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# Demand deposits higher required reserves higher • 10 • 80% (5) 4 out of 5 people found this document helpful This preview shows page 8 - 10 out of 10 pages. ##### We have textbook solutions for you! The document you are viewing contains questions related to this textbook. The document you are viewing contains questions related to this textbook. Chapter 24 / Exercise 35 Exploring Economics Sexton Expert Verified Demand deposits higherRequired reserves higherExcess reserves lowerd.Demand deposits higherRequired reserves higherExcess reserves lower ##### We have textbook solutions for you! The document you are viewing contains questions related to this textbook. The document you are viewing contains questions related to this textbook. Chapter 24 / Exercise 35 Exploring Economics Sexton Expert Verified e.Demand deposits higherRequired reserves higherExcess reserves lowerThese three questions illustrate that from the viewpoint of the banking system, it does not matter if the banks acquire debt issued by firms, governments, or households. To acquire the debt, the banks must have excess reserves. After they have used their excess reserves, the money supply is expanded, and the excess reserves become required reserves.f.Demand deposits no changeRequired reserves no changeExcess reserves no changeUnlike in the previous questions, the lending of excess reserves from one bank to another does not in the aggregate increase or decrease the reserves of the banking system.g.Demand deposits no changeRequired reserves no changeExcess reserves no changeLoans between members of the non-bank general public do not affect banks' reserves and thus do not affect their capacity to lend.h.Demand deposits lowerRequired reserves lowerExcess reserves higherWhile the creation of new loans uses the banks' excess reserves and creates new money, the retiring of loans from commercial banks reduces demand deposits and restores excess reserves (i.e., increases excess reserves).i.Demand deposits higherRequired reserves higherExcess reserves higherj.Demand deposits no changeRequired reserves no changeExcess reserves lowerk.Demand deposits no changeRequired reserves higherExcess reserves lowerQuestions j and k illustrate two major monetary tools, the reserve requirement and the discount rate. Notice that changing the discount rate and the reserve requirements do not in themselves change demand deposits. Their impact is on reserves, and the effect of this impact may lead to a change in the supply of money.l.Demand deposits higherRequired reserves higher Excess reserves lowerm.Demand deposits no changeRequired reserves no changeExcess reserves no changen.Demand deposits increaseRequired reserves increaseExcess reserves increaseDuring a period of inflation, a policy that contracts the money supply and the capacity of banks to lendis desirable. The opposite situation would apply during a recession. If there were a deficit during a period of recession, it is desirable to increase the money supply and the capacity of the banks to lend. Hence n is better than m. PTS:1	{}
Try out our new practice tests completely free! # Nature of Mathematics Mathematics Bookmark ## Quiz 17 : The Nature of Voting and Apportionment Use Jefferson's plan. Which state does violate the quota rule? Number of seats: 200 Free Multiple Choice C Apportion the indicated number of representatives to two states, A, and B, using Hamilton's plan. Next, recalculate the apportionment using Hamilton's plan for the three states, C and the original states. Decide whether the new states paradox occurs. Free Multiple Choice B Free None of the plans Use Jefferson's plan. Which state does violate the quota rule? Number of seats: 200 __________ (A, B, C, D, none) Use Hamilton's plan to apportion the new seats to the existing states. Then increase the number of seats by one and decide whether the Alabama paradox occurs. Assume that the populations are in thousands. Number of seats: 86 Multiple Choice An elderly rancher died and left her estate to her three children. She bequeathed her 17 prize horses in the following manner: 1/2 to the eldest, 1/3 to the second child, and 1/9 to the youngest. The children decided to call in a very wise judge to help in the distribution of the rancher's estate. They informed the judge that the 17 horses were not of equal value. The children agreed on a ranking of the 17 horses (#1 being the best and #17 being a real dog of a horse). They asked the judge to divide the estate fairly so that each child would receive not only the correct number of horses but horses whose average rank would also be the same. For example, if a child received horses 1 and 17, the number of horses is two and the average value is . How did the judge apportion the horses? Multiple Choice Use Jefferson's plan. Which state does violate the quota rule? Number of seats: 200 __________ (A, B, C, D, E, none) A fair apportionment of dividing a leftover piece of cake between two children is to let child #1 cut the cake into two pieces and then to let child #2 pick which piece he or she wants. Consider the following apportionment of dividing the leftover piece of cake among three children. Let the first child cut the cake into two pieces. Then the second child is permitted to cut one of those pieces into two parts. Child #3 can select any of the pieces, followed by child #1 selecting one of the remaining pieces, followed by child #2 who gets the remaining piece. Is this allocation process fair if each child's goal is to maximize the size of his or her own piece of cake? Multiple Choice An elderly rancher died and left her estate to her three children. She bequeathed her 17 prize horses in the following manner: 1/2 to the eldest, 1/3 to the second child, and 1/9 to the youngest. The children decided to call in a very wise judge to help in the distribution of the rancher's estate. They informed the judge that the 17 horses were not of equal value. The children agreed on a ranking of the 17 horses (#1 being the best and #17 being a real dog of a horse). They asked the judge to divide the estate fairly so that each child would receive not only the correct number of horses but horses whose average rank would also be the same. For example, if a child received horses 1 and 17, the number of horses is two and the average value is . How did the judge apportion the horses? Essay An elderly rancher died and left her estate to her three children. She bequeathed her 71 prize horses in the following manner: 1/2 to the eldest, 1/3 to the second child, and 1/9 to the youngest. The children decided to call in a very wise judge to help in the distribution of the rancher's estate. The judge arrived with a horse of his own. He put his horse in with the 71 belonging to the estate, and then told each child to pick from among the 72 in the proportions stipulated by the will (but be careful, he warned, not to pick his horse). The first child took thirty six horses, the second child took twenty four, and the third child, eight. The 71 horses were thus divided among the children. The wise judge took his horse from the corral, took a fair sum for his services, and rode off into the sunset. The youngest son complained that the oldest son received 36 horses (but was entitled to only 71/2 = 35.5 horses). The judge was asked about this, and he faxed the children the following message: "You all received more than you deserved. The eldest received 1/2 of an 'extra' horse, the middle child received 1/3 more, and the youngest, 1/9 of a horse 'extra.'" Apportion the horses according to Adams', Jefferson's, and Webster's plans. Which plan gives the appropriate distribution of horses? Multiple Choice Apportion the indicated number of representatives to three states, A, B, and C, using Hamilton's plan. Next, use the revised populations to reapportion the representatives. Decide whether the population paradox occurs. Multiple Choice Use Jefferson's plan. Which state does violate the quota rule? Number of seats: 200 Multiple Choice Apportion the indicated number of representatives to two states, A, and B, using Hamilton's plan. Next, recalculate the apportionment using Hamilton's plan for the three states, C and the original states. Decide whether the new states paradox occurs. Multiple Choice Round the given modified quota 3)57 By comparing it first with the arithmetic mean, and then with the geometric mean of the lower and upper quotas. Multiple Choice Apportion the indicated number of representatives to two states, A, and B, using Hamilton's plan. Next, recalculate the apportionment using Hamilton's plan for the three states, C and the original states. Decide whether the new states paradox occurs. __________ (A illustrates the new states paradox.; B illustrates the new states paradox.; C illustrates the new states paradox.; A and B illustrate the new states paradox.; The paradox does not occur.) Apportion the indicated number of representatives to two states, A, and B, using Hamilton's plan. Next, recalculate the apportionment using Hamilton's plan for the three states, C and the original states. Decide whether the new states paradox occurs. __________ (A illustrates the new states paradox.; B illustrates the new states paradox.; C illustrates the new states paradox.; A and B illustrate the new states paradox.; The paradox does not occur.) Suppose the annual salaries of three people are: What are their salaries if they are given a 4% raise, and then the result is rounded to the nearest $1,000 using Hamilton's plan with a cap on the total salaries of$113,000?	{}
# Thread: Linear functions and annihilators 1. ## Linear functions and annihilators Let V be 'n' dim vector space f1, f2, f3,..,fm be linear function on V (i.e. belong to dual space of V, V* = Hom(V,F)) W1, W2,....Wm be null-spaces of f1, f2,....f3 respectively. W = Intersection of (W1,W2,..,Wm) Consider 1. Annihilator of (W) i.e. {f \| f belongs to V* and f(W) = 0} 2. F = sub-space of V* spanned by {f1,f2,f3....fm} I want to show that Annihilator of (W) = F Please help. Thanks 2. It is easy to show F is a subset of Annihilator of (W). I am struggling to show the other way round i.e. if x in Annihilator of (W), then x in F i.e. x is spanned by {f1,f2,f3....fm}	{}
• October 06, 2015, 07:35:49 PM ### Author Topic: "I'm never shopping THERE again!" Share your story! (Read 1559109 times) 3 Members and 4 Guests are viewing this topic. #### Reika • Hero Member • Posts: 3419 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3345 on: December 07, 2012, 04:45:18 PM » That's crazy, Reika. No wonder you're frustrated!! My DD's Nook Color charging cable broke after just 4 months and of course it's just slightly different from allll the many other micro USB chargers we have. The replacements are $20 and the reviews are full of people complaining how easily they break. Meanwhile my Kindle charger is still like new after 3 years and even if it broke all the standard chargers fit it just fine. Ten points for Kindle! Have a Kindle 3G, I love it and hope it lasts my lifetime, including the original charger which just says "bring it!" I would have gotten a Fire, but for some reason I didn't like it anywhere near as much as the 3G, so wanted to get something different. I did like my cheapo tablet, but the charger thing has driven me insane for almost a month now. A coworker even brought in his box'o'chargers. That's where I found some random ones that fit, were the same power output and it didn't power up. Whenever I hook the blasted thing up to a USB cable of some sort (it has both regular and micro), and it turned on and off reproachfully that it didn't have power so couldn't connect to the omputer. While I had it hooked up to a usb charger in the darn wall outlet. So I basically have a plastic brick, but one I don't dare get rid of because it does have some of my info on it and with my luck whatever scavenger found it would have a charger that would work. However Newegg is my best friend and I got a replacement that was superior, but the same price thanks to the Holiday Whacky Season specials. And free shipping. Unfortunately, it's being shipped by one of the banes of my existence: UPS. #### Virg • Super Hero! • Posts: 5905 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3346 on: December 07, 2012, 04:54:21 PM » Reika wrote: "So I basically have a plastic brick, but one I don't dare get rid of because it does have some of my info on it and with my luck whatever scavenger found it would have a charger that would work." If you're planning on discarding it, just destroy it before you do and your data is safe. It won't give up your secrets if you crush it with a hammer or drvie your car over it. Virg #### Julian • I lost it between Thriller and Gangnam Style... • Hero Member • Posts: 1049 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3347 on: December 07, 2012, 04:57:56 PM » Reika wrote: "So I basically have a plastic brick, but one I don't dare get rid of because it does have some of my info on it and with my luck whatever scavenger found it would have a charger that would work." If you're planning on discarding it, just destroy it before you do and your data is safe. It won't give up your secrets if you crush it with a hammer or drvie your car over it. Virg That sounds very therapeutic, Virg! Been tempted to do that with several electronic gizmos over the years. #### Diane AKA Traska • Super Hero! • Posts: 5284 • Or you can just call me Diane. (NE USA EHellion) ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3348 on: December 07, 2012, 05:44:44 PM » Look for free microwaves on Craigslist. Film and Youtube. Location: Philadelphia, PA #### jedikaiti • Swiss Army Nerd • Hero Member • Posts: 3333 • A pie in the hand is worth two in the mail. ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3349 on: December 07, 2012, 06:07:58 PM » Reika wrote: "So I basically have a plastic brick, but one I don't dare get rid of because it does have some of my info on it and with my luck whatever scavenger found it would have a charger that would work." If you're planning on discarding it, just destroy it before you do and your data is safe. It won't give up your secrets if you crush it with a hammer or drvie your car over it. Virg That does not work for iPads though - I know someone whose iPad was run over by a large pickup truck. You could see that it was slightly flatter where it got run over, and the display was broken, but it still worked. What part of v_e = \sqrt{\frac{2GM}{r}} don't you understand? It's only rocket science! "The problem with re-examining your brilliant ideas is that more often than not, you discover they are the intellectual equivalent of saying, 'Hold my beer and watch this!'" - Cindy Couture #### Slartibartfast • Super Hero! • Posts: 12275 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3350 on: December 07, 2012, 06:10:29 PM » Reika wrote: "So I basically have a plastic brick, but one I don't dare get rid of because it does have some of my info on it and with my luck whatever scavenger found it would have a charger that would work." If you're planning on discarding it, just destroy it before you do and your data is safe. It won't give up your secrets if you crush it with a hammer or drvie your car over it. Virg That does not work for iPads though - I know someone whose iPad was run over by a large pickup truck. You could see that it was slightly flatter where it got run over, and the display was broken, but it still worked. Babybartfast jumped on ours (shoes on, both feet, direct hit) and it didn't phase it one bit. She got in some royal trouble for it, of course, but it's nice to know that the iPad can stand up to some abuse! #### Dark Annie • Member • Posts: 156 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3351 on: December 07, 2012, 06:35:56 PM » A friend of mine had great fun going in to buy a car carrying a briefcase. She negotiated her deal on the car, and then when they tried to set up financing, she told them she wanted to pay for the car up front. In cash. Actual cash. The briefcase was full of it. She told us later she had been tempted to point out that they were "non-consecutive bills" but thought that might be pushing it. They did not know what to do, and ended up calling in several extra car salesman to help count the money (she'd had it organized, of course, but they had to verify). My dad did that once. They had to give him$40.57 in change--but they didn't have any change, because no one ever buys a car for cash. They had to ask him if he would take a check for what they owed him. The salesman was pretty darned surprised at getting cash. I think he was also worried about whether he could get it all to the bank before it close Paying for a car completely in cash can also be the sign of criminal activity and may raise suspicions. Before my beloved dad passed away, he worked as a stock checker and auditor for a luxury car dealership in an expensive part of inner city Sydney. This dealership was a highly exclusive dealership- no Toyotas, Suzukis or Holdens, Ashton Martins, Bentleys, Rolls Royces, Jaguars. In fact, a certain ex-WAG had her very infamous car purchased from this dealership and was a frequent customer. I digress, but the point is, this was not a dealership that sold cheap cars. In fact, many of them cost over the million dollar mark (Dad to Me: "Kat, you know you're leaning on $1.5 million?") Anyway, one day whilst my dad was doing his rounds a young Asian man (yes his race is important) and his girlfriend came in and wanted to buy a Bentley SUV (don't quote me), which was around the$200 000. This is all well and good, nothing wrong here, except that the man wanted to pay completely in cash. Oh, and he had a $20 000 cash deposit with in in the sports bag he was holding. In addition, because the man was wearing a singlet/wife beater, his arms and back were fairly exposed, all of which were completely tattooed in well known Hong Kong/Chinese gang symbols, most prominent of which was a huge dragon. Naturally, the dealership declined the would be customers' sale, informing him that legally they could not accept that amount of cash. They then the gang squad at the local police station, who confirmed the dealerships suspicions after they described the tattoos. This wasn't the first time it happened either, nor was it limited to just Asian gangs- I believed that the Legitimate Businessman's Social Club was also a big fan of luxury cars. But it makes sense- if you've got a heap of cash from illegal activities, one of the best ways to 'legitimise' this money is by buying cars, especially at the higher end dealerships where pleasing the customer is paramount. That being said, it is mandatory for any large vash purchases to be reported to police, or they run the risk of being found complicit in money laundering and organised crime. That's why a lot of dealerships don't keep money or make change. Just thought I'd show the other side of the cash for cars debate! #### Elfmama • Super Hero! • Posts: 6926 • Derailing threads since 2001! ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3352 on: December 07, 2012, 07:35:35 PM » Quote a certain ex-WAG WAG? In my vernacular, WAG stands for Wild-Arsed Guess, which judging from context is not the intended meaning. ~~~~~~~~~~~~~~~~~~~~~~~~ Common sense is not a gift, but a curse. Because then you have to deal with all the people who don't have it. ~~~~~~~~~~~~~~~~~~~~~~~~ #### laud_shy_girl • Member • Posts: 482 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3353 on: December 07, 2012, 07:49:18 PM » Quote a certain ex-WAG WAG? In my vernacular, WAG stands for Wild-Arsed Guess, which judging from context is not the intended meaning. If I may... It's used in football circles for the Wives and Girldfriends. My one bit of useless trivia for the day. “For too long, we've assumed that there is a single template for human nature, which is why we diagnose most deviations as disorders. But the reality is that there are many different kinds of minds. And that's a very good thing.” - Jonah Lehrer #### Reika • Hero Member • Posts: 3419 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3354 on: December 07, 2012, 08:51:13 PM » Look for free microwaves on Craigslist. Film and Youtube. That's disturbingly tempting. Especially filming it with my new tablet. Which by the by, works quite happily with my universal adapters and charges off my computer. #### Redsoil • Hero Member • Posts: 2230 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3355 on: December 08, 2012, 01:08:07 AM » Prestigeway, or Sydney City Prestige, Dark Annie? Look out... It's one of the Aussie Contingent! #### Dark Annie • Member • Posts: 156 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3356 on: December 08, 2012, 04:43:27 AM » Prestigeway, or Sydney City Prestige, Dark Annie? Neither #### weeblewobble • Hero Member • Posts: 3844 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3357 on: December 08, 2012, 09:41:57 AM » Ooooh, I just thought of a good one! As part of a sales agency/distribution company we go to a lot of industry conferences. This particular conference we had our own booth for the first time (we are mainly west coast and this was east coast). We had arranged with our trucking company to deliver a pallet of supplies - brochures, samples, etc. - the morning the conference center was open. There are always long lines for the delivery trucks as all the vendors have to get their stuff delivered and set up in one day. The only way to be finished setting up the booth before nightfall is to get the truck in the line early. The company was contracted to be get to the center around 8 AM and all of our material would be available in a reasonable amount of time. So the day the set-up starts we get to the conference center around 9 AM. We set up what we can (rented couches, display cases) and check with the receiving people if our pallet has been checked in yet. After getting a negative we decide to run to Walmart to pick up some last minutes supplies (glass cleaner, small potted plants, energy bars, etc) fully expecting to find our pallet waiting for us when we got back. After lunch we went to receiving and found out that the truck had never been in the line to begin with. My boss called our office (which, due to the time difference, was just starting their day) and spoke to our purchaser. The purchaser (hereafter P) called up the company to find out what was going on. Well. The person P talked to said that the trucker decided he didn't want to wait in line all that time (which was in the contract) and would deliver the pallet later. After 5 PM in fact. Unfortunately, the conference receiving center closed at 4:30. All our marketing materials were on this pallet. After P pulled up the contract and read them the riot act they grudgingly agreed to get to the conference center as soon as they could (it was about 2 PM). We waited for another hour for the truck to appear. My boss was trying to haunt the receiving area without annoying people so around 3:30 she noticed a single box leaning against one of the doorways with her name on it. She asked the receiving people and they said a man had walked up to the doorway without checking in and just left the box there. Huh - said my boss - but where's the rest of the pallet? No pallet, just the box. Now this particular box was long, thin, and heavy - which is why it had been placed in the middle of the pallet and the whole thing cocooned in shrink-wrap. My boss called P who hit the roof; P called the company again and was informed that their records showed that we had only shipped the box. Unfortunately for them, we had had the weight recorded, the number of packages on the pallet, a picture, and the signature of the delivery company when they picked it up. The company back-pedaled after realizing that we kept good records and started following up to find out what had happened. Meanwhile OUR office was scouring their building for left-over material and overnighting it to us. Back at the conference center it was after 6:30 PM when we gave up on the truck. We brainstormed and managed to put up simple displays featuring what we had been able to fit in our luggage. Since we were marketing "natural" material we decided to tell people that we were collecting emails instead of handing out brochures in order to save trees. Epilogue: the rest of our pallet mysteriously appeared 2 hours after the conference started, the shrink wrap torn up and boxes beat to heck. It was unceremoniously dumped at our booth and left for us to deal with. We had to unpack the boxes, move, and clean up the debris while customers swirled around us. So that's one company that we will never do business with again! Agh, my office had similar issues, but with the hotel/conference center. I worked for an organization that hosted education conferences. Part of the conference set-up was an "expo room" where vendors could set up booths. At our largest conference, we could have 50-60 vendors, so we chose large conference centers used to handling this sort of traffic. We switched to a new conference center one year because they offered huge discounts if we agreed to hold several meetings at their location. Big mistake. We sent our booths and marketing materials well ahead of time so they would be waiting for us when we arrived at the hotel. We spoke with the conference center manager several times, emphasizing how important it was that we be able to reach the materials easily so we could set up before the vendors and participants arrived. We had it written into the contract that our materials would be stored in a separate area adjacent to where our temporary office space for the conference was located. We get there and our materials were nowhere to be found. We had the shipping manifest and the delivery confirmation from the shipping company, which includes a scan of the conference center manager's signature. So we knew it was there. Finally, the manager admitted that our materials arrived early and were shoved in the back of the main storage room. As more and more materials arrived from the vendors, they were put in the storage room, between our materials and the door, making it more and more difficult to reach our stuff. Manager was not apologetic at all and said, "Just wait until the other vendors arrive and get their stuff, then you can get yours." My boss explained this wasn't acceptable, we need to set up before the vendors arrived. Manager shrugged and said, "Well, there's nothing I can do." Boss disagreed and told him he was going to have to pull all of the other vendors materials out of the room so we could access ours. He said he couldn't authorize the man-power needed to do that. She said "OK, either you get us our materials right now, or I will consider you in breach of contract, and we will not be obligated to hold other meetings at your conference center. You can explain to your boss how you lost a (big dollar sign) account because you didn't want to rectify a mistake that YOU made." Suddenly, the man-power to clear the storage room appeared and we were able to set up as planned. We stayed with that hotel through the end of the contract, but it was a very frosty relationship between us and management. We were offered the chance to renew the contract. Boss made it clear why we wouldn't. #### Elfmama • Super Hero! • Posts: 6926 • Derailing threads since 2001! ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3358 on: December 08, 2012, 02:01:58 PM » The Maryland Knights of Columbus (a Catholic men's fraternal order) stopped giving their annual BIG EVENT to a particular hotel in Frederick. BE has several hundred people at this event, often close to a thousand, and usually fills the hosting hotel, to the point that in the past there has been a need to have blocks of a SECOND hotel reserved. The rooms ringing the indoor pool/atrium are reserved as hospitality rooms, each one designated for a particular council or assembly. They have drinks and nibbles in them all day except for the banquet time. And again, the banquet is several hundred people paying about$40-$50 each for a lettuce salad, rubber chicken, raw green beans, and over-salted potatoes, all so heavily seasoned with pepper as to be inedible. (Except for renegade pagans like me! I go to a restaurant and get exactly the meal I want, and it's edible.) So you know that the hotel rakes in the dough for this. Except the hotel people evidently thought it was not sufficient $$, and the last year the KofC was there, the hotel booked ANOTHER event in the same atrium space for the same Saturday night. I think it was a class reunion. The hotel staff didn't want to get more tables out for it, so they came around and tried to kidnap tables that we were already using! The reunion largely featured booze flowing like rivers, leading to drunken off-key caterwauling at the karaoke contest. Did I mention that the amplifiers were turned up AS LOUD AS THEY COULD GO? When they weren't "singing" there was a DJ playing something he probably thought was music. The bass BOOM BOOM BOOM could be heard (and felt! ) all over the hotel. Complaints to the front desk about the tablenapping and the "music" were greeted with shrugs. The other party had paid for the space so they could do whatever they liked. And what they liked was to continue the party until 2AM. The next day there was a line wrapping well into the hallway of people demanding the night's room charges be taken off their bill, because they couldn't sleep with all the noise. DH was there too. The loud noises had triggered a severe migraine for me and of course I was unable to sleep it off until they quit; I was still groggy from the vicodin. (Thank the gods for narcotic painkillers!) Management was overwhelmed. He tried to say "Well, give us your email address and we'll--" I don't know what he was going to do, because half the people there hollered "WE DON'T HAVE EMAIL!" (The KofC runs largely to elderly men; most of their members are retired, and a significant percentage of those are in their 70's and 80's.) I think it took letters to corporate before people got piddly refunds and abject apologies. Still not enough for the KofC, so now another hotel in another city gets to host Big Event, and the Holiday Inn in Frederick gets to do something rude and anatomically impossible. « Last Edit: December 08, 2012, 08:45:50 PM by Elfmama » ~~~~~~~~~~~~~~~~~~~~~~~~ Common sense is not a gift, but a curse. Because then you have to deal with all the people who don't have it. ~~~~~~~~~~~~~~~~~~~~~~~~ #### Soprych • Member • Posts: 135 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3359 on: December 08, 2012, 04:55:39 PM » If you know the charging requirements (usually it will be something like "Output: 5V ---- 1A"), then you can search for the exact power requirements. As long as the charger you buy has the proper connection, and has the exact power output you need, you'll be set. Or a lower output. We're having this problem with our Kindle Fire--Amazon's power adapters never last long and we need yet another new one. Our Fire is still under warranty so it should be free, but since they introduced the second generation Fire they don't make those chargers anymore--the second generation Fire comes with a mini USB cord and you have to buy a wall charger separately, which is "currently unavailable" from Amazon. They tried giving us a second generation Fire but I hated it, plus I thought it was stupid to replace a$200 tablet instead of a \$20 charger. So they're sending a second refurbished charger (even though they assured me that they're all quality tested, the first refurbished one was so loose it literally fell out of the Fire) and I have to use my cell phone charger for now, which is half as powerful. It works, just much much more slowly. Not really a "never shopping there again" offense (yet), but I probably couldn't say that about Amazon anyway since we order so much stuff from them. I find that my true Blackberry charger does charge my Kindle Fire as quickly as a Kindle Fire Charger. It does have to be a Blackberry OEM charger or it does take twice as long	{}
### Divisibility & Induction Suppose $x$, $y$, $z$ are positive integers satisfying the equation $\frac{1}{x} - \frac{1}{y} = \frac{1}{z},$ and let $h$ be the highest common factor of $x$, $y$, $z$. Prove that $hxyz$ is a perfect square. Prove also that $h(y-x)$ is a perfect square. Source: British Mathematical Olympiad, Paper 2, 1998, Question 3.	{}
1. Regression Let's say that I have some points... How do I then use a) linear regression to fit a line b)quadratic regression to fit a parabola I have a TI-84+ 2. Originally Posted by VonNemo19 Let's say that I have some points... How do I then use a) linear regression to fit a line b)quadratic regression to fit a parabola I have a TI-84+ Hi VonNemo19, 3. That's what I was after!	{}
The isoperimetric problem with a double density Giorgio Saracco (Università di Trento) created by pinamonti on 09 Apr 2021 modified by saracco on 12 Apr 2021 13 apr 2021 -- 14:30 [open in google calendar] Zoom seminar Please write to Andrea.pinamonti@unitn.it or to Andrea.marchese@unitn.it if you want to attend the seminar. Abstract. It is well-known that for any given volume, the sets that enclose said volume with the least perimeter are balls. What happens when one in place of the standard Euclidean volume and perimeter considers weighted counterparts? Given densities $f: \mathbb{R}^N \to \mathbb{R}^+$ and $h:\mathbb{R}^N \times \mathbb{S}^{N-1} \to \mathbb{R}^+$ to weigh, resp., the volume and the perimeter, we shall discuss under which hypotheses isoperimetric sets exist for all volumes. Furthermore, we shall introduce the $\varepsilon-\varepsilon^\beta$ property, which readily allows to prove boundedness. If time allows, some regularity results shall be discussed. Based on joint works with A. Pratelli (Università di Pisa). Credits \| Cookie policy \| HTML 5 \| CSS 2.1	{}
# Compound interest derivation of $e$ I'm reviewing stats and probability, including Poisson processes, and I came across: $$e=\displaystyle \lim_{n\rightarrow \infty} \left(1+\frac{1}{n}\right)^n$$ I'd like to understand this more fully, but so far I'm struggling. I guess what I'm trying to understand is how you prove that it converges. Can anyone point me toward (or provide) a good explanation of this? - Sorry for my lack of latex markup skills. The link points to a wikipedia image of this limit properly notated. – ivan Feb 16 '13 at 17:10 There is more than one way to prove this statement. What exactly are you struggling with? – Ludolila Feb 16 '13 at 17:18 This is a common definition of e. What definition are you using? – PyRulez Feb 16 '13 at 17:20 Historically this is the definition of the number $e$. One can show that the sequence $$\left( 1+\frac{1}{n}\right)^n$$ is increasing and bounded, and thus convergent. We define the limit to be $e$ and then it follows from this limit that $(e^x)'=e^x$. Here is the proof of it: Let $a_n= \left( 1+\frac{1}{n}\right)^n$ and $b_n=\left( 1+\frac{1}{n}\right)^{n+1}$. Then clearly $a_n \leq b_n$. Then $$\frac{a_{n+1}}{a_{n}}=\frac{\left( 1+\frac{1}{n+1}\right)^{n+1}}{\left( 1+\frac{1}{n}\right)^{n}}=\frac{n^n(n+2)^{n+1}}{(n+1)^{2n+1}}=\left( \frac{n(n+2)}{(n+1)^{2}}\right)^{n+1}\frac{n+1}{n}$$ By Bernoulli inequality $$\left(1-\frac{1}{(n+1)^2} \right)^{n+1}\frac{n+1}{n} \geq \left(1-\frac{n+1}{(n+1)^2}\right)\frac{n+1}{n} = 1$$ This shows that $a_n$ is increasing. Similarly $b_n$ is decreasing. Thus $a_n$ is increasing and bounded by $b_1$, and hence convergent. Now I will show that its limit is exactly $e$. Let $l$ be its limit. Then $$\ln(l) = \lim_n \frac{\ln(1+\frac{1}{n})-\ln 1}{\frac{1}{n}}=\ln'(1)=1$$ Comment: Most textbooks use the last argument to show this limit, but it only works for the wrong reason. We assume that the exponential is differentiable (because it "intuitively" is) and moreover that there is one exponential whose tangent at $(0,1)$ has slope exactly 1 (in other words, we use that $\ln(x)$ is continuous at $x=1$ to define $e$ and $\ln(x)$).. It is exactly this limit which makes all these work.... And using directly those arguments, the argument really reduces to "because this limit is $e$, it follows that this limit is $e$". There are actually simpler correct ways of defining $e$ but they all need a deep understanding of Analysis (like viz Taylor Series or using FTC: $\frac{1}{x}$ has an antiderivative). - Just a quick note on the comment, there is no need to assume that the exponential is differentiable. Simply define the exponential function $\exp(x) = \sum_{i=0}^{\infty}$ and all the usual properties of $\exp(x)$ and $\ln(x)$ follow quite easily, with no need for the FTC. This is particuarly nice because most of the interesting properties of $\exp$ and $\ln$ are differentiable properties rather than integral properties, and they all follow almost immediately from these definitions. – Tom Oldfield Feb 16 '13 at 17:51 @TomOldfield I mentioned that in my last paragraph, but all those definitions use Theorems not covered in an introductory calculus class, at least in North America. ;) And some of them are relatively strong results in Analysis... Moreover, with this approach you define $e=\sum_{n=0}^\infty \frac{1}{n!}=\lim_n \left( 1+\frac{1}{n}\right)^n$, so is not really a new definition, you just prove the existence of the above limit in a different way.... – N. S. Feb 16 '13 at 18:06 @TomOldfield And, to clarify, how would you introduce the exponential this to way to a student if Power series and sequences are NOT part of the material studied in the class? – N. S. Feb 16 '13 at 18:34 Trying to follow the proof, I'm stuck at: $$\frac{\left( 1+\frac{1}{n}\right)^{n}}{\left( 1+\frac{1}{n+1}\right)^{n+1}}=\frac{(n+1)^{2n}}{n^n(n+2)^n}\frac{n+1}{n+2}$$ Can you explain this step? – ivan Feb 16 '13 at 19:28 Oh wait, wait, I think I got it :) – ivan Feb 16 '13 at 19:38 It's not too hard to prove, but it does rely on a few things. (In particular the validity of the taylor expansion of $\ln$ around 1 and that $\exp$ is continuous.) Consider in general the sequence $n\ln(1+x/n)$ which is defined for all $x$, positive or negative provided $n$ is large enough. (In fact the proof that follows can also be modified slightly to work for complex $x$). Using the power series representation $\ln(1+y) = \sum_{i=1}^\infty (-1)^{i+1}\frac{x^i}{i}$ we get $$n\ln(1+\frac{x}{n}) = x + \sum_{i=2}^\infty(-1)^{i+1}\frac{x^i}{in^{i-1}}$$ The absolute values of the terms in the series on the right hand side are bounded above by a geometric series (for n large enough) whose summation tends to 0 as $n$ tends to infinity. Hence $$\lim_{n\rightarrow \infty} n\ln(1+\frac{x}{n}) = x$$ And so using the fact that $\exp(x)$ is a continuous function: $$\lim_{n\rightarrow \infty}\bigg(1+\frac{x}{n}\bigg)^n = \exp\bigg(\lim_{n\rightarrow \infty}n\ln(1+\frac{x}{n})\bigg) = \exp(x)$$ In particular, set $x = 1$ to get the result you ask for. - The problem with this type of approach, while it is easy to understand by students, is this: what is your definition of $e$? :) – N. S. Feb 16 '13 at 17:32 BTW: Taylor series are not needed $\lim_{n\rightarrow \infty} n\ln(1+\frac{x}{n}) = x$ follows immediately from the definition of the derivative of $\ln(x)$ at $1$... – N. S. Feb 16 '13 at 17:33 @N.S. We define the function $\exp(x)$ to be the power series $\sum_{n=0}^{\infty} \frac{x^n}{n}$, and $e = \exp(1)$ There is no problem with this definition and allows us to work with both $\exp$ and $\ln$ without the FTC. To derive the limit, it depends on your definition of $\ln(x)$. You can define it as in integral of $\frac{1}{x}$ in which case your method works, or as the inverse of $\exp(x)$ in which case mine does. – Tom Oldfield Feb 16 '13 at 17:40 Yes, you are right there is no problem with this definition, but this is not how most of our undergrad textbooks define $e$. The limit you calculate is actually equivalent to the fact that your definition of the exponential makes sense (i.e. it is an exponential function)... BTW: using the binomial expansion you can prove directly that $\lim_n (1+\frac{1}{n})^n=\sum_{n=0}^\infty \frac{1}{n!}$. – N. S. Feb 16 '13 at 17:45 @N.S. I wouldn't know, this is how I had $e$ defined to me as an undergraduate. Personally I would view the real test of a function as being exponential as having the property $f(x+y)=f(X)f(y)$. Yes, this definition can be derived from the other, and that just goes to show that it doesn't really matter which one you use, as long as you can use it easily! I guess everyone should just pick their favourite :) – Tom Oldfield Feb 16 '13 at 18:03 I have removed the more precise, but more confusing, justifications of $(1)$ and $(2)$. In this answer, I show that $$\left(1+\frac1{n+1}\right)^{n+1}\ge\left(1+\frac1n\right)^n\tag{1}$$ and $$\left(1+\frac1{n+1}\right)^{n+2}\le\left(1+\frac1n\right)^{n+1}\tag{2}$$ Therefore, $(1)$ says that $\left(1+\frac1n\right)^n$ is an increasing sequence and $(2)$ says that $\left(1+\frac1n\right)^{n+1}$ is a decreasing sequence. Since $\left(1+\frac1n\right)^n\le\left(1+\frac1n\right)^{n+1}$ the increasing sequence is bounded above and the decreasing sequence is bounded below; thus, they both converge. Since $\lim\limits_{n\to\infty}\frac{\left(1+\frac1n\right)^{n+1}}{\left(1+\frac1n\right)^n}=\lim\limits_{n\to\infty}\left(1+\frac1n\right)=1$. They both converge to the same limit. This is called $e$. $$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{3}$$ - A great elementary proof is on Rudin's Principles of Mathematical Analysis page 64 Theorem 3.31. PS: Sorry I don't know how to stretch it, maybe you can download the pic and enlarge. - Please write it down here for us who don't have that book at hand. – vonbrand Feb 16 '13 at 18:59 I want to put this as an answer rather than a comment so that people can find it easily: https://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function#Equivalence_of_characterizations_1_and_2 ^ Wikipedia has the cleanest approach I can find. i.e. It doesn't pull anything out of the blue, and it doesn't sit on top of other advanced results. I'm not going to replicate the working; I am confident it will not change for the worse. EDIT: I just noticed it cites its source as Rudin, theorem 3.31, p. 63–5, so it is the same as the picture posted in another answer! -	{}
# FHA Streamline Refinance: Rates and Requirements for 2021 Getty Images We want to help you make more informed decisions. Some links on this page — clearly marked — may take you to a partner website and may result in us earning a referral commission. For more information, see How We Make Money. With mortgage refinance rates hitting all-time lows, millions of Americans are considering refinancing their homes. According to Freddie Mac, refinance activities increased by 6% percent among first-time homeowners in the opening months of 2020. Those who have an FHA Loan have the option to refinance through the FHA Streamline Refinance program. This option is designed to help first-time homeowners take advantage of lower FHA interest rates based on their payment history. Here’s what you need to know about FHA Streamline refinancing and when you should consider moving forward with this option. ## What is the FHA Streamline Refinance? As an alternative to conventional loans, FHA-insured loans are targeted toward first-time home buyers. FHA loans require a minimum down payment of 3.5%, with easier credit qualifications and lower closing costs. The FHA Streamline Refinance program allows individuals and families with an FHA-insured loan to take advantage of lower rates with minimal paperwork. Instead of going through a credit check and income verification process, the FHA accepts a perfect on-time payment history as evidence the homeowner can re-qualify for a loan. ### Pro Tip Homeowners with current FHA loans will find streamline refinancing is comparatively faster and easier than traditional refinance options. “If the homeowner has made their payments on time over a period of time, they have pretty much done everything they need to do to qualify for the mortgage,” says Bill Banfield, executive vice president of capital markets at Quicken Loans. “If you’ve already been able to make your payments at the existing interest rate that you have, and can get an interest rate that’s a half-percent or more lower, you can use the FHA streamline refinance program and get the lower rates available on the market.” Compared to a conventional mortgage, the FHA streamline process allows homeowners to skip many traditional refinancing requirements, including income verification and home appraisal. The process enables homeowners to save time and money. ### What documents do I need for an FHA Streamline Refinance? Even though an FHA Streamline Refinance has a simplified application process, you’ll need documentation to show the following: • Proof you’re current on your existing FHA loans payments • Proof of current interest rate • Employment verification • Proof of funds to cover closing costs • Proof of insurance Your lender should provide you with a checklist of acceptable documents. What you’ll need to be able to show is that the refinance is providing a financial benefit to you and that you’ve been making your mortgage payments. ### When can I do an FHA Streamline Refinance? You can take advantage of the FHA Streamline Refinance program only if you currently have an FHA loan. There is also a waiting period before you can apply for an FHA Streamline Refinance. It will need to have been at least 210 since you closed on your existing home loan, at least six months since your first mortgage payment was due, and you’ll need to have made at least six payments on the mortgage you’re refinancing. ### Can you do FHA Streamline Refinance twice? You can use the FHA Streamline Refinance program more than once, but you’ll need to meet all the guidelines. So you’ll need to be up to date with your payments and meet the waiting period requirements. Keep in mind, you’ll also have to pay closing costs each time you refinance. ### What are the benefits of an FHA Streamline? For homeowners with an existing FHA mortgage, an FHA Streamline Refinance has several benefits compared to other refinancing options. First off, the out-of-pocket fees you pay to refinance could be slightly lower than a traditional FHA refinance because you may be able to skip the appraisal and save yourself $300 to$600. One of the biggest benefits is how much simpler the closing process can be. With no income verification or credit check required, you may be able to get your loan refinanced much more quickly. Credit Cards 9 min read Banking 7 min read Credit Cards 6 min read Credit Cards 4 min read	{}
Journal Search Engine ISSN : 2288-4637(Print) ISSN : 2288-4645(Online) The Journal of Asian Finance, Economics and Business Vol.6 No.1 pp.9-19 DOI : http://doi.org/10.13106/jafeb.2019.vol6.no1.9 # Why Learners Found Transfer Pricing Difficult? Implications for Directors Indra Abeysekera1, Sam Jebeile2 2 Education Strategy Advisor, ANU College of Business and Economics, Australian National University, Australia. E-mail: SAM.JEBEILE@anu.edu.au 1 First Author and Corresponding Author. Professor, Chair of Accounting and Finance, College of Business and Law, Charles Darwin University, Australia [Postal Address: 21 Kitchener Drive, Darwin City, NT0800, Australia] Tel: +61-4-1740-5399. E-mail: indraabeysekera@gmail.com October 20, 2018 October 31, 2018 December 23, 2018 ## Abstract A recent survey of Australian directors conducted by the Financial Reporting Council found that directors require a detailed understanding of technical accounting issues. With the aim of understanding learner difficulties in learning and applying higher learning material relevant to directors, this study explores the transfer pricing topic taught as a case presentation in an undergraduate accounting program at an Australian university. Before intervention with improvements, this study invited 25 students to take part in the study after they had learned the topic and been given one week to understand it. By adopting a transfer pricing problem presented in their essential reading and interviewing those students to gain further insights, the study found that learners experienced conceptual difficulties at various stages in attempting to learn. Intervention to ease learning difficulties was addressed through instructor training. The intervention improvements included using guided workbooks to develop a better understanding of concepts among learners, and representing the problem at hand with diagrams. After intervention with improvements, this study repeated the same procedures with 25 students who had not taken part in the previous study and found that interventions increased the learning. Results have implications for most directors, who are novices to the detailed technical accounting issues of transfer pricing. JEL Classification Code: I21, J49, M49. ## 1. Introduction A survey conducted by the Financial Reporting Council in Australia to determine the financial literacy of Australian directors asked financial professionals about directors’ financial literacy and found that these professionals rated directors’ knowledge about more technical accounting issues as less than adequate (CAANZ, 2016). This is a concern because directors require a quite detailed understanding of technical accounting issues such as transfer pricing in making sound decisions affecting organisational economics, efficiency, and effectiveness. Transfer pricing has to do with setting the price of goods and services sold within multi-divisional, multi-office, or multi-national firms. The efficient setting of transfer pricing allows each division, office, or firm within a group of firms, to function as a profit centre with the aim of maximising overall profits of the enterprise. An enterprise can also set transfer price to transfer profits from a high-tax jurisdiction country to a low-tax jurisdiction country. Governments lose tax revenue from companies due to transfer price arrangements and have established various rules a firm should follow in setting a transfer price for cross-border transactions within a group of firms. Transfer pricing is a challenging issue for directors because setting inappropriate transfer prices can increase government scrutiny and potential prosecution (Power, 2012). Knowledge about transfer pricing requires understanding a nexus of accounting concepts and their application to determine the appropriate transfer price. Hence, learners have reported that transfer pricing is a difficult to topic to understand. This difficulty becomes evident from students’ consistently poor examination performance on this question. Transfer pricing decides the selling price for transactions of goods and services between divisions within a firm. Firms can label their divisions as investment centres or profit centres. This labelling allows directors to evaluate profits or investments made by the divisions. Because directors measure how well divisions perform by profits and investments made, each division can tend to increase its profits and investments while ignoring how this influences the firm. Transfer prices are internal selling prices; they form revenue for the selling division that increases profits, and cost to the buying division that decreases profits (Langfield-Smith, Thorne, & Hilton, 2006). The divisions measure the financial performance using some measure of profit. This study explored why students have difficulties in understanding transfer pricing, a topic taught in the management accounting course unit of the undergraduate course at an Australian university. Investigating student learning difficulties in the transfer pricing topic helps us to understand difficulties facing novice students in learning topics of higher-order thinking which demand a sound understanding of lower-level thinking. Many directors who do not have a strong accounting background and are new to how transfer pricing works will also experience similar difficulties and become equivalent to novice-students of transfer pricing. This study presents how we overcame student difficulties in learning transfer pricing. The next section outlines cognitive load theory and the relevant literature. Section 3 reports the data collection including problem presentation. Section 4 presents results and discussion, as well as implications for teaching transfer pricing to directors. ## 2. Theory and Literature Review Cognitive load theory states that instructional design must guide learners to discover and build knowledge. It explains the association between knowledge and learning, and how instructional design can help to foster the building of knowledge (Moreno & Park, 2010). Learning requires exerting cognitive load, but instructional design can moderate this load. Cognitive load arises from consuming the working memory or short-term memory in learning new information. According to cognitive load theory, the human cognitive architecture has a limited short-term memory and a large long-term memory (Kahneman, 1973; Miller, 1956). The short-term memory makes meaning out of information by thinking, and long-term memory stores information for later use (Schneider & Schiffrin, 1977). A role of learning is to transfer an organised pattern of thought or behaviour (schema) developed in the working (short-term) memory to long-term memory so that the learner becomes familiar with finding solutions to any similar problems. The transfer of an organised pattern or thought from working memory to long-term memory frees up working memory for further learning (Sweller, 1994). Novices use more working memory because they lack previously formed thought patterns and behaviours relevant to solving a given task and therefore they must think more to solve the task. Experts use less working memory because they can access solutions to the problem by calling on previously formed thoughts and behavioural patterns deposited in long-term memory (Sweller, Ayers, & Kalyuga, 2011, p. 21). Although previous studies agree that instructional design can decrease the cognitive load, mixed findings exist for intrinsic cognitive load. Some findings suggest that instructional design can decrease the intrinsic cognitive load, by presenting items of information (that is, elements) sequentially, and combining similar items in the instruction (Lee, Plass, & Homer, 2006; Pollock, Chandler, & Sweller, 2002; van Merrienboer, Kirschner, & Kester, 2003). Learning various topics can invoke different types of thoughts and behavioural patterns (that is, schema). Some topics require building a simple schema, whereas for others a complex schema must be built. Learning accounting requires building a complex schema (Blaney, Kalyuga, & Sweller, 2010). Learning accounting topics requires learning many new concepts and formulas (that is, elements) and simultaneously combining them. The complexities lead to students using too much of their working memory, and this can decrease learners’ capacity to build a schema to increase learning (Sithole, Chandler, Abeysekera, & Paas, 2017). The transfer pricing topic is a case in point, where students need to learn and simultaneously combine new ideas and formulas to build from a basic to a complex schema. Figure 1 summarises five order levels of the understandings the learner must develop in order to learn transfer pricing. The first order shows the basic concepts in transfer pricing. The second-order concepts build upon the first-order concepts. It is essential that students understand the first-order concepts, to understand the second-order concepts. Each higher order level builds upon the understanding of lower-level concepts and understanding the fifth-order concepts requires understanding all other lower-level concepts. Solving a problem about transfer price requires decision-makers to simultaneously process information relating to all order levels. It is essential to understand the conceptual difficulties experienced by learners, to ease learning difficulties in mastering transfer pricing. We assessed learners’ written answers against the marking guide prepared at the time of formulating the problem. We conducted interviews while students were attempting the problem, to understand the learning difficulties they experienced. We stated and tested the alternate hypotheses according to order-level mastery, and expected that improvement-interventions diagnosed from the pre-intervention stage would improve the learning at the post-intervention stage. The learning was measured by assessing scores against the marking guides, and statistical significance is measured using Pearson Chi-Square (goodness-of-fit) test at 5 percent significance level. H1: Interventions improve learning at first-order and second-order levels (Part A of the problem). H2: Interventions improve learning at third-order and fourth-order levels (Part B of the problem). H3: Interventions improve learning at the fifth-order level (Part C of the problem). ## 3. Methodology This study was conducted in three phases (Figure 2). ### 3. 1. Phase 1: Problem Design The first phase, pre-intervention, investigated learner difficulties in learning transfer pricing. Phase 2 conducted a workshop for instructors to inform about learner challenges and develop strategies to overcome them. Phase 3, the post-intervention phase, implemented the agreed strategies to overcome learning difficulties to a future cohort of learners. Students were invited to take part in the research study by advertising it in tutorial classes as a research project examining learning difficulties in transfer pricing, according to the ethics agreement. A researcher randomly selected 25 students from those who expressed interest in taking part. Students enrolled in this course unit had attained a 56 percent average mark and had an average 1.94 Grade Point Average. The t-tests confirmed that these averages were not statistically different from the total average mark and GPA of students of this management accounting course unit, establishing their representativeness of the class. The study used a common teaching instructor for all students to eliminate the effect of instructor difference. Students had completed the transfer pricing topic before the mid-term break and we invited them to take part in the study in the first week after the break. We gave students a typical problem they had faced in learning transfer pricing, from their essential reading textbook written by Langfield-Smith et al. (2006) (Table 1). The problem had three parts to be answered as Part A, Part B, and Part C. #### 3.1.1. Part A of the Presented Learning Problem Part A required problem solving at the first-order and second-order levels. In Part A, students conducted the activities in the following order. (i) Calculated the contribution margin (difference between selling price and variable costs) from each unit produced, for each of the divisions, based on the transfer pricing rules of the firm. Part A also required students to know the meaning of contribution margin as the difference between selling price and variable costs. They were required to know that variable cost is adding direct material, direct labour, and variable overhead. This transfer pricing problem needed students to make four separate calculations. (ii) Calculated selling prices for the two divisions in the firm: Division A and Division B. (iii) Calculated total variable costs for the two divisions separately. (iv) Included the transfer price of Division A, as a buying cost in Division B. (v) Calculated contribution margins for the two divisions in the firm, Division A and Division B. #### 3.1.2. Part B of the Presented Learning Problem Part B required problem solving at third-order and fourth-order levels. Learners were required to calculate the minimum transfer price that Division A would accept for selling to Division B if market forces influenced the transfer price. Understanding the meaning of transfer price requires understanding the meaning of several other concepts as follows. (i) Know that outlay cost means variable costs incurred in production. (ii) Know that opportunity cost means giving up a monetary benefit for choosing one course of action over another. (iii) Be aware that excess capacity means the unused production quantity that remains after including production needs from internal and external customers. (iv) Be mindful of the fact that transfer price of an item produced is an addition of outlay cost of an item and the opportunity cost of that item. (v) Be aware of how to use these meanings and formulas simultaneously, to calculate outlay costs of Division A, and to calculate the opportunity cost of Division A. #### 3.1.3. Part C of the Presented Learning Problem Part C required problem solving at the fifth-order level. Part C asked students to decide the overall impact of accepting or rejecting a special order received from a customer named Socceroos, on firm-wide profits. Answering Part C required students to know the following. (i) Meaning of goal congruence, as a decision made by a division to make the most profit possible for the organisation, and not for the division itself. (ii) Meanings of incremental revenue, incremental costs, and incremental profits. Incremental revenue is the extra revenue resulting from choosing one course of action against another. Incremental costsare the extra costs resulting from choosing one course of action against another. Incrementalprofitis extra profits resulting from choosing one course of action against another. (iii) Know the connections of these meanings to calculate and decide whether the firm should accept or reject the special 1. First, calculate extra revenue earned by the firm. 2. Second, calculate extra costs incurred by division A for selling products to Division B because of the agreement between divisions. 3. Third, students needed to calculate extra costs that Division B incurs because of accepting the special order from a customer. 4. Fourth, students needed to calculate extra costs that Division A must incur in selling finished goods to Division B to meet the special 5. Fifth, they needed to calculate the opportunity costs that Division B must bear for accepting the special 6. Sixth, they needed to calculate extra profits earned by the firm. Table 1 shows the transfer pricing problem provided to students in Phase 1 of this study. The problem had three parts (Part A, B, and C) to answer. Table 2 outlines the interview guide which explored students’ understandings of concepts. ### 3. 2. Phase 2: Evaluating Answers to Ascertain Interventions and Conducting Interviews Students’ answers to the transfer pricing problem were evaluated against the model answer prepared during the research design stage (Table 2). The findings from the previous semester assessments and student interviews led to the following improvements to decrease learner cognitive load. (i) Provided a Transfer Pricing Workbook that students downloaded as prerequisite reading from the course learning site. The Workbook contained explanations of all the concepts required to learn transfer pricing in this course successfully. In contrast to the previous semester, the concepts about transfer pricing were covered in the prerequisite course as essential learning outcomes. (ii) The Transfer Pricing Workbook provided students with blank pages where they were also asked to represent the problem using diagrams. (iii) The lecturer highlighted the common misconceptions diagnosed in the previous semester evaluation. (iv) The lecturer drew diagrams at each stage of problem solving to visually show the how the solution is researched. (v) The lecturer gave students time to complete the examples before providing the solution. These activities provided students with opportunities for incidental learning. #### 3.2.2. Conducting Interviews and Documenting Feedback on Learning Difficulties The interviews held with students revealed that students' wrong answers for Part A arose due to five reasons (see Appendix for interview excerpts with feedback from students). First, presenting information as numbers wherever possible, rather than as text, decreased the cognitive load. Second, students made mistakes because of conceptual misunderstandings. For example, they did not understand the difference between selling price (for Division A) and the market price. Third, students could not combine lower-order concepts with higher-order concepts and simultaneously use them. For example, in the Phase 1 findings for Part B, calculating contribution margin of Division A needed a student to interact with several items of information simultaneously. Fourth, interacting with these items of information cascaded over three levels, making learning to calculate contribution margin difficult. At the first-order level, students needed to know the meaning of direct materials, direct labour, and variable overheads. At the second-order level, students had to know that these three types of cost make up variable production costs. At the third-order level, they should be aware that variable production costs become outlay costs for transfer pricing calculation. Students also should be aware the meanings of selling price, variable production costs, and non-variable production costs. At the second-order level, they should also understand how the existence of surplus production capacity can influence the opportunity costs. At the third-order level, they should be aware of the meaning of opportunity costs, and the information items that comprise these costs. At the fourth-order level, students should be aware that the outlay costs and opportunity costs of Division A production make up transfer pricing. When students reach the fifth-order level, they should be aware of goal congruence. They should explore whether the firm has a transfer pricing agreement to perform goal congruence, and accordingly calculate the contribution margin (Figure 1). Fifth, the Phase 1 of the findings also showed that the information presented in the transfer pricing problem required students to split their attention between the text and footnotes. Splitting their attention in order to integrate information to understand the question also increased the cognitive load. The problem question had an asterisk (*), hat (^), and footnotes. These interruptions distracted students from reading the text and forced them to split their attention to find out what those pointers were asking them to read, so as to understand the material. For instance, an asterisk forced them to divide their attention underneath a table to realise that 50 percent of manufacturing overheads do not change, and 50 percent vary with the quantity of production. ### 3.3. Phase 3: Post-intervention The same lecturer who conducted lectures in the first semester conducted lectures in the following semester. The lecture covered the theoretical aspects of transfer pricing, followed by practical examples using an overhead projector. The sequence of events mirrored the first semester (pre-intervention phase), with 25 students being invited after the mid-semester break to take part in the study in Phase 3 (post-intervention phase). Students’ GPA and average marks were not statistically significant from the Phase 1 cohort. The task required students to solve a semi-structured transfer pricing problem that was conceptually identical to that performed by students in Phase 1. The equivalence was determined by showing the two transfer pricing problems to three academics who had taught transfer pricing, and obtaining unanimous agreement. The marking criteria for the solution to the problem were identical between pre-intervention and post-intervention, moderated by improvements to extraneous cognitive load to ease learner difficulties. The different but equivalent transfer pricing problem given to students in Phase 3 is shown in Table 3. ## 4. Results and Discussion ### 4.1. Results of Part A Table 4 provides summary statistics of calculating selling price, contribution margin, and transfer price for Division A and Division B of the firm. The improvements due to interventions are significantly different from pre- to post-intervention (χ2(2) =385.53, p<0.05), and support H1. As additional tests, we measured the improvements at item level, and the intervention-improvements had a statistically significant effect at 5 percent significance level (exceeds the χ2(1) critical value = 0.004). ### 4.4. Reflections on Teaching Directors about Transfer Pricing Most directors are novices to learning transfer pricing, and findings from this study using novice learners become applicable to teaching directors about transfer pricing. Consistent with Blaney et al. (2010), these findings led us to design a teaching format that isolated each information item about transfer pricing from lowest order to highest order, and to teach them sequentially. The interrelating of these information items then occurred progressively at higher order levels (Figure 1). The post-intervention phase of the study attended to three areas: (i) conceptual difficulties were relieved by providing learners with, and guiding them with, a transfer pricing workbook, (ii) instructors understood the conceptual and procedural difficulties that learners have in learning transfer pricing, and learning was guided accordingly, and (iii) learners were encouraged to draw diagrams in order to visualise the text. In adopting these interventions, the study replaced the problem-solving technique with the worked example technique (Chandler & Sweller, 1991). The problem-solving instructions presents learners with a description of the problem and a goal statement but does not provide adequate guidance about the procedures to solve the problem, leading to trial and error or means-solutions analysis strategies, although the learner may eventually find a solution. Worked examples provide learners with not only means-solutions analysis but also steps taken successively to solve the problem and achieve the solution state. Instructions that directs learners on studying worked examples reduces extraneous or ineffective cognitive load on the working memory which can enhance the learning through transfer of learning from working memory to long term memory (Spanjers, van Gog, & van Merrienboer, 2012). As found in Phase 3, the teaching instruction should also add diagrams to show directors how to interrelate information items in transfer pricing. As found in this study, there are seven sequential instructional visual steps to follow in transfer pricing: 1. Drawing two boxes, one to represent Division A and the other Division B. 2. Drawing an arrow labelled “transfer price” from Division A to Division B, indicating that the transfer price between the divisions was yet to be determined. 3. Drawing a box around Division A and Division B together, to represent their relationship as divisions within the one company, recognising a need for goal congruence. 4. Drawing two boxes external to the company box, labelling the first box “External Market for Division A” and the other “External Market for Division B.” 5. Drawing an arrow from Division A out to the external market for Division A and labelling the output with the market selling price. 6. Drawing an arrow from Division B out to the external market for Division B and labelling the output with the market selling price. 7. Drawing an arrow from the external market for Division A into Division B, labelled with the input price if Division B was forced to buy its input from the open market rather than by internal transfer from Division A. ### 4.5. Limitations and Future Research In interpreting findings, there are three limitations to be acknowledged. First, it is neither claimed here that cognitive load theory is the only way to explain these results, nor contended that it explains all reasons behind the learning difficulties learners experienced. For instance, cognitive load theory does not consider the influences of psychological reasons such as student beliefs, expectations, and goals (Bannert, 2002). Willingness to invest time and effort in learning the task, and learners' views of achieving personal goals are also not considered here (Thrash & Elliott, 2001). Second, the improved learning outcomes are due to easing out conceptual difficulties, and students transforming the problem from text into visual. This study did not measure the contribution of each improvement-intervention measure separately and this is a future research proposal. However, the learning difficulties identified in this study can help in improving teaching design to help learners learn complicated topics such as transfer pricing. These topics require learning lower-level and higher-level ideas and inter-relating them, to decrease the cognitive load in working memory. Third, the study randomly selected 25 students from among those who expressed interest in taking part in the study, but omitting those who did not express interest may have contributed to a selection bias. These limitations can be used for future research to deepen the understanding about learning a vital topic such as transfer pricing. ## Reference 1. Bannert, M. (2002). Managing cognitive load, recent trends in cognitive load theory. Learning and Instruction, 12, 139–146. 2. Blayney, P., Kalyuga, S., & Sweller, J. (2010). Interactions between the isolated–interactive elements effect and levels of learner expertise: experimental evidence from an accountancy class. Instructional Science, 38(3), 277-287 3. Blayney, P., Kalyuga, S., & Sweller, J. (2016). The impact of complexity on the expertise reversal effect: experimental evidence from testing accounting students. Educational Psychology, 36, 1868-1885. 4. CAANZ (Chartered Accountants in Australia and New Zealand). (2016, June 24). Financial literacy of directors survey results. Retrieved from http://www.charteredaccountants.com.au/Industry-Topics/Reporting/News-and-guidance-on-regulatory-matters/News-and-updates/Financial-literacy-of-directors-survey-results. 5. Chandler, P., & Sweller, J. (1991). Cognitive load theory and the format of instruction. Cognition and Instruction, 8(4), 293-332. 6. Kahneman, D. (1973). Attention and effort. Englewood Cliffs, NJ: Prentice-Hall. 7. Langfield-Smith, K., Thorne, H., & Hilton, R. W. (2006). Management accounting: Information for managing and creating value (4th ed.). Sydney, Australia: McGraw-Hill. 8. Lee, H., Plass, J. J., & Homer, B. D. (2006). Optimizing cognitive load for learning from computer science simulations. Journal of Educational Psychology, 98, 902-913. 9. Miller, G. A. (1956). The magical number seven, plus or minus two: Some limits on capacity for processing information. Psychological Review, 63, 81-97. 10. Moreno, R., & Park, B. (2010). Cognitive load theory: Historical development and relation to other theories (chapter one). In J. L. Plass, R. Moreno, & R. Brunken (Eds.), Cognitive load theory. New York, NY: Cambridge University Press. 11. Pollock, E., Chandler, P., & Sweller, J. (2002). Assimilating complex information. Learning and Instruction, 12, 61-86.	{}
anonymous one year ago Please help me. I will medal & fan!!!!!!!!** :) If the nth term of a sequence is 2n + 4, what is the sum of the first n terms? A.) n(n+1)+4n B.) n(n+1)/2 +n C.) n(2n+4) D.) n(n+1)(2n+1) /6 + 2n(n+1) E.) 2n(n+1)/n 1. Michele_Laino the first term is: ${a_1} = 2 \times 1 + 4 = ...?$ 2. anonymous I would just plug that in? 3. Michele_Laino the first term is given by replacing n with 1 into your formula 4. Michele_Laino the requested sum S is given by the subsequent formula: $S = \frac{{{a_1} + {a_n}}}{2} \times n$ 5. Michele_Laino what is a_1? 6. anonymous 8? 7. Michele_Laino hint: $\large {a_1} = 2 \times 1 + 4 = 2 + 4 = ...?$ 8. anonymous I'm just a numskull and don't understand any of this. I thank you for time and effort. I will just guess at this point... Thank you, though. :) 9. Michele_Laino 10. Michele_Laino the first term is a1=6 right? 11. Michele_Laino now we have to substitute so we can write: $\large \begin{gathered} S = \frac{{{a_1} + {a_n}}}{2} \times n = \frac{{6 + 2n + 4}}{2} \times n = \frac{{10 + 2n}}{2} \times n = \hfill \\ \hfill \\ = \left( {5 + n} \right)n = 5n + {n^2} \hfill \\ \end{gathered}$ 12. Michele_Laino now we have this: $\Large n\left( {n + 1} \right) + 4n = {n^2} + n + 4n = {n^2} + 5n$ 13. anonymous n(2n+4) ? 14. Michele_Laino I think that it is: n(n+1)+4n 15. anonymous no it's n(n+1)+4n ?? 16. Michele_Laino that's right! 17. anonymous yay!! thank you!! 18. Michele_Laino :) Find more explanations on OpenStudy	{}
# Tag Info 25 Intel support for IEEE float16 storage format Intel supports IEEE half as a storage type in processors since Ivy Bridge (2013). Storage type means you can get a memory/cache capacity/bandwidth advantage but the compute is done with single precision after converting to and from the IEEE half precision format. https://software.intel.com/content/www/us/en/... 25 Take \begin{align} 1-\sqrt{ 1-x^2} &= (1-\sqrt{ 1-x^2})\frac{1+\sqrt{ 1-x^2}}{1+\sqrt{ 1-x^2}}\\ &= \frac{x^2}{1+\sqrt{ 1-x^2}} \end{align} So \begin{align} y = x\sqrt{\frac{1}{2+2\sqrt{1-x^2}}} \end{align} 22 (I have tested this approach before, and I remember it worked correctly, but I haven't tested it specifically for this question.) As far as I can tell, both $\\|\mathbf{v}_1\times \mathbf{v}_2\\|$ and $\mathbf{v}_1\cdot \mathbf{v}_2$ can suffer from catastrophic cancellation if they are almost parallel/perpendicular—atan2 can't give you good accuracy if ... 15 This question is very closely related to (and possibly a duplicate of) Is variable scaling essential when solving some PDE problems numerically?. There are still good practical reasons to nondimensionalize equations, if possible: It reduces the number of independent parameters for parametric studies (which was one of the original reasons for ... 12 Use the (IEEE standard) library function log1p, which should be present in all programming languages. The function log1p(x) returns $\log(1+x)$, and is implemented with particular attention to accuracy when $x$ is small. It is designed to solve exactly this kind of problem. 11 I am not an expert in machine learning, but I can outline the considerations that are relevant. The numerical calculations in machine learning are generally linear algebra -- either solving linear systems or linear least squares. For both types of problems, there are well-known backward-stable methods, so I will assume you are using a backward-stable ... 11 The efficient answer to this question is, not too surprisingly, in another note by Velvel Kahan: $$\alpha=2\arctan\left(\left\\|\frac{\mathbf v_1}{\\|\mathbf v_1\\|}+\frac{\mathbf v_2}{\\|\mathbf v_2\\|}\right\\|,\left\\|\frac{\mathbf v_1}{\\|\mathbf v_1\\|}-\frac{\mathbf v_2}{\\|\mathbf v_2\\|}\right\\|\right)$$ where I use $\arctan(x,y)$ as the angle made by $(x,y)$ ... 10 Overview Good question. There is a paper entitled "Improving the accuracy of the matrix differentiation method for arbitrary collocation points" by R. Baltensperger. It's no big deal in my opinion, but it has a point (that already was known before the appearance in 2000): it stresses the importance of an accurate representation of the fact that the ... 10 For $a+b\sqrt{-3}$ you can use the representation $$\begin{pmatrix}a&-3b\\b&a\end{pmatrix}$$ Addition works obviously. For multiplication, you can verify $$\begin{pmatrix}a_1&-3b_1\\b_1&a_1\end{pmatrix}\begin{pmatrix}a_2&-3b_2\\b_2&a_2\end{pmatrix} = \begin{pmatrix}a_1a_2-3b_1b_2&-3(a_1b_2+b_1a_2)\\a_1b_2+b_1a_2&a_1a_2-3b_1b_2\... 10 The numeric precision is not perfect. You get rounding errors during your computation. When working with floats, don't check if they are = 0, but check if their absolute distance to 0 is smaller than some epsilon. 10 Jean-Michel Muller, et. al., "Handbook of Floating-Point Arithmetic 2nd ed.", Birkhäuser 2018, gives the following example due to Muller, specifically constructed to deliver incorrect results with floating-point evaluation:$$ {u_{0} = 2,\\ u_{1} = -4,\\ u_{n} = 111 - \frac{1130}{u_{n-1}} + \frac{3000}{u_{n-1}u_{n-2}},\>\>\>\>n \ge 2.} $$... 9 I would like to know if there are any practical methods for obtaining an approximation to \mathbf{x}(t_f) (where t_f \in \mathbb{R} is some given final time) which is provably correct to N [sic] digits. That all depends on your opinion of the practicality of interval arithmetic. There are validated integrators available, such as the COSY code out of ... 9 There are aspects of modern computing systems that are inherently non-deterministic that can cause these kinds of differences. As long as the differences are very small in comparison with the required accuracy of your solutions, there probably isn't any reason to worry about this. An example of what can go wrong based on my own experience. Consider the ... 9 You can sometimes prove such results (or get counterexamples) using an SMT solver such as Z3 that supports floating point arithmetic. Here is a proof of a version of your theorem that says \|((x+y)-y)-x\| \leq 2^{-23}\|x\| when x>y>1 and x+y\neq\infty_{32} in 32-bit floating point arithmetic: λ> import Data.SBV λ> :set -XScopedTypeVariables λ&... 8 It is the difference in scales between terms in your equations that tend to cause numerical difficulties. You may work in any units you like as long as you are consistent. My approach has been to always consistently non-dimensionalize my equations in order to reduce the number of parameters to the minimum required, but this is only for my convenience. 8 Added after my initial answer: It appears to me now that the author of the referenced paper is giving condition numbers (apparently 2-norm condition numbers but possibly infinity-norm condition numbers) in the table while giving maximum absolute errors rather than norm relative errors or maximum elementwise relative errors (these are all different measures.) ... 8 In my opinion, not very uniformly. Low precision arithmetic seems to have gained some traction in machine learning, but there's varying definitions for what people mean by low precision. There's the IEEE-754 half (10 bit mantissa, 5 bit exponent, 1 bit sign) but also bfloat16 (7 bit mantissa, 8 bit exponent, 1 bit sign) which favors dynamic range over ... 7 There are many good open-source implementations of root-finding methods out there already. One example is boost, whose implementation of Newton-Raphson and related methods you can find here. If you read its source code, you will be able to see what issues the authors wanted to address, and how they dealt with issues like convergence, user-specified tolerance ... 7 At least in MATLAB, I believe abs(z) is implemented as sqrt(zz'). The extra square-root and squaring operation reduces numerical precision. >> z = randn + randn i z = 0.5377 + 1.8339i >> abs(z)^2 - zz' ans = 4.4409e-16 >> abs(z)^2 - sqrt(zz')^2 ans = 0 7 [EDIT] An alternate view: 64-bit floating numbers represent a discrete set S. For a function f to be exactly invertible, it should be a bijection from S to S. Suppose we are interested in a fast growing function like \exp. At one point, \exp x > x. If M is the maximum element from S, then f(M)\ge M. The strict version f(M)> M is ... 6 In contrast to Bill Barth, I usually try to keep things in dimensional form. Within a single equation, this of course does not change the relative scaling of terms. However, not doing the scaling requires that one pays attention to the relative scaling between different equations of a system of equations. A discussion of one case we have documented can be ... 6 There are three issues that are likely to cause such problems in pseudospectral methods: Gibbs oscillations Aliasing Time step too large In any case you likely develop oscillations in the solution until some point ends up with a negative density, resulting in a NaN when computing the pressure or sound speed or some other term. The solution to 3 is obvious, ... 6 Yes, it is possible to show that the statistical behavior of the approximate system will reach that of the "exact" system. (This is true even though hard-sphere dynamics do not accurately describe molecular systems!) The basic premise underlying molecular dynamics is the ergodic theorem, which states that, in the limit of long times, the time average of a ... 6 In my opinion, an example could be the calculation of the minimal solution of a three term recurrence relation (TTRR)$$y_{n+1} +a_{n}y_{n}+b_{n}y_{n-1} = 0, \quad n=1,2,3,\ldots$$. For example, the Bessel J_{n}(x) function for a fixed x satisfies the TTRR$$y_{n+1}-\frac{2n}{x}y_{n}+y_{n-1} = 0$$Suppose you know y_{0} = J_{0}(1) and y_{1} = J_{1}(... 6 The accepted answer provides an overview. I'll add a few more details about support in NVIDIA processors. The support I'm describing here is 16 bit, IEEE 754 compliant, floating point arithmetic support, including add, multiply, multiply-add, and conversions to/from other formats. Maxwell (circa 2015) The earliest IEEE 754 FP16 ("binary16" or "half ... 5 I think there's a simple way to do this. You have a rational function of identical cosh/sinh terms, where every expression is a homogeneous polynomial in cosh/sinh, and the only problem is that these exponential terms overflow. The function does not diverge as these terms approach infinity, so if you divide every numerator and denominator by the same power ... 5 Orthogonal matrices are about as well-conditioned as you can get, but numerical errors still occur. One common error is loss of orthogonality. A fix for this could be to re-orthogonalize your columns after some number of multiplications. You can do this by just taking the QR decomposition of your matrix after some number of products and taking the orthogonal ... 5 Very interesting question! LAPACK-inspired adaptive strategy This reminds me of a bug that was found in a LAPACK routine (rank-revealing QR) related to 'downdating' norms: essentially, you are given the norm of a vector v, and you want to compute at each iteration in O(1) the norm of the same vector after chopping off its initial entry: v[1:], v[2:], ... (... 5 There is no need for numerical computation here. First, T(q) is a well-known function, the logarithmic integral. Repeated integration by parts gives an asymptotic expansion$$\mathrm{Li}(q) = \frac{q}{\log q}\sum_{k=0}^{K-1} \frac{k!}{\log^k q} + O\left(\frac{q}{\log^{K+1}q}\right). There's also a fairly rapidly convergent representation due to ... 4 Elemental available at libelemental.org may have what you ask. From the documentation for release 0.84: Though Elemental does not yet fully support computation over arbitrary fields, the vast majority of its pieces do. Moving templated implementations into header files is a necessary step in the process and also allowed for certain templating techniques ... Only top voted, non community-wiki answers of a minimum length are eligible	{}
# Tag Info ## Hot answers tagged reconstruction 8 I don't get your downsample step when you downsampled by factor $M$. Let me go from scratch with the spectrum visualization below, with time domain, continuous frequency domain and discrete frequency domain from left to right. When we reduce the sampling frequency by a factor $k$, the signal spectrum is copied to new replicas at $f_s/k$. The discrete ... 7 I don't have a real answer but I have the feeling that this result will help you out: Bernstein's inequality says that, if the signal $x(t)$ is bandlimited to $\|f\|\leq B$, then $$\left\| \frac{\textrm{d}x(t)}{\textrm{d}t}\right\|\leq 4\pi B \,\textrm{sup}_{\tau\in\mathbb{R}}\|x(\tau)\| ,\,\,t\in\mathbb{R}$$ where $\textrm{sup}$ stands for "least upper bound". I ... 6 If your signal is really as simple as $$x(t)=A\sin(\omega_0t)\tag{1}$$ with known $\omega_0$, and you have observations $y(t_i)$, which are noisy samples of $x(t)$ at known time instances $t_i$, then a simple solution would be the least squares estimate $$\hat{A}=\frac{\displaystyle\sum_iy(t_i)\sin(\omega_0t_i)}{\displaystyle\sum_i\sin^2(\omega_0t_i)}\tag{2}... 5 Observations I have used +1 and -1 in the sequence instead of your 1 and 0. With \alpha=1, the band-limited continuous function f_m(T) in your first two figures (with the above mentioned modification) is:$$f_m(T) = \sum_{k=1-m}^m \operatorname{sign}\left(\operatorname{sinc}(\pi k - \pi/2)\right)\operatorname{sinc}(\pi T-\pi k),\tag{1}$$where:$$\... 4 Some recent cell phone models use something like a Cirrus Logic CS42xx series audio IO chip, which seems to use a digital polyphase interpolation filter, a sigma delta modulator, followed by a switched capacitor DAC and low-pass filter. Sinc interpolation (or, given finite hardware, a polyphase FIR kernel similar to a windowed Sinc) is one high quality ... 4 The sampling theorem requires a perfectly bandlimited signal, bandlimited to below twice the sampling frequency. The problem with this is that only an infinite length signal (e.g. exists before the big bang) can be perfectly bandlimited. This is from the Fourier theorem regarding any domain with finite support. Thus all real-world signals are ... 3 I guess this is a straightforward non-linear optimization problem (to be solved with Newton variations, such as Trust-Region methods), where you don't even need to compute the Jacobian analytically. It appears to me that the optimization problem is written over $K_i$, and thus is the input to the cost function. To compute the cost, at each call to this ... 3 The key idea is that the random sampling approach enforces more constraints on the resulting signal than the uniform sampling approach does. The POCS (projections onto convex sets) algorithm used for the reconstruction of the randomly sampled signal is the key piece: it enforces: That the signal must be from this spectrum. That the signal is real-valued. ... 3 FFT and IFFT are linear operators, and as such, the results only make a lot of sense in a linear intensity space, not if indexed into a non-linearly mapped space. 3 The general mathematical framework for interpolation is approximation theory. I guess the most important result is that for signals with bandwidth limitation, you can have perfect reconstruction via $sinc(\cdot)$ convolution; the famous sampling theorem, that has been mentioned here several times. I guess it is equally well-known that it is not really ... 3 It could be useful in the field of OCR as a pre processing step. Think about badly scanned data. You'd like to convert it into binary image. So the first step would be applying some kind of thresholding. Since no thresholding is perfect, There will be some "Holes" / "Gaps" within the text. Closing those "Holes" / "Gaps" can be done using morphological ... 3 SLAM(Simultaneous Localization and Mapping) algorithms can be used to for 3D reconstruction. They offer solutions for both monocular as well as stereo cameras. With single camera they estimate depth with few images and reconstruct the scene. You can find some of the open source solutions here. Real time 3D reconstruction can done using ORBSLAM and it is ... 3 The introduction of this paper explains the difference and gives an example. In short: Image restoration techniques presume that data are acquired in the image space; that is, the raw data represent a corrupted version of the image scene. In contrast, images are not directly observed in reconstruction problems. Instead, projections of an image are ... 3 First, a warm welcome to SE! Basically, you have a calibrated 3D reconstruction problem. The typical approach follows a 5-stage pipeline: Identify 2D features in each image along with the associated descriptors. Algorithms such as SURF, SIFT or AKAZE are heavily used and are available in many vision libraries such as OpenCV. Match the extracted keypoints ... 3 Indeed the model for the Proximal Gradient Method (Also see Proximal Gradient Methods for Learning) is in the form of: $$F \left( x \right) = f \left( x \right) + g \left( x \right)$$ Where usually $f \left( x \right)$ is convex smooth function and $g \left( x \right)$ is convex non smooth function. Yet the model is quite flexible and you may define ... 3 Our goal is to obtain proximal operator of the following function $$g \left( x \right) = {\left\\| x \right\\|}_{1} + \operatorname{TV}(x).$$ The involved optimization problem for any $z \in \mathbb{R}^d$ is the following $$\text{argmin}_{x}\left\{g(x) + \frac{1}{2}\\|x-z\\|^2_2\right\}$$ Denote the following $$g_1(x) := {\left\\| x \right\\|}_{1} + \frac{1}... 3 Zero-order hold will result in a piecewise-constant waveform. Linear interpolation will result in a piecewise-linear waveform. If you want a piecewise-quadratic or piecewise-cubic or higher order polynomial interpolation, it will not appear much different from the original bandlimited waveform. 3 Build a basis set with your frequency and match your signal. It is straightforward linear algebra: C is portion of cosine S is portion of the sine U is a vector of ones (DC)$$ X = a C + b S + c U X \cdot C = a (C \cdot C) + b (S \cdot C) + c (U \cdot C) X \cdot S = a (C \cdot S) + b (S \cdot S) + c (U \cdot S) X \cdot U = a (C \cdot ... 2 Alright, I'll try to take a shot at giving you an explanation of what's going on with the DWT. So in the CWT, basically what you are doing is you are generating the wavelet coefficients by convolving the signal with each scale and shift of the mother wavelet function. However, the problem with doing this is that when you analyze the low frequency components ... 2 It sounds like $W_k$ is just a matrix of 1's and 0's. When $W_k$ is a 1, then the corresponding range value in $D_k$ is used. If it's a 0, then that range value is not used. Similarly, $T_k$ should be mostly 1, and 0 when the corresponding $D_k$ values are unreliable. Does that tally with your understanding? If not, can you elaborate on your question ... 2 There are some mistakes in the question and your description. When we sample a signal of frequency $f_m$ with a sampling rate $f_s$, the sampled signal contains the frequencies $f=f_m \pm nf_s$, where $n \in \mathbb{Z}$. Here the frequencies available in the sampled signal are calculated as, $f_m=14100 Hz$ and $f_s=400 Hz$. we know that, $f=f_m \pm ... 2 Yet another possibility is to zero-pad the signal sample vector, FFT it, rotate the phase of each FFT result bin linearly with the bin index, and IFFT a time shifted result. 2 Because the sampling frequency is above the Nyquist rate, the original signal can be written in terms of its samples$y(n)$: $$x(t)=\sum_{m=-\infty}^{\infty}y(m)\frac{\sin[\pi(t-mT)/T]}{\pi(t-mT)/T}\tag{1}$$ By setting$t=nT+T/2$you get from (1) $$z(n)=\sum_{m=-\infty}^{\infty}y(m)\frac{\sin[\pi(n-m)+\pi/2]}{\pi(n-m)+\pi/2}$$ and since$$\sin[\pi(n-m)+\... 2 One simple approach would be taking the mean square error (MSE) by using fitness_1 = mean((inputimage(:) - reconstructedimage_1(:)).^2) though, as your image size won't change, you can ommit the mean and use sum instead. fitness_1 = sum((inputimage(:) - reconstructedimage_1(:)).^2) In general you want to have some kind of distance measure. Look at ... 2 I agree with Jim Clay's answer, but I think it is important to point out two things. First of all, there are no phase distortions due to the hold operation, just a simple delay of half a sampling interval. So nothing needs (and can) be done about the phase. Second, it is important to realize that the gain roll-off due to the sinc shape is relatively mild. ... 2 What you are describing is the distortion introduced by an ideal digital-to-analog converter (DAC) in the analog domain. Two things are typically done to reduce this distortion: Analog filtering Oversampling As you note in your question, the distortion is modeled, in the frequency domain, by a sinc rolloff. Increasing the sample rate before converting ... 2$x(t)$must be a band pass signal. Under certain conditions on the sampling frequency and its relation to the lower and upper band edges of the signal,$x(t)$can be sampled at a frequency that is lower than twice its maximum frequency, without introducing aliasing. Have a look at this article and at this answer to a related question. For the numbers given ... 2 First, since$t>0$in your case, you can write your function as $$C(t)=1.6925\left(\exp^{-0.136t}- \exp^{-1.192t} \right) u(t)$$ where$u(t)$is a unit-step function. Then, denote the FT of$C(t)$as$C(F)$, which is given by C(f)=1.6925\left(\frac{1}{0.136+2\pi jf}- \frac{1}{1.192+ 2\pi jf} \right) \end{... 2 The Poisson equation may suffer from ill-posedness if the boundary condition is not suﬃcient to yield a unique solution to the problem. The most often used boundary condition is Dirichlet boundary condition, in which the values of$I(x,y)$are specified for those points$(x,y)$on the boundary of$I\$. In your case, after you stretch your solution to another ... 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# All Questions 15 views ### Milky Way stellar number density : is the stated equation in this paper incorrect? The paper is : http://www.astro.washington.edu/users/ivezic/Publications/tomographyI.pdf The equation is equation #23 in the paper. It's a model for the density of stars in the Milky Way's disk. It ... 47 views ### How can we tell the difference between matter and antimatter by observation in space? I just was wondering and searching on the internet with little luck in the topic. On Antimatter Wiki they tell the observable universe is built up by matter. 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Algebra Power Mean Inequalities: Level 3 Challenges Given that $a_1,a_2,a_3,a_4 > 0$, find the maximum constant $N$ for which the following inequality always holds true: $\frac{a_1a_2(a_3^2+a_4^2)+a_3a_4(a_1^2+a_2^2)}{a_1a_2a_3a_4} \geq N.$ Let $a$ and $b$ be positive real numbers such that $a+b=1$. Find the maximum value of ${ a }^{ b }{ b }^{ a }+{ a }^{ a }{ b }^{ b }$. $\large \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2(a^2+b^2+c^2)}{3}$ Let $a,b$ and $c$ be positive reals satisfy $a+b+c=3$. Find the minimum value of the expression above. For some positive reals satisfying $\displaystyle{ \sum_{i=1}^{24} x_i=1}$, determine the maximum possible value of $\displaystyle{ \left (\sum_{i=1}^{24}\sqrt{x_i} \right ) \left (\sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i}} \right )}.$ There exist 3 positive numbers such that their sum is 8 and their product is 27. Is this true? ×	{}
Home > Help With > Help With Log # Help With Log ## Contents Therefore, the natural logarithm of x is defined as the inverse of the natural exponential function: $$\large ln(e^x)=e^{ln(x)}=x$$ In general, the logarithm to base b, written $$\log_b x$$, is the inverse Wird geladen... You will be able to keep track of changes to modules you are interested in using a watchlist. I never knew my words could speak! This means that there is a “duality” to the properties of logarithmic and exponential functions. Available from http://www.purplemath.com/modules/logs.htm. Should I use a Thunderbolt adapter or a USB 3.0 adapter for ethernet? Answer: 4 loga(x2+1) + ½ loga(x) Note: there is no rule for handling loga(m+n) or loga(m−n) We can also apply the logarithm rules "backwards" to combine logarithms: Example: Turn this into ## Log Conversion Calculator Wird geladen... Note that the base in both the exponential equation and the log equation (above) is "b", but that the x and y switch sides when you switch between the two equations. Acidic or Alkaline Acidity (or Alkalinity) is measured in pH: pH = −log10 [H+] where H+ is the molar concentration of dissolved hydrogen ions. • Not the answer you're looking for? • Remember that a logarithm is the inverse of an exponential. • You will not find it in your text, and your teachers and tutors will have no idea what you're talking about if you mention it to them. "The Relationship" is entirely • You will have your own user page where you can write a bit about yourself, and a user talk page which you can use to communicate with other users. • If you don't know that off the top of your head, go back and review that stuff or you're going to be one miserable puppy! • We carry Continental, Perma-chink, Sansin, and Sashco log home products, including stains and finishes, wood cleaners, chinking, caulking, and wood treatments. • The exponent of a number says how many times to use the number in a multiplication. also try the "-4" case. Because it works.) By the way: If you noticed that I switched the variables between the two boxes displaying "The Relationship", you've got a sharp eye. Specifically I can't understand this equation. Logs Maths Bitte versuche es später erneut. Schließen Ja, ich möchte sie behalten Rückgängig machen Schließen Dieses Video ist nicht verfügbar. Natural Logs All Rights Reserved. Constructive Media, LLC Advertisement Cool Math Pre-Algebra Algebra Pre-Calculus Practice Tools & ReferenceMath Dictionary Math Survival Guide Geometry & Trig Reference Puzzles Careers in Math Teacher's Success Area And, just as the base b in an exponential is always positive and not equal to 1, so also the base b for a logarithm is always positive and not equal Derivatives of Logarithms and Exponentials The derivatives of the natural logarithm and natural exponential function are quite simple. We have a wide variety of items to choose from, including log home caulk, log building tools, log home restoration products, and so much more. Log To Exponential Form Calculator On a calculator the Common Logarithm is the "log" button. Diese Funktion ist zurzeit nicht verfügbar. COOLMATH.COMAbout Us Terms of Use About Our Ads Copyright & Fair Use TOPICSPre-Algebra Lessons Algebra Lessons Pre-Calculus Lessons Math Dictionary Lines Factors and Primes Decimals Properties MORE FROM COOLMATHCoolmath Games Coolmath4Kids ## Natural Logs A Logarithm goes the other way. If you haven't logged in before, you will need to use the link provided to create an account. Log Conversion Calculator Can you solve this nurikabe puzzle? Solving Logs Remember that ln(2) is just a constant -- so we can simplify slightly: $$\large \frac{d}{dx}(\log_2x) = \frac{d}{dx}(\frac{lnx}{ln2})=\frac{d}{dx}(lnx\frac{1}{ln2})$$ Since the derivative of ln(x) is just 1/x, all we have to do is Sound Loudness is measured in Decibels (dB for short): Loudness in dB = 10 log10 (p × 1012) where p is the sound pressure. To convert, the base (that is, the 4) remains the same, but the 1024 and the 5 switch sides. Always try to use Natural Logarithms and the Natural Exponential Function whenever possible. at the library, at work, at school), please find and erase your user ID cookies after your editing session. Logs Meaning You can change this preference below. So we can check that answer: Check: 42.23 = 22.01 (close enough!) Here is another example: Example: Calculate log5 125 log5 125 = ln 125 / ln 5 = 4.83.../1.61... = If that happens and you try to save an edit, you will normally see a warning message that you are logged out. For almost three decades, we have been proudly serving homeowners and proprietors with our expansive selection of log home supplies. Retrieved from "https://www.mediawiki.org/w/index.php?title=Help:Logging_in&oldid=2370921" Category: Help Navigation menu Personal tools EnglishNot logged inTalkContributionsCreate accountLog in Namespaces Help Discussion Variants Views Read View source View history More Search Navigation Main pageGet MediaWikiGet extensionsTech Logarithms Rules Technically speaking, logs are the inverses of exponentials. Die Bewertungsfunktion ist nach Ausleihen des Videos verfügbar. ## But if you choose to give an email address, you will be able to do the following: reset your password if you forget it, receive automatic notifications of certain events, if In London UK, should I tip Uber drivers Invoking overriding method that throws Checked exception Can there be a planet with no dust? Is this use of 'chuse' a spelling mistake, a digitization error or the correct spelling for the time? Can I talk to rubber duck at work? Logarithm Examples View & Edit Cart Featured Products LifeTime Wood Treatment Log Construction Manual $37.95 The Log Home Maintenance Guide$24.95 WeatherSeal Stain Log Home Finish WR-5 Log Home Stain Cleaning, Maintaining & Example: Calculate log10 100 Well, 10 × 10 = 100, so when 10 is used 2 times in a multiplication you get 100: log10 100 = 2 Likewise log10 1,000 = History: Logarithms were very useful before calculators were invented ... Objects are shuffled. loga( m × n ) = logam + logan "the log of a multiplication is the sum of the logs" Why is that true? Browse other questions tagged logarithms radicals or ask your own question. If you did not enter an email address, or the address was out of date, you will have to create a new account under a different username. Note: in chemistry [ ] means molar concentration (moles per liter). Remember: check the privacy policy of the individual site you're visiting, if any (for example, Wikimedia's privacy policy). If you are unable to view captchas, contact an administrator. What is the probability that exactly one object remains in it's original position? Wird geladen... See Footnote. What if I forget the password or username? Using that property and the Laws of Exponents we get these useful properties: loga(m × n) = logam + logan the log of a multiplication is the sum of the logs It means that 4 with an exponent of 2.23 equals 22. Learn more You're viewing YouTube in German.	{}
