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https://www.physicsforums.com/threads/oppenheimer-snyder-model-of-star-collapse.651362/	1,531,856,661,000,000,000	text/html	crawl-data/CC-MAIN-2018-30/segments/1531676589892.87/warc/CC-MAIN-20180717183929-20180717203929-00253.warc.gz	930,840,559	26,126	Oppenheimer-Snyder model of star collapse 1. Nov 11, 2012 TrickyDicky This comes from this thread https://www.physicsforums.com/showthread.php?t=647627&page=7 discussion in posts #103,#104,#107 and #108. The Oppenheimer-Snyder model was mentioned by PeterDonis as a more plausible model than the Schwarzschild spacetime, well this has an element of subjectivity, but one reason I don't share this view is because the only way to relax the highly idealized conditions required by the O-S model is to recurr to the Kruskal-Szekeres diagram for the Schwarzschild solution as is shown in MTW sec. 32.5 second paragraph. So how can one consider more plausible a model than the one it owes its plausibility to? Also I have a few things to clarify from this model. As I understand it the O-S model basically joins the exterior Schwarzschild to a contracting FRW spatially spherical solution, (a pressureless isotropic and homogeneous dust). I usually interpret the exterior Schwarzschild solution to refer to the Schwarzschild metric, outside the Schwarzschild radius, or region I in the K-S diagram and this leads me to a second dependency of the O-S model on the maximally extended Schwarzschild solution, since in order to sy that the Schwarzschild exterior includes region II and the event horizon one must obviously rely on the K-S diagram (that didn't exist in 1939) to begin with. I'm still not convinced that it is commonly understood that the region inside the Schwarzschild radius is also considered an exterior region, since then, what is the interior region?, the singularity by itself? Here is the abstract from the original paper from O-S "When all thermonuclear sources of energy are exhausted a sufficiently heavy star will collapse. Unless fission due to rotation, the radiation of mass, or the blowing off of mass by radiation, reduce the star's mass to the order of that of the sun, this contraction will continue indefinitely. In the present paper we study the solutions of the gravitational field equations which describe this process. In I, general and qualitative arguments are given on the behavior of the metrical tensor as the contraction progresses: the radius of the star approaches asymptotically its gravitational radius; light from the surface of the star is progressively reddened, and can escape over a progressively narrower range of angles. In II, an analytic solution of the field equations confirming these general arguments is obtained for the case that the pressure within the star can be neglected. The total time of collapse for an observer comoving with the stellar matter is finite, and for this idealized case and typical stellar masses, of the order of a day; an external observer sees the star asymptotically shrinking to its gravitational radius." My bold: the first sentence I bolded lists some of the conditions required for the model to hold, have they all been theoretically and empirically ruled out? If so how? And I mean by other ways other than the K-S spacetime mathematical solution, that is considered not plausible due to its implying white holes. 2. Nov 11, 2012 Staff: Mentor I think you're misreading that paragraph. It only talks about relaxing one idealization, that of zero pressure inside the collapsing matter. It doesn't talk at all about relaxing spherical symmetry. The only reference to the K-S diagram is to show that, once the star has collapsed to R < 2M, no amount of pressure can stop it from collapsing, because inside the horizon, all timelike worldlines end in the singularity, not just geodesic ones. Pressure can make the worldlines of the infalling matter geodesic, but it can't make them not timelike. Also, note that the K-S diagram in MTW Figure 32.1b (the one that is referenced in the paragraph you refer to) is not a K-S diagram of the maximally extended spacetime. It's a K-S diagram of exactly the type of solution I described. The gray portion on the left is the region of spacetime occupied by the infalling matter; the white portion on the right (and below the singularity) is the vacuum region outside and inside the horizon (i.e., a portion of regions I and II of the maximally extended spacetime). The word "exterior" is used to mean two different things, which I agree is an unfortunate abuse of terminology. Sometimes it means "the vacuum region exterior to the horizon", and sometimes it means "the vacuum region exterior to the collapsing matter". In the O-S model the latter meaning is the one that's meant. As you can see from the K-S diagram that you referenced, the vacuum region includes portions of regions I and II of the maximally extended spacetime (see my comments above). No, of course not. The O-S model is a highly idealized model; nobody thinks otherwise. The problem with including all that other stuff is that nobody has an analytical solution that includes it. Numerical simulations, as referenced in MTW, still show the same qualitative behavior (formation of a horizon and collapse of the matter to form a singularity) when the other stuff is included. I'm not sure what you mean here. The O-S model uses a portion of the maximally extended Schwarzschild spacetime, which is what the "K-S spacetime mathematical solution" describes. There's no problem with doing that as long as you enforce the appropriate junction conditions at the boundary between the vacuum portion of the spacetime and the portion containing the collapsing matter. Yes. 3. Nov 11, 2012 TrickyDicky I know that paragraph only refers to the pressureless idealization, I only brought it up to show the dependence of the O-S model on the posterior K-S extended solution, because you were claiming the O-S model was physically more plausible than the latter. IMO this is a meaningless statement given the commented dependence of one model on the other i.e. the O-S model at least originally when it was first published, seems to refer to the collapsing star before the BH singularity is formed, and subsequently this paper has been interpreted in the light of the progress made more than 20 years later by Kruskal and others. 