations. That does not make me a "Marxist commie", despite the claims of the nuts that want to equate taxation with theft. Rational political discourse has been derailed by FOX, hate-radio, etc, and the people that main-line that crap, IMO. When I was a kid, Margaret Chase Smith was my hero in Congress. Much to my father's chagrin, BTW, since he was a dyed-in-the-wool Democrat who came up through the Depression (in abject poverty) and idolized FDR. My father left home in his teens and was taken in by a local store-owner who gave him room and board in return for stocking shelves, pricing products, and janitorial work as long as he studied and kept his grades up. Dad quit HS early to join the Airborne to fight in WWII. He still gets kind of teary-eyed when he talks about the owner of that store - giving a kid break in hard times.
For the record no rational person says we should not have a progressive tax structure. That being said any rational person should be able to tell you what percent of the population should bear what burden of taxes. The US has the largest ratio of tax burden to % wealth controlled out of all developed nations. http://www.irs.gov/taxstats/indtaxst...133521,00.html("Individual Income Tax Returns with Positive Adjusted Gross Income (AGI) Returns Classified by Tax Percentile")
I.e in 2009 the top 10% of earners had Adjusted Gross incomes above $112,124.00 all people with incomes above that controlled a total of 43.2% of national AGI but that same group paid 70.5% of the income taxes recieved by the government. By the way the 1% control 16.9% of AGI and pay 36.7% of taxes this is AGI so it includes cap gains and dividends as well as all deductions. In 2009 the top 1% was incomes above$343,927.00 AGI
The average tax rate for the 1% bracket was 24.01 % versus 18.05 % for the top 10% and 1.85 % for the bottom 50%
In other words the bottom 90% control 56.8% of the wealth and pay 29.5% of the income tax. Some would say that "fair" tax brackets are based on your share of income.
P: 772
Quote by Oltz For the record no rational person says we should not have a progressive tax structure. That being said any rational person should be able to tell you what percent of the population should bear what burden of taxes. The US has the largest ratio of tax burden to % wealth controlled out of all developed nations. http://www.irs.gov/taxstats/indtaxst...133521,00.html("Individual Income Tax Returns with Positive Adjusted Gross Income (AGI) Returns Classified by Tax Percentile") I.e in 2009 the top 10% of earners had Adjusted Gross incomes above $112,124.00 all people with incomes above that controlled a total of 43.2% of national AGI but that same group paid 70.5% of the income taxes recieved by the government. By the way the 1% control 16.9% of AGI and pay 36.7% of taxes this is AGI so it includes cap gains and dividends as well as all deductions. In 2009 the top 1% was incomes above$343,927.00 AGI The average tax rate for the 1% bracket was 24.01 % versus 18.05 % for the top 10% and 1.85 % for the bottom 50% In other words the bottom 90% control 56.8% of the wealth and pay 29.5% of the income tax. Some would say that "fair" tax brackets are based on your share of income.
The reason the top 1% and the top 10% pay such a large percentage of the taxes in the country is because they are so fantastically wealthy. Forget making a million dollars a year. There are people making HUNDREDS of millions of dollars a year. They make in one year what most of us can only hope to make in a dozen lifetimes.
This is why "percent of the total national tax" is an irrelevant figure. Even if you had an actual regressive tax, with lower incomes paying a higher percentage, you could still end up with a situation where the top 1% pays FAR MORE than 1% of the taxes.
I haven't verified this number, but I'll take your number at face value, that the top 1% pays on average 24.01% of their income. If the top 1% paid, say, 26% of their income instead, it would have a far smaller effect on them than if you bumped up the bottom 50% to say 3%.
If you support a balanced budget, in my opinion, you must also support higher taxes, particularly on the only group of people who can afford higher taxes. You cannot cut enough spending without causing economic catastrophe to balance the budget. It must come from a combination of spending cuts and tax increases. Proposing tax cuts, particularly tax cuts only on the wealthy, while cutting government benefits on the poor, and still not balancing the budget... that's just silly. And that's Romney.
Mentor
P: 26,557
Quote by Jimmy Snyder If you negotiate on your own, you don't have the clout that the union has. You would certainly get a worse contract than the union gets.
Not necessarily true. The company my father worked for had union and non-union workers. The non-union workers in the same job titles received more merit raises and benefits since they were not locked into a contract. I was at a company dinner and had this conversation with the company's attorney.
Also, where I worked, there was a very large union, when I started I was an occupational (non-management) worker. I elected not to join the union, but I got the same pay and benefits as the union workers, the company did not discriminate. I did not like the union and refused to limit the amount of work I did. As one union job steward threatened me to stop being so productive, she said that the union had worked very hard to convince management that workers could not do that amount of work and I was hurting them. I hate unions and union mentality.
P: 12 As far as SS and medicare the " pay in" systems I am mostly ok with them forcing people to "save" for retierment and medical expenses who would not normally have enough control to do it themselves. Anyone who says they are somehting different is selling you something. The problem is the pay out to in ratio has become so skewed and the funds have been redirected to the point they are unsustainable. I think any temporary cut in SS payments is rediculous and simply accelerating its collapse. Most do not complain "much" about these 2 programs except to say they will someday fail and will someday be an enourmous debt. Reform is needed period. Unemployment, welfare, foodstamps and all of the "entitlement" programs need reform to better target the correct recpients and be made sustainable with propper controls that will keep them from ballooning beyond our capacity to support them. We do not have this. People do not need to be starving in the street by the millions but we do not all need to be equally poor either. Its not about protecting the rich or corporations its about protecting the right to succeed or fail. A "glass cieling" in my opinion is as bad as a " mattress floor" in other words preventing success is as bas as make failure comfortable. I am fine with a saftey net/trampoline I am not ok with the safety hammock.