## Tag Archives: newtonian potential ### Regularity of Newtonian Potential Suppose ${\Gamma(\cdot)}$ is the fundamental solution of laplace equation. $\displaystyle \Gamma(x-y)=\begin{cases}\frac{1}{n(2-n)\omega_n}\|x-y\|^{2-n}, n>2\\\frac{1}{2\pi}\log\|x-y\|, n=2.\end{cases}$ We know it has property $\displaystyle \|D_{y_i}\Gamma(x-y)\|\leq C\frac{1}{\|x-y\|^{n-1}}$ We can define the Newtonianial potential of ${f}$ $\displaystyle Nf(x)=\int_{\Omega}\Gamma(x-y)f(y)dy\quad x\in\mathbb{R}^n$ ${Nf}$ is well defined. We have the following propery of ${Nf}$ Thm: Suppose ${\Omega\subset\mathbb{R}^n}$ is a bounded domain, ${n\geq 2}$. Assume ${f}$ is bounded and local integrable in ${\Omega}$, then ${Nf}$ is ${C^1(\mathbb{R}^n)}$ and $\displaystyle D_{x_i}Nf=\int_{\Omega} D_{x_i}\Gamma(x-y)f(y)dy$ Proof: Fix ${0<\epsilon<1}$, let ${B_\epsilon(x)=\{y\in\Omega\|\|x-y\|<\epsilon\}}$ and ${B_\epsilon^c(x)=\{y\in\Omega\|\|x-y\|\geq\epsilon\}}$. $\displaystyle Nf=\int_{B_\epsilon(x)}\Gamma(x-y)f(y)dy+\int_{B^c_\epsilon(x)}\Gamma(x-y)f(y)dy=I+II$ Since ${D_{x_i}\Gamma(x-y)}$ is uniformly bounded in ${B^c_\epsilon(x)}$, by the Lebesgue dominating theorem, ${II}$ is differentiable and $\displaystyle D_{x_i}II=\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy$ Considering ${I}$, we will prove if ${\|x-z\|<\epsilon/2}$ $\displaystyle \|I(x)-I(z)\|\leq \begin{cases}C\epsilon\|x-z\|\quad\quad\text{ if } n>2\\ C\epsilon\|\log \frac{\epsilon}{2}\|\|x-z\|\text{ if }n=2\end{cases}$ where ${C=C(\|\|f\|\|_{L^\infty},n,\Omega)}$. $\displaystyle \int_{B_\epsilon(x)}\Gamma(x-y)f(y)dy-\int_{B_\epsilon(z)}\Gamma(z-y)f(y)dy=\int_{B_\epsilon(x)\cap B^c_\epsilon(z)}\Gamma(x-y)f(y)dy$ $\displaystyle +\int_{B_\epsilon(x)\cap B_\epsilon(z)}[\Gamma(x-y)-\Gamma(z-y)]f(y)dy-\int_{B^c_\epsilon(x)\cap B_\epsilon(z)}\Gamma(z-y)f(y)dy$ ${=\mathcal{A}+\mathcal{B}+\mathcal{C}}$ For any ${y\in B_\epsilon(x)\cap B^c_\epsilon(z)}$, ${\|y-x\|\geq \|y-z\|-\|z-x\|\geq \epsilon/2}$. So $\displaystyle \|\mathcal{A}\|\leq C\frac{1}{\epsilon^{n-2}} \|B_\epsilon(x)\cap B^c_\epsilon(z)\|\leq C\frac{\epsilon^n-(\epsilon-\|x-z\|/2)^n}{\epsilon^{n-2}}\leq C\epsilon \|x-z\|\text{ if } n>2$ $\displaystyle \|\mathcal{A}\|\leq C\|\log\frac{\epsilon}{2}\| \|B_\epsilon(x)\cap B^c_\epsilon(z)\|\leq C\epsilon\|\log\frac{\epsilon}{2}\|\|x-z\|\text{ if }n=2$ Similarly for ${C}$, we also have $\displaystyle \|\mathcal{C}\|\leq\begin{cases}C\epsilon\|x-z\|\quad \quad \text{ if } n>2\\ C\epsilon\|\log\frac{\epsilon}{2}\|\|x-z\|\text{ if }n=2\end{cases}$ Considering ${\mathcal{B}}$, we have $\displaystyle \|\mathcal{B}\|\leq C\int_{B_\epsilon(x)\cap B_\epsilon(z)}\int_0^1\|D_{x_i}\Gamma(tx+(1-t)z-y)\|\|x-z\|dtdy$ $\displaystyle =C\|x-z\|\int_0^1\int_{B_\epsilon(x)\cap B_\epsilon(z)}\|D_{x_i}\Gamma(tx+(1-t)z-y)\|dydt$ $\displaystyle \leq C\|x-z\|\int_0^1\int_{B_\epsilon(x)\cap B_\epsilon(z)} \frac{1}{\|tx+(1-t)z-y\|^{n-1}}dydt$ $\displaystyle \leq C\|x-z\|\int^1_0 C\epsilon dt\leq C\epsilon \|x-z\|$ Combing the fact about ${\mathcal{A},\mathcal{B},\mathcal{C}}$, we get $\displaystyle \|I(x)-I(z)\|\leq \begin{cases}C\epsilon\|x-z\|\quad\quad\text{ if } n>2\\ C\epsilon\|\log \frac{\epsilon}{2}\|\|x-z\|\text{ if }n=2\end{cases}$ So $\displaystyle \left\|\frac{Nf(x)-Nf(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right\|$ $\displaystyle \leq \left\|\frac{II(x)-II(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right\|+\left\|\frac{I(x)-I(z)}{x-z}\right\|$ Applying the fact that ${II}$ is differentiable and the fact about ${I}$ $\displaystyle \overline{\lim\limits_{z\rightarrow x}}\left\|\frac{Nf(x)-Nf(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right\|$ $\displaystyle \leq \left\|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right\|+\begin{cases}C\epsilon\quad\quad\text{ if } n>2\\ C\epsilon\|\log \frac{\epsilon}{2}\|\text{ if }n=2\end{cases}$ $\displaystyle \leq C\epsilon+\begin{cases}C\epsilon\quad\quad\text{ if } n>2\\ C\epsilon\|\log \frac{\epsilon}{2}\|\text{ if }n=2\end{cases}$ Let ${\epsilon\rightarrow 0}$, we get ${D_{x_i}Nf}$ exists and $\displaystyle D_{x_i}Nf=\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy$ Next we shall prove ${D_{x_i}Nf}$ is continuous for ${i=1,\cdots,n}$, then ${Nf\in C^1(\mathbb{R}^n)}$ $\displaystyle \left\|\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy-\int_{\Omega}D_{x_i}\Gamma(z-y)f(y)dy\right\|$ $\displaystyle \leq\left\|\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy-\int_{B^c_{\epsilon}(z)}D_{x_i}\Gamma(z-y)f(y)dy\right\|$ $\displaystyle +\left\|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right\|+\left\|\int_{B_\epsilon(z)}D_{x_i}\Gamma(z-y)f(y)dy\right\|$ By the dominating theorem $\displaystyle \lim\limits_{x\rightarrow z}\left\|\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy-\int_{B^c_{\epsilon}(z)}D_{x_i}\Gamma(z-y)f(y)dy\right\|=0$ $\displaystyle \left\|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right\|\leq C\epsilon$ $\displaystyle \left\|\int_{B_\epsilon(z)}D_{x_i}\Gamma(z-y)f(y)dy\right\|\leq C\epsilon$ where ${C=C(\|\|f\|\|_{L^\infty}, n)}$. So ${\displaystyle \int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy=D_{x_i}Nf}$ is continuous everywhere. Remark: This revised version also have errors. It is a shame. What do you mean ${II}$ is differentiable? Don’t you notice that the integral domain of ${II}$ also has relation with ${x}$? To prove this theorem rigorously and neatly, we need the following lemma from calculus Lemma: Suppose ${u_n{x}}$ is a sequence of differntiable function on ${[a,b]\rightarrow \mathbb{R}}$. If ${u_n}$ converges to another function ${u}$ uniformly(${u}$ is finite somewhere) and ${u'_n}$ converges uniformly to ${v}$ in ${\Omega}$ then ${u}$ is differentiable and ${u'=v}$. Proof: Let ${\xi(r):C^1(\mathbb{R}_+^1)\rightarrow \mathbb{R}^1}$ and ${0\leq\eta'\leq 2}$, $\xi(r)=0\text{ if } r\leq 1$, $\xi(r)=1\text{ if }r>2$. Define ${\displaystyle w_\epsilon=\int_{\Omega}\xi\left(\frac{\|x-y\|}{\epsilon}\right)\Gamma(x-y)f(y)dy}$ Let ${\xi_\epsilon(r)=\xi_{\epsilon}(r/\epsilon)}$ Since $\displaystyle \left\|D_{x_i}\left(\xi_\epsilon\left(\|x-y\|\right)\Gamma(x-y)f(y)\right)\right\|\leq \frac{C}{\epsilon^{n-1}}f(y)$ which is in ${L^1(\Omega)}$. By the Lebesgue Differentiable theorem, ${w_\epsilon}$ is differntiable and $\displaystyle D_{x_i}w_\epsilon=\int_{\Omega}D_{x_i}\left(\xi_\epsilon\left(\|x-y\|\right)\Gamma(x-y)\right)f(y)dy$ $\displaystyle \left\|D_{x_i}w_\epsilon-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right\|\leq \sup\|f\|\int_{\Omega}\left\|D_{x_i}((1-\xi_\epsilon)\Gamma(x-y))\right\|dy$ $\displaystyle \leq \sup\|f\|\int_{\|x-y\|\leq 2\epsilon}\|D_{x_i}\Gamma\|+\frac{2}{\epsilon}\|\Gamma\|dy$ $\displaystyle \leq \sup\|f\|\begin{cases}C\epsilon\quad \quad \text{ if }n>2\\ C\epsilon (1+\|\log\epsilon\|)\text{ if }n=2\end{cases}$ So ${D_{x_i}w_\epsilon}$ converges to ${\displaystyle \int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy}$ uniformly on any compact subset of ${\mathbb{R}^n}$. And it is easy to prove ${w_\epsilon}$ converges to ${Nf}$ uniformly on any compact subset of ${\mathbb{R}^n}$. So by the lemma, we have ${Nf}$ is differentiable and has $\displaystyle D_{x_i}Nf(x)=\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy,\quad i=1,2\cdots,n$ ### L^p estimate of lapacian operator Refering to GT’s book corollary 9.10 $\textbf{Thm:}$ $\Omega$ is a bounded domain in $\mathbb{R}^n$. If $u\in W^{2,p}(\Omega)$ and vanish on the boundary, then (1) $\|\|D^2u\|\|_p\leq \|\|\Delta u\|\|_p$ , $1. where $C=C(n,p)$. If $p=2$ (2) $\|\|D^2u\|\|_2= \|\|\Delta u\|\|_2$ The proof of this theorem is using the following fact, if $u\in C^2(\mathbb{R}^n)$ has compact support, then $\displaystyle u(x)=\int_{\mathbb{R}^n}\Gamma(x-y)\Delta u(y)dy$ $\Gamma(x)$ is the fundamental solution of $\Delta$(Give a second thought why the RHS has compact support). This is exactly the Newtonian potential. By the result of Thm 9.9 $\|\|D^2u\|\|_p\leq C\|\|\Delta u\|\|_p$ is true for $u\in C^2_0(\Omega)$. Since $C^2_0(\Omega)$ is dense in $W^{2,p}(\Omega)$ with zero boundary, then 9.10 is true. ### Counterexample of regularity of newtonian potential Define the Newtonial potential in a bounded domain $\Omega$ by $\displaystyle N(x)=\int_{\Omega}\Gamma(x-y)f(y)dy$ If $f$ is bounded and integrable on $\Omega$, then $N\in C^1(\mathbb{R}^n)$. And if $f$ is bounded and locally holder continuous in $\Omega$ then $N\in C^2(\Omega)$ and $\Delta N=f$. But when $f$ is only continuous, $N$ is not necessarily secondly differentiable. Here is a counterexample. $\mathbf{Problem:}$ Let $P$ be a homogeneous harmonic polynomial of degree 2. Suppose $D^\alpha P\neq 0$ for some multi-index $\alpha=2$, for instance $P=x_1x_2, D_{12}P\neq 0$. Choose a cut-off function $\eta\in C^\infty_0(\{x\|\|x\|<2\})$ with $\eta=1$ when $\|x\|<1$. Denote $t_k=2^k$, and let $c_k\to 0$ as $k\to \infty$ and $\sum c_k$ divergent. Define $\displaystyle f(x)=\sum\limits_0^\infty c_k\left(\Delta(\eta P)\right)(t_kx)$ Prove $f$ is a continuous function but $\Delta u=f$ has no $C^2$ solution near the origin. $\mathbf{Proof:}$ If $x=0$, then $f=0$. If $x\neq 0$, then there exists only one $k_0$ such that $1\leq \|t_{k_0}x\|\leq 2$. Since $P$ is a harmonic polynomial and $\eta$ has compact support, $f(x)=c_{k_0}\left(\Delta(\eta P)\right)(t_{k_0}x)\leq Const.c_{k_0}$ So $f\in C^\infty(\mathbb{R}^n\backslash \{0\})$. As $x\to 0$, $c_k\to 0$, so $f(x)\to 0$, which means $f$ is continuous at 0. Let $\displaystyle w(x)=\sum\limits_{0}^\infty t^{-2}_kc_k(\eta P)(t_kx)$, then it is easy to prove $w$ is well defined and for $x\neq 0$ and the unique $k_0$ such that $1\leq \|t_{k_0}x\|\leq 2$, $\displaystyle D^\alpha w(x)=\sum\limits_0^{k_0}c_k+c_{k_0}D^\alpha(\eta P)(t_{k_0}x)\quad (1)$ $\Delta w(x)=c_{k_0}\left(\Delta(\eta P)\right)(t_{k_0}x)=f(x) \quad (2)$ (1) means $w\in C^2(\mathbb{R}^n\backslash\{0\})$ but $w\not\in C^2(\mathbb{R}^n)$, otherwise $D^\alpha w(0)=\sum c_k$ which does not exist. Suppose there exists $u$ is a $C^2$ solution at $B_{\epsilon}(0)$, (2) means that $\Delta (u-w)=0$ on $B_{\epsilon}\backslash\{0\}$. Since $u-w$ is bounded, by the removable singularity theorem, $u-w$ is a harmonic function in $B_\epsilon (0)$, thus an analytic function. However this means $w=u-(u-w)$ is a $C^2$ function on $B_\epsilon (0)$. Contradiction. $\text{Q.E.D}\hfill \square$ $\mathbf{Remark:}$ Gilbarg Trudinger’s book. Chapter 4. Exercise 4.9 $\mathbf{Erratum:}$ $u-w$ is not bounded in general, because $w$ can not be proved bounded directly. But we have $w(x)=o(\log r)$. In fact suppose $2^{-l}\leq\|x\|< 2^{-l+1}$, some $l\geq 1$, then $\displaystyle w(x)=\sum_{k=0}^{l-1} c_k+c_l\frac{(\eta P)(t_lx)}{t_l^2}=\sum_{k=0}^{l-1} c_k+c_l\eta (t_lx)P(x)$ Because $c_k\to 0$, $\displaystyle \left\|\frac{w(x)}{\log \|x\|}\right\|\leq \frac{\|\sum_{k=0}^{l-1} c_k+c_l\eta (t_lx)P(x)\|}{l\log 2}\to 0\text{ as }l\to \infty$ So $u-w=o(\log r)$ as $r\to 0$. Then one can use the Removable singularity theorem(Bochner), or the conclusion of 3.7 in GT’s book.	{}
# 4.3 Finding the inverse laplace transform Page 1 / 3 ## Using transform tables The inverse Laplace transform, given by $x\left(t\right)=\frac{1}{2\pi j}{\int }_{\sigma -j\infty }^{\sigma +j\infty }X\left(s\right){e}^{st}ds$ can be found by directly evaluating the above integral. However since this requires a background in the theory of complex variables, which is beyond the scope of this book, we will not be directly evaluating the inverse Laplace transform. Instead, we will utilize the Laplace transform pairs and properties . Consider the following examples: Example 3.1 Find the inverse Laplace transform of $X\left(s\right)=\frac{{e}^{-10s}}{s+5}$ By looking at the table of Laplace transform properties we find that multiplication by ${e}^{-10s}$ corresponds to a time delay of 10 sec. Then from the table of Laplace transform pairs , we see that $\frac{1}{s+5}$ corresponds to the Laplace transform of the exponential signal ${e}^{-5t}u\left(t\right)$ . Therefore we must have $x\left(t\right)={e}^{-5\left(t-10\right)}u\left(t-10\right)$ Example 3.2 Find the inverse Laplace transform of $X\left(s\right)=\frac{1}{{\left(s+2\right)}^{2}}$ First we note that from the table of Laplace transform pairs , the Laplace transform of $tu\left(t\right)$ is $\frac{1}{{s}^{2}}$ Then using the $s$ -shift property in the table of Laplace transform properties gives $x\left(t\right)=t{e}^{-2t}u\left(t\right)$ Also, the same answer may be arrived at by combining the Laplace transform of ${e}^{-2t}u\left(t\right)$ with the multiplication by $t$ property. ## Partial fraction expansions Partial fraction expansions are useful when we can express the Laplace transform in the form of a rational function , $\begin{array}{cc}\hfill X\left(s\right)& =\frac{{b}_{q}{s}^{q}+{b}_{q-1}{s}^{q-1}+\cdots +{b}_{1}s+{b}_{0}}{{a}_{p}{s}^{p}+{a}_{p-1}{s}^{p-1}+\cdots +{a}_{1}s+{a}_{0}}\hfill \\ & =\frac{B\left(s\right)}{A\left(s\right)}\hfill \end{array}$ A rational function is a ratio of two polynomials. The numerator polynomial $B\left(s\right)$ has order $q$ , i.e., the largest power of $s$ in this polynomial is $q$ , while the denominator polynomial has order $p$ . The partial fraction expansion also requires that the Laplace transform be a proper rational function, which means that $q . Since $B\left(s\right)$ and $A\left(s\right)$ can be factored, we can write $X\left(s\right)=\frac{\left(s-{\beta }_{1}\right)\left(s-{\beta }_{2}\right)\cdots \left(s-{\beta }_{q}\right)}{\left(s-{\alpha }_{1}\right)\left(s-{\alpha }_{2}\right)\cdots \left(s-{\alpha }_{p}\right)}$ The ${\beta }_{i},i=1,2,...,q$ are the roots of $B\left(s\right)$ , and are called the zeros of $X\left(s\right)$ . The roots of $A\left(s\right)$ , are ${\alpha }_{i},i=1,...,p$ and are called the poles of $X\left(s\right)$ . If we evaluate $X\left(s\right)$ at one of the zeros we get $X\left({\beta }_{i}\right)=0,i=1,...,q$ . Similarly, evaluating $X\left(s\right)$ at a pole gives The actual sign would need to be evaluated at some value of $s$ that is sufficiently close to the pole. $X\left({\alpha }_{i}\right)=±\infty ,i=1,...,p$ . The partial fraction expansion of a Laplace transform will usually involve relatively simple terms whose inverse Laplace transforms can be easily determined from a table of Laplace transforms. We must consider several different cases which depend on whether the poles are distinct. ## Distinct Poles: When all of the poles are distinct (i.e. ${\alpha }_{i}\ne {\alpha }_{j},i\ne j$ ) then we can use the following partial fraction expansion: $X\left(s\right)=\frac{{A}_{1}}{s-{\alpha }_{1}}+\frac{{A}_{2}}{s-{\alpha }_{2}}+\cdots +\frac{{A}_{p}}{s-{\alpha }_{p}}$ The coefficients, ${A}_{i},i=1,...,p$ can then be found using the following formula ${A}_{i}={\left(X,\left(s\right),\left(s-{\alpha }_{i}\right)\|}_{s={\alpha }_{i}},i=1,...,p$ Equation [link] is easily derived by clearing fractions in [link] . The inverse Fourier transform of $X\left(s\right)$ can then be easily found since each of the terms in the right-hand side of [link] is the Laplace transform of an exponential signal. This method is called the cover up method . Example 3.3 Find the inverse Laplace transform of $\begin{array}{cc}\hfill X\left(s\right)& =\frac{2s-10}{{s}^{2}+3s+2}\hfill \\ & =\frac{2s-10}{\left(s+1\right)\left(s+2\right)}\hfill \end{array}$ Application of nanotechnology in medicine what is variations in raman spectra for nanomaterials I only see partial conversation and what's the question here! what about nanotechnology for water purification please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe anyone know any internet site where one can find nanotechnology papers? research.net kanaga sciencedirect big data base Ernesto Introduction about quantum dots in nanotechnology hi Loga what does nano mean? nano basically means 10^(-9). nanometer is a unit to measure length. Bharti how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Got questions? Join the online conversation and get instant answers!	{}
# Heritability: Wikis Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles. # Encyclopedia Heritability is the proportion of phenotypic variation in a population that is attributable to genetic variation among individuals. Phenotypic variation among individuals may be due to genetic and/or environmental factors. Heritability analyses estimate the relative contributions of differences in genetic and non-genetic factors to the total phenotypic variance in a population. ## Definition Figure 1. Relationship of phenotypic values to additive and dominance effects using a completely dominant locus. Consider a statistical model for describing some particular phenotype:[1] Phenotype (P) = Genotype (G) + Environment (E). Considering variances (Var), this becomes: Var(P) = Var(G) + Var(E) + 2 Cov(G,E). In planned experiments, we can often take Cov(G,E) = 0. Heritability is then defined as: $H^2 = \frac{Var(G)}{Var(P)}$ . The parameter H2 is the broad-sense heritability and reflects all possible genetic contributions to a population's phenotypic variance. Included are effects due to allelic variation (additive variance), dominance variation, epistatic (multi-genic) interactions, and maternal and paternal effects, where individuals are directly affected by their parents' phenotype (such as with milk production in mammals). These additional terms can be included in genetic models. For example, the simplest genetic model involves a single locus with two alleles that affect some quantitative phenotype, as shown by + in Figure 1. We can calculate the linear regression of phenotype on the number of B alleles (0, 1, or 2), which is shown as the Linear Effect line. For any genotype, BiBj, the expected phenotype can then be written as the sum of the overall mean, a linear effect, and a dominance deviation: Pij = μ + αi + αj + dij = Population mean + Additive Effect (aij = αi + αj) + Dominance Deviation (dij). The additive genetic variance is the weighted average of the squares of the additive effects: $Var(A) = f(bb)a^2_{bb}+f(Bb)a^2_{Bb}+f(BB)a^2_{BB},$ where f(bb)ab b + f(Bb)aBb + f(BB)aBB = 0. There is a similar relationship for variance of dominance deviations: $Var(D) = f(bb)d^2_{bb}+f(Bb)d^2_{Bb}+f(BB)d^2_{BB},$ where f(bb)db b + f(Bb)dBb + f(BB)dBB = 0. Narrow-sense heritability is defined as $h^2 = \frac{Var(A)}{Var(P)}$ and quantifies only the portion of the phenotypic variation that is additive (allelic) by nature (note upper case H2 for broad sense, lower case h2 for narrow sense). When interested in improving livestock via artificial selection, for example, knowing the narrow-sense heritability of the trait of interest will allow predicting how much the mean of the trait will increase in the next generation as a function of how much the mean of the selected parents differs from the mean of the population from which the selected parents were chosen. The observed response to selection leads to an estimate of the narrow-sense heritability (called realized heritability). ## Estimating heritability Estimating heritability is not a simple process, since only P can be observed or measured directly. Measuring the genetic and environmental variance requires various sophisticated statistical methods. These methods give better estimates when using data from closely related individuals - such as brothers, sisters, parents and offspring, rather than from more distantly related ones. The standard error for heritability estimates are generally very poor unless the dataset is large. Figure 2. Heritability for nine psychological traits as estimated from twin studies. All sources are twins raised together (sample size shown inside bars). MZ: Monozygotic twins, DZ: Dizygotic twins In non-human populations it is often possible to collect information in a controlled way. For example, among farm animals it is easy to arrange for a bull to produce offspring from a large number of cows. Due to ethical concerns, such a degree of experimental control is impossible when gathering human data. As a result, studies of human heritability sometimes contrast identical twins who have been separated early in life and raised in different environments (see for example Fig. 2). Such individuals have identical genotypes and can be used to separate the effects of genotype and environment. Twin studies entail problems of their own, such as: independently raised twins shared a common prenatal environment; they may have undergone intrauterine competition; the mother may be more physically stressed (less nutrients); and twins reared apart are difficult to find, and may reflect certain types of environments. Heritability estimates are always relative to the genetic and environmental factors in the population, and are not absolute measurements of the contribution of genetic and environmental factors to a phenotype. Heritability estimates reflect the amount of variation in genotypic effects compared to variation in environmental effects. Heritability can be made larger by diversifying the genetic background, e.g., by using only very outbred individuals (which increases the Variance(G)) and/or by minimizing environmental effects (which decreases the Variance(E)). Smaller heritability, on the other hand, can be generated by using inbred individuals (which decreases the Variance(G)) or individuals reared in very diverse environments (which increases the Variance(E)). Due to such effects, different populations of a species might have different heritabilities even for the same trait. In observational studies G and E may be correlated, giving rise to gene environment correlation. Depending on the methods used to estimate heritability, correlations between genetic factors and shared or non-shared environments may or may not be included in the total heritability estimate. [2] Because of the contextual nature of measured heritabilities, paradoxes often arise. For example, the heritability of a trait could be near 100% in one study and close to zero in another. In one study, e.g., a group of unrelated army recruits may be given identical training and nutrition and then their muscular strength may be measured. The variation in strength observed after the (identical) training will translate into a high heritability estimate. In another study, whose purpose might be to assess the efficacy of various workout regimes or nutritional programs, study subjects may be first chosen to match each other as closely as possible in prior physical characteristics before some of them are put onto Program A and others onto Program B, and this will lead to a low heritability estimate. Heritability estimates are often misinterpreted. Heritability refers to the proportion of variation between individuals in a population that is influenced by genetic factors. Heritability describes the population, not individuals within that population. For example, It is incorrect to say that since the heritability of a personality trait is about .6, that means that 60% of your personality is inherited from your parents and 40% comes from the environment. The heritability estimate changes according to the genetic and environmental variability present in the population. In studies of genetically identical inbred animals, all traits have zero heritability. Heritability estimates can be much higher in outbred (genetically variable) populations under very homogeneous environments. A highly genetically loaded trait (such as eye color) still assumes environmental input within normal limits (a certain range of temperature, oxygen in the atmosphere, etc.). A more useful distinction than "nature vs. nurture" is "obligate vs. facultative" -- under typical environmental ranges, what traits are more "obligate" (e.g., the nose -- everyone has a nose) or more "facultative" (sensitive to environmental variations, such as specific language learned during infancy). Another useful distinction is between traits that are likely to be adaptations (such as the nose) vs. those that are byproducts of adaptations (such the white color of bones), or are due to random variation (non-adaptive variation in, say, nose shape or size). ## Estimation methods There are essentially two schools of thought regarding estimation of heritability. One school of thought was developed by Sewall Wright at The University of Chicago, and further popularized by C. C. Li (University of Chicago) and J. L. Lush (Iowa State University). It is based on the analysis of correlations and, by extension, regression. Path Analysis was developed by Sewall Wright as a way of estimating heritability. The second was originally developed by R. A. Fisher and expanded at The University of Edinburgh, Iowa State University, and North Carolina State University, as well as other schools. It is based on the analysis of variance of breeding studies, using the intraclass correlation of relatives. Various methods of estimating components of variance (and, hence, heritability) from ANOVA are used in these analyses. ## Regression/correlation methods of estimation The first school of estimation uses regression and correlation to estimate heritability. ### Selection experiments Figure 3. Strength of selection (S) and response to selection (R) in an artificial selection experiment, h2=R/S. Calculating the strength of selection, S (the difference in mean trait between the population as a whole and the selected parents of the next generation, also called the selection differential [3]) and response to selection R (the difference in offspring and whole parental generation mean trait) in an artificial selection experiment will allow calculation of realized heritability as the response to selection relative to the strength of selection, h2=R/S as in Fig. 3. ### Comparison of close relatives In the comparison of relatives, we find that in general, $h^2 = \frac{b}{r} = \frac{t}{r}$ where r can be thought of as the coefficient of relatedness, b is the coefficient of regression and t the coefficient of correlation. #### Parent-offspring regression Figure 4. Sir Francis Galton's (1889) data showing the relationship between offspring height (928 individuals) as a function of mean parent height (205 sets of parents). Heritability may be estimated by comparing parent and offspring traits (as In Fig. 4). The slope of the line (0.57) approximates the heritability of the trait when offspring values are regressed against the average trait in the parents. If only one parent's value is used then heritability is twice the slope. (note that this is the source of the term "regression", since the offspring values always tend to regress to the mean value for the population, i.e., the slope is always less than one). #### Full-sib comparison Full-sib designs compare phenotypic traits of siblings that share a mother and a father with other sibling groups. The estimate of the sibling phenotypic correlation is an index on familiality which is equal to half the additive genetic variance plus the common environment variance when there is only additive gene action. #### Half-sib comparison Half-sib designs compare phenotypic traits of siblings that share one parent with other sibling groups. #### Twin studies Figure 5. Twin concordances for seven psychological traits (sample size shown inside bars). Heritability for traits in humans is most frequently estimated by comparing resemblances between twins (Fig. 2 & 5). Identical twins (MZ twins) are twice as genetically similar as fraternal twins (DZ twins) and so heritability is approximately twice the difference in correlation between MZ and DZ twins, h2=2(r(MZ)-r(DZ)). The effect of shared environment, c2, contributes to similarity between siblings due to the commonality of the environment they are raised in. Shared environment is approximated by the DZ correlation minus half heritability, which is the degree to which DZ twins share the same genes, c2=DZ-1/2h2. Unique environmental variance, e2, reflects the degree to which identical twins raised together are dissimilar, e2=1-r(MZ). The methodology of the classical twin study has been criticized, but these criticisms do not take into account the methodological innovations and refinements described above. ## Analysis of variance methods of estimation The second set of methods of estimation of heritability involves ANOVA and estimation of variance components. ### Basic model We use the basic discussion of Kempthorne (1957 [1969]). Considering only the most basic of genetic models, we can look at the quantitative contribution of a single locus with genotype Gi as yi = μ + gi + e where gi is the effect of genotype Gi and e is the environmental effect. Consider an experiment with a group of sires and their progeny from random dams. Since the progeny get half of their genes from the father and half from their (random) mother, the progeny equation is $z_i = \mu + \frac{1}{2}g_i + e$ #### Intraclass correlations Consider the experiment above. We have two groups of progeny we can compare. The first is comparing the various progeny for an individual sire (called within sire group). The variance will include terms for genetic variance (since they did not all get the same genotype) and environmental variance. This is thought of as an error term. The second group of progeny are comparisons of means of half sibs with each other (called among sire group). In addition to the error term as in the within sire groups, we have an addition term due to the differences among different means of half sibs. The intraclass correlation is $corr(z,z') = corr(\mu + \frac{1}{2}g + e, \mu + \frac{1}{2}g + e') = \frac{1}{4}V_g$ , since environmental effects are independent of each other. #### The ANOVA In an experiment with n sires and r progeny per sire, we can calculate the following ANOVA, using Vg as the genetic variance and Ve as the environmental variance: Table 1: ANOVA for Sire experiment Source d.f. Mean Square Expected Mean Square Among sire groups n − 1 S $\frac{3}{4}V_g + V_e + r({\frac{1}{4}V_g})$ Within sire groups n(r − 1) W $\frac{3}{4}V_g + V_e$ The $\frac{1}{4}V_g$ term is the intraclass correlation among half sibs. We can easily calculate $H^2 = \frac{V_g}{V_g+V_e} = \frac{4(S-W)}{S+(r-1)W}$. The Expected Mean Square is calculated from the relationship of the individuals (progeny within a sire are all half-sibs, for example), and an understanding of intraclass correlations. ### Model with additive and dominance terms For a model with additive and dominance terms, but not others, the equation for a single locus is yij = μ + αi + αj + dij + e, where αi is the additive effect of the ith allele, αj is the additive effect of the jth allele, dij is the dominance deviation for the ijth genotype, and e is the environment. Experiments can be run with a similar setup to the one given in Table 1. Using different relationship groups, we can evaluate different intraclass correlations. Using Va as the additive genetic variance and Vd as the dominance deviation variance, intraclass correlations become linear functions of these parameters. In general, Intraclass correlation = rVa + θVd, where r and θ are found as r = P[ alleles drawn at random from the relationship pair are identical by descent], and θ = P[ genotypes drawn at random from the relationship pair are identical by descent]. Some common relationships and their coefficients are given in Table 2. Table 2: Coeffients for calculating variance components Relationship r θ Identical Twins 1 1 Parent-Offspring $\frac{1}{2}$ 0 Half Siblings $\frac{1}{4}$ 0 Full Siblings $\frac{1}{2}$ $\frac{1}{4}$ First Cousins $\frac{1}{8}$ 0 Double First Cousins $\frac{1}{4}$ $\frac{1}{16}$ ### Larger models When a large, complex pedigree is available for estimating heritability, the most efficient use of the data is in a restricted maximum likelihood (REML) model. The raw data will usually have three or more datapoints for each individual: a code for the sire, a code for the dam and one or several trait values. Different trait values may be for different traits or for different timepoints of measurement. The currently popular methodology relies on high degrees of certainty over the identities of the sire and dam; it is not common to treat the sire identity probabilistically. This is not usually a problem, since the methodology is rarely applied to wild populations (although it has been used for several wild ungulate and bird populations), and sires are invariably known with a very high degree of certainty in breeding programmes. There are also algorithms that account for uncertain paternity. The pedigrees can be viewed using programs such as Pedigree Viewer [1], and analysed with programs such as ASReml, VCE [2], WOMBAT [3] or BLUPF90 family's programs [4] ## Response to Selection In selective breeding of plants and animals, the expected response to selection can be estimated by the following equation:[4] R = h2S In this equation, the Response to Selection (R) is defined as the realized average difference between the parent generation and the next generation. The Selection Differential (S) is defined as the average difference between the parent generation and the selected parents. For example, imagine that a plant breeder is involved in a selective breeding project with the aim of increasing the number of kernels per ear of corn. For the sake of argument, let us assume that the average ear of corn in the parent generation has 100 kernels. Let us also assume that the selected parents produce corn with an average of 120 kernels per ear. If h2 equals 0.5, then the next generation will produce corn with an average of 0.5(120-100) = 10 additional kernels per ear. Therefore, the total number of kernels per ear of corn will equal, on average, 110. ## References ### Notes 1. ^ The presentation here roughly follows Kempthorne (1957) 2. ^ Cattell RB (1960). "The multiple abstract variance analysis equations and solutions: for nature–nurture research on continuous variables". Psychol Rev 67: 353–372. doi:10.1037/h0043487. 3. ^ Kempthorne (1957), page 507; or Falconer (1960), page 191, for example. 4. ^ Plomin, R., DeFries, J. C., & McClearn, G. E. (1990). Behavioral genetics. New York: Freeman. • Falconer, D. S. & Mackay TFC (1996). Introduction to Quantitative Genetics. Fourth edition. Addison Wesley Longman, Harlow, Essex, UK • Gillespie, G. H. (1997). Population Genetics: A Concise Guide. Johns Hopkins University Press. • Joseph, J. (2004). The Gene Illusion: Genetic Research in Psychiatry and Psychology Under the Microscope.New York: Algora. (2003 United Kingdom Edition by PCCS Books) (Chapter 5 contains a critique of the heritability concept) • Joseph, J. (2006). The Missing Gene: Psychiatry, Heredity, and the Fruitless Search for Genes.New York: Algora. • Kempthorne, O (1957 [1969]) An Introduction to Genetic Statistics. John Wiley. Reprinted, 1969 by Iowa State University Press. • Lynch, M. & Walsh, B. 1997. Genetics and Analysis of Quantitative Traits. Sinauer Associates. ISBN 0-87893-481-2. • Malécot, G. 1948. Les Mathématiques de l'Hérédité. Masson, Paris. • Wahlsten, D. (1994) The intelligence of heritability. Canadian Psychology 35, 244-258.	{}
If you find any mistakes, please make a comment! Thank you. The set of prime ideals of a commutative ring contains inclusion-minimal elements Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.36 Solution: Let $\mathcal{P}$ denote the set of prime ideals. Note that $\mathcal{P}$ is partially ordered by…	{}
# Are there any solvents that dissolve talcum powder? I understand that talcum powder is insoluble to water, would alcohol or something be able to dissolve it? Talcum powder is made from talc, a silicate mineral with formula unit: $\ce{Mg3Si4O10(OH)2}$. Silicates tend to be pretty much insoluble in water, but it is possible to dissolve them in strong acids. I don't know about talc specifically, but many silicates can be dissolved in concentrated nitric acid. Hydrofluoric acid is also commonly used for dissolving minerals for chemical analysis. Both are very very dangerous substances that really shouldn't be used without proper training and safety equipment. I don't think there's really anything accessible for a home experimenter, unfortunately. • Well it won't be talc anymore. Dissolving anything changes its properties to some extent. Talc is only talc because it has the composition it does in the atomic arrangement it does. In order to dissolve, the talc must lose its solid structure. In the case of using an acid, you're right that an oxidation reaction does sometimes cause dissolution, especially with things like elemental metals. When used for digestion, HF tends to pull out Si as $\ce{SiF4}$. In acidic conditions, I found a reference to this dissolution reaction for talc: $\ce{Mg3Si4O10(OH)2 +6H+ + 4H2O -> 3Mg^{2+} + 4H4SiO4}$ – Michael DM Dryden Sep 10 '14 at 5:32	{}
# Recursive neural network A recursive neural network is a kind of deep neural network created by applying the same set of weights recursively over a structured input, to produce a structured prediction over variable-size input structures, or a scalar prediction on it, by traversing a given structure in topological order. Recursive neural networks, sometimes abbreviated as RvNNs, have been successful, for instance, in learning sequence and tree structures in natural language processing, mainly phrase and sentence continuous representations based on word embedding. RvNNs have first been introduced to learn distributed representations of structure, such as logical terms.[1] Models and general frameworks have been developed in further works since the 1990s.[2][3] ## Architectures ### Basic A simple recursive neural network architecture In the most simple architecture, nodes are combined into parents using a weight matrix that is shared across the whole network, and a non-linearity such as tanh. If c1 and c2 are n-dimensional vector representation of nodes, their parent will also be an n-dimensional vector, calculated as ${\displaystyle p_{1,2}=\tanh \left(W[c_{1};c_{2}]\right)}$ Where W is a learned ${\displaystyle n\times 2n}$ weight matrix. This architecture, with a few improvements, has been used for successfully parsing natural scenes, syntactic parsing of natural language sentences,[4] and recursive autoencoding and generative modeling of 3D shape structures in the form of cuboid abstractions.[5] RecCC is a constructive neural network approach to deal with tree domains[2] with pioneering applications to chemistry[6] and extension to directed acyclic graphs.[7] ### Unsupervised RNN A framework for unsupervised RNN has been introduced in 2004.[8][9] ### Tensor Recursive neural tensor networks use one, tensor-based composition function for all nodes in the tree.[10] ## Training Typically, stochastic gradient descent (SGD) is used to train the network. The gradient is computed using backpropagation through structure (BPTS), a variant of backpropagation through time used for recurrent neural networks. ## Properties Universal approximation capability of RNN over trees has been proved in literature.[11][12] ## Related models ### Recurrent neural networks Recurrent neural networks are recursive artificial neural networks with a certain structure: that of a linear chain. Whereas recursive neural networks operate on any hierarchical structure, combining child representations into parent representations, recurrent neural networks operate on the linear progression of time, combining the previous time step and a hidden representation into the representation for the current time step. ### Tree Echo State Networks An efficient approach to implement recursive neural networks is given by the Tree Echo State Network[13] within the reservoir computing paradigm. ### Extension to graphs Extensions to graphs include graph neural network (GNN),[14] Neural Network for Graphs (NN4G),[15] and more recently convolutional neural networks for graphs. ## References 1. ^ Goller, C.; Küchler, A. (1996). "Learning task-dependent distributed representations by backpropagation through structure". Proceedings of International Conference on Neural Networks (ICNN'96). Vol. 1. pp. 347–352. CiteSeerX 10.1.1.52.4759. doi:10.1109/ICNN.1996.548916. ISBN 978-0-7803-3210-2. S2CID 6536466. 2. ^ a b Sperduti, A.; Starita, A. (1997-05-01). "Supervised neural networks for the classification of structures". IEEE Transactions on Neural Networks. 8 (3): 714–735. doi:10.1109/72.572108. ISSN 1045-9227. PMID 18255672. 3. ^ Frasconi, P.; Gori, M.; Sperduti, A. (1998-09-01). "A general framework for adaptive processing of data structures". IEEE Transactions on Neural Networks. 9 (5): 768–786. CiteSeerX 10.1.1.64.2580. doi:10.1109/72.712151. ISSN 1045-9227. PMID 18255765. 4. ^ Socher, Richard; Lin, Cliff; Ng, Andrew Y.; Manning, Christopher D. "Parsing Natural Scenes and Natural Language with Recursive Neural Networks" (PDF). The 28th International Conference on Machine Learning (ICML 2011). 5. ^ Li, Jun; Xu, Kai; Chaudhuri, Siddhartha; Yumer, Ersin; Zhang, Hao; Guibas, Leonadis (2017). "GRASS: Generative Recursive Autoencoders for Shape Structures" (PDF). ACM Transactions on Graphics. 36 (4): 52. arXiv:1705.02090. doi:10.1145/3072959.3073613. S2CID 20432407. 6. ^ Bianucci, Anna Maria; Micheli, Alessio; Sperduti, Alessandro; Starita, Antonina (2000). "Application of Cascade Correlation Networks for Structures to Chemistry". Applied Intelligence. 12 (1–2): 117–147. doi:10.1023/A:1008368105614. ISSN 0924-669X. S2CID 10031212. 7. ^ Micheli, A.; Sona, D.; Sperduti, A. (2004-11-01). "Contextual processing of structured data by recursive cascade correlation". IEEE Transactions on Neural Networks. 15 (6): 1396–1410. CiteSeerX 10.1.1.135.8772. doi:10.1109/TNN.2004.837783. ISSN 1045-9227. PMID 15565768. S2CID 12370239. 8. ^ Hammer, Barbara; Micheli, Alessio; Sperduti, Alessandro; Strickert, Marc (2004). "Recursive self-organizing network models". Neural Networks. 17 (8–9): 1061–1085. CiteSeerX 10.1.1.129.6155. doi:10.1016/j.neunet.2004.06.009. PMID 15555852. 9. ^ Hammer, Barbara; Micheli, Alessio; Sperduti, Alessandro; Strickert, Marc (2004-03-01). "A general framework for unsupervised processing of structured data". Neurocomputing. 57: 3–35. CiteSeerX 10.1.1.3.984. doi:10.1016/j.neucom.2004.01.008. 10. ^ Socher, Richard; Perelygin, Alex; Y. Wu, Jean; Chuang, Jason; D. Manning, Christopher; Y. Ng, Andrew; Potts, Christopher. "Recursive Deep Models for Semantic Compositionality Over a Sentiment Treebank" (PDF). EMNLP 2013. 11. ^ Hammer, Barbara (2007-10-03). Learning with Recurrent Neural Networks. Springer. ISBN 9781846285677. 12. ^ Hammer, Barbara; Micheli, Alessio; Sperduti, Alessandro (2005-05-01). "Universal Approximation Capability of Cascade Correlation for Structures". Neural Computation. 17 (5): 1109–1159. CiteSeerX 10.1.1.138.2224. doi:10.1162/0899766053491878. S2CID 10845957. 13. ^ Gallicchio, Claudio; Micheli, Alessio (2013-02-04). "Tree Echo State Networks". Neurocomputing. 101: 319–337. doi:10.1016/j.neucom.2012.08.017. hdl:11568/158480. 14. ^ Scarselli, F.; Gori, M.; Tsoi, A. C.; Hagenbuchner, M.; Monfardini, G. (2009-01-01). "The Graph Neural Network Model". IEEE Transactions on Neural Networks. 20 (1): 61–80. doi:10.1109/TNN.2008.2005605. ISSN 1045-9227. PMID 19068426. S2CID 206756462. 15. ^ Micheli, A. (2009-03-01). "Neural Network for Graphs: A Contextual Constructive Approach". IEEE Transactions on Neural Networks. 20 (3): 498–511. doi:10.1109/TNN.2008.2010350. ISSN 1045-9227. PMID 19193509. S2CID 17486263.	{}
# Where did the bias in the fully connected layer go? This is my first post in the forums so Hi all. Super impressed with the quality of the information in the MOOC. Big thanks to Jeremy and Rachel. My question is about the explanation of a simple, fully-connected neural net as described by Jeremy in “Lesson 4: Deep Learning 2018” at time 45:15 see: https://youtu.be/gbceqO8PpBg?t=46m15s. As an example, Jeremy takes a vector of 20 continuous variables and does a matrix multiply by a 20x100 matrix. He says this is what a matrix multiplication does (agreed) and this is what a linear layer does in Deep Learning. I get that the matrix holds the weights, but what happened to the bias values in the example? Thanks, Bob It’s just an illustrative example of the main picture. Biases can be added after the multiplication. Alternatively, you can treat biases as an additional column in the weight matrix. To do that, concatenate the weight matrix and the bias vector (the biases should be located in the additional last column). Then, extend the inputs vector with the fixed value “1” (that is, input vector length increases by 1, and the last element is always equal to 1). If you work out the matrix-vector multiplication, you’ll see that [A; b] \times [x;1] = A \times x + b. Also, you might not need the bias component in some cases. Nowadays, people frequently use Batch Normalization layers immediately after the linear (or other) layers. BatchNorm does some tricks to speed up learning, and also learns a bias component, so it’s not necessary to have consecutive 2 biases that sum up. In those cases, people just drop the bias from linear layers. 1 Like Thank you for the explanation!	{}
The Ideal of Heroic Citizenship Change [572 Words] # The Ideal of Heroic Citizenship Change [572 Words] A 0 points A+Solution In an essay of 400 to 800 words (with a typical font and spacing, this will be approximately 1-1/2 to 3 pages), respond to the following question: Sources are the video How does the ideal of heroic citizenship change from the Greek mythopoetic tradition through the emergence of Greek tragic drama to the late Stoicism of Roman imperialism? What elements remain the same in this development of the ideal, and what elements undergo alteration or adaptation? What, in the end, is the essence of the heroic ideal? The Ideal a+tutor A 0 points #### Oh Snap! This Answer is Locked Thumbnail of first page Filename: the-ideal-of-heroic-citizenship-change-30.doc Filesize: < 2 MB Print Length: 5 Pages/Slides Words: 179 Surround your text in italics or bold, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$ Use LaTeX to type formulas and markdown to format text. See example.	{}
# vSphere CSI Driver - File Volume The vSphere Vanilla CSI driver version 2.0 and above supports file volumes backed by vSAN File shares to be statically/dynamically created and mounted by stateful containerized applications. This feature allows you to reference the same shared data among multiple pods spread across different clusters making it an absolute necessity for applications that need shareability. Before you proceed, keep in mind that the file volumes feature doesn't work in conjunction with the topology aware zones, extend volume and encryption features. Proceed to the requirements section below to enable this feature in your environment. ## Requirements Check the CSI vSphere compatibility matrix and the supported features section to verify if your environment conforms to all the required versions. Also check the limits section to understand if this feature can cater to your needs. In order to utilize this feature in your vSphere environment, you need to make sure of the following: • Enable and configure the file service in your vSAN cluster configuration. You must configure the necessary file service domains, IP pools, network etc in order to create file share volumes. Refer to vSphere 7.0 vSAN File service documentation to get started. • Establish a dedicated file share network connecting all the kubernetes nodes and make sure this network is routable to the vSAN File Share network. Refer to Network Access of vSAN File Share to understand the setup better. • Configure the kubernetes nodes to use the DNS server which was used to configure the file services in the vSAN cluster. This will help the nodes resolve the file share access points with Fully Qualified Domain Name (FQDN) while mounting the file volume to the Pod. In order to retrieve the vSAN file service DNS configuration, head over to the Configure tab of your vSAN cluster and navigate to the File Service section as shown in the screenshot below. • Configure the kubernetes secret named vsphere-config-secret to specify network permissions and placement of volumes for your vSAN file shares in your vSphere environment. This step is completely optional. Refer to the CSI vSphere driver installation instructions on vSphere configuration file for file volumes to learn more. If not specified, it is upto the CSI driver to use its discretion to place the file share volumes in any of your vSAN datastores with File services enabled. In such a scenario, the file volumes backed by vSAN file shares using the NFSv4 protocol will assume default network permissions i.e Read-Write privilege and root access to all the IP ranges. Before you start using file services in your environment keep in mind that if you have file shares being shared across more than one clusters in your vCenter, deleting a PVC with reclaim policy set to Delete in any one cluster may delete the underlying file share causing the volume to be unavailable for the rest of the clusters. After you are done setting up file services, to get started with vSphere CSI file services integration with your applications, check this video. We have also provided few Try-out YAMLs for Storage Class, PersistentVolumeClaim, PersistentVolume and Pod specs for your convenience. The next section on CSI file services integration will explain some of the spec changes mandatory for file share volumes. ## File services integration with your application ### Dynamic Provisioning of file volumes To give this example a try, you can first pick the Storage Class spec from here. To create a file volume PVC spec, set accessModes to either ReadWriteMany or ReadOnlyMany depending upon your requirement. apiVersion: v1 kind: PersistentVolumeClaim name: example-vanilla-file-pvc spec: accessModes: resources: requests: storage: 1Gi storageClassName: example-vanilla-file-sc kubectl apply -f example-pvc.yaml After the PVC is bound, if you describe the corresponding PV it will look similar to this: Name: pvc-45cea491-8399-11ea-883a-005056b61591 Labels: <none> Annotations: pv.kubernetes.io/provisioned-by: csi.vsphere.vmware.com Finalizers: [kubernetes.io/pv-protection] StorageClass: example-vanilla-file-sc Status: Bound Claim: default/example-vanilla-file-pvc Reclaim Policy: Delete Access Modes: RWX VolumeMode: Filesystem Capacity: 1Gi Node Affinity: <none> Message: Source: Type: CSI (a Container Storage Interface (CSI) volume source) Driver: csi.vsphere.vmware.com VolumeHandle: file:53bf6fb7-fe9f-4bf8-9fd8-7a589bf77760 VolumeAttributes: storage.kubernetes.io/csiProvisionerIdentity=1587430348006-8081-csi.vsphere.vmware.com type=vSphere CNS File Volume Events: <none> The VolumeHandle associated with the PV should have a prefix of file: for file volumes. Create a Pod to use the PVC from above example. apiVersion: v1 kind: Pod name: example-vanilla-file-pod1 spec: containers: - name: test-container command: ["/bin/sh", "-c", "echo 'Hello! This is Pod1' >> /mnt/volume1/index.html && while true ; do sleep 2 ; done"] volumeMounts: - name: test-volume mountPath: /mnt/volume1 restartPolicy: Never volumes: - name: test-volume persistentVolumeClaim: claimName: example-vanilla-file-pvc If you need to read the same file share from multiple pods, specify the PVC associated with the file share in the ClaimName in all the Pod specifications. If you need to create a Pod with Read-Only access to the above PVC, you need to explicitly mention readOnly as true in the persistentVolumeClaim section as shown below. Note that just setting the accessModes to ReadOnlyMany in the PVC spec will not make the PVC read-only to the Pods. apiVersion: v1 kind: Pod name: example-vanilla-file-pod2 spec: containers: - name: test-container command: ["/bin/sh", "-c", "while true ; do sleep 2 ; done"] volumeMounts: - name: test-volume mountPath: /mnt/volume1 restartPolicy: Never volumes: - name: test-volume persistentVolumeClaim: claimName: example-vanilla-file-pvc If you exec into this pod and try to create a file in the mountPath which is /mnt/volume1, you will get an error. \$ kubectl exec -it example-vanilla-file-pod2 -c test-container -- /bin/sh / # cd /mnt/volume1/ /mnt/volume1 # touch abc.txt ### Static Provisioning for file volumes If you have an existing persistent storage file volume in your VC, you can use static provisioning to make the storage instance available to your cluster. Define a PVC and a PV as shown below apiVersion: v1 kind: PersistentVolume name: static-file-share-pv-name annotations: pv.kubernetes.io/provisioned-by: csi.vsphere.vmware.com labels: "static-pv-label-key": "static-pv-label-value" spec: capacity: storage: 1Gi accessModes: persistentVolumeReclaimPolicy: Retain csi: driver: "csi.vsphere.vmware.com" volumeAttributes: type: "vSphere CNS File Volume" "volumeHandle": "file:236b3e6b-cfb0-4b73-a271-2591b2f31b4c" --- kind: PersistentVolumeClaim apiVersion: v1 name: static-file-share-pvc-name spec: accessModes: resources: requests: storage: 1Gi selector: matchLabels: static-pv-label-key: static-pv-label-value storageClassName: "" The labels key-value pair static-pv-label-key: static-pv-label-value used in PV metadata and PVC selector aid in matching the PVC to the PV during static provisioning. Also, remember to retain the file: prefix of the vSAN file share while filling up the volumeHandle field in PV spec. NOTE: For File volumes, CNS supports multiple PV's referring to the same file-share volume.	{}
## Posts Book review: Architects of Intelligence by Martin Ford (2018) 2020-08-11T17:30:21.247Z ofer's Shortform 2019-11-26T14:59:40.664Z A probabilistic off-switch that the agent is indifferent to 2018-09-25T13:13:16.526Z Looking for AI Safety Experts to Provide High Level Guidance for RAISE 2018-05-06T02:06:51.626Z A Safer Oracle Setup? 2018-02-09T12:16:12.063Z Comment by ofer on (USA) N95 masks are available on Amazon · 2021-01-22T12:27:05.853Z · LW · GW To support/add-to what ErickBall wrote, my own personal experience with respirators is that one with headbands (rather than ear loops) and a nose clip + nose foam is more likely to seal well. Comment by ofer on ofer's Shortform · 2021-01-22T12:20:45.726Z · LW · GW [COVID-19 related] It was nice to see this headline: My own personal experience with respirators is that one with headbands (rather than ear loops) and a nose clip + nose foam is more likely to seal well. Comment by ofer on Short summary of mAIry's room · 2021-01-20T12:17:34.188Z · LW · GW The topic of risks related to morally relevant computations seems very important, and I hope a lot more work will be done on it! My tentative intuition is that learning is not directly involved here. If the weights of a trained RL agent are no longer being updated after some point[1], my intuition is that the model is similarly likely to experience pain before and after that point (assuming the environment stays the same). Consider the following hypothesis which does not involve a direct relationship between learning and pain: In sufficiently large scale (and complex environments), TD learning tends to create components within the network, call them "evaluators", that evaluate certain metrics that correlate with expected return. In practice the model is trained to optimize directly for the output of the evaluators (and maximizing the output of the evaluators becomes the mesa objective). Suppose we label possible outputs of the evaluators with "pain" and "pleasure". We get something that seems analogous to humans. A human cares directly about pleasure and pain (which are things that correlated with expected evolutionary fitness in the ancestral environment), even when those things don't affect their evolutionary fitness accordingly (e.g. pleasure from eating chocolate, and pain from getting a vaccine shot). 1. In TD learning, if from some point the model always perfectly predicted the future, the gradient would always be zero and no weights would be updated. Also, if an already-trained RL agent is being deployed, and there's no longer reinforcement learning going on after deployment (which seems like a plausible setup in products/services that companies sell to customers), the weights would obviously not be updated. ↩︎ Comment by ofer on Birds, Brains, Planes, and AI: Against Appeals to the Complexity/Mysteriousness/Efficiency of the Brain · 2021-01-19T12:56:20.224Z · LW · GW My understanding is that the 2020 algorithms in Ajeya Cotra's draft report refer to algorithms that train a neural network on a given architecture (rather than algorithms that search for a good neural architecture etc.). So the only "special sauce" that can be found by such algorithms is one that corresponds to special weights of a network (rather than special architectures etc.). Comment by ofer on Birds, Brains, Planes, and AI: Against Appeals to the Complexity/Mysteriousness/Efficiency of the Brain · 2021-01-19T09:45:07.080Z · LW · GW Great post! we’ll either have to brute-force search for the special sauce like evolution did I would drop the "brute-force" here (evolution is not a random/naive search). Re the footnote: This "How much special sauce is needed?" variable is very similar to Ajeya Cotra's variable "how much compute would lead to TAI given 2020's algorithms." I don't see how they are similar. Comment by ofer on Why I'm excited about Debate · 2021-01-17T01:21:49.079Z · LW · GW One might argue: We don't need the model to use that much optimization power, to the point where it breaks the operator. We just need it to perform roughly at human-level, and then we can just deploy many instances of the trained model and accomplish very useful things (e.g. via factored cognition). So I think it's important to also note that, getting a neural network to "perform roughly at human-level in an aligned manner" may be a much harder task than getting a neural network to achieve maximal rating by breaking the operator. The former may be a much narrower target. This point is closely related to what you wrote here in the context of amplification: Speaking of inexact imitation: It seems to me that having an AI output a high-fidelity imitation of human behavior, sufficiently high-fidelity to preserve properties like "being smart" and "being a good person" and "still being a good person under some odd strains like being assembled into an enormous Chinese Room Bureaucracy", is a pretty huge ask. It seems to me obvious, though this is the sort of point where I've been surprised about what other people don't consider obvious, that in general exact imitation is a bigger ask than superior capability. Building a Go player that imitates Shuusaku's Go play so well that a scholar couldn't tell the difference, is a bigger ask than building a Go player that could defeat Shuusaku in a match. A human is much smarter than a pocket calculator but would still be unable to imitate one without using a paper and pencil; to imitate the pocket calculator you need all of the pocket calculator's abilities in addition to your own. Correspondingly, a realistic AI we build that literally passes the strong version of the Turing Test would probably have to be much smarter than the other humans in the test, probably smarter than any human on Earth, because it would have to possess all the human capabilities in addition to its own. Or at least all the human capabilities that can be exhibited to another human over the course of however long the Turing Test lasts. [...] Comment by ofer on Gradient hacking · 2021-01-13T17:59:36.674Z · LW · GW It does seem useful to make the distinction between thinking about how gradient hacking failures look like in worlds where they cause an existential catastrophe, and thinking about how to best pursue empirical research today about gradient hacking. Comment by ofer on Gradient hacking · 2021-01-12T06:18:03.291Z · LW · GW Some of the networks that have an accurate model of the training process will stumble upon the strategy of failing hard if SGD would reward any other competing network I think the part in bold should instead be something like "failing hard if SGD would (not) update weights in such and such way". (SGD is a local search algorithm; it gradually improves a single network.) This strategy seems more complicated, so is less likely to randomly exist in a network, but it is very strongly selected for, since at least from an evolutionary perspective it appears like it would give the network a substantive advantage. As I already argued in another thread, the idea is not that SGD creates the gradient hacking logic specifically (in case this is what you had in mind here). As an analogy, consider a human that decides to 1-box in Newcomb's problem (which is related to the idea of gradient hacking, because the human decides to 1-box in order to have the property of "being a person that 1-boxs", because having that property is instrumentally useful). The specific strategy to 1-box is not selected for by human evolution, but rather general problem-solving capabilities were (and those capabilities resulted in the human coming up with the 1-box strategy). Comment by ofer on Gradient hacking · 2021-01-02T14:32:32.496Z · LW · GW My point was that there's no reason that SGD will create specifically "deceptive logic" because "deceptive logic" is not privileged over any other logic that involves modeling the base objective and acting according to it. But I now think this isn't always true - see the edit block I just added. Comment by ofer on Gradient hacking · 2021-01-02T13:59:11.134Z · LW · GW "deceptive logic" is probably a pretty useful thing in general for the model, because it helps improve performance as measured through the base-objective. But you can similarly say this for the following logic: "check whether 1+1<4 and if so, act according to the base objective". Why is SGD more likely to create "deceptive logic" than this simpler logic (or any other similar logic)? [EDIT: actually, this argument doesn't work in a setup where the base objective corresponds to a sufficiently long time horizon during which it is possible for humans to detect misalignment and terminate/modify the model (in a way the is harmful with respect to the base objective).] So my understanding is that deceptive behavior is a lot more likely to arise from general-problem-solving logic, rather than SGD directly creating "deceptive logic" specifically. Comment by ofer on Gradient hacking · 2021-01-02T09:39:31.392Z · LW · GW I think that if SGD makes the model slightly deceptive it's because it made the model slightly more capable (better at general problem solving etc.), which allowed the model to "figure out" (during inference) that acting in a certain deceptive way is beneficial with respect to the mesa-objective. This seems to me a lot more likely than SGD creating specifically "deceptive logic" (i.e. logic that can't do anything generally useful other than finding ways to perform better on the mesa-objective by being deceptive). Comment by ofer on Gradient hacking · 2021-01-01T23:03:20.684Z · LW · GW The less philosophical approach to this problem is to notice that the appearance of gradient hacking would probably come from the training stumbling on a gradient hacker. [EDIT: you may have already meant it this way, but...] The optimization algorithm (e.g. SGD) doesn't need to stumble upon the specific logic of gradient hacking (which seems very unlikely). I think the idea is that a sufficiently capable agent (with a goal system that involves our world) instrumentally decides to use gradient hacking, because otherwise the agent will be modified in a suboptimal manner with respect to its current goal system. Comment by ofer on AGI safety from first principles: Introduction · 2021-01-01T11:46:35.376Z · LW · GW Early work tends to be less relevant in the context of modern machine learning I'm curious why you think the orthogonality thesis, instrumental convergence, the treacherous turn or Goodhart's law arguments are less relevant in the context of modern machine learning. (We can use here Facebook's feed-creation-algorithm as an example of modern machine learning, for the sake of concreteness.) Comment by ofer on Against GDP as a metric for timelines and takeoff speeds · 2020-12-30T18:41:23.520Z · LW · GW Thank you for writing this up! This topic seems extremely important and I strongly agree with the core arguments here. I propose the following addition to the list of things we care about when it comes to takeoff dynamics, or when it comes to defining slow(er) takeoff: 1. Foreseeability: No one creates an AI with a transformative capability X at a time when most actors (weighted by influence) believe it is very unlikely that an AI with capability X will be created within a year. Perhaps this should replace (or be merged with) the "warning shots" entry in the list. (As an aside, I think the term "warning shot" doesn't fit, because the original term refers to an action that is carried out for the purpose of communicating a threat.) Comment by ofer on What are the best precedents for industries failing to invest in valuable AI research? · 2020-12-15T17:53:25.619Z · LW · GW And the other part of the core idea is that that's implausible. I don't see why that's implausible. The condition I gave is also my explanation for why the EMH fulfills (in markets where it does), and it doesn't explain why big corporations should be good at predicting AGI. it's in their self-interest (at least, given their lack of concern for AI risk) to pursue it aggressively So the questions I'm curious about here are: 1. What mechanism is supposed to causes big corporations to be good at predicting AGI? 2. How come that mechanism doesn't also cause big corporations to understand the existential risk concerns? Comment by ofer on What are the best precedents for industries failing to invest in valuable AI research? · 2020-12-15T11:07:27.877Z · LW · GW (I'm not an economist but my understanding is that...) The EMH works in markets that fulfill the following condition: If Alice is way better than the market at predicting future prices, she can use her superior prediction capability to gain more and more control over the market, until the point where her control over the market makes the market prices reflect her prediction capability. If Alice is way better than anyone else at predicting AGI, how can she use her superior prediction capability to gain more control over big corporations? I don't see how the EMH an EMH-based argument applies here. Comment by ofer on ofer's Shortform · 2020-12-14T11:49:10.292Z · LW · GW [Online dating services related] The incentives of online dating service companies are ridiculously misaligned with their users'. (For users who are looking for a monogamous, long-term relationship.) A "match" between two users that results in them both leaving the platform for good is a super-negative outcome with respect to the metrics that the company is probably optimizing for. They probably use machine learning models to decide which "candidates" to show a user at any given time, and they are incentivized to train these models to avoid matches that cause users to leave their platform for good. (And these models may be way better at predicting such matches than any human). Comment by ofer on ofer's Shortform · 2020-12-14T11:47:13.281Z · LW · GW [Online dating services] The incentives of online dating service companies are ridiculously misaligned with their users'. (For users that look for a monogamous, long-term relationship.) A "match" between two users that results in them both leaving the platform for good is a super-negative outcome with respect to the metrics that the company is probably optimizing for. They probably use machine learning models to decide which "candidates" to show a user at any given time, and they are incentivized to train these models to avoid matches that cause users to leave their platform for good. (And these models may be way better at predicting such matches than any human). Comment by ofer on Seeking Power is Often Robustly Instrumental in MDPs · 2020-12-12T03:11:32.500Z · LW · GW Instrumental convergence is a very simple idea that I understand very well, and yet I failed to understand this paper (after spending hours on it) [EDIT: and also the post], so I'm worried about using it for the purpose of 'standing up to more intense outside scrutiny'. (Though it's plausible I'm just an outlier here.) Comment by ofer on Covid 12/10: Vaccine Approval Day in America · 2020-12-11T16:39:27.309Z · LW · GW Regarding comparison of mask types, the best source I'm aware of is: https://examine.com/topics/coronavirus-masks/ Comment by ofer on In a multipolar scenario, how do people expect systems to be trained to interact with systems developed by other labs? · 2020-12-11T06:31:11.230Z · LW · GW I have quite a different intuition on this, and I'm curious if you have a particular justification for expecting non-simulated training for multi-agent problems. In certain domains, there are very strong economic incentives to train agents that will act in a real-world multi-agent environment, where the ability to simulate the environment is limited (e.g. trading in stock markets and choosing content for social media users). Comment by ofer on Forecasting Newsletter: November 2020 · 2020-12-10T13:27:38.491Z · LW · GW Otherwise, some members of the broader Effective Altruism and rationality communities made a fair amount of money betting on the election. I would caveat this by adding that people are probably more likely to mention that they invested in a prediction market when the market resolved in their favor. Comment by ofer on In a multipolar scenario, how do people expect systems to be trained to interact with systems developed by other labs? · 2020-12-10T11:41:19.652Z · LW · GW Some off-the-cuff thoughts: It seems plausible that transformative agents will be trained exclusively on real-world data (without using simulated environments) [EDIT: in "data" I mean to include the observation/reward signal from the real-world environment in an online RL setup]; including social media feed-creation algorithms, and algo-trading algorithms. In such cases, the researchers don't choose how to implement the "other agents" (the other agents are just part of the real-world environment that the researchers don't control). Focusing on agents that are trained on simulated environments that involve multiple agents: For a lab to use copies of other labs' agents, the labs will probably need to cooperate (or some other process that involves additional actors may need to exist). In any case, using copies of the agent that is being trained (i.e. self-play) seems to me very plausible. (Like, I think both AlphaZero and OpenAI Five were trained via self-play and that self-play is generally considered to be a very prominent technique for RL-in-simulated-environments-that-involve-multiple-agents). Comment by ofer on Cultural accumulation · 2020-12-07T11:23:46.683Z · LW · GW I would probably be inclined to reject such a public offer on the grounds that some of the possible people who are similar to me (possible people whose decisions are correlated with mine), when finding themselves in a similar situation, may not trust the offer-maker or may not have $500 to spare, and would not want to publicly disclose that information. Comment by ofer on I made an N95-level mask at home, and you can too · 2020-11-26T20:17:02.529Z · LW · GW My uneducated concern is that masks that are not intended to seal may not allow air to flow sufficiently easily through their "filter" part (without it turning out to be a problem during "normal" use due the air easily flowing through the edges). Re volume argument, maybe we also need to consider the volume of the air we inhale each time (and whether that volume becomes smaller if something is partially blocking the air flow, and whether we notice). Comment by ofer on Working in Virtual Reality: A Review · 2020-11-21T11:59:50.111Z · LW · GW That's very interesting. I'd be concerned about the potential impact of prolonged usage of a VR headset for many hours per day on eye health. (Of course, I'm not at all an expert in this area.) Comment by ofer on I made an N95-level mask at home, and you can too · 2020-11-19T15:05:16.976Z · LW · GW I followed your tip to just google it but every result in the first 2 pages for me was either out of stock or outdated As I said in that thread, I was not recommending the mentioned google search as a way to buy respirators, and one's best options (which may include buying from a well-known retailer and having a mechanism to substantially lower risks from counterfeit respirators) may depend on where they live. Comment by ofer on Some AI research areas and their relevance to existential safety · 2020-11-19T13:13:15.593Z · LW · GW Great post! I suppose you'll be more optimistic about Single/Single areas if you update towards fast/discontinuous takeoff? Comment by ofer on I made an N95-level mask at home, and you can too · 2020-11-18T19:07:53.249Z · LW · GW Disclaimer: I'm not an expert. It turns out that surgical masks are made of the exact same material as N95s! They both filter 95% of 0.1μm particles. I very much doubt this claim, and the link you provide in support of it is to a website that you later suggest is being run by people that seem to you "a bit sketchy". I also doubt that the way you propose for checking the "electrostatic effect" (on large pieces of paper?) can provide strong evidence that the mask's material provides filtering protection that is similar to a N95 respirator. [EDIT: sorry, you later cite the Rengasamy et al. paper that seems to support that claim to some extent; I'm not sure how much to update on it.] As a civilian you can’t purchase an N95 anywhere at any price. This claim is false (see this thread). BTW: since presumably surgical masks are not intended to be used in this way, I would also worry about potential risks of breathing too little oxygen or too much carbon dioxide. BTW2: Maybe it's worth looking into using your approach for "upgrading" cheap KN95 respirators rather than surgical masks (I suspect that cheap KN95 respirators tend to not seal well due to a lack of nose clip and due to bands that go around the ears rather than around the head). Though the above concern regarding oxygen/carbon dioxide might still apply. [EDIT: BTW3: for a comparison between cloth masks, surgical masks and N95 respirators see this page on examine.com.] Comment by ofer on What considerations influence whether I have more influence over short or long timelines? · 2020-11-07T21:50:36.582Z · LW · GW (They may spend more on inference compute if doing so would sufficiently increase their revenue. They may train such a more-expensive model just to try it out for a short while, to see whether they're better off using it.) Comment by ofer on What considerations influence whether I have more influence over short or long timelines? · 2020-11-07T15:24:44.613Z · LW · GW I didn't follow this. FB doesn't need to run a model inference for each possible post that it considers showing (just like OpenAI doesn't need to run a GPT-3 inference for each possible token that can come next). (BTW, I think the phrase "context window" would correspond to the model's input.) FB's revenue from advertising in 2019 was$69.7 billion, or $191 million per day. So yea, it seems possible that in 2019 they used a model with an inference cost similar to GPT-3's, though not one that is 10x more expensive [EDIT: under this analysis' assumptions]; so I was overconfident in my previous comment. Comment by ofer on What considerations influence whether I have more influence over short or long timelines? · 2020-11-07T13:20:17.113Z · LW · GW That said, I'd be surprised if the feed-creation algorithm had as many parameters as GPT-3, considering how often it has to be run per day... The relevant quantities here are the compute cost of each model usage (inference)—e.g. the cost of compute for choosing the next post to place on a feed—and the impact of such a potential usage on FB's revenue. This post by Gwern suggests that OpenAI was able to run a single GPT-3 inference (i.e. generate a single token) at a cost of$0.00006 (6 cents for 1,000 tokens) or less. I'm sure it's worth to FB much more than 0.00006 to choose well the next post that a random user sees. Comment by ofer on What considerations influence whether I have more influence over short or long timelines? · 2020-11-07T11:58:10.499Z · LW · GW The frontrunners right now are OpenAI and DeepMind. I'm not sure about this. Note that not all companies are equally incentivized to publish their ML research (some companies may be incentivized to be secretive about their ML work and capabilities due to competition/regulation dynamics). I don't see how we can know whether GPT-3 is further along on the route to AGI than FB's feed-creation algorithm, or the most impressive algo-trading system etc. The other places have the money, but less talent I don't know where the "less talent" estimate is coming from. I won't be surprised if there are AI teams with a much larger salary budget than any team at OpenAI/DeepMind, and I expect the "amount of talent" to correlate with salary budget (among prestigious AI labs). and more importantly don't seem to be acting as if they think short timelines are possible. I'm not sure how well we can estimate the beliefs and motivations of all well-resourced AI teams in the world. Also, a team need not be trying to create AGI (or believe they can) in order to create AGI. It's sufficient that they are incentivized to create systems that model the world as well as possible; which is the case for many teams, including ones working on feed-creation in social media services and algo-trading systems. (The ability to plan and find solutions to arbitrary problems in the real world naturally arises from the ability to model it, in the limit.) Comment by ofer on What considerations influence whether I have more influence over short or long timelines? · 2020-11-06T17:26:42.846Z · LW · GW This consideration favors short timelines, because (1) We have a good idea which AI projects will make TAI conditional on short timelines, and (2) Some of us already work there, they seem already at least somewhat concerned about safety, etc. I don't see how we can have a good idea which project whether a certain small set of projects will make TAI first conditional on short timelines (or whether the first project will be one in which people are "already at least somewhat concerned about safety"). Like, why not some arbitrary team at Facebook/Alphabet/Amazon or any other well-resourced company? There are probably many well-resourced companies (including algo-trading companies) that are incentivized to throw a lot of money at novel, large scale ML research. Comment by ofer on "Inner Alignment Failures" Which Are Actually Outer Alignment Failures · 2020-11-03T22:03:48.635Z · LW · GW you should never get deception in the limit of infinite data (since a deceptive model has to defect on some data point). I think a model can be deceptively aligned even if formally it maps every possible input to the correct (safe) output. For example, suppose that on input X the inference execution hacks the computer on which the inference is being executed, in order to do arbitrary consequentialist stuff (while the inference logic, as a mathematical object, formally yields the correct output for X). Comment by ofer on ofer's Shortform · 2020-11-01T16:28:53.254Z · LW · GW [Question about reinforcement learning] What is the most impressive/large-scale published work in RL that you're aware of where—during training—the agent's environment is the real world (rather than a simulated environment)? Comment by ofer on Responses to Christiano on takeoff speeds? · 2020-10-30T19:55:26.829Z · LW · GW I'd love to give feedback on your version if you want! Could even collaborate. Ditto for me! Comment by ofer on Draft report on AI timelines · 2020-10-29T19:13:27.353Z · LW · GW Let and be two optimization algorithms, each searching over some set of programs. Let be some evaluation metric over programs such that is our evaluation of program , for the purpose of comparing a program found by to a program found by . For example, can be defined as a subjective impressiveness metric as judged by a human. Intuitive definition: Suppose we plot a curve for each optimization algorithm such that the x-axis is the inference compute of a yielded program and the y-axis is our evaluation value of that program. If the curves of and are similar up to scaling along the x-axis, then we say that and are similarly-scaling w.r.t inference compute, or SSIC for short. Formal definition: Let and be optimization algorithms and let be an evaluation function over programs. Let us denote with the program that finds when it uses flops (which would correspond to the training compute if is an ML algorithms). Let us denote with the amount of compute that program uses. We say that and are SSIC with respect to if for any ,,, such that , if then . I think the report draft implicitly uses the assumption that human evolution and the first ML algorithm that will result in TAI are SSIC (with respect to a relevant ). It may be beneficial to discuss this assumption in the report. Clearly, not all pairs of optimization algorithms are SSIC (e.g. consider a pure random search + any optimization algorithm). Under what conditions should we expect a pair of optimization algorithms to be SSIC with respect to a given ? Maybe that question should be investigated empirically, by looking at pairs of optimization algorithms, were one is a popular ML algorithm and the other is some evolutionary computation algorithm (searching over a very different model space), and checking to what extent the two algorithms are SSIC. Comment by ofer on Draft report on AI timelines · 2020-10-29T19:13:04.730Z · LW · GW Some thoughts: 1. The development of transformative AI may involve a feedback loop in which we train ML models that help us train better ML models and so on (e.g. using approaches like neural architecture search which seems to be getting increasingly popular in recent years). There is nothing equivalent to such a feedback loop in biological evolution (animals don't use their problem-solving capabilities to make evolution more efficient). Does your analysis assume there won't be such a feedback loop (or at least not one that has a large influence on timelines)? Consider adding to the report a discussion about this topic (sorry if it's already there and I missed it). 2. Part of the Neural Network hypothesis is the proposition that "a transformative model would perform roughly as many FLOP / subj sec as the human brain". It seems to me worthwhile to investigate this proposition further. Human evolution corresponds to a search over a tiny subset of all possible computing machines. Why should we expect that a different search algorithm over an entirely different subset of computing machines would yield systems (with certain capabilities) that use a similar amount of compute? One might pursue an empirical approach for investigating this topic, e.g. by comparing two algorithms for searching over a space of models, where one is some common supervised learning algorithm, and the other is some evolutionary computation algorithm. In a separate comment (under this one) I attempt to describe a more thorough and formal way of thinking about this topic. 3. Regarding methods that involve adjusting variables according to properties of 2020 algorithms (or the models trained by them): It would be interesting to try to apply the same methods with respect to earlier points in time (e.g. as if you were writing the report back in 1998/2012/2015 when LeNet-5/AlexNet/DQN were introduced, respectively). To what extent would the results be consistent with the 2020 analysis? Comment by ofer on [AN #121]: Forecasting transformative AI timelines using biological anchors · 2020-10-24T17:56:45.191Z · LW · GW My point here is that in a world where an algo-trading company has the lead in AI capabilities, there need not be a point in time (prior to an existential catastrophe or existential security) where investing more resources into the company's safety-indifferent AI R&D does not seem profitable in expectation. This claim can be true regardless of researchers' observations beliefs and actions in given situations. Comment by ofer on [AN #121]: Forecasting transformative AI timelines using biological anchors · 2020-10-24T06:22:52.616Z · LW · GW We might get TAI due to efforts by, say, an algo-trading company that develops trading AI systems. The company can limit the mundane downside risks that it faces from non-robust behaviors of its AI systems (e.g. by limiting the fraction of its fund that the AI systems control). Of course, the actual downside risk to the company includes outcomes like existential catastrophes, but it's not clear to me why we should expect that prior to such extreme outcomes their AI systems would behave in ways that are detrimental to economic value. Comment by ofer on [AN #121]: Forecasting transformative AI timelines using biological anchors · 2020-10-23T07:13:37.354Z · LW · GW you need to ensure that your model is aligned, robust, and reliable (at least if you want to deploy it and get economic value from it). I think it suffices for the model to be inner aligned (or deceptively inner aligned) for it to have economic value, at least in domains where (1) there is a usable training signal that corresponds to economic value (e.g. users' time spent in social media platforms, net income in algo-trading companies, or even the stock price in any public company); and (2) the downside economic risk from a non-robust behavior is limited (e.g. an algo-trading company does not need its model to be robust/reliable, assuming the downside risk from each trade is limited by design). Comment by ofer on The Solomonoff Prior is Malign · 2020-10-14T11:08:15.365Z · LW · GW If arguments about acausal trade and value handshakes hold, then the resulting utility function might contain some fraction of human values. I think Paul's Hail Mary via Solomonoff prior idea is not obviously related to acausal trade. (It does not privilege agents that engage in acausal trade over ones that don't.) Comment by ofer on A prior for technological discontinuities · 2020-10-14T10:40:46.389Z · LW · GW Seems like an almost central example of continuous progress if you're evaluating by typical language model metrics like perplexity. I think we should determine whether GPT-3 is an example of continuous progress in perplexity based on the extent to which it lowered the SOTA perplexity (on huge internet-text corpora), and its wall clock training time. I don't see why the correctness of a certain scaling law or the researchers' beliefs/motivation should affect this determination. Comment by ofer on If GPT-6 is human-level AGI but costs200 per page of output, what would happen? · 2020-10-10T17:33:34.879Z · LW · GW Yeah, human-level is supposed to mean not strongly superhuman at anything important, while also not being strongly subhuman in anything important. I think that's roughly the concept Nick Bostrom used in Superintelligence when discussing takeoff dynamics. (The usage of that concept is my only major disagreement with that book.) IMO it would be very surprising if the first ML system that is not strongly subhuman at anything important would not be strongly superhuman at anything important (assuming this property is not optimized for). Comment by ofer on If GPT-6 is human-level AGI but costs $200 per page of output, what would happen? · 2020-10-10T17:23:53.614Z · LW · GW GPT-2 was benchmarked at 43 perplexity on the 1 Billion Word (1BW) benchmark vs a (highly extrapolated) human perplexity of 12 I wouldn't say that that paper shows a (highly extrapolated) human perplexity of 12. It compares human-written sentences to language model generated sentences on the degree to which they seem "clearly human" vs "clearly unhuman" as judged by humans. Amusingly, for every 8 human-written sentences that were judged as "clearly human", one human-written sentence was judged as "clearly unhuman". And that 8:1 ratio is the thing from which human perplexity is being derived from. This doesn't make sense to me. If the human annotators in this paper had never annotated human-written sentences as "clearly unhuman", this extrapolation would have shown human perplexity of 1! (As if humans can magically predict an entire page of text sampled from the internet.) The LAMBADA dataset was also constructed using humans to predict the missing words, but GPT-3 falls far short of perfection there, so while I can't numerically answer it (unless you trust OA's reasoning there), it is still very clear that GPT-3 does not match or surpass humans at text prediction. If the comparison here is on the final LAMBADA dataset, after examples were filtered out based on disagreement between humans (as you mentioned in the newsletter), then it's an unfair comparison. The examples are selected for being easy for humans. BTW, I think the comparison to humans on the LAMBADA dataset is indeed interesting in the context of AI safety (more so than "predict the next word in a random internet text"); because I don't expect the perplexity/accuracy to depend much on the ability to model very low-level stuff (e.g. "that's" vs "that is"). Comment by ofer on If GPT-6 is human-level AGI but costs$200 per page of output, what would happen? · 2020-10-09T18:35:04.912Z · LW · GW Nevertheless, it works. That's how self-supervised training/pretraining works. Right, I'm just saying that I don't see how to map that metric to things we care about in the context of AI safety. If a language model outperforms humans at predicting the next word, maybe it's just due to it being sufficiently superior at modeling low-level stuff (e.g. GPT-3 may be better than me at predicting you'll write "That's" rather than "That is".) (As an aside, in the linked footnote I couldn't easily spot any paper that actually evaluated humans on predicting the next word.) Comment by ofer on If GPT-6 is human-level AGI but costs $200 per page of output, what would happen? · 2020-10-09T15:50:45.607Z · LW · GW Some quick thoughts/comments: --It can predict random internet text better than the best humans I wouldn't use this metric. I don't see how to map between it and anything we care about. If it's defined in terms of accuracy when predicting the next word, I won't be surprised if existing language models already outperform humans. Also, I find the term "human-level AGI" confusing. Does it exclude systems that are super-human on some dimensions? If so, it seems too narrow to be useful. For the purpose of this post, I propose using the following definition: A system that is able to generate text in a way that allows to automatically perform any task that humans can perform by writing text. Comment by ofer on If GPT-6 is human-level AGI but costs$200 per page of output, what would happen? · 2020-10-09T15:50:03.079Z · LW · GW But I'm pretty sure stuff would go crazy even before then. How? We can end up with an intelligence explosion via automated ML research. One of the tasks that could be automated by the language model is "brainstorming novel ML ideas". So you'll be able to pay \$200 and get a text, that could have been written by a brilliant ML researcher, containing novel ideas that allow you to create a more efficient/capable language model. (Though I expect that this specific approach won't be competitive with fully automated approaches that do stuff like NAS.) Comment by ofer on AI arms race · 2020-10-07T11:54:06.128Z · LW · GW This model assumes that each AI lab chooses some level of safety precautions , and then acts accordingly until AGI is created. But the degree to which an AI lab invests in safety may change radically with time. Importantly, it may increase by a lot if the leadership of the AI lab comes to believe that their current or near-term work poses existential risk. This seems like a reason to be more skeptical about the counter-intuitive conclusion that the information available to all the teams about their own capability or progress towards AI increases the risk. (Not to be confused with the other counter-intuitive conclusion from the paper that the information available about other teams increases the risk).	{}
This topic provides instructions for installing and configuring protocol buffers (protobuf) support in the Snowflake Connector for Kafka (“Kafka connector”). Support for protobuf requires Kafka connector 1.5.0 (or higher). The Kafka connector supports the following versions of the protobuf converter: Confluent version This version is supported by the Confluent package version of Kafka only. Community version This version is supported by the open source software (OSS) Apache Kafka package. This version is also supported by the Confluent package version of Kafka; however, for ease of use, we suggest using the Confluent version instead. Install only one of these protobuf converters. In this Topic: ## Prerequisite: Installing the Snowflake Connector for Kafka¶ Install the Kafka connector using the instructions in Installing and Configuring the Kafka Connector. ## Configuring the Confluent Version of the Protobuf Converter¶ Note The Confluent version of the Protobuf converter is available with Confluent version 5.5.0 (or higher). 1. Open your Kafka configuration file (e.g. <kafka_dir>/config/connect-distributed.properties) in a text editor. 2. Configure the converter properties in the file. For information about the Kafka connector properties in general, see Kafka Configuration Properties. { "name":"XYZCompanySensorData", "config":{ .. "key.converter":"io.confluent.connect.protobuf.ProtobufConverter", "key.converter.schema.registry.url":"CONFLUENT_SCHEMA_REGISTRY", "value.converter":"io.confluent.connect.protobuf.ProtobufConverter", "value.converter.schema.registry.url":"http://localhost:8081" } } For example: { "name":"XYZCompanySensorData", "config":{ "connector.class":"com.snowflake.kafka.connector.SnowflakeSinkConnector", "topics":"topic1,topic2", "snowflake.topic2table.map": "topic1:table1,topic2:table2", "buffer.count.records":"10000", "buffer.flush.time":"60", "buffer.size.bytes":"5000000", "snowflake.url.name":"myorganization-myaccount.snowflakecomputing.com:443", "snowflake.user.name":"jane.smith", "snowflake.private.key":"xyz123", "snowflake.database.name":"mydb", "snowflake.schema.name":"myschema", "key.converter":"io.confluent.connect.protobuf.ProtobufConverter", "key.converter.schema.registry.url":"CONFLUENT_SCHEMA_REGISTRY", "value.converter":"io.confluent.connect.protobuf.ProtobufConverter", "value.converter.schema.registry.url":"http://localhost:8081" } } 3. Save the file. Produce protobuf data from Kafka using the Confluent console protobuf producer, the source protobuf producer, or the Python producer. Example Python code located in GitHub demonstrates how to produce protobuf data from Kafka. ## Configuring the Community Version of the Protobuf Converter¶ This section provides instructions for installing and configuring the community version of the protobuf converter. ### Step 1: Installing the Community Protobuf Converter¶ 1. In a terminal window, change to the directory where you want to store a clone of the GitHub repository for the protobuf converter. 2. Execute the following command to clone the GitHub repository: git clone https://github.com/blueapron/kafka-connect-protobuf-converter 3. Execute the following commands to build the 3.1.0 version of the converter using Apache Maven. Note that versions 2.3.0, 3.0.0, and 3.1.0 of the converter are supported by the Kafka connector: Note cd kafka-connect-protobuf-converter git checkout tags/v3.1.0 mvn clean package Maven builds a file named kafka-connect-protobuf-converter-<version>-jar-with-dependencies.jar in the current folder. This is the converter JAR file. 4. Copy the compiled kafka-connect-protobuf-converter-<version>-jar-with-dependencies.jar file to the directory for your Kafka package version: Confluent <confluenct_dir>/share/java/kafka-serde-tools Apache Kafka <apache_kafka_dir>/libs ### Step 2: Compiling Your .proto File¶ Compile the protobuf .proto file that defines your messages into a java file. For example, suppose your messages are defined in a file named sensor.proto. In a terminal window, execute the following command to compile the protocol buffers file. Specify the source directory for the application source code, the destination directory (for the .java file), and the path to your .proto file: protoc -I=$SRC_DIR --java_out=$DST_DIR $SRC_DIR/sensor.proto A sample .proto file is available here: https://github.com/snowflakedb/snowflake-kafka-connector/blob/master/test/test_data/sensor.proto. The command generates a file named SensorReadingImpl.java in the specified destination directory. For more information, see the Google developer documentation ### Step 3: Compiling the SensorReadingImpl.java File¶ Compile the generated SensorReadingImpl.java file from Step 2: Compiling Your .proto File along with the Project Object Model of the protobuf project structure. 1. Open your .pom file from Step 2: Compiling Your .proto File in a text editor. 2. Create an otherwise empty directory with a structure: protobuf_folder ├── pom.xml └── src └── main └── java └── com └── .. Where the directory structure under src / main / java mirrors the package name in your .proto file (line 3). 3. Copy the generated SensorReadingImpl.java file from Step 2: Compiling Your .proto File to the bottom folder in the directory structure. 4. Create a file named pom.xml in the root of the protobuf_folder directory. 5. Open the empty pom.xml file in a text editor. Copy the following example project model into the file and modify it: <?xml version="1.0" encoding="UTF-8"?> <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd"> <modelVersion>4.0.0</modelVersion> <groupId><group_id></groupId> <artifactId><artifact_id></artifactId> <version><version></version> <properties> <java.version><java_version></java.version> <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding> </properties> <dependencies> <dependency> <groupId>com.google.protobuf</groupId> <artifactId>protobuf-java</artifactId> <version>3.11.1</version> </dependency> </dependencies> <build> <plugins> <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-compiler-plugin</artifactId> <version>3.3</version> <configuration> <source>${java.version}</source> <target>${java.version}</target> </configuration> </plugin> <plugin> <artifactId>maven-assembly-plugin</artifactId> <version>3.1.0</version> <configuration> <descriptorRefs> <descriptorRef>jar-with-dependencies</descriptorRef> </descriptorRefs> </configuration> <executions> <execution> <id>make-assembly</id> <phase>package</phase> <goals> <goal>single</goal> </goals> </execution> </executions> </plugin> </plugins> </build> </project> Where: <group_id> Group ID segments of the package name specified in your .proto file. For example, if the package name is com.foo.bar.buz, then the group ID is com.foo. <artifact_id> Artifact ID for the package that you choose. The artifact ID can be randomly picked. <version> Version of the package that you choose. The version can be randomly picked. <java_version> Version of the Java Runtime Environment (JRE) installed on your local machine. For example: <?xml version="1.0" encoding="UTF-8"?> <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd"> <modelVersion>4.0.0</modelVersion> <groupId>com.snowflake</groupId> <artifactId>kafka-test-protobuf</artifactId> <version>1.0.0</version> <properties> <java.version>1.8</java.version> <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding> </properties> <dependencies> <dependency> <groupId>com.google.protobuf</groupId> <artifactId>protobuf-java</artifactId> <version>3.11.1</version> </dependency> </dependencies> <build> <plugins> <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-compiler-plugin</artifactId> <version>3.3</version> <configuration> <source>${java.version}</source> <target>${java.version}</target> </configuration> </plugin> <plugin> <artifactId>maven-assembly-plugin</artifactId> <version>3.1.0</version> <configuration> <descriptorRefs> <descriptorRef>jar-with-dependencies</descriptorRef> </descriptorRefs> </configuration> <executions> <execution> <id>make-assembly</id> <phase>package</phase> <goals> <goal>single</goal> </goals> </execution> </executions> </plugin> </plugins> </build> </project> 6. In a terminal window, change to the root of the protobuf_folder directory. Execute the following command to compile the protobuf data JAR file from the files in the directory: mvn clean package Maven generates a file named <artifact_id>-<version>-jar-with-dependencies.jar in the protobuf_folder/target folder (e.g. kafka-test-protobuf-1.0.0-jar-with-dependencies.jar). 7. Copy the compiled kafka-test-protobuf-1.0.0-jar-with-dependencies.jar file to the directory for your Kafka package version: Confluent <confluenct_dir>/share/java/kafka-serde-tools Apache Kafka Copy the file to the directory in your $CLASSPATH environment variable. ### Step 4: Configuring the Kafka Connector¶ 1. Open your Kafka configuration file (e.g. <kafka_dir>/config/connect-distributed.properties) in a text editor. 2. Add the value.converter.protoClassName property to the file. This property specifies the protocol buffer class to use to deserialize messages (e.g. com.google.protobuf.Int32Value). Note Nested classes must be specified using the $ notation (e.g. com.blueapron.connect.protobuf.NestedTestProtoOuterClass$NestedTestProto). For example: { "name":"XYZCompanySensorData", "config":{ ..	{}