4. Nov 11, 2012 Staff: Mentor The Schwarzschild spacetime would require any black hole to pre-date the big bang, and the OS spacetime does not. So calling it "more plausible" seems reasonable to me. 5. Nov 11, 2012 Staff: Mentor I don't agree that the O-S model "depends" on the "extended K-S solution". What you are calling the "extended K-S solution" is just the maximal analytic extension of the Schwarzschild geometry. It's a mathematical object. There is no physical principle that I'm aware of that makes the maximal analytic extension of a manifold logically prior to just using a portion of that manifold in a physical solution. Another way of looking at this is to observe that the maximally extended Schwarzschild geometry, as described by K-S, requires the entire spacetime to be vacuum. This immediately makes the full geometry unsuitable for a model where matter is present, such as the O-S model. But since the EFE is local, there's no requirement that we use the entire maximally extended manifold; in fact, looking at it that way gets things backwards. We don't pick the portions of the different manifolds (regions I and II of extended Schwarzschild, plus collapsing FRW) first; we solve the EFE first, and then discover which portions of what manifolds arise when we develop the global solution. Can you give actual quotes from the original paper that support this view? My impression from reading the abstract (which appears in MTW) is that the original O-S model already includes all three regions I referred to (FRW region containing collapsing matter, vacuum region outside the horizon, and vacuum region inside the horizon) plus the singularity. Certainly that's how MTW describe the model, and they don't give any impression that their description was something "interpreted" later that wasn't present in the original O-S model. 6. Nov 12, 2012 TrickyDicky Let me try and explain what I mean by the O-S model relying on the extended Schwarzschild mathematical solution, when thinking of it as a BH model. The FRW dust plus Schwarzschild exterior model only describes the situation of a collapsing star (not charged and not rotating) from the moment the contraction of the star starts , up to the instant previous to the singularity being the only entity inside the Schwarzschild radius of the star (so there is no longer FRW isotropic dust, and therefore no more O-S model). Right at that point the mathematical model of the maximally extended Schwarzschild spacetime takes the place. Further as commented above if one wants to relax the requirement of the O-S model concerning the star fluid being pressureless, one does it (or at least it is done in MTW) alluding to the fact that the causal logic, that is, the expected consequence of the O-S collapse model is the extended Schwarzschild spacetime wich doesn't care about the initial conditions of the collapse since it is an eternal exact solution of the EFE. So saying that the O-S model is just a local version that needs not rely on the global spacetime solution misses the causality of the collapse model, and if one wants to add physical plausability to it like not demanding exactly zero star pressure one also needs the extended mathematical model to account for the final result of the collapse. So I'm still finding hard to separate the physical plausability (or lack of) of one model from the other. Also historycally, if one looks at the timeline of the 1939 paper citations, one can see it was basically ignored until the beginning of the sixties when the mathematical models by Kruskal et al. were published. 7. Nov 12, 2012 Staff: Mentor Schwarzschild BH has existed forever, even before the big bang, OS BH has existed for a finite time. Existing for a finite time is more plausible than existing forever. What is so hard to understand about that? 8. Nov 12, 2012 Staff: Mentor Here is a geometric analogy: If I have a car with a 3 m long interior and I go to a lumber yard and they have only 4 m long pieces of lumber, then I might ask them to cut off 1 m of length and I might even ask them to taper it nicely. Then, regardless of the fact that the cut lumber was based on the long lumber, and regardless of the fact that beyond the taper the cut lumber has the same shape as the long lumber, and regardless of the fact that historically lumber yards ignored the sizes of cars from the development of tapered cuts in 1939 until the beginning of the sixties, it is clearly more plausible that the cut lumber will fit in my car. 9. Nov 12, 2012 Staff: Mentor Yes, that's the initial condition, that the star is at rest with some finite radius. No, this is not correct. The "model" includes the entire spacetime to the future of the start of the star's contraction. You don't "switch models" when the collapse forms the singularity. For one thing, "when the collapse forms the singularity" depends on how you choose your spatial slices. In the Penrose diagram of the O-S model, the singularity is as far in the future as it gets--spacelike slices that are cut arbitrarily close to the singularity also come arbitrarily close to future infinity. See the diagrams on Hamilton's web page: Even if we leave