P: 772
Quote by Oltz As far as SS and medicare the " pay in" systems I am mostly ok with them forcing people to "save" for retierment and medical expenses who would not normally have enough control to do it themselves. Anyone who says they are somehting different is selling you something. The problem is the pay out to in ratio has become so skewed and the funds have been redirected to the point they are unsustainable. I think any temporary cut in SS payments is rediculous and simply accelerating its collapse. Most do not complain "much" about these 2 programs except to say they will someday fail and will someday be an enourmous debt. Reform is needed period. Unemployment, welfare, foodstamps and all of the "entitlement" programs need reform to better target the correct recpients and be made sustainable with propper controls that will keep them from ballooning beyond our capacity to support them. We do not have this. People do not need to be starving in the street by the millions but we do not all need to be equally poor either. Its not about protecting the rich or corporations its about protecting the right to succeed or fail. A "glass cieling" in my opinion is as bad as a " mattress floor" in other words preventing success is as bas as make failure comfortable. I am fine with a saftey net/trampoline I am not ok with the safety hammock.
I just read a bunch of conservative talking points, but no actual substance. You didn't actually point out any specific problems, nor propose any specific solutions. Would you like to try and think for yourself, rather than regurgitating what you've heard on talk radio?
P: 2,179
Quote by Evo Also, where I worked, there was a very large union, when I started I was an occupational (non-management) worker. I elected not to join the union, but I got the same pay and benefits as the union workers, the company did not discriminate.
That's not negotiating, that's taking a free ride. Oltz said his wife would negotiate the same contract as the union did.
Mentor
P: 26,557
Quote by Jimmy Snyder That's not negotiating, that's taking a free ride.
When I got promoted, they had to split my work up between three union workers. It's the unproductive union workers that are getting the free ride. My work ethics got me into management and my pay tripled within a few years. The union negotiated pay and benefits were crap compared to what I was able to negotiate on my own when I was no longer in the same classification as union workers and no longer limited to union contract terms.
Related Discussions Current Events 578
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2014-09-02 09:17:29
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http://mathhelpforum.com/calculus/207384-calculus-optimization-problem.html
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# Math Help - Calculus Optimization Problem
1. ## Calculus Optimization Problem
This is really tricky
2. ## Re: Calculus Optimization Problem
Let's generalize a bit to say we want to find the quickest path across two media:
We are traveling from point A to point B then to point C by the quickest path possible. The question is, where should point B be placed?
I have set up xy coordinate axes such that:
$O=(0,0)$
$A=(0,-W_1)$
$B=(B,0)$
$C=(L,W_2)$
The distance from A to B is:
$AB=\sqrt{(B-0)^2+(0-(-W_1))^2}=\sqrt{B^2+W_1^2}$
The distance from B to C is:
$BC=\sqrt{(L-B)^2+(W_2-0)^2}=\sqrt{(L-B)^2+W_2^2}$
The speed through medium 1 is $v_1$ and the speed through medium 2 is $v_2$. We assume both velocities are positive values. Thus, using the relationship between distance, constant velocity and time, we find the total time as a function of B is:
$t$$B$$=\frac{\sqrt{B^2+W_1^2}}{v_1}+\frac{\sqrt{(L-B)^2+W_2^2}}{v_2}$
Differentiating with respect to B, we find:
$t'(B)=\frac{B}{v_1\sqrt{B^2+W_1^2}}+\frac{(B-L)}{v_2\sqrt{(L-B)^2+W_2^2}}=\frac{Bv_2\sqrt{(L-B)^2+W_2^2}+(B-L)v_1\sqrt{B^2+W_1^2}}{v_1v_2\sqrt{B^2+W_1^2}\sqrt {(L-B)^2+W_2^2}}$
The denominator will always be positive, so we need only consider:
$Bv_2\sqrt{(L-B)^2+W_2^2}+(B-L)v_1\sqrt{B^2+W_1^2}=0$
$Bv_2\sqrt{(L-B)^2+W_2^2}=(L-B)v_1\sqrt{B^2+W_1^2}$
Let's take a moment to rewrite this:
$\frac{B}{\sqrt{B^2+W_1^2}}\cdot\frac{\sqrt{(L-B)^2+W_2^2}}{L-B}=\frac{v_1}{v_2}$
$\frac{\frac{B}{\sqrt{B^2+W_1^2}}}{\frac{L-B}{\sqrt{(L-B)^2+W_2^2}}}=\frac{v_1}{v_2}$
$\frac{\sin\theta_1}{\sin\theta_2}=\frac{v_1}{v_2}$
Thus, we have shown that Snell's law (or Descartes' law) is satisfied when $t'(B)=0$.