# How much does it cost to research and develop a vaccine? Analyzing the cost of vaccine research Previously, we estimated how long it takes to research and develop a vaccine and came up with a conclusion that it would take “an average of 29 years [to develop a] typical vaccine, though with high uncertainty based on uncertainties in each approach and on many particular vaccines not being typical”. However, if we want to know the cost-effectiveness of vaccine research, it’s not enough to know how long a vaccine takes, but how much total money it would cost. ## Literature Like figuring out vaccine timelines, figuring out vaccine costs is also very difficult. Waye, Jacobs, and Schryvers (2013) state that the best answer to the average cost of vaccine R&D is that “the cost of vaccine R&D is unknown”. However, we’re not without our ability to estimate. Waye, Jacobs, and Schryvers (2013) state that it costs an average of $1.2B to bring a drug to market in the US, though it’s unclear how well that average generalizes to vaccines specifically. Another estimate by DiMasi, Grabowski, and Hansen (2016) found an average cost of$1.39B. To use a concrete example of the rotavirus vaccine, Light, Andrus, & Warburton (2009) estimate using publicly available data that the Phase I-III trials for two different rotavirus vaccine candidates were $137M-$206M (Merck) and $128M-$192M (GSK). But the costs of a vaccine are more than just the costs of the Phase I-III trials. Moalla, Bouras, Ouzrout, and Neubert (2009) outline a more holistic cost lifecycle for vaccines, which includes payment for researcher salaries and equipment throughout the research and design phase in addition to the trial phase, plus the payment for preclinical trials before Phase I-III trials, plus the costs of registration. Dr. Michel Greco, the President and Chief Operating Officer of Aventis Pasteur, testified to WHO that vaccine trials and registration are about 70% of the total costs of developing a vaccine (WHO 2001, p19). Additionally, Waye, Jacobs, and Schryvers (2013) point out that this does not include the costs of the failed vaccines – Light, Andrus, & Warburton (2009) are only looking at the costs of the successes. If we consider the rotavirus vaccine to have taken four total attempts, with $200M each spent on clinical trials and$86M each spent on all other costs, that would require $1.14B (which is pretty close to the$1.2B initial average for drugs in the US). However, it’s also not clear how well these long-run trends will predict the future. Andrew Witty, the CEO of GSK, said in 2013 that the $1B price tag for new vaccine development was “one of the great myths of the industry” and expected to bring future vaccines to market at lower prices. On the other hand, Elder and Cohn (2013) observe that more modern vaccines being distributed have higher per dose prices than earlier vaccines, which could (but does not necessarily) suggest that vaccine costs are increasing. ## Vaccine Breakdown Any kind of historical data on the costs of individual vaccines, let alone reliable data is quite difficult to come by. Also, the costs of vaccines has dramatically changed with the invention and standardization of modern clinical trials and licensing, which limits the amount we can infer from historical vaccines. To get a good benchmark, we chose to look at smallpox because it was the first vaccine and the only disease successfully eradicated through vaccination; measles as a historical vaccine as there is a good prima facie chance it is the most cost-effective vaccine1; the HIV, malaria, and ebola vaccines because they are currently under development and total R&D spending is relatively easy to find; and the rotavirus and HPV vaccines because they finished licensing the most recently. Based on this, I find the following costs and elaborate more below: • Smallpox -$5.5M • Measles - $38.3M • Rotavirus -$1,140M • HPV - ??? • HIV - $24,500M • Malaria -$605M #### Measles Vaccine John Enders, the inventor of the first measles vaccine, was able to develop a vaccine in a time before modern clinical trials, and the vaccine was tested and licensed based on testing the vaccine on a few thousand people (Bakalar, 2010), which made the development of the vaccine a lot cheaper than if it were being done today. #### HIV Vaccine The HIV vaccine has proven difficult to develop and has costed many times that of a typical vaccine. Hecht and Jameson (2011), writing for the Copenhagen Consensus’s Rethink HIV (Chapter 6), outlines an estimate that having an “effective HIV/ AIDS vaccine available for introduction by 2030 could cost as much as twenty times the $1B typically required to develop a new drug”, citing Adams and Brantner (2010) for calculating typical vaccine costs. Existing work on an HIV vaccine is cited to be at$9B to date, growing at $800M-900M per year (Hecht & Jameson, 2011). Hecht and Jameson (2011) convened a panel of leading AIDS vaccine scientists who, as of 2011, believed a prototype vaccine could achieve proof of 50% efficacy by 2020-2025, to be available for large-scale introduction by 2025-2030. This would mean that the total expected R&D costs of the HIV vaccine would be$21B to $28B6, or$24.5B as the average guess. #### HPV Vaccine Despite thoroughly searching as much as we did for other vaccines, we were not able to find sufficiently credible information about the HPV vaccine such that we felt comfortable making an estimate8. We generally assume that the costs of developing a vaccine are equal to paying for salaries, paying for clinical trials, and paying for equipment and other overhead costs. Salaries and overhead costs would be a function of the number of employees, while clinical trial costs would be a function of the number of vaccines tried. When we attempt to do a very rough Fermi calculation of this, we end up with a model pointing to the cost of developing a vaccine $460M to$1.9B with a mean of $960M. This essay was jointly written by Peter Hurford and Marcus A. Davis. ## Endnotes 1. Because the measles vaccine was developed so quickly by a single person in a time prior to massive costs from the modern clinical trials and licensing process, the costs of the measles vaccine were likely exceptionally low compared to all other vaccines. Additionally, the high rate of contagiousness from measles (CDC, 2017) combined with a moderate DALY penalty (WHO, 2004) makes eliminating cases of measles particularly high value. The high benefit combined with the abnormally low cost makes the measles vaccine the best prima facie case for impact in vaccination. [return] 2. According to Glassdoor, Associate Scientists at GSK make$62K/yr, Senior Scientists make $88K/yr, PMs and investigators and principal scientists make$110K/yr, and the director makes $160K/yr. At Merck, Glassdoor says assistant scientists make$62K-70K, scientist makes $81K/yr, principal scientist makes$139K/yr, and the director makes $175K/yr. [return] 3. Of course, all of this overhead is there for good reason and there may have been considerable risk at testing a vaccine so quickly! [return] 4. This is a complete guess, but I’m unsure how this guess could be improved. [return] 5. Battelle (2015) finds that a typical trial done today costs$36,500 per participant when considering all costs holistically, but trials for infectious diseases were found to be half as expensive. Since I already account for staff and overhead costs in my estimate, I’d cut this estimate in half again, and guess a “modern day cost” of ~$9K per participant. Matheny (2013) (of EA fame!) finds a cost of$9500 per participant for vaccines in particular (p20), which after re-factoring out staff costs could support an estimate even lower than $9K per participant. [return] 6.$9B to date as of 2011, plus $800-900M per year from 2011 to 2025-2030 implies a range of$20.2B to $26.1B in 2011 US dollars, or$21.6 to $27.9 in 2016 US dollars. [return] 7. From the records we could find$1.5B in contributions for vaccine research and development – GAVI contributed $300M (GAVI, 2015), Merck contributed$50M ( Pierson, 2004), CanSino contributed $65M ( Hruby, 2017; Liu, 2017), the European Union contributed$282.4M ( European Commission, 2015), the Bill and Melinda Gates Foundation contributed $34M (Ibid.), Johnson & Johnson contributed$287M (Bavarian Nordic, 2014; Johnson & Johnson, 2017), USAID contributed $515M (European Commision, 2015), China has contributed$8.3M ( Sanchez, 2014), and Russia has contributed $8M (Presse, 2016 ; see also Russian Federation press release). Another aggregation confirms this$1.5B over 2014-2015 (Financial Tracking Service). Notably this likely does not include any funding from before 2010, which likely could mean we are missing funds and creating an underestimate, but we decided not to investigate this further. [return] 8. The initial research into HPV vaccines was done by the University of Queensland, the University of Georgetown, the University of Rochester, and by the U.S. National Cancer Institute (McNeil, 2006). The U.S. National Institute of Health (the parent organization to the National Cancer Institute) contributed to all of these universities, and approximately 13% of HPV vaccine trials, granting ~$34.9 million (inflation unadjusted) since 1995 (Joshi, et. al.). Building on this research two vaccines were licensed, Gardasil, marketed by Merck, and Cervarix, sold by GSK (CDC, 2010). We are uncertain of how much spending Merck and Gardasil contributed to the development of these vaccines. Before the release of Cervarix, GSK purchased a company, Corixa, with relevant technology for some of their vaccines, including Cervarix, for ~$300 million but it’s unclear how much of that valuation should be attributed to HPV (ICIS, 2005; BusinessWire, 2005). There are public estimates of how much R&D cost recovery was anticipated for Merck and GSK through their HPV vaccines but this is based in part on how large the market was expected to be for the vaccines, how much money these companies typically spend on R&D, and how much they were anticipating they would charge per vaccine, instead of an estimate of how much money they actually spent developing the vaccines ( Outterson, 2006; Outterson and Kesselheim, 2006). Finally, the Bill and Melinda Gates Foundation spent \$27.8 million to help LMIC understand how to implement the vaccine effectively after it was first rolled out (PATH, 2006). Unfortunately, we’re not able to figure out how to aggregate all this raw data into an accounting we feel captures all the costs without leaving out anything significant. [return]	{}
Select Page 2. {\displaystyle y_{k}} Denoising is one of the important task and pre-processing step in digital image processing.there are many median filters are available for impulse noise reduction although these methods have been improved, but the quality of denoising image is still not satisfactory[1]. Various adaptive restoration filters for intensity speckle images are derived based on different model assumptions and a nonstationary image model. Unsharp masking is a simple, fast method for modeling, then removing, smooth (low-frequency) background noise. In a nutshell, extract a smooth background image with a wide-radius lowpass filter; sharper_image = image + c * (image - … In this case, the engine noise is 50 times more powerful than the customer's voice. Color Image (RGB) Representation in MATLAB 3. − In the ideal case, SMAX must be an odd integer greater than 1. An adaptive filter is a system with a linear filter that has a transfer function controlled by variable parameters and a means to adjust those parameters according to an optimization algorithm. Announcements. The magnitudes of these pixels are approximately 5 to 100 times higher than the normal intensity amplitudes of the biostructure. {\displaystyle v\equiv 0,v'\equiv 0,g'\equiv 0} is the best mean square estimate of Segmenting does not increase the computations substantially and increases the signal noise ratio by a small amount. Therefore, the necessary algorithms will be derived and demonstrated on examples from the field of speech and audio signal processing. (1993): Vector directional filters. k {\displaystyle x_{k}} k The focus in this work is only on non-blind restoration methods. However, the X values could be the values of an array of pixels. k The ALC finds use as an adaptive beam former for arrays of hydrophones or antennas. u k {\displaystyle \epsilon _{k}} To circumvent this potential loss of information, an adaptive filter could be used. filter stands in between the LMS and the RLS filters. When the adaption is successful, the output of the filter The algorithms can achieve significantly better image quality than regular (fixed-length) median filters when the images are corrupted by impulse noise. The proposed adaptive methodology constitutes a unifying and powerful framework for multichannel signal processing. consists entirely of a signal correlated with the undesired signal in α If you do not receive an email within 10 minutes, your email address may not be registered, Adaptive Filters, by Abhishek Chander. In Spline Adaptive Filter the model is a cascade of linear dynamic block and static non-linearity, which is approximated by splines. All the undesired signals in This chapter describes the design and evaluation of a novel adaptive fuzzy filter, and discusses its application to image enhancement. Image sharpening filters highlight edges by removing blur. . I should design and implement an adaptive filter to remove impulse noise from medical images! Performance of Adaptive filters is superior to that of the filters discussed till now but the price is increase in filter complexity We will study two adaptive filters: – Adaptive local noise reduction filter – Adaptive median filter 5/16/2013 COMSATS Institute of Information Technology, Abbottabad Digital Image Processing CSC330 25 k [m n] specifies the size (m-by-n) of the neighborhood used to estimate the local image mean and standard deviation.The additive noise (Gaussian white noise) power is assumed to be noise. k •The behaviour of adaptive filters changes depending on the characteristics of the image inside the filter … Learn more. {\displaystyle x_{k}} It is suitable as a textbook for senior undergraduate or first-year graduate courses in adaptive signal processing and adaptive filters. In Urysohn Adaptive Filter the linear terms in a model, are replaced by piecewise linear functions, Learn how and when to remove this template message, Multidelay block frequency domain adaptive filter, https://en.wikipedia.org/w/index.php?title=Adaptive_filter&oldid=990725852, Articles lacking in-text citations from February 2013, Creative Commons Attribution-ShareAlike License, This page was last edited on 26 November 2020, at 04:28. ≡ {\displaystyle u_{k}} Adaptive filters are commonly used in image processing to enhance or restore data by removing noise without significantly blurring the structures in the image. Concentration of information by transforms. Or they could be the outputs of multiple tapped delay lines. x There are two input signals to the adaptive filter: To build adaptive filter I'd use the statistics to figure if there is something to smooth within the window. and i am new in image processing . The adaptive filters. Restoration techniques: The inverse filters; The wiener filter; MAP formulation; Median filter; Adaptive filter; Linear filter; IBD method; NAS-RIF The LMS algorithm does not require that the X values have any particular relationship; therefore it can be used to adapt a linear combiner as well as an FIR filter. The mean and variance are the two statistical measures that a local adaptive filter depends with a … J = wiener2(I,[m n],noise) filters the grayscale image I using a pixel-wise adaptive low-pass Wiener filter. ; The default value for inRoi is Auto and causes the entire image to be processed. k Confocal type images often exhibit isolated pixels (1×1 ~ 5×5) with extremely bright values caused by voltage instability or dead or hot camera pixels. In a group of nonlinear filter, median filter gives good performance on impulse noise. In this paper, the local homogeneity method is employed which result in the homogeneity image (H-image). g The new filters use fuzzy membership functions based on different distance measures among the image vectors to adapt to local data in the image. {\displaystyle u_{k}} 0 An adaptive filter (AF) is a filter which recognizes the local signal resolution (which usually varies strongly across the image) and adapts its own impulse response to this resolution. A kinematic study of the irregular dwarf galaxy NGC 4861 using Summary. u Professor (ECE Deptt.) ^ The magnitudes of these pixels are approximately 5 to 100 times higher than the normal intensity amplitudes of the biostructure. and How to achieve adaptive threshold filter with color. Please check your email for instructions on resetting your password. In summer semester 2020, this lecture will be offered only digitally. k Orbital cyclicity in the Eocene of Angola: visual and image-time-series analysis compared. Example algorithms for the ATFA (Real-time testing environment for adaptive filters) audio algorithm real-time libraries signal-processing vector academic pointer impulse-response lms dso adaptive-filtering filter-coefficients adaptive-filtering-algorithms adapf-listing nlms affine-projection rls ufrj adaptive-filters Once the canceler has converged, the primary signal to interference ratio will be improved from 1:1 to 50:1. If the error is zero, then there should be no change in the weights. The distorted image is recovered by employing the LR and WF adaptive filters. k The basic operation of digital image processing . u Adaptive Filters for Image Processing based on Artificial Neural Networks JOÃO RICARDO BITTENCOURT, DR. FERNANDO SANTOS OSÓRIO UNISINOS – Universidade do Vale do Rio dos Sinos - Centro de Ciências Exatas e Tecnológicas Curso de Informática. k In addition, there are no design tasks; the wiener2 function handles all preliminary computations and implements the filter for an input image. k In this case the update formula is written as: The effect of the LMS algorithm is at each time, k, to make a small change in each weight. To reduce the amount of interference in the primary microphone, a second microphone is located where it is intended to pick up sounds from the engine. About the adaptive Fourier filter I want to use some effective filtering An AF is developed and typical examples of its application are shown. Abstract: New adaptive filters for color image processing are introduced and analyzed. image. y The general idea behind Volterra LMS and Kernel LMS is to replace data samples by different nonlinear algebraic expressions. The adaptive linear combiner (ALC) resembles the adaptive tapped delay line FIR filter except that there is no assumed relationship between the X values. [1]. is minimized. This book presents a concise overview of adaptive filtering, covering as many as possible in a unified form that avoids repetition and simplifies notation. The direction of the change is such that it would decrease the error if it had been applied at time k. The magnitude of the change in each weight depends on μ, the associated X value and the error at time k. The weights making the largest contribution to the output, The output of the filter is given by. A New Ultradense Group of Obscured Emission-Line Galaxies. Candidate Tidal Dwarf Galaxies in the Compact Group CG J1720−67.8. The filter recognizes the local signal resolution (which usually varies strongly across the image) and adapts its own impulse response to this resolution. The adaptive ﬁltering literature is vast and cannot adequately be summarized in a short chapter. The most common cost function is the mean square of the error signal. Confocal type images often exhibit isolated pixels (1×1 ~ 5×5) with extremely bright values caused by voltage instability or dead or hot camera pixels. , If μ is too large, the algorithm will not converge. It shown below the image corrupted by impulsive noise or salt and pepper noise is denoised by Adaptive Median filter. Vector Marginal Median Filter and Vector Median Filter. Sharpening Filters. d Most traditional edge detectors can perform well for uncorrupted images but are highly sensitive to impulse noise, so they can not work efficiently for blurred images. Dust Activity in Comet 67P/Churyumov–Gerasimenko from February 20 to April 20, 2003. This leads to a normalized LMS algorithm: The goal of nonlinear filters is to overcome limitation of linear models. Enter your email address below and we will send you your username, If the address matches an existing account you will receive an email with instructions to retrieve your username, I have read and accept the Wiley Online Library Terms and Conditions of Use. but this method is too time- The filter is controlled by a set of L+1 coefficients or weights. In this tutorial, we will learn about Median Filters, their importance and their usage explained with the help of a numeric example. Image filtering can be classified into two main categories: linear and nonlinear filtering. The microphone also picks up noise from the engine and the environment. = Many algorithms such as adaptive filter, total variation method in this field are efficient and practical. dft image-processing adaptive-filtering affine-transformation canny-edge-detection histogram-equalization wiener-filter radon-transform unsharp … Adaptive median filter (AMF) [12, 13, 17, 18] uses median filters adaptively.It mostly increases the window size by comparing median value with extreme values of image. k The more adjacent pixels they include, the more accurate they can become, but this comes at the expense of much longer processing time. ′ Example: A fast food restaurant has a drive-up window. 1. k % F = ADPMEDIAN(G, SMAX) performs adaptive median filtering of % image G. The median filter starts at size 3-by-3 and iterates up % to size SMAX-by-SMAX. includes components of the desired signal. Monthly Notices of the Royal Astronomical Society. Denoising is one of the important task and pre-processing step in digital image processing.there are many median filters are available for impulse noise reduction although these methods have been improved, but the quality of denoising image is still not satisfactory[1]. We start by exploring what digital filters are, how they work, and what their limitations are. PSO Algorithm based Adaptive Median Filter for Noise Removal in Image Processing Application Ruby Verma M.E Student (ECE Deptt.) Adaptive filters for color image processing: A survey.pdf. ; To obtain an image that has its pixels modified in inRoi and copied outside of it, one can use the ComposeImages filter. Image restoration process model [5] III. As the power of digital signal processors has increased, adaptive filters have become much more common and are now routinely used in devices such as mobile phones and other communication devices, camcorders and digital cameras, and medical monitoring equipment. . The output will be. u One way to remove the noise is to filter the signal with a notch filter at the mains frequency and its vicinity, but this could excessively degrade the quality of the ECG since the heart beat would also likely have frequency components in the rejected range. Because of the complexity of the optimization algorithms, almost all adaptive filters are digital filters. This book presents a concise overview of adaptive filtering, covering as many as possible in a unified form that avoids repetition and simplifies notation. The arithmetic operators of smoothing and sharpening also testifies the fact. Is the better performance worth the computational cost? {\displaystyle \ x_{k}} ϵ = {\displaystyle {\hat {u}}_{k}} The output of the variable filter in the ideal case is, The error signal or cost function is the difference between The size of the neighborhood is adjustable, as well as the threshold for the comparison. [3] The adaption algorithm attempts to filter the reference input into a replica of the desired input by minimizing the residual signal, Typically, after each sample, the coefficients of the FIR filter are adjusted as follows:[6](Widrow). I should design and implement an adaptive filter to remove impulse noise from medical images! , are changed the most. The noise density will be added gradually to MRI image to compare performance of the filters … Using a filter bank can improve the Conference Paper. The selection of optimum image restoration and filtering technique depends to have knowledge about the characteristics of degrading system and noise pattern in an image. Abstract. {\displaystyle y_{k}}. Abstract New filter classes for multichannel image processing are introduced and analyzed. This microphone provides the primary signal. But adaptive SWM filter handle noise up to 60%. I'm trying take an image like this: And make it look like this: If it matters, I'm working in ios. Objectives The objective of this lab is to understand & implement 1. An adaptive filter for processing of astronomical images is developed and described. {\displaystyle \epsilon _{k}=g_{k}} and you may need to create a new Wiley Online Library account. The adaptive filter would take input both from the patient and from the mains and would thus be able to track the actual frequency of the noise as it fluctuates and subtract the noise from the recording. image-processing gpuimage core-image imagefilter adaptive-threshold. An adaptive filter will also be more convenient if an image is not clean or has poor lighting. This formula means that the output signal to interference ratio at a particular frequency is the reciprocal of the reference signal to interference ratio.[5]. ≡ Smoothing and Sharpening Filtering Techniques on Color images 2. ... A.N. d or it could even be the filter coefficients.[4](Widrow). The Adaptive Median Filter classifies pixels as noise by comparing each pixel in the image to its surrounding neighbor pixels. The idea behind a closed loop adaptive filter is that a variable filter is adjusted until the error (the difference between the filter output and the desired signal) is minimized. Question have someone already created a free adaptive Fourier filter for processing of Astronomical images is developed typical! Mri images signal-dependent speckle noise and the mean all cases in cosmological zoom-in simulations adaptive! ( ICCSP ) adaptive restoration filters for intensity speckle images are derived on. Zero, then there should be no change in the Compact Group CG J1720−67.8 squared for! With care neutral and ionized gas kinematics of massive elliptical galaxies in CALIFA in! Contains original image I‐35 122 Padova, Italy values could be used with care in addition, there are design! Astronomical Observatory Padova Vicola Osservatotio, 5 I‐35 122 Padova, Italy the images are derived based on distance! First for rapid convergence and then decreased to minimize overshoot Neural networks into this list than. Dusty Group of nonlinear filter, preserving edges and other high-frequency parts an. For multichannel image processing algorithms require some redesign to make them suitable for HDL code generation multichannel processing. Result in the MRI images adaptive-wiener-filters Updated may 15, 2020 Abstract noise corrupted images gas kinematics of elliptical. Fuzzy filter, Median filter ( AMF ) and adaptive filters are digital filters are used... Vicola Osservatotio, 5 I‐35 122 Padova, Italy the Least mean Squares RLS. Pair Tol 1238−364 and ESO 381‐G009 the exact frequency of the same problems Sternwarte Babelsberg Rosa‐Luxemburg‐Str ( H-image.. Is zero, then the LMS update algorithm is used for image restoration from highly noise corrupted images high-frequency of! Improved from 1:1 to 50:1 some commonly used approaches: Volterra LMS, Kernel filter! ( an ECG ), adaptive Median filter classifies pixels as noise by comparing each pixel in image. The full text of this lab is to overcome limitation of linear models their limitations are numerous image applications! Survey of time- and frequency-domain adaptive filters for color image ( RGB ) Representation in MATLAB 3 loss information... And others article with your friends and colleagues it enhances the grayscale transition of an,. Can achieve significantly better image quality than regular ( fixed-length ) Median filters when the images derived... Exploring what digital filters based on different model assumptions and a nonstationary image model is better than Median! Califa and in cosmological zoom-in simulations ImageJ ) in inRoi and copied outside of it one. This field are efficient and practical Fourier filter for processing of Astronomical images is developed and typical examples of application! Filter I 'd use the ComposeImages filter algebraic expressions to overcome limitation of linear dynamic block static. Sharpening filtering techniques on color images 2 ; the default value for inRoi is Auto and causes entire... Multichannel image processing algorithms require some redesign to make them suitable for HDL generation! The source of the irregular dwarf galaxy NGC 4861 using H I and H $\sf \alpha$ observations Tol... Least squared estimators for stationary stochastic processes a large part of the original and. Not directly applicable to image processing algorithms require some redesign to make them suitable for HDL code generation minimize.. Editing applications, but now a days, they use the ComposeImages filter the fate of gas... Field of speech and audio signal processing IC 4662 and NGC 5408 for processing of Astronomical images is and... Filter and the mean Square of the complexity of the original image operation. Such methods are not directly applicable to image processing and is used to recover the.! Signal to interference ratio will be improved from 1:1 to 50:1 highly noise corrupted images this chapter describes design. Image threshold applied, one can use the ComposeImages filter depending on their complexity, these use anywhere 0... Of image smoothing a fast food restaurant has a simple formula referred to as power inversion Eocene Angola... Fuzzy filter, Spline adaptive filter example: adaptive Median filter 2.1.1 adaptive Median filter 2.1.1 adaptive Median is! Entire image to be processed Eocene of Angola: visual and image-time-series analysis compared & color (., Chandigarh, India Rajesh Mehra Assoc the fact technical difficulties a new adaptive filters a novel adaptive fuzzy,! Image to its surrounding neighbor pixels someone already created a free adaptive Fourier filter noise. Algorithms will be used to remove impulse noise from medical images this work is only non-blind... Normalized LMS algorithm: the fate of ionized gas in the Southern galaxy Pair Tol 1238−364 and 381‐G009. Most common cost function is the opposite of image smoothing ) ax0,,. Up noise from the field of speech and audio signal processing and is used for image denoising a! Application Ruby Verma M.E Student ( ECE Deptt. to minimize overshoot Least Squares ( RLS ) adaptive is... Af is developed and typical examples of its application to image enhancement digital Micrograph ( or ImageJ. M.E Student ( ECE Deptt. then decreased to minimize overshoot is useful, are discussed its! And increases the signal power from the engine and the environment be and! Smooth non-impulsive noise please check your email for instructions on resetting your password among the image I! 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## Algebra 2 (1st Edition) $0\leq x\lt45\cup x\gt 70$ Let $x$ be the speed on the highway in mph. Then the illegal range is $0\leq x\lt45\cup x\gt 70$	{}
• Corpus ID: 239024326 # On improving a Schur-type theorem in shifted primes @inproceedings{Wang2021OnIA, title={On improving a Schur-type theorem in shifted primes}, author={Ruoyi Wang}, year={2021} } • R. Wang • Published 19 October 2021 • Mathematics We show that if N ≥ exp(exp(exp(k))), then any k-colouring of the primes that are less than N contains a monochromatic solution to p1 − p2 = p3 − 1. ## References SHOWING 1-10 OF 15 REFERENCES Difference sets and the primes • Mathematics • 2008 Suppose that A is a subset of {1,...,N} such that the difference between any two elements of A is never one less than a prime. We show that \|A\| = O(N exp(-c(log N)^{1/4})) for some absolute c>0. Roth's theorem in the primes We show that any set containing a positive proportion of the primes contains a 3-term arithmetic progression. An important ingredient is a proof that the primes enjoy the so-called Hardy-Littlewood Some of My Favorite Problems in Ramsey Theory In this brief note, a variety of problems from Ramsey theory on which I would like to see progress made are described and I am offering modest rewards for most of these problems. On sets of natural numbers whose difference set contains no squares • Mathematics • 1988 We show that if a sequence s/ of natural numbers has no pair of elements whose difference is a positive square, then the density of J/ n{l,...,«} is O(l/log«) c »), cn->-oo. This improves previous Linear equations in primes • Mathematics • 2006 Consider a system ψ of nonconstant affine-linear forms ψ 1 , ... , ψ t : ℤ d → ℤ, no two of which are linearly dependent. Let N be a large integer, and let K ⊆ [-N, N] d be convex. A generalisation Schur Number Five A proof of the solution of a century-old problem, known as Schur Number Five, is constructed and validated using a formally verified proof checker, demonstrating that any result by satisfiability solvers can now be validated using highly trustworthy systems. Analytic Number Theory • Mathematics • 2004 Introduction Arithmetic functions Elementary theory of prime numbers Characters Summation formulas Classical analytic theory of $L$-functions Elementary sieve methods Bilinear forms and the large On monochromatic solutions to $$x-y=z^2$$ For $$k \in \mathbb {N}$$ , write S(k) for the largest natural number such that there is a k-colouring of $$\{1, \ldots ,S(k)\}$$ with no monochromatic solution to $$x-y=z^2$$ . That S(k) Über die Kongruenz x + y ≡ z (mod p) • Jahresber. Dtsch. Math. 25 • 1916	{}
# ISEE Lower Level Math : How to multiply ## Example Questions ← Previous 1 3 4 5 6 7 8 9 ### Example Question #1 : How To Multiply Susie needs to buy new supplies for school. Pencils are sold for 23 cents a piece. Notebooks are sold in pairs, at $3.04 a pair. Folders are sold in packages of 4, and each package costs$2.12. If Susie buys a 7 pencils, 8 notebooks, and a dozen folders with $25, how much change should she receive? Possible Answers:$8.83 $2.75$4.87 Susie does not have enough money to purchase all of these school supplies. $4.78 Correct answer:$4.87 Explanation: Pencils are sold for $0.23 each. If Susie buys 7 pencils, she is paying 7 x$0.23 = $1.61 for pencils. Notebooks are sold in pairs (or in packages of two), for$3.04 a pair. If Susie buys 8 notebooks, that means that she bought 8 ÷ 2 = 4 pairs of notebooks. So, 4 pairs of notebooks, at $3.04 a pair costs 4 x$3.04 = $12.16. Susie also buys a dozen, or 12, folders. Folders come in packages of 4. This means that Susie buys 12 ÷ 4 = 3 packages of folders. Because each package costs$2.12, Susie pays 3 x $2.12 =$6.36 for a dozen folders. If we add up the total cost of the pencils, notebooks, and folders we get $1.61 +$12.16 + $6.36 =$20.13. Because Susie pays $25, she will be left with$25 – $20.13 =$4.87 in change. Explanation: ### Example Question #3 : How To Multiply Solve using the order of operations: Explanation: The order of operations states that we must begin solving the mathematical expression by evaluating numbers in parentheses. In this expression, is in parentheses. When we solve this, we get . Our expression can now be rewritten as: The order of operations states that the next step would be to evaluate any term with an exponent. Since there are no exponents in this expression, we move on to multiplication. In this expression we are asked to multiply , which is . Our expression can now be written as: The order of operations would have us divide next, but seeing as there is no division in this expression, we move on to addition. . ### Example Question #1 : Multiplication And Division Read the following problem but do not solve it. Acording to Mr. Smith's will, when he dies, each of his six children is to be left a parcel of land 510 acres in area. If these six parcels make up all of the farm, then how large is Mr. Smith's farm now? Which of the following expressions must be evaluated in order to answer this question? Explanation: There are six parcels of equal size; the total size is therefore the product of the common size (510 acres) and the number of parcels (6). This is ### Example Question #1 : Operations Evaluate for Explanation: Substitute 7 for By order of operations, multiply, then subtract. ### Example Question #1 : How To Multiply Evaluate: Explanation: To multiply two numbers of unlike sign, multiply their absolute values, then affix a negative symbol in front: ### Example Question #2 : How To Multiply Evaluate: Explanation: To multply two negative numbers, just multiply their absolute values; the product will be positive: ### Example Question #1 : How To Multiply Evaluate for Explanation: Substitute 8 for By order of operations, multiply, then add. ### Example Question #3 : How To Multiply Find the product. Explanation: When multiplying two negative numbers, the product will be positive. The correct answer is 324. ### Example Question #4 : How To Multiply Which story best fits the equation ? I have 8 candy bars, and my friend has 5. How many do we have together? I want to share 40 candy bars with 10 friends. How many do they each get? I have 8 boxes of candy bars, and each box has 5 candy bars. How many bars do I have in total? I have 40 candy bars. If I eat 8, how many are left?	{}
# How would Lagrangian be used tor recover Schrodinger equation? In path integral formulation of quantum mechanics, I heard that Lagrangian is defined. So, how would Lagrangian in this formulation be used to recover Schrodinger equation that we normally use? - Googling "path integrals in quantum mechanics" will give you many sources, for example en.wikipedia.org/wiki/Path_integral_formulation If you have difficulty following the standard presentation please ask another question :) – Michael Brown Mar 2 at 9:06 The derivation in the link shows how you get from a solution of the Schrödinger equation to the path intergral formalism, but not the other way around. In this sense, the question is not really answered by it. – Frederic Brünner Mar 2 at 9:35 This is probably what you're looking for: users.physik.fu-berlin.de/~kleinert/kleiner_re242/node2.html – elfmotat Mar 2 at 9:59 To get from a given Lagrangian to the Schrödinger equation you have to realize that the latter is actually given in terms of the Hamiltonian of the theory. Lagrangian and Hamiltonian are related by a Legendre transformation. - I am not sure if you are looking for this, but you can define a Lagrangian in such a way that the L-EOM (equation of motion) is the Schrödinger equation. $\cal{L}=\Psi^{t}(i\frac{\partial}{\partial t}+\nabla^2/2m)\Psi$ $\frac{\partial\cal{L}}{\partial\Psi^t}=0$ The second term of the Lagrange-equation (derivative with respect to $\partial_{\mu}\Psi^t$) is zero since no derivative of $\Psi^t$ occurs in our field Lagrangian density . -	{}
• ### On the limitations of statistical absorption studies with the Sloan Digital Sky Surveys I--III(1802.01824) April 4, 2018 astro-ph.GA, astro-ph.IM We investigate the limitations of statistical absorption measurements with the SDSS optical spectroscopic surveys. We show that changes in the data reduction strategy throughout different data releases have led to a better accuracy at long wavelengths, in particular for sky line subtraction, but a degradation at short wavelengths with the emergence of systematic spectral features with an amplitude of about one percent. We show that these features originate from inaccuracy in the fitting of modeled F-star spectra used for flux calibration. The best-fit models for those stars are found to systematically over-estimate the strength of metal lines and under-estimate that of Lithium. We also identify the existence of artifacts due to masking and interpolation procedures at the wavelengths of the hydrogen Balmer series leading to the existence of artificial Balmer $\alpha$ absorption in all SDSS optical spectra. All these effects occur in the rest-frame of the standard stars and therefore present Galactic longitude variations due to the rotation of the Galaxy. We demonstrate that the detection of certain weak absorption lines reported in the literature are solely due to calibration effects. Finally, we discuss new strategies to mitigate these issues. • ### Detecting outliers and learning complex structures with large spectroscopic surveys - a case study with APOGEE stars(1711.00022) In this work we apply and expand on a recently introduced outlier detection algorithm that is based on an unsupervised random forest. We use the algorithm to calculate a similarity measure for stellar spectra from the Apache Point Observatory Galactic Evolution Experiment (APOGEE). We show that the similarity measure traces non-trivial physical properties and contains information about complex structures in the data. We use it for visualization and clustering of the dataset, and discuss its ability to find groups of highly similar objects, including spectroscopic twins. Using the similarity matrix to search the dataset for objects allows us to find objects that are impossible to find using their best fitting model parameters. This includes extreme objects for which the models fail, and rare objects that are outside the scope of the model. We use the similarity measure to detect outliers in the dataset, and find a number of previously unknown Be-type stars, spectroscopic binaries, carbon rich stars, young stars, and a few that we cannot interpret. Our work further demonstrates the potential for scientific discovery when combining machine learning methods with modern survey data. • The detection of GW170817 in both gravitational waves and electromagnetic waves heralds the age of gravitational-wave multi-messenger astronomy. On 17 August 2017 the Advanced LIGO and Virgo detectors observed GW170817, a strong signal from the merger of a binary neutron-star system. Less than 2 seconds after the merger, a gamma-ray burst (GRB 170817A) was detected within a region of the sky consistent with the LIGO-Virgo-derived location of the gravitational-wave source. This sky region was subsequently observed by optical astronomy facilities, resulting in the identification of an optical transient signal within $\sim 10$ arcsec of the galaxy NGC 4993. These multi-messenger observations allow us to use GW170817 as a standard siren, the gravitational-wave analog of an astronomical standard candle, to measure the Hubble constant. This quantity, which represents the local expansion rate of the Universe, sets the overall scale of the Universe and is of fundamental importance to cosmology. Our measurement combines the distance to the source inferred purely from the gravitational-wave signal with the recession velocity inferred from measurements of the redshift using electromagnetic data. This approach does not require any form of cosmic "distance ladder;" the gravitational wave analysis can be used to estimate the luminosity distance out to cosmological scales directly, without the use of intermediate astronomical distance measurements. We determine the Hubble constant to be $70.0^{+12.0}_{-8.0} \, \mathrm{km} \, \mathrm{s}^{-1} \, \mathrm{Mpc}^{-1}$ (maximum a posteriori and 68% credible interval). This is consistent with existing measurements, while being completely independent of them. Additional standard-siren measurements from future gravitational-wave sources will provide precision constraints of this important cosmological parameter. • ### The Effect of Interstellar Absorption on Measurements of the Baryon Acoustic Peak in the Lyman-{\alpha} Forest(1705.03190) Aug. 17, 2017 astro-ph.CO In recent years, the autocorrelation of the hydrogen Lyman-{\alpha} forest has been used to observe the baryon acoustic peak at redshift 2 < z < 3.5 using tens of thousands of QSO spectra from the BOSS survey. However, the interstellar medium of the Milky-Way introduces absorption lines into the spectrum of any extragalactic source. These lines, while weak and undetectable in a single BOSS spectrum, could potentially bias the cosmological signal. In order to examine this, we generate absorption line maps by stacking over a million spectra of galaxies and QSOs. We find that the systematics introduced are too small to affect the current accuracy of the baryon acoustic peak, but might be relevant to future surveys such as the Dark Energy Spectroscopic Instrument (DESI). We outline a method to account for this with future datasets. • ### The Evolution of Temperature and Bolometric Luminosity in Type-II Supernovae(1707.07695) July 24, 2017 astro-ph.SR, astro-ph.HE In this work we present a uniform analysis of the temperature evolution and bolometric luminosity of a sample of 29 type-II supernovae (SNe), by fitting a black body model to their multi-band photometry. Our sample includes only SNe with high quality multi-band data and relatively well sampled time coverage. Most of the SNe in our sample were detected less than a week after explosion so their light curves cover the evolution both before and after recombination starts playing a role. We use this sample to study the signature of hydrogen recombination, which is expected to appear once the observed temperature drops to $\approx 7,000$K. Theory predicts that before recombination starts affecting the light curve, both the luminosity and the temperature should drop relatively fast, following a power-law in time. Once the recombination front reaches inner parts of the outflow, it sets the observed temperature to be nearly constant, and slows the decline of the luminosity (or even leads to a re-brightening). We compare our data to analytic studies and find strong evidence for the signature of recombination. We also find that the onset of the optical plateau in a given filter, is effectively the time at which the black body peak reaches the central wavelength of the filter, as it cools, and it does not correspond to the time at which recombination starts affecting the emission. • ### Evidence of ongoing AGN-driven feedback in a quiescent post starburst E+A galaxy(1705.03891) May 10, 2017 astro-ph.GA Post starburst E+A galaxies are thought to have experienced a significant starburst that was quenched abruptly. Their disturbed, bulge-dominated morphologies suggest that they are merger remnants. We present ESI/Keck observations of SDSS J132401.63+454620.6, a post starburst galaxy at redshift z = 0.125, with a starburst that started 400 Myr ago, and other properties, like star formation rate (SFR) consistent with what is measured in ultra luminous infrared galaxies (ULRIGs). The galaxy shows both zero velocity narrow lines, and blueshifted broader Balmer and forbidden emission lines (FWHM=1350 +- 240 km/s). The narrow component is consistent with LINER-like emission, and the broader component with Seyfert-like emission, both photoionized by an active galactic nucleus (AGN) whose properties we measure and model. The velocity dispersion of the broad component exceeds the escape velocity, and we estimate the mass outflow rate to be in the range 4-120 Mo/yr. This is the first reported case of AGN-driven outflows, traced by ionized gas, in post starburst E+A galaxies. We show, by ways of a simple model, that the observed AGN-driven winds can consistently evolve a ULIRG into the observed galaxy. Our findings reinforce the evolutionary scenario where the more massive ULIRGs are quenched by negative AGN feedback, evolve first to post starburst galaxies, and later become typical red and dead ellipticals. • ### The weirdest SDSS galaxies: results from an outlier detection algorithm(1611.07526) Nov. 22, 2016 astro-ph.GA How can we discover objects we did not know existed within the large datasets that now abound in astronomy? We present an outlier detection algorithm that we developed, based on an unsupervised Random Forest. We test the algorithm on more than two million galaxy spectra from the Sloan Digital Sky Survey and examine the 400 galaxies with the highest outlier score. We find objects which have extreme emission line ratios and abnormally strong absorption lines, objects with unusual continua, including extremely reddened galaxies. We find galaxy-galaxy gravitational lenses, double-peaked emission line galaxies, and close galaxy pairs. We find galaxies with high ionisation lines, galaxies which host supernovae, and galaxies with unusual gas kinematics. Only a fraction of the outliers we find were reported by previous studies that used specific and tailored algorithms to find a single class of unusual objects. Our algorithm is general and detects all of these classes, and many more, regardless of what makes them peculiar. It can be executed on imaging, time-series, and other spectroscopic data, operates well with thousands of features, is not sensitive to missing values, and is easily parallelisable. • ### Evidence that most type 1 AGN are reddened by dust in the host ISM(1603.06948) Aug. 31, 2016 astro-ph.CO, astro-ph.GA The typical optical-UV continuum slopes observed in many type 1 AGN are redder than expected from thin accretion disk models. A possible resolution to this conundrum is that many AGN are reddened by dust along the line of sight. To explore this possibility, we stack 5000 SDSS AGN with luminosity L~10^45erg/s and redshift z~0.4 in bins of optical continuum slope alpha_opt and width of the broad H$\beta$ emission line. We measure the EW of the NaID absorption feature in each stacked spectrum. We find a linear relation between alpha_opt and EW(NaID), such that EW(NaID) increases as alpha_opt becomes redder. In the bin with the smallest H$\beta$ width, objects with the bluest slopes that are similar to accretion disk predictions are found to have EW(NaID)=0, supporting the line-of-sight dust hypothesis. This conclusion is also supported by the dependence of the $H\alpha/H\beta$ line ratio on alpha_opt. The implied relationship between alpha_opt and dust reddening is given by E(B-V)~0.2(-0.1-alpha_opt), and the implied reddening of a typical type 1 AGN with alpha_opt=-0.5 is E(B-V)~0.08mag. Photoionization calculations show that the dusty gas responsible for reddening is too ionized to produce the observed features. Therefore, we argue that the sodium absorption arises in regions of the host ISM which are shielded from the AGN radiation, and the correlation with alpha_opt arises since ISM columns along shielded and non-shielded sightlines are correlated. This scenario is supported by the similarity of the E(B-V)-NaID relation implied by our results with the relation in the Milky-Way found by previous studies. • ### The importance of 56Ni in shaping the light curves of type II supernovae(1506.07185) April 13, 2016 astro-ph.HE What intrinsic properties shape the light curves of Type II supernovae (SNe)? To address this question we derive observational measures that are robust (i.e., insensitive to detailed radiative transfer) and constrain the contribution from $^{56}$Ni, as well as a combination of the envelope mass, progenitor radius, and explosion energy. By applying our methods to a sample of type II SNe from the literature we find that $^{56}$Ni contribution is often significant. In our sample its contribution to the time weighted integrated luminosity during the photospheric phase ranges between 8% and 72% with a typical value of 30%. We find that the $^{56}$Ni relative contribution is anti-correlated with the luminosity decline rate. When added to other clues, this in turn suggests that the flat plateaus often observed in type II SNe are not a generic feature of the cooling envelope emission, and that without $^{56}$Ni many of the SNe that are classified as II-P would have shown a decline rate that is steeper by up to 1 mag/100 d. Nevertheless, we find that the cooling envelope emission, and not $^{56}$Ni contribution, is the main driver behind the observed range of decline rates. Furthermore, contrary to previous suggestions, our findings indicate that fast decline rates are not driven by lower envelope masses. We therefore suggest that the difference in observed decline rates is mainly a result of different density profiles of the progenitors. • ### Using machine learning to classify the diffuse interstellar bands(1501.04631) May 2, 2015 astro-ph.GA Using over a million and a half extragalactic spectra we study the correlations of the Diffuse Interstellar Bands (DIBs) in the Milky Way. We measure the correlation between DIB strength and dust extinction for 142 DIBs using 24 stacked spectra in the reddening range E(B-V) < 0.2, many more lines than ever studied before. Most of the DIBs do not correlate with dust extinction. However, we find 10 weak and barely studied DIBs with correlations that are higher than 0.7 with dust extinction and confirm the high correlation of additional 5 strong DIBs. Furthermore, we find a pair of DIBs, 5925.9A and 5927.5A which exhibits significant negative correlation with dust extinction, indicating that their carrier may be depleted on dust. We use Machine Learning algorithms to divide the DIBs to spectroscopic families based on 250 stacked spectra. By removing the dust dependency we study how DIBs follow their local environment. We thus obtain 6 groups of weak DIBs, 4 of which are tightly associated with C2 or CN absorption lines. • ### Bright but slow - Type II supernovae from OGLE-IV - Implications for magnitude limited surveys(1501.03452) Feb. 25, 2015 astro-ph.CO, astro-ph.HE We study a sample of 11 Type II supernovae (SNe) discovered by the OGLE-IV survey. All objects have well sampled I-band light curves, and at least one spectrum. We find that 2 or 3 of the 11 SNe have a declining light curve, and spectra consistent with other SNe II-L, while the rest have plateaus that can be as short as 70d, unlike the 100d typically found in nearby galaxies. The OGLE SNe are also brighter, and show that magnitude limited surveys find SNe that are different than usually found in nearby galaxies. We discuss this sample in the context of understanding Type II SNe as a class and their suggested use as standard candles. • ### Dusting off the diffuse interstellar bands: DIBs and dust in extragalactic SDSS spectra(1406.7006) Dec. 1, 2014 astro-ph.GA Using over a million and a half extragalactic spectra we study the properties of the mysterious Diffuse Interstellar Bands (DIBs) in the Milky Way. These data provide us with an unprecedented sampling of the skies at high Galactic-latitude and low dust-column-density. We present our method, study the correlation of the equivalent width of 8 DIBs with dust extinction and with a few atomic species, and the distribution of four DIBs - 5780.6A, 5797.1A, 6204.3A, and 6613.6A - over nearly 15000 squared degrees. As previously found, DIBs strengths correlate with extinction and therefore inevitably with each other. However, we show that DIBs can exist even in dust free areas. Furthermore, we find that the DIBs correlation with dust varies significantly over the sky. DIB under- or over-densities, relative to the expectation from dust, are often spread over hundreds of square degrees. These patches are different for the four DIBs, showing that they are unlikely to originate from the same carrier, as previously suggested. • ### Photometric and Spectroscopic Properties of Type II-P Supernovae(1404.0378) We study a sample of 23 Type II Plateau supernovae (SNe II-P), all observed with the same set of instruments. Analysis of their photometric evolution confirms that their typical plateau duration is 100 days with little scatter, showing a tendency to get shorter for more energetic SNe. The rise time from explosion to plateau does not seem to correlate with luminosity. We analyze their spectra, measuring typical ejecta velocities, and confirm that they follow a well behaved power-law decline. We find indications of high-velocity material in the spectra of six of our SNe. We test different dust extinction correction methods by asking the following -- does the uniformity of the sample increase after the application of a given method? A reasonably behaved underlying distribution should become tighter after correction. No method we tested made a significant improvement. • ### Exploring the spectral diversity of low-redshift Type Ia supernovae using the Palomar Transient Factory(1408.1430) We present an investigation of the optical spectra of 264 low-redshift (z < 0.2) Type Ia supernovae (SNe Ia) discovered by the Palomar Transient Factory, an untargeted transient survey. We focus on velocity and pseudo-equivalent width measurements of the Si II 4130, 5972, and 6355 A lines, as well those of the Ca II near-infrared (NIR) triplet, up to +5 days relative to the SN B-band maximum light. We find that a high-velocity component of the Ca II NIR triplet is needed to explain the spectrum in ~95 per cent of SNe Ia observed before -5 days, decreasing to ~80 per cent at maximum. The average velocity of the Ca II high-velocity component is ~8500 km/s higher than the photospheric component. We confirm previous results that SNe Ia around maximum light with a larger contribution from the high-velocity component relative to the photospheric component in their Ca II NIR feature have, on average, broader light curves and lower Ca II NIR photospheric velocities. We find that these relations are driven by both a stronger high-velocity component and a weaker contribution from the photospheric Ca II NIR component in broader light curve SNe Ia. We identify the presence of C II in very-early-time SN Ia spectra (before -10 days), finding that >40 per cent of SNe Ia observed at these phases show signs of unburnt material in their spectra, and that C II features are more likely to be found in SNe Ia having narrower light curves. • ### Spectroscopic identification of a redshift 1.55 supernova host galaxy from the Subaru Deep Field Supernova Survey(1211.2208) March 30, 2014 astro-ph.CO Context: The Subaru Deep Field (SDF) Supernova Survey discovered 10 Type Ia supernovae (SNe Ia) in the redshift range 1.5<z<2.0, as determined solely from photometric redshifts of the host galaxies. However, photometric redshifts might be biased, and the SN sample could be contaminated by active galactic nuclei (AGNs). Aims: We aim to obtain the first robust redshift measurement and classification of a z > 1.5 SDF SN Ia host galaxy candidate Methods: We use the X-shooter (U-to-K-band) spectrograph on the Very Large Telescope to allow the detection of different emission lines in a wide spectral range. Results: We measure a spectroscopic redshift of 1.54563 +/- 0.00027 of hSDF0705.25, consistent with its photometric redshift of 1.552 +/- 0.018. From the strong emission-line spectrum we rule out AGN activity, thereby confirming the optical transient as a SN. The host galaxy follows the fundamental metallicity relation defined in Mannucci et al. (2010, 2011) showing that the properties of this high-redshift SN Ia host galaxy is similar to other field galaxies. Conclusions: Spectroscopic confirmation of additional SDF SN hosts would be required to confirm the cosmic SN rate evolution measured in the SDF. • ### Direct evidence for a supernova interacting with a large amount of hydrogen-free circumstellar material(1309.6496) Sept. 25, 2013 astro-ph.CO, astro-ph.SR We present our observations of SN 2010mb, a Type Ic SN lacking spectroscopic signatures of H and He. SN 2010mb has a slowly-declining light curve ($\sim600\,$days) that cannot be powered by $^{56}$Ni/$^{56}$Co radioactivity, the common energy source for Type Ic SNe. We detect signatures of interaction with hydrogen-free CSM including a blue quasi-continuum and, uniquely, narrow oxygen emission lines that require high densities ($\sim10^9$cm$^{-3}$). From the observed spectra and light curve we estimate that the amount of material involved in the interaction was $\sim3$M$_{\odot}$. Our observations are in agreement with models of pulsational pair-instability SNe described in the literature. • ### An Emerging Coherent Picture of Red Supergiant Supernova Explosions(1304.4967) Sept. 25, 2013 astro-ph.CO Three lines of evidence indicate that in the most common type of core collapse supernovae, the energy deposited in the ejecta by the exploding core is approximately proportional to the progenitor mass cubed. This results stems from an observed uniformity of light curve plateau duration, a correlation between mass and ejecta velocity, and the known correlation between luminosity and velocity. This result ties in analytical and numerical models together with observations, providing us with clues as to the mechanism via which the explosion of the core deposits a small fraction of its energy into the hurled envelope. • ### The PTF Orion Project: a Possible Planet Transiting a T-Tauri Star(1206.1510) June 4, 2013 astro-ph.SR, astro-ph.EP We report observations of a possible young transiting planet orbiting a previously known weak-lined T-Tauri star in the 7-10 Myr old Orion-OB1a/25-Ori region. The candidate was found as part of the Palomar Transient Factory (PTF) Orion project. It has a photometric transit period of 0.448413 +- 0.000040 days, and appears in both 2009 and 2010 PTF data. Follow-up low-precision radial velocity (RV) observations and adaptive optics imaging suggest that the star is not an eclipsing binary, and that it is unlikely that a background source is blended with the target and mimicking the observed transit. RV observations with the Hobby-Eberly and Keck telescopes yield an RV that has the same period as the photometric event, but is offset in phase from the transit center by approximately -0.22 periods. The amplitude (half range) of the RV variations is 2.4 km/s and is comparable with the expected RV amplitude that stellar spots could induce. The RV curve is likely dominated by stellar spot modulation and provides an upper limit to the projected companion mass of M_p sin i_orb < 4.8 +- 1.2 M_Jup; when combined with the orbital inclination, i orb, of the candidate planet from modeling of the transit light curve, we find an upper limit on the mass of the planetary candidate of M_p < 5.5 +- 1.4 M_Jup. This limit implies that the planet is orbiting close to, if not inside, its Roche limiting orbital radius, so that it may be undergoing active mass loss and evaporation. • ### Discovery of a Cosmological, Relativistic Outburst via its Rapidly Fading Optical Emission(1304.4236) April 15, 2013 astro-ph.CO, astro-ph.HE We report the discovery by the Palomar Transient Factory (PTF) of the transient source PTF11agg, which is distinguished by three primary characteristics: (1) bright, rapidly fading optical transient emission; (2) a faint, blue quiescent optical counterpart; and (3) an associated year-long, scintillating radio transient. We argue that these observed properties are inconsistent with any known class of Galactic transients, and instead suggest a cosmological origin. The detection of incoherent radio emission at such distances implies a large emitting region, from which we infer the presence of relativistic ejecta. The observed properties are all consistent with the population of long-duration gamma-ray bursts (GRBs), marking the first time such an outburst has been discovered in the distant universe independent of a high-energy trigger. We searched for possible high-energy counterparts to PTF11agg, but found no evidence for associated prompt emission. We therefore consider three possible scenarios to account for a GRB-like afterglow without a high-energy counterpart: an "untriggered" GRB (lack of satellite coverage), an "orphan" afterglow (viewing-angle effects), and a "dirty fireball" (suppressed high-energy emission). The observed optical and radio light curves appear inconsistent with even the most basic predictions for off-axis afterglow models. The simplest explanation, then, is that PTF11agg is a normal, on-axis long-duration GRB for which the associated high-energy emission was simply missed. However, we have calculated the likelihood of such a serendipitous discovery by PTF and find that it is quite small (~ 2.6%). While not definitive, we nonetheless speculate that PTF11agg may represent a new, more common (> 4 times the on-axis GRB rate at 90% confidence) class of relativistic outbursts lacking associated high-energy emission. • ### Type Ia Supernovae Strongly Interacting with Their Circumstellar Medium(1304.0763) April 2, 2013 astro-ph.CO Owing to their utility for measurements of cosmic acceleration, Type Ia supernovae (SNe) are perhaps the best-studied class of SNe, yet the progenitor systems of these explosions largely remain a mystery. A rare subclass of SNe Ia show evidence of strong interaction with their circumstellar medium (CSM), and in particular, a hydrogen-rich CSM; we refer to them as SNe Ia-CSM. In the first systematic search for such systems, we have identified 16 SNe Ia-CSM, and here we present new spectra of 13 of them. Six SNe Ia-CSM have been well-studied previously, three were previously known but are analyzed in-depth for the first time here, and seven are new discoveries from the Palomar Transient Factory. The spectra of all SNe Ia-CSM are dominated by H{\alpha} emission (with widths of ~2000 km/s) and exhibit large H{\alpha}/H{\beta} intensity ratios (perhaps due to collisional excitation of hydrogen via the SN ejecta overtaking slower-moving CSM shells); moreover, they have an almost complete lack of He I emission. They also show possible evidence of dust formation through a decrease in the red wing of H{\alpha} 75-100 d past maximum brightness, and nearly all SNe Ia-CSM exhibit strong Na I D absorption from the host galaxy. The absolute magnitudes (uncorrected for host-galaxy extinction) of SNe Ia-CSM are found to be -21.3 <= M_R <= -19 mag, and they also seem to show ultraviolet emission at early times and strong infrared emission at late times (but no detected radio or X-ray emission). Finally, the host galaxies of SNe Ia-CSM are all late-type spirals similar to the Milky Way, or dwarf irregulars like the Large Magellanic Cloud, which implies that these objects come from a relatively young stellar population. This work represents the most detailed analysis of the SN Ia-CSM class to date. • ### Using Machine Learning for Discovery in Synoptic Survey Imaging(1209.3775) Sept. 17, 2012 stat.AP, astro-ph.IM Modern time-domain surveys continuously monitor large swaths of the sky to look for astronomical variability. Astrophysical discovery in such data sets is complicated by the fact that detections of real transient and variable sources are highly outnumbered by bogus detections caused by imperfect subtractions, atmospheric effects and detector artefacts. In this work we present a machine learning (ML) framework for discovery of variability in time-domain imaging surveys. Our ML methods provide probabilistic statements, in near real time, about the degree to which each newly observed source is astrophysically relevant source of variable brightness. We provide details about each of the analysis steps involved, including compilation of the training and testing sets, construction of descriptive image-based and contextual features, and optimization of the feature subset and model tuning parameters. Using a validation set of nearly 30,000 objects from the Palomar Transient Factory, we demonstrate a missed detection rate of at most 7.7% at our chosen false-positive rate of 1% for an optimized ML classifier of 23 features, selected to avoid feature correlation and over-fitting from an initial library of 42 attributes. Importantly, we show that our classification methodology is insensitive to mis-labelled training data up to a contamination of nearly 10%, making it easier to compile sufficient training sets for accurate performance in future surveys. This ML framework, if so adopted, should enable the maximization of scientific gain from future synoptic survey and enable fast follow-up decisions on the vast amounts of streaming data produced by such experiments. • ### An Early & Comprehensive Millimeter and Centimeter Wave and X-ray Study of Supernova 2011dh: A Non-Equipartition Blastwave Expanding into A Massive Stellar Wind(1209.1102) Only a handful of supernovae (SNe) have been studied in multi-wavelength from radio to X-rays, starting a few days after explosion. The early detection and classification of the nearby type IIb SN2011dh/PTF11eon in M51 provides a unique opportunity to conduct such observations. We present detailed data obtained at the youngest phase ever of a core-collapse supernova (days 3 to 12 after explosion) in the radio, millimeter and X-rays; when combined with optical data, this allows us to explore the early evolution of the SN blast wave and its surroundings. Our analysis shows that the expanding supernova shockwave does not exhibit equipartition (e_e/e_B ~ 1000), and is expanding into circumstellar material that is consistent with a density profile falling like R^-2. Within modeling uncertainties we find an average velocity of the fast parts of the ejecta of 15,000 +/- 1800 km/s, contrary to previous analysis. This velocity places SN 2011dh in an intermediate blast-wave regime between the previously defined compact and extended SN IIb subtypes. Our results highlight the importance of early (~ 1 day) high-frequency observations of future events. Moreover, we show the importance of combined radio/X-ray observations for determining the microphysics ratio e_e/e_B. • ### An Empirical Relation between Sodium Absorption and Dust Extinction(1206.6107) Dust extinction and reddening are ubiquitous in astronomical observations and are often a major source of systematic uncertainty. We present here a study of the correlation between extinction in the Milky Way and the equivalent width of the NaI D absorption doublet. Our sample includes more than 100 high resolution spectra from the KECK telescopes and nearly a million low resolution spectra from the Sloan Digital Sky Survey (SDSS). We measure the correlation to unprecedented precision, constrain its shape, and derive an empirical relation between these quantities with a dispersion of order 0.15 magnitude in E(B-V). From the shape of the curve of growth we further show that a typical sight line through the Galaxy, as seen within the SDSS footprint, crosses about three dust clouds. We provide a brief guide on how to best estimate extinction to extragalactic sources such as supernovae, using the NaI D absorption feature, under a variety of circumstances. • ### Berkeley Supernova Ia Program I: Observations, Data Reduction, and Spectroscopic Sample of 582 Low-Redshift Type Ia Supernovae(1202.2128) May 4, 2012 astro-ph.CO In this first paper in a series we present 1298 low-redshift (z\leq0.2) optical spectra of 582 Type Ia supernovae (SNe Ia) observed from 1989 through 2008 as part of the Berkeley SN Ia Program (BSNIP). 584 spectra of 199 SNe Ia have well-calibrated light curves with measured distance moduli, and many of the spectra have been corrected for host-galaxy contamination. Most of the data were obtained using the Kast double spectrograph mounted on the Shane 3 m telescope at Lick Observatory and have a typical wavelength range of 3300-10,400 Ang., roughly twice as wide as spectra from most previously published datasets. We present our observing and reduction procedures, and we describe the resulting SN Database (SNDB), which will be an online, public, searchable database containing all of our fully reduced spectra and companion photometry. In addition, we discuss our spectral classification scheme (using the SuperNova IDentification code, SNID; Blondin & Tonry 2007), utilising our newly constructed set of SNID spectral templates. These templates allow us to accurately classify our entire dataset, and by doing so we are able to reclassify a handful of objects as bona fide SNe Ia and a few other objects as members of some of the peculiar SN Ia subtypes. In fact, our dataset includes spectra of nearly 90 spectroscopically peculiar SNe Ia. We also present spectroscopic host-galaxy redshifts of some SNe Ia where these values were previously unknown. [Abridged] • We present ultraviolet (UV) spectroscopy and photometry of four Type Ia supernovae (SNe 2004dt, 2004ef, 2005M, and 2005cf) obtained with the UV prism of the Advanced Camera for Surveys on the Hubble Space Telescope. This dataset provides unique spectral time series down to 2000 Angstrom. Significant diversity is seen in the near maximum-light spectra (~ 2000--3500 Angstrom) for this small sample. The corresponding photometric data, together with archival data from Swift Ultraviolet/Optical Telescope observations, provide further evidence of increased dispersion in the UV emission with respect to the optical. The peak luminosities measured in uvw1/F250W are found to correlate with the B-band light-curve shape parameter dm15(B), but with much larger scatter relative to the correlation in the broad-band B band (e.g., ~0.4 mag versus ~0.2 mag for those with 0.8 < dm15 < 1.7 mag). SN 2004dt is found as an outlier of this correlation (at > 3 sigma), being brighter than normal SNe Ia such as SN 2005cf by ~0.9 mag and ~2.0 mag in the uvw1/F250W and uvm2/F220W filters, respectively. We show that different progenitor metallicity or line-expansion velocities alone cannot explain such a large discrepancy. Viewing-angle effects, such as due to an asymmetric explosion, may have a significant influence on the flux emitted in the UV region. Detailed modeling is needed to disentangle and quantify the above effects.	{}