out the issues about "right at that point", this is not correct; if this were true, then a white hole would magically appear in the past instead of the collapsing star. The O-S model is a model of the entire spacetime to the future of the start of the star's collapse. The spacetime to the past of that point is not included in the model, but it certainly is not a white hole; the simplest assumption (which of course is not realistic) would be that the star simply sat there statically for an infinite time into the past. A more realistic model would include the rest of the universe, all originating from the Big Bang. In no case would we have a white hole or a maximally extended Schwarzschild spacetime; that spacetime, as I've said before, assumes that there is vacuum everywhere, and there can't be vacuum everywhere with a collapsing star present. First of all, the "expected consequence" still depends on exact spherical symmetry; in any real case there is not exact spherical symmetry, so Birkhoff's theorem doesn't apply and we can't say that the vacuum region will be Schwarzschild. Second, you're misunderstanding what I said about the EFE being local. See below. I didn't say the model was local, I said the EFE was local. That means that when I am putting together a global model for a spacetime, I don't have to use only one single solution; i.e., I don't have to use the entirety of one particular spacetime (one particular mathematical geometry). I can stitch together pieces of different geometries, as long as I satisfy the appropriate junction conditions when I do the stitching. Here's a simpler example to illustrate what's going on. A 2-sphere is a particular mathematical geometry. So is a cylinder that extends infinitely far in the direction along its axis. Each one can be described very simply in terms of coordinates on it. But I can also form a third geometry by taking half of the sphere and stitching it together with half of the cylinder; as long as I do the stitching right (I have to match up the radius of the 2-sphere with the radius of the cylinder, and orient the junction so the tangent vectors of the two surfaces match up at the boundary), meeting the appropriate "junction conditions", the resulting surface will be continuous and differentiable (I'm hand-waving on terminology a bit here, hopefully you can see what I mean), and will therefore be just as much of a legitimate mathematical geometry as the sphere and the cylinder. I won't be able to describe it quite as simply using coordinates on it, but it is still a perfectly good geometry, and it is perfectly self-contained; nothing in my description of it will have to take into account the "existence" of the other half of the sphere or the other half of the cylinder. Similarly, to form the O-S model, I take the maximally extended Schwarzschild spacetime and "cut" it along the boundary where the surface of the collapsing star is going to be, and use only the portion to the future of that boundary. I then stitch that portion together with a collapsing FRW spacetime, making sure that things match up along the boundary. (And, if I want to have a complete solution, I also stitch in something to the past of the initial spacelike surface where the star's collapse begins, so my complete model includes the entire past history of the star and its vacuum exterior. That will still only include a further portion of region I of the extended Schwarzschild spacetime, i.e., I will still be "cutting" that spacetime and only using a portion of it in my model.) The final solution therefore only contains a piece of the extended Schwarzschild spacetime, a piece comprised of a portion of region I and a portion of region II. The rest of the maximally extended spacetime is simply not there in the model, just as half of the 2-sphere and half of the cylinder were simply not there in the object I made by stitching a half-sphere and half-cylinder together. I don't understand what you're saying here. Causality just means the local light cone structure is continuous throughout the spacetime. As long as the junction conditions are satisfied, this holds when I stitch together pieces of different geometries. That may be so (I don't know enough about the citation history to know, perhaps you have a link?). So what? Last edited: Nov 12, 2012 10. Nov 12, 2012 TrickyDicky I'm not saying you have to "swith models" necessarily, just highlighting the domain of appliccation of each model. Penrose diagrams didn't even exist for more than 20 years after the paper that describes the O-S model. The O-S model predicts that "the total time of collapse for an observer comoving with the stellar matter is finite", what does the O-S model say about what comes inmediately after that finite time? IOW, I'm only referring to the future direction after finite time for the comoving observer in the O-S model. AFAIK this is not correct, numerical and perturbation methods alows us to use the extended Schwarzschild spacetime in the absence of perfect spherical symmetry. Note you are all the time using the extended Schwarzschild spacetime as a template in wich to cut the O-S model. That's the kind of dependency I'm referring to. 11. Nov 12, 2012 TrickyDicky OS collapsing star is a static object. It's exterior geometry is Schwarzschild, so it has existed forever before its contraction started ("even before the big bang") too. What has existed for a finite time is its collapsing process for the comoving observer POV. 12. Nov 12, 2012 Staff: Mentor So what? They're a legitimate slicing of the spacetime; whether that slicing was known when O-S wrote their paper is irrelevant. By which they mean "proper time as experienced by an observer comoving with the stellar matter". Nothing "comes after" it; the