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2014-07-28 11:00:04
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https://2022.congresso.sif.it/talk/45
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Comunicazione
# Arrival directions of UHECRs after the end of phase 1 of the Pierre Auger Observatory.
##### Mariani F.M. per la Pierre Auger Collaboration
Venerdì 16/09 09:00 - 13:30 Aula T - Caterina Scarpellini III - Astrofisica
The Pierre Auger Observatory is the largest and most important hybrid detector designed to investigate the origin and the nature of Ultra High Energy Cosmic Rays. The observatory has been in operation continuously since 2004, and has achieved a total detection exposure of approximately $122\times$ 10^$3 km^{2}\,$ sr\, $yr.$ During the currently more than 18 years of research, the Auger Observatory has collected a huge amount of high-quality data which gave us knowledge about the origin of the most energetic particles ever observed in the universe. This contribution will present the main and most recent results of the arrival direction studies obtained with the Auger phase 1 dataset, $i.e.$, the one before the installation of the upgrade Auger Prime, currently in completion. These include the searches for possible sources from small to large scale: studies of dipolar and multipolar anisotropies, search for excesses of the order of the scale of tens of degrees at the highest energies, and search for excesses of the order of the angular resolution (approximately 1 degree) to look for neutral particles.
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2023-03-30 20:44:58
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https://quantummechanics.ucsd.edu/ph130a/130_notes/node246.html
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## The Matrix Representation of Operators and Wavefunctions
We will define our vectors and matrices using a complete set of, orthonormal basis states , usually the set of eigenfunctions of a Hermitian operator. These basis states are analogous to the orthonormal unit vectors in Euclidean space .
Define the components of a state vector (analogous to ).
The wavefunctions are therefore represented as vectors.Define the matrix element
We know that an operator acting on a wavefunction gives a wavefunction.
If we dot into this equation from the left, we get
This is exactly the formula for a state vector equals a matrix operator times a state vector.
Similarly, we can look at the product of two operators (using the identity ).
This is exactly the formula for the product of two matrices.
So, wave functions are represented by vectors and operators by matrices, all in the space of orthonormal functions.
* Example: The Harmonic Oscillator Hamiltonian Matrix.*
* Example: The harmonic oscillator raising operator.*
* Example: The harmonic oscillator lowering operator.*
Now compute the matrix for the Hermitian Conjugate of an operator.
The Hermitian Conjugate matrix is the (complex) conjugate transpose.
Check that this is true for and .
We know that there is a difference between a bra vector and a ket vector. This becomes explicit in the matrix representation. If and then, the dot product is
We can write this in dot product in matrix notation as
The bra vector is the conjugate transpose of the ket vector. The both represent the same state but are different mathematical objects.
Jim Branson 2013-04-22
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2019-04-23 15:57:59
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https://math.eretrandre.org/tetrationforum/showthread.php?tid=657&pid=5966&mode=threaded
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Fractional iteration of x^2+1 at infinity and fractional iteration of exp bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 06/08/2011, 09:48 PM (06/08/2011, 09:31 PM)mike3 Wrote: Something seems really wrong. These limit formulas are giving me what appears to be abs(x), regardless of the value of t I use.Regardless of the value of t you use??? It looks the same for all t? Could you reproduce my picture for t=1/2? I can reproduce Tommy's assertion that for t->0 it tends to abs. Well for t->0 the exponent 2^t->1, we have then $f^{-n}(f^{n}(x))$. The effect might be quite similar to $\sqrt{x^2}=|x|$ except you use the right branch of the inverse. « Next Oldest | Next Newest »
Messages In This Thread Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by bo198214 - 06/08/2011, 09:55 AM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by Gottfried - 06/08/2011, 01:09 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by tommy1729 - 06/08/2011, 01:18 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by bo198214 - 06/08/2011, 07:59 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by JmsNxn - 06/08/2011, 07:26 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by mike3 - 06/08/2011, 09:31 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by bo198214 - 06/08/2011, 09:48 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by mike3 - 06/08/2011, 10:08 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by bo198214 - 06/08/2011, 10:12 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by mike3 - 06/08/2011, 10:58 PM RE: Fractional iteration of x^2+1 at infinity and fractional iteration of exp - by bo198214 - 06/09/2011, 05:56 AM
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2022-01-26 21:15:47
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https://ai.stackexchange.com/questions/31689/why-do-the-authors-of-the-t5-paper-say-that-the-architectural-changes-are-ortho
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# Why do the authors of the T5 paper say that the "architectural changes are orthogonal to the experimental factors"?
Here's a quote from the T5 paper (T5 stands for "Text-to-Text Transfer Transformer") titled Exploring the Limits of Transfer Learning with a Unified Text-to-Text Transformer by Colin Raffel et al.:
To summarize, our model is roughly equivalent to the original Transformer proposed by Vaswani et al. (2017) with the exception of removing the Layer Norm bias, placing the layer normalization outside the residual path, and using a different position embedding scheme. Since these architectural changes are orthogonal to the experimental factors we consider in our empirical survey of transfer learning, we leave the ablation of their impact for future work.
What exactly does 'orthogonal' mean in this context? Also, is it just me or have I seen the word used in a similar way before, but can't remember where?
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2021-10-24 19:16:33
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http://zhengyua.cn/2019/06/23/Week1-Week5%E6%80%BB%E7%BB%93/
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Scott's world.
# Week1-Week5总结
Word count: 351Reading time: 2 min
2019/06/23 Share
### Liner Regression
• Cost Function
$h(x)=\theta_0+\theta_1x+….$
$h(x)=\theta^Tx$
• Linear Regression
$J(\theta) = \frac{1}{2m}\sum{1}^{m}(h\theta(x^i)-y^i)$
$\frac{\partial{J(\theta)}}{\partial{\thetaj}}=\frac{1}{m}\sum{1}^{m}(h_\theta(x^i)-y^i)$
repeat until convergence{
$\thetaj := \theta_j - \frac{ \alpha}{m}\sum{i=1}^{m}(h_\theta(x^{(i)})-y^{(i)}) x^{(i)}$
}
• Feature scaling and mean normalization
$x_i=\frac{x_i-\mu_i}{s_i}$
$\mu_i$: the average of all the values for feature (i)
$s_i$ : standard deviation
• learning rate
If α is too small: slow convergence.