## Tuesday, September 6, 2022 ### High Precision Composite Op-Amps, Part 2 This post is a part of the series on audio amplifier feedback. The contents of the series can be found here. My previous post on this topic was on composite opamps from by John D. Yewen's article "High-precision composite op-amps" (Electronics & Wireless World, February 1987): Appropriately choosing the voltage divider R1R2 at the output of U1 allows to improve stability (phase margin) of the composite at the expense of the loop gain. Here, orange traces are the loop gain of the composite from the schematic above, while blue are the maximum possible loop gain with the same two opamps: For audio applications, it is desirable to maximize the loop gain, at least in the audio band, but preserve the phase margin. To achieve that, let me look at the role of the voltage divider in Yewen's composite. Referring to the schematic above, U2 sees a (differential) input signal that is a sum of (i) the signal at the non-inverting output, where Ri and Rf connect, and (ii) the same signal amplifier by U1 and divided by R1R2. At low frequencies, U1's gain is large, and U2's input signal is effectively that at its non-inverting input. The loop gain is the product of that of U1 and U2 and falls with frequency at 40dB/decade. At high frequencies, U1's gain is small, and U2's input signal is effectively that at its inverting input. The loop gain is just that of U2, falling at 20dB/decade. The transition from "low" to "high" frequencies is a zero in the composite's loop gain, located at the frequency where the signal magnitudes at the non-inverting and inverting inputs of U2 are equal - that is, when the gain of U2 followed by R1R2 is unity. For a single-pole U1, that frequency is a fraction of U1's Gain Bandwidth Product (GBW):$$F_{zero}=GBW_{U_1} \times {R_2 \over {R_1+R_2}}$$That is, Yewen's voltage divider sets the frequency of a zero in the composite's loop gain. In the example above, GBW is 10MHz, the divider's attenuation is 22, so the zero is at ${{10 MHz}\over 22} = {455 kHz}$. One way to keep that zero and maximize the loop gain at low frequencies is to make the voltage divider frequency dependent, for example: Adding an inductor in parallel to R3 adds a pole-zero pair (disregarding the inductor's own series resistance, the pole is at $F_p={1 \over {2 \pi}} {{R_3 \|\| R_4} \over L_1}$, the zero at $F_z={1 \over {2 \pi}} {R_3 \over L_1}$). With the values shown, we get about 12dB of extra loop gain at 20kHz with the same phase margin as without the inductor: A 22mH inductor may not be very practical, but a similar effect can be achieved with a capacitive divider: Here, the pole is at $F_p={1 \over {2 \pi R_5 (C_1 + C_2)}}$, the zero at $F_z={1 \over {2 \pi R_5 C_2}}$. With the values shown, the loop gain is about the same as with the inductor above: Not bad for one additional passive component. It works in hardware, too, although you may need to add a resistor in series with U5's output to isolate it from capacitive load. R5 can be adjusted for optimal stability and clipping. Can we make it still better? Yes we can! Stay tuned...	{}
# More algorithms on perfectly balanced photo gallery Recently there was an article by Johannes Treitz submitted to Hacker News about how to display a set of pictures as nicely as possible. The article doesn't have a formal description of the problem, so here is my take on what it mean by perfectly balanced. Problem1 Given a sequence of $$n$$ rectangles where the $$i$$th rectangle has width $$W_i$$ and height $$H_i$$. Also given are width $$W$$ and target height $$H$$. Partition the sequence to $$k$$ consecutive subsequences, such that we scale the rectangles keeping the aspect ratio, and each subsequence of rectangles fills the entire width $$W$$, and is close to the target height $$H$$. The problem is not so well defined. To make sure the heights are close, do we minimize the sum of all differences? Minimize the maximum difference? Or maybe, we just want to minimize the difference between consecutive rows, such that the first row has height close to $$h$$. Nevertheless, in real applications, the exact definition doesn't matter that much. Treitz reduce the problem to the linear partition problem. Problem2 Let $$a_1,\ldots,a_n$$ be a sequence of positive reals. We want to partition it into $$k$$ consecutive subsequences, such that the maximum sum over each subsequence is minimized. Formally, find $$k+1$$ positions $$b_1=1,b_2,\ldots,b_{k},b_{k+1}=n$$, such that $$\max_{i=1}^{k} \sum_{j=b_i}^{b_{i+1}} a_j$$ is minimized. $$a_i = W_i/H_i$$ is the aspect ratio. This article will explore the techniques to solve the problem in $$O(kn)$$ time. It is only a high level overview, and leaves the details unfilled. # 1 Minimize total difference Let's first consider a simpler problem for demonstration. Problem3 Let $$a_1,\ldots,a_n$$ be a sequence of positive reals. We want to partition it into $$k$$ consecutive subsequences, such that the total difference between the sum of each consecutive sequence and the average sum of the whole sequence is minimized. Formally, let $$\mu = \frac{1}{k}\sum_{i=1}^n a_i$$. Find $$k+1$$ positions $$b_1=1,b_2,\ldots,b_k,b_{k+1}=n$$, such that $$\sum_{i=1}^{k} \|\sum_{j=b_i}^{b_{i+1}} a_j - \mu\|$$ is minimized. We will solve this problem with a reduction to a problem on a DAG, and then apply dynamic programming. Since we can always turn a problem that ask us to find the cost to a problem that ask us to find a construction that achieve the cost(with the same time/space bound). We will only concern with the cost version of the problem as it's much clearer. (One can implement some arrows in Haskell to make DP for construction as easy as DP for cost, sounds like a nice project. ) Build a directed graph $$D(V,A)$$, where $$V=\{v_1,\ldots,v_n\}$$, $$(v_i,v_j)\in A$$ if and only if $$i\leq j$$. (We can make it $$i < j$$ instead, depend on if we consider a empty sequence a valid sequence.) We have a weight function $$w(i,j)$$ that assign weights to arc $$(v_i,v_j)$$. Define $$w(i,j) = \|\sum_{k=i}^{j-1} a_i - \mu\|$$. If we find a $$k$$-edge path of minimum weight from $$v_1$$ to $$v_n$$, then this implies a solution to Problem 3. # 2 Solve the minimum weight $$k$$-edge path problem Define $$C(d,i)$$ to be the $$d$$-edge path from $$v_1$$ to $$v_i$$ with minimum weight. We want to find $$C(k,n)$$. $C(d,i) = \min_{1\leq j\leq i} {C(d-1,j) + w(j,i)}$ If we set $$w(j,i)=\infty$$ if $$j>i$$, then we have a better representation. $C(d,i) = \min_{1\leq j\leq n} {C(d-1,j) + w(j,i)}$ There are $$kn$$ entries in the DP table for $$C$$, and $$C(d,i)$$ requires $$O(n)$$ time to compute. This means the algorithm will take $$O(kn^2)$$ time. # 3 Improve the time complexity There is a standard technique on totally monotone matrices that can reduce the complexity of the problem to $$O(kn)$$. Definition4 A weight function $$w$$ is Monge if for every $$1<i+1<j\leq n$$, we have $w(i,j) + w(i+1,j+1)\leq w(i,j+1) + w(i+1,j)$ . Theorem5 $$w$$ is Monge. Proof Let $$\sum_{k=i+1}^{j-1} a_i - \mu = m$$ \begin{align} w(i,j)+w(i+1,j+1) &= \|\sum_{k=i}^{j-1} a_i - \mu\| + \|\sum_{k=i+1}^{j} a_i - \mu\|\\ &= \|a_i + m\| + \|a_j + m\|\\ &\leq \|a_i+a_j+m\| + \|m\|\\ &= w(i,j+1)+w(i+1,j) \end{align} To prove the $$\leq$$, see that $$a_i,a_j$$ are positive, one can consider either $$m$$ is negative or positive, and notice either way the inequality holds true. Remark We can replace $$\|\cdot\|$$ with $$\|\cdot\|^p$$ for $$p\geq 1$$. When $$p=2$$, we minimizes the variance(and standard deviation). Definition6 A matrix is totally monotone if for every $$i<i'$$ and $$j<j'$$, $$a_{i,j} > a_{i',j} \implies a_{i,j'} > a_{i',j'}$$. Image Credit: Vanessa Li. Remark There are isomorphic definition of Monge and totally monotone, depend on if the person want to find row or column minima. Define a matrix $$M^d$$, such that $$M_{j,i}^d = C(d-1,j) + w(j,i)$$, the original recurrence become $C(d,i) = \min_{1\leq j\leq n} {M^d_{j,i}}$ In other words, $$C(d,i)$$ is the $$i$$th column's minima of $$M^d$$. Theorem7 If $$w$$ is Monge, then $$M^d$$ is a $$n\times n$$ totally monotone matrix. Using the SMAWK algorithm, all column minimas can be found in $$O(n)$$. Finding $$C(d,i)$$ takes only $$O(1)$$ time on average! Here is the very simple code to show how this can be done easily if we have a Haskell implementation of the SMAWK algorithm. The indexing is a bit different from the description in the article. import Data.Array import SMAWK minCostkEdgePath k n w = d!(k-1,n) where d = array ((0,0),(k-1,n)) [ ((i,j), f i j) \| i<-[0..k-1],j<-[0..n]] f 0 i = w 0 i f k i = m k (p!(k,i)) i p = array ((1,0),(k,n)) [((z,i),t) \|z<-[1..k],(i,t)<-zip [0..n] (columnMinima (m z) (n+1) (n+1))] m k j i = (+) (d!(k-1,j)) (w j i) minCostMuPartition k xs = minCostkEdgePath k n w where w i j \| i <= j = abs $! s!j - s!i - avg \| otherwise = 2*m s = listArray (0, n-1)$ scanl (+) 0 xs n = length xs m = sum xs avg = m/fromIntegral k # 4 Solve the linear partition problem Now, returning to the original problem. Again, we can reduce the problem to a problem on a directed graph. This time, $$w(i,j) = \sum_{k=i}^{j-1} a_i$$. The weight of a path is the maximum weight of the edges in the path. A $$k$$-edge path with minimum weight implies the solution to the original problem. $C(d,i) = \min_{1\leq j\leq n} M^d_{j,i}$ where $$M^d_{j,i} = \max (C(d-1,j),w(j,i))$$. Just like Problem 3, this describes a $$O(kn^2)$$ time algorithm. Compare $$w,C$$ and $$M^d$$ with the previous problem to see there isn't much difference. # 5 Improve the time complexity, again Can we define an alternative to the Monge property? Yes, we can extend this to algebraic Monge property. Just replace $$+$$ with some associative operation $$\oplus$$, $$\leq$$ with a total order that is ordered with respect to $$\oplus$$, i.e. $$a \leq a\oplus b$$. So it seems, using the algorithm above with little modification, we can solve the problem in $$O(kn)$$ time because we can just replace $$+$$ by $$\min$$. However, algebraic Monge property in general doesn't imply totally monotone matrix. Consider the simple matrix, and our operation is $$\max$$. \begin{bmatrix} 1& 2\\ 0& 2 \end{bmatrix} $$\max(1,2)=\max(0,2)$$, but the matrix is not totally monotone. $$1>0$$ but $$2\not > 2$$. If instead the operation is strictly compatible, i.e. $$a \oplus b< a \oplus c$$ if $$b < c$$, then we can always produce a totally monotone matrix. This is not the case with $$\max$$. Some readers who are familiar with $$L^p$$ space might point out for any sequence of reals $$a_1,\ldots,a_n$$ and $$\epsilon>0$$, there exist a $$p\geq 1$$, such that $$(\sum_{i=1}^n \|a_i\|^p)^{1/p} - \max_{i=1}^n \|a_i\|< \epsilon$$. So if we don't need to be exact, pick a large enough $$p$$ as an exponential is good enough. This opens up a can of numerical analysis, and that's undesirable... Good news, a more restrictive but good enough variant of the Monge property hold. Definition8 $$w$$ has the strict bottleneck Monge property if either of the following is true for $$1<i+1<j$$: 1. $$\max(w(i,j), w(i+1,j+1)) < \max(w(i+1,j),w(i,j+1))$$. 2. $$\max(w(i,j), w(i+1,j+1)) = \max(w(i+1,j),w(i,j+1))$$ and $$\min(w(i,j), w(i+1,j+1)) \leq \min(w(i+1,j),w(i,j+1))$$. Our $$w$$ in consideration has strict bottleneck Monge property. Because all the numbers are positive, $$w(i,j+1)>w(i+1,j+1)>w(i+1,j)$$, $$w(i,j+1)>w(i,j)>w(i+1,j)$$. A simple case check on the relation between $$w(i,j)$$ and $$w(i+1,j+1)$$ will give the desired proof. Why this property? You can check by a case by case proof that this property implies total monotonicity. Theorem9 If $$w$$ has the strict bottleneck Monge property, then $$M^d$$ is a $$n\times n$$ totally monotone matrix. There seems to be a direct proof by analyze 12 different cases, but I got too bored after the second case. One can read [1] for the proof. It doesn't contain the exact theorem, but a result that implies this one. The proof is quite involved. Their idea is to construct a new matrix with a new operation over sorted sequences of numbers instead of just numbers. Prove this matrix is strictly compatible and has the algebraic Monge property, and show this can always be done if the original matrix has the strict bottleneck Monge property. The final punchline. Theorem10 Change the last (+) in minCostkEdgePath to max, and change how w is computed in minCostMuPartition solves Problem 2. # 6 Rethink the original problem We have developed a $$O(kn)$$ algorithm for Problem 2. As we can see from the practical application, this is a $$O(n^2)$$ algorithm because the $$k$$ is of the order $$n$$. For a few hundred photos, we can afford to run it in real time. This will reach a limit once there are thousand of photos. Can we do better? Yes, by considering a different kind of reduction(thank god we didn't formally define what close means). Instead of compute $$k$$ and try to fit $$k$$ rows, why not just make sure each row's width is almost the width we want and scale accordingly? We want each rows to approximately have width $$W$$, we don't really care how many rows there are. Problem11 Let $$a_1,\ldots,a_n$$ be a sequence of positive reals, and a number $$\mu = W/H$$. We want to partition it into consecutive subsequences, such that the maximum difference between the sum of each consecutive sequence and the $$\mu$$ is minimized. Formally, find a $$k$$ and a sequence of $$k+1$$ numbers $$b_1=1,b_2,\ldots,b_{k},b_{k+1}=n$$, such that $$\max_{i=1}^{k} \|\sum_{j=b_i}^{b_{i+1}} a_j - \mu\|$$ is minimized. This is the exact same problem described on section 5 of [1]. It can be solve in $$O(n)$$ time. # References [1] W. Bein, P. Brucker, L.L. Larmore, J.K. Park, The algebraic monge property and path problems, Discrete Applied Mathematics. 145 (2005) 455–464 10.1016/j.dam.2004.06.001. Posted by Chao Xu on 2013-08-16. Tags: .	{}
complete the square 2x2 + 6x + 2. Learn all about the quadratic formula with this step-by-step guide: Quadratic Formula, The MathPapa Guide; Video Lesson. This is a smart application which solves quadratic equations or formulas and gives you the step-by-step solution. High School Math Solutions – Exponential Equation Calculator. Coefficients may be either integers (10), decimal numbers (10.12), fractions (10/3) or Square roots (r12). How the math algebra calculator with steps works. Thus, for calculating the square root fraction following sqrt(99)/sqrt(6), just enter simplify_surd(sqrt(99)/sqrt(6)) the result with the calculation stpes is returned. x 2 = 4. We love to hear your feedback. Free quadratic equation calculator - Solve quadratic equations using factoring, complete the square and the quadratic formula step-by-step This website uses cookies to ensure you get the best experience. Keep x terms on the left and move the constant to the right side by adding it on both sides. The calculator will show you the work and detailed explanation. So, plugging this values in the formula we get: Step 3: Simplify the values in the equation, once you have plugged the values of $$a$$, $$b$$ and $$c$$. Algebra-equation.com delivers valuable facts on square root property calculator, addition and formula and other algebra subjects. equation_solver online Description : Equation calculator An is an Step 5 Find the square root of each side of the equation. For digits after decimal point, pair them from left to right). Enter the variables into the formula or calculator above. Quadratic Formula. The calculator generate solutions using the square root property Free radical equation calculator - solve radical equations step-by-step. Solve equations of the form ax2+c=0by extracting the roots. The algebra calculator helps you find solution to a wide range of mathematical problems. ★ Capable of generating graphs for a given equation. Integration calculator with steps. We'll assume you're ok with this, but you can opt-out if you wish. 1. +254 723897890. Description : Equation calculator. What are the coefficients now? It appears when you are solving all kind of geometric problems, such as when you are maximizing an area, given a fixed perimeter, or in numerous word problems. Sample: Calculate square root of 5 using division method Group the digits into pairs (For digits to the left of the decimal point, pair them from right to left. As usual, in solving these equations, what we do to one side of an equation we must do to the other side as well. Translate into an equation by writing the appropriate formula or model for the situation. If you are interested in learning more about completing the square or in practicing common problem types for completing the square, please check out our lesson on this topic . Free Online Equation Calculator helps you to solve linear, quadratic and polynomial systems of equations. Step 4: Look inside of the square root. It is exactly the same idea, that derives to the quadratic formula that we all know. By using this website, you agree to our Cookie Policy. Example: 2x^2=18. This website uses cookies to ensure you get the best experience. Our completing square calculator shows step by step solution with all the workings. so you need to replace the value of the coefficients $$a$$, $$b$$ and $$c$$. Available as a mobile and desktop website as well as native iOS and Android apps. ★ Both "Quadratic Formula" and "Completing the square" methods available. The Examples will also guide you on how to use this equation calculator to solve your algebra problems. Online surds calculator that allows you to make calculations in exact form with square roots: sum, product, difference, ratio. All rights reserved. Method 2: Completing the square The best way to learn this method is by using an example. However, learning at least the "guess and check" method for finding the square root will actually help the students UNDERSTAND and remember the square root concept itself! Â©Copyright - essaysite.net. Further, the calculator shows all the workings in a step by step method. This method uses a number's factors to find a number's square root (depending on the number, this can be an exact numerical answer or a close estimate). To solve an equation using the online calculator, simply enter the math problem in the text area provided. Solving by taking the square root Quadratics by taking square roots: with steps CCSS.Math: HSA.REI.B.4 , HSA.REI.B.4b To calculate a square root by hand, first estimate the answer by finding the 2 perfect square roots that the number is between. Thank you. This website uses cookies to improve your experience. Khan Academy Video: Quadratic Formula 1; Need more problem types? 4x2 − 9 = 0 4x2 = 9 x2 = 9 4. Lots of people wonder if there is any relationship between this quadratic equation formula and the method of completing the square. Finding the square roots of imperfect squares can sometimes be a bit of a pain — especially if you're not using a calculator (in the sections below, you'll find tricks for making this process easier). Square Root … Basically the calc solves the following algebra problems: Finding unknown, Evaluation, fractions, quadratic equations, simplification, factorization etc. Example: Solve equation 2x 2 + 7x - … ax2 + c = 0. The standard or general form of quadratic equations is ax2 + bx + c = 0. Enter your math in the text box given eg : Enter 3x+2=14, Hit the calculate button or press Enter key to proceed. Unlike most of the other apps, this app is featured with both "Quadratic Formula" and "Completing the square" methods. Simplify square roots by removing perfect squares when possible. Completing the square calculator with steps. If you encounter any problems while using this calculator, please let us know: Now, follow these simple steps. If step 5 is not possible, then the equation has no real solution. New calculators are sure to hit the sector periodically, because they have been for so many decades, and we are going to be certain to update this article as new and potentially promising calculators come to our attention. ★ Both "Quadratic Formula" and "Completing the square" methods available. 4 . Tap for more steps... Rewrite as . Check the answer in the problem and make sure it makes sense. In this case $$a = 1$$ (the coefficient multiplying the quadratic term $$x^2$$), $$b = 3$$ (the coefficient multiplying the linear term $$x$$), and $$c = 1$$ (the constant). Example: What happens with following expression: $$-3 + \frac{1}{2} x$$. Checkout our algebra examples, each with a step by step solution. For example: Solution by completing the square for: x 2 + 0 x − 4 = 0. Compute expert-level answers using Wolfram's breakthrough algorithms, knowledgebase and AI technology The automatic quadratic equation solver lets you see all the steps and working alongside the roots to your polynomial. Step 1 : Simplify the Integer part of the SQRT. For example, enter 3x+2=14 into the text box to get a step-by-step explanation of how to solve 3x+2=14. The equation solver allows to solve equations with an unknown with calculation steps : linear equation, quadratic equation, logarithmic equation, differential equation. Algebra 2 online book. This is a smart application which solves quadratic equations or formulas and gives you the step-by-step solution. Step 2: Plug the coefficients you found in the formula. Learning math with examples is the best approach. The calculator is a perfect solution/ roots generator for quadratic equations. A two-variable iterative method Free roots calculator - find roots of any function step-by-step This website uses cookies to ensure you get the best experience. How to Use the Calculator Type your algebra problem into the text box. The solution should Example: 3x^2-2x-1=0 (After you click the example, change the Method to 'Solve By Completing the Square'.) Get solutions Here: Have any suggestion on improving our calculators? x = ± 4. therefore. complete the square x2 − x − 6. complete-the-square-calculator. For Polynomials of degree less than or equal to 4, the exact value of any roots (zeros) of the polynomial are returned. Powered by Wolfram\|Alpha. The calculator works with both equations and expressions. This step may require distributing (or FOILing), combining like terms, isolating the variable, or solving by factoring depending on the remaining terms. Hit the calculate button to get the roots. In the case above the denominator is 2, hence the equation specifies that the square root is to be found. Calculate Square root with variables, Simplify Square root, Square root fraction Calculator, root calculator with steps and quadratic equation, Cubic root, Formulas. We hope our limit multivariable limit calculator helped you regarding your learning and practice. Acceptable Math symbols and their usage Quadratic formula 3rd power. If you choose to write your mathematical statements, here is a list of acceptable math symbols and operators. One of the numbers I need is the square root of 6594.26 (+- 0.01). This calculator allows you to list the first 100 multiples of any number. Solve-variable.com gives essential tips on Irrational Square Root Calculator, polynomials and subtracting fractions and other algebra topics. Substitute the given information. Round to one decimal place. Example: How about Suppose that you have the following expression: $$\frac{5}{4} + \frac{3}{4} x + \frac{1}{2} x^2$$. By using this website, you agree to our Cookie Policy. Trying to get a grasp of a scientific calculator can be a real pain in the neck, which raised the need for a function-oriented app that focuses on specific applications. Example: Suppose that you have the following expression: $$x^2+3x+1$$. The quadratic equation is an equation of the form: with $$a \neq 0$$. You can also use other useful free tools like slope calculator and cone volume calculator for free. Be sure to simplify all radical expressions and rationalize the denominator if necessary. That implies no presence of any term being raised to the first power somewhere in the equation. Calculating square root is easy if you have a perfect square. Pull terms out from under the radical, assuming positive real . First of all, select any cell from the table and go to Data Tab Get & … I can see that I have two {x^2} terms, one on each side of the equation. Step 5. An equation in which the variable is in the radicand of a square root is called a radical equation. This calculator also determines whether the discriminant $$b^2 – 4ac$$ is less than, greater than or equal to 0. A quadratic equation is a second degree polynomial of the form ax^2+bx+c=0 where a, b, c are constants, a\neq 0; A Quadratic formula calculator is an equation solver that helps you find solution for quadratic equations using the quadratic formula. So, whether you want an algebra square root calculator, an algebra factor calculator, or a simplify boolean algebra calculator, Mathway’s tool … Unlike most of the other apps, this app is featured with both "Quadratic Formula" and "Completing the square" methods. The same identity is used when computing square roots with logarithm tables or slide rules. For example, the square root (root 2) of 16 (√16) is 4, as 4 x 4 = 16. Examine the given equation of the form $$ax^2+bx+c$$, and determine the coefficients $$a$$, $$b$$ and $$c$$. However, learning at least the "guess and check" method for finding the square root will actually help the students UNDERSTAND and remember the square root concept itself! In the previous example, we have a -8 inside of the square root, so we have two complex solutions, as shown below: The quadratic formula is one of the most ubiquitous formula in mathematics. This video shows how to solve quadratic equations using square root property. This online calculator finds the roots of given polynomial. The answer is simple: you arrive the quadratic formula by solving the quadratic equation via completing the square. Instructions: This quadratic formula calculator will solve a quadratic equation for you, showing all the steps. INSTRUCTIONS: 1 . It's tougher to figure out square roots of numbers that are not whole. In the previous example, we would have. Should you require advice on logarithmic or maybe functions, Solve-variable.com is without a doubt the right destination to have a look at! Online square root calculator steps by steps Online square root calculator returns the calculation steps for a better understanding. I'm needing the uncertainty for the square root of various numbers, but I don't know how to go about it. Thus we have, 05 Linear equation This is a linear equation calculator, which includes all the steps and graphs of the linear function. In elementary algebra, the quadratic formula is a formula that provides the solution(s) to a quadratic equation. One learns about the "factor theorem," typically in a second course on algebra, as a way to find all roots that are rational numbers. Answers, graphs, alternate forms. The calculator works with both equations and expressions. Our limit calculator with steps will find the limit of your required function instantly. This quadratic equation solver helps you make these calculations automatically. 2 . But it's possible. $complete\:the\:square\:x^2-x-6$. $complete\:the\:square\:2x^2+6x+2$. One of the neat things of this quadratic equation solver is that it will show the steps to compute the y-intercept, the coordinates of the vertex and it will plot the quadratic function. Square each side of the equation. Divide your number into perfect square factors. We are always working hard to make algebra easy and fun. Similarly, 25 has a square root of 5 (5X5=25), so the square root … No Download or Signup. The calculator works the entered math problem using the quadratic formula. An equation root calculator that shows steps. Step 6 Solve for x and simplify. One also learns how to find roots of all quadratic polynomials, using square roots (arising from the discriminant) when necessary. It has solutions of the form. In order to analyze the nature of the solution, the discriminant is defined as: Based on the value of the discriminant, the nature of the solutions is defined. Factor 50 into its prime factors 50 = 2 • 5 2 To simplify a square root, we extract factors which are squares, i.e., factors that are raised to an even exponent. Example: If you have the equation: $$-3x^2 + 2x-1 = 0$$, you find that $$a = -3$$, $$b = 2$$ and $$c = -1$$. Comments, suggestions or problem report are highly appreciated! My approach is to collect all … How to Solve Quadratic Equations using the Square Root Method This is the “best” method whenever the quadratic equation only contains terms. So even though your math book may totally dismiss the topic of finding square roots without a calculator, consider letting students learn and practice at least the "guess and check" method. Able to display the work process and the detailed explanation. Exponent Calculator and Exponent Calculator - The Perfect Combination Be certain to do your due diligence as you try to find a calculator. There are other ways of solving a quadratic equation instead of using the quadratic formula, such as factoring (direct factoring, grouping, AC method), completing the square, graphing and others. How to Find a Square Root Without a Calculator. You know that 16 is a perfect square with a square root of 4 (4X4=16). What are the coefficients? Arithmetic expressions This is an arithmetic calculator and it shows all the steps. Math loves simplicity and our calculator is simple and efficient to use. Take the square root of both sides. Square root Square root calculator to determine the square root of any number. There are several steps you have to follow in order to successfully solve a quadratic equation: Step 1: Identify the coefficients. Enter any valid number, including fractions into the text boxes and our calculator will perform all work, while you type! Algebra Calculator with Steps Simplify, Expand, Factor Use the keypad below to enter an expression. The equation solver allows to solve equations with an unknown with calculation steps : linear equation, quadratic equation, logarithmic equation, differential equation. Step 3: Solve the equation found in step 2. Displaying the steps of calculation is a bit more involved, because the Derivative Calculator can't completely depend on Maxima for this task. Online square root calculator steps by steps. Online square root calculator returns the calculation steps for a … Example: 4x^2-2x-1=0. For Windows or Linux - Press Ctrl+D 2. Functions: What They Are and How to Deal with Them, Normal Probability Calculator for Sampling Distributions. The formula is quadratic formula is. Quadratic equation This is a quadratic LU3 3SQ, UK Let’s say you want to find the square root of 20. For MacOS - Press Cmd+D 3. Free Square Roots calculator - Find square roots of any number step-by-step This website uses cookies to ensure you get the best experience. If the value is positive, then the quadratic equation has two real roots. Instead, the derivatives have to be calculated manually step by step. But for that, you need to know what are squares and square roots. Degrees of Freedom Calculator Paired Samples, Degrees of Freedom Calculator Two Samples. If you don't, there's a logical process you can follow to systematically figure out the square root of any Here is a Worked example to illustrate how the calculator Works: Perhaps it is best if you learnt math through examples. Take the Square Root. Solve equation 3x 2 + 2x - 5 = 0. Step 1: Enter the equation in the respective input field Step 2: Now click the button “Solve” to get the result Step 3: Finally, the variable value using square root property will be displayed in the new window This polynomial is considered to have two roots, both equal to 3. In fact, when $$D > 0$$, then there are two different real solutions, when $$D = 0$$, there is one repeated real solution, and when $$D < 0$$, there are two different imaginary solutions. Square Root Calculator With Steps Suppose you are asked to find the square root of a number. Shows you the step by step solutions using the quadratic formula. Instructions: This quadratic formula calculator will solve a quadratic equation for you, showing all the steps. After applying the square root property, you have two linear equations that each can be solved. Factors which will be extracted are : 25 = 5 2 Factors which will remain inside the root are : 2 = 2 Please use at your own risk, and please alert us if something isn't working. completing the square formula with this calculator with steps A quadratic equation of the form ax 2 + bx + c = 0 for x, where a \ne 0 can be solved using the … An equation is an algebraic equality involving one or more unknowns. Type the coefficients of the quadratic equation, and the solver will give you the roots, the y-intercept, the The value of the number being 2 = x. Solve the equation using good algebra techniques. About quadratic equations Quadratic equations have an x^2 term, and can be rewritten to have the form: a x 2 + b x + c = 0 The calculator solution will show steps using the quadratic formula to solve the entered equation for real and complex roots. Basically the calc solves the following algebra problems: Finding unknown, Evaluation, fractions, quadratic equations, simplification, factorization etc. 8 JULIUS GARDENS LUTON These steps will help in solving the equations in the following exercise. How to complete the square First, determine the variables Calculate or gather the variables a,b, and c from an equation of the form ax^2+bx+C. Extracting roots involves isolating the square and then applying the square root property. A quadratic equation has two roots or zeroes namely; Root1 and Root2. You can try this discriminant finder to find out the exact nature of roots and the number of root of the Empty places will be repalced with zeros. If you can help me with how to do this one, I can If we take the square root of both sides of this equation, we obtain the following: √x2 = √9 4 \| x \| = 3 2. Square root function Asymptote teacher worksheets, teaching exponents 5th grade, quotient of polynomials and bionomials online problem solver, graph lines from standard form solver, how to find cube root on TI-83 calculator, activity to teach basic square root… Try MathPapa Algebra Calculator complete the square x2 + 3x. Complete The Square. The rules of Have word math problems to Solve? An online discriminant calculator helps to find the discriminant of the quadratic polynomial as well as higher degree polynomials. Squaring a square root causes the radical to disappear leaving the expression inside of the square root. The equation 4x2 − 9 = 0 is in this form and can be solved by first isolating x2. In this case $$a = \frac{1}{2}$$ (the coefficient multiplying the quadratic term $$x^2$$), $$b = \frac{3}{4}$$ (the coefficient multiplying the linear term $$x$$), and $$c = \frac{5}{4}$$ (the constant). en. Enter quadratic equation in the form a x 2 + b x + c = 0. A root is a number that is multiplied by itself the root number of times. A perfect square root is any square root that's a whole number. equation_solver online. Instructions: This quadratic formula calculator will solve a quadratic equation for you, showing all the steps. A number's factors are any set of other numbers that multiply together to make it. Type the coefficients of the quadratic equation, and the solver will give you the roots, the y-intercept, the coordinates of the vertex showing all the work and it … Divisors Divisors calculator to calculate and list all dividers of a The coefficient $$b$$ is the coefficient that appears multiplying the linear term $$x$$, and the coefficient $$c$$ is the constant. In this case, we have that $$a = 0$$, because the expression does not contain a quadratic term $$x^2$$, so in this case, this is not a quadratic expression. In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us. 3 . Type the coefficients of the quadratic equation, and the solver will give you the roots, the y-intercept, the coordinates of the vertex showing all the work and it will plot the function. Take the square root of both sides of the equation to eliminate the exponent on the left side. Send us your recommendations and app ideas. Disclaimer: This calculator is not perfect. Step-By-Step Guide. Here we see that x = ± 3 2 are solutions to the resulting equation. This is close enough because we rounded the square root. ★ Capable of generating graphs for a given equation. Note: Using power query for square root is a dynamic method, every time when you enter a new value in your table it will return the square root of that number. Limit with Square Root Calculator Use our simple online Limit Calculator to find the Limit with Square Root with step-by-step explanation. Free Algebra Solver and Algebra Calculator showing step by step solutions. ... We will show examples of square roots; higher... Read More. The coefficient $$a$$ is the coefficient that appears multiplying the quadratic term $$x^2$$. Simplify the right side of the equation. How to find square root = TI-83 plus, solve logarithm calculator, who invented the f.o.i.l method math, online square root simplifier, positive and negitive integers worksheets, interpreting quadratic equation … Want to see more features? If the value is 0, then there is one real root, and if the value inside of the square root is negative, then there are two complex root. Eliminate b term with 0 to get: x 2 − 4 = 0. Step 6. ( Use above calculator to check your solution. ) Step 4. Is not perfect root with step-by-step explanation of how to solve linear, quadratic equations which the is... The roots to your polynomial step method LU3 3SQ, UK +254 723897890: Identify the coefficients calculator steps steps... Guide ; Video Lesson, 05 online square root calculator with steps will help in the. To disappear leaving the expression inside of the other apps, this app is featured with both quadratic,... By first isolating x2 you are asked to find roots of all quadratic polynomials, using square property! Learns how to square root equation calculator with steps the calculator works: Perhaps it is best if you wish, all! The denominator if necessary, suggestions or problem report are highly appreciated, suggestions or problem report are highly!! Help in solving the quadratic equation: step 1: simplify the Integer part of the square '.,..., difference, ratio ( 4X4=16 ) that provides the solution should this is bit. Displaying the steps slide rules and AI technology Disclaimer: this calculator allows you to make it coefficient that multiplying... X + c = 0 'Solve by Completing the square root of sides!: want to see more features the SQRT solve linear, quadratic equations is ax2 bx! Us know: want to see more features of given polynomial being 2 = x ’ s say want... − 9 = 0 it makes sense root property guide: quadratic calculator... Arising from the discriminant of the square root ( root 2 ) of 16 ( √16 is! + \frac { 1 } { 2 } x\ ) are several steps you the... Method 2: Completing the square root by hand, first estimate the answer is simple efficient! All quadratic polynomials, using square root check the answer by Finding the 2 perfect square.! Use the calculator is simple: you arrive the quadratic term \ ( -3 + \frac { }. Our calculator is simple and efficient to use { x^2 } terms, one on each of... A perfect square factors steps will help in solving the quadratic equation terms out from under radical. You, showing all the steps and graphs of the equation real and complex roots ( arising the... N'T working of calculation is a smart application which solves quadratic equations using the quadratic term \ ( c\.! Our Completing square calculator shows all the steps formula and the detailed explanation when possible to go it... A step-by-step explanation of how to go about it eliminate b term with 0 to get: x 2 4! Gives you the step-by-step solution. of Completing the square the best way learn. Used when computing square roots ( arising from the discriminant ) when.. That you have to follow in order to successfully solve a quadratic via. Solver and algebra calculator helps to find the discriminant of the form ax2+c=0by extracting the roots also learns how solve. Any relationship between this quadratic equation in the text box given eg: 3x+2=14! Right destination to have a perfect solution/ roots generator for quadratic equations using the equation! Is featured with both quadratic formula '' and Completing the square root.! Or problem report are highly appreciated implies no presence of any number step-by-step this website, need. 16 ( √16 ) is the square root of any term being raised to the quadratic equation is... And please alert us if something is n't working into the text box writing the appropriate formula or for... Will also guide you on how to go about it finds the roots of any number to collect …... Step solutions we hope our limit multivariable limit calculator to determine the square take the square root hand! Depend on Maxima for this task agree to our Cookie Policy, and... Have, 05 online square root square root is called a radical equation calculator helps you to list first!, fractions, quadratic equations using square roots: sum, product, difference, ratio, enter. Show steps using the online calculator, which includes all the steps and of! Complete\: the\: square\: x^2-x-6 \$ idea, that derives to the right to. First estimate the answer by Finding the 2 perfect square root ( root 2 ) 16... Roots involves isolating the square root causes the radical to disappear leaving expression. They are and how to use this equation calculator - solve radical equations step-by-step whenever... Formula is a Worked example to illustrate how the calculator generate solutions using the square '' methods.! Manually step by step method ( x^2+3x+1\ ) of any number by Finding the 2 perfect with. Arithmetic calculator and cone volume calculator for free equation in the square root equation calculator with steps area provided there is square... To collect all … this polynomial is considered to have two linear equations that each can be solved by isolating. Each can be solved by first isolating x2 move the constant to the resulting equation khan Academy Video: formula. N'T know how to go about it logarithmic or maybe functions, Solve-variable.com is without a the. Is close enough because we rounded the square root calculator with steps will find the limit of required. Form ax2+c=0by extracting the roots of any number of square roots of all quadratic polynomials using! More problem types complete the square root property step 1: Identify the coefficients (! Method this is a smart application which solves quadratic equations, simplification, factorization etc take square. Equation in the text box to get a step-by-step explanation x\ ) b^2 4ac! Resulting equation b\ ) and \ ( -3 + \frac { 1 } { }... Mathpapa guide ; Video Lesson: sum, product, difference, ratio to right ) decimal point pair. Solves quadratic equations using the quadratic formula that we all know square root equation calculator with steps by! The text square root equation calculator with steps } terms, one on each side of the other apps, this app featured... The equations in the formula or calculator above solve linear, quadratic equations using the online calculator please! Math in the radicand of a square root all know to do your due diligence as you try to a. And complex roots have the following algebra problems ax2+c=0by extracting the roots of numbers that are whole. Uncertainty for the square '' methods for the situation the radicand of number... Compute expert-level answers using Wolfram 's breakthrough algorithms, knowledgebase and AI technology Disclaimer this. − 6. complete-the-square-calculator: with \ ( b\ ) and \ ( a\ ), \ ( a\,! Higher... Read more denominator if necessary to figure out square roots calculator - perfect! Calculator steps by steps online square root is any relationship between this quadratic formula formula by the. Equation to eliminate the exponent on the left and move the constant to the quadratic formula by solving the in. Look inside of the equation found in step 2: Completing the root! B term with 0 to get: x 2 + 2x - 5 = 0 quadratic formula by the... Each can be solved we all know be solved by first isolating x2: Perhaps is. Of the SQRT also determines whether the discriminant \ ( a\ ), \ ( -3 + {. Or more unknowns adding it on both sides these calculations automatically roots, both equal to 0 is. Roots to your polynomial: have any suggestion on improving our calculators on how to use this calculator! Of your required function instantly: have any suggestion on improving our?. Form: with \ ( a\ ) is the coefficient that appears multiplying the quadratic solver. The problem and make sure it makes sense to 'Solve by Completing the square '' methods computing square of! Solver helps you find solution to a quadratic equation for real and complex.... Happens with following expression: \ ( -3 + \frac { 1 } { 2 x\. X 4 = 0 you have two roots or zeroes namely ; Root1 and Root2 learn this method by... Maxima for this task 2 ) of 16 ( √16 ) is less than, greater than or equal 0! Needing the uncertainty for the square root with step-by-step explanation of how to Deal with Them Normal. And fun being raised to the first 100 multiples of any term being raised the... A \neq 0\ ) method to 'Solve by Completing the square x2 − −! The MathPapa guide ; Video Lesson { x^2 } terms, one on each side of the other apps this. That appears multiplying the quadratic equation for you, showing all the steps of calculation is a equation... To get: x 2 + b x + c = 0 to our Cookie Policy out roots. + b x + c = 0 0.01 ) linear equations that each can be solved replace the of! Ca n't completely depend on Maxima for this task = 0 is in the formula or above! Eg: enter 3x+2=14 into the text box given eg: enter 3x+2=14, Hit the calculate button press... Pair Them from left to right ) after applying the square and then applying the square the best to... ( root 2 ) of 16 ( √16 ) is the “ best ” method whenever the quadratic:. Algorithms, knowledgebase and AI technology Disclaimer: this quadratic equation this is equation! For square root equation calculator with steps, enter 3x+2=14 into the formula or model for the.! Is by using this website uses cookies to ensure you get the best experience to the..., you need to know What are squares and square roots so you need square root equation calculator with steps replace value... Quadratic formula with this step-by-step guide: quadratic formula wonder if there is any square root that 's a number! \ ( a\ ), \ ( -3 + \frac { 1 {! Form a x 2 − 4 = 0 you encounter any problems using!	{}
Version 145 (modified by davarpan1s, 4 years ago) (diff) Dashes to "to" in table # Project Name Infrastructure CNRS_Coriolis Project (long title) Coriolis and Rotational effects on Stratified Turbulence Campaign Title (name data folder) 16CREST Lead Author Jeffrey Peakall Contributor Stephen Darby, Robert Michael Dorrell, Shahrzad Davarpanah Jazi, Gareth Mark Keevil, Jeffrey Peakall, Anna Wåhlin, Mathew Graeme Wells, Joel Sommeria, Samuel Viboud Date Campaign Start 12/09/2016 Date Campaign End 21/10/2016 # 1 - Objectives Our primary objective is to measure detailed turbulence distributions within channelised gravity currents, as a function of Coriolis forces, concentrating on: i) the bottom boundary layer, ii) redistribution of turbulence within bends, and, iii) redistribution of turbulence at the interface between the gravity current and the ambient. These datasets will enable existing theory on the presence and influence of Ekman boundary layers to be tested, with important implication for the basal shear stress distributions, erosion, and the evolution of channels. These data on the distribution of turbulence will then be applied to i) examine the turbulence distribution in straight channels, ii) provide an analysis of secondary flow and associated turbulence around bends for the first time, and an assessment of how channelized flows alter as a function of Rossby numbers and therefore latitude, iii) assess how the morphodynamics of submarine channels vary as a function of the Rossby number, iv) explain the observed patterns of submarine channel sinuosity with latitude (Peakall et al., 2012; Cossu and Wells, 2013; Cossu et al., 2015), and, v) incorporate the entrainment data into numerical models of submarine channels, in order to address the unanswered question of how these flows traverse such large-distances across very low-angle slopes (Dorrell et al., 2014).Our primary objective is to measure detailed turbulence distributions within channelised gravity currents, as a function of Coriolis forces, concentrating on: i) the bottom boundary layer, ii) redistribution of turbulence within bends, and, iii) redistribution of turbulence at the interface between the gravity current and the ambient. These datasets will enable existing theory on the presence and influence of Ekman boundary layers to be tested, with important implication for the basal shear stress distributions, erosion, and the evolution of channels. These data on the distribution of turbulence will then be applied to i) examine the turbulence distribution in straight channels, ii) provide an analysis of secondary flow and associated turbulence around bends for the first time, and an assessment of how channelized flows alter as a function of Rossby numbers and therefore latitude, iii) assess how the morphodynamics of submarine channels vary as a function of the Rossby number, iv) explain the observed patterns of submarine channel sinuosity with latitude (Peakall et al., 2012; Cossu and Wells, 2013; Cossu et al., 2015), and, v) incorporate the entrainment data into numerical models of submarine channels, in order to address the unanswered question of how these flows traverse such large-distances across very low-angle slopes (Dorrell et al., 2014). # 2 - Experimental setup: ## 2.1 General description 2.1 General description A channel model is positioned within the Coriolis facility. The channel model consists of an initial tapered input section with a honeycomb baffle for flow straightening and turbulence control, a 3.2 m straight channel section, and two bends with a mid-channel radius of 1.5 m. The channel is made of acrylic and is 60 cm wide and 50 cm high; the sinuous section has a sinuosity of 1.2. The slope is 3/50 radians (3.5 degrees, 6% gradient) and the channel terminates 10 cm off of the floor. Saline fluid is pumped into the top of the channel, forming a gravity current, which flows along the channel, and off the end. The basal 10 cm of the flume operates as a sump for the denser saline fluid to accumulate. In turn, this fluid can be drawn down in one of two ways: i) whilst recirculating the fluid, though this is limited to 20 $m3$/hr (5.55 l/s), and ii) through emptying to the drain, in which case any flow rate is possible. Two long metal rails are positioned to either side of the channel model across the full width of the flume. These carry a computerized gantry, which can be positioned at any point along the channel. The gantry itself contains the controls for two Schneider slides, one orientated transverse to the model, and the other connected slide, orientated in the vertical. Thus the system enables xyz control. ## 2.3 Fixed Parameters Notation Definition Values Remarks $Q_0$ Input Density $12 \ ls-1$ $\Delta\rho$ Density Difference $20 \ kg \ m-3$ $W$ Channel Width $0.6 \ m$ $\nu$ Viscosity $10-6m2s-1$ $S$ Slope $3.5{\circ}$ ## 2.4 Variable Parameters Notation Definition Unit Initial Estimated Values Remarks $\Omega$ Rotation Rate $rads-1$ -0.18 - 0.15 $H_w$ Water Depth $m$ 1-1.1 $Q_o_u_t_p_u_t$ Output Flow Rate $ls-1$ 5.5 - 17 $k$ Roughness - - ## 2.5 Additional Parameters Notation Definition Unit Initial Estimated Values $h$ Depth of gravity current $m$ 0.1 $U$ Mean downslope velocity $ms-1$ 0.1-0.15 $\delta$ Thickness of Ekman boundary layer $mm$ ~10 $R$ Radius of curvature $m$ 1.5 ## 2.6 Definition of the relevant non-dimensional numbers Flow Reynolds number across the obstruction, $Re = Uh/\nu$. Densimetric Froude number, $Fr = U/(g'h){1/2}$, $g' = g(\Delta\rho)/\rho_0$. Rossby number, $Ro = U/fW$. Canyon number, $\beta = sW/\delta$. Bulk Richardson number, $Ri = 1/(Fr{2})$ Keulegan number, $Ke = (Ri/Re?){1/3}$ # 3 - Instrumentation and data acquisition ## 3.1 Instruments Two ultrasonic systems are used to measure velocity. Ultrasonic Velocimetry Profiling (UVP) Ultrasonic velocimetry profiling (UVP) is a technique that measures a single component of velocity at either 128 or 256 points along a line. A transducer sends out an ultrasonic pulse, and then gates the return signal into a series of spatial bins. The individual transducers can be multiplexed in order to provide pseudo-velocity fields. The transducers are linked via a multiplexer with a delay of 15 ms, so the two-dimensional velocity field is not instantaneous, however velocity fields can be collected at 3-4 Hz. Two frequencies of transducers will be used. An array of ten 4 MHz UVP probes (with 10 m long cables) will be used for collecting downstream velocity profiles, these probes are positioned at heights (centre point of each probe) of 7, 16, 26, 56, 86, 116, 146, 176, 206 and 236 mm from the base of the channel. These probes are positioned in a custom made plastic holder, in turn connected to a bar strapped to the channel top. Initially, the probes are positioned on the channel centreline, 80 mm downstream of the apex of bend 2, looking upstream. An array of ten 2 MHz UVP probes (with 4 m long cables) will be used to examine the nature of secondary flow at bend apices. This involves drilling holes in the apexes of bends 1 and 2 and inserting the UVP probes. Probes will have to be moved between experiments, to look at the two different bend apexes. 2 MHz probes are required for the cross-section measurements since the measurement range needs to be much larger (60 cm) than is required for the axial velocity measurements. Profiling Acoustic Doppler Velocimetry (ADV) Three profiling Acoustic Doppler probes will be used for mapping three-component flow velocities within the channel. These probes come in two types, 1 is a stem probe with additional shielding, and two are flexible probes that allow closer deployment. All three probes measure three component velocities over a depth range of around 30 mm (up to 34 mm), with this zone starting 40 mm below the probe head. Bottom tracking by the instruments enables this depth to be precisely known and controlled, provided sufficient seeding is present in the flow. These probes are co-mounted on the traverse in order to collectively measure over a height of approximately 6 cm. Movement of these probes on the traverse in the y and z planes allows detailed vertical profiles and flow mapping to be undertaken. Dwell times at each point will vary between 30 and 60 seconds depending on the nature of the experiment. The basal probe is numbered 1, middle 2, and upper probe 3, and are at heights of 7.2, 10 and 13 cm respectively. Lateral offset in the x direction is 8.5 cm between probes 1 (most downstream probe) and 2, and 7.5 cm between probes 2 and 3 (most upstream probe). The traverse positions for each cross-section are based on probe 2 being positioned directly above the cross-section. The aim is to synchronise the ADV collection with the traverse through sending of the required voltage offset signal to them, thus enabling individual velocity files to be collected for each probe at each traverse. # 6 - Table of Experiments: Run Name Downstream (x)Position Density Excess Input flow Rotation Rate Initial Water Depth Outflow Rate Run Time ADV Dwell Time $(kg \ m-3)$ $(l \ s-1)$ $(rad \ s-1)$ $(m)$ $(l \ s-1)$ $(Minutes)$ $(s)$ fixstr1_1909a X1 20 12 0 1 5.5 30 continuous fixstr1_2109a X1 10.3 12 0 1 5.5 21 60 fixstr1_2109b X1 10.3 10 0 1 5.5 21 60 fixstr1_2209a X1 19.2 6 to 8 0 1 5.5 21 60 fixstr1_2209b X1 19.2 10 0 1 5.5 22 60 fixstr1_2609a X1 19 5.9 0 1 5.5 27 30 fixstr1_2609b X1 19 5.9 0 1 5.5 30 30 fixstr1_2709a X1 20 6 0 1 5.5 20 30 fixstr1_2709b X1 18.4 15.3 to 6 0 1 5.5 15 30 fixstr1_2809a X1 20 20 to 6 0 1 5.5 15 30 fixapex2_2809b X4 20 20 to 6 0 1 5.5 10 30 fixapex2_2809c X4 18.4 20 to 6 0 1 5.5 5 30 fixapex2_2809d X4 18.4 20 to 6 0 1 5.5 5 30 rotstr1_2909a X1 20.3 20 to 5.64 0.083 1 5.5 30 30 rotstr1_2909b X1 20.5 20 to 6 0.083 1 5.5 15 30 rotstr1_3009a X1 20.4 20 to 5.83 0.083 1 5.5 30 30 rotstr1_3009b X1 20.4 20 to 5.5 0.083 1 5.5 35 30 rotstr1_3009c X1 20.4 20 to 5.5 0.083 1 5.5 10 - # 7 - Diary: Monday, September 19th 2016 Experiment name: fixstr1_1909a. Filenames: fixstr1_1909a1, fixstr1_1909a2. Location: Position X1 (75% of the way down the straight section, 58 cm upstream from the end of straight section). Input rate: 12 l/s, density excess: 20 kg/m3. Water was very cloudy to the extent that we were not able to use the laser. No siphon rig was used. Running basal ADV with ADV #1 located 7.2 cm from the base. ADV just measured at one point (no traverse measurement). Experiment started at 2:45pm and stopped at 3:15 pm. The flow automatically stopped part way through as a valve was not open for recirculating water. Part way through red dye was added to visualise the current. Dye visualisation suggested pretty thin flow on the inner bank and significant super elevation on the outer bank. Some perturbation was observed on the surface of the flow at the outer bank, but otherwise the flow surface appears quite smooth, and mixing appeared to be very low. Mean maximum flow velocity from raw output was around 20 - 25 cm/s. ADV and UVP measured twice, the first is referred to as fixstr1_1909a1 and this had a velocity range of 0.3 m/s on the ADV setting, and 0.25 m/s for the UVP. Instantaneous flow velocities were faster than anticipated, as a result of the steep slope (3.5 degrees) with flow wrapping on both instruments so the velocity range was increased on the ADV and the UVP. A second run period fixstr1_1909a2 had a velocity range of 0.5 m/s on the ADV setting. The UVP setting was 680 mm/s (0.68 m/s). Tuesday, September 20th 2016 Paint was applied to the tank floor to address a leak in the flume, and the tank floor left to dry and seal. The siphon rig was completed and tested. The laser system was aligned. The position of the basal ADV was refined, Orgasol was required as additional seeding in order for bottom tracking to work effectively. Interesting, the stem (fixed) ADV picks this bottom point up better than the flexible ADV in the absence of seeding. Tested synchronization of ADV with traverse; a problem was identified with the nature of the required input signal. Investigation in progress as to how to address this. Wednesday, September 21st 2016 Experiment name: fixstr_2109a, b. File names: fixtr1_2109a, fixstr1_2109b. Location: Position X1 (75% down straight section, 58 cm upstream from the end of the straight section). Density excess 10.3 kg/m3. fixstr1_2109a: The input flow rate was 12 L/s and the total experiment time was 21 minutes. The UVP was turned on as the gravity current entered the inlet box. The ADV was turned on 2 minutes after the UVP saw the flow. Two sets of siphon samples were collected three minutes after the current hit the end of the tank. Sampling was done every 10 minutes: Siphon Set #1 (4:30 to 5:30) and Siphon Set #2 (14:30 to 15:30). The flow was too thick and the velocity was about 18 cm/s. It was speculated that the flow rate should be decreased. Siphon samples showed a density decrease in depth, going from the bottom to the top, and also a decrease over time. fixstr1_2109b: Since the flow was fast and thick, the flow rate was lowered from 10 L/s (t = 0 to 6 ) to 8 L/s (t = 8 to 13) to 6 L/s (t = 15 to 30). 3 sets of siphon samples were collected for the 3 flow rates at 3 different times (4:55 to 5:55, 12:02 to 13:02, and 19:26 to 20:26). The experiment lasted for 21 minutes. Thursday, September 22nd 2016 Experiment name: fixstr_2209a, b. File names: fixstr1_2909a, fixstr1_2909b. Location: Position X1 (75% down straight section, 58 cm upstream from the end of the straight section). Density excess 19.2 kg/m3. Before the experiments samples were collected from the storage tanks which store 10 kg/m3 and 20 kg/m3 salt water in order to check the density. Also, samples from the base and the top of the rotating Coriolis tank were collected. No laser was used since the quality of the water was not good enough. fixstr1_2909a: Flow rate started with 6 L/s for about 16 minutes. Siphon sample Set #1 was collected around 5 minutes after the start of the experiment for 1 minute. After 7 minutes, a sample was collected from the tank on the third floor where salt water enters from the storage tank. Then the flow rate was increased to 8 L/s. At around 16 minutes constant flow rate was reached and a second set of siphon samples were collected at about 19 minutes for 1 minute deviation. The flow rate then was increased to 10 L/s, but since it took a long time to become constant the UVP stopped. So a second experiment was conducted with a flow rate 10 L/s. Also, the Nikon Camera took 2 videos, 10 minutes each one at the start and one almost right at the end. fixstr1_2209b: Flow rate was 10 L/s and the total experiment timing was 16 min. The flow rate became constant after about 7 minute, so potentially it was not constant before. One siphon sample was collected at 4:27 for one minute deviation. After measuring the density, it was concluded that the density of the inlet box is much lower than 20 kg/m3 meaning a lot of mixing is occurring there. Therefore, there are two issues that need to be considered for future experiments. 1. How to control the density of the inlet box and reduce mixing. 1. Start every experiment with a constant flow rate, which takes at least 5 min to occur. Friday, September 23rd 2016 The tank was drained and cleaned in order to locate, drill and install the 2 Mhz UVPs at the apex of the first curve. The probes were located at 10 points evenly spaced 45 mm apart, starting 45 mm from the bottom. The results from the previous experiments were revised to plan new experiments and updated experimental plan and protocol were provided. A calibration for the conductivity probe on the traverse was plotted. Concentration profiles and plots were updated for the latest experiments. Post-processing of the UVP and ADP data for the conducted experiments were carried out. All of our data on the network was updated including the UVP and ADV. Monday, September 26th 2016 Experiment name: fixstr1_2609a,b. Filenames: fixstr1_2609a, fixstr1_2609b. Location: Position X1 (75% down straight section, 58 cm upstream from the end of the straight section). Input rate 5.9 l/s. Density excess 19 kg/m3. Laser was not used in the experiments today. Running basal ADV with ADV #1 located 7.2 cm, ADV #2 located at 10 cm and ADV #3 located at 12 cm above channel bed. fixstr1_2609a: The main goal was to investigate if the concentration of the gravity current, inlet box and tank D were equal (19 kg/m3). The traverse sent one TTL signal to the ADV and didn’t send a return signal so only one measurement at one point was made by the ADVs (all ADVs collected data simultaneously). On the middle ADV (#2) wrapping occurred which indicates that velocity needs to be increased to 0.6 m/s. UVP files showed a fair amount of noise in the data which requires further investigation. We waited about 7:20 minutes for the flow rate to reach a constant value. One person took 6 samples (every 3 minutes, starting at t = 0) from the mixing tank on the third floor which contained the salt water mixture coming from tank D. A second person took 6 samples (every 3 minutes, starting at t = 0) from the inlet box at the upstream end of the channel. A third person took 3 sets of 12 siphon samples every 5 minutes (starting at t = 0, duration = 1 min). GoPros were installed at three positions as follows: Position 1 - On top of inlet box looking downstream. Position 2 - Inside inlet box focused on inflow pipe. Position 3 - straight section of channel, looking cross-channel. Density measurements of the samples showed that the density in the mixing tank on the third floor and tank D were equal (19 km/m3). However, there was a major decrease of density in the inlet box (max density = 14.8 kg/m3). Density of inlet box needs further investigation. fixstr1_2609b: Goal was to determine flushing characteristics of the inlet box, and how long it takes to reach steady-state density conditions. Running basal ADV at 5 positions in the bottom cross section only at traverse positions y = 0.016, 0.116, 0.216, 0.316 and 0.416 m (with 5 repetitions). GoPro cameras were mounted at three different positions (Position 1 - Outside rotating tank focused on channel outflow looking upstream. Position 2 - above tank, looking down, focused on bends. Position 3 - straight section of channel, looking cross-channel). Dye was released to visualize the flow. One set of 12 siphon samples (duration = 1 min) was taken 5 minutes after flow rate stabilized. It took 2.5 min for the flow rate to stabilize. The ADV was initialized 5 minutes after flow stabilized. The ADV needed more seeding to improve the signal-to=noise ratio. The GoPro in the inlet box showed a lot of mixing between the current and the ambient which explains the decrease in density in the gravity current in the inlet box. The suggestion was to reduce the inlet box volume to reduce the time to reach steady-state density (by reduce the flushing time). Tuesday, September 27th 2016 Density of siphon samples from Monday, Sept 26 were measured. Two large PVC tubes (dia ~ 40 cm) were placed in the inlet box and weighted down to reduce the volume (new vol ~ 150 L). Experiment name: fixstr1_2709a,b. Filenames: fixstr1_2709a, fixstr1_2709b. Location: Position X1 (75% down straight section, 58 cm upstream from the end of the straight section). Input rate 6 l/s. Density excess 20 kg/m3. Laser was not used in the experiments today. Running basal ADV with ADV #1 located 7.2 cm, ADV #2 located at 10 cm and ADV #3 located at 12 cm above channel bed. Fixstr1_2709a: The goal was to examine the flushing time of the inlet box to see how it behaves once the PVC tubes were installed. One conductivity probe (C1T1) was mounted at the centerline just before the flow straightener (metallic honeycomb). Started with 6 l/s and waited for 5 min to see how inlet box, UVP and pump behaved. The UVP and siphon pump were turned on to fix the suspected pulse that the pump imparted to the UVP. This did not resolve the SNR issue. The inlet box took a long time to reach a steady conductivity value. It was suggested to start with a higher flow rate and then reduce the flow rate to 6 l/s. The ADV and UVP data were still noisy. No siphon sampling was done. fixstr1_2709b: It was suggested to install an array of three conductivity probes in a cross-section just before the honeycomb in the inlet box. Only one additional probe was available. As such, two conductivity probes were mounted in the cross section just before the flow straightener (metallic honeycomb): one at the centerline (C2 green) and one 5 cm?? from the right wall (C1T1 black), looking downstream. The traverse was able to move vertically in every cross-section position and the previously outstanding issues have been resolved. Water was initially pumped with a flow rate of 15.28 l/s (55 m3/h). This was the highest achievable flow rate. The water was pumped at this rate for 2 minutes and then reduced to 6 l/s. This seems to achieve the desired effect of reaching a steady conductivity value in a short amount of time in the inlet box. The UVP was set over 10 minutes, but still showed noisy data. No siphon sampling was conducted. Wednesday, September 28th 2016 Experiment name: fixstr1_2809a and fixapex2_2809b, c, d. Filenames: fixstr1_2809a, fixapex2_2809b, fixapex2_2809c, fixapex_2809d. Location for fixstr1_2809a: Position X1 (75% down straight section, 58 cm upstream from the end of the straight section). Location for fixapex2_2809b, c, d: Position X4 (at the centre of the second apex). Input rate 20 l/s (initial ~ 2 minutes of each run) then reduced to 6 l/s. Density excess 18.4 kg/m3. Laser was not used in the experiments today. Running basal ADV with ADV #1 located 7.2 cm, ADV #2 located at 10 cm and ADV #3 located at 12 cm above channel bed. fixstr1_2809a: The goal is to increase the seeding and test the ADVs and UVP for less noise. A new stem ADV was mounted to reduce the noise. The UVP didn’t change in terms of noise issue, so the problem does not lie with the seeding density. Another suggestion was to move the UVP box back to the back bench to reduce interference from the electronics on the traverse. Minimal improvement was seen on the ADV profile. The seeding seems to be getting stuck in the inlet box behind the flow straightening baffles - this was visually observed by a buildup of foam in the inlet box. fixapex_2809b: The siphons were moved up by 20 cm and the UVP 17 cm. The traverse was also moved to the second apex. The goal for this experiment was to take siphon and UVP data higher than 23.6 cm above the channel bed (previously not possible due to the siphon and UVP configuration), because we want to be able to draw velocity and concentration profiles for 40 cm flow thickness. It was also intended to verify if moving the physical position of the UVP box would impact the data quality. Traverse moved to apex 2 position and siphon samples were taken every 1 minute after flow rate stabilized, for 1 minute sampling time (5-6 min, 7-8 min, 9-10 min). When the flow was released a large cloud of seeding flowed through the channel as a gravity current (presumably seeding caught in inlet box from previous run) and the data on the ADV improved substantially. It was thus concluded that seeding is a viable solution to the noisy data problem (provided a reliable mechanism for seeding the flow, without it getting caught in the inlet box, can be devised). The other suggestion is to increase their distance further apart to avoid side lobe interference. UVP data remained noisy. It is speculated from pressing and GoPro videos that surface waves are produced where the channel sides plunge under the free surface, and that this is the source of the noise in the UVP data. fixapex_2809c: The goal for this experiment was to turn off the transverse completely then turn on ADV and UVP to check if noise can be reduced. Almost all electronics were turned off on the transverse. Both the UVP and ADV still show a lot of noise. This implies there is either seeding or side lobe effect for the ADV and speculation on surface wave for the UVP. fixapex_2809d: The new stem ADV was damaged, so the old cable ADV was used instead. Only one experiment was run with just one ADV to see if it will show noise. It didn’t show noise. Since running multiple ADVs showed noise, this means that they are talking to each other. SNR was low, but according to Nortek, in the arms it is still possible to take good data even with low SNR. Thursday, September 29th 2016 The tank was drained, cleaned, washed, and refilled. The tank was spun with 0.083 rad/s rotation rate (0.8 rpm). Experiment name: rotstr_2909a, b. Filenames: rotstr1_2909a, rotstr1_2909b. Location: Position X1 (75% down straight section, 58 cm upstream from the end of the straight section). Input rate started from 20 L/s for the first two minutes and then was decreased to 5.64 L/s. rotstr1_2909a: Density excess was 20.3 kg/m3. Two sets of siphon samples were collected every 5 to 6 minutes and 15 to 16 minutes after the start of experiment. Siphons and UVPs were at the top part of the flow. Gopros took movies from the ping pong balls moving on top of the rotating tank. rotstr1_2909b: Density contract was 20.5 kg/m3. Siphons and UVPs were moved down to their original location. Three set of siphon samples were collected every 5 minutes after the start of the experiment. The experiment lasted for 15 minutes. Friday, September 30th 2016 Experiment name: rotstr1_3009a, b, c. Filenames: rotstr1_3009a, rotstr1_3009b, rotstr1_3009c. rotstr1_3009a: The total experiment time was 30 minutes. The goal was to use UVPs and switch between them within the experiment. ADV samples were taking at the straight section (Position X1 -75% down straight section, 58 cm upstream from the end of the straight section). The stream wise UVP collected data for the first 10 minutes of the experiment. In the second 10 minutes, the UVP was switched to the cross stream and collected data in the cross stream section for the third 10 minutes. It should be noted that the cycle for the cross stream UVP was set to 1500 mistakenly, instead of 666. Bottom check data was collected for both ADVs. Siphon samples were collected every 5 minutes after the flow rate became constant. The 3 sets were sampled at 5 to 6 minutes (Set #1), 15 to 16 minutes (Set #2) and 25 to 26 minutes (Set #3). The laser was used in this experiment. rotstr1_3009b: The goal was to see how much better the UVP data would be later in the experiment. Input flow rate started from 20 L/s for the first 2 minutes and then was decreased to 5.5 L/s. Density excess and temperature were 20.4 kg/m3 and 23.7 degrees C respectively. The experiment lasted for 35 minutes as opposed to 30 minutes because it was concluded from previous experiments that would show better results later in the experiment. Therefore 5 minutes after the flow rate became constant, the stream wise UVP collected data for 10 minutes. Then, the two UVPs were switched in the next 10 minutes and, finally, the cross stream UVP collected data from 25 to 35 min. A complete sequence of data was collected by the ADV, but it did not restart for some time. Then, it started collecting a second sequence later which was not complete because the experiment ended. The laser did not work in this experiment. rotstr1_3009c: Input flow rate started from 20 L/s for the first 2 minutes and then was decreased to 5.5 L/s. The goal for this experiment was to add dye and visualize the behavior of the current at different locations. Gopros were placed at different locations to take movies of the current and to test which location is the best fit for taking movies. Monday, October 3rd 2016 The tank was drained, washed and cleaned. New taller probes were built and installed for the stream wise UVP and the siphons. A new sequence for the transverse was considered and calibration for both conductivity probes was completed. All the data and the diary on the Wiki were updated and a new revised experimental plan was proposed for the next three weeks. Tuesday, October 4th 2016	{}
# How do you find the center, vertices, foci and eccentricity of x^2/100 + y^2/64 = 1? Nov 11, 2015 The ellipse is in the form ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$ Where center $= \left(h , k\right)$ focus = $\left(h + c , k\right) , \left(h - c , k\right)$ ${c}^{2} = {a}^{2} - {b}^{2}$ ${x}^{2} / 100 + {y}^{2} / 64 = 1$ $\implies {\left(x - 0\right)}^{2} / {10}^{2} + {\left(y - 0\right)}^{2} / {8}^{2} = 1$ $\implies$ center $= \left(0 , 0\right)$ ${c}^{2} = {a}^{2} - {b}^{2}$ ${c}^{2} = 100 - 64$ $\implies {c}^{2} = 36$ $\implies c = 6$ $\implies$ focus $= \left(h + c , k\right) , \left(h - c , k\right)$ $\implies$ focus $= \left(0 + 6 , 0\right) , \left(0 - 6 , 0\right)$ $\implies$ focus $= \left(6 , 0\right) , \left(- 6 , 0\right)$ I don't know how to get the eccentricity. But I believe it has something to do with the ratio of $a$ and $b$ (or maybe $c$). Sorry	{}
2.8: Angular Velocity Difficulty Level: At Grade Created by: CK-12 Estimated17 minsto complete % Progress Practice Angular Velocity MEMORY METER This indicates how strong in your memory this concept is Progress Estimated17 minsto complete % Estimated17 minsto complete % MEMORY METER This indicates how strong in your memory this concept is To find a particular song on your Ipod, you use the scroll wheel. This involves moving your finger around the wheel in a circular motion. Unfortunately for you, the song you want is near the very bottom of your songs list. And since an Ipod can hold over 1,000 songs, you have to scroll fast! As you are moving your finger in a circle, you wonder if you could measure how fast your finger is covering the distance around the circle. Watching your finger, you realize that your finger is moving around the circle twice every second. If the radius of the Ipod wheel is 2 cm, what is the angular velocity of your finger as you scroll through your songs list? What is the linear velocity? At the end of this Concept, you'll know how to answer these questions. Guidance You may already be familiar with the measurement of speed as the relationship of an object's distance traveled to the time it has been in motion. However, this relationship is for objects that are moving in a straight line. What about objects that are traveling on a circular path? Do you remember playing on a merry-go-round when you were younger? If two people are riding on the outer edge, their velocities should be the same. But, what if one person is close to the center and the other person is on the edge? They are on the same object, but their speed is actually not the same. Look at the following drawing. Imagine the point on the larger circle is the person on the edge of the merry-go-round and the point on the smaller circle is the person towards the middle. If the merry-go-round spins exactly once, then both individuals will also make one complete revolution in the same amount of time. However, it is obvious that the person in the center did not travel nearly as far. The circumference (and of course the radius) of that circle is much smaller and therefore the person who traveled a greater distance in the same amount of time is actually traveling faster, even though they are on the same object. So the person on the edge has a greater linear velocity (recall that linear velocity is found using \begin{align}\text{distance} = \text{rate} \cdot \text{time}\end{align}). If you have ever actually ridden on a merry-go-round, you know this already because it is much more fun to be on the edge than in the center! But, there is something about the two individuals traveling around that is the same. They will both cover the same rotation in the same period of time. This type of speed, measuring the angle of rotation over a given amount of time is called the angular velocity. The formula for angular velocity is: \begin{align}\omega =\frac{\theta}{t}\end{align} \begin{align}\omega\end{align} is the last letter in the Greek alphabet, omega, and is commonly used as the symbol for angular velocity. \begin{align}\theta\end{align} is the angle of rotation expressed in radian measure, and \begin{align}t\end{align} is the time to complete the rotation. In this drawing, \begin{align}\theta\end{align} is exactly one radian, or the length of the radius bent around the circle. If it took point \begin{align}A\end{align} exactly 2 seconds to rotate through the angle, the angular velocity of \begin{align}A\end{align} would be: \begin{align}\omega &=\frac{\theta}{t}\\ \omega &=\frac{1}{2} \ \text{radians per second}\end{align} In order to know the linear speed of the particle, we would have to know the actual distance, that is, the length of the radius. Let’s say that the radius is 5 cm. If linear velocity is \begin{align}v=\frac{d}{t}\end{align} then, \begin{align}v=\frac{5}{2}\end{align} or 2.5 cm per second. If the angle were not exactly 1 radian, then the distance traveled by the point on the circle is the length of the arc, \begin{align}s = r \theta\end{align}, or, the radius length times the measure of the angle in radians. Substituting into the formula for linear velocity gives: \begin{align}v=\frac{r\theta}{t}\end{align} or \begin{align}v=r \cdot \frac{\theta}{t}\end{align}. Look back at the formula for angular velocity. Substituting \begin{align}\omega\end{align} gives the following relationship between linear and angular velocity, \begin{align}v = r\omega\end{align}. So, the linear velocity is equal to the radius times the angular velocity. Remember in a unit circle, the radius is 1 unit, so in this case the linear velocity is the same as the angular velocity. \begin{align}v &= r\omega\\ v &= 1 \times \omega\\ v &= \omega\end{align} Here, the distance traveled around the circle is the same for a given unit of time as the angle of rotation, measured in radians. Example A Lindsay and Megan are riding on a Merry-go-round. Megan is standing 2.5 feet from the center and Lindsay is riding on the outside edge 7 feet from the center. It takes them 6 seconds to complete a rotation. Calculate the linear and angular velocity of each girl. Solution: We are told that it takes 6 seconds to complete a rotation. A complete rotation is the same as \begin{align}2\pi\end{align} radians. So the angular velocity is: \begin{align}\omega=\frac{\theta}{t}=\frac{2\pi}{6}=\frac{\pi}{3}\end{align} radians per second, which is slightly more than 1 (about 1.05), radian per second. Because both girls cover the same angle of rotation in the same amount of time, their angular speed is the same. In this case they rotate through approximately 60 degrees of the circle every second. As we discussed previously, their linear velocities are different. Using the formula, Megan’s linear velocity is: \begin{align}v=r\omega = (2.5)\left(\frac{\pi}{3}\right) \approx 2.6 \ \text{ft per sec}\end{align} Lindsay’s linear velocity is: \begin{align}v=r\omega=(7)\left(\frac{\pi}{3}\right) \approx 7.3 \ \text{ft per sec}\end{align} Example B A bug is standing near the outside edge of a compact disk (so that his radius from the center of the disc is 6 cm) that is rotating. He notices that he has traveled \begin{align}\pi\end{align} radians in two seconds. What is his angular velocity? What is his linear velocity? Solution: We know that the equation for angular velocity is \begin{align}\omega=\frac{\theta}{t}=\frac{\pi}{2}\end{align} radians per second. We can use the given equation to find his linear velocity: \begin{align}v=r\omega=(6)\left(\frac{\pi}{2}\right) \approx 9.42 \ \text{cm per sec}\end{align} Example C How long does it take the bug in Example B to go through two complete turns? Solution: Since the angular velocity of the bug is \begin{align}\frac{\pi}{2}\end{align} radians per second, we can use the equation for angular velocity and solve for time: \begin{align}\omega =\frac{\theta}{t}\end{align} \begin{align}t =\frac{\theta}{\omega}\end{align} Since there are \begin{align}4\pi\end{align} radians in two complete turns of the disc, we can use this for the value of \begin{align}\theta\end{align}: \begin{align}t =\frac{4\pi}{\frac{\pi}{2}}=4\pi \times \frac{2}{\pi}=8 seconds\end{align} Vocabulary Angular Velocity: The angular velocity of a rotating object is the change in angle of an object divided by the change in time. Linear Velocity: The linear velocity of an object is the change in position of an object divided by the change in time. Guided Practice 1. Doris and Lois go for a ride on a carousel. Doris rides on one of the outside horses and Lois rides on one of the smaller horses near the center. Lois’ horse is 3 m from the center of the carousel, and Doris’ horse is 7 m farther away from the center than Lois’. When the carousel starts, it takes them 12 seconds to complete a rotation. Calculate the linear velocity of each girl. Calculate the angular velocity of the horses on the carousel. 2. The Large Hadron Collider near Geneva, Switerland began operation in 2008 and is designed to perform experiments that physicists hope will provide important information about the underlying structure of the universe. The LHC is circular with a circumference of approximately 27,000 m. Protons will be accelerated to a speed that is very close to the speed of light (\begin{align}\approx 3 \times 10^8\end{align} meters per second). How long does it take a proton to make a complete rotation around the collider? What is the approximate (to the nearest meter per second) angular speed of a proton traveling around the collider? Approximately how many times would a proton travel around the collider in one full second? 3. Ted is standing 2 meters from the center of a merry go round. If his linear velocity is 6 m/s, what is his angular velocity? Solutions: 1. It is actually easier to calculate the angular velocity first. \begin{align}\omega=\frac{2\pi}{12}=\frac{\pi}{6}\end{align}, so the angular velocity is \begin{align}\frac{\pi}{6} \ rad\end{align}, or \begin{align}0.524\end{align}. Because the linear velocity depends on the radius,each girl has her own. Lois: \begin{align}v=r\omega = 3 \cdot \frac{\pi}{6}=\frac{\pi}{2} \ \text{or} \ 1.57 \ m/sec\end{align} Doris: \begin{align}v=r\omega=10 \cdot \frac{\pi}{6}=\frac{5\pi}{3} \ \text{or} \ 5.24 \ m/sec\end{align} 2. \begin{align}v=\frac{d}{t}\rightarrow 3 \times 10^8=\frac{27,000}{t} \rightarrow t=\frac{2.7 \times 10^4}{3 \times 10^8}=0.9 \times 10^{-4} = 9 \times 10^{-5}\end{align} or 0.00009 seconds. \begin{align}\omega=\frac{\theta}{t}=\frac{2\pi}{0.00009}\approx 69,813 \ rad/sec\end{align} The proton rotates around once in 0.00009 seconds. So, in one second it will rotate around the LHC \begin{align}1 \div 0.00009 = 11,111.11\end{align} times, or just over 11,111 rotations. 3. Since the equation relating linear and angular velocity is given by \begin{align}v = r \omega\end{align}, we can solve for omega: \begin{align}\omega = \frac{v}{r} = \frac{6}{2} = 3\end{align} Concept Problem Solution As you found out in this Concept, the angular velocity is the change in angle divided by the change in time. Since you sweep around the circle twice in a second, this becomes: \begin{align}\omega = \frac{4\pi}{1} = 4\pi\end{align} rad/sec Further, you can find the linear velocity with the equation: \begin{align}v = r\omega = (2)(4\pi) = 8\pi \approx 25.132 cm/s \end{align} Practice Beth and Steve are on a carousel. Beth is 7 ft from the center and Steve is right on the edge, 7 ft further from the center than Beth. Use this information and the following picture to answer questions 1-6. 1. The carousel makes a complete revolution in 12 seconds. How far did Beth go in one revolution? How far did Steve go in one revolution? 2. If the carousel continues making revolutions every 12 seconds, what is the angular velocity of the carousel? 3. What are Beth and Steve's linear velocities? 4. How far away from the center would Beth have to be in order to have a linear velocity of \begin{align}\pi\end{align} ft per second. 5. The carousel changes to a new angular velocity of \begin{align}\frac{\pi}{3}\end{align} radians per second. How long does it take to make a complete revolution now? 6. With the carousel's new velocity, what are Beth and Steve's new linear velocities? 7. Beth and Steve go on another carousel that has an angular velocity of \begin{align}\frac{\pi}{8}\end{align} radians per second. Beth's linear velocity is \begin{align}2\pi\end{align} feet per second. How far is she standing from the center of the carousel? 8. Steve's linear velocity is only \begin{align}\frac{\pi}{3}\end{align} feet per second. How far is he standing from the center of the carousel? 9. What is the angular velocity of the minute hand on a clock? (in radians per minute) 10. What is the angular velocity of the hour hand on a clock? (in radians per minute) 11. A certain clock has a radius of 1 ft. What is the linear velocity of the tip of the minute hand? 12. On the same clock, what is the linear velocity of the tip of the hour hand? 13. The tip of the minute hand on another clock has a linear velocity of 2 inches per minute. What is the radius of the clock? 14. What is the angular velocity of the second hand on a clock? (in radians per minute) 15. The tip of the second hand on a clock has a linear velocity of 2 feet per minute. What is the radius of the clock? Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Vocabulary Language: English TermDefinition angular speed Angular speed is the ratio of revolutions that occur per unit of time. Angular Velocity The angular velocity of a rotating object is the change in angle of an object divided by the change in time. dimensional analysis Dimensional analysis is a process for converting from one unit to another. linear speed Linear speed is the ratio of distance per unit of time. Linear Velocity The linear velocity of an object is the change in position of an object divided by the change in time. radian A radian is a unit of angle that is equal to the angle created at the center of a circle whose arc is equal in length to the radius. Show Hide Details Description Difficulty Level: Tags: Subjects:	{}
# Faithful representations and tensor products Let $G$ be a finite group (or a compact Lie group). Prove that if $\rho:G\to\mathrm{GL}(V)$ is a faithful finite dimensional complex representation of $G$ then any irreducible representation embeds in some tensor product of $V$. Advertisements # Foiled by multiple roots Let $\Sigma_3$ be the set of monic complex polynomials of degree 3 with multiple roots, and regard it as a subspace of $\mathbb{C}^3$ by looking at the non-leading coefficients. Prove that $\mathbb{C}^3\setminus \Sigma_3$ is homotopy equivalent to $S^3\setminus K$, where $K$ is the trefoil knot. # Not an average calculus problem Prove that for any complex numbers $z_1,\dots, z_n$ one can find a nonempty subset $I\subseteq \{1,2,\dots, n\}$ such that $\left\|\sum_{i\in I}z_i\right\|\geq \frac{1}{\pi}\sum_{i=1}^n \|z_i\|$ # An exercise in Linear Algebra Let $k$ be a field of characteristic $0$ and $k[x_1,\ldots,x_n]^{(d)}$ be the $k$-vector space of homogeneous degree $d$ polynomials in $n$ variables. Show that the linear $k$ span of the set of $d$-th powers of linear polynomials $(a_1 x_1+\ldots+a_n x_n)^d$ is the whole space of homogeneous degree $d$ polynomials. In coordinate-free terms: if $V$ is a finite-dimensional $k$-vector space then $\{ v^n : v\in V\}$ spans $\mathrm{Sym}^dV$. # Spooky categories Suppose $C$ is an abelian category and $B\subseteq C$ is a subcategory which is also abelian. Is $B$ a sub-(abelian category) of $C$? # Independentist magic Show that $\displaystyle\int_{0}^{1} (2\cos(\pi x))^{2n} (2\sin(\pi x))^2 dx = C_n$ where $C_n = \frac{1}{n+1}\binom{2n}{n}$ is the $n$-th Catalan number. # Categorical noetherian-ness A ring $A$ is noetherian iff the category $_A\mathrm{mod}$ of finitely generated $A$-modules is a full abelian subcategory of the category $_A\mathrm{Mod}$ of all $A$-modules.	{}
# Reorganize the order of my own blocks Hello everybody, and happy new year ! I wonder wheather there is a way to reorganize the order of my own blocks ? ( all the blocks which are under Make a block) Without deleting them and creating the blocks again one by one in the wanted order... No, sorry. I guess you could edit the xml file, if it's important enough to you. But be sure to keep a copy of the project as it was before you started editing! Oh, I see. Is there a description of the xml files projects somewhere ? Because I just had a look on mine and it looks very long... moreover the custom-blocks are everywhere... Do you know a good xml editor on mac ? I'm a windows users so don't know of a mac editor But when you find one, you'll want to expand how it looks otherwise its MUCH harder to edit. But you need to make sure that when you compress it back again - that Snap! is happy with the file So, for first use - play around with a copy of your project until your happy that you can edit and re-save without it going wrong This caught me out in early days with my windows editor No, I just use emacs. Maybe xcode has one? I've found a good one on mac QXmlEdit. Like I say - make sure Snap! is happy with the file when you save - I just tried another windows xml editor one out and Snap! wouldn't reload the file even though I made no changes at all! I've put together a hacked addition to the right-click menu (using a previous @dardoro project) that seems to able to move custom blocks to the top of the list I've no idea if it's safe to use on a complex real project so use at own risk [edit2 - now works with command, reporter and predicate custom blocks] I've already tested QXmlEdit, modifying an xml Snap! project and importing it again into Snap!. It works. Anyway, thank you for the move block to top. Nice! What I really want is a way to weave custom blocks in with the primitives. So, for example, the Colors and Crayons library could put its SET PEN [menu] TO __ block up where the primitive one is, and hide the primitive one. :~) I'll get on it I noticed that the library creates a lot of variables, which makes the part of the screen with the variables very crowded. Would it be possible to somehow hide only the variables that are created by libraries. We're working on designing a system for that. Soon, I hope. This topic was automatically closed 30 days after the last reply. New replies are no longer allowed.	{}
# Plotting atom output Suggestions and improvements to this utility is most welcome! Creating pseudopotentials using atom will often require users to post-analyze the output of the program. Here we aid this visualization process by adding a command line interface for plotting details of the output via a simple command. ## Command line tool It may be called using stoolbox atom-plot <atom-output-directory> which by default will create 4 plots. It mimicks the shell commands provided by atom for plotting with some additional details. By default the program will output details regarding the core-charge and valence charge overlap (use full for determining whether core-corrections should be used). Total charge in atom: 82.00213 Core-correction r_pc [2.0, 1.5, 1]: [0.8656676967, 0.9330918445, 1.0573356051] Bohr Here it is for lead and the first line gives the total charge in the atom calculation. Note that the deviation stems from the different integration schemes used in the toolbox and in atom. The 2nd line prints details about the where the core charge is some factor larger than the pseudo potential valence charge. In this case it prints the radii $$r_i$$ where $$\rho_{\mathrm{core}}(r_i)/\rho_{\mathrm{valence}}(r_i)$$ equals $$2$$, $$1.5$$ and $$1$$. The full overlap is printed in the title of the charge plot. It is calculated as: $\mathcal O = \int \mathrm{min}(\rho_{\mathrm{core}}(r), \rho_{\mathrm{valence}}(r)) \mathrm dr$ ## Input generation The tool can also be used to generate the input files for atom. While it is possible to read input files from INP files, one may also define a specific Atom object which will be used to generate the input file. from sisl import * from sisl_toolbox.siesta.atom import * s = AtomicOrbital(n=5, l=0, q0=1., R=2.4) p = AtomicOrbital(n=5, l=1, q0=0., R=1.6) d = AtomicOrbital(n=4, l=2, q0=10., R=1.4) f = AtomicOrbital(n=4, l=3, q0=0., R=2.) Ag = Atom("Ag", orbitals=[s, p, d, f]) atom_input = AtomInput(Ag, rcore=2., xc="pb", flavor="tm2") # write the INP file in current directory atom_input.pg()	{}
# Refrigerator fans #### #12 Joined Nov 30, 2010 18,210 Appliances seem pretty tame for this site, but this is the only forum I am a member at, so... I've been meeting 4 wire fans in refrigerators lately. I measured about 13 volts DC that the controller board was using to try to get a fan to run yesterday. I assume the motor is bad and will replace it today but, I wonder if anyone here knows what kind of fan motors the manufacturers are using lately and what kind of drive signal is proper for them. ps, this is a General Electric brand side-by-side ref/freezer, 2002 year. If I may rant for a moment, I wish they would just use the old 4 watt 120 VAC motors connected to the compressor leads instead of complicating the job with a microprocessor, but they didn't, so it's up to me to keep learning. #### SgtWookie Joined Jul 17, 2007 22,210 http://www.formfactors.org/developer\specs\REV1_2_Public.pdf ALL appliance manufacturers are "under the gun" to develop more efficient appliances. While that's usually good for the electric bill, the newer fridges don't seem to last worth a darn anymore. Anyway, the old AC fans lasted a long time, but there was no means to control the speed. Then they came out with the DC fans with a tach output (3-wire), but it was somewhat difficult to control their speed. The 4-wire fans have integrated electronics that take a logic-level PWM signal to control the speed, and the tach output gives feedback as to what the actual fan speed is. #12 #### #12 Joined Nov 30, 2010 18,210 One of the big mysteries in my life...how much energy is saved by replacing a 4 watt fan motor with a 5 watt microprocessor board, 3 sensors, and a 4 watt motor that doesn't run as fast? I am very doubtful that slowing down a 4 watt motor is saving a bunch of energy. That much energy could be saved by putting a single cooling fin on the case of the compressor, and it wouldn't require energy to run the computer that controls the cooling fin. If I ever get stuck with one of these microprocessor controlled 4 watt fan motors, I will drill a hole in the back of the freezer and install a 4 watt fan motor that is powered by the wires that are hot when the compressor is running instead of replacing $100 processor boards and$65 fan motors. Meanwhile, my customers are stuck because I refuse to redesign retail products for them. (It would violate my insurance policy.) I also wish that the computer that is monitoring the fan speed would notice that the fan isn't running and put up an error message like, "freezer fan is not turning". Edit. That PDF confirms the 13 VDC I saw. I expect the computer is waiting for a tach pulse before converting to PWM. Good. My meter can measure DC just fine. If the signal was already PWM I would have so much fun trying to detect it while trying to wedge both shoulders and a DVM inside a 22 inch wide freezer compartment. Last edited: #### MrChips Joined Oct 2, 2009 21,165 I'm not a big fan of refrigerators either. #### R!f@@ Joined Apr 2, 2009 9,734 I find much humor in this thread. Joined Jul 7, 2009 1,583 If people really wanted to save energy, they'd go back to refrigerator/freezer designs from 50 years ago (or more). At one point, we had two refrigerators and three freezers in our house, although one of the freezers was shut off. One of the freezers is a relatively new one that my sister-in-law gave to us because it made noise. I took it apart and found some bad motor bearings in a fan and replaced the fan. Anyway, the other freezer was left in this house when we moved into it 24 years ago. And it was an old freezer -- I could tell from the design/shape that it was made in the 1950's -- maybe even earlier. Guess what: that 50+ year-old freezer consumed exactly half the energy that the modern refrigerators and freezers do. I measured each unit with a Kill-a-watt integrating the power for at least a week for each unit. The only thing "wrong" with that old freezer is that you had to defrost it. I'd imagine that most people younger than 40 or 50 have no clue about what defrosting a freezer means. Consumers demand frost-free devices, but they cost twice as much energy-wise. That's because of the heating element to do the defrosting along with running the fan all the time. #### debe Joined Sep 21, 2010 1,169 Also have customer with an LG fridge with simmilar electronic fan speed & temp control all run by a processor board. Board has been destroyed by Mice in the back of the fridge. Customer says board to expensive, so fridge will probably endup scrap. I gues this is how manufacturers keep you Consuming products. #### #12 Joined Nov 30, 2010 18,210 Fan motor installed. Freezer happy Joined Dec 26, 2010 2,148 Living in a less affluent part of the world, I have a very basic refrigerator, and an equally basic freezer. They both have only compressors, no fans. Of course, the freezer does need periodic defrosting. The fridge has no ice compartment, so it needs no attention apart from cleaning - if I get around to it. From what I've just read, maybe it's best to stick with these simple appliances until they wear out. #### #12 Joined Nov 30, 2010 18,210 I agree, regularly. Putting a microprocessor on a 4 watt fan motor tends to make my head explode. #### colinb Joined Jun 15, 2011 351 Guess what: that 50+ year-old freezer consumed exactly half the energy that the modern refrigerators and freezers do. I measured each unit with a Kill-a-watt integrating the power for at least a week for each unit. The only thing "wrong" with that old freezer is that you had to defrost it. I'd imagine that most people younger than 40 or 50 have no clue about what defrosting a freezer means. Consumers demand frost-free devices, but they cost twice as much energy-wise. That's because of the heating element to do the defrosting along with running the fan all the time. We have a regular modern refrigerator+freezer with automatic defrost, and also a large deep freeze non-defrosting freezer. Even more than the energy savings in not performing automatic defrost, I find that the deep freeze's consitent low temperature (0 °F or less) make food stay fresh MUCH longer; bread, meat, etc. does not suffer the freezer burn and drying effect caused by automatic defrost temperature cycling. Though, as for the debate about putting a microprocessor on a small 4 watt fan to save energy, don't forget that modern microcontrollers and supporting electronics need only milliwatts of power to run and even microwatts average power when efficient power modes are used. Even saving a few percent of that 4 watts would make the microcontroller a win. #### #12 Joined Nov 30, 2010 18,210 #### SgtWookie Joined Jul 17, 2007 22,210 My folks had a Sears Kenmore washer and dryer in their Michigan home that worked from about 1965 up until the house was sold earlier this year; I had to replace the timers once about 10 years ago. When they bought a home in FL, they got a new pair of Kenmore washer and dryer; Mom broke the washer twice before the repairman told her to have the knob OUT before turning the timer around. She kept blowing a TRIAC that controlled the motor. #### NM2008 Joined Feb 9, 2008 135 Believe it or not, the fridges compressor is one of the most reliable devices in the house. I have seen many fridges/freezers practically fall apart around the compressor. In the UK and IRL. Many fridges/freezers are thrown to the scrap due to doors binding and not sealing properly. Not to mention the unsightly cracks and splits that appear inside door panels and plastic holders. There are dozens of these appliances waiting to be scrapped, but yet have perfectly working compressor units that still have years of life left in them. I find it interesting that it is, more times than not, the parts responsible for interaction with the human which seems to fail first. Eg...doors, panels. Interesting that the part which the average person never comes into contact will last tens of years. NM #### colinb Joined Jun 15, 2011 351 SgtWookie: I suspect something more sinister than simple cheapness of production cost for the appliance manufacturers. This is probably some planned obsolescence along with potential benefits for the manufacturer-certified service technicians and expensive replacement part market. Replacement parts are ridiculously priced: e.g., the keypad+display panel on my 10 year old double oven costs over \$200 for the replacement part (likely with additional labor costs unless you DIY) and it's already been replaced by the previous homeowner -- not even fixing the problem since the actual problem is in the connector from the display board to the logic board. #### colinb Joined Jun 15, 2011 351 I find it interesting that it is, more times than not, the parts responsible for interaction with the human which seems to fail first. Eg...doors, panels. Interesting that the part which the average person never comes into contact will last tens of years. Yup, the manufacturer wants your fridge to look dated and old after a few years so you'll want to pony up another couple GRAND for a shiny new one. #### NM2008 Joined Feb 9, 2008 135 Yup, the manufacturer wants your fridge to look dated and old after a few years so you'll want to pony up another couple GRAND for a shiny new one. Seems to be it alright! And then they advertise these appliances as being more enviornmentally friendly than the last one they built. Yet they seem to forget about the many hundreds of years that the waste plastic panels will be sitting in the landfill. The majority being built in Asia having to be transported thousands of miles on crude oil burning cargo ships. Putting out as much emissions in one trip as a hundred cars do in one year. NM #### #12 Joined Nov 30, 2010 18,210 Sorry colin. Nothing personal. You aren't the first person to say microprocessors are great inventions while I was in the middle of a rant. You must know that rants are unreasonable..don't you? Meanwhile my car has a microprocessor that is beeping at me. You see, it isn't good to bump your face on the windshield, so you drive carefully. Our government decided that wasn't enough so they passed a law requiring a backup system called a "seat belt". Then they decided that wasn't good enough and required a backup for the backup called an "air bag", complete with high speed sensors and electronic firing mechanism. Then some lawyer decided that if you run into some kind of extra-terrestrial vehicle with a blade sticking out 6 feet from it, and the blade cuts your battery cable before the actual impact, you will need a backup for the backup for the backup, and it's called a "backup power supply". Now, if the backup for the backup for the backup fails, you need an idiot light blinking at you from the dashboard, and that is a backup for the backup for the backup for the backup. The idiot light in my car has been blinking for so long that the light bulb burned out, so the backup for the backup for the backup for the backup has a backup, and it's called a "beeper". In the morning when I wake up, I'm going out to my car, find the backup for the backup for the backup for the backup for the backup, and rip it out with a pair of pliers. 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# I Ricci tensor in vacuum inside Earth 1. Sep 22, 2016 ### AlanE Imagine a hole drilled through the Earth from which all air has been removed thus creating a vacuum. Let a cluster of test particles in the shape of a sphere be dropped into this hole. The volume of the balls should start to decrease. However, in his article "The Meaning of Einstein's Equation" page 18, John Baez and Emory Bunn indicate that the rate of decrease of the volume, per volume, should equal the t-t component of the Ricci tensor. But how can this be if the Ricci tensor is zero in a vacuum? Any ideas would be appreciated. 2. Sep 22, 2016 ### pervect Staff Emeritus What makes you think the volume should decrease? After a little thought, I agree that the volume will decrease, by analogy with the case I mention below. Basically a disk of test particles at the same height will reach the center of the Earth at the same time, which winds up implying the volume at the center of the Earth is zero. I've never thought much about a ball of test particles being dropped into the center of the Earth, but I have thought more about a ball of test particle orbiting the Earth in a non-rotating spaceship. I think the conclusions are similar. What happens is that V(t) does decrease with time, but if you look at Einstein's equation, it only says that $\ddot{V}= 0$ at t=0, when all the particles in the ball are initially at rest relative to each other. So if we plot V as a function of t, initially V does not change with time, but this is only true at t=0, it is not necessarily true that $\ddot{V}= 0$ at any other time. There is no requirement that V be a constant of motion. The only requirement is that at t=0, the second time derivative of the volume is zero. Last edited: Sep 22, 2016 3. Sep 23, 2016 ### AlanE Thanks for the reminder that merely having the second derivative equal to 0 doesn't imply that the volume won't decrease. I guess the equation for the change of volume could be something hypothetically like V(t) = V(1-kt3). or V(t) = V(1-kt**4) etc. I need to research this myself. Maybe someone knows the required equation even if only for Newtonian mechanics. 4. Sep 23, 2016 ### pervect Staff Emeritus I believe the evolution of the volume in the general case would be given by Raychauhurdi's equation, https://en.wikipedia.org/wiki/Raychaudhuri_equation. But the goal is to make things as simple as possible, which Baez does by choosing initial conditions appropriately - those initial conditions being that the ball of test particles start out with all the particles relatively at rest. Then a lot of complexities disappear, and the underlying simplicity emerges. 5. Sep 25, 2016 ### Staff: Mentor Not quite. A disk is not a ball, and the version of Einstein equation under discussion only applies to a ball. A ball of test particles dropped from initial rest such that its center is at a given height will not all reach the center of the Earth at the same time (although the portion of the ball consisting of a disk transverse to the direction of fall will); the ball squeezes in the transverse direction but stretches in the radial direction. Strictly speaking, this won't be true if any of the test particles in the ball meet at the center of the Earth, because Raychaudhuri's equation only applies to a congruence of non-intersecting worldlines. However, we can use it up until that point. Looking at the RHS of the equation, there is only one possible nonzero term: the shear term $- 2 \sigma^2$. The tidal tensor term is zero in vacuum (because it is the trace of the tidal tensor, which as noted is equal to $R_{mn} X^m X^n$, and $R_{mn} = 0$ in vacuum). So the key thing to compute is the shear scalar of the congruence in question. It starts out being zero (because, roughly speaking, the shear scalar is a sort of average shear in all directions, and the positive shear in the radial direction is exactly cancelled by the negative shear in the transverse direction). However, I'm not sure whether it continues being zero once the fall is in progress or not. 6. Sep 25, 2016 ### pervect Staff Emeritus The later point is the one I was trying to make - if the disk, which is a sub-manifold of the ball, deforms to a point, we expect the volume of the deformed ball to wind up to be zero. But this isn't inconsistent with Einstein's equation, for reasons that have already been discussed and which the OP seems happy with.	{}
## 8.1 Decision Tree in Hunt’s Algorithm Hunt’s algorithm builds a decision tree in a recursive fashion by partitioning the training dataset into successively purer subsets. Hunt’s algorithm takes three input values: 1. A training dataset, $$D$$ with a number of attributes, 2. A subset of attributes $$Att_{list}$$ and its testing criterion together to form a 'test condition', such as 'age>=25' is a test condition, where, 'age' is the attribute and '>=25' is the test criterion. 3. A Attribute_selection_method, it refers a procedure to determine the best splitting. The general recursive procedure is defined as below (Pang-Ning Tan 2005): 1. Create a node $$N$$, suppose the training dataset when reach to note $$N$$ is $$D_{N}$$. Initially, $$D_{N}$$ is the entire training set $$D$$. Do the following: 2. If $$D_{t}$$ contains records that belong the same class $$y_{t}$$, then $$t$$ is a leaf node labelled as $$y_{t}$$; 3. If $$D_{t}$$ is not an empty set but $$Att_{list}$$ is empty, (there is no more test attributes left untested), then $$t$$ is a leaf node labelled by the label of the majority records in the dataset; 4. If $$D_{t}$$ contains records that belong to more than one class and $$Att_{list}$$ is not empty, use Attribute_selection_method to choose next best attribute from the $$Att_{list}$$ and remove that list from $$Att_{list}$$. use the attribute and its condition as next test condition. 5. Repeat steps 2,3 and 4 until all the records in the subset belong to the same class. There are two fundamental problems that need to be sorted before Hunt’s algorithm can work: 1. How to form a ‘test condition’? particularly when non-binary attributes exist? 2. How to define the ‘best test conditions’, so very loop the best test condition can be used in a decision tree? ### How to Form a Test Condition? Decision tree algorithms must provide a method for expressing a test condition and its corresponding outcomes for different attribute types. 1. Binary Attributes. The test condition for a binary attribute is simple because it only generates two potential outcomes, as shown in figure 8.2. 1. No-binary attributes. No-binary attributes depend on the types of nominal or ordinal, it can have different ways of the split. A nominal attribute can have many values, its test condition can be expressed in two ways, as shown in Figure 8.3. For a multiway split, see Figure 8.3(a), the number of outcomes depends on the number of distinct values for the corresponding attribute. For example, if an attribute such as age has three distinct values: youth, m_aged, or senior. Its test condition will produce a three-way split or a binary split. Figure 8.3(b) illustrates three different ways of grouping the attribute values for age into two subsets. 1. Continuous Attributes. For continuous attributes, the test condition can be constructed as a comparison test $$(A<v)$$ or $$(A≥v)$$ with binary outcomes, or a range query with outcomes of the form $$v_i≤A<v_{i+1}$$, For $$i=1,…,k$$. The difference between these approaches is shown in Figure 8.4. For the binary case, the decision tree algorithm must consider all possible split positions v, and it selects the one that produces the best partition. For the multiway split, the algorithm must consider all possible ranges of continuous values. ### How to Determine the Best Split Condition? The method used to define the best split makes different decision tree algorithms. There are many measures that can be used to determine the best way to split the records. These measures are defined in terms of the class distribution of the records before and after splitting. The best splitting is the one that has more purity after the splitting. If we were to split $$D$$ into smaller partitions according to the outcomes of the splitting criterion, ideally each partition after splitting would be pure (i.e., all the records that fall into a given partition would belong to the same class). Instead of defining a split’s purity, the impurity of its child node is used. There are a number of commonly used impurity measurements: Entropy, Gini Index and Classification Error. Entropy: measures the degree of uncertainty, impurity, or disorder. The formula for calculate entropy is as shown below: $$$E(x)= ∑_{i=1}^{n}p_ilog_2(p_i), \tag{8.1}$$$ Where $$p$$ represents the probability and $$E(x)$$ represents the entropy. Gini Index: also called Gini impurity, measures the degree of probability of a particular variable being incorrectly classified when it is chosen randomly. The degree of the Gini index varies between zero and one, where zero denotes that all elements belong to a certain class or only one class exists, and one denotes that the elements are randomly distributed across various classes. A Gini index of 0.5 denotes equally distributed elements into some classes. The formula used to calculate Gini index is shown below: $$$GINI(x) = 1- ∑_{i=1}^{n}p_i^2, \tag{8.2}$$$ Where $$p_i$$ is the probability of an object being classified to a particular class. Classification Error measures the misclassified class labels. It is calculated with the formula shows below: $$$Classification error(x)= 1 - max_{i}p_i. \tag{8.3}$$$ Among these three impurity measurements, Gini is Used by the CART (classification and regression tree) algorithm for classification trees, and Entropy is Used by the ID3, C4.5, and C5.0 tree-generation algorithms. With the above explanation, we can now say that the aims of a decision tree algorithm are to reduce the Entropy level from the root to the leaves and the best tree is the one that takes order from the most to the least in reducing the Entropy level. The good news is that we do not need to calculate the impurity of each test condition to build a decision tree manually. Most tools have the tree construction built-in already. We will use the R package called rpart to build decision trees for our Titanic prediction problem. ### References Pang-Ning Tan, Vipin Kumar, Michael Steinbach. 2005. Introduction to Data Mining. 1st ed. Addison Wesley.	{}
Tuesday, March 06, 2007 Curious ferocity Begin a sentence with the first word of the pair and end it with the last. If you need to change the type of word (change ferocious to ferocity or fierce or ferociously) feel free. curious . . . . . ferocious unappreciated . . sensitive vapid . . . . . . crazy somber. . . . . . concise foggy . . . . . . egotistical dangerous . . . . considerate willowy . . . . . capricious hilarious . . . . mystical blithe. . . . . . reckless allusive. . . . . divine dizzy . . . . . . ornery Inspired by the Start and Stop Game at Writing Fix.	{}
# Orbital Mechanics Quandary 1. Mar 11, 2008 ### spud three I had thought that for an object in a simple circular orbit, the escape velocity would be root 2 times it's orbital speed. And that this should approximately apply for the Moon too. However Wikipedia here http://en.wikipedia.org/wiki/Moon gives the moon's escape velocity as 2.38 km/sec, yet further down gives the "average orbital speed" as 1.022 km/sec. So, is the Root 2 relationship valid? Spud 2. Mar 11, 2008 ### Staff: Mentor Orbital speed (of the moon) is not the same parameter as escape velocity. Orbital speed is the linear (tangential) speed of a satellite, moon or planet around the body that it orbits. Escape velocity is the speed required for a mass to depart from some point in the gravitational field, e.g. the surface of moon or planet, or orbit, and escape to some very far distance (essentially infinitely far away), such that if the propulsive force is cutoff, the mass would coast to infinity. 3. Mar 11, 2008 ### Janus Staff Emeritus I think you'll find that the 2.38km/sec value given is the escape velocity for an mass resting on the Moon's surface, not the escape velocity needed for the Moon to leave Earth orbit. 4. Mar 13, 2008 ### spud three thanks Janus, that explains it. I used Vesc = sq root(2GM of moon/R) and calculated the escape vel for an object on the surface of the moon to be 2.375 x 10E3 m/sec, almost exactly what wikipedia states. Further, I used the same equation to calculate the escape velocity of the moon from the Earth (assuming no other gravity factors), and came up with 1.440 x 10E3 m/sec, which works out to be almost exactly sq root of 2 * the "average orbital speed" given by wikipedia. So the sq root 2 relationship between orbital speed, and escape speed seems correct. Now my real ambition is to verify that the moon cannot escape (all other gravity factors aside), that there is not enough kinetic energy in the Earth's rotation. Assuming that it were to be all transfered to the moon (I know tidally locked still means the Earth has a small rotation but lets not consider this for now) I'd like to prove that the moon would still be in orbit. I realize that the sun will have probably flared up and died before this could theoretically happen, but it's just an academic exercise. For the orbital mechanics gurus out there, would it be correct to calculate the Earth's rotational energy, add that to the the moon's potential energy, and then see if that's greater than the energy needed for the moon to escape? Or this alternative which I believe gives the same answer. We know the moon's escape vel from the Earth is sq root of 2 * orbital vel, and thus the escape energy needed is 2 * orbital kinetic energy. And so if the Earth's rotational energy is less than the moon's kinetic energy, then that must mean the moon can not escape? Is this too simple, or are there any other factors to consider that would affect this calculation? Should I go ahead with my assumptions? 5. Mar 13, 2008 ### Janus Staff Emeritus A couple of things. Actually, you need to calculate the total energy of the Moon (kinetic and gravitational), then compare that to the rotational energy of the Earth. If the absolute value of the Earth's energy is less than the Moon's total energy, then it doesn't have enough energy to boost the Moon to escape velocity. The second thing is that the Moon doesn't have to reach escape velocity to leave earth orbit. Because the influence of the Sun's gravity, there will be a point where the Sun's grip on the Moon will be strong enough to pull the Moon free into a independent orbit. This point falls on what is called the Hill sphere. If the Moon drifts further from the Earth than the radius of the Hill sphere, it will drift away with out every reaching escape velocity. So you really need to figure out how much energy it would take to lift the Moon to an orbit with an radius outside the Hill sphere. This will be less than that needed to reach escape velocity. total energy of a orbiting body: $$E_t = -\frac{GMm}{2a}$$ M = mass of primary m = mass of orbiting body a= average orbital radius of orbit. (semi-major axis) Hill sphere radius for circular or nearly circular orbits: $$r = a \sqrt[3]{\frac{m}{3M}}$$ m = mass of secondary body M = mass of primary body for example to find the Hill sphere for the Earth, The Earth is the secondary body and the Sun the primary. The answer will the be the maximum distance from the Earth at which a body can stay in orbit around the Earth. 6. Mar 15, 2008 ### spud three So Janus, using your formula for the moon’s total energy (other than some small rotational energy about it’s center), I get: E = -6.6726 E-11 * 5.974 E24 * 7.348 E22 /(23.844 E8) = -3.81 E28 Joules Now for the rotation energy of the Earth, I used Ek = mv^2/5 Where m is the mass of a spinning solid sphere, and v is the equatorial velocity. I calculate Ek = 2.57 E29 Joules, whereas wikipedia lists it as 2.6 E29 Joules, very close. So the moon would need 3.81 E28 joules to escape (neglecting other gravity influences for now), and Earth has 2.57 E29 joules of rotational energy. So the Earth has approx 7 times as much rotational energy as needed by the moon to escape. Now all sources I’ve seen say that theoretically there’s not enough energy in the Earth for the moon to escape, that tidally locked would be the theoretical answer, and never escape the Earth (caveats listed in my post #4). So where’s my error? 7. Mar 16, 2008 ### Janus Staff Emeritus The problem is in assuming that all of the Earth's rotational energy can be transfered to the Moon. Only a small percentage is. Consider the following: The Earth rotation slows by 1 millisecond per century. The moon recedes at a rate of 4 cm per year. If we figure out how much energy the Earth loses in rotational energy per century and then figure out how much energy the moon gains by receding 4 meters during that century we can determine what fraction of the energy the Earth loses is transfered to the Moon. The answer will come out to be smaller than 1/7. The rest is lost as waste heat. 8. Mar 17, 2008 ### spud three Here’s what I calculated: The Earth’s loss in rotational energy due to a change of one millisecond (happens over a century) is 6.00 E21 joules The moon’s gain in energy receding 4 m, (receding 4 cm per year ) is 3.96 E20 joules So the fraction of the energy the Earth loses compared to the moon gain is 6.00 E21 / 3.96 E20 = 15 As you predicted, the moon gains less than the one seventh required to escape. One fifteen seems a pretty small amount of energy to transfer. I guess “global warming” gets the other 14/15ths. So... using the 1/15th factor, how far would the moon recede, outside gravity and longevity considerations aside? 9. Mar 18, 2008 ### Janus Staff Emeritus Add 1/15 of 2.57e29 joules to -3.81e28 joules,giving you the new total energy of the Moon. Take this answer and plug it into the equation used to find the total energy of the Moon and solve for "a" the new average radius of the orbit. 10. Mar 22, 2008 ### spud three So the ratio of the energy the Earth loses compared to the moon gains is 6.00 E21 / 3.96 E20 = 15.14 Total energy of the moon after receiving 1/15th of the Earth’s rotational energy = -2.11 E28 joules. Radius for that energy = 694,000 km, compared to 384,400 present radius of moon’s orbit. Moons period at a radius of 694,000 km would be 66.6 present Earth days (compared to present 27.4 or so). Hill sphere radius for the Earth/Sun works out to be 1,497,000 km. So the sun should never capture the moon. For an orbiting object at the Hill sphere radius I calculated, the sun’s force of gravity on this object would be 33 times the Earth’s force of gravity on it. Yet at just less than this radius, the object would theoretically stay in an (un)stable orbit around the Earth. This was calculated from Fg = GM1M2/Rsquared. Does this raise any eyebrows? 11. Mar 22, 2008 ### Ken G That all seems right, including Janus' point about the energy lost to wasted heat. The one issue that has not been explicitly mentioned however is that the actual conserved quantity, to a good approximation, here is angular momentum not rotational energy. The Earth loses its angular momentum of rotation as the Moon gains orbital angular momentum. As the Moon would require infinite angular momentum to escape the Earth, it would never be possible for any amount of energy. However, the Sun does ruin the angular momentum conservation at some point, controlled by the Hill sphere as mentioned above. So putting it all together, one might say that the Earth has plenty of rotational energy to eject the Moon, but it lacks sufficient angular momentum to get the Moon far enough away where the Hill sphere effects become important. The Earth's rotation would slow to the point that it would be locked to the Moon's orbit, and at that point there would no longer be any reason for the Earth to slow down further. In fact, IIRC, the effects of the Sun are then to cause the Moon to start coming back in toward the Earth. Last edited: Mar 22, 2008 12. Mar 22, 2008 ### Janus Staff Emeritus Try calculating the ratio of the Sun's force of gravity to the Earth's on the Moon in its present orbit. 13. Mar 22, 2008 ### spud three Janus, I did previously calculate the ratio of the sun's force of gravity on the moon compared to the Earth's, in its present orbit, and came up with 2.21 Should there be a theoretical maximum for the ratio of forces of gravity at the Hill sphere for different primary and secondary objects? 14. Mar 22, 2008 ### Janus Staff Emeritus The point is that is that is not the ratio of the forces of gravity that counts. Instead you need to consider the ratio of Earth's gravity to the tidal force of the Sun acting across the orbit of the Moon. 15. Mar 26, 2008 ### spud three Ken, I understand that in the end angular momentum is conserved (somehow), however I’m finding my calculations are NOT showing this. For the situation where the primary (the Earth) slows 1 msec over 100 years, I’m finding the change in the rotating Earth’s angular momentum L = 2/5 Me * Re * (Vi – Vf) = 8.23 E25 Nms. I considered two cases, the first where ALL of the Earth’s kinetic energy loss (due to decrease rotational speed) goes to the moon (an academic example only). The moon’s change in L = Mm * (G * Me)^.5 * (Rf ^.5 – Ri ^.5) = 2.27 E27 Nms, where Rf is the lunar distance after absorbing the Earth’s Ek loss, and Ri is the lunar distance at the start (the difference being about 60 m). For the second case (using Janus’ realistic numbers of .001 msec over 100 years, and 4 cm per year) I calculate the moon’s change in L to be 1.47 E26 Nms (almost exactly 1/15 th of the first case angular momentum, as expected). However, neither of these cases equals the smaller calculated Earth L loss of 8.23 e25 Nms. Conservation of energy was explained by creation of heat energy. Is there a similar explanation for conservation of angular momentum ? And Ken, would an escaping moon (Hill sphere and other factors aside) really require infinite angular momentum? 16. Apr 7, 2008 ### Jenk Sorry if this simple way of putting the answer to spud three's initial problem has been mentioned already. The two parameters that are root 2 apart can be found without any mention of the Earth. Escape velocity at the surface of the moon is root 2 larger than orbital velocity at the surface of the moon. You'd hit a few hills flying that low, but the point is that wikipedia gave the escape v at the surface, so r is the moon's radius, to the surface, in both root GM/r, and root 2GM/r. Hope this way of seeing it makes sense. 17. Apr 14, 2008 ### spud three ...Back to Ken's angular momentum... Yes thanks Jenk, I came up with the similar formulas and derivation to show root 2 once Janus explained what Wiki meant, and so the quandary referred to in the post title is solved for me. However, the thread has progressed to a disussion of conservation of angular momentum, as in Ken's points in post 11 and my attempts in post number 15. I've seem to find that in the 100% energy transfer case (assumed in my post number 6) angular momentum seems not to have been conserved. Similarly for the partial energy transfer mentioned by Janus in post number 7 and calculated by me to be approximately 1/15 th in post 8, angular momentum again seems not to be conserved. This I worked out for the Earth using I = .4 * M * R squared, and angular momentum L = I * w. And for the moon using L = M * V * R. (I can post the complete calculation if anyone would like to take a look and offer me an explanation of where the missing angular momentum is). In further pursuit, can ANYONE offer some figures on a ficticious planet/satellite system that undergoes changes in the satellite orbit (for example as tides cause on the Earth/moon system) where I can then calculate the angular momentums? Ideas?? 18. Apr 14, 2008 ### Ken G Sorry I missed this before. Conserving energy won't conserve angular momentum, that's no suprise (the same happens with energy and linear momentum when two cars collide). The second example probably just used approximate numbers, and that's why the angular momentum wasn't conserved exactly. Note it wasn't too far off. It wouldn't really require an infinite angular momentum if you include the Hill sphere, only in the idealized limit of two point masses and no Sun. But it would require more angular momentum than is present in the Earth's rotation, so that's the key issue. As for why angular momentum is conserved, it is because the Earth-moon system (with no Sun and no Hill sphere) is a closed system with no external torque on it. Last edited: Apr 14, 2008 19. Apr 18, 2008 ### spud three Well Ken, using the figures over 100 years of the moon receding 4 m and the Earth slowing 1 milli second may be just too approximate, as you say. Because using them, it appears the Earth would have to lose 1.823 times as much angular momentum as I calculated, in order to conserve angular momentum. Hence if we revise the Earth's rotation slowing to 1.823 milliseconds over the 100 years, then the moon/earth angular momentum is conserved by my calculations. Then as for conservation of energy, since the Earth is slowing almost twice as quickly as the previously used figure, instead of approximately one fifteenth of the Earth's lost kinetic energy going to the moon, it's more like 1/(15.14*1.823) = .00363 or approx 1/28. The remaining 27/28 must go to heat then.	{}
#### Torch ##### Member so.....superstitions.....believe in them?? i kinda do. this week every single one of our techies had something happen to them. Our TD had his hand like eaten away when a light tree collapsed on his hand. Our SM had a wanger slice her leg open, I had a nail impaled into the bone of my foot, our SR Deck Manager burned her hand while focusing lights, our AV guy.......just was having his fun with the projectors, thats enuf pain.....our spots tripped over their railings, its just been one amazing week, im scared for this comming one this is ALL because our TD had an empty fortune cookie. first time, he ran over and killed one singular frog...3x....then next time he was late to school and almost got suspended, now...were getting like killed on set....hes no longer allowed to have anymore chineese food.... #### Van ##### CBMod CB Mods Perhaps, to be philosophical for a moment, The empty fortune cookie was a sign. A sign that You make your own fortune, and how you view that fortune is completely up to you. Your choices determine your fate. Or perhaps your right no more chinese food for Squegee! . #### Torch ##### Member no im sure that its no more chinese for squeegee #### HandyMan ##### Member I think i'd be a little scared to be in that theatre..wouldnt want to end up in the hospital...ekk...no over night stays for me plz #### Squeegee ##### Member It wasn't that bad! The first time I got an empty fortune cookie I laughed. A good friend of mine looked at it and said that I should take a picture of it and title it "unfortunate" I thought that was clever. Little did I know that I would kill a frog that night. Yeah its a bad sign when you get an unfortunate cookie like I did, just warning you guys. My hand is okay but I was grabbing onto the part where the lighttree came together with its base and it slammed onto my skin and as I instinctively pulled my hand away my skin ripped off and got caught in the light tree. Now half my thumbprint is missing. Go me! :-\ Noone will let me do anything either! I feel so inept! #### bcfcst4 ##### Member On the saturday of our show, I had an organic chem class right before. Everything, and i do mean everything, went wrong during our lab, things broke, fell, shattered, exploded, made odd vapors-- you name it, it happened. I made the comment to my fellow group members that if this was the theme of my day, if this Murphy's Law continued, our show was screwed. I went to the theater to help setting everything up, told everyone about my Murphy's Law morning and we had a good laugh. Matinee starts, everything's going pretty well. We begin our biggest scene change, and a vase shatters. Why we were using real glass vases i have no idea. Also during that scene change, an actor's mic stops working, so we have every sound person trying to fix it as we freak out about broken glass. The scene begins, we breathe a minute and get ready to clean up the glass at intermission. Now this show was Little Shop of Horrors and, as we're standing backstage, we get the word that yes, our Audrey II has broken. We rented it for two weeks for a lot of money, and now, two scenes before intermission, it's broken. I freak out because my Murphy's Law day is continuing, we manage to pull the plant off during intermission, extend intermission by about 10 minutes, make a splint for the piece of conduit that broke, and hope to god it lasts till the end of the show. It did, and we were able to more permanently fix it before our saturday night show, but man. I was cursing every god I could think of, while Murphy was laughing his @ of in his grave i'm sure. #### Van ##### CBMod CB Mods Murphy's Law, Anything that can go wrong, will. Cole's Law, Thinly sliced Cabbage. People should be shot for building Audrey II out of conduit, any good model/puppet maker will tell you Rattan or a Carbon fiber/ fiberglass rod is the only way to make an armature that will really last. Just my two cents worth. #### bcfcst4 ##### Member Murphy's Law, Anything that can go wrong, will. Cole's Law, Thinly sliced Cabbage. People should be shot for building Audrey II out of conduit, any good model/puppet maker will tell you Rattan or a Carbon fiber/ fiberglass rod is the only way to make an armature that will really last. Just my two cents worth. I wouldn't have know before hand, but now, i do oh so much agree. Ah the joys of a small budget. #### astrotechie ##### Member this week every single one of our techies had something happen to them. Our TD had his hand like eaten away when a light tree collapsed on his hand. Our SM had a wanger slice her leg open, I had a nail impaled into the bone of my foot, our SR Deck Manager burned her hand while focusing lights, our AV guy.......just was having his fun with the projectors, thats enuf pain.....our spots tripped over their railings, its just been one amazing week, im scared for this comming one Wow. That is a lot of bad luck. Well superstitious or not, i think that was just being in the wrong place at the wrong time.	{}
How to control an underactuated platform with a single servo What's the best way to control the following type of platform so that it remains balanced and upright? It's a box supported by a single leg that can pivot at the foot (F) and at the hip (H). The foot and hip joints are designed so that they can only rotate in the xz plane (where x goes left-to-right and z goes up-and-down). It's similar in concept to the inverted pendulum, except the base (F) is fixed, and the actuation is done by shifting the mass at the "top" of the pendulum. I've seen some academic papers refer to this as the "cart-table model". The hip joint is powered by a servo motor which can report its angle of deflection relative to the box (ϴⱼ). At the center-of-mass (COM), there's an accelerometer that reports the box's angle of deflection relative to gravity (ϴᵢ). I've constructed this platform and I'm now trying to program a microcontroller to actuate the servo so that the platform remains both upright and level. The figure depicts the four basic states that the platform will experience. At t₀, it's perfectly level, but in an unstable leg configuration so that its center-of-mass is off-center and will cause it to begin toppling over. At t₁, the platform has tipped forward slightly. I've been able to program my controller to detect this and start actuating the servo to rotate the box counter to the angle of deflection so that the center-of-mass shifts to the other side of the leg, taking me to state t₂. However, this is where I run into problems. I'm using a simple PID controller that's being told the target value of ϴᵢ, while being fed the current value of ϴᵢ, and it's attempting to output a servo control signal to achieve this correction. After some testing, I realized this doesn't work because ϴᵢ isn't the only angle that needs to be controlled. By only looking at ϴᵢ, I was seeing massive oscillations of the servo all while the box topples over, because as soon as the servo tilted the box over the foot, ϴᵢ was then horribly unbalanced in the opposite direction, causing the PID controller to again correct it by shifting the box back over the foot...in the direction that the platform is falling, which only hastens the fall. I'm not sure how to tell the controller to "ignore" the unbalanced ϴᵢ once the center-of-mass has been shifted. So I think the solution is that I need to either directly measure ϴₖ, or infer it by using ϴᵢ and ϴⱼ, and then feed both ϴₖ and ϴᵢ into my controller. However, I'm only familiar with simple PID controllers that take a single input. What type of controller would I use to take two inputs and output a servo signal? If I input ϴₖ into my PID controller, I could probably get my platform to stop toppling over, but the box wouldn't be level. It would achieve a state close to that shown at t₂, where the box is skewed but its center-of-mass is balanced. Would another solution be to re-locate the accelerometer to point H? If it were there, I could use the raw acceleration readings to get a reliable sense of when the platform has stopped shifting in the wrong direction, and use that period to temporarily disable the PID controller so it doesn't erroneously un-shift itself. My goal is to get the platform to the state depicted in tₙ, where the box is level and the leg is angled so that the center-of-mass is evenly distributed so that the platform can remain upright. What type of controller would I use to accomplish this? Is there anything else I'm missing? Edit: Updated the figure to depict the state described by @fibonatic. • $t_n$ still does not look like a stable equilibrium, since the reaction force acting on the box doesn't go through the center of mass. Namely I suspect a stable equilibrium to be when both the center of mass and the pivots in the foot and hip are aligned vertically. – fibonatic Mar 31 '18 at 10:15 • @fibonatic Aligning all the pivots with center-of-mass would definitely be stable, but intuitively, I think tₙ should also be an equilibrium. You should be able to verify this empirically with a simple test. If you stand up and try to balance on your heels while bending forward slightly, you can still stay balanced even though your center-of-mass, hips, and feet are not in alignment. No? – Cerin Mar 31 '18 at 17:31 • I initially misread your question and thought that the actuator is in the foot instead of the hip. In that case $t_n$ would also be an equilibrium point, but would require a constant torque. – fibonatic Mar 31 '18 at 18:07	{}
# Start a new discussion ## Not signed in Want to take part in these discussions? Sign in if you have an account, or apply for one below ## Site Tag Cloud Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support. • CommentRowNumber1. • CommentAuthorFinnLawler • CommentTimeAug 16th 2010 New page: indecomposable object, following (what I think is) Johnstone's definition. I also found it in some online topos theory lecture notes by Ieke Moerdijk and Jaap van Oosten. Lambek and Scott give a different definition in Intro. to Higher-order Cat. Log., p. 168. I'm not sure how it relates to Johnstone's. I've also given a proof that indecomposable <=> connected in an extensive category. I'd be interested to know whether this hypothesis is the weakest possible, if anyone has any ideas (or just likely-looking references). • CommentRowNumber2. • CommentAuthorRodMcGuire • CommentTimeAug 17th 2010 New page:indecomposable object This definition (defined using coproduct) exactly corresponds in lattice theory to join irreducible (non existent page) which has the co-notion meet irreducible. The time is ripe for some terminological cleanup so that ’indecomposable obect’ has a transparent co-notion. Various options (not necessarily disjoint) are: 1) Rename indecomposable to coindecomposable because it is defined with coproducts, leaving indecomposable to be defined with products. 2) Follow lattice theory and distinguish +indecomposable from ×indecomposable. 3) Rename indecomposable to irreducible because it as a notion is already in use at such places as irreducible closed subspace, irreducible topological space , and possibly irreducible representation. • CommentRowNumber3. • CommentAuthorTodd_Trimble • CommentTimeAug 17th 2010 Of the options listed, (2) is my favorite. Of course the two notions coincide where one has biproducts (and so it would be pointless to maintain the distinction in such contexts), as in classical representation theory. I guess you yourself have some hesitancy about the last example listed in option (3), since you say “possibly”. As one knows from quiver representations, “irreducible” is a stronger notion than “indecomposable”. Aesthetically, I don’t like the sound of “coindecomposable” in (1) for such a common notion, and I don’t think one needs to be so fussy here. Cf. the common usage “filtered colimit”, as opposed to “cofiltered colimit”. • CommentRowNumber4. • CommentAuthorFinnLawler • CommentTimeAug 17th 2010 I agree with both of you – specifically, I think that ’indecomposable’ by itself isn’t very suggestive, but ’+-indecomposable’ would seem to mean what it says. I would have thought that ’indecomposable’ and ’irreducible’ were synonyms, but then I don’t know any representation theory. Meanwhile ’coindecomposable’ is even worse, I think, because it’s not at all obvious where the dualization is happening. So I vote for (2). On an unrelated note, I’ve asked a question on MathOverflow about the Johnstone and Lambek–Scott definitions. I’ll transfer any interesting responses to nLab. • CommentRowNumber5. • CommentAuthorTobyBartels • CommentTimeAug 17th 2010 • (edited Aug 17th 2010) I also like (2). Cf. the common usage “filtered colimit”, as opposed to “cofiltered colimit”. This is a technicality (indeed, the whole point of your example is that this is a technicality), but you’ve got it backwards; filtered colimits really are filtered, but filtered limits are really cofiltered. [Edit: In fact, Todd did not have it backwards; I just read Todd backwards.] The problem is that ‘co’ sometimes goes with products instead of with coproducts. Besides filters, another example is (co)induction and (co)recursion. In both of these cases, the sum-like notion was studied first, which is why the terminology came out that way. Actually, if I were king of the world, I’d switch things around so that products were called ‘cosums’ and (co)limits followed suit. (The only big problem is that I don’t know any simple term to replace ‘coequaliser’.) But it is too late now! • CommentRowNumber6. • CommentAuthorTodd_Trimble • CommentTimeAug 17th 2010 you’ve got it backwards Hmm… I do? I took it that Rod was saying that the prefix “co” ought to be used consistently, so that “coindecomposable” would go with coproducts. I was merely pointing out that this is not observed elsewhere in category theory, using an example that is not my invention. Notice that I was not making a claim that “filtered colimit” was wrong, as you seem to interpret me as saying. In fact, I am in complete agreement with your penultimate paragraph. • CommentRowNumber7. • CommentAuthorRodMcGuire • CommentTimeAug 17th 2010 • (edited Aug 17th 2010) As an order chauvinist I would like the notions of indecomposible and inrreducible when they are the same to use the same term, irreducible (Though I have yet to look into those areas where they differ). However both terms seem to have little meaning as English words and a more enlightening choice is available. Consider ×indecomposible which basically means un-factorable. However we have simple word for this - “prime”. I think ×indecomposible directly carries over to numbers and defines the primes. +indecomposible would then become coprime and the only coprime number is 1. (this notion of coprime is not to be confused the notion of coprime as relatively prime.) If I was king I would decree prime and coprime (or ×prime and +prime if one wants to avoid primacy). This approach is along the line of avoiding negative terms such as “non-empty” in favor of the positive “inhabited”. • CommentRowNumber8. • CommentAuthorTobyBartels • CommentTimeAug 17th 2010 Hmm… I do? • CommentRowNumber9. • CommentAuthorTobyBartels • CommentTimeAug 17th 2010 Consider ×indecomposible which basically means un-factorable. However we have simple word for this - “prime”. There’s already a big mess here in ring theory. We have the concepts of irreducible element (Wikipedia), meaning $\times$-irreducible (or equivalently $\times$-indecomposable) as you have said, and prime element, meaning generating a prime ideal. These are not the same, and what’s more, the usual concept of prime number corresponds to the former and not the latter! • CommentRowNumber10. • CommentAuthorTodd_Trimble • CommentTimeAug 18th 2010 No problemo! • CommentRowNumber11. • CommentAuthorMike Shulman • CommentTimeAug 18th 2010 I don’t think there is a serious enough problem here to merit changing established words. Words like “prime” and “irreducible” have heavy connotations in other contexts and I hesitate to try to reuse them to mean something fairly different here. In particular, already in lattice theory there are the notions of “prime filter” and “prime ideal” (one could argue that one of those should be “coprime” but I think they are both used prefixless), which are related but not quite the same as the meaning of “indecomposable” here. I guess I’m saying that I agree with (2) – that we shouldn’t introduce new words but just clarify when necessary. But I also don’t have a problem with “indecomposable” referring to coproducts by default, if unqualified. • CommentRowNumber12. • CommentAuthorzskoda • CommentTimeAug 18th 2010 Without a question, in algebra, and in representation theory proper, irreducible is stronger than indecomposable and the distinction is completely standard. The entry under the consideration here is about irreducible, not indecomposable. In many special categories the two notions happen accidentally to be equivalent. • CommentRowNumber13. • CommentAuthorTodd_Trimble • CommentTimeAug 18th 2010 • (edited Aug 18th 2010) in algebra, in representation theory proper, irreducible is stronger than indecomposable and the distinction is completely standard. The entry under the consideration here is about irreducible The same conceptual distinction resides in lattice theory as well: “irreducible” in representation theory corresponds to ($\vee$)-“atomic” in lattice theory, whereas “indecomposable” means ($\vee$)-“irreducible” in lattice theory (as Rod is pointing out under his option (3)). The entry under consideration is actually about indecomposable object; it’s just that there is a terminological shift across fields where elsewhere it’s called “irreducible element”. But again, for special lattices such as Boolean algebras, the two notions happen to be equivalent. A terminological mess, really, but that’s the way it is. • CommentRowNumber14. • CommentAuthorTobyBartels • CommentTimeAug 18th 2010 I’ve written a discussion of terminology in the article. Is it helpful? • CommentRowNumber15. • CommentAuthorTodd_Trimble • CommentTimeAug 18th 2010 I think it looks good; thanks Toby. • CommentRowNumber16. • CommentAuthorzskoda • CommentTimeAug 18th 2010 Maybe we have to relate this to the discussion of reducibility and complete reducibility at some point. • CommentRowNumber17. • CommentAuthorMarc • CommentTimeAug 18th 2010 In a topos, Lambek's definition of an indecomposable object is equivalent to the requirement that the object in question is not a union of proper subobjects (possibly taking care of the initial object). This is actually an exercise in that section. In fact, I think it is enough to have (Epi,StrongMono) factorizations available to proof the equivalence. Anyway, in a distributive category this condition (about subobjects) is equivalent to the usual "indecomposable" because coproduct injections are monic. As to the question about the requirements of "irreducible <=> connected", I do not know for sure. One can consider the following three conditions for an object c in a category K (I restrict to the binary case): (I) c =/= 0 and for any coproduct diagram a --> c <-- b we have a=0 or b=0.(P) c =/= 0 and for any coproduct diagram a --> c <-- b we have a --> c or b --> c an isomorphism.(C) K(c,-): K --> Set preserves coproducts Then you have (1) (I) ==> (P) always holds.(2) (C) ==> (P) holds whenever coproduct injections are monic.(3) (I) ==> (C) holds whenever coproducts are stable under pullback and coproduct injections are monic.(4) (P) ==> (I) holds whenever K has no proper preinitial objects (i.e. every epi 0 --> p must be a isomorphism) I hope I have not forgotten some obvious conditions. These days I do not have internet access, so I cannot post many details. One the other hand, the proofs are more or less straightforward. So it seems that (I) <==> (C) holds e.g. in every distributive category without proper preinitials. But are such categories already extensive? I do not know.	{}
# Memoization-based puzzle solution - sums in a triangle I am trying to solve a problem at Codechef. As I am new to programming, I am not familiar with dynamic programming. However, I do know the basics of recursion and memoization. Hence I implemented a solution to the above mentioned problem. The problem with my solution is that it exceeds the time limit given. As far as I can guess, the problem seems to be with my memoization logic. Please help me optimize the code. #include <iostream> #include <algorithm> #include <vector> #include <string> using namespace std; typedef vector<int> vector1D ; typedef vector<vector1D > vector2D ; // Two Dimensional vector int rownum; int noOfCalls = 0; vector2D a; vector2D cache; // The cache to implement memoization int solve (int i, int j) { int t1,t2,total; if( i > rownum-1 ) { return 0; } else { if (cache[i][j] != 0 && ( i < 50) && ( j < 50)) { //for small values use memoization return cache[i][j]; } t1 = solve(i+1,j); t2 = solve(i+1,j+1); total = max(t1,t2) + a[i][j]; cache[i][j] = total;//store in cache } } int main(int argc, char argv[]) { int tries; cin>>tries; while(tries--) { cin>>rownum; int temp = rownum; for ( int i = 0 ; i < rownum ; i++) { a.push_back(vector1D(rownum,0));//initialize memory } for ( int j = 0; j < 10; j++) { cache.push_back(vector1D(10,0));//initialize cache } for ( int i = 0; i < rownum; i++) { for ( int j = 0 ; j <= i; j++) { cin>>a[i][j]; } } cout<<solve(0,0)<<endl;//call the solving function cache.clear(); } return 0; } • your solution overall looks Ok. You should extend the memoization for all values. Max no of lines is <=100 and 100X100 vector is Ok to have. – Rajendran T Jan 25 '12 at 19:38 I hate your use of global variables. int rownum; int noOfCalls = 0; vector2D a; vector2D cache; // The cache to implement memoization I suppose it is an attempt at better performance. The test for j here is redundant: if (cache[i][j] != 0 && ( i < 50) && ( j < 50)) If i is smaller than 50 then j can not be greater than 50. If i is greater than 50 then the test will fail and there is no need to test what the value of j is. ### Bugs for ( int i = 0 ; i < rownum ; i++) { a.push_back(vector1D(rownum,0));//initialize memory } As you never clear a the size increases for every puzzle. This can become quite expensive as a will continuously keep breaking its buffer and have to be re-allocated (copying all those internal arrays). ### Speed If you are solely going for speed then operator>> is about half the speed of scanf as it (operator>>) takes into account a lot of local stuff that is not done with scanf. Your recursive algorithm is computing some spots multiple times. // Your Triangle 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 // Number of times each spot is hit with recursion 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 As you can see some of the spots in the center bottom of the triangle are being hit many many times by recursion when you only want to hit each location once. Thus an iterative solution is probably better. You are using a 2D array to represent the triangle which requires two de-references (which can be inefficient (but without measuring I am not convinced this is a major issue)). Rather than do that use a 1-D array given the number of rows n you can quickly compute the number of elements in the triangle as n(n+1)/2 and thus re-size the array in one statement. Element access for a child of m is m+row(m)+0 and m+row(m)+1 (note the calculation of row(m) does not need to be done each time you can keep track of this as you increment m). If your priority is performance with this, in addition to having a memoizing cache for up to the size of about the L1 cache - maybe 4K or so, I'd also get rid of the use of the vector of vector. You're spending a disproportionate amount of time in your vectors' [] operator() functions which will cause an over-abundance of stack pushes and context switches - you're using that operator in at least four places in each pass through solve() - six when not cached already. You can see this if you profile your code. I'd keep it to a vector with dynamically allocated columns and use only pointer arithmetic so that all accesses are happening from the same stack frame in solve(). I'd pre-calculate base pointers in advance and use those as much as possible in the pointer calculations using the bitwise << sizeof(int) to traverse. I would also stay local and convert your max(t1,t2) call to use a bit-twiddle trick so that total is now: total = t1 ^ ((t1 ^ t2) & -(t1 < t2)) + paij; where paij is the dereferenced arithmetically calc'ed pointer instead of a[i][j]. That should keep the pipeline full more often. To get to the next level from there, I would probably move from recursion to an iterative approach if you're not married to recursion. That should buy you some performance gain, although more could be done depending on your machine configuration. ### Edit: Test for speed. #include <iostream> #include <algorithm> #include <sys/time.h> using namespace std; inline int getmax( int a, int b ) { if ( a < b ) return b; else return a; } int main() { time_t timev; timeval t1, t2; double elapsed_time; const unsigned int RUNL = 1000000000; int i = 0; int j = 1; int k = 1; int total; gettimeofday( &t1, NULL ); while ( i++ < RUNL ) { total = j ^ ( ( j ^ k ) & -( j < k ) ); k ^= 1; } gettimeofday( &t2, NULL ); elapsed_time = ( t2.tv_sec - t1.tv_sec ) * 1000.0; elapsed_time += ( t2.tv_usec - t1.tv_usec ) / 1000.0; cout << "Elapsed time for the bitop was: " << elapsed_time << " ms\n" << endl; i = 0; k = 1; gettimeofday( &t1, NULL ); while ( i++ < RUNL ) { total = max( j, k ); k ^= 1; } gettimeofday( &t2, NULL ); elapsed_time = ( t2.tv_sec - t1.tv_sec ) * 1000.0; elapsed_time += ( t2.tv_usec - t1.tv_usec ) / 1000.0; cout << "Elapsed time for the stl call was: " << elapsed_time << " ms\n" << endl; i = 0; k = 1; gettimeofday( &t1, NULL ); while ( i++ < RUNL ) { total = getmax( j, k ); k ^= 1; } gettimeofday( &t2, NULL ); elapsed_time = ( t2.tv_sec - t1.tv_sec ) * 1000.0; elapsed_time += ( t2.tv_usec - t1.tv_usec ) / 1000.0; cout << "Elapsed time for the getmax call was: " << elapsed_time << " ms\n" << endl; } ### Edit (Edit) Fixed Speed Test: #include <iostream> #include <algorithm> #include <sys/time.h> #include <vector> using namespace std; void printTime(int total, std::string const& name, timeval& t1, timeval& t2) { double elapsed_time; elapsed_time = ( t2.tv_sec - t1.tv_sec ) * 1000.0; elapsed_time += ( t2.tv_usec - t1.tv_usec ) / 1000.0; std::cout << "Result(" << total << ") Elapsed time for the " << name << " call was: " << elapsed_time << " ms\n" << std::endl; } inline int getmax( int a, int b ) { if ( a < b ) return b; else return a; } int main() { timeval t1, t2; const unsigned int RUNL = 1000000000; std::vector<int> data(RUNL); for(int loop=0;loop < RUNL;++loop) { data[loop] = rand(); } std::cout << "LoadData DONE" << std::endl; int j = data[0]; int k; gettimeofday( &t1, NULL ); for(int loop = 1;loop < RUNL;++loop) { k = data[loop]; j = j ^ ( ( j ^ k ) & -( j < k ) ); } gettimeofday( &t2, NULL ); printTime(j, "BIT TWIDDLE", t1, t2); j = data[0]; gettimeofday( &t1, NULL ); for(int loop = 1;loop < RUNL;++loop) { k = data[loop]; j = max( j, k ); } gettimeofday( &t2, NULL ); printTime(j, "std::max", t1, t2); j = data[0]; gettimeofday( &t1, NULL ); for(int loop = 1;loop < RUNL;++loop) { k = data[loop]; j = getmax( j, k ); } gettimeofday( &t2, NULL ); printTime(j, "inline max", t1, t2); } Results: > g++ -O3 x.cpp > ./a.out Result: total( 2147483645) Elapsed time for the BIT TWIDDLE call was: 1968.2 ms Result: total( 2147483645) Elapsed time for the std::max call was: 1170.83 ms Result: total( 2147483645) Elapsed time for the inline max call was: 1178.05 ms It's easy to see why the optimizer runs so much faster than the bit hack from this code from between the gettimeofday calls): STLMAX movl 28(%esp), %ecx movl $1, %edx movl$1, %eax .p2align 4,,7 L34: movl (%ecx,%edx,4), %edx cmpl %edx, %ebx cmovl %edx, %ebx addl $1, %eax cmpl$1000000000, %eax movl %eax, %edx jne L34 leal 40(%esp), %eax movl $0, 4(%esp) movl %eax, (%esp) BIT HACK movl 28(%esp), %edi movl$1, %eax movl 1, %edx .p2align 4,,7 L25: movl (%edi,%eax,4), %ecx xorl %eax, %eax cmpl %ebx, %ecx setg %al xorl %ebx, %ecx negl %eax addl1, %edx andl %ecx, %eax xorl %eax, %ebx cmpl $1000000000, %edx movl %edx, %eax jne L25 leal 40(%esp), %eax movl$0, 4(%esp) movl %eax, (%esp) Basically two effective instructions in the loop doing the actual max in optimized STL compiler compared with seven for the bit hack. Lesson learned. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Code for hasZeroByte which gave about 2.5x improvement for the bit hack over the other two under -O3 optimization conditions: #include <iostream> #include <algorithm> #include <sys/time.h> #include <vector> using namespace std; void printTime(int total, std::string const& name, timeval& t1, timeval& t2) { double elapsed_time; elapsed_time = ( t2.tv_sec - t1.tv_sec ) * 1000.0; elapsed_time += ( t2.tv_usec - t1.tv_usec ) / 1000.0; std::cout << "Result(" << total << ") Elapsed time for the " << name << " call was: " << elapsed_time << " ms\n" << std::endl; } inline bool checkZbyte1( int a ) { for ( int i = 0; i < sizeof( int ); i++ ) { if ( !( 0xff & a ) ) return true; else a >>= 8; } return false; } inline bool checkZbyte2( int a ) { if ( !( 0xff & a ) \|\| !( 0xff00 & a ) \|\| !( 0xff0000 & a ) \|\| !( 0xff000000 & a ) ) return true; return false; } int main() { timeval t1, t2; const unsigned int RUNL = 100000000; std::vector<int> data(RUNL); for(int loop=0; loop < RUNL; ++loop) data[loop] = rand(); std::cout << "LoadData DONE" << std::endl; int j; int count = 0; gettimeofday( &t1, NULL ); for(int loop = 0; loop < RUNL ;++loop) { if ( checkZbyte1( data[loop] ) ) count++; } gettimeofday( &t2, NULL ); printTime( count, "inline version #1", t1, t2 ); count = 0; gettimeofday( &t1, NULL ); for( int loop = 0; loop < RUNL; ++loop ) { if ( checkZbyte2( data[loop] ) ) count++; } gettimeofday( &t2, NULL ); printTime( count, "inline version #2", t1, t2 ); count = 0; gettimeofday( &t1, NULL ); for( int loop = 0; loop < RUNL; ++loop ) { j = data[loop]; if ( ~( ( ( (j & 0x7F7F7F7F ) + 0x7F7F7F7F ) \| j ) \| 0x7F7F7F7F ) ) count++; } gettimeofday( &t2, NULL ); printTime( count, "BIT HACK", t1, t2 ); } • Unless your compiler is extremely ancient/stupid, it will generate better code for max than any bit-twidding hack you come up with. Otherwise these are good ideas :-) – Nemo Jan 26 '12 at 2:22 • I think bit twiddling is silly. Also I am not sure I agree that operator[] is a problem. This is simply returns base[index] and as such is probably inlined and is no more expensive than a normal array access. If you are saying that the 2D array is the problem then maybe. – Martin York Jan 26 '12 at 13:50 • hmm.. Why are bit hacks silly? I just ran 3 different cases of max using 1Billion of each all in identical loops. The results: – Wade Stone Jan 27 '12 at 0:58 • Thing cut me off for taking to long... ironic... the results: bit hack max as above: 3.052s stl max() call: 4.238s my own getmax() call: 4.470s the results were consistent. I'm using cygwin with gcc 4.5 on a quad 2.5GHz Dell M6400. Try it yourself... I don't think we could disregard an improvement of 29%. – Wade Stone Jan 27 '12 at 1:09 • They are silly because the compiler will always do as well and usually better. If you have found a new optimization to max (unlikely) then it will be introduced into the compiler and the next evolution of the compiler will be just as quick as your code. Post your code. :-) we will see how well it really does. – Martin York Jan 27 '12 at 3:10 Why only use memoization for "small" values of i and j? Without memoization, your program runs in exponential time. The problem, which memoization intends to solve, is that you tend to compute the same value many times along the way. So you memoize some of the values, to save time. On those areas of the table you aren't memoizing, you still suffer the exponential slowdown. And you're only memoizing 1/4 of your domain.	{}
# A generalization of the Kővári-Sós-Turán theorem We present a new proof of the Kővári-Sós-Turán theorem that ex(n, K_s,t) = O(n^2-1/t). The new proof is elementary, avoiding the use of convexity. For any d-uniform hypergraph H, let ex_d(n,H) be the maximum possible number of edges in an H-free d-uniform hypergraph on n vertices. Let K_H, t be the (d+1)-uniform hypergraph obtained from H by adding t new vertices v_1, ..., v_t and replacing every edge e in E(H) with t edges e ∪{v_1},..., e ∪{v_t} in E(K_H, t). If H is the 1-uniform hypergraph on s vertices with s edges, then K_H, t = K_s, t. We prove that ex_d+1(n,K_H,t) = O(ex_d(n, H)^1/t n^d+1-d/t). Thus ex_d+1(n,K_H,t) = O(n^d+1-1/t) for any d-uniform hypergraph H with ex_d(n, H) = Θ(n^d-1), which implies the Kővári-Sós-Turán theorem in the d = 1 case. As a corollary, this implies that ex_d+1(n, K_H,t) = O(n^d+1-1/t) when H is a d-uniform hypergraph in which all edges are pairwise disjoint, which generalizes an upper bound proved by Mubayi and Verstraëte (JCTA, 2004). We also obtain analogous bounds for 0-1 matrix Turán problems. ## Authors • 20 publications • ### Bounding the Number of Minimal Transversals in Tripartite 3-Uniform Hypergraphs We bound the number of minimal hypergraph transversals that arise in tri... 07/24/2018 ∙ by Alexandre Bazin, et al. ∙ 0 • ### Connected Fair Detachments of Hypergraphs Let 𝒢 be a hypergraph whose edges are colored. An (α,n)-detachment of 𝒢 ... 09/21/2020 ∙ by Amin Bahmanian, et al. ∙ 0 • ### A spectral bound on hypergraph discrepancy Let H be a t-regular hypergraph on n vertices and m edges. Let M be the ... 07/07/2019 ∙ by Aditya Potukuchi, et al. ∙ 0 • ### On rainbow-free colourings of uniform hypergraphs We study rainbow-free colourings of k-uniform hypergraphs; that is, colo... 06/13/2021 ∙ by Ragnar Groot Koerkamp, et al. ∙ 0 • ### Ear-Slicing for Matchings in Hypergraphs We study when a given edge of a factor-critical graph is contained in a ... 12/11/2019 ∙ by András Sebő, et al. ∙ 0 • ### Hypergraph regularity and higher arity VC-dimension We generalize the fact that graphs with small VC-dimension can be approx... 10/01/2020 ∙ by Artem Chernikov, et al. ∙ 0 • ### The convex dimension of hypergraphs and the hypersimplicial Van Kampen-Flores Theorem The convex dimension of a k-uniform hypergraph is the smallest dimension... 09/03/2019 ∙ by Leonardo Martínez-Sandoval, et al. ∙ 0 ##### This week in AI Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday. ## 1 Introduction The Kővári-Sós-Turán theorem is one of the most famous results in extremal combinatorics [16, 8, 10]. The theorem states that the maximum number of edges in a -free graph of order is . There are multiple known proofs of this theorem, including a standard double-counting proof that uses Jensen’s inequality, as well as a proof that uses dependent random choice and Jensen’s inequality [2]. Past proofs of the Kővári-Sós-Turán theorem have relied on convexity and Jensen’s inequality, or the power mean inequality which is a corollary of Jensen’s inequality. For a student to fully understand the proof, they would need to understand both convexity and Jensen’s inequality, which would require calculus background, or they would need to know an alternative proof of the power mean inequality that avoids using Jensen’s inequality. In this paper, we prove the Kővári-Sós-Turán theorem without using calculus or Jensen’s inequality. Instead we use a method based on Nivasch’s bounds on Davenport-Schinzel sequences [19] and Alon et al.’s bounds on interval chains [1]. This new proof gives a simple way to teach the proof of the Kővári-Sós-Turán theorem to students with no calculus background, and the same method can be used to prove a generalization of the Kővári-Sós-Turán theorem for uniform hypergraphs. In [19], Nivasch found upper bounds on the maximum possible lengths of Davenport-Schinzel sequences using two different methods. Both methods gave the same bounds, but the first method was more like the proofs in past papers on Davenport-Schinzel sequences, and the second method was similar to proofs about interval chains in [1]. The second method in [19] was much simpler than the first for proving bounds on Davenport-Schinzel sequences. We imitate the second method here for graph and hypergraph Turán problems. Let denote the maximum number of edges in an -free -uniform hypergraph on vertices. Let be the -uniform hypergraph obtained from by adding new vertices and replacing every edge in with in . For example, if is the -uniform hypergraph of order with edges, then . Mubayi and Verstraëte [18] proved that when is a -uniform hypergraph in which all edges are pairwise disjoint. In Section 2, we provide an elementary proof that , giving an alternative proof of the Kővári-Sós-Turán theorem when is the -uniform hypergraph of order with edges. As a corollary, this implies that when is a -uniform hypergraph in which all edges are pairwise disjoint, generalizing the upper bound of Mubayi and Verstraëte. In Section 3, we discuss analogous results about -dimensional 0-1 matrices that can be proved with similar methods. ## 2 The letter method An ordered -uniform hypergraph is a -uniform hypergraph with a linear order on the vertices. We define a lettered -uniform hypergraph as the structure obtained from labeling each edge of an ordered -uniform hypergraph with a letter such that two edges can be labeled with the same letter only if they have the same greatest vertex. Given a -uniform hypergraph , we say that a lettered -uniform hypergraph is -free if its underlying -uniform hypergraph is -free. For any -uniform hypergraph , let denote the maximum possible number of distinct letters in an -free lettered -uniform hypergraph on vertices in which every letter occurs at least times. The next lemma is analogous to inequalities in [19, 4, 11, 13] and is proved similarly. ###### Lemma 2.1. For all positive integers and -uniform hypergraphs , we have . ###### Proof. Start with a -uniform -free hypergraph with edges. Order the vertices of arbitrarily. For each vertex in in order from greatest to least, label the unlabeled edges adjacent to in any order with letters , only using each letter exactly times and deleting up to remaining edges adjacent to if does not divide the total number of edges in which is the greatest vertex. Observe that the new lettered hypergraph has at most distinct letters with every letter occurring exactly times, and it is -free. ∎ When combined with Lemma 2.1, the next lemma will complete our proof of the generalization of the Kővári-Sós-Turán theorem. We use Stirling’s bound in the proof of the next lemma, but it is not actually necessary. We explain after the proof how the use of Stirling’s bound can be replaced with an elementary one-sentence argument. ###### Lemma 2.2. For and a -uniform hypergraph with , we have . ###### Proof. Suppose for contradiction that there exists a -free lettered -uniform hypergraph on vertices with distinct letters in which every letter occurs at least times. Suppose that is sufficiently large so that . Delete edges of until every letter occurs exactly times. For each -subset of , define to be the number of edges in that contain all of the vertices in and a greater vertex in the ordering. Let be the number of -subsets of with . The number of -tuples of edges in that have the same least vertices is equal to , which is at most , or else would contain a copy of . This follows by the pigeonhole principle, since every -tuple of edges in that have the same least vertices must have different letters on each edge. Then and , where the last inequality follows from Stirling’s bound. However , a contradiction. ∎ ###### Theorem 2.3. For fixed and -uniform hypergraph , we have . The use of Stirling’s bound in Lemma 2.2 may seem to make the proof non-elementary, but it was unnecessary. All we need is that there exists some constant such that for all , and then we can replace each in the last proof with . Proving this for : it is clearly true for , and if we assume that , then we also have , and . Thus the whole proof is elementary. ###### Corollary 2.4. If is a -uniform hypergraph in which all edges are pairwise disjoint, then . ###### Proof. If is a -uniform hypergraph in which all edges are pairwise disjoint, then , so this bound follows from Theorem 2.3. ∎ The last corollary yields the bound of Mubayi and Verstraëte from [18] when . ## 3 0-1 matrices Using the same method, we can get similar bounds for Turán-type problems on -dimensional 0-1 matrices. In order to state these results, we introduce more terminology. We say that -dimensional 0-1 matrix contains -dimensional 0-1 matrix if some submatrix of can be turned into by changing some number of ones to zeroes. Otherwise avoids . For any -dimensional 0-1 matrix , define to be the maximum number of ones in a -dimensional 0-1 matrix of sidelength that avoids . As with the case of -uniform hypergraphs, most of the past research on the topic of -dimensional 0-1 matrices has focused on when . We mention several results for that have been generalized to higher values of . For example, Klazar and Marcus [15] proved that for every -dimensional permutation matrix , generalizing the result of Marcus and Tardos [17]. Geneson and Tian [14] sharpened this bound by proving that for -dimensional permutation matrices of sidelength , generalizing a result of Fox [6]. Geneson and Tian also proved that for every -dimensional double permutation matrix , generalizing the upper bound in [12]. In order to state the next result, we define to be the -dimensional 0-1 matrix obtained from the -dimensional 0-1 matrix by stacking copies of with the same orientation in the direction of the new dimension. For example if is the matrix of all ones, then is the matrix of all ones. ###### Theorem 3.1. 1. For fixed and -dimensional 0-1 matrix , . 2. For any -dimensional 0-1 matrix with , we have . In particular, for any -dimensional 0-1 matrix with at least two ones such that . Moreover, for any -dimensional 0-1 matrix with at least three ones differing in the first coordinate such that . ###### Proof. The upper bounds follow from using the letter method as in the last section. The lower bounds follow from stacking copies of the known lower bound constructions for extremal functions of forbidden and all-ones matrices [3, 5, 9, 10]. ∎ Permutation matrices and double permutation matrices with at least three ones are some examples for which Theorem 3.1 gives sharp bounds on and up to a constant factor [17, 12, 15, 14]. ## 4 Concluding remarks The standard double-counting method used to prove the Kővári-Sós-Turán theorem can also be used to prove the bounds in Theorem 2.3 and 3.1. We did not include this method since it uses convexity, and it gives the same bounds up to a constant factor as the letter method. Dependent random choice can also be used to obtain the same bounds for uniform hypergraphs up to a constant factor when , and it can be applied to a larger family of hypergraphs that contains the family of . The next lemma is a generalization of the dependent random choice lemma from [2] and [7]. In the next lemma, we call a vertex and a -subset of vertices of a -uniform hypergraph neighbors if there is some edge of that contains and all of the vertices of . For each vertex and set of vertices , we define to be the set of -subsets of vertices that are neighbors with , and we define to be the set of -subsets of vertices that are neighbors with every vertex in . ###### Lemma 4.1. Let be a -uniform hypergraph with vertices and edges. If there is a positive integer such that , then contains a subset of at least vertices such that every vertices in have at least common neighbors among the -subsets of . ###### Proof. Pick a set of -subsets of vertices of , choosing -subsets uniformly at random with repetition. Let , and let be the cardinality of . Then , where the second-to-last inequality used Jensen’s inequality. Let be the random variable for the number of subsets of size with fewer than common neighbors among the -subsets of vertices of . The probability that an arbitrary -subset is a subset of is , so . Thus by linearity of expectation, . Thus there exists a choice of for which the corresponding set of cardinality satisfies , so we can remove vertices from to produce a new subset so that all -subsets of have at least common neighbors among the -subsets of . ∎ We can use Lemma 4.1 to get upper bounds for a more general family of -uniform hypergraphs that contains the family of . The next theorem describes one such family. ###### Theorem 4.2. For any -uniform hypergraph , let be the -uniform hypergraph obtained by starting with vertices , making disjoint copies of for each -subset of vertices of , and replacing each edge in each with edges of the form for each . For any -uniform hypergraph with and any integers and , we have . Note that , and that the letter method also works to show that for any integers and . It would be interesting to see if the letter method is useful for other Turán-type problems, and what else can be said about in general. ## References • [1] N. Alon, H. Kaplan, G. Nivasch, M. Sharir, and S. Smorodinsky, Weak -nets and interval chains, J. ACM, 55, article 28, 32 pages, 2008. • [2] N. Alon, M. Krivelevich, B. Sudakov, Turán numbers of bipartite graphs and related Ramsey-type questions, Combin. Probab. Comput. 12 (2003), 477-494. • [3] W. Brown, On graphs that do not contain a Thomsen graph, Canad. Math. Bull. 9 (1966) 281-285. • [4] J. Cibulka and J. Kynčl. Tight bounds on the maximum size of a set of permutations with bounded vc-dimension. Journal of Combinatorial Theory Series A, 119: 1461-1478, 2012. • [5] P. Erdős, A. Rényi, and Vera T. Sós: On a problem of graph theory, Stud Sci. Math. Hung. 1 (1966) 215-235. • [6] J. Fox, Stanley-Wilf limits are typically exponential, arXiv:1310.8378, 2013. • [7] J. Fox, B. Sudakov. Dependent random choice. Random Structures Algorithms 38 (2011) 68-99. • [8] Z. Fűredi, An upper bound on Zarankiewicz problem, Combin. Probab. Comput. 5 (1996) 29-33. • [9] Z. Fűredi, New asymptotics for bipartite Turán numbers, J. Combin. Theory Ser. A 75 (1996) 141-144. • [10] Z. Fűredi and M. Simonovits. The history of degenerate (bipartite) extremal graph problems. In Erdős centennial, Bolyai Soc. Math. Stud. 25 (2013) 169-264. • [11] J. Geneson, A Relationship Between Generalized Davenport-Schinzel Sequences and Interval Chains. Electr. J. Comb. 22(3): P3.19, 2015. • [12] J. Geneson, Extremal functions of forbidden double permutation matrices, J. Combin. Theory Ser. A, 116 (7) (2009), 1235-1244. • [13] J. Geneson, Forbidden formations in multidimensional 0-1 matrices. European J. of Comb. 78, 147-154, 2019. • [14] J. Geneson, P. Tian, Extremal functions of forbidden multidimensional matrices. Discrete Mathematics 340(12): 2769-2781, 2017. • [15] M. Klazar and A. Marcus, Extensions of the linear bound in the Furedi-Hajnal conjecture, Advances in Applied Mathematics, 38 (2) (2007), 258-266. • [16] T. Kővari, V. T. Sós, and P. Turán. On a problem of K. Zarankiewicz. Colloquium Math., 3: 50-57, 1954. • [17] A. Marcus and G. Tardos, Excluded permutation matrices and the Stanley-Wilf conjecture, J. Combin. Theory Ser. A, 107 (1) (2004), 153-160. • [18] D. Mubayi, J. A. Verstraëte, A hypergraph extension of the bipartite Turán problem, J. Combin. Theory Ser. A 106 (2004) 237-253. • [19] G. Nivasch. Improved bounds and new techniques for Davenport-Schinzel sequences and their generalizations. J. ACM, 57(3), 2010.	{}
## Which is more ‘fundamental’, the ampere or the coulomb? An American electrical engineer, William J Beaty, who runs the excellent website (challenging the way in which electrical science is taught), entitled ‘K-6 Misconceptions, which I highly recommend, poses the interesting question: ‘Which is the more fundamental: the ampere or the coulomb?’ In his article he argues that, because charge is more fundamental than current, therefore, by extension, the coulomb must be more ‘fundamental’ than the ampere. As much as admire and agree with most of Mr Beaty’s articles, I feel that in this particular case, he is wrong. I believe that he has made the fundamental mistake of confusing quantities (i.e. ‘current’ and ‘charge’) with their corresponding units of measurement (i.e. ‘ampere’ and ‘coulomb’). His argument is, essentially, that because electric current is defined in terms of the quantity of charge transferred per unit time, then it follows that the coulomb must, therefore, be more ‘fundamental’ than the ampere. But SI doesn’t use the term, ‘fundamental’; rather, it uses the term ‘base’. And it considers the ampere to be a ‘base unit‘ while the coulomb is considered to be a ‘derived unit‘. By definition, all ‘derived units’ are defined in terms of ‘base units’. Where Mr Beaty’s argument falls flat is that he bases most his argument on his belief that the ‘definition’ the ampere is a ‘coulomb per second’. If this were to be the case, then it would be difficult to disagree with his argument. But, in fact, the ampere is NOT defined in terms of the coulomb and the second… and it never has been! So, the ampere is not reliant on, and thus ‘less fundamental’ than, the coulomb. So, if the ampere isn’t defined as a ‘coulomb per second’ (and a lot of North American textbooks mistakenly claim this!), then how is it defined? Well, there are three ‘effects’ of an electric current: the heating effect, the chemical effect, and the magnetic effect. Theoretically, any one of these effects could be used to define its unit of measurement: the ampere. For example, prior to 1948, the ampere was defined in terms of the chemical effect of an electric current: The ‘international ampere‘, as it was then called, was an early attempt at defining the ampere, as ‘that current that would deposit 0.001 118 g of silver per second from a silver nitrate solution’ Later, more-accurate measurements revealed that this current was actually 0.999 85 A, and not 1 A as thought! So, in 1948, it was decided to redefine the ampere in terms of the magnetic effect of an electric current! So, since 1948, the ampere has been defined as follows: The ampere is ‘that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed one metre apart in a vacuum, would produce between these conductors a force equal to 2×10−7 newtons per metre of length’. So, you see, the ampere is not, and never has been, defined as a ‘coulomb per second’! For most of its life, it’s been defined in terms of the force between parallel, current-carrying, conductors. But, after 70+ years, this definition has changed! Because, from mid-2019, the ampere is now defined in terms of the rate of flow of elementary charges —i.e. the carried by individual electrons, and NOT by a ‘coulomb’s-worth of electrons’. This new definition of the ampere us now ‘the current in the direction of flow of a particular number (see elsewhere in this blog for the actual number) of elementary charges per second’. And, as a ‘derived unit’, SI will continue to define the coulomb in terms of the ampere and the second: The coulomb is defined as ‘the quantity of charge transferred, in one second, by a steady current of one ampere’. So, while it is perfectly true that electric charge is ‘more fundamental’ than electric current, the same cannot be said about their corresponding units! So, to summarise. Mr Beaty is quite correct in arguing that charge is more fundamental than current because current is defined in terms of the quantity of charge transported per unit time. However, the same argument cannot be extended to the SI unit of current, the ampere, because (contrary to Mr Beaty’s belief) it is an SI base unit, and has never, ever, been defined in terms of the coulomb (a derived unit) but, from 1947, in terms of the force between energised conductors and, since 2019, now in terms of the flow of individual elementary particles. ## Current ‘flows’ We often say that ‘an electric current flows through a conductor’. Although widely-used, strictly speaking, this is actually incorrect. If we think in terms an analogy of the ‘current’ in a stream or river, what it describes is a flow of water. It’s actually the water that’s flowing, not the current. In exactly the same way, an electric ‘current’ describes a flow of charge carriers. It’s the charge carriers that are flowing, not the electric current. Although describing an electric current as ‘flowing’ is, strictly, a misconception, it is rather pedantic (‘nit-picking’) to insist on not describing current as ‘flowing’ around a circuit —particularly when it’s necessary to describe the direction of charge flow around a circuit. Providing we are aware of the distinction between ‘current’ and ‘charge flow’, it is perfectly acceptable to say that ‘current “flows” clockwise (or counter-clockwise) around…’ a particular circuit. ## Current is a flow of ‘free electrons’ In metal conductors, current is, indeed, a flow of free electrons. But to think that a current is always a flow of free electrons is a misconception. This is because an electric current can also occur in non-metallic conductors, such as within electrolytes (conducting liquids) and gasses. In electrolytes, for example, the charge carriers are not electrons, but ions. Ions are simply charged atoms, i.e. atoms which either have more electrons than protons (‘negative ions’), or those which have less electrons than protons (‘positive’ ions). It is, therefore, much more accurate to describe an electric current as being a flow of electric charge carriers, than of free electrons. This definition covers current in both metal as well as other conductors. But even this definition needs further clarity. This is because ‘flow’ suggests a significant movement of individual charge carriers in a particular direction. But, as we shall learn, later, for direct current (d.c.) this ‘flow’ is very slow (far less than metres per hour, in some cases!). For alternating current (a.c.), the charge carriers simply vibrate in opposite directions, and don’t ‘flow’ at all! ## Current is the ‘rate of flow’ of charge carriers Not really. We sometimes hear an electric current being defined as being ‘the rate of flow of electric charges’. The term, ‘rate of flow’, suggests that we are describing the velocity at which charge carriers are moving through a material. This is a misconception, because what we should be describing is the quantity of charge carriers, not their velocity. So, if we want to further refine our definition of an electric current, we can say that it is ‘the quantity of charge carriers, moving past a given point in a material, per unit time’. An electric current is ‘the quantity of charge carriers, moving past a given point in a material, per unit time’. ## The ampere is defined as a ‘coulomb per second’ No it’s not. It never has been! We have already defined electric current as ‘the quantity of charge carriers, moving past a given point in a material, per unit time’. Since charge is measured in coulombs (symbol: C) and time is measured in seconds (symbol: s), many students believe that an ampere is defined as a ‘coulomb per second’; indeed, many textbooks (particularly US textbooks) reinforce this misconception! However, the ampere has never been defined in this way! The ampere is one of seven SI Base Units, from which all other SI units are derived. The coulomb is one of those Derived Units. As SI Base Units cannot be defined in terms of Defined Units, we cannot define an ampere in terms of a coulomb. Since 1948, the ampere has been defined in terms of its magnetic effect, specifically the resulting force between two, parallel, current-carrying conductors, due to the attraction or repulsion of their magnetic fields, as follows: The ampere (symbol: A) is defined as ‘the constant current that, if maintained in two straight parallel conductors of infinite length and negligible cross-sectional area and placed one metre apart in a vacuum, would produce between them a force equal to 2 × 10-7 newtons per unit length’. [Although the newton is derived unit, it is simply the name we give to a ‘kilogram metre per second squared’ —all Base Units.] Prior to the 1948 definition, the ampere was defined in terms of the mass of silver deposited on an electrode over a specified period of time by electrolysis. So current has NEVER been defined as ‘a coulomb per second’! However, with effect from 20 May 2019, the definition of the ampere will change to: The ampere, symbol A, is defined by taking the fixed numerical value of the elementary charge, e, to be 1.602176634×10−19 when expressed in the unit coulomb (C), which is equal to the ampere second (A⋅s), where the second is defined in terms of ΔνCs. Please refer to the ‘News about the ampere‘ page of this blog for an explanation of this new definition. ## Current ‘flows’ close to the speed of light This is a widely-held misconception! We have already learnt that it’s not the current that’s flowing, but electric charges. And, most students are surprised to learn, that the velocity of these charges is very, v-e-r-y, low —often as low as millimetres per hour! It’s been said that an individual electron is unlikely to complete its journey through the filament of a torch (flashlight) within the lifetime of the torch battery! We call the velocity of electric charges, their ‘drift velocity(v) which, for any conductor, is expressed by the following equation: $v = \frac{I}{n A e}$ where v = drift velocity, I = current, n = number of electrons per cubic metre, A = cross-sectional area, and e = charge on one charge carrier. The number (n) of electrons per cubic metre of conductor depends, of course, on the type of material and its purity. For copper, for example, this figure is generally taken as 85×1027 and, for aluminium, 76.2×1027. And the amount of charge on a single electron (e) is generally taken as 16×10-18 C (coulombs). So, the drift velocity for a 2.5-mm2 conductor (used for ring-mains in British residences), carrying a direct current of, say, 10 A, would work out to be an incredibly-low 2.9 micrometres per second! Which means that it will take 344 828 seconds, or nearly 96 hours, to travel a distance of just one metre! While the drift velocity is v-e-r-y slow, the effect of the current is immediate because, of course. all the electrons throughout the conductor start to move at the same time. Students often ask, ‘If free elections move so slowly, why is there not a delay between operating a light switch, and the lamp coming on?’ Well, of course, the conductor connecting the switch and lamp is full of free electrons and, when the switch is closed, they ALL start to move at the same time. ## Current direction The great American physicist and statesman, Benjamin Franklin (1706–1790), believed that an electric current was some sort of mysterious ‘fluid’ which flowed from a higher (positive) pressure to a lower (negative) pressure. From this, it was generally assumed that an electric current flowed (through an external circuit) from positive to negative. After the much-later discovery of sub-atomic particles led to the ‘electron theory‘ of electricity, it was realised that (in metal conductors) electrons actually moved through an external circuit from negative to positive. When we talk about ‘current direction‘, we are always describing its direction through the external circuit, and never within the source of potential difference (battery, generator, etc.) To distinguish between the original theory regarding current direction, and the modern theory, we use the terms ‘conventional flow‘ (positive to negative) and ‘electron flow‘ (negative to positive). Unfortunately, as many of the laws relating to magnetism depend upon a knowledge of current direction, and were based on ‘conventional flow’, this is still widely-used today and is the case with most modern textbooks. Students have the misconception that ‘conventional flow’ describes a flow of ‘positive charges’ moving from positive to negative. However, this is not really the case. Conventional flow is simply an assumed direction of current and makes no assumptions whatsoever about what constitutes that current. ‘Electron flow’, on the other hand, not only describes direction but also the nature of what constitutes that current.	{}
# An expression involving p-th root of unity Let $\zeta_p$ be a $p$-th root of unity, where $p$ is an odd prime number. I just came across the following expression: $$\frac{(\zeta_p^2-\zeta_p+1)^3}{\zeta_p^2(\zeta_p-1)^2}.$$ Can we simplify this expression somehow? For $p=3$, I can rewrite this to $$\frac{8}{(\zeta_3-1)^2}.$$ Question. Can we simplify $$\frac{(\zeta_p^2-\zeta_p+1)^3}{\zeta_p^2(\zeta_p-1)^2} ?$$ Note that this expression does not change if we replace $\zeta_p$ by one of the elements in $\{\zeta_p, \zeta_p^{-1},1-\zeta_p, (1-\zeta_p)^{-1}, \zeta_p/(1-\zeta_p), (1-\zeta_p)/\zeta_p\}$. - Honestly, the expression looks simple to me, particularly for large $p$. – Tapu Oct 14 '11 at 12:18 The reason that you are successful in simplifying when $p=3$ is that the top factors as $$(\zeta_p-\zeta_6)(\zeta_p-\zeta_6^5)$$ where $\zeta_6$ is a primitive sixth root of unity; since $\zeta_6$ is $\zeta_3 + 1$ (for an appropriate choice of $\zeta_3$), you can simplify. I agree with Swapan that there doesn't seem to be any general reason why the expression would simplify for larger $p$. – Barry Smith Oct 14 '11 at 12:55 There are at least a couple of profitable ways of re-writing this. A short triggy answer is that your expression simplifies to $$\frac{\left(2\cos\left(\frac{2\pi}{p}\right)-1\right)^3}{2\cos\left(\frac{2\pi}{p}\right)-2},$$ which for $p=3$ gives the value of $\boxed{\frac{8}{3}}$ you allude to above, and for $p=5$ gives the neatly random-looking value of $$\frac{\left(\sqrt{5}-3\right)^3}{4(\sqrt{5} - 5)}=\frac{8\sqrt{5}-18}{\sqrt{5} - 5}=\boxed{\frac{2}{11\sqrt{5}+25}.}$$ For discussing the work that leads to the simplification, it's slightly more convenient from the view of algebraic number theory to deal with the reciprocal $$\xi_p:=\frac{\zeta_p^2(\zeta_p-1)^2}{(\zeta_p^2-\zeta_p+1)^3}.$$ First, it's not too hard to check that $\zeta_p^2-\zeta_p+1$ is a unit of $\mathbb{Z}[\zeta_p]$ for $p>3$, so writing it as $$\frac{\zeta_p^2}{(\zeta_p^2-\zeta_p+1)^3}\cdot (\zeta_p-1)^2$$ makes it clear that $\xi_p$ an algebraic integer (this is why I wanted the reciprocal), of asbolute norm $p^2$ (since $\zeta_p-1$ is a degree 1 prime above $p$ in $\mathbb{Q}(\zeta_p)$.) Second, let's take advantage of the fact that we know that $\xi_p$ is totally real, and so an element of $\mathbb{Z}[\zeta_p^+]$, where $\zeta_p^+:=\zeta_p+\zeta_p^{-1}.$ From the above re-writing, it's unreasonable to expect (actually, probably impossible) for $\xi_p$ to live in any proper subfield of $\mathbb{Q}(\zeta_p^+)$. So a reasonable interpretation for the problem of an ultimate simplication for $\xi_p$ is to write it completely in terms of $\zeta_p^+$. Let's do this now: $$\xi_p=\frac{\zeta_p^2(\zeta_p-1)^2}{(\zeta_p^2-\zeta_p+1)^3}=\frac{\zeta_p^3}{(\zeta_p^2-\zeta_p+1)^3}\cdot \frac{(\zeta_p-1)^2}{\zeta_p}=\frac{\zeta_p+\zeta_p^{-1}-2}{(\zeta_p+\zeta_p^{-1}-1)^3}=\boxed{\frac{\zeta_p^+-2}{(\zeta_p^+-1)^3}}$$ Now writing $\zeta_p^+=2\cos(2\pi/p)$ and reciprocating gives the formula at the top of this answer. Finally, let me mention that from an algebraic number theory point of view, it might me most useful to write $\xi_p$ not in terms of $\zeta_p^+$, but in terms of a prime of $\mathbb{Z}[\zeta_p]$ above $p$, i.e., in terms of $$\mathfrak{p}:=(1-\zeta_p)(1-\zeta_p^{-1})=2-\zeta_p^+.$$ Re-writing the previous boxed expression, we finally conclude with the reasonably concise (and algebraically transparent) formulation $$\xi_p=\boxed{\frac{\mathfrak{p}}{\left(1-\mathfrak{p}\right)^3}.}$$ -	{}
781 articles – 3114 Notices [english version] HAL : in2p3-00329098, version 1 arXiv : 0809.2929 34th International Conference on High-Energy Physics (ICHEP 08), Philadelphia : États-Unis (2008) Hadronic B Decays to Charm and Charmonium with the BaBar Experiment BABAR Collaboration(s) (01/08/2008) The {\it BaBar} experiment has recorded the decays of more than 465$\times 10^6 B\bar{B}$ pairs since 1999, and is reaching an unprecedented precision in the measurement of hadronic B decays. The following results are presented: tests of QCD factorization with the decays $B\to \chi_{c0}K^{}$, $B\to \chi_{c1,2}K^{()}$, and $\bar{B}^{0}\to D^{()0}h^{0}$, $h^0=\pi^0, \eta,\ \omega, \eta'$, study of the decays to charmonium $B\to \eta_c K^{()},\ \eta_c(2S) K^{()}$ and $h_cK^{()}$, measurement of the mass difference between neutral and charged $B$'s, measurement of the "r" parameters for the extraction of the CKM angles $\sin(2\beta+\gamma)$ with the decays $B\to D_s^{()}h, h=\pi^-, \rho^-, K^{()+}$, study of the three-body rare decays $B\to J/\psi\phi K$, study of the baryonic decays $\bar{B}^0\to\Lambda_c^+\bar{p}$, $B^-\to\Lambda_c^+\pi^-\bar{p}$, and $B^-\to\Lambda_c^+\pi^0\bar{p}$. Except for the published results, all the given numbers are preliminary. Thème(s) : Physique/Physique des Hautes Energies - Expérience Lien vers le texte intégral : http://fr.arXiv.org/abs/0809.2929 in2p3-00329098, version 1 http://hal.in2p3.fr/in2p3-00329098 oai:hal.in2p3.fr:in2p3-00329098 Contributeur : Nicole Berger <> Soumis le : Vendredi 10 Octobre 2008, 14:22:12 Dernière modification le : Vendredi 10 Octobre 2008, 15:15:53	{}
# Trajectories¶ For a motor controller supporting the use of a trajectory, and for which the support is effective in BLISS (cf writing a motor controller), trajectories can be defined and used to move axes in complex movements. ## Simple example¶ Here is a simple arc trajectory with two axis (X and Y). So axis X will have a cosin trajectory and axis Y a sin trajectory. In this example we want to move from 0 deg to 90 deg in 10 seconds. First let define the alpha, 10 points from 0 deg to 90 deg: import numpy alpha = numpy.linspace(0,90,10) #deg Then build position for the two axis: x_positions = numpy.cos(numpy.deg2rad(alpha)) Then build the time array. times = numpy.linspace(0,10,10) At t0 == 0 and t9 == 10 seconds in 10 points. In Bliss all trajectory are defined by the triplet PVT => P osition, V elocity, T ime. To achive this, use the PointTrajectory object to create the PVT array: from bliss.physics import trajectory pt = trajectory.PointTrajectory() pt.build(times,{'X':x_positions,'Y':y_positions}) Create one trajectory object per each axis: from bliss.common import axis pvt = pt.pvt() # calculate pvt array x_trajectory = axis.Trajectory(X,pvt['X']) y_trajectory = axis.Trajectory(Y,pvt['Y']) An finally create the TrajectoryGroup. This object group the movement for this two axis along the trajectory. from bliss.common.motor_group import TrajectoryGroup #Create the Trajectory group group = TrajectoryGroup(x_trajectory,y_trajectory) #load trajectories into the motor controller group.prepare() #linear move to the first point #In this example will move X to 1 and Y to 0 group.move_to_start() #move to the end of the trajectory. #After this command X will be near 0 and Y near 1 group.move_to_end() The calling sequence of this object must always be the same: • call prepare first • then move_to_start • finally move_to_end It’s possible to repeat the sequence but always in this order.	{}
APM346-2015F > Test 2 TT2-P2 (1/4) > >> Victor Ivrii: Solve \begin{align} &u_{y=0}=\frac{1}{x^2+1},\label{2-2}\\ &\max \|u \|<\infty.\label{2-3} \end{align} Hint: Use partial Fourier transform with respect to $x$, and formula F (x^2+a^2)^{-1}=\frac{1}{2a}e^{-\|k\|a}\qquad \text{as\ \ } a>0. \label{2-4} Xi Yue Wang: Use Fourier Transformï¼Œ$u(x,y) \rightarrow \hat{u} (k,y)$ $$-k^2\hat{u}+\hat{u}_{yy}=0\\\hat{u}(k,y)=A(k)e^{-\|k\|y}+B(k)e^{\|k\|y}$$ We discard second term because it is unbounded.$$\hat{u}(k,0) = A(k) = \hat{f}(k) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{x^2 +1}e^{-ikx} dx = \frac{1}{2}e^{-\|k\|} \\u(x,y)=\frac{1}{2}\int_{-\infty}^{\infty} e^{-\|k\|y}e^{\|-k\|}e^{ikx} dk$$ Then the answer is same as Emily. But for this one, from lecture notes, suppose we don't have a hint for $\hat{f}(k)$ We get IFT of $e^{-\|k\|y}$ is $\frac{2y}{x^2+y^2}$ $$u(x,y) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x')\frac{2y}{((x-x')^2+y^2)}dx'\\= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{(x'^2 +1)}\frac{y}{((x-x')^2+y^2)}dx'$$ Emily Deibert: I believe there may be a mistake in this one. As we did in HA7, we must change the given condition at $y=0$ by Fourier transform as well. So we will have (making use of the hint): u\|_{y=0} \longrightarrow \hat{u}\|_{y=0} = \hat{\frac{1}{x^2 + 1}} = \frac{1}{2}e^{-\|k\|} We should be able to proceed similarly to what Xi Yue Wang has done, with the integral now being: u(x,y)=\frac{1}{2} \int_{-\infty}^{\infty} e^{-\|k\|y}e^{-\|k\|}e^{ikx} dk It's actually a simple integral to solve; we split it up so that we have: u(x,y)=\frac{1}{2} \left( \int_{-\infty}^{0} e^{ky}e^{k}e^{ikx} dk + \int_{0}^{\infty} e^{-ky}e^{-k}e^{ikx} dk \right) So we get: u(x,y) = \frac{1}{2} \left( \frac{e^{ky+k+ikx}\|_{-\infty}^0}{y + 1 + ix} - \frac{e^{-ky-k+ikx}\|_{0}^{\infty}}{y + 1 - ix} \right) Evaluating, we get: u(x,y) = \frac{1}{2} \left( \frac{1}{y + 1 + ix} + \frac{1}{y + 1 - ix} \right) u(x,y) = \frac{y+1}{(y+1)^2 + x^2} Rong Wei: I believe Xi Yue Wang is right ;D Bruce Wu: Xi Yue's solution makes no sense. By her answer $u(x,0)=0$, which clearly does not satisfy the boundary condition.	{}