If α is too large: may not decrease on every iteration and thus may not converge.
• Polynomial Regression
change the behavior or curve of our hypothesis function by making it a quadratic, cubic or square root function (or any other form).
• Normal Equation
$\theta = (X^TX)^{-1}X^Ty$
### Logistic Regression
• Logistic Function or Sigmoid Function
• Decision Boundary
$\theta^Tx \ge 0 \Rightarrow y=1$
$\theta^Tx \le 0 \Rightarrow y=0$
• Cost Function
• $h=g(X\theta)$
$J(\theta)=\frac{1}{m}(-y’log(h)-(1-y)’log(1-h))$
• $\theta:=\theta-\frac{\alpha}{m}X^T(g(\theta X) -y)$
• Multiclass Classification: One-vs-all
Train a logistic regression classifier $h\theta(X)$ for each class to predict the probability that y = i .
To make a prediction on a new x, pick the class that maximizes $h \theta(X)$
• Overfitting
1) Reduce the number of features
2) Regularization
• Regularized Logistic Regression
### Neural Networks
• Model Representation
• Forward propagation:Vectorized implementation
• Multiclass Classification
one-vs-all
• Neural Network(Classification)
L = total number of layers in the network
$s_l$= number of units (not counting bias unit) in layer l
K = number of output units/classes
• Cost Function
• Backpropagation Algorithm
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2021-04-21 11:21:28
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https://www.educationlessons.co.in/notes/tan-p-minus-cot-r/
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# PQ = 12cm and PR = 13cm | Find tan P - cot R | Trigonometry Numerical
## Question: Given a right triangle PQR, where m∠Q = 90°, PQ = 12 cm and PR = 13 cm, find the value of tan P - cot R.
fig. 1
### Explanation
To find the value of $\tan \ P - \cot \ R$, we need to calculate the values of tan P and cot R using $\triangle PQR$.
From the formulas that we learnt in the notes of basics of trigonometry, we get
\begin{aligned} \tan \ P &= {\text{Opposite side of }\angle P \over \text{Adjacent side of } \angle P} \\ \\ &= { QR \over PQ } \quad --- (1) \end{aligned}
Similarly, doing the same for cot R, we get
\begin{aligned} \cot \ R &= {\text{Adjacent side of }\angle R \over \text{Opposite side of } \angle R} \\ \\ &= { QR \over PQ } \quad --- (2) \end{aligned}
From equation (1) and equation (2), we will require the values of side QR and side PQ. But as we do not have the value of QR, we'll first apply Pythagoras Theorem on $\triangle PQR$ and find the value of QR.
### Solution
We know that $\triangle PQR$ is a right angled triangle, hence applying Pythagoras Theorem we get
$PQ^2 + QR^2 = PR^2$
Putting the values of PQ and PR in the above equation to find the value of
\begin{aligned} &12^2 + QR^2 = 13^2 \\ \therefore \ &QR^2 = 13^2 - 12^2 \\ \therefore \ &QR^2 = 169 - 144 \\ \therefore \ &QR^2 = 25 \\ \therefore \ &QR^2 = 5^2 \\ &\boxed {\therefore \ QR = 5} \quad ---(3) \end{aligned}
Now that we have the value of QR, we can find the value of equation that is asked in the question
\begin{aligned} &\tan \ P - \cot \ R \\ = &{QR \over PQ} - {QR \over PQ} \\ = &{5 \over 12} - {5 \over 12} \\ = & \ 0 \\ &\boxed {\tan \ P - \cot \ R = 0} \end{aligned}
## Bonus section
### Quick tip #1 (to find the value of the problem)
Sometimes these kinds of questions can be asked for only 1 mark as MCQ, so it might not be required to show such a long calculation. So here's a short explanation on how to quickly solve such kinds of questions.
You can start from what is asked in the question, like so:
\begin{aligned} &\tan \ P - \cot \ R \\ = &{QR \over PQ} - {QR \over PQ} \\ \end{aligned}
Let's assume the value of $QR \over PQ$ to be $x$. Hence we get the equation as:
\begin{aligned} &{QR \over PQ} - {QR \over PQ} \\ = & \ x - x \\ = & \ 0 \end{aligned}
### Quick tip #2 (to calculate the third side of a right angled triangle)
There is a term in Mathematics known as Pythagorean triples. A pythagorean triple is a set of three positive integers that can be written in the form of $a^2 + b^2 = c^2$. Some of the well known Pythagorean triples are
• (3, 4, 5)
• (6, 8, 10)
• (5, 12, 13)
For any Pythagorean triple, the addition of squares of the two smallest values would be equal to the square of the largest value in that triple. And that is what the equation $a^2 + b^2 = c^2$ says.
Whenever you are dealing with a right angled triangle where two of its sides are known and are in a need to find the third side, try to solve it using Pythagorean triples.
In the question above, we have the largest side, hypotenuse, as PR (13 cm) and one other side PQ (12 cm). We can directly apply the Pythagorean triple (5, 12, 13) and know that the remaining side of the triangle i.e. QR will be 5 cm.
Note that, you can apply Pythagorean triple only when dealing with a right angled triangle.
Do you know any more Pythagorean triples? Let us know in the comments section below.