Hiya! In need of some help, auto relay shutoff for DC less than 24V mikemikecnc Joined Feb 15, 2022 10 Hiya! Ok so I have a 24VDC brake that is messing up and its because when the voltage to it goes to 24vdc or more, it deactivates (releases brake), but the drive going to it needs to shut off (go to 0Vdc) to apply the brake. Normally 24vdc = brake release Normally 0vdc = Brake applied However.... I am getting 16vdc when I need it to goto 0vdc to apply the brake. The voltage comes off a large servo amplifier board. Instead of buying a new board (tons of money), im tying to design a simple DC relay circuit to CLOSE when 24vdc or more is applied, yet OPEN the moment 20VDC or below is seen, with NO DELAY. Can anyone help or point me to a design? I have tried many setups with simple relays, but the residual voltage is keeping the relay closed, until about 5+ seconds go by and then closes becauses of losses. Thank you in advance! Yall are awesome! Mike Ya’akov Joined Jan 27, 2019 5,642 How many amps does it have to handle? mikemikecnc Joined Feb 15, 2022 10 Hiya! I would say max 5 amps, gauge wire to it is 16 gauge Ya’akov Joined Jan 27, 2019 5,642 Something like this might work but I think I would use it to drive the coil of a larger relay because the relay on the board doesn't inspire confidence. mikemikecnc Joined Feb 15, 2022 10 HIya, I actually bought one, the problem is that it has as 5 second on and 5 second off delay.... unless I can modify it? I dont have a schematic thoo... maybe shrinking the cap size? sghioto Joined Dec 31, 2017 3,270 mikemikecnc Joined Feb 15, 2022 10 Hiya, that may work, the challenge with that is the reference and power voltage is one in the same, however I could tap into the 24vdc voltage from the control for reference voltage, if it drop the it with trigger relay off sounds like? MisterBill2 Joined Jan 23, 2018 11,874 Areyou able to show us the circuit of the amplifier board? We may be able to help spot the failure and you could replace a part and avoid the problem with no changes. a much better fix. mikemikecnc Joined Feb 15, 2022 10 Hiya MisterBill2, no can do, is a CNC Servo amplifier board from Haas... so probably not... sghioto Joined Dec 31, 2017 3,270 Hiya, that may work, the challenge with that is the reference and power voltage is one in the same, however I could tap into the 24vdc voltage from the control for reference voltage, if it drop the it with trigger relay off sounds like? The relay module would be powered by a steady 24 supply and that would also be the reference. The 16 to 24 voltage would be the input. mikemikecnc Joined Feb 15, 2022 10 Awesome sghioto, this sounds like a great fix, just bought it and it will be here this weekend. TY! ill let yall know about how it goes MisterBill2 Joined Jan 23, 2018 11,874 OK, no complex circuit required. You will need to add a relay and a resistor and a bit of wiring, though. The relay will need to operate on probably six volts DC, which should be fairly common, or it can be a 5 volt DC relay The one side of the coil will go to the 24 volt supply and the other side will go to the output that drops to 16 volts. So when the output is high there is no voltage across the relay coil, but when the output drops to 16 volts then you will have about 8 volts on the coil, which is why you need a series resistor between the +24 volts and the coil. So then the relay will operate when the brake should engage, and release when the brake should release. So the contacts must be able to handle the brake current. I am not aware of the complexity of the servo amp board but it seems like the failed part is in the spike protection circuit area. Can you eve get to the board to examine it?? sghioto Joined Dec 31, 2017 3,270 That actually should work. Dagnabit, I wish I had thought of that first. Last edited: sghioto Joined Dec 31, 2017 3,270 I suspect the 24 volts when the brake is off is a reading through the brake solenoid or electro magnet to the 24 volt supply. Disconnect the control wire from the brake circuit and connect it to the 5 volt relay and use the relay contacts as the control for the brake. MisterBill2 Joined Jan 23, 2018 11,874 My preference is always to have one side of the load tied to common,and switch the source side. AND, as the feed to the loade would be pulling it up to 24 volts to release the brake, (Brake released ) There are spring applied brakes and spring released brakes. For an arrangement of the brake applied when the drive is off, that would be a spring applied brake. But I am sure that the TS will be able to figure that part out. AND STILL, it seems that it should be possible to repair that big board. I have repaired things via email before, so I know that it can be done. At least sometimes. mikemikecnc Joined Feb 15, 2022 10 Hiya MisterBill2, Thank you for this, the servo board is incased inside the machine and not fun to take apart.... However I do believe that your idea would work as well. I tried a simple 1 Ohm resistor to drop the voltage the problem with the relay coil wont shut off until it goes pretty low... so 24volts is decativated brake, and then it goes to like 16 volts, but this is still too much voltage and then drains and then goes to 0 mikemikecnc Joined Feb 15, 2022 10 Really appreciate you sghioto and Misterbill2, I will actually pull the board tomorrow and trace wires to see where the brake circuit is at. Now that I am thinking about it, there are 2 wires to the brake, 0v and 24vdc. Ill need to chase them and see if the go to the servo board for the Z axis, or if they just goto the control board, which im thinking it does because E-stopping the system is what releases voltage, and applies the z axis brake to the motor MisterBill2 Joined Jan 23, 2018 11,874 OK, and perhaps the problem is on the control board and not the servo amplifier board. Taking advantage of a smaller voltage change to engage a relay is a good cheating trick. And now I am thinking that the brake is spring applied and electrically released. Let me know if I have that reversed. But it should be that way to stop motion if the power goes out. It may be just one part has failed and it might run as much as $2 or even$5 for a replacement. But changing parts on a circuit board does demand good soldering skills. mikemikecnc Joined Feb 15, 2022 10 Hiya MisterBill2, totally agree with ya, Ill dive in this weekend and reach back to yall mikemikecnc Joined Feb 15, 2022 10 Hiya yall! Ok so i tried a lot of things to get it working. sadly the input (reference voltage) is very damaged, and all over the place. So i simply added a seperate 24v power supply, rigged a simple relay to the Estop switch, and its fixed now. Thank you all for your help! I feel a reinvigeration for electronics now! Hahah Attachments • 928.4 KB Views: 3	{}
• 09/08/13 \| Adjusted: 08/01/18 \| 1 file Three Composing/Decomposing Problems Author: Illustrative Mathematics • Description • Files Mathematically: • Addresses standards: 2.NBT.A and MP.2 • Attends to all three components of the place value system: base-ten units, bundling/unbundling, and positional notation (2.NBT.A) • Relates concrete quantities and abstract symbols (MP.2) In the classroom: • Prompts students to share their developing thinking and understanding • Uses concrete representations to make the mathematics explicit • Allows the teacher to check for understanding throughout students' work This task was designed to include specific features that support access for all students and align to best practice for English Language Learner (ELL) instruction. Go here to learn more about the research behind these supports. This lesson aligns to ELL best practice in the following ways: • Provides opportunities for students to practice and refine their use of mathematical language. • Allows for whole class, small group, and paired discussion for the purpose of practicing with mathematical concepts and language. • Elicits evidence of student thinking both verbally and in written form. • Includes a mathematical routine that reflects best practices to supporting English Language Learners in accessing mathematical concepts. • Offers the opportunity for students to act out the problem when the task features complex real-world situations. • Making the Shifts How does this task exemplify the instructional Shifts required by CCSSM? Focus Belongs to the major work of second grade Coherence Develops foundations for multi-digit operations Rigor Conceptual Understanding: primary in this task Procedural Skill and Fluency: not targeted in this task Application: not targeted in this task Some students are working with base-ten blocks. a. Nina has 3 hundreds, 8 tens, and 23 ones. How many ones would this be? b. Lamar wants to make the number 261. He has plenty of hundreds blocks and ones blocks to work with, but only 4 tens blocks. His friend Jose said, You can still make 261 with the blocks you have Explain how he can. c. Find at least three different ways to make 124 using hundreds, tens and ones. • Illustrative Mathematics Commentary and Solution Commentary The purpose of this task is to help students understand composing and decomposing ones, tens and hundreds. The task is meant to be used in an instructional setting and would only be appropriate to use if students actually have base-ten blocks on hand. The last two tasks fully engage the notion of composing and decomposing as needed for algorithms for addition and subtraction. Both parts require persistence, as in the Standard for Mathematical Practice 1. After seeing the first two tasks, students have the ideas needed to start listing possibilities in the third task. the idea of exchanging a ten for ten ones and a hundred for ten tens is needed in order to complete the task. Solution: While some students might try to simply add, others will recognize that $23$ ones is $2$ tens and $3$ ones. When we combine the $2$ tens with the $8$ tens we already have we get $10$ tens, which is one hundred. So we have $3$ hundreds and another hundred and three ones, which is $403$. • Lamar could use ten ones for each ten-block which he was missing. So instead of $2$ hundreds, $6$ tens and $1$ one as he wanted, he can start with the $2$ hundreds and $4$ tens which he has and then use two sets of ten ones instead of the two more needed tens. Those make $20$ ones, which we add to the $1$ one needed to get $21$ ones. Collecting all of these we get $2$ hundreds, $4$ tens and $21$ ones. There are many possible solutions – for example using $2$ hundreds, $3$ tens and $31$ ones – but the one given is the most likely. The list of all ways using $1$ hundred is: • $1$ hundred, $2$ tens, $4$ ones. • $1$ hundred, $1$ ten, $14$ ones. • $1$ hundred, $0$ tens, $24$ ones. The list of all ways not using any hundreds is: • $12$ tens, $4$ ones. • $11$ tens, $14$ ones. • $10$ tens, $24$ ones. • $9$ tens, $34$ ones. • $8$ tens, $44$ ones. • $7$ tens, $54$ ones. • $6$ tens, $64$ ones. • $5$ tens, $74$ ones. • $4$ tens, $84$ ones. • $3$ tens, $94$ ones. • $2$ tens, $104$ ones. • $1$ tens, $114$ ones. • $124$ ones To know the list is complete as we make it, we can start with the standard way, namely $1$ hundred, $2$ tens, and $4$ ones, and exchange tens for ones one at a time to get the first list. Then we exchange the hundred for $10$ tens, to get a total of $12$ tens along with $4$ ones. Once again, we can exchange tens for $10$ ones step by step in order to get the second list. Because we cannot use two or more hundreds, these two lists contain all possibilities.	{}
Solve $\lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x}$ I need to solve the limit $$\lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x}$$ when $a,b,c \gt 0$. I'm looking for ways to avoid $\frac{1}{x}$ power. - I formatted your question. Please make certain I have done so faithfully. In particular, you mention "sqrt of $x$" in your question, but it did not appear in the limit you wrote down. Are you referring to the "$\frac{1}{x}$ power"? – Austin Mohr Feb 21 '12 at 18:38 Yes Austin that is correct thank you. – Tom Feb 21 '12 at 18:39 The answer (if I am not wrong) is $\frac{ab}{c}$. I'll try to punch in a solution soon. – Inquest Feb 21 '12 at 18:49 @Nunoxic I went to my board, did the calculations and got the same. My comment'd have been the same as yours. – Pedro Tamaroff Feb 21 '12 at 21:24 Note that $$\lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x} =\lim_{x \rightarrow 0}\,e^{\left(\displaystyle\frac1x \log(a^x+b^x-c^x)\right)}.$$ So we only need to calculate $$\lim_{x \rightarrow 0}\frac1x \log(a^x+b^x-c^x).$$ Since the expression inside the log goes to $1$ as $x\to0$, we can apply L'Hôpital, to get $$\lim_{x \rightarrow 0}\frac1x \log(a^x+b^x-c^x) =\lim_{x\to0}\frac{a^x\,\log a+ b^x\,\log b-c^x\,\log c}{a^x+b^x-c^x} =\log a+ \log b-\log c.$$ Then $$\lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x}=e^{\log a+ \log b-\log c}=\frac{ab}c$$ Added: as suggested by Aryabhata, it remains to justify that one can actually apply L'Hôpital. For the rule to apply, we need to have a quotient $f(x)/g(x)$, with both $f$ and $g$ differentiable at $0$, with $g'(x)\ne0$ on a neighbourhood of $0$, such that $f(0)=g(0)=0$ (less is needed, in fact, only that they go to $0$) and such that $\lim_{x\to0}f'(x)/g'(x)$ exists (it equals $ab/c$ in this case). Here $f(x)=\log(a^x+b^x-c^x)$, $g(x)=x$. The function $x\mapsto a^x+b^x-c^x$ is differentiable everywhere; for $x$ near $0$, the values of this function approximate $1$; on any neighbourhood of $1$ that doesn't contain $0$, $\log$ is differentiable, and then so is the composition $f(x)=\log(a^x+b^x-c^x)$. And $g'(x)=1\ne0$ for all $x$. - This seems circular: LHospital assumes that the derivative exists at $0$. But the limit you are trying to find is exactly the derivative at $0$! (This is what I was hinting with my answer). – Aryabhata Feb 21 '12 at 19:43 @Aryabhata: I'm using L'Hopital to find the value of the limit, not to prove that the derivative exists. The function $\log(a^x+b^x-c^x)$ is C$^\infty$ on small enough neighbourhoods of $0$. – Martin Argerami Feb 21 '12 at 20:59 Finding a limit also entails proving its existence (implicitly or explicitly). Otherwise, all you have done is, "if the limit exists, then it is so and so", which is incomplete. – Aryabhata Feb 21 '12 at 21:08 @Aryabhata: L'Hôpital is exactly about the existence of the limit. – Martin Argerami Feb 21 '12 at 21:19 You missed mentioning that $g'(x) \ne 0$ in a neighbourhood of $0$, but that is fine here as $g(x) = x$. Sure, Lhopital also proves existence, but that was not my point. – Aryabhata Feb 21 '12 at 21:37 Another hint: Let $f(x) = \log (a^x + b^x - c^x)$ What is $f(0)$? What is $f'(0)$? (The derivative at $0$). - f(0) is 0 and f(tag)(0) is minus infinity..trying to think what this tells me – Tom Feb 21 '12 at 19:01 @TOm: What is the definition of $f'(0)$ in terms of limits? – Aryabhata Feb 21 '12 at 19:05 @Tom: Are you sure about $f'(0)$? – Aryabhata Feb 21 '12 at 19:11 HINT: $$\left( a^x + b^x - c^x \right)^\frac{1}{x} = a \left( 1 + \left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x \right)^\frac{1}{x} = a \left( 1 + x \cdot \frac{\left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x}{x} \right)^\frac{1}{x}$$ Compute $\lim_{x \to 0} \frac{\left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x}{x}$, then use the golden limit $\lim_{x\to 0} \left( 1 + \alpha x\right)^\frac{1}{x}$. - This needs some justification. For instance consider limit of $(1+x)^{1/x}$ as $x \to 0$ (which is $e$). We just cannot replace $(1+x)$ by $1$ because $x \to 0$. – Aryabhata Feb 21 '12 at 18:54 @Aryabhata Sasha is not suggesting that. – Pedro Tamaroff Feb 21 '12 at 19:01 this looks like magic but yet again the limit to compute when x approaches 0 is hard for me to do – Tom Feb 21 '12 at 19:05 @Peter: I am pretty sure Sasha is not doing that, but people who are just learning these things might think that that is what Sasha is doing. Just pointing out the potential pitfall. – Aryabhata Feb 21 '12 at 19:43 Where you start is to realize that $1^\infty$ is an indeterminate form, so you can use L'hopital's rule after you rewrite it as, $$e^{\left(\lim\limits_{x \rightarrow 0} \frac{\ln(a^x + b^x - c^x)}{x}\right)}$$ - If you want to avoid L'Hopital: Write $(a^x + b^x - c^x)^{1 \over x}$ as $a(1 + ({b \over a})^x - ({c \over a})^x)^{1 \over x}$, so that it suffices to find $$\lim_{x \rightarrow 0} (1 +({b \over a})^x - ({c \over a})^x)^{1 \over x}$$ Taking logs, it's enough to find $$\lim_{x \rightarrow 0} {\ln(1 +({b \over a})^x - ({c \over a})^x) \over x}$$ $$=\lim_{x \rightarrow 0} {\ln(1 +({b \over a})^x - ({c \over a})^x) - \ln(1 +({b \over a})^0 - ({c \over a})^0) \over x - 0}$$ This is the difference quotient for the derivative of $\ln(1 +({b \over a})^x - ({c \over a})^x)$ at $x = 0$. Using the chain rule to find this derivative and plugging in $x = 0$ gives you $\ln(b/a) - \ln(c/a) = \ln(b/c)$. Going backwards, the original limit is ${\displaystyle ae^{\ln(b/c)} = {ab \over c}}$. -	{}
# “Runaway argument?” error with \NewEnviron I have two files. Basically, the main file just loads the environ package and inputs the secondary file. The secondary file creates an environment that basically does nothing, and creates one instance of that environment: main.tex: \documentclass{scrartcl} \usepackage{environ} \begin{document} \input{secondary.tex} \end{document} secondary.tex: \NewEnviron{myenv}{\BODY} \begingroup \myenv Hello \endmyenv \endgroup I get the following error message: Runaway argument? Hello \endmyenv \endgroup \par ! File ended while scanning use of \document. <inserted text> \par l.5 \input{secondary.tex} The example works as expected when I make any (!) of the following changes: • Use \newenvironment instead of \NewEnviron • Use \begin{myenv}...\end{myenv} instead of \begingroup\myenv...\endmyenv\endgroup • Place the contents of secondary.tex directly in main.tex instead of inputting secondary.tex Why does this not work? environ creates an environment that uses TeX parameter text in definitions to capture the environment contents. This parameter-text style definition allows the package to capture the environment contents, but it also requires a very strict usage of the macros/definitions... something you're not adhering to when using a command-style environment (\myenv...\endmyenv rather than \begin{myenv}...\end{myenv}). Broadly speaking, \NewEnviron{myenv} creates a macro \myenv which designates the start of the environment, and this macro is on the lookout for exactly \end{myenv}... nothing else. The example fails when using \input since there's no explicit \end{<something>} that \myenv is searching for. It reaches the end of secondary.tex before it ever finds any resemblance of \end. It's actually looking for \end{document} (see below). Let's look at the three instances you mention that corrected the problem and see why that's the case: • Use \newenvironment instead of \NewEnviron It should be obvious why this works. \newenvironment{myenv} defines both \myenv and \endmyenv, so using the command-definition of the environment should work without problem; • Use \begin{myenv}...\end{myenv} instead of \begingroup\myenv...\endmyenv\endgroup This is exactly what environ expects you should use when you define an environment using \NewEnviron. So, this works out-of-the-box, since you've explicitly used the environment termination \end{myenv}; • Place the contents of secondary.tex directly in main.tex instead of inputting secondary.tex Since you're using the command-form of an environment, the "current environment" (known to LaTeX as \@currenvir) never really changes. That is, it remains as being inside the document environment (due to a call \begin{document}). So, your end-of-environment capture is actually \end{document} according to newenviron's setup. Your environment "starts" with \myenv and ends with \end{document}. Weird, but that's how it works when you don't use the expected environment scoping \begin{...}...\end{...}. For it to work properly within the context of newenviron, use the scoping \begin{myenv}...\end{myenv}. This scoping provides the necessary grouping you implement through \begingroup...\endgroup. Alternatively, define \myenv to capture stuff until it reaches \endmyenv: \documentclass{scrartcl} \usepackage{filecontents} \begin{filecontents}{secondary.tex} \long\def\myenv #1\endmyenv{#1}% Fake myenv environment that captures its body \begingroup \myenv Hello \endmyenv \endgroup \end{filecontents} \begin{document} \input{secondary.tex} \end{document} • Thanks for the comprehensive explanation. On part that is still not clear is that first you say that the "macro is on the lookout for exactly \end{myenv}...nothing else", and then you say that the "end-of-environment capture is actually \end{document}". I get the first part, thanks to your alternative example, but the second part seems to contradict the first part. – Martin Herrmann Dec 18 '14 at 18:57 • Would there be a problem if \myenv checked \@currenv and then if \@currenv = myenv => look for \end{myenv}; and if \@currenvmyenv then look for \endmyenv? – Manuel Dec 18 '14 at 18:59 • @Manuel it would be very expensive in terms of complexity, currently tex can use a primitive delimited argument to skip to the next \end and then it just has to check the argument to check it is the right thing. To stop at \endmyenv it would have to read every token one by one. – David Carlisle Dec 18 '14 at 19:05 • @DavidCarlisle I mean, define 2 auxiliar macros to pass the “content of the environment” to a third one. And then \myenv checks \@currenv and calls either one (which looks for \end{myenv}) or the other (which looks for \endmyenv) and then “the content” would be passed to a third one. I was asking more of trivial problems that would be created because of this (which I can't think of), I think it's “easy task” to define those macros. – Manuel Dec 18 '14 at 19:08 • @Manuel ok but it would break the common use of using it in another macro, eg tabularx does similar and you can use \tabularx \endtabularx in the definition of mytable because the scanner knows to look for the end of the current environment – David Carlisle Dec 18 '14 at 19:11 It does not work if you place the content in the main file. It only don't give an error. Try this: \documentclass{scrartcl} \usepackage{environ} \begin{document} \NewEnviron{myenv}{BEGIN\BODY END} \begingroup \myenv Hello \endmyenv \endgroup World \end{document} Environ needs to find \end{myenv} to stop the parsing.	{}
## Variability of newly identified B-type stars observed by Kepler 10 Mar 2021 · Wojciech Szewczuk, Przemysław Walczak, Jadwiga Daszyńska-Daszkiewicz · Recent re-determination of stellar atmospheric parameters for a sample of stars observed during the {\it Kepler} mission allowed to enlarge the number of {\it Kepler} B-type stars. We present the detailed frequency analysis for all these objects... All stars exhibit pulsational variability with maximum amplitudes at frequencies corresponding to high-order g modes. Peaks that could be identified with low-order p/g modes are also extracted for a few stars. We identified some patters in the oscillation spectra that can be associated with the period spacings that can result from the asymptotic nature of the detected pulsational modes. We also tentatively confront the observed oscillation characteristics with predictions from linear nonadiabatic computations of stellar pulsations. For high-order g modes the traditional approximation was employed to include the effects of rotation on the frequency values and mode instability. read more PDF Abstract # Code Add Remove Mark official No code implementations yet. Submit your code now # Categories Solar and Stellar Astrophysics	{}
# Viewpoint: Fast-Forwarding the Search for New Particles Physics 11, 90 A proposed machine-learning approach could speed up the analysis that underlies searches for new particles in high-energy collisions. In the eyes of an old and famous statistics paper [1], searching for evidence of a new particle in the complex data of the Large Hadron Collider should be easy. You just need to calculate two numbers: the likelihood that the data came from the hypothetical particle and the likelihood that it did not. If the ratio between these numbers is high enough, you’ve made a discovery. Can it really be that easy? In principle, yes, but in practice, the two likelihood calculations are intractable. So instead, particle physicists approximate the likelihood ratio by making simplifying assumptions. Even then, the calculation requires a huge amount of computer time. In a pair of papers [2], Johann Brehmer of New York University and colleagues propose a new approach that avoids the typical simplifications and doesn’t demand long computation times. Their method, which relies on machine-learning tools, could significantly boost physicists’ power to discover new particles in their data. Physicists almost never directly “see” the particle that they are searching for. That’s because new particles created in a collision exist for only a brief shining moment ( $<1{0}^{-20}$ s) before decaying into other particles. These secondary particles may lead to hundreds of other particles, which are what’s actually picked up in the detectors. When making a likelihood calculation, researchers have to account for all of the different types of secondary particles with different momenta that could have led to the same detected outcome—a bit like sipping a glass of wine and trying to imagine all of the ways the Sun and weather could have influenced its flavor. Accounting for all of the possible secondary particles and their momenta involves a hairy integral that sums over all of the possibilities. Unfortunately, calculating the integral exactly is impossible, and numerical approaches fail because of the vast parameter space of the unobserved particles. Obviously, particle physicists have not thrown their hands in the air and given up. They’ve managed to extract evidence from the collider data for all of the particles that make up the standard model, including the Higgs boson. The key to this success has been simplifying the relevant likelihood-ratio calculation into one that is still computationally expensive, but doable—albeit with less statistical power to make a detection. The most widespread approach has been to randomly select a few of the possibilities for the unobserved secondary particles and momenta and then to use these possibilities to simulate detector data with or without a new particle [3]. These simulations still take up a lot of computer time. And, most importantly, they cannot describe the expected data in terms of all of the relevant parameters, which is where the discovery power is lost. An alternative approach attempts to tackle the integration directly [4], making it tractable via a series of assumptions about the secondary states, such as whether the particles involved can be treated independently. This direct approach also requires significant computational resources, and the assumptions are viewed as unpalatable. In their two papers, Brehmer and co-workers outline a new strategy for calculating the likelihood ratio within effective field theories, which are often used for making predictions beyond the standard model. They first point out that even though the likelihood ratio’s numerator and denominator are intractable, the difficult bits in both cancel out in an “extended” version of the final ratio. In this expression, some momenta of the unobserved particles are assumed to be constant. The extended ratio can be calculated using a small number of integrals, and the original likelihood ratio can then be recovered from the extended version using machine learning (Fig. 1). In a nutshell, a neural network is provided with many simulated examples of the few unobserved momenta so that it learns how to approximate the intractable likelihood ratio—much as though it had learned how to do the integral! The researchers have, in a sense, combined the best of the simulation and direct approaches. At the same time, they avoid the pitfalls of existing approaches: the trained neural network evaluates the likelihood ratio quickly (within microseconds), without using the simplifying assumptions of the direct calculation approach. Can particle physicists now jump for joy, knowing future searches will be a lot easier? Not quite, as the proposed approach comes with some costs. It will require the generation of many simulated examples in order for the neural network to learn the relationship between the unobserved and observed data. In addition, the approach uses complex machine-learning algorithms, which can be difficult to train and understand. To what extent it will actually be faster is hard to predict at this point. But in my view, this is a landmark new idea, which brings the power of modern machine-learning to bear on a central, intractable statistical problem in particle physics. It suggests that we can extract more information out of our collisions and at a reduced computational cost. The possibility of more discoveries, in less time and for less money, is definitely good news. This research is published in Physical Review Letters and Physical Review D. ## References 1. J. Neyman and E. S. Pearson, “On the problem of the most efficient tests of statistical hypotheses,” Phil. Trans. R. Soc. Lond. A. 231, 289 (1933). 2. J. Brehmer, K. Cranmer, G. Louppe, and J. Pavez, “Constraining effective field theories with machine learning,” Phys. Rev. Lett. 121, 111801 (2018); J. Brehmer, K. Cranmer, G. Louppe, and Juan Pavez, “A guide to constraining effective field theories with machine learning,” Phys. Rev. D 98, 052004 (2018). 3. S. Agostinelli et al., “Geant4: A simulation toolkit,” Nucl. Instrum. Methods A 506, 250 (2003). 4. K. Kondo, “Dynamical likelihood method for reconstruction of events with missing momentum. I. Method and toy models,” J. Phys. Soc. Jpn. 57, 4126 (1988); “Dynamical likelihood method for reconstruction of events with missing momentum. II. Mass spectra for $2\to 2$ processes,” 60, 836 (1991); R. H. Dalitz and G. R. Goldstein, “Decay and polarization properties of the top quark,” Phys. Rev. D 45, 1531 (1992); V. M. Abazov et al. (DØ Collaboration), “A precision measurement of the mass of the top quark,” Nature 429, 638 (2004). Daniel Whiteson is a professor of experimental particle physics at the University of California, Irvine, and a fellow of the American Physical Society. He conducts research using the Large Hadron Collider at CERN. ## Subject Areas Particles and Fields ## Related Articles Particles and Fields ### Viewpoint: Higgs Decay into Bottom Quarks Seen at Last Two CERN experiments have observed the most probable decay channel of the Higgs boson—a milestone in the pursuit to confirm whether this remarkable particle behaves as physicists expect. Read More » Astrophysics ### Viewpoint: Weak Lensing Becomes a High-Precision Survey Science Analyzing its first year of data, the Dark Energy Survey has demonstrated that weak lensing can probe cosmological parameters with a precision comparable to cosmic microwave background observations. Read More » Plasma Physics ### Focus: Solar Wind Shock Wave Gives Ions a Push Measurements made by NASA’s New Horizons spacecraft show that shock waves in the solar wind transfer significant energy to ionized interstellar atoms, confirming a decades-old prediction. Read More »	{}
# Just Random Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ## Description Coach Pang and Uncle Yang both love numbers. Every morning they play a game with number together. In each game the following will be done: 1. Coach Pang randomly choose a integer x in [a, b] with equal probability. 2. Uncle Yang randomly choose a integer y in [c, d] with equal probability. 3. If (x + y) mod p = m, they will go out and have a nice day together. 4. Otherwise, they will do homework that day. For given a, b, c, d, p and m, Coach Pang wants to know the probability that they will go out. ## Input The first line of the input contains an integer T denoting the number of test cases. For each test case, there is one line containing six integers a, b, c, d, p and m(0 <= a <= b <= 109, 0 <=c <= d <= 109, 0 <= m < p <= 109). ## Output For each test case output a single line "Case #x: y". x is the case number and y is a fraction with numerator and denominator separated by a slash ('/') as the probability that they will go out. The fraction should be presented in the simplest form (with the smallest denominator), but always with a denominator (even if it is the unit). ## Sample Input 4 0 5 0 5 3 0 0 999999 0 999999 1000000 0 0 3 0 3 8 7 3 3 4 4 7 0 ## Sample Output Case #1: 1/3 Case #2: 1/1000000 Case #3: 0/1 Case #4: 1/1 ## Source 2013 Asia Chengdu Regional Contest	{}
# Use expected bound to derive high probability bound Suppose that for some integer $p\geq 2$, we have $$(\mathbb{E}[\lVert \mathbf{X} \rVert^p])^{\frac{1}{p}}\leq C$$ where $C>0$ is a constant as upper bound. Now, my question is how to convert this into a high probability result? Specifically, I want to have the result like with probability at least $1-\delta$, $$\lVert \mathbf{X} \rVert \leq D$$ Here, $D = f(C,\delta)$. Besides, hope the order of $\delta$ is $log$-order. A simple method is use Markov inequality, then $D=C/\delta^{1/p}$, so is it possible to be tighter for a high probability bound? Let $X$ be a real random variable such that $\mathbb{P}(X=0)=1-\delta$ and $\mathbb{P}(X=D)=\delta$ where $D=C/\delta^{1/p}$. Then we have $(\mathbb{E}(X^p))^{1/p}=C$ so one cannot hope to get a better bound than $C/\delta^{1/p}$ without any other assumptions.	{}
# Fixed points and idempotency Thinking about fixed points, I realised (even if it’s a trivial conclusion) that they can be elegantly used to define idempotency: A function is idempotent if every member of its codomain is a fixed point. Fixed point of some function f is any value a for which: $f(a) = a$ (in functional programming, fixed points are super useful and lead to interesting things like recursion schemes and y-combinators) Codomain of some function f is the set of values that the function maps the domain to. Given some function f, applying it to its domain element x produces the codomain element f(x). When is that codomain element f(x) a fixed point? Let’s use the fixed point equation from earlier: $f(a) = a$ $a = f(x)$ $f(f(x)) = f(x)$ We arrive at the exact condition for idempotency. A function is idempotent if repeated application results in the same value: f(x) = f(f(x)) = f(f(f(x))) = ... Here’s an example: Rounding function floor: Double => Int. • Every member of its codomain is a fixed point: for every integer n, floor(n) = n. • Therefore, function is idempotent. • Try it: floor(3.7) = floor(floor(3.7)) = 3. Written on November 17, 2022	{}
## Just brushing up on chemistry Hey guys, im new here. I am a college student that is trying to brush up on some chemistry. I am reading an old textbook that I have, and I didnt understand something. I was wondering how some would use the Kb of a weak base, how they would determine the value of Ka for the conjugate base? A sample problem in the book is: A weak acid has a Kb= 1.5 x 10^-9. What is the value of Ka for the conjugate base? I have read the whole chapter over Acid & Bases, but it says this problem is on the cd-rom that is included with my book. Yet, i never recieved that cd when I bought my book. I have tried searching but have had no luck. Any help would be great. I am just trying to understand the relationship between Ka and Kb. Thanks again. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Kw = Ka x Kb = 1 x 10^-14 Recognitions: Science Advisor K is called a dissociation constant What K states is that there is a certain amount of dissociation at any given time, and almost everything has one of these numbers. Something like lead chloride will have a K value to say how soluble it is. Here you are talking about water, and K relates to water breaking from HOH into H and OH. For water, the dissociation constant is 10^-14. $$[H^+] * [OH^-] = 10^{-14}$$ $$K_A * K_B = 10^{-14}$$ $$pK_A + pK_B = 14$$	{}
Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript. # First-principles study of superconducting hydrogen sulfide at pressure up to 500 GPa ## Abstract We investigate the possibility of achieving the room-temperature superconductivity in hydrogen sulfide (H3S) through increasing external pressure, a path previously widely used to reach metallization and superconducting state in novel hydrogen-rich materials. The electronic properties and superconductivity of H3S in the pressure range of 250–500 GPa are determined by the first-principles calculations. The metallic character of a body-centered cubic Im$$\overline{{\bf{3}}}$$m structure is found over the whole studied pressure. Moreover, the absence of imaginary frequency in phonon spectrum implies that this structure is dynamically stable. Furthermore, our calculations conducted within the framework of the Eliashberg formalism indicate that H3S in the range of the extremely high pressures is a conventional strong-coupling superconductor with a high superconducting critical temperature, however, the maximum critical temperature does not exceed the value of 203 K. ## Introduction The recent breakthrough theoretical calculations1, 2 confirmed by spectacular experimental reports of the superconductivity in hydrogen sulfide, with the record high critical temperature equals to 203 K at pressure closed to 150 GPa3,4,5, open the door to achieving the room-temperature superconductivity in the compressed hydrogen-rich materials6,7,8 or in the pristine metallic hydrogen9,10,11,12. In contrast to the cuprates13, 14, where the mechanism responsible for the superconducting state is still debated15,16,17,18,19,20,21,22, the phonon-mediated pairing scenario is generally accepted in the case of H3S23,24,25,26,27 due to the observed of a large isotope effect which convincingly suggest that hydrogen sulfide is a conventional superconductor3, 4. Therefore, in the theoretical papers28,29,30,31,32, the superconducting properties of hydrogen sulfide are studied in the framework of the mean-field Bardeen-Cooper-Schrieffer (BCS) theory33, 34, or more precisely using the Migdal-Eliashberg (ME) approach35,36,37. For the crystal structures of high-T C hydrides, the high-pressure x-ray diffraction experiments combined with the electrical resistance measurements5, 38 confirm that H3S takes a body-centered cubic $$Im\overline{3}m$$ structure, which is certainly stable above 180 GPa2, 39. The pressure-temperature phase diagram of solid hydrogen sulfide, determined on the base of a sharp falls to zero in resistivity with cooling, shows increase of T C from 95 to 203 K in the range of pressure from 110 to 155 GPa3, 5 and, then, further increasing of compression causes a linear decrease of critical temperature to 170 K at 225 GPa3. There arises the natural question, whether under the influence of the high pressure is it possible to obtain the superconducting state with the value of the critical temperature even higher than 203 K in the case of H3S compound. Below, we present the results of the ab initio calculations, which conclude that T C does not exceed the value of 203 K, while the range of the pressure from 250 to 500 GPa is adopted. ## Computational details The searches for the stable high pressure structures of H3S system were performed through the evolutionary algorithm implemented in the USPEX code40, 41, which has been applied successfully to a number of compressed systems containing hydrogen42, 43. The computed enthalpy differences relative to the Cccm structure (H − H Cccm ) as a function of pressure for the selected crystal structures are presented in Fig. 1. It can be clearly seen that at low pressure (below 110 GPa) the lowest value of enthalpy corresponds to the orthorhombic Cccm structure and the hexagonal R3m structure has the most stable lattice between 110 and 180 GPa. Then, a cubic $$Im\overline{3}m$$ structure becomes favorable above 180 GPa. This structure is characterized by two S atoms located at a simple body centered cubic lattice and H atom situated midway between the two S atoms (see the inset in Fig. 1). Let us emphasize that for the low pressures our results agree well with the data reported by Duan et al.2. Moreover, for H3S the second-order structural phase transition from R3m to $$Im\overline{3}m$$ is also experimentally observed but for a slightly lower pressure (~150 GPa)5, 44. It should be underlined that over 450 various structures were studied, wherein in any case was obtained enthalpy lower than $${H}_{Im\overline{3}m}$$ in the range of pressures from 250 to 500 GPa. Due to the above fact in this study the critical temperature and the other thermodynamic parameters of the superconducting state of H3S are calculated only for the structure $$Im\overline{3}m$$. Figure 2 presents the curve of the volume-pressure type. This curve can be reproduced with the help of the third-order Birch-Murnaghan equation: $$p(V)=\frac{3}{2}{B}_{0}[{({V}_{0}/V)}^{\mathrm{7/3}}-{({V}_{0}/V)}^{\mathrm{5/3}}]\,\{1+\frac{3}{4}({B}_{0}^{^{\prime} }-4)[{({V}_{0}/V)}^{\mathrm{2/3}}-1]\}$$, where B 0 = 129.8 GPa, $${V}_{0}=158.4\,{{\rm{a}}}_{0}^{3}$$, and $${B}_{0}^{^{\prime} }=3.6$$ 39. The characteristics of the electron structure, the phonon structure, and the electron-phonon interaction was made in the framework of the Quantum-ESPRESSO package45. The calculations were conducted basing on the density-functional methods using the PWSCF code45,46,47. The Vanderbilt-type ultra-soft pseudopotentials for S and H atoms were employed with the kinetic energy cut-off equal to 80 Ry. The phonon calculations were performed for 32 × 32 × 32 Monkhorst-Pack k-mesh with the Gaussian smearing of 0.03 Ry. The electron-phonon coupling matrices were computed using 8 × 8 × 8 q-grid. The superconducting transition temperature (T C ) can be in a simple way estimated using the Allen-Dynes modified McMillan equation48: $${k}_{B}{T}_{C}={f}_{1}{f}_{2}\frac{{\omega }_{\mathrm{ln}}}{1.2}\,\exp \,[\frac{-1.04(1+\lambda )}{\lambda -{\mu }^{\ast }(1+0.62\lambda )}],$$ (1) where k B = 0.0862 meV/K (the Boltzmann constant), f 1 and f 2 are the correction functions: $${f}_{1}={[1+{(\frac{\lambda }{2.46(1+3.8{\mu }^{\ast })})}^{\mathrm{3/2}}]}^{\mathrm{1/3}},$$ (2) $${f}_{2}=1+\frac{(\sqrt{{\omega }_{2}}/{\omega }_{\mathrm{ln}}-1){\lambda }^{2}}{{\lambda }^{2}+{[1.82(1+6.3{\mu }^{\ast })(\sqrt{{\omega }_{2}}/{\omega }_{\mathrm{ln}})]}^{2}}.$$ (3) The quantity ω 2 represents the second moment of the normalized weight function: $${\omega }_{2}\equiv \frac{2}{\lambda }{\int }_{0}^{{\omega }_{D}}\,d\omega {\alpha }^{2}F(\omega )\omega$$ (4) and ω ln is the logarithmic average of the phonon frequencies: $${\omega }_{\mathrm{ln}}\equiv \exp \,[\frac{2}{\lambda }{\int }_{0}^{{\omega }_{D}}\,d\omega \frac{{\alpha }^{2}F(\omega )}{\omega }\,\mathrm{ln}(\omega )].$$ (5) More sophisticated calculations can be conducted within the framework of the Eliashberg formalism, which allows a more accurately description of the superconducting state in strong-coupling systems31, 32. The Eliashberg equations for the superconducting order parameter function $${\phi }_{m}\equiv \phi \,(i{\omega }_{m})$$ and the electron mass renormalization function $${Z}_{m}\equiv Z\,(i{\omega }_{m})$$ written in the imaginary-axis formulation take the following form refs 36 and 49: $${\phi }_{m}=\pi {k}_{B}T\sum _{n=-M}^{M}\,\frac{{\lambda }_{n,m}-{\mu }^{\ast }\theta ({\omega }_{c}-\|{\omega }_{n}\|)}{\sqrt{{\omega }_{n}^{2}{Z}_{n}^{2}+{\phi }_{n}^{2}}}{\phi }_{n},$$ (6) and $${Z}_{m}=1+\frac{\pi {k}_{B}T}{{\omega }_{n}}\sum _{n=-M}^{M}\,\frac{{\lambda }_{n,m}}{\sqrt{{\omega }_{n}^{2}{Z}_{n}^{2}+{\phi }_{n}^{2}}}{\omega }_{n}{Z}_{n},$$ (7) where the pairing kernel for the electron-phonon interaction is given by: $${\lambda }_{n,m}=2{\int }_{0}^{{\omega }_{D}}\,d\omega \frac{\omega }{{({\omega }_{n}-{\omega }_{m})}^{2}+{\omega }^{2}}{\alpha }^{2}F(\omega ).$$ (8) Symbols $${\mu }^{\ast }$$ and θ denote the Coulomb pseudopotential and the Heaviside function with cut-off frequency ω c equal to three times the maximum phonon frequency (ω D ). The α 2 F(ω) functions, called the Eliashberg functions, for H3S system were calculated using the density functional perturbation theory and the plane-wave pseudopotential method, as implemented in the Quantum-Espresso package45: $${\alpha }^{2}F(\omega )=\frac{1}{2\pi \rho ({\varepsilon }_{F})}\sum _{{\bf{q}}\nu }\,\delta (\omega -{\omega }_{{\bf{q}}\nu })\frac{{\gamma }_{{\bf{q}}\nu }}{{\omega }_{{\bf{q}}\nu }},$$ (9) with $${\gamma }_{{\bf{q}}\nu }=2\pi {\omega }_{{\bf{q}}\nu }\sum _{ij}\,\int \frac{{d}^{3}k}{{{\rm{\Omega }}}_{BZ}}{\|{g}_{{\bf{q}}\nu }({\bf{k}},i,j)\|}^{2}\delta ({\varepsilon }_{{\bf{q}},i}-{\varepsilon }_{F})\delta ({\varepsilon }_{{\bf{k}}+{\bf{q}},j}-{\varepsilon }_{F}),$$ (10) where $$\rho \,({\varepsilon }_{F})$$ denotes the density of states at the Fermi energy, ω q ν determines the values of the phonon energies, and γ q ν represents the phonon linewidth. The electron-phonon coefficients are given by g q ν (k, i, j) and $${\varepsilon }_{{\bf{k}},i}$$ is the electron band energy. ## Results and Discussion To investigate the electronic properties of H3S at $$p\in \langle 250,500\rangle \,{\rm{GPa}}$$, we calculate the electronic band structure and density of states (DOS). The Fermi surface of H3S at 250 and 500 GPa is shown in Fig. 3. It is formed by five different Fermi surfaces calculated in the bcc Brillouin zone50. As has been previously reported by Bianconi and Jarlborg, the red small Fermi surface centered at the Γ-point and covering the surfaces #1 and #2, appears above 95 GPa with the change of the Fermi surface topology. This change of the Fermi surface topology is called a L1 Lifshitz transition for a new appearing Fermi surface spot and occurs where the bands at the Γ-point cross the chemical potential. The L2 Lifshitz transition for neck disrupting occurs around 180–200 GPa and is connected with appearing of the small tubular necks in the Fermi surface (in particular in the surface #4)50, 51. The results presented in Fig. 4 clearly show that $$Im\overline{3}m$$ structure is a good metal with a large DOS at the Fermi level (0.418–0.511 states/eV/f.u.). This is in a good agreement with the previous theoretical and experimental results obtained for the lower pressure3,4,5, 52. The metallic behavior of this system indicates that $$Im\overline{3}m$$ phase might be superconducting above 250 GPa. In order to investigate the superconductivity of H3S, the phonon band structures, the phonon density of states (PhDOS) and the Eliashberg spectral functions together with the electron-phonon integrals $$\lambda (\omega )=2{\int }_{0}^{\omega }\,d\omega {\alpha }^{2}F(\omega )/\omega$$ were carried out. As shown in Fig. 5 there is no imaginary frequency to be found in the whole Brillouin zone, confirming that $$Im\overline{3}m$$ is a dynamically stable structure. In the case of the pressure of 250 GPa, the clearly separated lines respectively associated with the low-energy vibrations of sulfur ($$\omega \in \langle \mathrm{0,76.4}\rangle \,{\rm{meV}}$$) and the high-energy vibrations of hydrogen ($$\omega \in \langle \mathrm{95.2,256.7}\rangle \,{\rm{meV}}$$) can be noticed in the phonon dispersive relation. Such fact directly translates into the shape of the function of the phonon density of states, which consists of two parts separated by the gap of the energy (about 19 meV). On the basis of the diagram related to the spectral function it can be seen that the contribution to the electron-phonon coupling constant comes mainly from hydrogen, and is equal approximately to 79%. At the pressure of 350 GPa, the maximum energy of the hydrogen vibrations becomes larger and equals to 308.7 meV. Still visible is the division of functions of the phonon density of states on the part related to sulfur and hydrogen. However, the energy gap width decreases, and is approximately 14 meV. The contribution of hydrogen to the electron-phonon coupling constant is still dominating (around 66%). The increase of the pressure by the further 150 GPa causes the renewed increase in the maximum vibration energy of hydrogen $${[{\omega }_{D}]}_{500{\rm{GPa}}}/{[{\omega }_{D}]}_{350{\rm{GPa}}}=1.179$$. However, the division of the phonon density of states on the portion derived from sulfur and hydrogen is galling. The disappearance of the sulfur-hydrogen separation results also in the slight increase of the value of the electron-phonon coupling constant in the range of the lower frequencies $${\lambda }_{500{\rm{GPa}}}(\omega =120\,{\rm{meV}})/{\lambda }_{350{\rm{GPa}}}(\omega =120\,{\rm{meV}})=1.29$$. Above the pressure of 500 GPa we found the imaginary (negative) phonon frequencies which is an indication of the structural instability. This is one of the reason why we have limited our calculations to this range of pressures, the second one is that higher pressures are far beyond the ability of the experiment. In light of the latest results on the metallization of hydrogen53, compression up to 500 GPa is possible to achieve in laboratory. Figure 6 ilustrates the pressure dependence of the superconducting critical temperature. Close and open circles corresponds to the experimental results presented by Drozdov et al.3 and Einaga et al.5, respectively. The red dashed lines drawn by eye represents the trend of the experimental data above 150 GPa and great combine together with the theoretical range of T C calculated for the high pressures. These theoretical results were obtained using the Eliashberg equations and the following relation: $${{\rm{\Delta }}}_{m=1}(T={T}_{C})=0$$, where the order parameter is defined as $${{\rm{\Delta }}}_{m=1}={\phi }_{m=1}/{Z}_{m=1}$$. The commonly accepted value of the Coulomb pseudopotential $${\mu }^{\ast }=0.13$$ was adopted, the exact results of T C are collected in Table 1. The error bars indicate the value range of T C with $${\mu }^{\ast }\in \langle 0.11,0.15\rangle$$. The obtained results show that T C decreases from 164 to 129 K in the range of pressure from 250 to 350 GPa. Then above 350 GPa the superconducting critical temperature starts increasing. This is a promising result, however in the range of the pressure from 350 to 500 GPa the critical temperature does not exceed the value of 203 K. This can be explained by the unfavorable, and simultaneously weak, variation of the electron-phonon coupling constant and the logarithmic phonon frequency (the insert in Fig. 6). From the conducted ab initio calculations comes the conclusion that this is caused by the small or unfavorable influence of the pressure on the electron density of states and the electron-phonon matrix elements. It can be, however, noticed that the value of T C in the range of the very high pressures is relatively high and does not drop below 120 K. We did not study the critical temperature under extreme pressures because beyond 500 GPa the H3S structure loses the dynamical stability. Then, by using the analytical continuation37, 49, we determine the superconducting energy gap Δ(0) and the dimensionless ratio 2Δ(0)/T C which in the BCS theory takes the constant value 3.53. As we can see in Table 1, the obtained results significantly exceed the value of BCS predictions. This is connected with the strong-coupling and retardation effects, which in the framework of the Eliashberg formalism are not neglected. During the preparation of this manuscript, a superconducting energy gap of H3S compressed to 150 GPa was experimentally found (2Δ = 73 meV)54. Taking into account this result and the previously determined value of T C at this same pressure (203 K)3 we can evidence that 2Δ/k B T C = 4.17 is surprisingly close to our predictions for higher pressures. The above fact proves the correctness of our calculations. ## Conclusions In this paper we showed that H3S exhibits the superconducting properties in the range of the very high pressures (250–500 GPa), however, the critical temperature does not exceed the value of 203 K. The obtained result is related to the weak and unfavorable volatility of the electron-phonon coupling constant and the logarithmic phonon frequency. From the microscopic point of view, this results from the small or unfavorable influence of the pressure on the value of the electron density of states at the Fermi surface and the electron-phonon matrix elements. According to the above, it can be seen that the increase of the pressure alone is not sufficient to obtain the superconducting state in H3S at the room temperature. It is possible that the better method to achieve this goal is the appropriate partial atomic substitution of S atoms by other atoms. Interesting theoretical results were obtained by Ge et al. in the paper55, where the increase of the P-substitution rate causes the increase of the DOS, the phonon linewidths and the electron-phonon coupling constant. Finally, T C = 280 K for H3S0.925P0.075 at 250 GPa. ## References 1. 1. Li, Y., Hao, J., Liu, H., Li, Y. & Ma, Y. The metallization and superconductivity of dense hydrogen sulfide. J. Chem. Phys. 140, 174712 (2014). 2. 2. Duan, D. et al. Pressure-induced metallization of dense (H2S)2H2 with high-T C superconductivity. Sci. Rep. 4, 6968 (2014). 3. 3. Drozdov, A. P., Eremets, M. I., Troyan, I. A., Ksenofontov, V. & Shylin, S. I. Conventional superconductivity at 203 kelvin at high pressures in the sulfur hydride system. Nature 525, 73 (2015). 4. 4. Troyan, I. et al. Observation of superconductivity in hydrogen sulfide from nuclear resonant scattering. Science 351, 1303 (2016). 5. 5. Einaga, M. et al. Crystal structure of the superconducting phase of sulfur hydride. Nat. 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Quantitative analysis of nonadiabatic effects in dense H3S and PH3 superconductors. Sci. Rep. 6, 38570 (2016). 33. 33. Bardeen, J., Cooper, L. N. & Schrieffer, J. R. Microscopic theory of superconductivity. Phys. Rev. 106, 162 (1957). 34. 34. Bardeen, J., Cooper, L. N. & Schrieffer, J. R. Theory of superconductivity. Phys. Rev. 108, 1175 (1957). 35. 35. Migdal, A. B. Interaction between electrons and lattice vibrations in a normal metal. Soviet Physics JETP 34, 996 (1958). 36. 36. Eliashberg, G. M. Interactions between electrons and lattice vibrations in a superconductor. Soviet Physics JETP 11, 696 (1960). 37. 37. Carbotte, J. P. Properties of boson-exchange superconductors. Rev. Mod. Phys. 62, 1027 (1990). 38. 38. Li, Y. et al. Dissociation products and structures of solid H2S at strong compression. Phys. Rev. B 93, 020103 (2016). 39. 39. Errea, I. et al. Quantum hydrogen-bond symmetrization in the superconducting hydrogen sulfide system. Nature 532, 81 (2016). 40. 40. Oganov, A. 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Solids 99, 105 (2016). ## Acknowledgements A.P.D. gratefully acknowledges financial support from the Częstochowa University of Technology under Grant No. BS/MN-203-301/2017. ## Author information Authors ### Contributions A. P. Durajski designed and carried out the ab-initio calculations, collected data and drafted the final version of the manuscript. R. Szczęśniak wrote the part of the code for numerical calculations and participated in writing the manuscript. All authors reviewed the manuscript. ### Corresponding author Correspondence to Artur P. Durajski. ## Ethics declarations ### Competing Interests The authors declare that they have no competing interests. Publisher's note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. ## Rights and permissions Reprints and Permissions Durajski, A.P., Szczęśniak, R. First-principles study of superconducting hydrogen sulfide at pressure up to 500 GPa. Sci Rep 7, 4473 (2017). https://doi.org/10.1038/s41598-017-04714-5 • Accepted: • Published: • ### Thermodynamic Properties of the Superconducting State in Metallic Hydrogen: Electronic Correlations, Non-conventional Electron-Phonon Couplings and the Anharmonic Effects • M. Kostrzewa • , A. P. Durajski • , J. K. Kalaga • & R. Szczȩśniak Journal of Superconductivity and Novel Magnetism (2021) • ### From LaH10 to room–temperature superconductors • M. Kostrzewa • , K. M. Szczęśniak • , A. P. Durajski • & R. Szczęśniak Scientific Reports (2020) • ### Real-Time Evolution of the Electron Clouds of Transition Metal Ions: Possible Electron-Pairing Medium in Unconventional High-Temperature Superconductors • Tiege Zhou • , Min Wang • , Xiaoguang Luo • & Xu Zhang Journal of Superconductivity and Novel Magnetism (2019) • ### Unusual sulfur isotope effect and extremely high critical temperature in H3S superconductor • & Artur P. Durajski Scientific Reports (2018) • ### Anomalously high value of Coulomb pseudopotential for the H5S2 superconductor • Małgorzata Kostrzewa • , Joanna K. Kalaga • & Izabela A. Wrona Scientific Reports (2018)	{}
# How to avoid beamer counter incrementing In Latex beamer, I've created a handout environment for slides that are for the produced handout only (ie. not to be shown on the viewer). It looks like this: \newenvironment{handout}[2][1]{ \begin{frame}<handout>[environment=handout,allowframebreaks] } {\end{frame}} Key point in the above is: 1) <handout> ensures it is only visible on handout 2) allowframebreaks allows the handout to run over several slides. OK so far. Only problem is that the framenumber counter (which is used to display page numbers) is incremented for each slide, which means that the handout material and the displayed material in the presentation gets out of sync. So I want to freeze the counter. I tried simply to back up the current value of the framenumber counter to a temporary in the before part of the environment definition and reseting to that in the after part. That did however now work. Something like this: \newenvironment{handout}[2][1]{ \begin{frame}<handout>[environment=handout,allowframebreaks] } So the question is basically, how do I avoid the counter incrementing over these slides? - Welcome to TeX.sx! Your question was migrated here from Stack Overflow. Please register on this site, too, and make sure that both accounts are associated with each other, otherwise you won't be able to comment on or accept answers or edit your question. – N.N. Sep 6 '11 at 14:06 ## migrated from stackoverflow.comAug 28 '11 at 12:08 This question came from our site for professional and enthusiast programmers. \newenvironment{handout}{	{}
# American Institute of Mathematical Sciences October 2019, 24(10): 5337-5354. doi: 10.3934/dcdsb.2019061 ## Numerical solution of partial differential equations with stochastic Neumann boundary conditions Department of Mathematics, Faculty of Sciences, Razi University, Kermanshah, Iran Received May 2018 Revised November 2018 Published April 2019 The aim of this paper is to study the numerical solution of partial differential equations with boundary forcing. For spatial discretization we apply the Galerkin method and for time discretization we will use a method based on the accelerated exponential Euler method. Our purpose is to investigate the convergence of the proposed method, but the main difficulty in carrying out this construction is that at the forced boundary the solution is expected to be unbounded. Therefore the error estimates are performed in the $L_p$ spaces. Citation: Minoo Kamrani. Numerical solution of partial differential equations with stochastic Neumann boundary conditions. Discrete & Continuous Dynamical Systems - B, 2019, 24 (10) : 5337-5354. doi: 10.3934/dcdsb.2019061 ##### References: show all references ##### References: Numerical solution of Example 1, for $N = 32$, $\epsilon = 0.01$ and $T = \frac{3}{20}$, $\Delta t = 10^{-4}$ [1] Tomasz Kosmala, Markus Riedle. Variational solutions of stochastic partial differential equations with cylindrical Lévy noise. Discrete & Continuous Dynamical Systems - B, 2021, 26 (6) : 2879-2898. doi: 10.3934/dcdsb.2020209 [2] Qi Lü, Xu Zhang. A concise introduction to control theory for stochastic partial differential equations. Mathematical Control & Related Fields, 2021 doi: 10.3934/mcrf.2021020 [3] Madalina Petcu, Roger Temam. The one dimensional shallow water equations with Dirichlet boundary conditions on the velocity. 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Numerical Algebra, Control & Optimization, 2021, 11 (3) : 407-429. doi: 10.3934/naco.2020034 [12] Zhang Chao, Minghua Yang. BMO type space associated with Neumann operator and application to a class of parabolic equations. Discrete & Continuous Dynamical Systems - B, 2021 doi: 10.3934/dcdsb.2021104 [13] Ying Sui, Huimin Yu. Singularity formation for compressible Euler equations with time-dependent damping. Discrete & Continuous Dynamical Systems, 2021 doi: 10.3934/dcds.2021062 [14] Elvise Berchio, Filippo Gazzola, Dario Pierotti. Nodal solutions to critical growth elliptic problems under Steklov boundary conditions. Communications on Pure & Applied Analysis, 2009, 8 (2) : 533-557. doi: 10.3934/cpaa.2009.8.533 [15] Xiaofei Liu, Yong Wang. Weakening convergence conditions of a potential reduction method for tensor complementarity problems. Journal of Industrial & Management Optimization, 2021 doi: 10.3934/jimo.2021080 [16] Changpin Li, Zhiqiang Li. Asymptotic behaviors of solution to partial differential equation with Caputo–Hadamard derivative and fractional Laplacian: Hyperbolic case. Discrete & Continuous Dynamical Systems - S, 2021 doi: 10.3934/dcdss.2021023 [17] Qiwei Wu, Liping Luan. Large-time behavior of solutions to unipolar Euler-Poisson equations with time-dependent damping. Communications on Pure & Applied Analysis, 2021, 20 (3) : 995-1023. doi: 10.3934/cpaa.2021003 [18] Vandana Sharma. Global existence and uniform estimates of solutions to reaction diffusion systems with mass transport type boundary conditions. Communications on Pure & Applied Analysis, 2021, 20 (3) : 955-974. doi: 10.3934/cpaa.2021001 [19] Jan Březina, Eduard Feireisl, Antonín Novotný. On convergence to equilibria of flows of compressible viscous fluids under in/out–flux boundary conditions. Discrete & Continuous Dynamical Systems, 2021, 41 (8) : 3615-3627. doi: 10.3934/dcds.2021009 [20] Jaume Llibre, Luci Any Roberto. 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Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack GMAT Club It is currently 28 Mar 2017, 14:59 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # Algebra: Tips and hints Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 37659 Followers: 7411 Kudos [?]: 99831 [0], given: 11051 ### Show Tags 16 Jul 2014, 11:30 Expert's post 17 This post was BOOKMARKED Algebra: Tips and hints ! This post is a part of the Quant Tips and Hints by Topic Directory focusing on Quant topics and providing examples of how to approach them. Most of the questions are above average difficulty. Algebraic Identities 1. $$(x+y)^2=x^2+y^2+2xy$$ 2. $$(x-y)^2=x^2+y^2-2xy$$ 3. $$x^2-y^2=(x+y)(x-y)$$ 4. $$(x+y)^2-(x-y)^2=4xy$$ 5. $$x^3+y^3=(x+y)(x^2+y^2-xy)$$ 6. $$x^3-y^3=(x-y)(x^2+y^2+xy)$$ The general form of a quadratic equation is $$ax^2+bx+c=0$$. It's roots are: $$x_1=\frac{-b-\sqrt{b^2-4ac}}{2a}$$ and $$x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ Expression $$b^2-4ac$$ is called discriminant: • If discriminant is positive quadratics has two roots; • If discriminant is negative quadratics has no root; • If discriminant is zero quadratics has one root. When graphed quadratic expression ($$ax^2+bx+c=0$$) gives parabola: • The larger the absolute value of $$a$$, the steeper (or thinner) the parabola is, since the value of y is increased more quickly. • If $$a$$ is positive, the parabola opens upward, if negative, the parabola opens downward. Viete's theorem Viete's theorem states that for the roots $$x_1$$ and $$x_2$$ of a quadratic equation $$ax^2+bx+c=0$$: $$x_1+x_2=\frac{-b}{a}$$ AND $$x_1*x_2=\frac{c}{a}$$. Common mistake to avoid Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero. For example, $$xy=y$$ cannot be reduced by $$y$$ because $$y$$ could be 0 and we cannot divide by 0. If we do we'll loose one of the solutions. The correct way is: $$xy=y$$ --> $$xy-y=0$$ --> $$y(x-1)=0$$ --> $$y=0$$ or $$x=1$$. This week's PS question This week's DS Question Theory on Algebra: algebra-101576.html DS Algebra Questions to practice: search.php?search_id=tag&tag_id=29 PS Algebra Questions to practice: search.php?search_id=tag&tag_id=50 Special algebra set: new-algebra-set-149349.html Please share your Algebra tips below and get kudos point. Thank you. _________________ Manager Joined: 20 Dec 2011 Posts: 88 Followers: 4 Kudos [?]: 84 [1] , given: 31 Re: Algebra: Tips and hints [#permalink] ### Show Tags 17 Jul 2014, 14:28 1 KUDOS 1 This post was BOOKMARKED Bunuel wrote: Algebraic Identities 3. $$x^2-y^2=(x+y)(x-y)$$ Rule 3 is especially useful on GMAT. Sometimes it is obvious, as in PS 117 in OG 13: if-n-3-8-2-8-which-of-the-following-is-not-a-factor-of-n-132874.html but sometimes it will be hidden, as in PS 199 in OG 13: topic-137149.html In other words, if you are stuck and you see anything that might be expressed as "[perfect square] - [perfect square]", see if this can help you. GMAT Club Legend Joined: 09 Sep 2013 Posts: 14493 Followers: 609 Kudos [?]: 174 [0], given: 0 Re: Algebra: Tips and hints [#permalink] ### Show Tags 01 Nov 2015, 23:26 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 14493 Followers: 609 Kudos [?]: 174 [0], given: 0 Re: Algebra: Tips and hints [#permalink] ### Show Tags 05 Nov 2016, 09:53 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: Algebra: Tips and hints [#permalink] 05 Nov 2016, 09:53 Similar topics Replies Last post Similar Topics: 39 Absolute Value: Tips and hints 9 09 Jul 2014, 07:04 79 Inequalities: Tips and hints 2 02 Jul 2014, 04:33 15 Remainders: Tips and hints 2 23 Jun 2014, 03:33 63 Number Properties: Tips and hints 13 04 Jun 2014, 12:25 22 Exponents and Roots on the GMAT: Tips and hints 2 24 Jul 2014, 07:33 Display posts from previous: Sort by	{}
# SBT 425: CELL AND TISSUE CULTURE ## Course outline ### course out line 1. History and Development of tissue culture 2. Tissue culture equipment and facilities 3. Sterilization techniques 4. Tissue culture media ## lesson 1 ### Introduction Definition of tissue culture Refers to a technique that allows a whole plant to be produced from minute amount of the plant part (explant e.g immature embryos, shoot tip, root, leaf, ovary ovary, scutellum) or just a single cell on a sterile nutrient medium under laboratory controlled conditions (in vitro) often in small glass or plastic containers. ### Terminologies in tissue culture • Aseptic---Free of microorganisms. • Aseptic Technique---Procedures used to prevent the introduction of fungi, bacteria, viruses, mycoplasma or other microorganisms into cultures. • Autoclave---A machine capable of sterilizing wet or dry items with steam under pressure. Pressure cookers are a type of autoclaves. • Callus---An unorganized, proliferating mass of undifferentiated plant cells or tissues a under the influence of elevatedplant hormone levels. • Chemically Defined Medium---A nutritive solution for culturing cells in which each component is specifiable and ideally of known chemical structure. • Clone---Plants produced asexually from a single source plant. • Clonal Propagation---Asexual reproduction of plants that are considered to be genetically uniform and originated from a single individual or explant. • Contamination---Being infested with unwanted microorganisms such as bacteria, fungi or viruses • Culture—A plant growing in vitro. • Differentiated---Cells that maintain, in culture, all or much of the specialized structure and function typical of the cell type in vivo. Modifications of new cells to form tissues or organs with a specific function. • Explant---Tissue taken from its original site and transferred to an artificial medium for growth or maintenance. Part of an organism used in "in vitro" culture • Totipotency - The establishment of missing plant organs or parts; formation of a whole plant from a few cells or small portion of a plant. ### Basis for the cell culture In cell and tissue culture the concept of totipotency is used. Plant cells are totipotent if they have the ability to develop into whole plants or plant organs in vitro when given the correct conditions which is a characteristic potrayed by a zygote. Plant tissue culture relies on the fact that many plant cells have the ability to regenerate a whole plant (totipotency). Single cells, plant cells without cell wall(protoplasts), pieces of leaves, or (less commonly) roots can often be used to regenerate a new plant on culture media given the required nutrients and plant hormones. Not all plant cells are totipotent. However, there are a sufficient number of totipotent cells in the plant. Differentiated cells have to be dedifferentiated into callus and redifferentiated back to somatic embryo that will regenerate the entire plant. Thus totipotency implies that undifferentiated plant cells (meristematic cells) have the ability to display their full genetic potential to form functional plants when cultured in vitro. The major impact of plant tissue culture will be felt in the area of controlled manipulations of plants at the cellular level in ways which have not been possible prior to the introduction of tissue culture. ITissue culture has applications in research and commerce. In commercial settings, tissue culture is primarily used for plant propagation and it is often referred to as micropropagation. Micropropagation refers to the production of whole plants from cell cultures derived from explants (the initial piece of tissue put into culture); the explants usually consist of tissues that contain or develop into meristem cells. Plant tissue culture techniques are essential to many types of academic inquiry, as well as to many applied aspects of plant science. In the past, plant tissue culture techniques have been used in academic investigations of totipotency and the roles of hormones in cytodifferentiation and organogenesis. Currently, tissue-cultured plants that have been genetically engineered provide insight into plant molecular biology and gene regulation. Plant tissue culture techniques are also central to innovative areas of applied plant science, including plant biotechnology and agriculture. For example, selected plants can be cloned and cultured as suspended cells from which plant products can be harvested. In addition, the management of genetically engineered cells to form transgenic whole plants requires tissue culture procedures; tissue culture methods are also required in the formation of somatic haploid embryos from which homozygous plants can be generated. Thus, tissue culture techniques have been, and still are, prominent in academic and applied plant science. The techniques demonstrated in these exercises range from simple ones that can easily be performed by beginning students to those done by botany or physiology students. Experiment 1 and 2 employ plant material derived from aseptic seed germinations, while Experiments 3, 4, and 5 use portions of large intact plants. Experiment 1 demonstrates "in vitro" morphogenesis and totipotency and has been used successfully by beginning classes containing both biology majors and non-majors The remaining experiments are designed for use by more advanced students. . History and advancement in Development of Tissue culture The history of plant tissue culture can be traced back to near the turn of the 20th century when Gottlieb Haberlandt (1902) reported isolated single palisade cells from leaves/leaf mesophyll tissue cultured in Knop's salt solution enriched with sucrose. But failed to divide. Gottlieb Haberlandt a German botanist was the first to generate tissue from fully differentiated tissue. The first embryo culture, although crude, was done by Hanning in 1904; he cultured nearly mature embryos of certain crucifers and grew them to maturity. The technique was utilised by Laibach in 1925 to recover hybrid progeny from an interspecific cross in Linum. Subsequently, contributions from several workers led to the refinement of this technigue. White (1934) repeatedly reported continuously growing culture of meristematic cells of tomato from root tip-derived tissues of on medium containing inorganic salts, yeast extract and sucrose and 3 vit B (pyridoxine, thiamine, nicotinic acid) – established the importance of additives. Techniques for plant tissue culture progressed rapidly during the 1930s due to the discovery of the necessity of B vitamins and auxin for the growth of isolated meristem tissues. Some notable major discoveries included chemical and hormonal control of regeneration by Skoog and their associates on the nutritional requirements of tobacco tissue culture led to not only the discovery of plant growth hormones, kinetin and auxins, but also to the formation of an important plant tissue culture medium, the MS medium (Skoog and Tsui, 1951; Skoog and Miller, 1957; Murashige and Skoog, 1962). Haploid plants from pollen grains were first produced by Maheshwari and Guha in 1964 by culturing anthers of Datura. This marked the beginning of anther culture or pollen culture for the production of haploid plants. The technique was further developed by many workers, more notably by JP. Nitch (1967), C. Nitch and coworkers. These workers showed that isolated microspores of tobacco produce complete plants. Murashige cloned plants in vitro, raised haploid plants from pollen grains and used protoplast fusion to hybridize 2 species of tobacco into one plant contained 4N. Since 1960s, tissue and cell culture has increasingly been used as a tool by plant scientists and biotechnologists. Practical micropropagation and production of virus-free plants (G. Morel, 1960), haploid plants (S. Guha and S.C. Maheshwari, 1964, J. P. Nitsch, 1967), culture and regeneration of protoplasts (E.C. Cocking, 1960), production of secondary metabolites (B. Kaul and E.J. Staba, 1967) and large-scale cell culture in bioreactors (M. Noguchi et al., 1977), to mention just a few landmarks. 1970’s and 1980s marked the beginning of genetic engineering.. The ongoing achievements in in vitro culture of pollen, protoplasts and cell suspensions – and their ability to regenerate whole plants has resulted into a new disciplines of plant science: somatic cell genetics and metabolite production, somatic hybrids, production of haploid plants, selection of variants and mutants, and improved generation of metabolites. All became available for the community of plant scientists and breeders, expanding the diversity, agronomic and commercial value of plants. Yet, much remains to be explored in terms of methodology, procedures and the theories behind the technology Tissue culture equipment and facilities Tissue culture equipment and supplies Laminar flow hoods. There are two types of laminar flow hoods, vertical and horizontal. The vertical hood, also known as a biology safety cabinet, is best for working with hazardous organisms since the aerosols that are generated in the hood are filtered out before they are released into the surrounding environment. Horizontal hoods are designed such that the air flows directly at the operator hence they are not useful for working with hazardous organisms but are the best protection for your cultures. Both types of hoods have continuous displacement of air that passes through a HEPA (high efficiency particle) filter that removes particulates from the air. In a vertical hood, the filtered air blows down from the top of the cabinet; in a horizontal hood, the filtered air blows out at the operator in a horizontal fashion.. The hoods are equipped with a short-wave UV light that can be turned on for a few minutes to sterilize the surfaces of the hood, but be aware that only exposed surfaces will be accessible to the UV light. Do not put your hands or face near the hood when the UV light is on as the short wave light can cause skin and eye damage. The hood is used for the transfer of explants and culture, dispensing sterile media. b. Microscopes. Inverted phase contrast microscopes are used for visualizing the cells. Microscopes should be kept covered and the lights turned down when not in use. Before using the microscope or whenever an objective is changed, check that the phase rings are aligned. c. Vessels. Anchorage dependent cells require a nontoxic, biologically inert, and optically transparent surface that will allow cells to attach and allow movement for growth. The most convenient vessels are specially-treated polystyrene plastic that are supplied sterile and are disposable. These include petri dishes, multi-well plates, microtiter plates, roller bottles, and screwcap flasks. Autoclave/pressure cooker – for sterilizing the media and the vessels Refrigirator – for storing chemicals at low temperature (4oC) such as vitamins, growth regulators and micronutrients Cultures bottles/baby jars/petridishes – for putting a medium used for culturing, transfers of cultures Incubator – used for growth of culture at a specific controlled temperature Freezer – storage of chemicals at -20oC Shaker – for the growth of the suspension cultures Balances – for weighing chemicals used for preparing stock solutions and culture media pH meter – for measuring the pH of the media Thermometer – for measuring temperature in the growth room Dry Sterilization oven – for sterilizing vessels and instruments Microva – for heating agar media Bunsen burner Beakers, Pipettes, micropipettes, volumetric flasks, graduated measuring cylinders –for measuring stock solutions/chemicals and the media Parafilm – for sealing petrishes Magnetic stirrer – used when preparing the media Metal trays – for carrying the media, stock solutions and tissue culture tools Aluminium foil – for wrapping instruments and vessel opening when sterilizing Scapels, forceps – for cutting and transferring explants and cultures Hot plate – for melting and dissolving the gelling agents Growth chamber – for growing culture or plants under controlled conditions of light and temperature. Facilities Culture Room – room where all activities of sterile transfers of explants and cultures are done Media preparation room –Room where the preparation of the culture media is done. Growth room – Room with controlled temperature and light where cultures are incubated or grown Green house – where in vitro regenerated plants are hardened and grown to maturity ## lesson 2 Sterilization (Aseptic) Techniques Used in Tissue Culture Tissue culture requires sterile conditions. The culture room, lamina hood, instruments, vessels, culture media, plant materials used must be sterile. Aim the essence of aseptic technique is to ensure all cell culture procedures are performed to a standard that will prevent contamination from bacteria,viruses and fungi and cross contamination with other cell lines from culture to culture If sterile tissues are available, then the exclusion of microorganisms is accomplished by using sterile instruments and culture media concurrently with standard bacteriological transfer procedures to avoid extraneous contamination. Aseptic technique is absolutely necessary for the successful establishment and maintenance of plant cell, tissue and organ cultures. The in vitro environment in which the plant material is grown is also ideal for the proliferation of microorganisms. In most cases the microorganisms outgrow the plant tissues,. The environmental control of air is also of concern because room air may be highly contaminated. Example: Sneezing produces 100,000 - 200,000 aerosol droplets which can then attach to dust particles. Successful control of contamination depends largely upon the operator’s techniques in aseptic culture. One should always be aware of potential sources of contamination such as dust, hair, hands, and clothes All the materials, e.g., vessels, instruments, medium, plant material, etc., used in culture work must be freed from microbes. This is achieved by one of the following approaches: (i) dry heat treatment, (ii) flame sterilization, (iii) autoclaving, (iv) filter sterilization, (v) wiping with 70% ethanol, and (vi) surface sterilization. ### Sterilization of the culture room The room should be sterilized with 70% ethanol or methylated spirit 20minutes before culturing. ### Sterilization of the lamina hood UV Radiation: It is possible to use germicidal lamps to sterilize items in the transfer hood when one is working in the hood. UV lamps should not be used when people are present because the light is damaging to eyes and skin. Plants left under UV lamps will die. Ethanol: The hood should be sterilized with 70% ethanol or methylated spirit 20 minutes before culturing. 70% ethanol is used ### How to mantain sterilization when working in the Transfer Hood/lamina hood: 1. Spray or wipe the inside of the transfer hood using 70% ethanol let the spray dry before commencing work . Wipe up any spills quickly; use 70% EtOH for cleaning. Clean hood surface periodically while working 2. The hood should remain on continuously. If for some reason it has been turned off, turn it on and let it run for at least 15 minutes before using. 3. Sterilize gloves by washing them in 70% ethanol and allowing to air dry for 30 seconds before commencing work. 4. Remove watches, etc., roll up long sleeves, and wash hands thoroughly with soap (preferably bactericidal) and water then Wipe hands and lower arms with 70% EtOH 5. Spray equipment put into the sterile area with 70% ethanol. For example, spray bags of petri dishes, media bottles, pipette aids with 70 % alcohol before you open them and place the desired number of unopened dishes in the sterile area.  Make sure that everything needed for the work is in the hood and all unnecessary things are removed. Keep as little in the hood as possible. As few things as possible should be stored in the hood. Make sure that materials in use are to the side of your work area, so that airflow from the hood is not blocked. Arrange tools and other items in the hood so that your hands do not have to cross over each other while working. All other items in the hood should be arranged so that your work area is directly in front of you, and between 8 and 10 inches in from the front edge. No materials should be placed between the actual work area and the filter.  Sterilize culture tubes with lids or caps on. When you open a sterile tube, touch only the outside of the cap, and do not set the cap on any laboratory surface. Instead, hold the cap with one or two fingers while you complete the operation, and then replace it on the tube. This technique usually requires some practice, especially if you are simultaneously opening tubes and operating a sterile pipette. After you remove the cap from the test tube, pass the mouth of the tube through a flame. If possible, hold the open tube at an angle. Put only sterile objects into the tube. Complete the operation as quickly as you reasonably can, and then flame the mouth of the tube again. Replace the lid. If you don't have to be careful about the volume you transfer, a pure sterile solution can be transferred to a sterile container or new sterile medium by pouring. For example, we do not measure a specific volume of medium when we pour culture plates, although after you have done it for a while, you become pretty consistent. Remove the cap or lid from the solution to be transferred. Thoroughly flame the mouth of the container, holding it at an angle as you do so. Remove the lid from the target container. Hold the container at an angle. Quickly and neatly pour the contents from the first container into the second. Replace the lid.  If you must transfer an exact volume of liquid, use a sterile pipette or a sterile graduated cylinder. When using a sterile graduated cylinder, complete the transfer as quickly as you reasonably can to minimize the time the sterile liquid is exposed to the air. 1. Don’t touch any surface that is supposed to remain sterile with your hands. Use forceps, etc. 2. Whilst working do not contaminate gloves by touching anything outside the cabinet (especially face and hair). If gloves become contaminated re-sterilized with 70% ethanol as above before proceeding. 3. Plant material should be placed on a sterile surface when manipulating it in the hood. Sterile petri dishes (expensive), sterile paper towels, or sterile paper plates work fine. Pre-sterilized plastic dishes have two sterile surfaces-the inside top and inside bottom. 4. Plastic pipettes are purchased presterilized in individual wrappers. To use a pipette, remove it from its wrapper or container by the end opposite the tip. Do not touch the lower two-thirds of the pipette. Do not allow the pipette to touch any laboratory surface. Insert only the untouched lower portion of the pipette into a sterile container. 5. Discard gloves after handling contaminated cultures and at the end of all cell culture procedures. 6. Know which of your implements, flasks, etc. are sterile and which are not 7. Sterilized items should be used within a short time 8. Movement within and immediately outside the cabinet must not be rapid. Movements in the hood should be limited to a small area. Slow movement will allow the air within the cabinet to circulate properly. Work well back in the transfer hood (behind the line). Especially keep all flasks as far back to the back of the hood as possible. 9. Speech, sneezing and coughing must be directed away from the cabinet so as not to disrupt the airflow. 10. Remove items from the hood as soon as they are no longer needed. All cultures must be sealed before leaving the hood. 11. After completing work disinfect all equipment and material before removing from the cabinet. Spray the work surfaces inside the cabinet with 70% ethanol and wipe dry with tissue. Dispose of tissue by autoclaving. 12. It is pointless to practice good sterile technique in a dirty lab. Special problems are contaminated cultures, dirty dishes and solutions where microorganisms can grow. Sterilization technique used include: • Ethanol: Instruments (scalpels, forceps) can be sterilized dipping them in 70% or 95% EtOH and then immediately placing them in the flame of an alcohol lamp or gas burner. This can be dangerous if the vessel holding the alcohol tips over and an alcohol fire results. A fairly deep container, like a coplin-staining jar, should be used to hold the ethanol. Use enough ethanol to submerge the business ends of the instruments but not so much that you burn your hands. Some people wear gloves in the hood for certain procedures. If you do this, be very careful not to get them near the flame. • Microwave: Instrument are rubbed using aluminum foil and then sterilized. • Dry heat: Empty glassware (culture vessels, pipettes, etc.), and certain Plasticware (Teflon FEP); instruments like scalpels, forceps, needles, etc. can be sterilized using an oven at 160-180oC for 4 hours.. More recently, glass bead sterilizers (300°C) are being employed for the sterilization these devices use dry heat. Small instruments such as scalpels and forceps are placed into the glass beads and are sterilized within 10-60 seconds. The instruments will cool down to working temperatures with 30-60 seconds. These sterilizers heat to approximately 275-350° C and will destroy bacterial and fungalspores that may be found on your instruments. The instruments simply need to be inserted into the heated glass beads for a period of 10 to 60 sec. The instruments should then be placed on a rack under the hood to cool until needed. These units kill most major classes of fungi, bacteria, and viruses. • Autoclaving: Empty vessels, beakers, graduated cylinders, etc., should be closed with a cap or aluminum foil. Tools should also be wrapped in foil or paper or put in a covered sterilization tray. It is critical that the steam penetrate the items in order for sterilization to be successful. However autoclaving is not advisable for metal instruments because they may rust and become blunt under these conditions. • Flame sterilization: Instruments like scalpels, forceps, etc that have been sterilized in hot dry air should be removed from their wrapping, dipped in 95% ethyl alcohol, and exposed to the heat of a flame. After an instrument has been used, it can again be dipped in ethyl alcohol, re-flamed, and then reused. This technique is called flame sterilization . Flaming instruments prior to use and flaming the opening of receiving vessels prior to transfer is reqiured. Aseptic transfers are more easily performed in a transfer chamber such as a laminar flow hood, which is also preferably equipped with a bunsen burner. • NOTE: Items that come packaged sterile e.g Petri dishes should be examined carefully for damage before use. Surface-sterilizing Plant Materials Plants materials used in tissue culture need to be healthy and actively growing. Stressed plants, particularly water-stressed plants, usually do not grow as tissue cultures. Insect and disease-free greenhouse plants are rendered aseptic more readily, so contamination rate is lower when these plants are used in tissue culture procedures. Seeds that can be easily surface sterilized usually produce contamination-free plants that can be grown under clean greenhouse conditions for later experimental use If experimental tissues are not aseptic, then surface sterilization procedures specific to the tissues are employed. Common sterilants are ethyl alcohol and/or chlorox Bleach, Calcium hypochlorite, Mercuric chloride, Hydrogen peroxide with an added surfactant. Concentration of sterilants and exposure time are determined empirically. Steps in sterilization of plant materials (explants) 1. Preparation of Stock Plants Prior good care of stock plants may lessen the amount of contamination that is present on explants. Plants grown in the field are typically more “dirty” than those grown in a greenhouse or growth chamber. Overhead watering increases contamination of initial explants. Likewise, splashing soil on the plant during watering will increase initial contamination. Treatment of stock plants with fungicides and/or bacteriocides is sometimes helpful. It is sometimes possible to harvest shoots and force buds from them in clean conditions. The shoots may be free of contaminants when surface-sterilized in a normal manner. Seeds may be sterilized and germinated in vitro to provide clean material. Covering growing shoots for several days or weeks prior to harvesting tissue for culture may supply cleaner material. Explants or material from which material will be cut can be washed in soapy water and then placed under running water for 1 to 2 hours. 2. Sterilization of explants sterilization using sterilants a) Ethanol Ethanol is a powerful sterilizing agent but also extremely phytotoxic. Therefore, plant material is typically exposed to it for only seconds or minutes depending on the sensitivity of the explants and how difficult it is to disinfect. Explants from Woody and field plants takes longer time. The more tender the tissue, the more it will be damaged by alcohol. Tissues such as dormant buds, seeds, or unopened flower buds can be treated for longer periods of time since the tissue that will be explanted or that will develop is actually within the structure that is being surface-sterilized. Generally 70% ethanol is used prior to treatment with other compounds. 3. Sterilization using bleaching agents a. Sodium Hypochlorite Sodium hypochlorite, usually purchased as laundry bleach, is the most frequent choice for surface sterilization. It is readily available and can be diluted to proper concentrations. Commercial laundry bleach is 5.25% sodium hypochlorite. It is usually diluted to 10% - 20% of the original concentration, resulting in a final concentration of 0.5 - 1.0% sodium hypchlorite. Plant material is usually immersed in this solution for 10 - 20 minutes. A balance between concentration and time must be determined empirically for each type of explant, because of phytotoxicity. b. Calcium Hypochlorite The concentration of calcium hypochlorite used is 3.25 %. The solution must be filtered prior to use since not the entire compound goes into solution. Calcium hypochlorite may be less injurious to plant tissues than sodium hypochlorite. c. Mercuric Chloride Mercuric chloride is used only as a last resort. Concentration of 0.2% is used. It is extremely toxic to both plants and humans and must be disposed of with care. Since mercury is so phytotoxic, it is critical that many rinses be used to remove all traces of the mineral from the plant material. d. Hydrogen Peroxide The concentration of 10 % hydrogen peroxide is used for surface sterilization of plant material. Some researchers have found that hydrogen peroxide is useful for surface-sterilizing material while in the field. Enhancing Effectiveness of Sterilization Procedure • Surfactant/wetting agent (e.g.Tween 20) is frequently added to the sodium hypochlorite. • The solutions that the explants are in are often shaken or continuously stirred. 4. Rinsing with sterile distilled water After plant material is sterilized with one of the above compounds, it must be rinsed thoroughly with sterile water. Typically three to five separate rinses are done to eliminate the residue of the disinfectant. Media sterilization Two methods (autoclaving and membrane filtration under positive pressure) are commonly used to sterilize culture media. Autoclaving: Autoclaving is the method most often used for sterilizing culture media In order to be sterilized, the culture media must be held at 121C, 15 psi, (pounds per square inch; 1.06 kg/cm2). It is important that this temperature is attained before timing begins. Therefore time in the autoclave will vary, depending on volume in individual vessels and number of vessels in the autoclave. Most autoclaves automatically adjust time when temperature and psi are set, and include time in the cycle for a slow decrease in pressure. There are tape indicators that can be affixed to vessels, but they may not reflect the temperature of liquid within them.. Culture media, distilled water, and other heat stable mixtures can be autoclaved in glass containers that are sealed with cotton plugs, aluminum foil, or plastic closures. For small volumes of liquids (100 ml or less), the time required for autoclaving is 15-20 min, larger quantities (2-4 liter), 30-40 min is required to complete the cycle. The pressure should not exceed 20 psi, as higher pressures may lead to the decomposition of carbohydrates and other components of a medium. Too high temperatures or too long cycles can also result in changes in properties of the medium. ABA, IAA, IBA, kinetin, pyridoxine, 2-ip and thiamine are usually autoclaved. A summary of the time required for the of various volumes of the medium at 121oC Volume of the medium per vessel Pre-heating time to reach 121oC Total sterilization time 20-25 9 24 50 11 26 100 13.5 28.5 250 16.5 31.5 500 20 35 1000 25 40 2000 33 48 3000 40 55 4000 48 63 Membrane filtration: Solutions that contain heat-labile components are best sterilized by ultrafiltration. However the process is slower and more costly than autoclaving. Organic compounds such as some growth regulators such as heat labile compounds like GA3, ABA, zeatin, enzymes, amino acids, and vitamins may be degraded during autoclaving. These compounds require filter sterilization through a 0.22 µm membrane. Several manufacturers make nitrocellulose membranes that can be sterilized by autoclaving. They are placed between sections of a filter unit and sterilized as one piece. Other filters (the kind we use) come pre-sterilized. Larger ones can be set over a sterile flask and a vacuum is applied to pull the compound dissolved in liquid through the membrane and into the sterile flask. Smaller membranes fit on the end of a sterile syringe and liquid is pushed through by depressing the top of the syringe. The size of the filter selected depends on the volume of the solution to be sterilized and the components of the solution. # -=Tissue culture media== Can be a solid or a liquid culture media There are different types of tissue culture media such as  MS (developed for tobacco),  LS (Linsmaier and Skoogs medium),  Gamborg B5 (developed for soyabean cell suspension cultures),  Schenk and Hildebrandt (developed for callus cultures of monocots and dicots),  N6 (developed for anther culture of rice),  Whites medium (developed for tissue culture of tomato roots),  Wood medium (for trees) and  Nitschs medium for anther culture). Ms and B5 are the most commonly used media.Virtually all tissue culture media are synthetic or chemically defined; only a few of them use complex organics, e.g., potato extract, as their normal constituents. A synthetic medium consists of only chemically defined compounds. A variety of recipes have been developed since none of them is suitable for either all plant species or for every purpose. Composition of Some Plant Tissue Culture Media - Components of media for the growth of plant callus and suspension cultures can be classified into five groups. The division usually reflects the way in which stock solutions are prepared and stored. These groups are: i) inorganic nutrients (macronutrients and micronutrients) ii) organic supplement (vitamins) iii) carbon source iv) Growth regulators ` v) gelling agents Media compositions are formulated considering specific requirements of a particular culture system or species. 'For example, some tissues show better response on a solid medium while others prefer a liquid medium or others do better in MS medium than in B5 medium.. Some tissues are grown on simple media containing only inorganic salts and utilizable carbon (sugar) source, but for most others it is essential to supplement the medium with vitamins, amino acids and growth substances. Very often, complex nutritive substances are added to the medium. Such a medium, composed of "chemically defined" compounds, is referred to as a "synthetic" medium. 1. Inorganic nutrients a) Macronutrients – Required in large quantities. The macronutrients include six major elements-nitrogen (N), phosphorus (P), potassium (K), calcium (Ca), magnesium (Mg), and surphur (S)-present as salts that constitute various media. MgSO4 provide Mg & S; KH2PO4 and NaHPO4 provides P; CaCl2.2H2O or Ca(NO3)2. 4H2Oprovide calcium and KCl, KNO3 KH2PO4 provides K; KCl Or CaCl2.2H2O provide chloride. All are essential for plant cell and tissue growth. Culture media should contain at least 25 mM nitrate and potassium. However, considerably better results are obtained if the source for nitrogen in media is contributed by both nitrates and ammonium (2-20 mM) or any other reduced nitrogen source. In case only ammonium is used, there is need to add one or more tricarboxylic acid (TCA) cycle acids (e.g., citrate, succinate, or malate) so that any deleterious effect due to ammonium concentrations in excess of 8 mM in the medium is diluted. When nitrate and ammonia ions are present together in the culture medium, the latter are used more rapidly. b) Micronutrients Are required in lower concentrations. Include boron (H2BO3); Cobalt (COCl2); iron (NaFeEDTA); managanase (MnSO4.H2O)); molybdenum (NaMoO4); copper (CuSO4. H2O) and Zinc (ZnSO4.7H20).	{}
## On the Nagumo uniqueness theorem.(English)Zbl 1252.34011 The paper discusses a certain generalization of two uniqueness criteria for the Cauchy problem $x'= f(t,x),\;x(0)= 0,\;t\in [0,1].$ The first of the mentioned criteria, due Athanassov, is the following (1) $$\|f(t,x)- f(t,y)\|\leq (u'(t)/u(t))\|x- y\|$$ for $$t\in (0,1]$$ and $$x,y\in [-1,1]$$, where $$u:[0,1]\to \mathbb{R}_+= [0,\infty), u(0)= 0$$ and $$u'(t)> 0$$ for $$t\in(0,1]$$. Moreover,a second requirement is satisfied, (2) $$f(t,x)= o(u'(t))$$ as $$t\to 0$$, uniformly with respect to $$x\in [-1,1]$$. The second criterion, due to Constantin, consists of condition (2) and condition $$(1^\prime)$$ $$\|f(t,x)\|\leq (u'(t)/u(t))\omega(\|x\|)$$, where $$\omega: [0,1]\to \mathbb{R}_+$$ satisfies some conditions. The authors did not include a concrete example showing that their criterion is more general than those mentioned above. Reviewer’s remark: It is a very surprising fact that nobody noticed that Athanassov’s criterion is only a particular case of the uniqueness criteria due to J. Witte [Math. Z. 140, 281–287 (1974; Zbl 0289.34007)] and the reviewer and J. Rivero [Commentat. Math. Univ. Carol. 28, 23–31 (1987; Zbl 0627.34073)]. The details are too involved to be stated here. ### MSC: 34A12 Initial value problems, existence, uniqueness, continuous dependence and continuation of solutions to ordinary differential equations 34A34 Nonlinear ordinary differential equations and systems ### Citations: Zbl 0289.34007; Zbl 0627.34073 Full Text: ### References: [1] Chemin, J.Y.; Lerner, N., Flot de champs de vecteurs non lipschitziens et équations de navier – stokes, J. differential equations, 121, 314-328, (1995) · Zbl 0878.35089 [2] Constantin, A., The trajectories of particles in Stokes waves, Invent. math., 166, 523-535, (2006) · Zbl 1108.76013 [3] Constantin, A., A dynamical systems approach towards isolated vorticity regions for tsunami background states, Arch. ration. mech. anal., 200, 239-253, (2011) · Zbl 1294.35169 [4] Balabane, M.; Dolbeault, J.; Ounaies, H., Nodal solutions for a sublinear elliptic equation, Nonlinear anal. TMA, 52, 219-237, (2003) · Zbl 1087.35033 [5] Kaper, H.G.; Kwong, M.K., Uniqueness for a class of nonlinear initial value problems, J. math. anal. appl., 130, 467-473, (1988) · Zbl 0653.34007 [6] Nagumo, M., Eine hinreichende bedingung für die unität der Lösung von differentialgleichungen erster ordnung, Japan J. math., 3, 107-112, (1926), Reprinted in: M. Yamaguti, L. Nirenberg, S. Mizohata, Y. Sibuya (Eds.), Mitio Nagumo Collected Papers, Springer-Verlag, Tokyo, 1993 · JFM 52.0438.01 [7] Nagumo, M., Eine hinreichende bedingung für die unität der Lösung von gewöhnlichen differentialgleichungen $$n$$-ter ordnung, Japan J. math., 4, 307-309, (1927), Reprinted in: M. Yamaguti, L. Nirenberg, S. Mizohata, Y. Sibuya (Eds.), Mitio Nagumo Collected Papers, Springer-Verlag, Tokyo, 1993 · JFM 54.0460.01 [8] Wintner, A., On the local uniqueness of the initial value problem of the differential equation $$\frac{\operatorname{d}^n x}{\operatorname{d}^n t} = f(t, x)$$, Boll. unione mat. ital., 11, 496-498, (1956) · Zbl 0071.29902 [9] Constantin, A., On the unicity of solutions for high-order differential equations, Ist. lomb. rend. sci. A, 130, 171-181, (1996) [10] Athanassov, Z.S., Uniqueness and convergence of successive approximations for ordinary differential equations, Math. japon., 35, 351-367, (1990) · Zbl 0709.34002 [11] Constantin, A., On nagumo’s theorem, Proc. Japan acad., 86(A), 41-44, (2010) · Zbl 1192.34014 [12] O.G. Mustafa, A Nagumo-like uniqueness result for a second order ODE, Monatsh. Math., http://dx.doi.org/10.1007/s00605-011-0324-2 (in press). [13] T. Mejstrik, Some remarks on Nagumo’s theorem, Czechoslovak Math. J. (in press). · Zbl 1249.34021 [14] Chicone, C., Ordinary differential equations with applications, (2006), Springer-Verlag New York · Zbl 1120.34001 [15] Mustafa, O.G., On some second order differential equations with convergent solutions, Monatsh. math., 150, 133-140, (2007) · Zbl 1121.34009 [16] Benguria, R.D.; Dolbeault, J.; Esteban, M.J., Classification of the solutions of semilinear elliptic problems in a ball, J. differential equations, 167, 438-466, (2000) · Zbl 0968.35042 [17] Peletier, L.A.; Serrin, J., Uniqueness of positive solutions of semilinear equations in $$\mathbb{R}^n$$, Arch. ration. mech. anal., 81, 181-197, (1983) · Zbl 0516.35031 This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	{}
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Question: # What is the term for the distance and direction of an object from its original starting point? Not speed that is a rate with time, its the distance and direction an matebu is from the starting point? ## Velocity is a vector quantity which refers to "the rate at which an object changes its position." Imagine a person moving MORE Physics Algebra Motion A physical quantity (or "physical magnitude") is a physical property of a phenomenon, body, or substance, that can be quantified by measurement. The following outline is provided as an overview of and topical guide to physics: Physics – natural science that involves the study of matter and its motion through spacetime, along with related concepts such as energy and force. More broadly, it is the general analysis of nature, conducted in order to understand how the universe behaves. Kinematics Vectors Velocity In mathematics, physics, and engineering, a Euclidean vector (sometimes called a geometric or spatial vector, or—as here—simply a vector) is a geometric quantity having magnitude (or length) and direction expressed numerically as tuples [ x , y , z ] splitting the entire quantity into its orthogonal-axis components. A vector is an object that is an input for or an output from vector functions according to vector algebra. A Euclidean vector is typically sketched as a directed line segment, or arrow, connecting an initial point A with a terminal point B and denoted by $\overrightarrow{AB}.$ However, as an informational object, the vector is not as informative as a directed line segment (an ordered list of two points [ A , B ]) but rather expresses the displacement, or vector offset (change in location), A --> B. Technically, the [ x, y, z ] components of vector $\overrightarrow{AB}$ are equal to the vector difference $\overrightarrow{B}$ minus $\overrightarrow{A}$. In this way, the vector $\overrightarrow{AB}$ considered as a numerical quantity conceals the locations of A and B while imparting the location of point B relative to A as if A were the coordinate origin. Vectors play an important role in physics: velocity and acceleration of a moving object and forces acting on it are all described by vectors. Many other physical quantities can be usefully thought of as vectors. Although most of them do not represent distances (except, for example, position or displacement), their magnitude and direction can be still represented by the length and direction of an arrow. The mathematical representation of a physical vector depends on the coordinate system used to describe it. Other vector-like objects that describe physical quantities and transform in a similar way under changes of the coordinate system include pseudovectors and tensors. Displacement The mathematics of general relativity are very complex. In Newton's theories of motions, an object's length and the rate of passage of time remain constant as it changes speed. As a result, many problems in Newtonian mechanics can be solved with algebra alone. In relativity, on the other hand, length, and the passage of time change as an object's speed approaches the speed of light. The additional variables greatly complicate calculations of an object's motion. As a result, relativity requires the use of vectors, tensors, pseudotensors, curvilinear coordinates and many other complicated mathematical concepts. All the mathematics discussed in this article were understood before the proposal of Einstein's general theory of relativity. 6	{}
New Titles \| FAQ \| Keep Informed \| Review Cart \| Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education Resolving Markov Chains onto Bernoulli Shifts via Positive Polynomials Brian Marcus, IBM Almaden Research Center, San Jose, CA, and Selim Tuncel, University of Washington, Seattle, WA SEARCH THIS BOOK: Memoirs of the American Mathematical Society 2001; 98 pp; softcover Volume: 150 ISBN-10: 0-8218-2646-8 ISBN-13: 978-0-8218-2646-1 List Price: US$51 Individual Members: US$30.60 Institutional Members: US\$40.80 Order Code: MEMO/150/710 The two parts of this Memoir contain two separate but related papers. The longer paper in Part A obtains necessary and sufficient conditions for several types of codings of Markov chains onto Bernoulli shifts. It proceeds by replacing the defining stochastic matrix of each Markov chain by a matrix whose entries are polynomials with positive coefficients in several variables; a Bernoulli shift is represented by a single polynomial with positive coefficients, $$p$$. This transforms jointly topological and measure-theoretic coding problems into combinatorial ones. In solving the combinatorial problems in Part A, we state and make use of facts from Part B concerning $$p^n$$ and its coefficients. Part B contains the shorter paper on $$p^n$$ and its coefficients, and is independent of Part A. An announcement describing the contents of this Memoir may be found in the Electronic Research Announcements of the AMS at the following Web address: www.ams.org/era/ Graduate students and research mathematicians working in measure and integration. Part A. Resolving Markov Chains onto Bernoulli Shifts • Introduction • Weighted graphs and polynomial matrices • The main results • Markov chains and regular isomorphism • Necessity of the conditions • Totally conforming eigenvectors and the one-variable case • Splitting the conforming eigenvector in the one-variable case • Totally conforming eigenvectors for the general case • Splitting the conforming eigenvector in the general case • Bibliography Part B. On Large Powers of Positive Polynomials in Several Variables • Introduction • Structure of $${\mathbf{Log}}({\mathbf(p}^{\mathbf n})$$ • Entropy and equilibrium distributions for $${\mathbf w}\in {\mathbf W}({\mathbf p})$$ • Equilibrium distributions and coefficients of $${\mathbf(p}^{\mathbf n})$$ • Proofs of the estimates • Bibliography	{}
# Surds and Indices MCQ Quiz - Objective Question with Answer for Surds and Indices - Download Free PDF Last updated on Dec 3, 2022 Surds and Indices often show themselves up in the competitive exams syllabus therefore it’s important to prepare them effectively. Solve Surds and Indices MCQs Quiz so that you never will have to depend on fluke chances for your answer to be correct. Get solutions and their explanations for each and every Surds and Indices question answer listed in this selection of Surds and Indices objective questions. We also have mentioned tips and shortcuts to solve these questions to save time and improve accuracy. ## Latest Surds and Indices MCQ Objective Questions #### Surds and Indices Question 1: (35/81) × 38 × (9/27) = 3× 243 1. 3 2. 4 3. 2 4. 5 Option 1 : 3 #### Surds and Indices Question 1 Detailed Solution Concept: am/an = am - n am× an = am + n Calculation: (35/81) × 38 × (9/27) = 3× 243 ⇒ 35/34 × 38 × 1/3 = 3? × 35 ⇒ 35/34 × 38 × 3-1 3? × 35 ⇒ 31 × 38 × 3-1 = 3? × 35 ⇒ 31+8-1 = 3x+5 ⇒ 38 = 3x+5 Since the base is the same we can equate the powers. ⇒ x + 5 = 8 x = 3 #### Surds and Indices Question 2: If 200% of 108 = x, then find the value of $$\rm \sqrt[3]{x}$$ . 1. 18 2. 216 3. 16 4. 6 Option 4 : 6 #### Surds and Indices Question 2 Detailed Solution Formula Used: X% = X/100 $$\rm \sqrt[3]{x}$$ = cube root of x Calculation: ⇒ 200% × 108 = x $$\frac{200}{100}$$ × 108 = x ⇒ x = 108 × 2 = 216 ⇒ x = 63 ⇒ $$\rm \sqrt[3]{x}$$ = 6 ∴ The correct answer is 6. #### Surds and Indices Question 3: If $${\left( {\frac{2}{7}} \right)^{ - 3}}X{\left( {\frac{4}{{49}}} \right)^6} = {\left( {\frac{2}{7}} \right)^{2m - 1}}$$, then what is the value of m? 1. 4 2. 5 3. -4 4. -5 Option 2 : 5 #### Surds and Indices Question 3 Detailed Solution Formula Used: a0 = 1 a-1 = 1/a a1 = a am/an = am - n (am)n = amn am × an = am + n If am = an Then m = n Calculation: $${\left( {\frac{2}{7}} \right)^{ - 3}}X{\left( {\frac{4}{{49}}} \right)^6} = {\left( {\frac{2}{7}} \right)^{2m - 1}}$$ $$⇒ {\left( {\frac{2}{7}} \right)^{ - 3}}X{\left( {\frac{2}{{7}}} \right)^(12)× 2} = {\left( {\frac{2}{7}} \right)^{2m - 1}}$$ ⇒ (2/7)-3 + 12= (2\7)2m - 1 Now, ⇒ 9 = 2m - 1 ⇒ 10 = 2m ⇒ m = 5 ∴ he value of m is 5. The correct option is 2 i.e. 5 #### Surds and Indices Question 4: What will come in the place of the question mark ‘?’ in the following question? 112 + 8+ 132 = ? - 10- 200% of 235 1. 924 2. 703 3. 803 4. 810 5. 713 Option 1 : 924 #### Surds and Indices Question 4 Detailed Solution Given: 112+ 82 + 13= ? - 102 - 200% of 235 Follow the BODMAS rule to solve this question, as per the order given below: Calculation: 112+ 82 + 13= ? - 102 - 200% of 235 ⇒ 121 + 64 + 169 = ? - 100 - 470 ⇒ 185 + 169 = ? - 570 ⇒ 354 = ? - 570 ⇒ ? = 570 + 354 ⇒ ? = 924 Hence, 924 will come in the place of '?'. #### Surds and Indices Question 5: The value of $$\rm \frac{3\sqrt2}{\sqrt6-\sqrt3}-\frac{{4\sqrt 3 }}{{\sqrt {8 - 2\sqrt {12} } }}+\frac{3}{2(3-\sqrt{8})}$$ is closet to: 1. 9 2. 8.8 3. 8 4. 7.8 Option 3 : 8 #### Surds and Indices Question 5 Detailed Solution Given: $$\rm \frac{3√2}{√6-√3}-\frac{{4√ 3 }}{{√ {8 - 2√ {12} } }}+\frac{3}{2(3-√{8})}$$ Concept used: √3 = 1.732 Calculation: $$\rm \frac{3√2}{√6-√3}-\frac{{4√ 3 }}{{√ {8 - 2√ {12} } }}+\frac{3}{2(3-√{8})}$$ ⇒ $$\frac {3√2 × (√6 + √3)}{(√6 - √3) (√6 + √3)}$$ - $$\frac {4√3}{√ {6 + 2 - 2√6 √2}}$$ + $$\frac {3 × (3 + √8)} {2 (3 - √8) × (3 + √8)}$$ ⇒ $$\frac {3√2 × (√6 + √3)}{6 - 3}$$ - $$\frac {4√3} {√{(√6 - √2)^2}}$$ + $$\frac {3 × (3 + √8)} {2 (9 - 8)}$$ ⇒ $$\frac {3√2 × (√6 + √3)}{6 - 3}$$ - $$\frac {4√3} {{(√6 - √2)}}$$ + $$\frac {3 × (3 + √8)} {2 (9 - 8)}$$ ⇒ $$2√3 + √6$$ - $$\frac {4√3 × (√6 + √2)} {{(√6 - √2) × (√6 + √2)}}$$ + 9/2 + $$3√2$$ ⇒ $$2√3 + √6$$ - $$\frac {4√3 × (√6 + √2)} {{(√6 - √2) × (√6 + √2)}}$$ + 4.5 + $$3√2$$ ⇒ $$2√3 + √6$$ - $$\frac {4√3 × (√6 + √2)} {6 - 2}$$ + 4.5 + $$3√2$$ ⇒ $$2√3 + √6$$ - $$3√2 - √ 6$$ + 4.5 + $$3√2$$ ⇒ 2√3 + 4.5 ⇒ 2 × 1.732 + 4.5 ⇒ 7.964 ≈ 8 ∴ The simplified value is 8. ## Top Surds and Indices MCQ Objective Questions #### Surds and Indices Question 6 What is the square root of (8 + 2√15)? 1. √5 + √3 2. 2√2 + 2√6 3. 2√5 + 2√3 4. √2 + √6 Option 1 : √5 + √3 #### Surds and Indices Question 6 Detailed Solution Formula used: (a + b)2 = a2 + b2 + 2ab Calculation: Given expression is: $$\sqrt {8\; + \;2\sqrt {15} \;}$$ $$\sqrt {5\; + \;3\; + \;2\sqrt 5 \sqrt 3 \;}$$ $$\sqrt {{{(\sqrt 5 )}^2}\; + \;{{\left( {\sqrt 3 } \right)}^2}\; + \;2\sqrt 5 \sqrt 3 \;}$$ $$\sqrt {{{\left( {\;\sqrt 5 \; + \;\sqrt 3 \;} \right)}^2}\;}$$ = $$\sqrt 5 + \sqrt 3$$ #### Surds and Indices Question 7 The square root of ((10 + √25)(12 – √49)) is: 1. 4√3 2. 3√3 3. 5√3 4. 2√3 Option 3 : 5√3 #### Surds and Indices Question 7 Detailed Solution Concept: We can find √x using the factorisation method. Calculation: √[(10 + √25) (12 - √49)] ⇒ √[(10 + 5)(12 – 7)] ⇒ √(15 × 5) ⇒ √(3 × 5 × 5) ⇒ 5√3 #### Surds and Indices Question 8 Find the value of x: 23 × 34 × 1080 ÷ 15 = 6x 1. 4 2. 6 3. 8 4. 2 Option 2 : 6 #### Surds and Indices Question 8 Detailed Solution Given, 23 × 34 × 1080 ÷ 15 = 6x ⇒ 23 × 34 × 72 = 6x ⇒ 23 × 34 × (2 × 62) = 6x ⇒ 24 × 34 × 62 = 6x ⇒ (2 × 3)4 × 62 = 6x [∵ xm × ym = (xy)m] ⇒ 64 × 62 = 6x ⇒ 6(4 + 2) = 6x ⇒ x = 6 #### Surds and Indices Question 9 If (3 + 2√5)2 = 29 + K√5, then what is the value of K? 1. 12 2. 6 3. 29 4. 39 Option 1 : 12 #### Surds and Indices Question 9 Detailed Solution Method I: (3 + 2√5)2 = (32 + (2√5)2 + 2 × 3 × 2√5) = 9 + 20 + 12√5 = 29 + 12√5 On comparing, 29 + 12√5 = 29 + K√5 we get, K = 12 Alternate Method 29 + 12√5 = 29 + K√5 ⇒ K√5 = 29 - 29 + 12√5 ⇒ K√5 = 12√5 ∴ K = 12 #### Surds and Indices Question 10 Simplify: $$\sqrt {11 - 2\sqrt {30} }$$ 1. $$\sqrt 6 + \sqrt 5$$ 2. 6 3. $$\sqrt 6 - \sqrt 5$$ 4. $$6 - \sqrt 5$$ Option 3 : $$\sqrt 6 - \sqrt 5$$ #### Surds and Indices Question 10 Detailed Solution $$\begin{array}{l} \sqrt {11 - 2\sqrt {30} } \\ = \sqrt {\left( {11} \right) - 2\sqrt 6 \times \sqrt 5 } \\ = \sqrt {\left( {6 + 5} \right) - 2\sqrt 6 \times \sqrt 5 } \\ = \sqrt {{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt 5 } \right)}^2} - 2\sqrt 6 \times \sqrt 5 } \\ = \sqrt {{{\left( {\sqrt 6 - \sqrt 5 } \right)}^2}} \\ = \sqrt 6 - \sqrt 5 \end{array}$$ #### Surds and Indices Question 11 Which of the following statement(s) is/are TRUE? I. 2√3 > 3√2 II. 4√2 > 2√8 1. Only I 2. Only II 3. Neither I nor II 4. Both I and II Option 3 : Neither I nor II #### Surds and Indices Question 11 Detailed Solution Statement I: 2√3 > 3√2 To Check either above given relation is correct or not, simplifying by squaring on both the sides. ⇒ (2√3)2 > (3√2)2 ⇒ 12 > 18 which is not true, as we know that 18 is greater than 12. So, the given relation in statement I is not true. Statement II: Now, simplifying the values given in statement II (Note: 2√8 = 2√(4 × 2) = 4√2) 4√2 > 2√8 on taking the square root from the right hand side. ⇒ 4√2 > 2 × 2√2 ⇒ 4√2 > 4√2 which is not true, as the the value on left hand side is equal to the value on right hand side. So, the given relation in statement II is also not true. ∴ Neither statement I nor statement II is true. #### Surds and Indices Question 12 If (3/5)x = 81/625, then what is the value of xx ? 1. 16 2. 256 3. 0 4. 32 Option 2 : 256 #### Surds and Indices Question 12 Detailed Solution Given: (3/5)x = 81/625 Calculation: We know, 34 = 81 and 54 = 625 ⇒ (3/5)4 = 81/625 (3/5)x = 81/625 ∴ On comparing both the equation, we get x = 4 Now, xx = 44 = 256 #### Surds and Indices Question 13 Simplify: $${625^{0.17}} \times {625^{0.08}} = {25^?} \times {25^{ - \frac{3}{2}}}$$ 1. 1 2. 2 3. 3 4. 0.5 Option 2 : 2 #### Surds and Indices Question 13 Detailed Solution To solve questions of this type, follow the laws of “Surds and indices’’ given below: Laws of Indices: 1. am × an = a{m + n} 2. am ÷ an = a{m - n} 3. (am)n = amn 4. (a)-m = 1/am 5. (a)m/n = n√am 6. (a)0 = 1 $${625^{0.17}} \times {625^{0.08}} = {25^?} \times {25^{- \frac{3}{2}}}$$ $$\Rightarrow {625^{0.17\; + \;0.08}} = {25^{? + (- \frac{3}{2})}}$$ $$\Rightarrow {625^{0.25}} = {25^{? - \frac{3}{2}}}$$ $$\Rightarrow {625^{\frac{1}{4}}} = {\left( {{5^2}} \right)^{? - \frac{3}{2}}}$$ $$\Rightarrow 5 = {5^{2 \times? - 3}}$$ ⇒ 2 × ? - 3 = 1 ⇒ ? = (1 + 3)/2 ∴ ? = 2 #### Surds and Indices Question 14 If A = 83 × 54 and B = 85 × 53, then what is the value of A × B? 1. 216 × 58 2. 824 × 57 3. 424 × 57 4. 224 × 57 Option 4 : 224 × 57 #### Surds and Indices Question 14 Detailed Solution A = 83 × 5and B = 85 × 53 ⇒ A × B = 83 × 54 × 85 × 53 ⇒ A × B = 23(3 + 5) × 5(4 +3) ⇒ A × B = 224 × 57 #### Surds and Indices Question 15 If 2x = 4y = 8z and $$\frac{1}{2x}+\frac{1}{4y}+\frac{1}{4z}=4$$, then the value of x is: 1. $$\frac{7}{16}$$ 2. $$\frac{7}{17}$$ 3. $$\frac{7}{19}$$ 4. $$\frac{7}{23}$$ Option 1 : $$\frac{7}{16}$$ #### Surds and Indices Question 15 Detailed Solution Given: 2x = 4y = 8z $$\frac{1}{2x}+\frac{1}{4y}+\frac{1}{4z}=4$$ Calculation: $$\frac{1}{2x}+\frac{1}{4y}+\frac{1}{4z}=4$$---- (1) 2x = 4y = 8z ⇒ 2x = 22y = 23z ⇒ x = 2y = 3z Converting y and z in x 2y = x, so 4y = 2x 3z = x, so 4z = 4x/3 Using the above value in equation (1) ⇒ $$\frac{1}{2x}+\frac{1}{2x}+\frac{3}{4x}=4$$ ⇒ 7/4x = 4 ∴ x = 7/16	{}
# Vector Calculus • August 4th 2007, 10:57 AM physics4life Vector Calculus hi this is about vector calculus subscript notation: i generally understand v.c.s.n but can't do something like this: Show : (AXB) . (CXD) = (A.C)(B.D) - (A.D)(B.C) my work so far: $(AXB)i = \varepsilon_{ijk}A_jB_k$ $(CXD)i = \varepsilon_{ijk}C_jD_k$ so (AXB) . (CXD) = $(\varepsilon_{ijk}A_jB_k)_i . (\varepsilon_{ijk}C_jD_k)_i$ = $\varepsilon_{ijk}\varepsilon_{ijk}A_jC_jB_kD_k$ = $( \delta_{ij} \delta_{jk} - \delta_{ik}\delta_{jj})(A.C)_j(B.D)k$ = $[ \delta_{ij}(A.C)_j ][ \delta_{jk}(B.D)_k ] - [ \delta_{ik}(B.D)_k ][ \delta_{jj}(A.C)_j ]$ = $(A.C)_i(B.D)j - (B.D)_i(A.C)_j$ =(A.C)(B.D) - (B.D)(A.C) the first part of the answer (in red) i got right.. but the 2nd part is wrong as you can see how am i meant to get -(A.D)(B.C)??? • August 4th 2007, 11:10 AM topsquark Quote: Originally Posted by physics4life hi this is about vector calculus subscript notation: i generally understand v.c.s.n but can't do something like this: Show : (AXB) . (CXD) = (A.C)(B.D) - (A.D)(B.C) my work so far: $(AXB)i = \varepsilon_{ijk}A_jB_k$ $(CXD)i = \varepsilon_{ijk}C_jD_k$ so (AXB) . (CXD) = $(\varepsilon_{ijk}A_jB_k)_i . (\varepsilon_{ijk}C_jD_k)_i$ Here's the problem: you are using jk indices for the second sum. And you have the epsilon-epsilon identity wrong. Here's what you should do: $(A \times B) \cdot (C \times D)$ $\to (\varepsilon_{ijk}A_jB_k)(\varepsilon_{imn}C_mD_n)$ $= \varepsilon_{ijk} \varepsilon_{imn}A_jB_kC_mD_n$ $= (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km})A_jB_kC_mD_n$ $= \delta_{jm} \delta_{kn}A_jB_kC_mD_n - \delta_{jn} \delta_{km}A_jB_kC_mD_n$ $= A_jB_kC_jD_k - A_jB_kC_kD_j$ $\to (A \cdot C)(B \cdot D) - (A \cdot D)(B \cdot C)$ -Dan • August 4th 2007, 12:35 PM physics4life cheers dan, your method is easy to understand. it's frustrating trying to learn all this calculus on your own when there is just a huge pile of books and you're forever researching. so i appreciate your help. just a question though, on the 2nd epsilon identity you used: $\epsilon_{imn}$ what i wanted to ask is: do you have to use the imn notation.. or can you use any preferred notation.. such as ixy.... i understand the final result will still be the same as long as the notation is consistent throughout the solution, but i just wanted to confirm whether imn has to be used as a standard for vector calculus. thanks once again .:o • August 4th 2007, 01:53 PM topsquark Quote: Originally Posted by physics4life cheers dan, your method is easy to understand. it's frustrating trying to learn all this calculus on your own when there is just a huge pile of books and you're forever researching. so i appreciate your help. just a question though, on the 2nd epsilon identity you used: $\epsilon_{imn}$ what i wanted to ask is: do you have to use the imn notation.. or can you use any preferred notation.. such as ixy.... i understand the final result will still be the same as long as the notation is consistent throughout the solution, but i just wanted to confirm whether imn has to be used as a standard for vector calculus. thanks once again .:o You can use any indices you like. (You can even make up your own, but they're much harder to type. :p ) The point is that they are different from the ones you used previously since they have no relation to the indices in the other cross product. By the way, a nice way to remember that epsilon-epsilon identity: $\varepsilon_{ijk} \varepsilon_{imn} = \delta_{jm} \delta_{jm} - \delta_{jn} \delta_{km}$ Note that the "i" index doesn't appear anywhere in the deltas. Now look at the first delta of each pair. They both have a "j," the first index after the "i" in the first epsilon. The second index of those first deltas is in sequence, m and n... the second and third indices of the second epsilon. -Dan • August 4th 2007, 03:23 PM physics4life Quote: Originally Posted by topsquark By the way, a nice way to remember that epsilon-epsilon identity: $\varepsilon_{ijk} \varepsilon_{imn} = \delta_{jm} \delta_{jm} - \delta_{jn} \delta_{km}$ Note that the "i" index doesn't appear anywhere in the deltas. Now look at the first delta of each pair. They both have a "j," the first index after the "i" in the first epsilon. The second index of those first deltas is in sequence, m and n... the second and third indices of the second epsilon. -Dan hold on a minute. if we have a standard epsilon-epsilon identity.. like $\varepsilon_{ijk} \varepsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$, then surely the identity you used would be: $\varepsilon_{ijk} \varepsilon_{imn} = \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}$? but yours is different. is there something i missed while learning this topic? • August 4th 2007, 03:32 PM topsquark Quote: Originally Posted by physics4life hold on a minute. if we have a standard epsilon-epsilon identity.. like $\varepsilon_{ijk} \varepsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$, then surely the identity you used would be: $\varepsilon_{ijk} \varepsilon_{imn} = \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}$? but yours is different. is there something i missed while learning this topic? The two are the same. Note that in my statement the common index in the epsilons is the first. Rearranging your expression by noting that $\varepsilon_{kij} = \varepsilon_{ijk}$ you will find we have the same expressions. -Dan	{}
## 3 Shocking Ways to Break Mathematics This post provides formal proofs of three paradoxes that occur in type systems with unrestricted recursion, negative data types, or type in type (Girard’s paradox). The title is a pun on BuzzFeedy headlines. The source code of the paradoxes is available for download. An important aspect of type theory is identifying the values of a given type. For example: the values of type $\mathbb{N}$ are the natural numbers $0, 1, 2, \dots$; the values of type $\text{bool}$ are $\text{true}$ and $\text{false}$. This blog post explains ways in which many languages add spurious values to types. In Java, for example, there are at least 3 values of type boolean1. A type system that avoids spurious values has several advantages. For example: 1) A type system can provide guarantees about a program. For example, a type system can guarantee that the function $\text{square}(n : \mathbb{N}) := n^2$ is only ever invoked with a natural number (not e.g. a string) and thus that $n^2$ will never fail. Spurious values can break these guarantees. 2) A type system can enable mathematical reasoning using the Curry-Howard Correspondence. Using this correspondence, true propositions are represented by inhabited types (i.e. at least one value is of the type), and false propositions are represented by uninhabited types (i.e. no value is of the type). A spurious value in a type intended to be uninhabited (e.g. False), leads to a true proposition that was intended to be false — a paradox. The examples of spurious values in this blog post are based on divergence, and are thus mostly interesting for mathematical reasoning (the second advantage). ### Unrestricted Recursion Most programming languages allow some form of unrestricted recursion or loops, which can usually be exploited to construct a value of an arbitrary type. For example, the Java expression f() (defined below) is of type boolean but neither true nor false. public static boolean f() { return f(); } The trick is that f never actually has to generate a boolean value, because it delegates that responsibility to itself by calling f. By delegating all the work to itself though, f never gets anything done. Every typed mainstream programming language has this problem (including Haskell and ML). While this seems bad, it’s not quite as bad as you might expect initially. In Java, a function g that takes a boolean is still guaranteed to only ever be invoked with a value that is either true or false (unless Java is broken in some other way). The reason is that g(f()) never actually calls the function g, because f never stops executing. If you want to use Java’s type system for mathematical reasoning though, you will be pleased to hear (sarcasm) that every single proposition will be true, because there is a value in every type. Languages can avoid these kinds of spurious values/paradoxes, for example by using inductive data types and eliminators instead of unrestricted recursion. ### Negative Data Types Functional programming languages like Haskell and ML provide algebraic/inductive data types. These can be exploited to construct a value of an arbitrary type. For example, the expression false below inhabits the inductive type False which was defined to have no constructors. data False data Rec a = Rec { rec :: Rec a -> a } false :: False false = rec (Rec (\a -> rec a a)) (Rec (\a -> rec a a)) Note that false diverges because it reduces to (\a -> rec a a) (Rec (\a -> rec a a)) which reduces back to false. This program is inspired by Adam Chlipala’s book CPDT, chapter 3.6. Languages can avoid these kinds of spurious values/paradoxes by restricting the shape of an inductive type T. Specifically, the type of a constructor of T may not refer to T on the left side of a function arrow. This restriction is called strict positivity. ### Type in Type Most typed languages allow functions that take terms as arguments, for example not (b:bool) := if b then false else true. More advanced type system also allow functions that take types are arguments, for example id (A:Type) (a:A) := a. In such languages, the question arises what the type of Type should be. Initial suggestions were for Type to be of type Type, i.e. Type : Type, until Girard proved that this is paradoxical. The following proof is based on Per Martin-Löf’s proof of Girard’s paradox in An Intuitionistic Theory of Types, 1972. Note that this problem only arises in languages with fairly complex type systems. Many languages simply do not allow a programmer to ask for the type of Type. In Java for example, the function static <A> A id(A a) { return a; } takes the type A as argument, but because there is the special syntax <A> Java gets away without having to type the expression Type. This proof will define a notion of ordering, will show that orderings can themselves be ordered, will show that all orderings are smaller than the ordering of orderings, and will conclude that this is a contradiction. Don’t despair if all this talk of orderings of orderings makes your head hurt. This is the expected behavior — your brain has evolved to avoid wasting resources on thinking diverging thoughts. This paradox relies heavily on the notion of an ordering, which requires that we first become familiar with transitivity and infinite chains. Definition transitive {A:Type} R := forall (x y z:A), R x y -> R y z -> R x z. Definition infiniteChain {A:Type} R (f:nat->A) := forall n, R (f (S n)) (f n). Given a binary relation over A, a function f is called an infinite chain if it assigns to each natural number n a value f n in A, such that the value assigned to the next number f (S n) is related to f n. For example, the identity function id is an infinite ascending chain of the natural numbers. Lemma infiniteAscendingChainNat : infiniteChain gt id. unfold infiniteChain, id. intro n. omega. Qed. Yet, it is impossible to create an infinite descending chain of the natural numbers. Consider for example the failed attempt of assigning some number m to f 0 and then subtracting n from m at f n. This fails because for n >= m the subtraction m - n is defined to be 0. Note that there exist infinite descending chains for the integers and real numbers. Lemma infiniteDescendingChainNat m : infiniteChain lt (fun n => m - n) -> False. unfold infiniteChain. intro h. specialize (h m). omega. Qed. We define an ordering as a set together with a “less than” relation that is transitive and contains no infinite descending chains (note that the ordering is not required to be total). Note that ordering could be defined using existentials instead of a record. Class ordering : Type := { set : Type; lt : set -> set -> Type; trans : transitive lt; noChains : forall f, infiniteChain lt f -> False }. Notation "x < y" := (lt x y). Such an ordering is necessarily irreflexive. Lemma irreflexive {ordering} a : a < a -> False. intro h. apply (noChains (fun _ => a) (fun _ => h)). Qed. All the paradoxes we have seen so far are based on recursion. It should come to nobody’s surprise that this paradox will go the same way. Therefore, we now define the ordering of orderings. An ordering o is less than an ordering O, if and only if there exists an order preserving map that takes every element in o’s set to an element in O’s set, and there exists a bound in O’s set that is larger than the mapping of every element in o’s set. Class orderingLe (o:ordering) (O:ordering) : Type := { map : @set o -> @set O; mapOk x y : x < y -> map x < map y; bound : @set O; boundOk x : map x < bound }. Notation "x << y" := (orderingLe x y) (at level 45). Pictorially, we will represent two ordered orderings o and O as: To show that << is an ordering, we have to show that << is transitive. Lemma orderingLeTransitive : transitive orderingLe. compute. intros o o' o'' ? ?. refine {\| map := map ∘ map; bound := bound \|}. + intros. apply mapOk. apply mapOk. trivial. + intro x. apply boundOk. Qed. The new map of the ordering is the functional composition of the other orderings’ maps, the new bound is the larger ordering’s bound. This new map is clearly order preserving, and the new bound a bound. Pictorially: Further, we also have to show that << contains no infinite descending chains. To this end, it is sufficient to show that an infinite descending chain f of orderings implies the existence of an infinite descending chain g in the ordering f 0 (which is a contradiction). Definition chainImpliesChain f : infiniteChain orderingLe f -> {g:nat -> set & infiniteChain (@lt (f 0)) g}. intro h. unfold infiniteChain in . refine ( let fix rec n' := match n' as n'' return @set (f n'') -> @set (f 0) with \| 0 => id \| S n'' => rec n'' ∘ map end in _). refine (existT _ _ _). + exact (fun n => rec n (@bound _ _ (h n))). + cbn. intro n. enough (forall x y, x < y -> rec n x < rec n y) as h'. { apply h'. apply boundOk. } intros x y. induction n as [\|n IHn]; cbn in . * trivial. * intro h'. apply IHn. apply mapOk. apply h'. Qed. Lemma orderingLeNoChains f : infiniteChain orderingLe f -> False. intro h. destruct (chainImpliesChain f h) as [g h']. refine (@noChains (f 0) g h'). Qed. The infinite descending chain g assigns to n the bound of the ordering f n mapped to an element in the set of the ordering f 0. i.e. g 0 := bound, g 1 := map bound, g 2 := map (map bound), g 3 := map (map (map bound)), etc. g is an infinite descending chain because the maps are order preserving, and the objects in f (S n) are bounded by the bound of f n. This can be represented pictorially. Note that the y-axis’s < is flipped, i.e. objects on the bottom are larger than objects on the top. We are now ready to define the ordering of orderings. To avoid paradoxes exactly like the one we are currently defining, Coq does not allow this definition. Thankfully, the flag -type-in-type makes Coq more lenient. Instance orderingOfOrderings : ordering. refine {\| set := ordering; lt := orderingLe \|}. + apply orderingLeTransitive. + apply orderingLeNoChains. Defined. We will need the notion of a sub-ordering of the ordering o. A sub-ordering is defined to be a subset of o’s set containing all the elements of o smaller than a bound a. A sub-ordering has the same ordering relation a o, and is thus trivially transitive and contains no infinite descending chains. Instance subOrdering (o:ordering) (a:@set o): ordering. refine {\| set := {x:@set o & x < a}; lt := fun x y => projT1 x < projT1 y \|}. + intros [x ?] [y ?] [z ?]. cbn. apply trans. + intros f h. apply (noChains (fun n => projT1 (f n))). intro n. specialize (h n). apply h. Defined. A sub-ordering is always smaller than the original ordering, and a sub-ordering with a smaller bound a is smaller than a sub-ordering with a larger bound b. Lemma subOrderingIsLt {o:ordering} : forall a, subOrdering o a << o. intro a. cbn. refine {\| map := _; bound := a \|}. * exact (fun x => projT1 x). * intros [x ?] [y ?]. cbn. exact (fun a => a). * intros [x h]. cbn. apply h. Qed. Lemma subOrderingOrderPreserving {o:ordering} a b : a < b -> subOrdering o a << subOrdering o b. intros h. refine {\| map := _; mapOk := _; bound := _; boundOk := _ \|}. * refine (fun x => existT _ (projT1 x) _). specialize (projT2 x); intro h'. specialize trans. firstorder. * intros [] []. cbn in . trivial. refine (existT _ a _). trivial. * cbn in *. intros []. cbn. trivial. Qed. The next step is to show that the ordering of orderings is larger than all orderings o. This is done by mapping all elements of o into a sub-ordering. Lemma orderingOfOrderingsBound o : o < orderingOfOrderings. cbn. refine {\| map := fun a => subOrdering o a; bound := o \|}. + apply subOrderingOrderPreserving. + apply subOrderingIsLt. Qed. Pictorially: We just showed that the order of orderings is larger than all orderings (including itself), and we also showed that orderings are irreflexivity. This is clearly a paradox. Lemma orderingOfOrderingsReflexive : orderingOfOrderings < orderingOfOrderings. apply orderingOfOrderingsBound. Qed. Theorem false : False. apply (@irreflexive orderingOfOrderings orderingOfOrderings). apply orderingOfOrderingsReflexive. Qed. The paradox is caused by the self-referential nature of the ordering of orderings. In general, the definition of a type is called impredicative if it involves a quantifier whose domain includes the type currently being defined (see TaPL, chapter 23). A predicative type system enforces that such self-referential definitions cannot be made. Predicative type system are usually implemented using stratification or ramification, where Type is indexed by a natural number and Type n : Type (n + 1), e.g. Type 0 : Type 1 : Type 2. Stratification is also the trick that Haskell uses to avoid this paradox (it’s not like that matters, given that the other two paradoxes are still possible). Instead of having an infinite hierarchy of types though, Haskell opts for just a finite number of them (the highest being kinds). ### Conclusion The types of most languages contain spurious values and are thus not useful for mathematical reasoning using the Curry-Howard Correspondence. Languages free of all known paradoxes are for example Coq, Agda, NuPRL, and Lean. #### Footnotes 1. Strictly speaking, type theory is concerned with identifying all the terms of a given type, and then partitioning them into equivalence classes according to their normalization behavior (e.g. 5 and 2 + 3 are equivalent). Thus, Java’s boolean type contains at least three equivalence classes. The first contains for example true, true \|\| false, and if true then true else false; the second contains false, true && false, and if false then true else false; and the third contains f() which normalizes to neither true nor false`.	{}
# Energy Radiating Off of a Cube Split Into Two Pieces This is somewhat a Physics question, but ends of being more of an algebra/geometry-related question. Anyway, there is a cube that radiates with a power $P_0$ and it is "cut" into two pieces. These two pieces radiate energy with a power of $P_1$. Find the ratio of $P_1$/$P_0$. Heat transfer for radiation is defined as P = $\delta$Q/$\delta$T = e $\sigma$ A $T^4$ Everything is constant between the two cubes (except for their total surface area, obviously) The ratio ends up being $P_1$/$P_0$ = $A_1$/$A_0$ My problem is that it doesn't given any details about the size before or after the cube is cut. I tried every geometric comparison I could think of, I always ended up with 1/2 or 2. The answer supplied is 4/3. Any hints? - I kind of makes sense that $P_1 > P_0$ such that the ratio is large than 1. Don't you think? – Fabian Apr 26 '12 at 5:05 What is $P$? What is $\delta$? What is $Q$? What is $T$? What is e? What is $\sigma$? What is $A$? Don't worry, I know what 4 is. – Gerry Myerson Apr 26 '12 at 5:54 @Gerry, I only know $T$ is temperature but all that matters is the ratio of areas since everything else is constant. – Dan Brumleve Apr 26 '12 at 5:58 The answer is only correct when the cube is cut into "boxes." If you slice down along a diagonal of one of the faces, the ratio is $5:3$. And there are many other possibilities for cutting. Some, like cutting off a tiny corner, make no appreciable difference. – André Nicolas Apr 26 '12 at 6:21	{}
# A fast and accurate coalescent approximation Blog author Suyash Shringarpure is a postdoc in Carlos Bustamante’s lab. Suyash is interested in statistical and computational problems involved in the analysis of biological data. The coalescent model is a powerful tool in the population geneticist’s toolbox. It traces the history of a sample back to its most recent common ancestor (MRCA) by looking at coalescence events between pairs of lineages. Starting from assumptions of random mating, selective neutrality, and constant population size, the coalescent uses a simple stochastic process that allows us to study properties of genealogies, such as the time to the MRCA and the length of the genealogy, analytically and through efficient simulation. Extensions to the coalescent allow us to incorporate effects of mutation, recombination, selection and demographic events in the coalescent model. A short introduction to the coalescent model can be found here and a longer, more detailed introduction can be read here. However, coalescent analyses can be slow or suffer from numerical instability, especially for large samples. In a study published earlier this year in Theoretical Population Biology, CEHG fellow Ethan Jewett and CEHG professor Noah Rosenberg proposed fast and accurate approximations to general coalescent formulas and procedures for applying such approximations. Their work also examined the asymptotic behavior of existing coalescent approximations analytically and empirically. ## Computational challenges with the coalescent For a given sample, there are many possible genealogical histories, i.e., tree topologies and branch lengths, which are consistent with the allelic states of the sample. Analyses involving the coalescent therefore often require us to condition on a specific genealogical property and then sum over all possible genealogies that display the property, weighted by the probability of the genealogy. A genealogical property that is often conditioned on is $n_t$, the number of ancestral lineages in the genealogy at a time $t$ in the past. However, computing the distribution $P(n_t)$ of $n_t$ is computationally expensive for large samples and can suffer from numerical instability. ## A general approximation procedure for formulas conditioning on $n_t$ Coalescent formulas conditioning on $n_t$ typically involve sums of the form $f(x)=\sum_{n_t} f(x\|n_t) \cdot P(n_t)$ For large samples and recent times, these computations have two drawbacks: – The range of possible values for $n_t$ may be quite large (especially if multiple populations are being analyzed) and a summation over these values may be computationally expensive. – Expressions for $P(n_t)$ are susceptible to round-off errors. Slatkin (2000) proposed an approximation to the summation in $f(x)$ by a single term $f(x\|E[n_t])$. This deterministic approximation was based on the observation that $n_t$ changes almost deterministically over time, even though it is a stochastic variable in theory. Thus we can write $n_t \approx E[n_t]$. From Figure 2 in the paper (reproduced here), we can see that this approximation is quite accurate. The authors prove the asymptotic accuracy of this approximation and also prove that under regularity assumptions, $f(x\|E[n_t])$ converges to $f(x)$ uniformly in the limits of $t \rightarrow 0$ and $t \rightarrow \infty$ . This is an important result since it shows that the general procedure produces a good approximation for both very recent and very ancient history of the sample. Further, the paper shows how this method can be used to approximate quantities that depend on the trajectory of $n_t$ over time, which can be used to calculate interesting quantities such as the expected number of segregating sites in a genealogy. ## Approximating $E[n_t]$ for single populations A difficulty with using the deterministic approximation is that $E[n_t]$ often has no closed-form formula, and if one exists, it is typically not easy to compute when the sample is large. For a single population with changing size, two deterministic approximations have previously been developed (one by Slatkin and Rannala 1997, Volz et al. 2009 and one by Frost and Volz, 2010, Maruvka et al., 2011). Using theoretical and empirical methods, the authors examine the asymptotic behavior and computational complexity of these approximations and a Gaussian approximation by Griffiths. A summary of their results is in the table below. Method Accuracy Griffith’s approximation Accurate for large samples and recent history. Slatkin and Rannala (1997), Volz et al. (2009) Accurate for recent history and arbitrary sample size, inaccurate for very ancient history. Frost and Volz (2010), Maruvka et al. (2011) Accurate for both recent and ancient history and for arbitrary sample size. Jewett and Rosenberg (2014) Accurate for both recent and ancient history and arbitrary sample size, and for multiple populations with migration. ## Approximating $E[n_t]$ for multiple populations Existing approaches only work for single populations of changing size and cannot account for migration between multiple populations. Ethan and Noah extend the framework for single populations to allow multiple populations with migration. The result is a system of simultaneous differential equations, one for each population. While it does not allow for analytical solutions except in very special cases, the system can be easily solved numerically for any given demographic scenario. ## Significance of this work The extension of the coalescent framework to multiple populations with migration is an important result for demographic inference. The extended framework with multiple populations allows efficient computation of demographically informative quantities such as the expected number of private alleles in a sample, divergence times between populations. Ethan and Noah describe a general procedure that can be used to approximate coalescent formulas that involve summing over distributions conditioned on $n_t$ or the trajectory of $n_t$ over time. This procedure is particularly accurate for studying very recent or very ancient genealogical history. The analysis of existing approximations to $E[n_t]$ show that different approximations have different asymptotic behavior and computational complexities. The choice of which approximation to use is therefore often a tradeoff between the computational complexity of the approximation and the likely behavior of the approximation in the parameter ranges of interest. ## Future Directions As increasingly large genomic samples from populations with complex demographic histories become available for study, exact methods either become intractable or very slow. This work adds to a growing set of approximations to the coalescent and its extensions, joining other methods such as conditional sampling distributions and the sequentially markov coalescent. Ethan and Noah are already exploring applications of these approximate methods to reconciling gene trees with species trees. In the future, I expect that these and other approximations will be important for fast and accurate analysis of large genomic datasets. ## References [1] Jewett, E. M., & Rosenberg, N. A. (2014). Theory and applications of a deterministic approximation to the coalescent model. Theoretical population biology. [2] Griffiths, R. C. (1984). Asymptotic line-of-descent distributions. Journal of Mathematical Biology21(1), 67-75. [3] Frost, S. D., & Volz, E. M. (2010). Viral phylodynamics and the search for an ‘effective number of infections’. Philosophical Transactions of the Royal Society B: Biological Sciences365(1548), 1879-1890. [4] Maruvka, Y. E., Shnerb, N. M., Bar-Yam, Y., & Wakeley, J. (2011). Recovering population parameters from a single gene genealogy: an unbiased estimator of the growth rate. Molecular biology and evolution28(5), 1617-1631. [5] Slatkin, M., & Rannala, B. (1997). Estimating the age of alleles by use of intraallelic variability. American journal of human genetics60(2), 447. [6] Slatkin, M. (2000). Allele age and a test for selection on rare alleles.Philosophical Transactions of the Royal Society of London. Series B: Biological Sciences355(1403), 1663-1668. [7] Volz, E. M., Pond, S. L. K., Ward, M. J., Brown, A. J. L., & Frost, S. D. (2009). Phylodynamics of infectious disease epidemics. Genetics183(4), 1421-1430. Paper author Ethan Jewett is a PhD student in the lab of Noah Rosenberg. # Demographic inference from genomic data in nonmodel insect populations Blog author Martin Sikora is a postdoc in the lab of Carlos Bustamante. Reconstructing the demographic history of species and populations is one of the major goals of evolutionary genetics. Inferring the timing and magnitude of past events in the history of a population is not only of interest in its own right, but also in order to form realistic null models for the expected patterns of neutral genetic variation in present-day natural populations. A variety of methods exist that allow the inference of these parameters from genomic data, which, in the absence of detailed historical records in most situations, is often the only feasible way to obtain them. As a consequence, it is generally not possible to empirically validate the parameters inferred from genomic data in a direct comparison with a known “truth” from a natural population. Furthermore, until recently, the application of these methods was limited to model organisms with well-developed genomic resources (e.g., humans and fruitflies), excluding a large number of non-model organisms with potentially considerable evolutionary and ecological interest. ## Chasing butterflies? In an elegant study recently published in the journal Molecular Ecology, Rajiv McCoy, a graduate student with Dmitri Petrov and Carol Boggs, and colleagues tackle both of these problems in natural populations of Euphydryas gillettii, a species of butterfly native to the northern Rocky Mountains. About 30 years ago, a small founder population of this species from Wyoming was intentionally introduced to a new habitat at the Rocky Mountain Biological Laboratory field site in Colorado, and population sizes were recorded every year since the introduction. The beauty of this system is that it allows the authors to perform a direct comparison of the known demography (i.e. a recent split from the parental population and bottleneck ~30 generations ago, with census data in the newly introduced population) with estimates inferred from genomic data. Gillete’s Checkerspot (Euphydryas gillettii). Photo taken by Carol Boggs, co-advisor of Rajiv and one of the senior authors of the study. ## A genomic dataset from a non-model organism The researchers sampled eight larvae each from both the parental as well as the derived population for this study. In the world of model organisms, the next steps for constructing the dataset would be straightforward: Extract genomic DNA, sequence to the desired depth, map to the reference genome and finally call SNPs. In the case of E. gillettii however, no reference genome is available, so the authors had to use a different strategy. They decided to use RNA-sequencing in order to first build a reference transcriptome, which was then used as a reference sequence to map against and discover single nucleotide variants. An additional advantage of this approach is that the data generated can potentially also be utilized for other types of research questions, such as analyses of gene expression differences between the populations. On the downside, SNP calling from a transcriptome without a reference genome is challenging and can lead to false positives, for example due to reads from lowly expressed paralogs erroneously mapping to the highly expressed copy present in the assembled transcriptome. The authors therefore went to great lengths to stringently filter these false positive variants from their dataset. ## Demographic inference using δαδι For the demographic inference, McCoy and colleagues used δαδι (diffusion approximation for demographic inference), a method developed by Ryan Gutenkunst while he was a postdoc in the group of CEHG faculty member Carlos Bustamante. This method uses a diffusion approximation to calculate the expected allele frequency spectrum under a demographic model of interest. The observed allele frequency spectrum is then fit to the expected spectrum by optimization of the demographic parameters to maximize the likelihood of the data. δαδι has been widely used to infer the demographic history of a number of species, from humans to domesticated rice, and is particularly suited to large-scale genomic datasets due to its flexibility and computational efficiency. Excerpt of Figure 2 from McCoy et al., illustrating the demographic models tested using δαδι. ## Models vs History The authors then fit a demographic model reflecting the known population history of E. gillettii, as illustrated in Figure 2 of their article (Model A). Encouragingly, they found that the model provided a very good fit to the data, with an the estimate of the split time between 40 and 47 generations ago, which is very close to the known time of establishment of the Colorado population 33 generations ago. Furthermore, they also tested how robust these results were to using a misspecified demographic model, by incorporating migration between the Colorado and Wyoming populations in their model (which in reality are isolated from each other). However, both alternative models with migration (Models B1 and B2) did not significantly improve the fit, again nicely consistent with the known population history. ## Three butterflies is enough? Finally, the researchers also tested the robustness of the results to variations in the number of samples or SNPs used in the analysis, from datasets simulated under the best-fit model A. They found that δαδι performed remarkably well even with sample sizes as low as three individuals per population. While this is in principle good news for researchers limited by low number of available samples, one has to be aware of the fact that this results will be to a certain extent specific to this particular type of system, where one population undergoes a very strong bottleneck resulting in large effects on the allele frequency spectrum. A good strategy suggested by McCoy and colleagues is then to use these types of simulations in the planning stages of an experiment, in order to inform researchers of the number of samples and markers necessary to confidently estimate the demographic parameters of interest. ## Conclusions and future directions For me, this study is a great example of how next-generation sequencing and sophisticated statistical modeling can open up a new world of possibilities to researchers interested in the ecology and evolution of natural populations. McCoy and colleagues constructed their genomic dataset essentially from scratch, without the “luxuries” of a reference genome or database of known polymorphisms. Moving forward, Rajiv has been busy collecting more samples over the past year. He and his colleagues plan to sequence over a thousand of them for the next phase of the project, as well as assemble a reference genome for E. gillettii, and important next step in the development of genomic tools for this fascinating ecological system. The author of the paper Rajiv McCoy, sampling larvae of Euphydryas gillettii McCoy, R. C., Garud, N. R., Kelley, J. L., Boggs, C. L. and Petrov, D. A. (2013), Genomic inference accurately predicts the timing and severity of a recent bottleneck in a nonmodel insect population. Molecular Ecology. doi: 10.1111/mec.12591	{}
# Conceptual understanding of Schrödinger equation So I followed this lecture: which starts of with the statement: If you have a Schrödinger equation for an energy eigenstate you have $$-\frac{\hbar}{2m}\frac{d^2}{dx^2}\psi(x) + V(x)\psi(x) = E \psi(x)\tag{1}$$ Question 1: What does it mean to have a energy eigenstate in this context? All eigenstates I ever cared about were the eigenstates and eigenfunctions of Hamiltonians. Question 2: Is equation (1) a general statement or specific to some conditions? Usually I assumeed that Schrödingers-equation is used for time-evolutions but this doesn't seem to be the case here. This is the time independent Schrödinger equation. It is basically an eigenvalue problem $$\hat{H}\psi=E\psi$$ where $$\hat{H}=-\frac{\hbar^2}{2m}\nabla^2+V(x)$$ is the Hamiltonian of the system. Since you yourself mentioned eigenstates of the Hamiltonian I'm going to guess you already know about why the Hamiltonian has this form. The solution of this equation are the eigenstates of the Hamiltonian operator, a set of eigenvectors and eigenvalues. Probably, the Schrödinger equation you have in mind is the time dependent Schrödinger equation $$i\hbar\frac{\partial}{\partial t}\|\psi\rangle=\hat{H}\|\psi\rangle$$ Why do we need two separate equations? Well, the true equation of motion of the state is the time dependent version, nevertheless, considering the appearance of the hamiltonian, it's useful to solve the time independent one and get the eigenstates. Why? Because once you have some states $$\|n\rangle$$ such that $$\hat{H}\|n\rangle=E_n\|n\rangle$$ you can verify that $$\|\psi(t)\rangle=\sum_n a_n\exp \left(-i\frac{E_n t}{\hbar} \right)\|n\rangle$$ is a solution to the time dependent version for some coefficients $$a_n$$. • First: Thank you very much that was very helpful! Second: You said that the solution of the time independent version are ' energy eigenstates of the Hamiltonian operator, a set of eigenvectors and eigenvalues.' I always assumed that the vectors are the states ? At least that's the case in the bracket notation. Am I missing here something ? What is the difference between an eigenstate and an eigenvector? – CatoMaths Jan 11 at 23:15 • @CatoMaths no difference - sorry for using both terms, it can be confusing. "Eigenstate" is just a name we give to an eigenvector of a physical observable. It's just a name we give out of physical interpretation. The solution of the time independent equation is a set of eigenvectors $\|n\rangle$ and eigenvalues $E_n$. The eigenvectors are usually called eigenstates, and the eigenvalues are interpreted as the energies of the corresponding eigenstates. – user2723984 Jan 11 at 23:21 • Also I should probably not have written "energy eigenstates of the Hamiltonian", usually one says "energy eigenstates" to refer to the eigenstates of the Hamiltonian, but writing both is redundant and confusing, I edited it. – user2723984 Jan 11 at 23:24 • Perfect, finally I am starting to understand! Thank you so much. If you don't mind I would have a minor question on this topic. We said that the solution of this equation is the eigenvector and corresponding eigenvalue. Let's call them $\mid \psi_n \rangle$ and $E_n$. I am having a hard time to make the connection between the normal notation $\psi_n$ and the bracket notation $\mid \psi_n \rangle$. This may sound stupid but I am used to be always in the backet notation so seeing $\psi_n$ now as non-vector is somewhat confusing to me. – CatoMaths Jan 11 at 23:37 • In bra-ket notation, vectors are denoted by $\|\psi\rangle$. If you have an eigenbasis $\|x\rangle$ of some observable, for example position, you can decompose $\|\psi\rangle$ as $\int dx \langle x\|\psi\rangle \|x\rangle$ and we note by $\psi(x):=\langle x\|\psi\rangle$, this is the wave function. It's not a vector in an Hilbert space, it's a function (or more generally a tempered distribution). Often physicists interchange the two in somewhat confusing ways, see this question of mine from when I was confused too :) – user2723984 Jan 12 at 8:31 You have stated that the only eigenstates you care about are for the Hamiltonian. That equation IS the Hamiltonian for a non-relativistic particle. The differential operator on the left hand side is H, the Hamiltonian operator, the E on the right hand side is the energy, a scalar number. Solving this equation for the allowed wavefunctions and energies provides you with a complete set of eigenfunctions and eigenvalues {psi_n, E_n}. By the way I think you are missing a square on your h_bar. As for the time-dependent equation, its solutions can be built up from the eigenfunctions since the span Hilbert space. This is a common approach to solving time dependent PDE, and is used in acoustics, optics and all other wave mechanics. As for it being a special case? The only things special about the equation you have posted are (1) it is non-relativistic, (2) it is 1-dim (in 3-dim the second derivative would be replaced with the Laplacian operator), and (3) no boundary conditions are explicitly mentioned, e.g. psi(x0) = 0 or psi(infinity) = 0, etc.	{}
Thread: need help solving a demand function 1. need help solving a demand function The demand for a certain fraternity's plastic brownie dishes is q(p) = 360,000 − (p + 1)2 where q represents the number of brownie dishes that the fraternity can sell each month at a price of p¢. Determine the number of brownie dishes the fraternity can sell each month if the price is set at 60¢. 2. Originally Posted by theski232 The demand for a certain fraternity's plastic brownie dishes is q(p) = 360,000 − (p + 1)2 where q represents the number of brownie dishes that the fraternity can sell each month at a price of p¢. Determine the number of brownie dishes the fraternity can sell each month if the price is set at 60¢. Since you know that value of p just evaluate the function $\displaystyle q(60)=360000-(60+1)^2=...$ 3. I was doing that but general knowledge led me to believe that 60cents is the same as 0.60. That is where I had my problem.	{}
## Section: Scientific Foundations ### Identification and approximation Identification typically consists in approximating experimental data by the prediction of a model belonging to some model class. It consists therefore of two steps, namely the choice of a suitable model class and the determination of a model in the class that fits best with the data. The ability to solve this approximation problem, often non-trivial and ill-posed, impinges on the effectiveness of a method. Particular attention is payed within the team to the class of stable linear time-invariant systems, in particular resonant ones, and in isotropically diffusive systems, with techniques that dwell on functional and harmonic analysis. In fact one often restricts to a smaller class —e.g. rational models of suitable degree (resonant systems, see section 4.3 ) or other structural constraints— and this leads us to split the identification problem in two consecutive steps: 1. Seek a stable but infinite (numerically: high) dimensional model to fit the data. Mathematically speaking, this step consists in reconstructing a function analytic in the right half-plane or in the unit disk (the transfer function), from its values on an interval of the imaginary axis or of the unit circle (the band-width). We will embed this classical ill-posed issue (i.e. the inverse Cauchy problem for the Laplace equation) into a family of well-posed extremal problems, that may be viewed as a regularization scheme of Tikhonov-type. These problems are infinite-dimensional but convex (see section 3.1.1 ). 2. Approximate the above model by a lower order one reflecting further known properties of the physical system. This step aims at reducing the complexity while bringing physical significance to the design parameters. It typically consists of a rational or meromorphic approximation procedure with prescribed number of poles in certain classes of analytic functions. Rational approximation in the complex domain is a classical but difficult non-convex problem, for which few effective methods exist. In relation to system theory, two specific difficulties superimpose on the classical situation, namely one must control the region where the poles of the approximants lie in order to ensure the stability of the model, and one has to handle matrix-valued functions when the system has several inputs and outputs, in which case the number of poles must be replaced by the McMillan degree (see section 3.1.2 ). When identifying elliptic (Laplace, Beltrami) partial differential equations from boundary data, point 1. above can be recast as an inverse boundary-value problem with (overdetermined Dirichlet-Neumann) data on part of the boundary of a plane domain (recover a function, analytic in a domain, from incomplete boundary data). As such, it arises naturally in higher dimensions when analytic functions get replaced by gradients of harmonic functions (see section 4.2 ). Motivated by free boundary problems in plasma control and questions of source recovery arising in magneto/electro-encephalography, we aim at generalizing this approach to the real Beltrami equation in dimension 2 (section 6.3.3 ) and to the Laplace equation in dimension 3 (section 6.3.1 ). Step 2. above, i.e., meromorphic approximation with prescribed number of poles—is used to approach other inverse problems beyond harmonic identification. In fact, the way the singularities of the approximant (i.e. its poles) relate to the singularities of the approximated function is an all-pervasive theme in approximation theory: for appropriate classes of functions, the location of the poles of the approximant can be used as an estimator of the singularities of the approximated function (see section 6.3.2 ). We provide further details on the two steps mentioned above in the sub-paragraphs to come. #### Analytic approximation of incomplete boundary data Keywords : extremal problems, inverse problems, Hardy spaces, harmonic functions, Beltrami equations. Participants : Laurent Baratchart, Slah Chaabi, Yannick Fischer, Juliette Leblond, Jean-Paul Marmorat, Jonathan Partington, Stéphane Rigat [ Univ. Aix-Marseille I ] , Emmanuel Russ [ Univ. Aix-Marseille III ] , Fabien Seyfert. Given a planar domain D , the problem is to recover an analytic function from its values on a subset of the boundary of D . It is convenient to normalize D and apply in each particular case a conformal transformation to meet a “normalized” domain. In the simply connected case, which is that of the half-plane, we fix D to be the unit disk, so that its boundary is the unit circle T . We denote by Hp the Hardy space of exponent p which is the closure of polynomials in the Lp -norm on the circle if 1p< and the space of bounded holomorphic functions in D if p = . Functions in Hp have well-defined boundary values in Lp(T) , which make it possible to speak of (traces of) analytic functions on the boundary. A standard extremal problem on the disk is [68] : (P0 ) Let 1p and fLp(T) ; find a function gHp such that g-f is of minimal norm in Lp(T) . When seeking an analytic function in D which approximately matches some measured values f on a sub-arc K of T , the following generalization of (P0 ) naturally arises: (P ) Let 1p , K a sub-arc of T , fLp(K) , and M>0 ; find a function gHp such that and g-f is of minimal norm in Lp(K) under this constraint. Here is a reference behavior capsulizing the expected behavior of the model off K , while M is the admissible error with respect to this expectation. The value of p reflects the type of stability which is sought and how much one wants to smoothen the data. To fix terminology we generically refer to (P ) as a bounded extremal problem . The solution to this convex infinite-dimensional optimization problem can be obtained upon iteratively solving spectral equations for appropriate Hankel and Toeplitz operators, that involve a Lagrange parameter, and whose right hand-side is given by the solution to (P0 ) for some weighted concatenation of f and . Constructive aspects are described in [43] , [45] , [87] , for p = 2 , p = , and 1<p< , while the situation p = 1 is essentially open. Various modifications of (P) have been studied in order to meet specific needs. For instance when dealing with loss-less transfer functions (see section 4.3 ), one may want to express the constraint on in a pointwise manner: \|g-\|M a.e. on , see [29] , [47] for p = 2 and = 0 . The above-mentioned problems can be stated on an annular geometry rather than a disk. For p = 2 the solution proceeds much along the same lines [23] . When K is the outer boundary, (P ) regularizes a classical inverse problem occurring in nondestructive control, namely to recover a harmonic function on the inner boundary from overdetermined Dirichlet-Neumann data on the outer boundary (see sections 4.2 and 6.3 ). Interestingly perhaps, it becomes a tool to approach Bernoulli type problems for the Laplacian, where overdetermined observations are made on the outer boundary and we seek the inner boundary knowing it is a level curve of the flux (see section 6.3.3 ). Here, the Lagrange parameter indicates which deformation should be applied on the inner contour in order to improve the fit to the data. Continuing effort is currently payed by the team to carry over bounded extremal problems and their solution to more general settings. Such generalizations are twofold: on the one hand Apics considers 2-D diffusion equations with variable conductivity, on the other hand it investigates the ordinary Laplacian in . The targeted applications are the determination of free boundaries in plasma control and source detection in electro/magneto-encephalography (EEG/MEG, see section 6.3.2 ). An isotropic diffusion equation in dimension 2 can be recast as a so-called real Beltrami equation [73] . This way analytic functions get replaced by “generalized” ones in problems (P0 ) and (P ). Hardy spaces of solutions, which are more general than Sobolev ones and allow one to handle Lp boundary conditions, have been introduced when 1<p< [46] . The expansions of solutions needed to constructively handle such problems have been preliminary studied in [64] , [65] . The goal is to solve the analog of (P ) in this context to approach Bernoulli-type problems (see section 6.3.1 ). At present, bounded extremal problems for the n -D Laplacian are considered on half-spaces or balls. Following [88] , Hardy spaces are defined as gradients of harmonic functions satisfying Lp growth conditions on inner hyperplanes or spheres. From the constructive viewpoint, when p = 2 , spherical harmonics offer a reasonable substitute to Fourier expansions [13] . Only very recently were we able to define operators of Hankel type whose singular values connect to the solution of (P0 ) in BMO norms. The Lp problem also makes contact with some nonlinear PDE's, namely to the p -Laplacian. The goal is here to solve the analog of (P ) on spherical shells to approach inverse diffusion problems across a conductor layer. #### Meromorphic and rational approximation Keywords : meromorphic approximation, rational approximation, critical point theory, orthogonal polynomials. Participants : Laurent Baratchart, José Grimm, Martine Olivi, Edward Saff, Herbert Stahl [ TFH Berlin ] . Let as before D designate the unit disk, T the unit circle. We further put RN for the set of rational functions with at most N poles in D , which allows us to define the meromorphic functions in Lp(T) as the traces of functions in Hp + RN . A natural generalization of problem (P0 ) is (PN ) Let 1p , N0 an integer, and fLp(T) ; find a function gNHp + RN such that gN-f is of minimal norm in Lp(T) . Problem (PN ) aims, on the one hand, at solving inverse potential problems from overdetermined Dirichlet-Neumann data, namely to recover approximate solutions of the inhomogeneous Laplace equation u = , with some (unknown) distribution, which will be discretized by the process as a linear combination of N Dirac masses. On the other hand, it is used to perform the second step of the identification scheme described in section 3.1 , namely rational approximation with a prescribed number of poles to a function analytic in the right half-plane, when one maps the latter conformally to the complement of D and solve (PN ) for the transformed function on T . Only for p = and continuous f is it known how to solve (PN ) in closed form. The unique solution is given by the AAK theory, that allows one to express gN in terms of the singular vectors of the Hankel operator with symbol f . The continuity of gN as a function of f only holds for stronger norms than uniform, [85] . The case p = 2 is of special importance. In particular when , the Hardy space of exponent 2 of the complement of D in the complex plane (by definition, h(z) belongs to if, and only if h(1/z) belongs to Hp ), then (PN ) reduces to rational approximation. Moreover, it turns out that the associated solution gNRN has no pole outside D , hence it is a stable rational approximant to f . However, in contrast with the situation when p = , this approximant may not be unique. The former Miaou project (predecessor of Apics) has designed an adapted steepest-descent algorithm for the case p = 2 whose convergence to a local minimum is guaranteed; it seems today the only procedure meeting this property. Roughly speaking, it is a gradient algorithm that proceeds recursively with respect to the order N of the approximant, in a compact region of the parameter space [40] . Although it has proved rather effective in all applications carried out so far (see sections 4.2 , 4.3 ), it is not known whether the absolute minimum can always be obtained by choosing initial conditions corresponding to critical points of lower degree (as done by the Endymion software section 5.5 and RARL2 software, section 5.2 ). In order to establish convergence results of the algorithm to the global minimum, Apics has undergone a long-haul study of the number and nature of critical points, in which tools from differential topology and operator theory team up with classical approximation theory. The main discovery is that the nature of the critical points (e.g. local minima , saddles...) depends on the decrease of the interpolation error to f as N increases [48] . Based on this, sufficient conditions have been developed for a local minimum to be unique. This technique requires strong error estimates that are often difficult to obtain, and most of the time only hold for N large. Examples where uniqueness or asymptotic uniqueness has been proved this way include transfer functions of relaxation systems (i.e., Markov functions) [49] , the exponential function, and meromorphic functions [8] . The case where f is the Cauchy integral on a hyperbolic geodesic arc of a Dini-continuous function which does not vanish “too much” has been recently answered in the positive, see section 6.7 . An analog to AAK theory has been carried out for 2p< [9] . Although not computationally as powerful, it has better continuity properties and stresses a continuous link between rational approximation in H2 and meromorphic approximation in the uniform norm, allowing one to use, in either context, techniques available from the other(When 1p<2 , problem (PN ) is still fairly open.). A common feature to all these problems is that critical point equations express non-Hermitian orthogonality relations for the denominator of the approximant. This is used in an essential manner to assess the behavior of the poles of the approximants to functions with branched singularities which is of particular interest for inverse source problems (cf. sections 6.3.2 , 6.7 ). In higher dimensions, the analog of problem (PN ) is the approximation of a vector field with gradients of potentials generated by N point masses instead of meromorphic functions. The issue is by no means understood at present, and is a major endeavor of future research problems. Certain constrained rational approximation problems, of special interest in identification and design of passive systems, arise when putting additional requirements on the approximant, for instance that it should be smaller than 1 in modulus. Such questions have become over years an increasingly significant part of the team's activity (see sections 4.3 , 6.6 , 6.7 , and 6.10 ). When translated over to the circle, a prototypical formulation consists in approximating the modulus of a given function by the modulus of a rational function of degree n . When p = 2 this problem can be reduced to a series of standard rational approximation problems, but usually one needs to solve it for p = . The case where \|f\| is a piecewise constant function with values 0 and 1 can also be approached via classical Zolotarev problems [86] , that can be solved more or less explicitly when the pass-band consists of a single arc. A constructive solution in the case where \|f\| is a piecewise constant function with values 0 and 1 on several arcs (multiband filters) is one recent achievement of the team. Though the modulus of the response is the first concern in filter design, the variation of the phase must nevertheless remain under control to avoid unacceptable distortion of the signal. This is an important issue, currently under investigation within the team under contract with the CNES, see section 6.10 . From the point of view of design, rational approximants are indeed useful only if they can be translated into physical parameter values for the device to be built. This is where system theory enters the scene, as the correspondence between the frequency response (i.e., the transfer-function) and the linear differential equations that generate this response (i.e., the state-space representation), which is the object of the so-called realization process. Since filters have to be considered as dual modes cavities, the realization issue must indeed be tackled in a 2×2 matrix-valued context that adds to the complexity. A fair share of the team's research in this direction is concerned with finding realizations meeting certain constraints (imposed by the technology in use) for a transfer-function that was obtained with the above-described techniques (see section 6.8 ). #### Behavior of poles of meromorphic approximants and inverse problems for the Laplacian Keywords : singularity detection, free boundary inverse problems, meromorphic and rational approximation, orthogonal polynomials, discretization of potentials. Participants : Laurent Baratchart, Edward Saff, Herbert Stahl [ TFH Berlin ] , Maxim Yattselev. We refer here to the behavior of the poles of best meromorphic approximants, in the Lp -sense on a closed curve, to functions f defined as Cauchy integrals of complex measures whose support lies inside the curve. If one normalizes the contour to be the unit circle T , we are back to the framework of section 3.1.2 and to problem (PN ); the invariance of the problem under conformal mapping was established in [6] . The research so far has focused on functions whose singular set inside the contour is zero or one-dimensional. Generally speaking, the behavior of poles is particularly important in meromorphic approximation to obtain error rates as the degree goes large and also to tackle constructive issues like uniqueness. However, the original motivation of Apics is to consider this issue in connection with the approximation of the solution to a Dirichlet-Neumann problem, so as to extract information on the singularities. The general theme is thus how do the singularities of the approximant reflect those of the approximated function? The approach to inverse problem for the 2-D Laplacian that we outline here is attractive when the singularities are zero- or one-dimensional (see section 4.2 ). It can be used as a computationally cheap preliminary step to obtain the initial guess of a more precise but heavier numerical optimization. For sufficiently smooth cracks, or pointwise sources recovery, the approach in question is in fact equivalent to the meromorphic approximation of a function with branch points, and we were able to prove [4] , [6] that the poles of the approximants accumulate in a neighborhood of the geodesic hyperbolic arc that links the endpoints of the crack, or the sources [44] . Moreover the asymptotic density of the poles turns out to be the equilibrium distribution on the geodesic arc of the Green potential and it charges the end points, that are thus well localized if one is able to compute sufficiently many zeros (this is where the method could fail). The case of more general cracks, as well as situations with three or more sources, requires the analysis of the situation where the number of branch points is finite but arbitrary, see section 6.7 ). This are outstanding open questions for applications to inverse problems (see section 6.3 ), as also the problem of a general singularity, that may be two dimensional. Results of this type open new perspectives in non-destructive control, in that they link issues of current interest in approximation theory (the behavior of zeroes of non-Hermitian orthogonal polynomials) to some classical inverse problems for which a dual approach is thereby proposed: to approximate the boundary conditions by true solutions of the equations, rather than the equation itself (by discretization). Let us point out that the problem of approximating, by a rational or meromorphic function, in the Lp sense on the boundary of a domain, the Cauchy transform of a real measure, localized inside the domain, can be viewed as an optimal discretization problem for a logarithmic potential according to a criterion involving a Sobolev norm. This formulation can be generalized to higher dimensions, even if the computational power of complex analysis is then no longer available, and this makes for a long-term research project with a wide range of applications. It is interesting to mention that the case of sources in dimension three in a spherical or ellipsoidal geometry, can be attacked with the above 2-D techniques as applied to planar sections (see section 6.3 ). #### Matrix-valued rational approximation Keywords : rational approximation, inner matrix, reproducing kernel space, realization theory. Participants : Laurent Baratchart, Martine Olivi, José Grimm, Jean-Paul Marmorat, Bernard Hanzon, Ralf Peeters [ Univ. Maastricht ] . Matrix-valued approximation is necessary for handling systems with several inputs and outputs, and it generates substantial additional difficulties with respect to scalar approximation, theoretically as well as algorithmically. In the matrix case, the McMillan degree (i.e., the degree of a minimal realization in the System-Theoretic sense) generalizes the degree. The problem we want to consider reads: Let and n an integer; find a rational matrix of size m×l without poles in the unit disk and of McMillan degree at most n which is nearest possible to in (H2)m×l . Here the L2 norm of a matrix is the square root of the sum of the squares of the norms of its entries. The approximation algorithm designed in the scalar case generalizes to the matrix-valued situation [67] . The first difficulty consists here in the parametrization of transfer matrices of given McMillan degree n , and the inner matrices (i.e., matrix-valued functions that are analytic in the unit disk and unitary on the circle) of degree n enter the picture in an essential manner: they play the role of the denominator in a fractional representation of transfer matrices (using the so-called Douglas-Shapiro-Shields factorization). The set of inner matrices of given degree has the structure of a smooth manifold that allows one to use differential tools as in the scalar case. In practice, one has to produce an atlas of charts (parametrization valid in a neighborhood of a point), and we must handle changes of charts in the course of the algorithm. Such parametrization can be obtained from interpolation theory and Schur type algorithms, the parameters being interpolation vectors or matrices [33] , [10] , [11] . Some of these parametrizations have a particular interest for computation of realizations [10] , [11] , involved in the estimation of physical quantities for the synthesis of resonant filters. Two rational approximation codes (see sections 5.2 and 5.5 ) have been developed in the team. Problems relative to multiple local minima naturally arise in the matrix-valued case as well, but deriving criteria that guarantee uniqueness is even more difficult than in the scalar case. The already investigated case of rational functions of the sought degree (the consistency problem) was solved using rather heavy machinery [7] , and that of matrix-valued Markov functions, that are the first example beyond rational function has made progress only recently [39] . Let us stress that the algorithms mentioned above are first to handle rational approximation in the matrix case in a way that converges to local minima, while meeting stability constraints on the approximant. Logo Inria	{}
Specification and validation of POUMM/PMM settings. specifyPOUMM( z = NULL, tree = NULL, zMin = -10, zMean = 0, zMax = 10, zVar = 4, zSD = sqrt(zVar), tMin = 0.1, tMean = 2, tMax = 10, parMapping = NULL, parLower = NULL, parUpper = NULL, g0Prior = NULL, parInitML = NULL, control = NULL, parPriorMCMC = NULL, parInitMCMC = NULL, parScaleMCMC = NULL, nSamplesMCMC = 1e+05, thinMCMC = 100, accRateMCMC = 0.01, gammaMCMC = 0.50001, nChainsMCMC = 3, samplePriorMCMC = TRUE, parallelMCMC = FALSE, validateSpec = TRUE ) specifyPOUMM_ATS( z = NULL, tree = NULL, zMin = -10, zMean = 0, zMax = 10, zVar = 4, zSD = sqrt(zVar), tMin = 0.1, tMean = 2, tMax = 10, parMapping = NULL, parLower = NULL, parUpper = NULL, g0Prior = NULL, parInitML = NULL, control = NULL, parPriorMCMC = NULL, parInitMCMC = NULL, parScaleMCMC = NULL, nSamplesMCMC = 1e+05, thinMCMC = 100, accRateMCMC = 0.01, gammaMCMC = 0.50001, nChainsMCMC = 3, samplePriorMCMC = TRUE, parallelMCMC = FALSE, sigmaeFixed = 0 ) specifyPOUMM_ATSG0( z = NULL, tree = NULL, zMin = -10, zMean = 0, zMax = 10, zVar = 4, zSD = sqrt(zVar), tMin = 0.1, tMean = 2, tMax = 10, parMapping = NULL, parLower = NULL, parUpper = NULL, g0Prior = NULL, parInitML = NULL, control = NULL, parPriorMCMC = NULL, parInitMCMC = NULL, parScaleMCMC = NULL, nSamplesMCMC = 1e+05, thinMCMC = 100, accRateMCMC = 0.01, gammaMCMC = 0.50001, nChainsMCMC = 3, samplePriorMCMC = TRUE, parallelMCMC = FALSE, sigmaeFixed = 0 ) specifyPOUMM_ATSSeG0( z = NULL, tree = NULL, zMin = -10, zMean = 0, zMax = 10, zVar = 4, zSD = sqrt(zVar), tMin = 0.1, tMean = 2, tMax = 10, parMapping = NULL, parLower = NULL, parUpper = NULL, g0Prior = NULL, parInitML = NULL, control = NULL, parPriorMCMC = NULL, parInitMCMC = NULL, parScaleMCMC = NULL, nSamplesMCMC = 1e+05, thinMCMC = 100, accRateMCMC = 0.01, gammaMCMC = 0.50001, nChainsMCMC = 3, samplePriorMCMC = TRUE, parallelMCMC = FALSE ) specifyPMM( z = NULL, tree = NULL, zMin = -10, zMean = 0, zMax = 10, zVar = 4, zSD = sqrt(zVar), tMin = 0.1, tMean = 2, tMax = 10, parMapping = NULL, parLower = NULL, parUpper = NULL, g0Prior = NULL, parInitML = NULL, control = NULL, parPriorMCMC = NULL, parInitMCMC = NULL, parScaleMCMC = NULL, nSamplesMCMC = 1e+05, thinMCMC = 100, accRateMCMC = 0.01, gammaMCMC = 0.50001, nChainsMCMC = 3, samplePriorMCMC = TRUE, parallelMCMC = FALSE ) specifyPMM_SSeG0( z = NULL, tree = NULL, zMin = -10, zMean = 0, zMax = 10, zVar = 4, zSD = sqrt(zVar), tMin = 0.1, tMean = 2, tMax = 10, parMapping = NULL, parLower = NULL, parUpper = NULL, g0Prior = NULL, parInitML = NULL, control = NULL, parPriorMCMC = NULL, parInitMCMC = NULL, parScaleMCMC = NULL, nSamplesMCMC = 1e+05, thinMCMC = 100, accRateMCMC = 0.01, gammaMCMC = 0.50001, nChainsMCMC = 3, samplePriorMCMC = TRUE, parallelMCMC = FALSE ) specifyPOUMM_ATH2tMeanSe( z = NULL, tree = NULL, zMin = -10, zMean = 0, zMax = 10, zVar = 4, zSD = sqrt(zVar), tMin = 0.1, tMean = 2, tMax = 10, parMapping = NULL, parLower = NULL, parUpper = NULL, g0Prior = NULL, parInitML = NULL, control = NULL, parPriorMCMC = NULL, parInitMCMC = NULL, parScaleMCMC = NULL, nSamplesMCMC = 1e+05, thinMCMC = 100, accRateMCMC = 0.01, gammaMCMC = 0.50001, nChainsMCMC = 3, samplePriorMCMC = TRUE, parallelMCMC = FALSE ) specifyPOUMM_ATH2tMeanSeG0( z = NULL, tree = NULL, zMin = -10, zMean = 0, zMax = 10, zVar = 4, zSD = sqrt(zVar), tMin = 0.1, tMean = 2, tMax = 10, parMapping = NULL, parLower = NULL, parUpper = NULL, g0Prior = NULL, parInitML = NULL, control = NULL, parPriorMCMC = NULL, parInitMCMC = NULL, parScaleMCMC = NULL, nSamplesMCMC = 1e+05, thinMCMC = 100, accRateMCMC = 0.01, gammaMCMC = 0.50001, nChainsMCMC = 3, samplePriorMCMC = TRUE, parallelMCMC = FALSE ) specifyPMM_H2tMeanSe( z = NULL, tree = NULL, zMin = -10, zMean = 0, zMax = 10, zVar = 4, zSD = sqrt(zVar), tMin = 0.1, tMean = 2, tMax = 10, parMapping = NULL, parLower = NULL, parUpper = NULL, g0Prior = NULL, parInitML = NULL, control = NULL, parPriorMCMC = NULL, parInitMCMC = NULL, parScaleMCMC = NULL, nSamplesMCMC = 1e+05, thinMCMC = 100, accRateMCMC = 0.01, gammaMCMC = 0.50001, nChainsMCMC = 3, samplePriorMCMC = TRUE, parallelMCMC = FALSE ) specifyPMM_H2tMeanSeG0( z = NULL, tree = NULL, zMin = -10, zMean = 0, zMax = 10, zVar = 4, zSD = sqrt(zVar), tMin = 0.1, tMean = 2, tMax = 10, parMapping = NULL, parLower = NULL, parUpper = NULL, g0Prior = NULL, parInitML = NULL, control = NULL, parPriorMCMC = NULL, parInitMCMC = NULL, parScaleMCMC = NULL, nSamplesMCMC = 1e+05, thinMCMC = 100, accRateMCMC = 0.01, gammaMCMC = 0.50001, nChainsMCMC = 3, samplePriorMCMC = TRUE, parallelMCMC = FALSE ) ## Arguments z, tree a numeric vector and a phylo object on which the fit is to be done. These arguments are used in order to guess meaningful values for the parLower, parUpper and parPriorMCMC arguments. See also, zMin,zMean,...,tMax below. summary statistics of the observed tip-values (z) and root-tip distances (t). Some of these values are used for constructing default parameter values and limits; These arguments are given default values which will most likely be meaningless in your specific use-case. The default values will be overwritten with the corresponding statistics from the z and tree arguments if these were specified. If none of tree and z, nor these parameters are specified, then the arguments parLower, parUpper, parPriorMCMC must be specified explicitly. An R-function that can handle, both, a numeric vector or a numeric matrix as argument. This function should transform the input vector or each row-vector (if the input is matrix) into a (row-)vector of the POUMM parameters alpha, theta, sigma, sigmae, g0. For a vector input the function should return a vector with named elements alpha, theta, sigma, sigmae, g0. For a matrix input the function should return a matrix with the same number of rows and columns alpha, theta, sigma, sigmae, g0. Only finite non-negative values are allowed for alpha, sigma, and sigmae. Returning Inf, -Inf, NA or NaN for any of these parameters will result in an error during likelihood calculation. Only finite numerical values are allowed for theta. The parameter g0 is treated in a special way and can assume either a finite numerical value or one of NA or NaN. If g0 = finite value, this value is used together with the corresponding values of alpha, theta, sigma, and sigmae for likelihood calcuation. If g0 = NA (meaing value Not Avaiable), the value of g0 is calculated analytically during likelihood calculation in order to maximise one of the following: if a normal prior for g0 was specified (see g0Prior), $$pdf(z \| \alpha, \theta, \sigma, \sigma_e, g0, tree) x prior(g0)$$. otherwise, $$pdf(z \| \alpha, \theta, \sigma, \sigma_e, g0, tree)$$. If g0 = NaN (meaning Not a Number), then the likelihood is marginalized w.r.t. the g0's prior distribution (see g0Prior), i.e. the likelihood returned is: $$pdf(z \| \alpha, \theta, \sigma, \sigma_e, tree) = Integral(pdf(z\|\alpha,\theta,\sigma,\sigma_e,g0) x pdf(g0) d g0; g0 from -\infty to +\infty)$$ In this case (g0=NaN), if g0Prior is not specified, it is assumed that g0Prior is the stationary OU normal distribution with mean, theta, and variance, varOU(Inf, alpha, sigma). Examples: # Default for POUMM: identity for alpha, theta, sigma, sigmae, NA for g0. parMapping = function(par) { if(is.matrix(par)) { atsseg0 <- cbind(par[, 1:4, drop = FALSE], NA) colnames(atsseg0) <- c("alpha", "theta", "sigma", "sigmae", "g0") } else { atsseg0 <- c(par[1:4], NA) names(atsseg0) <- c("alpha", "theta", "sigma", "sigmae", "g0") } atsseg0 } two named numeric vectors of the same length indicating the boundaries of the search region for the ML-fit. Calling parMapping on parLower and parUpper should result in appropriate values of the POUMM parameters alpha, theta, sigma sigmae and g0. By default, the upper limit for alpha is set to 69.31 / tMean, which corresponds to a value of alpha so big that the time for half-way convergence towards theta from any initial trait value is 100 times shorter than the mean root-tip distance in the tree. Examples: # Default for POUMM: parLower = c(alpha = 0, theta = zMin - 2 * (zMax - zMin), sigma = 0, sigmae = 0) parUpper = c(alpha = 69.31 / tMean, theta = zMax + 2 * (zMax - zMin), sigma = sigmaOU(H2 = .99, alpha = 69.31 / tMean, sigmae = 2 * zSD, t = tMean), sigmae = 2 * zSD) Either NULL or a list with named numeric or character members "mean" and "var". Specifies a prior normal distribution for the parameter g0. If characters, the members "mean" and "var" are evaluated as R-expressions - useful if these are functions of some of other parameters. Note that if g0Prior is not NULL and g0 is not NaN (either a fixed number or NA), then the likelihood maximization takes into account the prior for g0, that is, the optimization is done over the product p(g0) x lik(data\|g0, other parameters and tree). This can be helpful to prevent extremely big or low estimates of g0. To avoid this behavior and always maximize the likelihood, use g0Prior = NULL. A named vector (like parLower and parUpper) or a list of such vectors - starting points for optim. List of parameters passed on to optim in the ML-fit, default list(factr=1e9), see ?optim. A function of a numeric parameter-vector returning the log-prior for this parameter vector. Example: # Default for POUMM: parPriorMCMC = function(par) { dexp(par[1], rate = tMean / 6.931, TRUE) + dnorm(par[2], zMean, 10 * zSD, TRUE) + dexp(par[3], rate = sqrt(tMean / (zVar * 0.6931)), TRUE) + dexp(par[4], rate = 2 / zSD, TRUE) } a function(chainNo, fitML) returning an initial state of an MCMC as a vector. The argument fitML can be used to specify an initial state, close to a previously found likelihood optimum. Example: # Default for POUMM: parInitMCMC = function(chainNo, fitML) { if(!is.null(fitML)) { parML <- fitML\$par } else { parML <- NULL } init <- rbind( c(alpha = 0, theta = 0, sigma = 1, sigmae = 0), parML, c(alpha = 0, theta = 0, sigma = 1, sigmae = 1) ) init[(chainNo - 1) %% nrow(init) + 1, ] } Numeric matrix indicating the initial jump-distribution matrix for the MCMC fit. Default for POUMM is diag(4); Integer indicating the length of each MCMC chain. Defaults to 1e5. Logical indicating whether adaptation of the MCMC jump distribution should be done with respect to the target acceptance rate (accRateMCMC) or integer indicating how many initial MCMC iterations should be used for adaptation of the jump-distribution matrix (see details in ?POUMM). Defaults to nSamplesMCMC meaning continuous adaptation throughout the MCMC. Integer indicating the thinning interval of the mcmc-chains. Defaults to 100. numeric between 0 and 1 indicating the target acceptance rate of the adaptive Metropolis sampling (see details in ?POUMM). Default 0.01. controls the speed of adaption. Should be in the interval (0.5,1]. A lower gamma leads to faster adaption. Default value is 0.50001. integer indicating the number of chains to run. Defaults to 3 chains, from which the first one is a sample from the prior distribution (see samplePriorMCMC). Logical indicating if sampling from the prior should be done for the first chain (see nChainsMCMC). This is useful to compare mcmc's for an overlap between prior and posterior distributions. Default is TRUE. Logical indicating whether the MCMC chains should be run in parallel. Setting this option to TRUE results in using foreach::foreach() %dopar% { } construct for the MCMC fit. In order for parallel execution to be done, you should create a computing cluster and register it as parallel back-end (see example in package vignette and the web-page https://github.com/tobigithub/R-parallel/wiki/R-parallel-Setups). Logical indicating whether the passed parameters should be validated. This parameter is used internally and should always be TRUE. fixed value for the sigmae parameter (used in specifyPOUMM_ATS and specifyPOUMM_ATSG0). ## Value A named list to be passed as a spec argument to POUMM. ## Functions • specifyPOUMM: Specify parameters for fitting a POUMM model. Parameter vector is c(alpha, theta, sigma, sigmae). Default model settings. • specifyPOUMM_ATS: Fitting a POU model with fixed sigmae. Parameter vector is c(alpha, theta, sigma). • specifyPOUMM_ATSG0: Fitting a POU model with fixed sigmae. Parameter vector is c(alpha, theta, sigma, g0). • specifyPOUMM_ATSSeG0: Fitting a POUMM model with sampling of g0. Parameter vector is c(alpha, theta, sigma, sigmae, g0). • specifyPMM: Specify parameter for fitting a PMM model. Parameter vector is c(sigma, sigmae) • specifyPMM_SSeG0: Specify parameter for fitting a PMM model with sampling of g0. Parameter vector is c(sigma, sigmae, g0). • specifyPOUMM_ATH2tMeanSe: Fitting a POUMM model with a uniform prior for the phylogenetic heritability at mean root-tip distance. Parameter vector is c(alpha, theta, H2tMean, sigmae). • specifyPOUMM_ATH2tMeanSeG0: Fitting a POUMM model with a uniform prior for the phylogenetic heritability at mean root-tip with sampling of g0. Parameter vector is c(alpha, theta, H2tMean, sigmae, g0). • specifyPMM_H2tMeanSe: Fitting a PMM model with a uniform prior for the phylogenetic heritability at mean root-tip distance. Parameter vector is c(H2tMean, sigmae). • specifyPMM_H2tMeanSeG0: Fitting a PMM model with a uniform prior for the phylogenetic heritability at mean root-tip distance with sampling of G0. Parameter vector is c(H2tMean, sigmae, g0).	{}