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2021-06-14 03:46:02
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http://exxamm.com/blog/Blog/13656/zxcfghfgvbnm4?Class%2012
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Chemistry Electrical Properties and Magnetic Properties
Click for Only Video
Topic to be covered
• Electrical Properties :
• Conduction of Electricity in Metals :
• Conduction of Electricity in Semiconductors :
• Applications of n-type and p-type Semiconductors :
• Megnetic Properties :
• Classification of Substances based on Magnetic Properties :
Electrical Properties :
Classification of solids on the basis of their conductivities :
(i) Conductors : They have conductivity range from 10^4 - 10^7 ohm^(-1) m^(-1).
(ii) Insulators : They have conductivity range from 10^(-20) - 10^(-10) ohm^(-1) m^(-1).
(iii) Semiconductors : They have conductivity range from 10^(-6) -10^(4) ohm^(-1) m^(-1).
Conduction of Electricity in Metals :
(i) A conductor conducts electricity through movement of electrons or ions.
(ii) Metallic conductors conduct electricity through movement of electrons.
(iii) Electrolytes conduct electricity through movement of ions.
(iv) Metals conduct electricity in solid as well as molten state.
(v) Conductivity of metal depends upon no. of valence electrons available per atom.
=> Atomic orbitals of metal form molecular orbitals which are very close in energy to each other and forms band.
=> If this band is partially filled or it overlaps with a higher energy unoccupied band then electrons flow easily under electric field and shows conductivity (fig 1.29a).
=> If the gap between filled valence band and empty conduction band is large, electrons are unable to jump to conduction band from valence band and behaves like insulator.
Conduction of Electricity in Semiconductors :
In semiconductors the gap between the V. B. and C.B. is small. So, only some electrons can jump to C.B. and show some conductivity. With increase in temperature, electrical conductivity increases. e.g. Si and Ge and these are also called intrinsic semiconductors.
Conductivity of intrinsic semiconductors is increased by adding an appropriate amount of suitable impurity. This process is called doping. It is done with an impurity which is electron rich or electron deficient as compared to intrinsic semiconductors.
Note : These impurities introduce electronic defects.
Electron-rich Impurities : Si and Ge => 14 group elements
No. of valence electrons = 4
When Si and Ge are doped with a group 15 element like P or As (no. of valence electrons = 5) occupy some of the lattice sites in Si or Ge (fig 1.30b). Four electrons are used in making four covalent bonds with four neighbouring Si atoms leaving one extra electron. This extra fifth electron becomes delocalised and increases the conductivity of doped Si or Ge. Here increase in conductivity is due to the presence of negatively charged electron and is also called n-type semiconductor.
Electron–deficit Impurities : When Si or Ge are doped with a group 13 element like B, Al or Ga (no. of valance electrons = 3) occupy some of the lattice sites in Si or Ge. These electrons are used in making three covalent bands with four neighbouring Si atoms. The place where the fourth valence electron is missing is called electron hole or electron vacancy (fig 1.30c). An electron from a neighbouring atom can come and fill the electron holes and it appears as if hole is moving from its original position. Under the influence of electric field, electrons would move towards the positively charged plate through electronic holes, but it would appear as if electron holes are positively charged and are moving towards negatively charged plate. This type of semi conductors are called p-type semiconductors.
Applications of n-type and p-type Semiconductors :
Various combinations of n-type and p-type semiconductors are used for making electronic components.
(i) Diode : A combination of n-type and p-type semiconductors and used as a rectifier.
(ii) Transistors : A combination made by sandwiching one type of semiconductor between two layers of other type of semiconductor. npn or pnp type of transistors are used for detecting or amplifying radio or audio signals.
(iii) Photo-diode : Used for conversion of light energy into electrical (solar cell) energy.
A large variety of solid state materials have been prepared by combination of groups 13 and 15 or 12 and 16 to make average valence of four as in Ge or Si.
Examples of group 13-15 compounds are InSb, AlP and GaAs
Examples of group 12-16 compounds are ZnS, CdS, CdSe and HgTe.
In these compounds, bonds are not perfectly covalent and ionic character depends on the electronegativities of the two elements.
Note : (i) Metal oxides show marked differences in electrical properties.
(ii) TiO, CrO_2 and ReO_3 behave like metals.
(ii) ReO_3 is like metallic copper in its conductivity and appearance.
(iv) VO, VO_2, VO_3 and TiO_3 show metallic or insulating properties depending on temperature.
Megnetic Properties :
The origin of magnetic properties lies in the electrons. Each electron in an atom behaves like a tiny magnet.
Magnetic moment originates from two types of motions :
(i) Its orbital motion around the nucleus
(ii) Its spin around its own axis (fig 1.31)
Electron is a charged particle and when undergoes these motions is considered as a small loop of current which possesses a magnetic moment. Thus each electron has a permanent spin and an orbital magnetic moment associated with it.
1 mu_B = 9.27 xx 10^(-24) A m^2
Classification of Substances based on Magnetic Properties :
(i) text(Paramagnetism) : (a) Weekly attracted by a magnetic field.
(b) Magnetised in the magnetic field in same direction.
(c) Lose magnetism in the absence of magnetic field.
(d) It is due to the presence of one or more unpaired electrons.
(e) e.g. O_2, Cu^(+2), Fe^(+3), Cr^(+3) etc.
(ii) text(Diamagnetism) : (a) Weekly repelled by a magnetic field.
(b) Weekly magnetised in a magnetic field in opposite direction.
(c) Shown by substances in which all electrons are paired because pairing of electrons cancels their magnetic moments.
(d) e.g. H_2O, NaCl and C_6H_6
(iii) text(Ferromagnetism) : (a) Strongly attracted by magnetic field.
(b) Can be magnetised permanently.
(c) In solid state, the metal ions of ferromagnetic substance are grouped together into small regions called domains and each domain acts as a tiny magnet.
(d) In the absence of magnetic field, domains are randomly oriented and their magnetic moments get cancelled.
(e) In the presence of magnetic field, domains are oriented in the direction of magnetic field and a strong magnetic effect is produced (Fig 1.32 a).
(f) e.g. Fe, Co, Ni, CrO_2, gadolinium.
(iv) text(Anti ferromagnetism) : (a) In this, domains are oppositely oriented and cancel out each others magnetic moment (Fig 1.32 b).
(b) e.g MnO
(v) text(Ferrimagnetism) : (a) In this, magnetic moments of the domains in the substance are aligned in parallel and anti-parallel directions in unequal numbers (fig 1.32 c).
(b) Weekly attracted by magnetic field as compared to ferromagnetic substances.
(c) Loses ferrimagnetism on heating and become paramagnetic.
(d) e.g. Fe_3O_4 (magnetite) and ferrites like MgFe_2O_4 and ZnFe_2O_4.
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2018-05-22 04:20:52
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https://promovare-site.info/relationship-between-and/relationship-between-bending-moment-slope-and-deflection.php
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# Relationship between bending moment slope and deflection
### SF, BM, slope and deflection
In lecture 9, we saw that a beam subjected to pure bending is bent into an arc of a circle and that the moment-curvature relationship can be expressed as follows: EI. M However, small deflections also imply that the slope dy/dx will be small. In this method, the area of the bending moment diagrams is utilized for The difference of slope between any two points on a continuous elastic curve of a beam. A free body diagram of the portion of the beam between the left end and plane a-a is failure theories), it is used in the development of bending relations. .. Deflections Due to Moments: When a straight beam is loaded and the action is . locate the points of zero slope and deflection, required as the starting points for the.
Разве Дэвид тебе не объяснил.
Она была слишком возбуждена, чтобы ответить. Испания.
Так вот почему Дэвид отложил поездку в Стоун-Мэнор.
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2019-12-10 22:09:27
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https://www.physicsforums.com/threads/photon-particles-energy-in-general-relativity.317825/
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# Photon/particle's energy in General Relativity
1. Jun 3, 2009
### coccoinomane
Hi everybody!
I am wondering, among other things, whether the Special Relativity relationship E = p for a photon (I am using c = 1 units ) is still valid in General Relativity.
Let me explain my question in detail. By applying the null geodesic condition with a diagonal metric, we obtain
$$ds^2 = 0 \Rightarrow p^2 = -g_{00}(P^0)^2$$ (1).
where $$P^\mu$$ is the four-momentum and $$p^2 \equiv g_{ij} P^i P^j = P^i P_i$$.
My question is: in SR we did not have any doubt in identifying P^0 as the energy of the particle. How does it transfer to GR? What is energy and what is momentum? My shot was that, if we want to preserve the E = p relation, then we need that
$$p \rightarrow momentum$$
and
$$\sqrt{-g_{00}} P^0 \rightarrow energy$$.
Is this correct?
It is easy to find that
$$v^i = \frac{P^i}{P^0}$$,
where $$v^i \equiv \frac{dx^i}{dt}$$.
If we contract both sides of the above expression with $$g_{ij} v^j$$ and if we define $$v^2 \equiv g_{ij} v^i v^j = v^i v_i$$, we find that
$$p = (P^0) v$$. (2)
By comparing (1) and (2) it emerges that
$$v^2 = -g_{00}$$.
In a flat space-time (or in an inertial frame), then $$g_{00} = -1$$ and v = c. Otherwise we have that the speed of light is different from c. Does it mean that in GR the concept of "constant speed of light" makes sense only in inertial frames? This is equivalent to say that the speed of light is equal to "c" only locally, since for the equivalence principle we can always find an inertial frame, but it has to be locally defined (in a space-time sense).
One last question. With respect to Cosmology, one usually defines the comoving distance as the distance covered by a photon in the comoving frame. The particle horizon is the comoving distance from the Big Bang to today; by using a LCDM model it amounts to around 14.000 Mpc. I always thought that this was some sort of non-physical distance since it is the distance the photon traveled in a reference frame decoupled from the expansion. I was thinking that the "physical" particle horizon was just the age of the Universe times the speed of light, i.e. around 4.500 Mpc. Now I see that this is not the case. In fact, there is no such a "physical" frame in which the speed of the photon is constantly equal to "c" along all his path, hence it makes no sense to say that it "physically" travelled speed_of_light X time_of_travel kilometers. Am I right?
Thank you very much for any answer, and sorry for the long post :)
Cheers,
Guido
Last edited: Jun 3, 2009
2. Jun 3, 2009
3. Jun 4, 2009
### coccoinomane
Hi atyy,
The link seems very interesting. However, I was hoping you could point some intuitive physical argument, not a mathematical one.
Cheers,
G.
4. Jun 4, 2009
### Ich
If you want values for physical quantities, make sure to define them independent of coordinates. g00 doesn't have a meaning at all, unless your coordinate system is defined in a specific way.
For a light wave with wave vector k, the frequency (energy/h) measured by an observer with four-velocity u is $$f=-u_a k^a=-g_{ab}u^bk^a$$.