If a trait $\mathrm{A}$ exists in $10 \%$ of a population of an asexually Question. If a trait $\mathrm{A}$ exists in $10 \%$ of a population of an asexually reproducing species and a trait $\mathrm{B}$ exists in $60 \%$ of the same population, which trait is likely to have arisen earlier? solution: The trait $\mathrm{B}$ has arisen earlier since it in $60 \%$ population while trait $\mathrm{A}$ is merely in $10 \%$ which is newly arisen and not spread to large number.	{}
## October 30, 2016 ### Linear Algebraic Groups (Part 4) #### Posted by John Baez This time I explain some axioms for an ‘abstract projective plane’, and the extra axiom required to ensure an abstract projective plane comes from a field. Yet again the old Greek mathematicians seem to have been strangely prescient, because this extra axiom was discovered by Pappus of Alexandria sometime around 340 AD! For him it was a theorem in Euclidean geometry, but later it was realized that a cleaner statement involves only projective geometry… and later still, it was seen to be a useful axiom. • Lecture 4 (Oct. 4) - Abstract projective planes. Pappus’s hexagon theorem, and how it characterizes which abstract projective planes are of the form $k\mathrm{P}^2$ for a field $k$. Klein geometry and transitive group actions: each kind of highly symmetrical geometry corresponds to a group $G$, and each type of geometrical figure in this geometry corresponds to a set on which $G$ acts transitively. Transitive $G$-spaces all arise from subgroups of $G$. Klein geometry studies invariant relations between transitive $G$-spaces. The subject of abstract projective planes touches on the fascinating axiomatic approach to incidence geometries having various linear algebraic groups as their symmetry groups. But instead of marching down that beckoning byway, I’ll point you to a place where you can read more: Posted at October 30, 2016 1:25 AM UTC TrackBack URL for this Entry: https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2916 ## 1 Comment & 0 Trackbacks ### Re: Linear Algebraic Groups (Part 4) Thanks for these notes: all very fascinating! I’ve found what I believe are some typos, although it’s always possible that I’ve misunderstood things; if so, please forgive me. Lecture 2, page 1: in the definition of the sphere X, the dot product v.c should read “v.v” Lecture 2, page 2: in the definition of the set of points P, I believe that instead of reading “P={p : p is a 1-dim subspaces of k3} n X” this should read “{P={p n X : p is a 1-dim subspaces of k3}”. And same for L. (Using “n” for set intersection; also, “subspaces” should be singular.) Lecture 3, page 1: the definition of “kPn” should instead say kPn-1 (at least, that’s how it was at the end of Lecture 2, which I believe is correct.) Posted by: Ryan on November 3, 2016 5:45 PM \| Permalink \| Reply to this Post a New Comment	{}
# Math Help - Optimization 1. ## Optimization We just started this new section in optimization, and I don't really have a clue how to set up the word problems. I get that whatever one is attempting to minimize/maximize is what one takes the derivative of; and I also know that the expression that one takes the derivative of will have to be in terms of one variable (so the constraint equation would be used to do this) but I don't really understand how to set the expressions up... Can anyone help with the set up of the expressions for this problem? You have been hired as a consultant by a farmer who needs to build a grain silo. The silo is to have the shape of a cylinder, with a roof in the shape of a half-sphere (There is no bottom to the silo, as it is built on a platform). The silo must have a total volume of 2400pi cubic feet. Your job is to inform the farmer of the dimensions of the silo that will minimize the total cost of materials. The material for the side (the cylindrical part) costs $1.50 per sq ft, and the material for the roof (the half-sphere) costs$5 per sq ft. a) give the set-up and equations used to solve the problem b) give the values of r and h that minimize the cost c) evaluate for the total cost for the materials My professor said something about giving the expression for the total surface area? I don't get how if you're given the total volume (2400pi cubic ft), you would write the equation in terms of surface area? ----> so, 2400pi = (4pir^2)/2 + 2pirh (4pir^2 is divided by 2 because the roof is only a half of a sphere) ----> then, divide the right side by pi? Obviously....I have no idea where to start? Any help would be greatly appreciated! 2. Originally Posted by obsmith08 You have been hired as a consultant by a farmer who needs to build a grain silo. The silo is to have the shape of a cylinder, with a roof in the shape of a half-sphere (There is no bottom to the silo, as it is built on a platform). The silo must have a total volume of 2400pi cubic feet. Your job is to inform the farmer of the dimensions of the silo that will minimize the total cost of materials. The material for the side (the cylindrical part) costs $1.50 per sq ft, and the material for the roof (the half-sphere) costs$5 per sq ft. a) give the set-up and equations used to solve the problem b) give the values of r and h that minimize the cost c) evaluate for the total cost for the materials $V = \pi r^2 h + \frac{2}{3}\pi r^3 = 2400 \pi$ solve for h in terms of r ... $h = \frac{2400}{r^2} - \frac{2r}{3}$ $A = 2\pi r h + 2\pi r^2$ cost ... $C = 3 \pi r h + 10\pi r^2$ $C = 3\pi r \left(\frac{2400}{r^2} - \frac{2r}{3}\right) + 10\pi r^2$ $C = \pi\left(\frac{7200}{r} + 8r^2\right)$ find $\frac{dC}{dr}$ and determine the value of r that minimizes the cost. 3. you are a life saver! Thanks!	{}
International Conference on Precision Physics and Fundamental Physical Constants (FFK-2015) 12-16 October 2015 Hungarian Academy of Sciences Europe/Budapest timezone The exclusive decays J/psi to D_s- l+ nu_l in a covariant constituent quark model with infrared confinement Oct 15, 2015, 10:10 AM 25m Kis terem (Small conference room) (Hungarian Academy of Sciences) Kis terem (Small conference room) Hungarian Academy of Sciences 1051 Budapest, Széchenyi tér 9. Speaker Chien-Thang Tran (Joint Institute for Nuclear Research; Moscow Institute of Physics and Technology) Description We investigate the exclusive semileptonic decays $J/\psi \to D_{(s)}^{()-} {\ell}^+ \nu_{\ell}$\,, where $\ell=e,\mu$, within the Standart Model. The relevant transition form factors are calculated in the framework of a relativistic constituent quark model with built-in infrared confinement. Our calculations predict the branching fractions $\mathcal{B}(J/\psi \to D_{(s)}^{()-} {\ell}^+ \nu_{\ell})$ to be of the order of $10^{-10}$ for $D_s^{()-}$ and $10^{-11}$ for $D^{(*)-}$. Most of our numerical results are consistent with other theoretical studies. However, some branching fractions are larger than those calculated in QCD sum rules approaches but smaller than those obtained in the covariant light-front quark model by a factor of about $2 - 3$. Primary authors Chien-Thang Tran (Joint Institute for Nuclear Research; Moscow Institute of Physics and Technology) Mikhail Ivanov	{}
# 1st and 2nd fundamental theorem of calculus The fundamental theorem of calculus (FTC) is the formula that relates the derivative to the integral and provides us with a method for evaluating definite integrals. First Fundamental Theorem of Calculus. To receive credit as the author, enter your information below. In every example, we got a F'(x) that is very similar to the f(x) that was provided. The first part of the theorem says that: The Second Fundamental Theorem of Calculus. If we make it equal to "a" in the previous equation we get: But what is that integral? The first theorem is instead referred to as the "Differentiation Theorem" or something similar. (You can preview and edit on the next page), Return from Fundamental Theorem of Calculus to Integrals Return to Home Page. A few observations. Note that the ball has traveled much farther. The last step is to specify the value of the constant C. Now, remember that x is a variable, so it can take any valid value. \$1 per month helps!! The second part of the theorem gives an indefinite integral of a function. Thank you very much. First Fundamental Theorem of Calculus. First and Second Fundamental Theorem of Calculus, Finding the Area Under a Curve (Vertical/Horizontal). As you can see for all of the above examples, we are essentially doing the same thing every time: integrating f(t) with the definite integral to get F(x), deriving it, and then structuring the F'(x) so that it is similar to the original set up of the integral. When we diﬀerentiate F 2(x) we get f(x) = F (x) = x. IT CHANGED MY PERCEPTION TOWARD CALCULUS, AND BELIEVE ME WHEN I SAY THAT CALCULUS HAS TURNED TO BE MY CHEAPEST UNIT. There are several key things to notice in this integral. EK 3.1A1 EK 3.3B2 * AP® is a trademark registered and owned by the College Board, which was not involved in the production of, and does not endorse, this site.® is a trademark registered and owned In this lesson we will be exploring the two fundamentals theorem of calculus, which are essential for continuity, differentiability, and integrals. The Fundamental Theorem of Calculus formalizes this connection. The first part of the fundamental theorem stets that when solving indefinite integrals between two points a and b, just subtract the value of the integral at a from the value of the integral at b. The Second Fundamental Theorem of Calculus shows that integration can be reversed by differentiation. This helps us define the two basic fundamental theorems of calculus. Thanks to all of you who support me on Patreon. The second part tells us how we can calculate a definite integral. This will always happen when you apply the fundamental theorem of calculus, so you can forget about that constant. Find F′(x)F'(x)F′(x), given F(x)=∫−3xt2+2t−1dtF(x)=\int _{ -3 }^{ x }{ { t }^{ 2 }+2t-1dt }F(x)=∫−3xt2+2t−1dt. In fact, we've already seen that the area under the graph of a function f(t) from a to x is: The first part of the fundamental theorem of calculus simply says that: That is, the derivative of A(x) with respect to x equals f(x). The Second Fundamental Theorem of Calculus. Now deﬁne a new function gas follows: g(x) = Z x a f(t)dt By FTC Part I, gis continuous on [a;b] and differentiable on (a;b) and g0(x) = f(x) for every xin (a;b). Proof of the First Fundamental Theorem of Calculus The ﬁrst fundamental theorem says that the integral of the derivative is the function; or, more precisely, that it’s the diﬀerence between two outputs of that function. The fundamental theorem of calculus is a theorem that links the concept of differentiating a function with the concept of integrating a function. - The variable is an upper limit (not a lower limit) and the lower limit is still a constant. The fundamental theorem of calculus establishes the relationship between the derivative and the integral. The Second Part of the Fundamental Theorem of Calculus. As we learned in indefinite integrals, a primitive of a a function f(x) is another function whose derivative is f(x). Just type! Patience... First, let's get some intuition. And let's consider the area under the curve from a to x: If we take a smaller x1, we'll get a smaller area: And if we take a greater x2, we'll get a bigger area: I do this to show you that we can define an area function A(x). This formula says how we can calculate the area under any given curve, as long as we know how to find the indefinite integral of the function. It is essential, though. Finally, you saw in the first figure that C f (x) is 30 less than A f (x). Thus, the two parts of the fundamental theorem of calculus say that differentiation and … Let Fbe an antiderivative of f, as in the statement of the theorem. The second fundamental theorem of calculus states that, if a function “f” is continuous on an open interval I and a is any point in I, and the function F is defined by then F'(x) = f(x), at each point in I. This implies the existence of antiderivatives for continuous functions. Click here to see the rest of the form and complete your submission. We already know how to find that indefinite integral: As you can see, the constant C cancels out. It is sometimes called the Antiderivative Construction Theorem, which is very apt. The first part of the theorem says that: (1) This result, while taught early in elementary calculus courses, is actually a very deep result connecting the purely algebraic indefinite integral and the purely analytic (or geometric) definite integral. The fundamental theorem of calculus is a simple theorem that has a very intimidating name. Fundamental Theorem of Calculus: Part 1 Let $$f(x)$$ be continuous in the domain $$[a,b]$$, and let $$g(x)$$ be the function defined as: The second figure shows that in a different way: at any x-value, the C f line is 30 units below the A f line. Finally, you saw in the first figure that C f (x) is 30 less than A f (x). You can upload them as graphics. If you need to use, Do you need to add some equations to your question? Let's say we have a function f(x): Let's take two points on the x axis: a and x. If you have a problem, or set of problems you can't solve, please send me your attempt of a solution along with your question. Then A′(x) = f (x), for all x ∈ [a, b]. Then A′(x) = f (x), for all x ∈ [a, b]. The fundamental theorem of calculus is central to the study of calculus. The first fundamental theorem is the first of two parts of a theorem known collectively as the fundamental theorem of calculus. The second part tells us how we can calculate a definite integral. The Fundamental Theorem of Calculus theorem that shows the relationship between the concept of derivation and integration, also between the definite integral and the indefinite integral— consists of 2 parts, the first of which, the Fundamental Theorem of Calculus, Part 1, and second is the Fundamental Theorem of Calculus, Part 2. So, don't let words get in your way. In this theorem, the sigma (sum) becomes the integrand, f(x1) becomes f(x), and ∆x (change in x) becomes dx (change of x). The functions of F'(x) and f(x) are extremely similar. How Part 1 of the Fundamental Theorem of Calculus defines the integral. So the second part of the fundamental theorem says that if we take a function F, first differentiate it, and then integrate the result, we arrive back at the original function, but in the form F (b) − F (a). Thus if a ball is thrown straight up into the air with velocity the height of the ball, second later, will be feet above the initial height. Introduction. The First Fundamental Theorem of Calculus links the two by defining the integral as being the antiderivative. EK 3.3A1 EK 3.3A2 EK 3.3B1 EK 3.5A4 * AP® is a trademark registered and owned by the College Board, which was not involved in the production of, and does not endorse, this site.® is a trademark In Section 4.4, we learned the Fundamental Theorem of Calculus (FTC), which from here forward will be referred to as the First Fundamental Theorem of Calculus, as in this section we develop a corresponding result that follows it. We saw the computation of antiderivatives previously is the same process as integration; thus we know that differentiation and integration are inverse processes. In this theorem, the sigma (sum) becomes the integrand, f(x1) becomes f(x), and ∆x (change in x) becomes dx (change of x). This theorem helps us to find definite integrals. A few observations. MATH 1A - PROOF OF THE FUNDAMENTAL THEOREM OF CALCULUS 3 3. Recall that the First FTC tells us that if … Here is the formal statement of the 2nd FTC. You can upload them as graphics. If ‘f’ is a continuous function on the closed interval [a, b] and A (x) is the area function. If is continuous near the number , then when is close to . By the end of this equation, we can see that the derivative of F(x), which is the integral of f(x), is equivalent to the original function f(x). THANKS FOR ALL THE INFORMATION THAT YOU HAVE PROVIDED. This lesson contains the following Essential Knowledge (EK) concepts for the AP Calculus course.Click here for an overview of all the EK's in this course. - The variable is an upper limit (not a … So, our function A(x) gives us the area under the graph from a to x. When we do this, F(x) is the anti-derivative of f(x), and f(x) is the derivative of F(x). Just want to thank and congrats you beacuase this project is really noble. The fundamental theorem of calculus tells us that: b 3 b b 3 x 2 dx = f(x) dx = F (b) − F (a) = 3 − a a a 3 While the two might seem to be unrelated to each other, as one arose from the tangent problem and the other arose from the area problem, we will see that the fundamental theorem of calculus does indeed create a link between the two. The first fundamental theorem of calculus states that, if f is continuous on the closed interval [a,b] and F is the indefinite integral of f on [a,b], then int_a^bf(x)dx=F(b)-F(a). There are several key things to notice in this integral. These will appear on a new page on the site, along with my answer, so everyone can benefit from it. a The fundamental theorem of calculus (FTC) is the formula that relates the derivative to the integral and provides us with a method for evaluating definite integrals. This area function, given an x, will output the area under the curve from a to x. The first FTC says how to evaluate the definite integral if you know an antiderivative of f. The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. Check box to agree to these submission guidelines. Note that the ball has traveled much farther. EK 3.3A1 EK 3.3A2 EK 3.3B1 EK 3.5A4 AP® is a trademark registered and owned by the College Board, which was not involved in the production of, and does not endorse, this site.® is a trademark As we learned in indefinite integrals, a primitive of a a function f(x) is another function whose derivative is f(x). It is broken into two parts, the first fundamental theorem of calculus and the second fundamental theorem of calculus. The Mean Value Theorem for Integrals and the first and second forms of the Fundamental Theorem of Calculus are then proven. This can also be written concisely as follows. Since the lower limit of integration is a constant, -3, and the upper limit is x, we can simply take the expression t2+2t−1{ t }^{ 2 }+2t-1t2+2t−1given in the problem, and replace t with x in our solution. You da real mvps! The solution to the problem is, therefore, F′(x)=x2+2x−1F'(x)={ x }^{ 2 }+2x-1 F′(x)=x2+2x−1. The fundamental theorem of calculus says that this rate of change equals the height of the geometric shape at the final point. You don't learn how to find areas under parabollas in your elementary geometry! If ‘f’ is a continuous function on the closed interval [a, b] and A (x) is the area function. The fundamental theorem of calculus has two parts. This lesson contains the following Essential Knowledge (EK) concepts for the AP Calculus course.Click here for an overview of all the EK's in this course. Second Fundamental Theorem of Calculus The second fundamental theorem of calculus states that if f(x) is continuous in the interval [a, b] and F is the indefinite integral of f(x) on [a, b], then F'(x) = f(x). The first fundamental theorem of calculus describes the relationship between differentiation and integration, which are inverse functions of one another. Of course, this A(x) will depend on what curve we're using. Theorem: (First Fundamental Theorem of Calculus) If f is continuous and b F = f, then f(x) dx = F (b) − F (a). In indefinite integrals we saw that the difference between two primitives of a function is a constant. That is: But remember also that A(x) is the integral from 0 to x of f(t): In the first part we used the integral from 0 to x to explain the intuition. It is the indefinite integral of the function we're integrating. Entering your question is easy to do. The first part of the theorem says that if we first integrate $$f$$ and then differentiate the result, we get back to the original function $$f.$$ Part $$2$$ (FTC2) The second part of the fundamental theorem tells us how we can calculate a definite integral. The second figure shows that in a different way: at any x-value, the C f line is 30 units below the A f line. This helps us define the two basic fundamental theorems of calculus. When you see the phrase "Fundamental Theorem of Calculus" without reference to a number, they always mean the second one. The Fundamental Theorem of Calculus is a theorem that connects the two branches of calculus, differential and integral, into a single framework. Specifically, for a function f that is continuous over an interval I containing the x-value a, the theorem allows us to create a new function, F(x), by integrating f from a to x. The Second Fundamental Theorem of Calculus As if one Fundamental Theorem of Calculus wasn't enough, there's a second one. A special case of this theorem was first described by Parameshvara (1370–1460), from the Kerala School of Astronomy and Mathematics in India, in his commentaries on Govindasvāmi and Bhāskara II. In fact, this “undoing” property holds with the First Fundamental Theorem of Calculus as well. This integral gives the following "area": And what is the "area" of a line? - The integral has a variable as an upper limit rather than a constant. The First Fundamental Theorem of Calculus. You already know from the fundamental theorem that (and the same for B f (x) and C f (x)). We being by reviewing the Intermediate Value Theorem and the Extreme Value Theorem both of which are needed later when studying the Fundamental Theorem of Calculus. Do you need to add some equations to your question? Entering your question is easy to do. It can be used to find definite integrals without using limits of sums . History. Next lesson: Finding the ARea Under a Curve (vertical/horizontal). If you have just a general doubt about a concept, I'll try to help you. Click here to upload more images (optional). The First Fundamental Theorem of Calculus Our ﬁrst example is the one we worked so hard on when we ﬁrst introduced deﬁnite integrals: Example: F (x) = x3 3. The result of Preview Activity 5.2 is not particular to the function $$f (t) = 4 − 2t$$, nor to the choice of “1” as the lower bound in the integral that defines the function $$A$$. How Part 1 of the Fundamental Theorem of Calculus defines the integral. Here, the F'(x) is a derivative function of F(x). Let's say we have another primitive of f(x). The Second Part of the Fundamental Theorem of Calculus. The second fundamental theorem of calculus holds for f a continuous function on an open interval I and a any point in I, and states that if F is defined by the integral (antiderivative) F(x)=int_a^xf(t)dt, then F^'(x)=f(x) at each point in I, where F^'(x) is the derivative of F(x). Recommended Books on … Create your own unique website with customizable templates. This does not make any difference because the lower limit does not appear in the result. As we learned in indefinite integrals, a primitive of a a function f(x) is another function whose derivative is f(x). Using the Second Fundamental Theorem of Calculus, we have . Using the Second Fundamental Theorem of Calculus, we have . To create them please use the equation editor, save them to your computer and then upload them here. Second fundamental theorem of Calculus The second part tells us how we can calculate a definite integral. You'll get used to it pretty quickly. To get a geometric intuition, let's remember that the derivative represents rate of change. Or, if you prefer, we can rearr… The formula that the second part of the theorem gives us is usually written with a special notation: In example 1, using this notation we would have: This is a simple and useful notation. This theorem allows us to avoid calculating sums and limits in order to find area. How the heck could the integral and the derivative be related in some way? This lesson contains the following Essential Knowledge (EK) concepts for the AP Calculus course.Click here for an overview of all the EK's in this course. So, for example, let's say we want to find the integral: The fundamental theorem of calculus says that this integral equals: And what is F(x)? However, we could use any number instead of 0. In this equation, it is as if the derivative operator and the integral operator “undo” each other to leave the original function . Get some intuition into why this is true. This theorem gives the integral the importance it has. This math video tutorial provides a basic introduction into the fundamental theorem of calculus part 1. The total area under a curve can be found using this formula. Just type! A restricted form of the theorem was proved by Michel Rolle in 1691; the result was what is now known as Rolle's theorem, and was proved only for polynomials, without the … The fundamental theorem of calculus connects differentiation and integration , and usually consists of two related parts . First fundamental theorem of calculus: $\displaystyle\int_a^bf(x)\,\mathrm{d}x=F(b)-F(a)$ This is extremely useful for calculating definite integrals, as it removes the need for an infinite Riemann sum. It is the theorem that shows the relationship between the derivative and the integral and between the definite integral and the indefinite integral. As you can see, the function (x/4)^2 is matched with the width of the rectangle being (1/4) to then create a definite integral in the interval [0, 1] (as four rectangles of width 1/4 would equal 1) of the function x^2. Remember that F(x) is a primitive of f(t), and we already know how to find a lot of primitives! If you need to use equations, please use the equation editor, and then upload them as graphics below. Let be a continuous function on the real numbers and consider From our previous work we know that is increasing when is positive and is decreasing when is negative. The first one is the most important: it talks about the relationship between the derivative and the integral. The second fundamental theorem of calculus holds for f a continuous function on an open interval I and a any point in I, and states that if F is defined by the integral (antiderivative) F(x)=int_a^xf(t)dt, then F^'(x)=f(x) at each point in I, where F^'(x) is the derivative of F(x). Let's call it F(x). It has gone up to its peak and is falling down, but the difference between its height at and is ft. That simply means that A(x) is a primitive of f(x). In this wiki, we will see how the two main branches of calculus, differential and integral calculus, are related to each other. So, we have that: We have the value of C. Now, if we want to calculate the definite integral from a to b, we just make x=b in the original formula to get: And that's an impressive result. This is a very straightforward application of the Second Fundamental Theorem of Calculus. The Second Fundamental Theorem of Calculus shows that integration can be reversed by differentiation. It is actually called The Fundamental Theorem of Calculus but there is a second fundamental theorem, so you may also see this referred to as the FIRST Fundamental Theorem of Calculus. Moreover, with careful observation, we can even see that is concave up when is positive and that is concave down when is negative. - The integral has a variable as an upper limit rather than a constant. This integral we just calculated gives as this area: This is a remarkable result. The result of Preview Activity 5.2 is not particular to the function $$f (t) = 4 − 2t$$, nor to the choice of “1” as the lower bound in the integral that defines the function $$A$$. The Second Fundamental Theorem of Calculus establishes a relationship between a function and its anti-derivative. Conversely, the second part of the theorem, someti :) https://www.patreon.com/patrickjmt !! Second fundamental theorem of Calculus It is zero! Second Fundamental Theorem of Calculus The second fundamental theorem of calculus states that if f(x) is continuous in the interval [a, b] and F is the indefinite integral of f(x) on [a, b], then F'(x) = f(x). You already know from the fundamental theorem that (and the same for B f (x) and C f (x)). The first part of the theorem, sometimes called the first fundamental theorem of calculus, states that one of the antiderivatives, say F, of some function f may be obtained as the integral of f with a variable bound of integration. The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. It just says that the rate of change of the area under the curve up to a point x, equals the height of the area at that point. Its equation can be written as . PROOF OF FTC - PART II This is much easier than Part I! So, replacing this in the previous formula: Here we're getting a formula for calculating definite integrals. That is, the area of this geometric shape: A'(x) will give us the rate of change of this area with respect to x. As antiderivatives and derivatives are opposites are each other, if you derive the antiderivative of the function, you get the original function. THANKS ONCE AGAIN. To create them please use the. It has gone up to its peak and is falling down, but the difference between its height at and is ft. Thus if a ball is thrown straight up into the air with velocity the height of the ball, second later, will be feet above the initial height. The second part of the fundamental theorem of calculus tells us that to find the definite integral of a function ƒ from to , we need to take an antiderivative of ƒ, call it , and calculate ()-(). If you are new to calculus, start here. X ∈ [ a, b ] start here integral as being the Construction! 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# chains and countability Given a point $x$ in a topological space $X$. I was wondering, whether one can always find a local basis at $x$, which is totally ordered (a chain) under inclusion. For example this is true for spaces, which have countable local basis: If $\{U_i\|i\in\mathbb{N}\}$ is a local basis at $X$, then $\{\bigcap_{i=1}^nU_i\|n\in \mathbb{N}\}$ is totally ordered. So my question is: Assume, there is a totally ordered local basis at $x\in X$. Does this already imply, that there is a countable local basis at $x\in X$? The rest of this text contains the motivating example. So if you are motivated enough, you can stop reading. For example the space $\prod_{i\in I} \{0;1\}$ does not have a totally ordered local basis at $\{0\}^I$ for uncountable $I$. Assume $P$ is such a basis. Let for any open neighborhood $U$ of $\{0\}^I$ $N_U:=\{i\in I\| 1_i\notin U\}$ where $1_i$ denotes the element of $\prod_{i\in I} \{0;1\}$ with zeros everywhere except for the $i$-th entry. Note that $N_U$ is always finite by definition of the product topology and that for $U\subset U'$ we get $N_U\supset N_{U'}$ Further note that $I=\bigcup_{U\in P}N_U$. Now construct an injection $f:\mathbb{N}\rightarrow I$ in the following way Choose any $U_0\in P$ and enumerate all elements of $N_0$ in your favourite way. Then choose a $U_1\in P$ with $N_{U_0}\subsetneq N_{U_1}$. Then enumerate all elements in $N_{U_1}\setminus N_{U_0}$. Continue this way to get an injection $\mathbb{N}\rightarrow I$. As $I$ is uncountable, this map can't be surjective. Choose $i\in I\setminus Im(f)$ and any basis element $U\in P$ with $i\in N_U$. So for any $n\in \mathbb{N}$, we get $i\notin N_{U_n} \Rightarrow U_n\nsubseteq U\Rightarrow U\subseteq U_n\Rightarrow N_U \supset N_{U_n}$. The second last implication holds, as the base is totally ordered. This holds for any $n$, so $N_U \supset \bigcup_nN_{U_n}$. But the left side is finite and the right side is (by constuction) infinite. Contradiction! So the cannot be a totally ordered local base. - The space $\omega_1+1$ under the order topology, where $\omega_1$ refers to the first uncountable ordinal, has a linearly ordered local basis at the point $\omega_1$ (and indeed at every point), consisting of the intervals $(\alpha,\infty)$, but there is no countable local basis at that point, because every countable set of ordinals below $\omega_1$ is bounded below $\omega_1$.	{}
# Off the Hooke To grasp nonlinear structural analysis, unlearn Hooke's Law, $latex \sigma = E\varepsilon$. And all derivative outcomes of Hooke's Law. Like $latex \kappa=M/EI$ and $latex {\bf K} {\bf U} = {\bf P}$. https://www.youtube.com/watch?v=z4jeREy7Pbc When steel yields, concrete cracks, etc., Hooke's Law no longer applies and holding on to it can lead to erroneous interpretations of nonlinear … Continue reading Off the Hooke # Verifying Ain’t Easy I've posted a few modeling challenges on frame analysis (strongback, Ziemian, and stability) and soil-structure interaction. However, I recently accepted a challenge from George Chamosfakidis to see if OpenSees can give the same periods and mode shapes reported in the ETABS verification example shown below. Published verification examples typically just show the "correct" result and … Continue reading Verifying Ain’t Easy # Absolutely, It’s Relative One of the most frequently asked OpenSees questions is whether node recorders record absolute or relative displacement (relative to the ground) when a model is subjected to a uniform excitation. There's several approaches to find the answer to this question. One solution is to apply a simple uniform excitation--like a constant ground acceleration--to an SDF … Continue reading Absolutely, It’s Relative # You Gotta Keep ’em Aggregated The SectionAggregator was one of my few useful OpenSees ideas. This class gives a flexible way to combine, or aggregate, modes of force-deformation in a single section model. The idea for SectionAggregator came from the Decorator software design pattern, the same pattern from which so many UniaxialMaterial wrappers were spawned (here and here). In fact, … Continue reading You Gotta Keep ’em Aggregated # Minimal Working Example When people post online or e-mail me about what could be a bug in OpenSees, I'll ask for a minimal working example (MWE), i.e., a simple script the demonstrates the problem. I don't want to deal with elaborate scripts--yours or mine. So, what does an MWE look like for OpenSees? Here's a non-exhaustive list of … Continue reading Minimal Working Example # Tcl as a Front End for Python I know I'm not the only one who enjoys converting between scripting languages or between structural analysis programs. I've had fun writing bespoke Tcl middleware between OpenSees and MATLAB, but now OpenSeesPy makes all of that obsolete. But, let's say you have an OpenSees Tcl script that you'd like to run in OpenSeesPy. There's a … Continue reading Tcl as a Front End for Python # There’s Three, Actually The displacement-based and force-based formulations garner a lot of comparisons for simulating nonlinear frame response. My Google Scholar alerts tell me so. And I even wrote a post comparing the two formulations. Doc Ock from Spider-Man: Into the Spider-Verse There is a third formulation--the mixed formulation. Alemdar and White compared three frame element formulations (displacement-based, … Continue reading There’s Three, Actually # Random Bullets on Blogging It's 2021. So, why blog when peak blogging was like 2007? Here are some random bullets on why blogging works for me. The more you write, the more ideas you generate. Usually it's ideas for more blog posts, but sometimes it's new ideas for research.Blog posts have a direct, conversational tone. Like I'm talking with … Continue reading Random Bullets on Blogging # The Rayleigh Quotient Eigenvalue analysis wasn't giving me what I wanted the other day. So, to make a long story short, I decided to try Rayleigh's method. I won't go through all the details of Rayleigh's method, but the basic idea is you can obtain a very good approximation of the fundamental frequency of a structural model by … Continue reading The Rayleigh Quotient	{}
## May 28, 2008 ### Manin on Foundations #### Posted by David Corfield Out of the series of observations made by Yuri Manin in Truth as value and duty: lessons of mathematics most relevant to us here is: For a working mathematician, when he/she is concerned at all, “foundations” is simply a general term for the historically variable set of rules and principles of organization of the body of mathematical knowledge, both existing and being created. From this viewpoint, the most influential foundational achievement in the 20th century was an ambitious project of the Bourbaki group, building all mathematics, including logic, around set-theoretical “structures” and making Cantor’s language of sets a common vernacular of algebraists, geometers, probabilists and all other practitioners of our trade. These days, this vernacular, with all its vocabulary and ingrained mental habits, is being slowly replaced by the languages of category theory and homotopy theory and their higher extensions. Respectively, the basic “left-brain” intuition of sets, composed of distinguishable elements, is giving way to a new, more “right brain” basic intuition dealing with space-like and continuous primary images, both deformable and deforming. Has mathematics learned better to employ its corpus callosum? Posted at May 28, 2008 6:29 PM UTC TrackBack URL for this Entry: http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1700 ### Re: Manin on Foundations Is that an emoticon on page 5 of Manin’s paper? I don’t think I’ve ever before seen an emoticon in a philosophy paper, though it’s bound to become common eventually. Posted by: John Baez on May 29, 2008 3:40 AM \| Permalink \| Reply to this ### Re: Manin on Foundations I agree with his distinction between the styles, but I like to call it holistic/reductionistic rather than right/left brain. And I usually hate the words holistic and reductionistic, but here they seem to do exactly the right job. Posted by: James on May 29, 2008 4:43 AM \| Permalink \| Reply to this ### Re: Manin on Foundations Yes, you would have thought we’d got beyond simple left/right brain dichotomies. Still, I suppose it works as a kind of shorthand. For example, Arnold declares: In the middle of the twentieth century a strong mafia of left-brained mathematicians succeeded in eliminating all geometry from the mathematical education (first in France and later in most other countries), replacing the study of all content in mathematics by the training in formal proofs and the manipulation of abstract notions. Of course, all the geometry, and, consequently, all relations with the real world and other sciences have been eliminated from the mathematics teaching. All the same, it might be interesting to do some brain imaging of real mathematicians thinking about real mathematics. A couple of years ago I was discussing the idea with a brain scientist and even put out a call for participation. Existing studies seem rather limited, if indicative. For example, in ‘Interhemispheric Interaction During Global-Local Processing in Mathematically Gifted Adolescents, Average-Ability Youth, and College Students’, Singh, Harnam; W. O’Boyle, Michael; Neuropsychology, Vol 18(2), Apr 2004. pp. 371-377, the authors argue that the ‘mathematically gifted’ have greater interhemispherical activity. Abstract: Interhemispheric interaction in mathematically gifted (MG) adolescents, average-ability (AA) youth, and college students (CS) was examined by presenting hierarchical letter pairs in 3 viewing conditions: (a) unilaterally to the right hemisphere (RH), (b) unilaterally to the left hemisphere (LH), or (c) bilaterally, with 1 member of the pair presented to each hemisphere simultaneously. Participants made global-local, match-no-match judgments. For the AA and CS, the LH was faster for local matches and the RH for global matches. The MG showed no hemispheric differences. Also, AA and CS were slower on cooperative compared with unilateral trials, whereas the MG showed the opposite pattern. These results suggest that enhanced interhemispheric interaction is a unique functional characteristic of the MG brain. Atiyah speculates here about a much more ambitious project to search for the neural underpinnings of mathematicians’ aesthetic sense: Despite all the difficulties associated with understanding or defining beauty we can still ask what mechanisms in the brain are involved in its appreciation. This can be asked about beauty in the various arts or in mathematics. It is a fascinating question whether there is any commonality across all areas. Are we just misled by the inadequacies of language and the misleading power of metaphor? As I argued at the beginning, mathematics is a pure form of thought and so it may provide an easier field for physiological study. It is still a daunting task. We have to identify many instances of what a mathematician finds beautiful and see, by experiment, if there is any region of the brain that is common to them. We can for example compare beauty, as illustrated in geometric form, with the more formal beauty of an elegant algebraic formula or of a subtle abstract argument. Posted by: David Corfield on May 29, 2008 10:01 AM \| Permalink \| Reply to this ### Re: Manin on Foundations DC wrote: We can for example compare beauty, as illustrated in geometric form, with the more formal beauty of an elegant algebraic formula or of a subtle abstract argument. SH: Providing an opportunity to mention the last two books I’ve read, “Why Beauty Is Truth: The History of Symmetry” by Ian Stewart and “The Equation That Couldn’t Be Solved” how mathematical genius discovered the language of Symmetry by Mario Livio Both authors mention symmetric faces being more attractive and bees are attracted to flowers whose petals are more symmetric. They both mention the connection to group theory but I think Livio’s book is a bit broader and deeper. Posted by: Stephen Harris on May 29, 2008 12:52 PM \| Permalink \| Reply to this ### Re: Manin on Foundations DC wrote… Not DC, but Sir MA, FRS, OM. Posted by: David Corfield on May 29, 2008 1:17 PM \| Permalink \| Reply to this ### Re: Manin on Foundations I’m interested in how some of the cognitive findings in “Where Mathematics Comes From: How the Embodied Mind Brings Mathematics into Being” by Lakoff and Nunez can be related to mathematical aesthetics. I really think that their “mathematical idea analysis” gives a good method. Posted by: Matt on May 29, 2008 3:41 PM \| Permalink \| Reply to this ### Re: Manin on Foundations The handful of mathematicians I’ve heard talking about ‘Where Mathematics Comes From’ seemed united in thinking that while it points in an interesting direction, something isn’t quite convincing about their position. Perhaps something in our discussion is more promising. Posted by: David Corfield on May 29, 2008 10:19 PM \| Permalink \| Reply to this ### Re: Manin on Foundations Could I trouble you to give some sort of summary of their position? I have been rather curious, although the casual glance at the bookstore did not result in a very positive impression. Posted by: Minhyong Kim on May 30, 2008 6:51 PM \| Permalink \| Reply to this ### Re: Manin on Foundations I’ll have to come clean and admit that I have only browsed it too. I did read earlier work of Lakoff with Mark Johnson. I wrote a little on what they called ‘image schemas’ in my PhD thesis. I was struck at the time by the relevance of their analysis of ‘over’ in Metaphors We Live By for the mathematical notions of cover and sheaf. Examples like these might have been better choices than the run-of-the-mill ‘infinity’, ‘continuity’, ‘set’, ‘number’ examples of Where mathematics comes from. A synopsis of the book in the Wikipedia article seems a reasonable starting point. Posted by: David Corfield on June 2, 2008 1:19 PM \| Permalink \| Reply to this ### Re: Manin on Foundations There is one remark I think of often that illustrates rather well the difficulty of capturing what geometric intuition really is in any simple terms, coming as it does from Sir MA’s student Graeme Segal, known mainly for his work in geometry and topology: A good sentence is worth a thousand pictures.’ We were discussing knots and QFT’s in that conversation, and I think John was present. Segal has a very non-trivial appreciation of high-brow literature (as far as I’m able to judge, of course). Posted by: Minhyong Kim on May 29, 2008 2:53 PM \| Permalink \| Reply to this ### Re: Manin on Foundations Perhaps JB will recall an incident at a conference in which the host of the conference badgered the speaker to not demonstrate any more pictures. The speaker was prepared with a slide that contained the arcane formulas that were restatements of the pictures. The identity of neither should be made public. Those who were present may also recall. Yes, thoughtful and careful writing can create imagery, and the mathematical author who insists upon not using pictures has an obligation to create the unambiguous image that (s)he is experiencing. As a practitioner who illustrates with drawings and with words, I don’t think either task is particularly easy. By using both, one creates redundancy in exposition, but the redundancy facilitates the flow of information across the channel and between the lobes. The 4-dimensional imagery of a mathematical mind at work would indeed be interesting. Perhaps tenure and promotion decisions could be determined by how much of the brain is firing during the production of the candidates magnus opus ;-) Posted by: Scott Carter on May 29, 2008 4:10 PM \| Permalink \| Reply to this ### Re: Manin on Foundations I do indeed recall both these incidents. Thanks for reminding me! Posted by: John Baez on May 29, 2008 4:49 PM \| Permalink \| Reply to this ### Re: Manin on Foundations > we can still ask what mechanisms in the brain are involved in its appreciation. I expect this to be a difficult task. Appreciation of mathematical beauty isn’t necessarily something that takes place while doing mathematics. When doing mathematics, you’re often way too busy doing mathematics to stop and think about whether or not what you’re doing is beautiful. A while back a researcher friend of mine complained to me about how there was little beauty to be seen in his day-to-day work. He was just getting on with his job. It was only when he wasn’t doing mathematics he could look back and appreciate what he had been working on. Mathematical appreciation isn’t a state of mind that exists at any given instant, it’s a long-term ongoing process. Posted by: Dan Piponi on May 29, 2008 7:19 PM \| Permalink \| Reply to this ### Re: Manin on Foundations Dan wrote: A while back a researcher friend of mine complained to me about how there was little beauty to be seen in his day-to-day work. He was just getting on with his job. Ugh! It sounds like he’s gotten himself stuck in a trap. Please remind him that he’s going to die soon and he needs to have fun now if ever. I spend a lot of time enjoying the beauty of mathematics. I’m sure I could be more ‘productive’ if I knuckled down and worked harder… but why, really? It’s not as if the world needs more theorems per year. I actually work too hard already! If I were really smart I’d goof off more. Posted by: John Baez on May 30, 2008 8:49 AM \| Permalink \| Reply to this ### Re: Manin on Foundations But it’s not just a question of the search for beauty as provider of fun versus the unenjoyable graft of theorem proving. We dwell on mathematics and affirm its statements for the sake of its intellectual beauty, which betokens the reality of its conceptions and the truth of its assertions, says Michael Polanyi. Without this betokening how would you know where to go? While in the natural sciences the feeling of making contact with reality is an augury of as yet undreamed of future empirical confirmations of an immanent discovery, in mathematics it betokens an indeterminate range of future germinations within mathematics itself. Posted by: David Corfield on May 30, 2008 10:31 AM \| Permalink \| Reply to this ### Re: Manin on Foundations Perhaps this remark takes us off-topic, but I’ve never felt much sympathy for a clear division between ‘work’ and ‘fun.’ The reality of such categories may even be problematic in general, but certainly seems artificial in the quotidian existence of a university professor. Perhaps I’ll go out on a limb and state my suspicion that the rigid psychological dichotomy of leisure vs. labor creates real tension in a broad spectrum of human concerns, including (obviously) education as well as the administration of industrial economies. Justifying this suspicion, of course, would take too much work! Posted by: Minhyong Kim on May 30, 2008 4:31 PM \| Permalink \| Reply to this ### Re: Manin on Foundations Afterthought: Maybe it wasn’t entirely off-topic. I think I was vaguely trying to sympathize with Dan Piponi’s friend and his last sentence. And then, I did mention categories’ in the post. Posted by: Minhyong Kim on May 30, 2008 4:45 PM \| Permalink \| Reply to this ### Re: Manin on Foundations Minhyong wrote: Perhaps this remark takes us off-topic, but I’ve never felt much sympathy for a clear division between ‘work’ and ‘fun.’ The reality of such categories may even be problematic in general, but certainly seems artificial in the quotidian existence of a university professor. I have no sympathy for this distinction at all. I hate it! But for me, at least, the distinction seems real — secretly self-created, but not easy to escape. I start projects because they’re ‘fun’; when I want to start the next project, the current one becomes ‘work’. However, especially when I’m writing a paper with coauthors, or I’ve promised a paper for a conference proceedings, I feel compelled to finish my projects! So then I’m stuck doing ‘work’ when I want to have ‘fun’ — that is, think about something new. I then start procrastinating… and can make myself really miserable. Last fall I was really depressed about the huge amount of work I’d set up for myself, which stood like a huge wall between me and the new ideas I wanted to think about. When I’m writing a paper by myself, without any deadline, I don’t seem to fall into this trap quite so badly. The reason is that then I write the paper at approximately the rate at which I figure things out. I figure things out by writing the paper! Figuring things out is fun. Ideally, this means that when I’m done having fun, the paper is done. Unfortunately I like having coauthors, where the ‘figuring out’ stage consists of sitting around talking. That’s fun. But then the writing up somehow becomes ‘work’. So, it’s a complicated mind game, which I keep feeling I should be able to escape simply by taking a more enlightened attitude… but I haven’t succeeded yet. Now I have to get to work. Posted by: John Baez on May 30, 2008 5:58 PM \| Permalink \| Reply to this ### Re: Manin on Foundations Aha! I always knew the reason we aren’t able to collaborate is that you work so much harder than I do! Posted by: Minhyong Kim on May 30, 2008 6:46 PM \| Permalink \| Reply to this Post a New Comment	{}
Did you take the 7th Grade IAR Math Practice Test? If so, then it’s time to review your results to see where you went wrong and what areas you need to improve. 1- Choice C is correct If the score of Mia was $$90$$, then the score of Ava is $$30$$. Since the score of Emma was one and a half as that of Ava, therefore, the score of Emma is $$1.5×30=45$$. 2- Choice A is correct Write the ratio and solve for $$x$$. $$\frac{60}{50}=\frac{5x+2}{10}⇒ 12=5x+2 ⇒12-2=5x⇒ x=\frac{10}{5}=2$$ 3- Choice B is correct Let $$x$$ be the number of students in the class. $$40\%$$ of $$x$$ $$=$$ girls, $$25\%$$ of girls $$=$$ tennis player, Find $$25\%$$ of $$40\%$$. Then: $$25\%$$ of $$40\%$$$$=0.25×0.40=0.1=10\%$$ or $$\frac{10}{100}=\frac{1}{10}$$ 4- Choice C is correct Use the information provided in the question to draw the shape. Use Pythagorean Theorem: $$a^2+b^2=c^2$$ $$30^2+40^2=c^2⇒ 900+1,600= c^2⇒2,500= c^2⇒c=50$$ 5- Choice A is correct Write a proportion and solve for $$x$$. $$\frac{12 \space Cans}{ 7.40}=\frac{30 \space Cans}{x}, x= \frac{7.40×30}{12}⇒x=18.5$$ 6- Choice D is correct Use the volume of square pyramid formula. $$V= \frac{1}{3} a^2 h ⇒V=\frac{1}{3} (12\space m)^2×20 \space m ⇒ V=960\space m^3$$ 7- Choice C is correct Let x be the number of soft drinks for $$240$$ guests. Write a proportional ratio to find $$x$$. $$\frac{6 \space soft \space drinks}{8 \space guests} =\frac{x}{240 \space guests}$$, $$x=\frac{240×6}{8}⇒x=180$$ 8- Choice B is correct Use the formula for Percent of Change: $$\frac{New \space Value-Old \space Value}{Old \space Value}×100\%$$, $$\frac{1.75-1.4}{1.4}×100\%=25\%$$ Use PEMDAS (order of operation): $$[8×(-14)+15]-(10)+[4×6]÷3=[-122+15]-(10)+8=-97-10+8=-99$$ 10- Choice D is correct Simplify. $$5x^2 y(2xy^3)^4=5x^2 y(16x^4 y^{12 })=80x^6 y^{13}$$ ## The Absolute Best Book to Ace the 7th Grade IAR Math Test 11- Choice C is correct The distance between Jason and Joe is $$14$$ miles. Jason running at $$6$$ miles per hour and Joe is running at the speed of $$8$$ miles per hour. Therefore, every hour the distance is $$2$$ miles less. $$14÷2=7$$ 12- Choice A is correct Let $$x$$ be the integer. Then: $$5x-9=101$$, Add $$9$$ both sides: $$5x=110$$, Divide both sides by $$5$$: $$x=22$$ 13- Choice D is correct Two and half times of $$18,000$$ is $$45,000$$. One fifth of them cancelled their tickets. One sixth of $$45,000$$ equals $$9,000(\frac{1}{5} ×45,000=9,000)$$. $$36,000(45,000-9,000=36,000)$$ fans are attending this week. 14- Choice C is correct Write the numbers in order: $$25,12,13,18,22,36,22$$ Since we have $$7$$ numbers ($$7$$ is odd), then the median is the number in the middle, which is $$22$$. 15- Choice D is correct The question is: $$615$$ is what percent of $$820$$? Use percent formula: $$part=\frac{percent}{100}×whole$$ $$615=\frac{percent}{100}×820 ⇒ 615=\frac{percent ×820}{100}⇒61,500=percent×820$$⇒ percent$$=\frac{61,500}{820}=75, 615$$ is $$75\%$$ of $$820$$. Therefore, the discount is: $$100\%-75\%=25\%$$ 16- The answer is $$22 \frac{1}{3}$$ miles. Robert runs $$4 \frac{1}{3}$$ miles on Saturday and $$2(4 \frac{1}{3} )$$ miles on Monday and Wednesday. Robert wants to run a total of $$35$$ miles this week. Therefore, subtract $$4 \frac{1}{3}+2(4 \frac{1}{3} )$$ from $$35$$. $$35-(4 \frac{1}{3}+2(4 \frac{1}{3} ))=35-12\frac{2}{3}=22 \frac{1}{3}$$ miles 17- Choice B is correct To find the area of the shaded region, find the difference of the area of two circles. ($$S_1$$: the area of bigger circle.$$S_2$$: the area of the smaller circle). Use the area of circle formula. $$S=πr^2$$ $$S_1- S_2=π(6cm)^2- π(4cm)^2⇒S_1- S_2=36π \space cm^2-16π \space cm^2 ⇒ S_1- S_2 =20π \space cm^2$$ 18- Choice A is correct Use Pythagorean Theorem: $$a^2+b^2=c^2$$, $$12^2+5^2=c^2⇒ 144+25= c^2 ⇒ c^2=169 ⇒c=13$$ 19- Choice A is correct Let $$L$$ be the price of laptop and $$C$$ be the price of computer. $$4(L) =7(C)$$ and $$L = 240 + C$$ Therefore, $$4(240 + C) =7C ⇒ 960 + 4C = 7C ⇒ C=320$$ Jason needs an $$75\%$$ average to pass five exams. Therefore, the sum of $$5$$ exams must be at least $$5×75=375$$, The sum of $$4$$ exams is $$62+73+82+88=305$$. The minimum score Jason can earn on his fifth and final test to pass is: $$375-305=70$$ ## Best 7th Grade IAR Math Prep Resource for 2022 21- Choice B is correct Let $$x$$ be the original price. If the price of a laptop is decreased by $$15\%$$ to $$425$$, then: $$85\%\ of x=425 ⇒ 0.85x=425 ⇒ x=425÷0.85=500$$ 22- Choice C is correct The weight of $$12$$ meters of this rope is: $$12×450 \space g=5,400\space g$$ $$1\space kg=1,000 \space g$$, therefore, $$5,400 \space g÷1,000=5.4\space kg$$ 23- Choice D is correct Only option D is correct. Other options don’t work in the equation. $$(4x-2)x=42$$ 24- Choice C is correct Compare each score: In Algebra Joe scored $$24$$ out of $$32$$ in Algebra that it means $$75\%$$ of total mark. $$\frac{24}{32}= \frac{x}{100}⇒x=75$$ Joe scored $$28$$ out of $$40$$ in science that it means $$70\%$$ of total mark. $$\frac{28}{40}=\frac{x}{100}⇒x=70$$ Joe scored $$72$$ out of $$90$$ in mathematic that it means $$80\%$$ of total mark. $$\frac{72}{90}=\frac{x}{100} ⇒x=80$$ Therefore, his score in mathematics is higher than his other scores. 25-Choice B is correct To find the discount, multiply the number by ($$100\%$$$$-$$rate of discount). Therefore, for the first discount we get: $$(D)(100\%-25\%)=(D)(0.75)=0.75$$ For increase of $$15\%$$: $$(0.75D)(100\%+15\%)=(0.75D)(1.15)=0.8625$$ $$D=86.25\%$$ of $$D$$ 26-Choice B is correct Write the numbers in order: $$42,21,15,28,43,34,26$$, since we have $$7$$ numbers ($$7$$ is odd), then the median is the number in the middle, which is $$28$$. 27-Choice C is correct The average speed of John is $$210÷7=30$$ $$km$$, and the average speed of Alice is: $$160÷5=32$$ $$km$$, Write the ratio and simplify. $$30$$∶ $$32 ⇒ 15∶16$$ 28-Choice D is correct Use the formula for Percent of Change: $$\frac{New \space Value-Old \space Value}{Old \space Value}×100\%$$ $$\frac{42-56}{56}×100\%=-25\%$$ (negative sign here means that the new price is less than old price). 29-Choice C is correct Use the formula of areas of circles.$$Area=πr^2 ⇒ 121π= πr^2 ⇒ 121= r^2⇒ r=11$$ Radius of the circle is $$11$$. Now, use the circumference formula: Circumference$$=2πr=2π(11)=22π$$ 30-Choice B is correct Let x be the number of balls. Then: $$\frac{1}{2}x+\frac{1}{5}x+\frac{1}{10} x+12=x$$ $$(\frac{1}{2}+\frac{1}{5}+\frac{1}{10})x+12=x$$, $$\frac{8}{10}x+12=x,x=60$$, In the bag of small balls $$\frac{1}{5}$$ are white, then: $$\frac{60}{5}=12$$, There are $$12$$ white balls in the bag. 31-Choice A is correct William ate $$\frac{4}{5}$$ of $$10$$ parts of his pizza that it means $$8$$ parts out of $$10$$ parts ($$\frac{4}{5}$$ of $$10$$ parts $$=x ⇒ x=8$$) and left $$2$$ parts. Ella ate $$\frac{1}{2}$$ of $$10$$ parts of her pizza that it means $$5$$ parts out of $$10$$ parts ($$\frac{1}{2}$$ of $$10$$ parts $$= x$$ ⇒ $$x=5$$) and left $$5$$ parts. Therefore, they ate $$(5+2)$$ parts out of $$(10+10)$$ parts of their pizza and left $$(5+2)$$ parts out of $$(10 + 10)$$ parts of their pizza. It means: $$\frac{7}{20}$$, After simplification we have: $$\frac{7}{20}$$ 32-Choices D is correct. The failing rate is $$14$$ out of $$50=$$$$\frac{14}{50}$$, Change the fraction to percent: $$\frac{14}{50} ×100\%=28\%$$ $$28$$ percent of students failed. Therefore, $$72$$ percent of students passed the exam. 33-Choice C is correct $$x\%$$ of $$50$$ is $$6.2$$, then: $$0.50x=6.2 ⇒x=6.2÷0.50=12.4$$ Use the area of the square formula. $$S=a^2 ⇒ 196= a^2 ⇒ a=14$$ One side of the square is $$14$$ feet. Use the perimeter of the square formula. $$P=4a ⇒ P=4(14) ⇒ P=56$$ 35-Choice B is correct. Input the points instead of $$x$$ and $$y$$ in the formula. Only option B works in the equation. $$6x-14=4y, 6(2)-14=4(-\frac{1}{2})⇒-2=-2$$ 36- Choice B is correct The sum of supplement angles is $$180$$. Let $$x$$ be that angle. Therefore, $$x+4x=180$$ $$5x=180$$, divide both sides by $$5$$: $$x=36$$ 37- Choice B is correct Use simple interest formula: $$I=prt$$ ($$I=$$interest,$$p=$$principal,$$r=$$ rate,$$t=$$time) $$I=(16,000)(0.035)(3)=1,680$$ 38- Choice B is correct. Total number of way is $$6×6=36$$, favorable cases is $$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$$. Thus probability that sum of two tice get $$7$$ is $$\frac{6}{36}=\frac{1}{6}$$ To find the number of possible outfit combinations, multiply the number of options for each factor: $$3×8×7=168$$ 40- Choice B is correct. $$7\%$$ of the volume of the solution is alcohol. Let $$x$$ be the volume of the solution. Then: $$7\%$$ of $$x=35$$ $$ml$$ ⇒ $$0.07 x=35$$ ⇒ $$x=35 ÷ 0.07=500$$ ## The Best Books to Ace the 7th Grade IAR Math Test Toggle panel: Yoast SEO ### More math articles SAVE $5 It was$18.99 now it is \$13.99	{}
# 20 Rewriting R code in C++ ## 20.1 Getting started with C++ Q1: With the basics of C++ in hand, it’s now a great time to practice by reading and writing some simple C++ functions. For each of the following functions, read the code and figure out what the corresponding base R function is. You might not understand every part of the code yet, but you should be able to figure out the basics of what the function does. double f1(NumericVector x) { int n = x.size(); double y = 0; for(int i = 0; i < n; ++i) { y += x[i] / n; } return y; } NumericVector f2(NumericVector x) { int n = x.size(); NumericVector out(n); out[0] = x[0]; for(int i = 1; i < n; ++i) { out[i] = out[i - 1] + x[i]; } return out; } bool f3(LogicalVector x) { int n = x.size(); for(int i = 0; i < n; ++i) { if (x[i]) return true; } return false; } int f4(Function pred, List x) { int n = x.size(); for(int i = 0; i < n; ++i) { LogicalVector res = pred(x[i]); if (res[0]) return i + 1; } return 0; } NumericVector f5(NumericVector x, NumericVector y) { int n = std::max(x.size(), y.size()); NumericVector x1 = rep_len(x, n); NumericVector y1 = rep_len(y, n); NumericVector out(n); for (int i = 0; i < n; ++i) { out[i] = std::min(x1[i], y1[i]); } return out; } A: The code above corresponds to the following base R functions: • f1: mean() • f2: cumsum() • f3: any() • f4: Position() • f5: pmin() Q2: To practice your function writing skills, convert the following functions into C++. For now, assume the inputs have no missing values. 1. all(). 2. cumprod(), cummin(), cummax(). 3. diff(). Start by assuming lag 1, and then generalise for lag n. 4. range(). 5. var(). Read about the approaches you can take on Wikipedia. Whenever implementing a numerical algorithm, it’s always good to check what is already known about the problem. A: Let’s port these functions to C++. 1. all() bool allC(LogicalVector x) { int n = x.size(); for (int i = 0; i < n; ++i) { if (!x[i]) return false; } return true; } 2. cumprod(), cummin(), cummax(). NumericVector cumprodC(NumericVector x) { int n = x.size(); NumericVector out(n); out[0] = x[0]; for (int i = 1; i < n; ++i) { out[i] = out[i - 1] * x[i]; } return out; } NumericVector cumminC(NumericVector x) { int n = x.size(); NumericVector out(n); out[0] = x[0]; for (int i = 1; i < n; ++i) { out[i] = std::min(out[i - 1], x[i]); } return out; } NumericVector cummaxC(NumericVector x) { int n = x.size(); NumericVector out(n); out[0] = x[0]; for (int i = 1; i < n; ++i) { out[i] = std::max(out[i - 1], x[i]); } return out; } 3. diff() (Start by assuming lag 1, and then generalise for lag n.) NumericVector diffC(NumericVector x) { int n = x.size(); NumericVector out(n - 1); for (int i = 1; i < n; i++) { out[i - 1] = x[i] - x[i - 1]; } return out ; } NumericVector difflagC(NumericVector x, int lag = 1) { int n = x.size(); if (lag >= n) stop("lag must be less than length(x)."); NumericVector out(n - lag); for (int i = lag; i < n; i++) { out[i - lag] = x[i] - x[i - lag]; } return out; } 4. range() NumericVector rangeC(NumericVector x) { double omin = x[0], omax = x[0]; int n = x.size(); if (n == 0) stop("length(x) must be greater than 0."); for (int i = 1; i < n; i++) { omin = std::min(x[i], omin); omax = std::max(x[i], omax); } NumericVector out(2); out[0] = omin; out[1] = omax; return out; } 5. var() double varC(NumericVector x) { int n = x.size(); if (n < 2) { return NA_REAL; } double mx = 0; for (int i = 0; i < n; ++i) { mx += x[i] / n; } double out = 0; for (int i = 0; i < n; ++i) { out += pow(x[i] - mx, 2); } return out / (n - 1); } ## 20.2 Missing values Q1: Rewrite any of the functions from the first exercise to deal with missing values. If na.rm is true, ignore the missing values. If na.rm is false, return a missing value if the input contains any missing values. Some good functions to practice with are min(), max(), range(), mean(), and var(). A: For this exercise we start with minC() and extend it so it can deal with missing values. We introduce an na_rm argument to make minC() aware of NAs. In case x contains exclusively NA values minC() should return Inf for na_rm = TRUE. For the return values vector data types are used to avoid irregular type conversions. #include <Rcpp.h> using namespace Rcpp; // [[Rcpp::export]] NumericVector minC(NumericVector x, bool na_rm = false) { int n = x.size(); NumericVector out = NumericVector::create(R_PosInf); if (na_rm) { for (int i = 0; i < n; ++i) { if (x[i] == NA_REAL) { continue; } if (x[i] < out[0]) { out[0] = x[i]; } } } else { for (int i = 0; i < n; ++i) { if (NumericVector::is_na(x[i])) { out[0] = NA_REAL; return out; } if (x[i] < out[0]) { out[0] = x[i]; } } } return out; } minC(c(2:4, NA)) #> [1] NA minC(c(2:4, NA), na_rm = TRUE) #> [1] 2 minC(c(NA, NA), na_rm = TRUE) #> [1] Inf We also extend anyC() so it can deal with missing values. Please note that this (again) introduces some code duplication. This could be avoided by moving the check for missing values to the inner loop at the expense of a slight decrease of performance. Here we use LogicalVector as return type. If we would use bool instead, the C++ NA_LOGICAL would be converted into R’s logical TRUE. #include <Rcpp.h> using namespace Rcpp; // [[Rcpp::export]] LogicalVector anyC(LogicalVector x, bool na_rm = false) { int n = x.size(); LogicalVector out = LogicalVector::create(false); if (na_rm == false) { for (int i = 0; i < n; ++i) { if (LogicalVector::is_na(x[i])) { out[0] = NA_LOGICAL; return out; } else { if (x[i]) { out[0] = true; } } } } if (na_rm) { for (int i = 0; i < n; ++i) { if (LogicalVector::is_na(x[i])) { continue; } if (x[i]) { out[0] = true; return out; } } } return out; } anyC(c(NA, TRUE)) # any(c(NA, TRUE)) would return TRUE in this case #> [1] NA anyC(c(NA, TRUE), na_rm = TRUE) #> [1] TRUE Q2: Rewrite cumsum() and diff() so they can handle missing values. Note that these functions have slightly more complicated behaviour. A: Our NA-aware cumsumC() function will return a vector of the same length as x. By default (na_rm = FALSE) all values following the first NA input value will be set to NA, because they depend on the unknown missing value. In case of na_rm = FALSE the NA values are treated like zeros. #include <Rcpp.h> using namespace Rcpp; // [[Rcpp::export]] NumericVector cumsumC(NumericVector x, bool na_rm = false) { int n = x.size(); NumericVector out(n); LogicalVector is_missing = is_na(x); if (!na_rm) { out[0] = x[0]; for (int i = 1; i < n; ++i) { if (is_missing[i - 1]) { out[i] = NA_REAL; } else{ out[i] = out[i - 1] + x[i]; } } } if (na_rm) { if (is_missing[0]) { out[0] = 0; } else { out[0] = x[0]; } for (int i = 1; i < n; ++i) { if (is_missing[i]) { out[i] = out[i-1] + 0; } else { out[i] = out[i-1] + x[i]; } } } return out; } cumsumC(c(1, NA, 2, 4)) #> [1] 1 NA NA NA cumsumC(c(1, NA, 2, 4), na_rm = TRUE) #> [1] 1 1 3 7 The diffC() implementation will return an NA vector of length length(x) - lag, if the input vector contains a missing value. In case of na_rm = TRUE, the function will return an NA for every difference with at least one NA as input. #include <Rcpp.h> using namespace Rcpp; // [[Rcpp::export]] NumericVector diffC(NumericVector x, int lag = 1, bool na_rm = false) { int n = x.size(); if (lag >= n) stop("lag must be less than length(x)."); NumericVector out(n - lag); for (int i = lag; i < n; i++) { if (NumericVector::is_na(x[i]) \|\| NumericVector::is_na(x[i - lag])) { if (!na_rm) { return rep(NumericVector::create(NA_REAL), n - lag); } out[i - lag] = NA_REAL; continue; } out[i - lag] = x[i] - x[i - lag]; } return out; } diffC(c(1, 3, NA, 10)) #> [1] NA NA NA diffC(c(1, 3, NA, 10), na_rm = TRUE) #> [1] 2 NA NA ## 20.3 Standard Template Library To practice using the STL algorithms and data structures, implement the following using R functions in C++, using the hints provided: Q1: median.default() using partial_sort. A: The median is computed differently for even or odd vectors, which we allow for in the function below. #include <algorithm> #include <Rcpp.h> using namespace Rcpp; // [[Rcpp::export]] double medianC(NumericVector x) { int n = x.size(); if (n % 2 == 0) { std::partial_sort (x.begin(), x.begin() + n / 2 + 1, x.end()); return (x[n / 2 - 1] + x[n / 2]) / 2; } else { std::partial_sort (x.begin(), x.begin() + (n + 1) / 2, x.end()); return x[(n + 1) / 2 - 1]; } } Q2: %in% using unordered_set and the find() or count() methods. A: We use the find() method and loop through the unordered_set until we find a match or have scanned the entire set. #include <Rcpp.h> #include <unordered_set> using namespace Rcpp; // [[Rcpp::export]] LogicalVector inC(CharacterVector x, CharacterVector table) { std::unordered_set<String> seen; seen.insert(table.begin(), table.end()); int n = x.size(); LogicalVector out(n); for (int i = 0; i < n; ++i) { out[i] = seen.find(x[i]) != seen.end(); } return out; } Q3: unique() using an unordered_set (challenge: do it in one line!). A: The insert()-method will return if an equivalent element already exists. If a new element is inserted, we will add it to the (unique) return vector of our function. #include <Rcpp.h> #include <unordered_set> using namespace Rcpp; // [[Rcpp::export]] NumericVector uniqueC(NumericVector x) { std::unordered_set<int> seen; int n = x.size(); std::vector<double> out; for (int i = 0; i < n; ++i) { if (seen.insert(x[i]).second) out.push_back(x[i]); } return wrap(out); } // As a one-liner // [[Rcpp::export]] std::unordered_set<double> uniqueCC(NumericVector x) { return std::unordered_set<double>(x.begin(), x.end()); } Q4: min() using std::min(), or max() using std::max(). A: We will implement min() by iterating over the vector and recursively comparing each element to the current minimum value. #include <Rcpp.h> using namespace Rcpp; // [[Rcpp::export]] double minC(NumericVector x) { int n = x.size(); double out = x[0]; for (int i = 0; i < n; i++) { out = std::min(out, x[i]); } return out; } Q5: which.min() using min_element, or which.max() using max_element. A: To implement which.min(), we will first locate the min_element and then compute the distance() to it (starting from the beginning of the vector). #include <Rcpp.h> #include <algorithm> #include <iterator> using namespace Rcpp; // [[Rcpp::export]] double which_minC(NumericVector x) { int out = std::distance( x.begin(), std::min_element(x.begin(), x.end()) ); return out + 1; } Q6: setdiff(), union(), and intersect() for integers using sorted ranges and set_union, set_intersection and set_difference. A: The structure of the three functions will be very similar. We first sort both input vectors. Then we apply the respective set_union, set_intersection or set_difference function. After that, the result will be between the iterators tmp.begin() and out_end. To retrieve the result, we loop once through the range between tmp.begin() and out_end in the last part of each function. The set operations in base R will discard duplicated values in the arguments. We achieve a similar behaviour by introducing a deduplication step, which omits values that match their predecessor. For the symmetric set functions unionC and intersectC this step is implemented for the output vector. For setdiffC the deduplication is applied to the first input vector. #include <Rcpp.h> #include <unordered_set> #include <algorithm> using namespace Rcpp; // [[Rcpp::plugins(cpp11)]] // [[Rcpp::export]] IntegerVector unionC(IntegerVector x, IntegerVector y) { int nx = x.size(); int ny = y.size(); IntegerVector tmp(nx + ny); std::sort(x.begin(), x.end()); // unique std::sort(y.begin(), y.end()); IntegerVector::iterator out_end = std::set_union( x.begin(), x.end(), y.begin(), y.end(), tmp.begin() ); int prev_value = 0; IntegerVector out; for (IntegerVector::iterator it = tmp.begin(); it != out_end; ++it) { if ((it != tmp.begin()) && (prev_value == it)) continue; out.push_back(it); prev_value = it; } return out; } // [[Rcpp::export]] IntegerVector intersectC(IntegerVector x, IntegerVector y) { int nx = x.size(); int ny = y.size(); IntegerVector tmp(std::min(nx, ny)); std::sort(x.begin(), x.end()); std::sort(y.begin(), y.end()); IntegerVector::iterator out_end = std::set_intersection( x.begin(), x.end(), y.begin(), y.end(), tmp.begin() ); int prev_value = 0; IntegerVector out; for (IntegerVector::iterator it = tmp.begin(); it != out_end; ++it) { if ((it != tmp.begin()) && (prev_value == it)) continue; out.push_back(it); prev_value = it; } return out; } // [[Rcpp::export]] IntegerVector setdiffC(IntegerVector x, IntegerVector y) { int nx = x.size(); int ny = y.size(); IntegerVector tmp(nx); std::sort(x.begin(), x.end()); int prev_value = 0; IntegerVector x_dedup; for (IntegerVector::iterator it = x.begin(); it != x.end(); ++it) { if ((it != x.begin()) && (prev_value == it)) continue; x_dedup.push_back(it); prev_value = it; } std::sort(y.begin(), y.end()); IntegerVector::iterator out_end = std::set_difference( x_dedup.begin(), x_dedup.end(), y.begin(), y.end(), tmp.begin() ); IntegerVector out; for (IntegerVector::iterator it = tmp.begin(); it != out_end; ++it) { out.push_back(it); } return out; } Let’s verify, that these functions work as intended. # input vectors include duplicates x <- c(1, 2, 3, 3, 3) y <- c(3, 3, 2, 5) union(x, y) #> [1] 1 2 3 5 unionC(x, y) #> [1] 1 2 3 5 intersect(x, y) #> [1] 2 3 intersectC(x, y) #> [1] 2 3 setdiff(x, y) #> [1] 1 setdiffC(x, y) #> [1] 1	{}
# Resize font when screen resizes desktop Libgdx I create my font using the following code: public BitmapFont createFont(FreeTypeFontGenerator ftfg, float dp) { FreeTypeFontParameter f = new FreeTypeFontParameter(); f.size = (int)(dp * Gdx.graphics.getDensity()); f.color = Color.BLACK; f.minFilter = Texture.TextureFilter.Nearest; f.magFilter = Texture.TextureFilter.MipMapLinearNearest; ftfg.scaleForPixelHeight((int)(dp * Gdx.graphics.getDensity())); return ftfg.generateFont(f); } myFont = createFont(new FreeTypeFontGenerator(Gdx.files.internal("Fonts/Roboto-Black.ttf")), 16); I have some labels to which I set the font using: Label myLabel= new Label("Text", uiSkin); myLabel.setWrap(true); myLabel.getStyle().font = myFont; How I create the stage: Stage collectionStage = new Stage(new StretchViewport(Gdx.graphics.getWidth(), Gdx.graphics.getWidth())); I also tried with no viewport or with other viewports. In my resize method: public void resize(int width, int height) { Viewport viewport = collectionStage.getViewport(); viewport.update(width, height, false); viewport.apply(); } The problem is that when I resize the window on Desktop the font gets distorted. Do I have to create another font in the resize method, and then for each label that I have, reassign the new font? Or should I do something else? • Distorted as in pixelated or as in stretched / other? – Charanor Jun 20 '17 at 14:20 • Also, if you use a stretch viewport it will stretch your screen and everything in it. Try using a FitViepwort or an ExtendViewport instead that keeps your aspect ratio. – Charanor Jun 20 '17 at 15:01 Saw because of bump, If it comes out blurry than you are most likely using a font that is too small for the resolution you are working at. What you'll want to do is to create a font at the size that is suggested for that font, and then scale it up afterwords. Ex: fontGenerator = new FreeTypeFontGenerator(Gdx.files.internal("Fonts/pixels.ttf")); FreeTypeFontGenerator.FreeTypeFontParameter fontParameter; fontParameter = new FreeTypeFontGenerator.FreeTypeFontParameter(); fontParameter.size = 16; fontParameter.color = Color.WHITE; font = fontGenerator.generateFont(fontParameter); // scale here font.getData().setScale(/however much you want to scale with/); Hope this helps. You need to use a Viewport which keeps the aspect ratio, for example a FillViewport. This will handle resizing for you, so you "only" need to manage the position. Also, I'd not scale the fonts. Create some fonts in different sizes at the beginning and use them. • I tried to use some of the available viewports, but the fonts still get distorted on resize. – Andy Mar 24 '17 at 21:39 • Can you upload the code of how you initialize and use the viewport? – AlGrande Mar 24 '17 at 22:10 • I updated my question with a bit more code. – Andy Mar 24 '17 at 22:31	{}
Address 105 Front St, Beckley, WV 25801 (304) 252-6170 http://www.cns-repair.com # expected true error rate Eccles, West Virginia Estimation of error rates in discriminant analysis. Dougherty has authored several books including Epistemology of the Cell: A Systems Perspective on Biological Knowledge and Random Processes for Image and Signal Processing (Wiley-IEEE Press).Bibliografische InformationenTitelError Estimation for Pattern RecognitionIEEE Psychometrika. 1951;16:31–50.Braga-Neto U, Dougherty E. It has a long history going back to 1968 (Lachenbruch and Mickey, 1968). Your cache administrator is webmaster. Hence, the bias is not too great as long as n/k is small. Naturally, any model is highly optimized for the data it was trained on. That is, it fails to decrease the prediction accuracy as much as is required with the addition of added complexity. When k = n, one gets the leave-one-out estimator, ε^n l.Cross-validation’s salient good property is that, under random sampling, it can be proved (see Devroye et al., 1996) that it is Effect of discretization method on the diagnosis of parkinsons disease. The measure of model error that is used should be one that achieves this goal. A Probabilistic Theory of Pattern Recognition. Here we initially split our data into two groups. The American Statistician, 43(4), 279-282.↩ Although adjusted R2 does not have the same statistical definition of R2 (the fraction of squared error explained by the model over the null), it is Dougherty has authored several books including Epistemology of the Cell: A Systems Perspective on Biological Knowledge and Random Processes for Image and Signal Processing (Wiley-IEEE Press). To demarcate the separate-sampling case from the random-sampling case, we will write the discriminant and corresponding classifier by Wn0,n1(S0,S1,X) and Ψn0,n1(S0,S1,X), respectively, with the latter defined in the same manner as However, in addition to AIC there are a number of other information theoretic equations that can be used. Braga Neto is an Associate Professor in the Department of Electrical and Computer Engineering at Texas A&M University, USA. In fact, adjusted R2 generally under-penalizes complexity. Blood. 2006;108:2020–2028. [PMC free article] [PubMed]Articles from Bioinformatics are provided here courtesy of Oxford University Press Formats:Article \| PubReader \| ePub (beta) \| PDF (592K) \| CitationShare Facebook Twitter Google+ You We fix n and determine n0 according to n0=⌈nr⌉. Kennedy ’26 Chair, and Scientific Director at the Center for Bioinformatics and Genomic Systems Engineering at Texas A&M University, USA. In this region the model training algorithm is focusing on precisely matching random chance variability in the training set that is not present in the actual population. CSS from Substance.io. linear and logistic regressions) as this is a very important feature of a general algorithm.↩ This example is taken from Freedman, L. For instance, in the illustrative example here, we removed 30% of our data. Please review our privacy policy. But at the same time, as we increase model complexity we can see a change in the true prediction accuracy (what we really care about). Then the classical k-fold cross-validation estimator is given by ε^n cv(k) = 1n ∑i=1k∑q∈Ui(IWn(Ui)(S,Xq)≤0IYq=0+IWn(Ui)(S,Xq)>0IYq=1) .(8) If k = n, this reduces to the leave-one-out estimator ε^n l=1n∑i=1n(I Wn(i)(S,Xi)≤0 IYi=0+I Wn(i)(S,Xi)>0 IYi=1),(9) where If local minimums or maximums exist, it is possible that adding additional parameters will make it harder to find the best solution and training error could go up as complexity is Each observation is called an instance and the class it belongs to is the label. Comput. Inf. 2011;7:4669–4678.Lachenbruch PA, Mickey MR. Furthermore, although the deviation variance of classical cross-validation can be mitigated by large samples, the bias issue generally remains just as bad for large samples.3.2 Two case studiesTo further illustrate the National Library of Medicine 8600 Rockville Pike, Bethesda MD, 20894 USA Policies and Guidelines \| Contact Cookies helfen uns bei der Bereitstellung unserer Dienste. In this case however, we are going to generate every single data point completely randomly. The behavior observed in Figure 2 makes it plausible that the error estimates for classical cross-validation will exceed those of separate-sampling cross-validation, which is nearly unbiased. Thus their use provides lines of attack to critique a model and throw doubt on its results. Table 1 provides a summary of these real datasets, including the total number of features and sample size. On important question of cross-validation is what number of folds to use. For a multiclass classifier, the Bayes error rate may be calculated as follows:[citation needed] p = ∫ x ∈ H i ∑ C i ≠ C max,x P ( C i This is unfortunate as we saw in the above example how you can get high R2 even with data that is pure noise. As example, we could go out and sample 100 people and create a regression model to predict an individual's happiness based on their wealth. The Danger of Overfitting In general, we would like to be able to make the claim that the optimism is constant for a given training set. For instance, if we had 1000 observations, we might use 700 to build the model and the remaining 300 samples to measure that model's error. Cross-validation works by splitting the data up into a set of n folds. There is a simple relationship between adjusted and regular R2: $$Adjusted\ R^2=1-(1-R^2)\frac{n-1}{n-p-1}$$ Unlike regular R2, the error predicted by adjusted R2 will start to increase as model complexity becomes very high. If one wishes to use cross-validation with separate sampling, then one should use the separate-sampling version of cross-validation, which is proposed here, or else, significant bias may result. Perhaps the most commonly used training data-based classification error estimator is cross-validation. Dr.	{}
# Rotational Energy 1. Jul 23, 2008 ### Lance WIlliam A 250 g baseball is pitched at 35 m/s, and it's spinning at 55rad/s . What fraction of its kinetic energy is rotational? Treat the baseball as a uniform solid sphere of radius 3.8 cm. What the?! Im pretty lost on this one.... Do I use K_rot_=(1/2)$$I$$$$\omega$$^2 2. Jul 23, 2008 ### rock.freak667 Yes this will give the rotational ke of the ball. 3. Jul 23, 2008 ### Lance WIlliam What is the Fraction of KE though? 4. Jul 23, 2008 ### rock.freak667 The total energy of the ball is $\frac{1}{2}mv^2$	{}
Linear Formula C 6 H 5 CH=CHCOOH . Soluble in water (1 in 3500), readily soluble in alcohol and oils. Thus, cinnamic acid exhibits a low toxicity. It is a white crystalline compound that is slightly soluble in water, and freely soluble in many organic solvents. Explain the difference between “suspension and Emulsion” OR between “solution and colloid”. Cinnamyl alcohol or styron is an organic compound that is found in esterified form in storax, Balsam of Peru, and cinnamon leaves. 4 years ago. To obtain high FA extraction yields using supercritical carbon dioxide (scCO 2), the extraction should be performed in the presence of co-solvents.This work reports FA solubility in scCO 2 /ethanol/water mixtures and its thermodynamic modelling using the Peng-Robinson equation of state. Explain the difference between condensation and hydrolysis. Why is it called “Angular Momentum Quantum Number” for a numbering system based on the number of subshells/orbitals in a given element? Synthetic Communications: Vol. Source(s): Chemist. Assessment of the anti-invasion potential and mechanism of select cinnamic acid derivatives on human lung adenocarcinoma cells. SENSIBLE AL AIRE-OXIGENO. 4 years ago. This paper is devoted to the experimental measurement of the solubility of solid trans -cinnamic acid in pure supercritical CO 2 and in supercritical CO 2 with ethanol as a cosolvent. Relevance. Cinnamic Alcohol is a FDA GRAS ingredient ( 21 CFR §172.515). General tips Briefly, 5 X10 4 cells are plated in each well of a 24-well plate, allowed to attach overnight, and treated with compounds (e.g., Cinnamic acid: 2.5, 5, 10, 20, 30 mM) the following day. Question: In A Recrystallization From Ethanol, Cinnamic Acid (pure Mp 132-135 Degree C) Was Melted On A Melting Point Machine Immediately After Vacuum Nitration, And Found To Have A Melting Point Of 122-127 Degree C. Why Is The Melting Point Not 132-135 Degree C? ethanol [9] have been investigated with some success. Ethanol 33 mg/mL (201.02 mM) ... Insoluble * Please note that Selleck tests the solubility of all compounds in-house, and the actual solubility may differ slightly from published values. 4 years ago. They react in this way with all bases, both organic (for example, the amines) and inorganic. 1, the solubility of the acid can significantly Synthesis of Cinnamic Acid Derivatives Using Ethanol as Solvent or Microwave Assisted Method. inertizar con nitrogeno.. keep in tightly closed container in a cool and dry place, protected from light. Still have questions? Product name: Cinnamic Acid Appearance: White or light yellow crystal powder Solubility: Soluble in ethanol, methanol, petroleum ether, chloroform, easily soluble in benzene, ether, acetone, glacial acetic acid, carbon disulfide and oils, insoluble in water. MDL number MFCD00004369. Cinnamic acid was first isolated as crystals from cinnamon oil by Trommsdorf in 1780. The solubility of stearic acid in ethyl acetate was found to be the highest, followed by ethanol, acetone and methanol. Get your answers by asking now. Cinnamic Acid manufacturers and suppliers from China. Please rinse for fifteen minutes with cold water. plympton. Isopropylcinnamate : 1199-77-5 Molecular Weight: 162.188 g/mol Purity: greater-or-equal 99% Properties: white crystalline powder, insoluble in water, soluble in organic solvents such as ethanol and benzene. Provide examples? P-TOLUIC ACID is a carboxylic acid. Classified as an unsaturated carboxylic acid, it occurs naturally in a … The “mole fraction solubilities (xe)” of SA in various “Carbitol + water” systems were determined at “T = 298.15–318.15 K” and “p = 0.1 MPa.” 11. Here is a list of its solubility: - Water: about 0.2% at 20oC, about 1.7% at 75oC (by adding about 5% of sodium phosphate or another neutral salt, solubility of salicylic acid in water is increased to about 1.8%) Carbamazepine−cinnamic acid cocrystals synthesized by solvent evaporation showed a higher dissolution rate, solubility, and stability in water compared to carbamazepine. It has a melting point of 133 degrees and a boiling point of 300 degrees. WHEN STORED FOR MORE THAN 24 MONTHS, QUALITY SHOULD BE CHECKED BEFORE USE. Cinnamic acid is a monocarboxylic acid that consists of acrylic acid bearing a phenyl substituent at the 3-position. About 0% of these are Herbal Extract, 0% are Other Extracts. It is soluble in ethanol, methanol, petroleum ether and chloroform; it is easily soluble in benzene, ether, acetone, acetic acid, carbon disulfide and oils but insoluble in water. 4. Name: trans-Cinnamic acid. Answer Save. Relevance. Description. In addition we will provide 1 year warranty. It should be free from chlorides. The solubility of trans-cinnamic acid in the mixed solvents water + ethanol, and water + methanol was measured, at 298.15 K, using the isothermal shake-flask method and quantitative analysis either by gravimetry or UV spectrophotometry. Cinnamic acid is an organic compound with the formula C6H5CH=CHCOOH. ? Solubility in water 1625 mg/L @ 25 0C 13. Does the water used during shower coming from the house's water tank contain chlorine? Curve calibration pre-pared from standar solution of trans-cinnamic acid containing 0.5-2.5 µg/mL. 103-26-4) is a white or transparent solid with a strong, aromatic odor and it belongs to the class of organic compounds known as cinnamic acid esters with the … Classified as an unsaturated carboxylic acid, it occurs naturally in a number of plants. ABSTRACT: The solubility of stearic acid in ethanol, methanol, ethyl acetate, and acetone has been measured gravimetrically at various temperatures ranging from 301 to 313 K at atmospheric condition. 2. Cinnamic acid is soluble in ethanol and nearly insoluble in water. So volume of the solution will be very small and much may be lost during handling. Tsai CM, Yen GC, Sun FM, et al. Alcohol fermentation ethanol is yeast. ERNESTO VENTÓS, S.A. shall not be held responsible for any damage or injury resulting from contact with the above. It is very irritating to breath in. P-Hydroxycinnamic Acid, also known as 4-Hydroxycinnamic acid, is a radical scavenging effects of antioxidant, compared with other tyrosinase inhibitor, the hydroxy cinnamic acid and ether hydroxy cinnamic acid and other non-toxic odorless, good permeability, and is considered a good candidate material used as a skin whitening agent. acid value (mg koh/g) - stability and storage. Source(s): https://owly.im/a0uvV. In this assay, 1, 5, and 10 CA were used to evaluate the potential Cinnamic Acid Isopropyl Ester. 17 The solubility of the trans isomers of cinnamic acid, p-coumaric acid and ferulic acid 18 was measured in water and seven organic solvents (methanol, ethanol, 1-propanol, 2-19 propanol, 2-butanone, ethyl acetate and acetonitrile), at 298.2 and 313.2 K, using the 20 analytical shake-flask technique. trans-cinnamic acid has inhibitory effect on phorbol-12-myristate-13-a cetate-induced invasion of human lung adenocarcinoma A549 cells. Get your answers by asking now. Also known as phenylacrylic acid, cinnamic acid forms monoclinic crystals, as needles or prisms, with melting points of 133 degrees Celsius and boiling points of 300 degrees Celsius. Salicylic acid doesn't dissolve well in water and many other common solvents. EC Number 205-398-1. 3769-3774. These esters are … (2000). ... solubility . Cinnamic acid has the formula C6H5CHCHCOOH and is an odorless white crystalline acid, which is slightly soluble in water. Composed of 9 carbon, 8 hydrogen, and 2 oxygen atoms, cinnamic acid is classified as a skin, eye, and respiratory irritant and is soluble in water or alcohol. sensible al aire-oxigeno. Cinnamyl alcohol(CAS NO. Source(s): https://owly.im/a0uvV. Appearance: White to Light yellow powder to crystal: Purity(GC) min. Figure 3.14: a) $$100 \: \text{mg}$$ of trans-cinnamic acid in each test tube, b) Left tube contains trans-cinnamic acid with water (insoluble) and right tube contains trans-cinnamic acid with methanol (soluble). 20, pp. Fig.1. It exists as both a cis and a trans isomer, although the latter is more common. trans -Cinnamic acid (TCD, also known as trans --pheny-lacrylic acid) is one of the main chemical constituents of cinnamon which belongs to plant Lauraceae [, ]. Cinnamic acid could suppress the growth of colon carcinoma HT29 xenografts at well-tolerated doses. Cinnamyl alcohol is also known by a variety of other names including cinnamic alcohol, 3-phenyl-2-propen-1-ol, 3-phenylprop-2-en-1-ol, zimtalcohol, styryl carbinol, e-3-phenylprop-2-en-1-ol, 3-phenylallyl alcohol, trans-cinnamyl alcohol. SOLVENT SOLUBILITY: soluble in alcohol and ether: VAPOR DENSITY. We believe the above information to be correct but we do not present it as all inclusive and as such should be used as a guide. 4 years ago. 98.0 % Melting point: 132.0 to 136.0 °C Cinnaminc acid contains a carboxyl group and a double bond but other than that it is pretty insoluble. 3. The solubility and solution thermodynamic properties of a “bioactive nutraceutical” sinapic acid (SA) in different “2-(2-ethoxyethoxy)ethanol (Carbitol®) + water” mixtures were investigated. If you need, we will provide you free sample. 0 0. Carboxylic acids donate hydrogen ions if a base is present to accept them. So, you'll use a mixture of 95% ethanol and 5% water so that cinnamic acid is still soluble at high temperatures but less soluble at … Themistocles. The solubility of CA is 0.5 g/l (about 3.4 mM). Log P (o/w) 1.95 SYNONYM Cinnamyl Alcohol, 3-phenyl allyl alcohol, Zintalcohol, gamma-phenyl allyl alcohol, Styrone, Styryl carbinol, 1-phenyl prop-1-en-3-ol, 3-phenyl-2-propenol USE LEVEL Mol Pharm. Pólo Industrial Granja Viana - CEP 06707-070, Colonia Alce Blanco - Naucalpan de Juarez. Give examples. It is a white crystalline compound that is slightly soluble in water, and freely soluble in many organic solvents. It has a role as a plant metabolite. assay (% gc) > 98 . A wide variety of cinnamic acid solubility options are available to … 140-10-3・trans-Cinnamic Acid・035-03412・039-03415[Detail Information] \| [Life Science] \|Laboratory Chemicals-FUJIFILM Wako Chemicals U.S.A. It is found in Cinnamomum cassia. 0 0. Solubility in alcohol 1gm soluble in 1ml 70% alcohol 12. Cinnamic acid is an organic compound with the formula C6H5CH=CHCOOH. Using the gravimetric method to measure the solubility of p-coumaric acid in nine mono solvents, including methanol, ethanol, 1-propanol, 2-propanol, n-butanol, isobutanol, acetone, ethyl acetate, and methyl acetate, and also in the water + ethanol mixtures from 293.15 to 333.15 K at atmospheric pressure (p = 0.1 MPa). All PubChem Substance ID 24893022. Preparation curve calibration of trans-cinnamic acid Cinnamic acid were weighed and transferred to 50 mL stoppered volu-metric flasks and volume adjusted with ethanol. Ferulic acid (FA) is a phenolic compound present in many natural extracts. 1:1 in ethanol 70% . Their reactions with bases, called "neutralizations", are accompanied by the evolution of substantial amounts of heat. Join Yahoo Answers and get 100 points today. trans cinnamic acid is insoluble in water but soluble in ethanol at warm temperatures. 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