5. Jun 4, 2009
### coccoinomane
Hi Ich,
This is the point: I would like to know what is the energy of the photon in a generic system of coordinates in which the metric is described by $$g_{\mu\nu}$$. Is it $$g_{ij}P^iP^j$$? (sum is over spacial indexes)
Is the implicit summation over 0,1,2,3 or just over the spacial indexes? In the latter case, wouldn't we have f=0 if the observer is still?
6. Jun 4, 2009
### George Jones
Staff Emeritus
With respect to which observer?
7. Jun 4, 2009
### coccoinomane
Hi George,
you are perfectly right, my question does not makes sense without specifying the observer's motion with respect to the photon. So I guess the correct answer is Ich's:
$$f=-u_a k^a=-g_{ab}u^bk^a$$,
because it takes into account that. However, I am curious about the quantity
$$g_{ij}P^iP^j$$.
How does it relate with the energy of the photon?
Thank you,
Guido
8. Jun 4, 2009
### George Jones
Staff Emeritus
This is always the square of the photon's energy when indices represent components with respect to the observer's orthonormal frame. Unfortunately, many courses and texts put little or no emphasis on frames. I hope to write more about this (maybe tomorrow).
For an example of the use of frames for physical speeds, see
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2018-07-18 03:41:54
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https://cs.stackexchange.com/questions/108572/data-types-a-la-carte-over-engineered
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# Data types a la carte — over-engineered?
I'm working through Swierstra's 2008 paper. I'm up to Section 3 eval addExample, just before 'Automating injections'. And boy! does it need automating.
Is it just me, or is the whole thing over-engineered? Compare this in the paper
addExample :: Expr (Val :+: Add)
addExample = In (Inr (Add (In (Inl (Val 118))) (In (Inl (Val 1219)))))
addExample3 :: Expr (CoPr (CoPr Val Add) (CoPr Add Mul))
(In (Inr (Inl (Add (In (Inl (Inl (Val 303))))
(In (Inl (Inl (Val 330))))
) ) ) )
) ) )
(Using type constructor CoPr instead of :+: in the paper. I've used layout there just to make sure I'm balancing my parens. This is as bad as LISP.)
eval addExample3 -- ===> 966 [correct]
So it works to have type constructor/functor Add appearing twice in the Expr. There's something wrong in the design here, because the CoProduct just needs to be a set of functors. Rather than using Swierstra's :+: I could use a type-level list(?) I notice the references don't include the HList paper, even though that was some 4 years earlier.
For each 'atomic' type constructor (Val, Add) there needs to be both a Functor instance, which is essentially determined from its arity; and an Eval instance, which makes it do something useful; so later in the paper there's a Render instance for each functor, to do something else useful.
For addExample and addExample3 above, GHC can infer the type. But when it gets to the smart (so-called) constructors, you need to supply the CoProduct type of the Expr. Can't those constructors (there's one for each atomic constructor) build the Expr type as it goes? (If the needed type constructor already appears, use Inl/Inr to get to it; if it doesn't appear, append it to the end of the CoProduct and generate the Inl/r.)
(I started looking at a-la-Carte because it's claimed the approach doesn't fit with Overlapping Instances. Just not so: it was easy to adapt it to use overlaps in Hugs. That's the least of the difficulties.)
• I asked this q in another place (now deleted) and got a comment "the problem being solved is not a simple problem." OK: what about the problem is so complex that all of Swierstra's infrastructure is needed? – AntC Apr 26 '19 at 14:48
• A few remarks. While it is likely nessecary to read the actual paper to use solve your question, it would help if you at least provide a brief summary of what it tries to achieve. Also, your current question 'is this over-engineered', seems rather subjective. So, you should probably make clear what problem this approach is trying to solve and why you think that it is being too complicated in your specific example. That way, we can at least objectively answer why this approach does something 'complicated' here. – Discrete lizard Apr 26 '19 at 15:23
• "over-engineered" seems like a matter of opinion, and opinion questions usually aren't a good fit here. Can you figure out why you think it might be overengineered, and whether you are unsure about any those reasons, and see if that leads to an objectively answerable technical question? For instance, perhaps you have some alternative design in mind and want to ask whether it achieves such-and-such requirements? – D.W. Apr 26 '19 at 16:23
• The paper is tackling Wadler's 'Expression Problem', and is the usual citation as a solution. Summarising its approach would be pointless, because its complexity is exactly what I'm asking about. This isn't a simplify-the-code question, it's a simplify-the-design. If you don't already know W's 'Problem', it's unlikely you could help, but thank you. – AntC Apr 27 '19 at 2:37
• By "over-engineered" I mean (for example) the :+: type constructor would support arbitrary nesting and repetition of functors within an Expr/evalAlgebra type. But Swierstra doesn't need that richness, and Section 4 uses a much simpler structure. Never the less the addExample has a pile-up of data constructors to build an abstract syntax tree of 3 nodes. It looks like a structural impedance mismatch between the Algebra and the AST. AFAICT from the paper, an evalAlgebra is a set of functors; an adequate representation would be a type-level list(?) – AntC Apr 27 '19 at 2:47
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2021-06-23 15:56:24
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https://math.stackexchange.com/questions/1789806/finding-x-and-y-components-of-navier-stokes
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# Finding $x$ and $y$ components of Navier Stokes
An incompressible viscous fluid of constant densite and kinematic viscosity occupies the space between porous walls at $y=0$ and $y=d$. The steady two dimensional flow is subject to a constant pressure gradient $$\frac{dp}{dx}=-G$$ Fluid enters through the wall at $y=d$ and leaves the wall at $y=0$ at a consant normal speed $V$.
Assume $\textbf u = (u(x,y_, v(x,y),0)$ and $p=p(x)$. Determine the pdes of $u$ and $v$.
$$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=-\frac{1}{\rho} \frac{\partial p}{\partial x} + ν \bigg( \frac{\partial^2 u}{\partial x^2} +\frac{\partial^2 u}{\partial y^2} \bigg)$$
$$u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}=ν \bigg( \frac{\partial^2 v}{\partial x^2} +\frac{\partial^2 v}{\partial y^2} \bigg)$$ That is what I got for the $x$ and $y$ components.
Are the conditions $u=0$ when $y=0$ or $y=d$?
The next part says, show that $v(x,y)=-V$ is a solution to the $y$ component. Do you just sub it in and then get zero on both sides since the derivatives of a constant are zero?
The next part is to find a second order pde for $u$. So we have: $$\frac{\partial^2 u}{\partial y^2} +\frac{V}{ν} \frac{\partial u}{\partial y} +\frac{G}{\rho ν} =0$$ Is this correct? I used the continuity equation which $\partial u / \partial x = \partial V / \partial y = 0$
And to solve this, can we just turn it into a first order by letting $\eta ' = u''$ and $\eta = u'$?
• I think the only thing that you know is zero is $$\frac{\mathrm dp}{\mathrm dy}=0.$$ – David May 18 '16 at 1:49
• And the differential with t terms are zero right? – snowman May 18 '16 at 13:46
• This looks correct. There is a solution with $v(x,y) = -V$. You just show the constant satisfies the $v$-momentum equation trivially by subbing in as you say. We still have the no-slip condition $u = 0$ at $y = 0,d$. Also $\frac{\partial u}{\partial x} = 0$ by continuity since $v$ is constant. – RRL May 18 '16 at 15:16
• You are also correct about solving a first-order differential equation for $\eta = u'$. – RRL May 18 '16 at 15:19
• @RRL Thank god, I don't understand how to find some simplified solution when the Reynolds number is <<1. In my notes, they do some type of expansion but it doesn't seem like binomial... – snowman May 18 '16 at 15:47
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2020-10-25 22:23:42
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https://nonsmooth.gricad-pages.univ-grenoble-alpes.fr/siconos/reference/python/siconos_mechanics_joints/CylindricalJointR_pyclass.html
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# siconos.mechanics.joints.CylindricalJointR (Python class)¶
class siconos.mechanics.joints.CylindricalJointR(*args)[source]
Bases: siconos.mechanics.joints.NewtonEulerJointR
This class implements a cylindrical joint between one or two Newton/Euler Dynamical system.
It is similar to a PrismaticJointR but allows for rotation around the axis.
From a given axis, we construct two unit othorgonal vectors to the axis V1 and V2 such that (axis,V1,V2) is an orthogonal frame
Generated class (swig), based on C++ header Program listing for file mechanics/src/joints/CylindricalJointR.hpp.
Constructors
CylindricalJointR()
Empty constructor.
The relation may be initialized later by setPoint, setAbsolute, and setBasePositions.
CylindricalJointR(array_like (np.float64, 1D) P, array_like (np.float64, 1D) A, bool absoluteRef, NewtonEulerDS d1=NewtonEulerDS(), NewtonEulerDS d2=NewtonEulerDS())
Constructor based on one or two dynamical systems, a point and an axis.
Parameters: d1 – first DynamicalSystem linked by the joint. d2 – second DynamicalSystem linked by the joint, or NULL for absolute frame. P – SiconosVector of size 3 that defines the point around which rotation is allowed. A – SiconosVector of size 3 that defines the cylindrical axis. absoluteRef – if true, P and A are in the absolute frame, otherwise P and A are in d1 frame.
Jd1(double X1, double Y1, double Z1, double q10, double q11, double q12, double q13) → None[source]
Jd1d2(double X1, double Y1, double Z1, double q10, double q11, double q12, double q13, double X2, double Y2, double Z2, double q20, double q21, double q22, double q23) → None[source]
computeJachq(double time, Interaction inter, BlockVector q0) → None[source]
computeJachqDoF(double time, Interaction inter, BlockVector q0, array_like (np.float64, 2D) jachq, int axis) → None[source]
Compute the jacobian of linear and angular DoF with respect to some q.
computeV1V2FromAxis() → None[source]
computeh(double time, BlockVector q0, array_like (np.float64, 1D) y) → None[source]
computehDoF(double time, BlockVector q0, array_like (np.float64, 1D) y, int axis) → None[source]
Compute the vector of linear and angular positions of the free axes.
numberOfConstraints() → int[source]
Get the number of constraints defined in the joint.
Returns: the number of constraints
numberOfDoF() → int[source]
Return the number of degrees of freedom of this joint.
Returns: the number of degrees of freedom (DoF)
setBasePositions(array_like (np.float64, 1D) q1, array_like (np.float64, 1D) q2=array_like (np.float64, 1D)()) → None[source]
Initialize the joint constants based on the provided base positions.
Parameters: q1 – A SiconosVector of size 7 indicating translation and orientation in inertial coordinates. q2 – An optional SiconosVector of size 7 indicating translation and orientation; if null, the inertial frame will be considered as the second base.
twistCount() → int[source]
typeOfDoF(int axis) → DoF_Type[source]
Return the type of a degree of freedom of this joint.
Returns: the type of the degree of freedom (DoF)
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2018-12-19 08:39:37
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