infalling observer gets destroyed in the singularity along with the stellar matter. Their worldlines simply end at the singularity; there is nothing "after" it. Remember that the singularity is spacelike; that is, it is "an instant of time", not "a place in space". My point about different slicings is simply that there is a slicing of the spacetime according to which the singularity is the instant of time "at the end of time", i.e., there is no time that is "after" the singularity. If you don't have perfect spherical symmetry, then the spacetime you're working with is only approximately Schwarzschild; how good the approximation is depends on how close you are to spherical symmetry. But that's irrelevant to the question of what portion of the maximally extended spacetime is actually used in the physical model. That's one way of describing what I'm doing, yes. But it's not the only way. Here's another: I start with the assumption of perfect spherical symmetry and solve the vacuum EFE on an "initial value" spacelike slice. I do the solution locally, starting at spatial infinity and working inwards. Eventually I reach the surface of the matter, which I assume is at rest in this initial spacelike slice; at that point my solution is no longer vacuum, but I can ensure that the switch is smooth by imposing appropriate junction conditions at the boundary. Once I'm inside the matter, I continue to assume spherical symmetry, and I also assume homogeneity because it's the only assumption that's simple enough to allow me to find an analytical solution, locally, to the EFE. I continue working inwards until I reach r = 0, the center of the collapsing matter. I now have a description of a spacelike slice on which the matter is instantaneously at rest. I then work the solution forward in time, using the EFE to evolve things from one spacelike slice to the next. I find that the matter is collapsing inward; then I find that a horizon forms; then I find that the matter collapses to r = 0 and forms a singularity. If I try to iterate further "forward" in time, I find that the singularity is actually spacelike; depending on exactly how I cut my spacelike slices, it may even be that the singularity is the single spacelike slice that I am at when the matter reaches r = 0, so there is nothing "after" it--it is the future endpoint of my solution. In any case, I can obtain a complete spacetime geometry to the future of the initial spacelike slice I started with. Notice that nowhere did I do any "cutting" of anything out of anything else; I simply used the EFE to build my solution point by point. After I have done all this, of course, I can discover that my solution is isometric to what I described in previous posts: a portion of regions I and II of the extended Schwarzschild spacetime, joined to a collapsing FRW geometry with appropriate junction conditions. But I didn't have to assume that, or start with that, or construct my solution from those pieces. The description of the solution in terms of those pieces joined together is just a helpful aid to visualizing what's going on; it is not logically or physically essential to deriving the actual solution. The knowledge that a solution could be constructed out of those pieces may have helped in arriving at the O-S model, but that's not the same as saying it is necessary to the O-S model. (There is also, of course, the question of what lies to the past of the initial spacelike slice. That's a separate question from how we construct the solution; it depends on what assumptions we make about what the star was doing before it started to collapse.) 13. Nov 12, 2012 TrickyDicky Thanks, that was all I meant when saying "up to the instant previous to the singularity being the only entity inside the Schwarzschild radius of the star (so there is no longer FRW isotropic dust, and therefore no more O-S model)." to wich you replied: "No, this is not correct." Once this is clarified, it is probably not very useful to keep debating your claim about what is more or less physically plausible, it depends on the subjective priorities one might want to consider. In fact all the EFE solutions even the most practical, empirically or computationally have tremendous idealization that are far from what is usually considered physical. BTW here's the citation history of the OS paper, curious to say the least. It was ignored until 1964, but it wasn't really until the last 5 years that it took off. http://libra.msra.cn/Publication/19921235/on-continued-gravitational-contraction 14. Nov 12, 2012 Staff: Mentor This is not correct, at least as far as how I understand it. To my knowledge, neither the original dust cloud, nor the singularity, nor the event horizon has existed forever in the OS spacetime, which starts with the collapse. I.e. the OS manifold only covers back in time until the initiation of collapse. By contrast, in the Schwarzschild solution the exterior of the horizon, the horizon itself, the interior of the horizon, and the singularity are all considered to have existed forever. I should think it is obvious to any reasonable person that that is much less plausible. 15. Nov 12, 2012 Staff: Mentor I thought you were saying that there must be something "after" the singularity forms. If you agree that there are slicings in which there is nothing "after" the singularity forms, then yes, we're in agreement. All of the analytical solutions, yes. Numerical solutions can be far more realistic. Thanks! I'll take a look. 16. Nov 12, 2012 Staff: Mentor This is my understanding too, at least based on all the presentations of the O-S model that I've seen; none of them talk about what happens before the initial instant of time at which the dust cloud is instantaneously at rest. However, this obviously leaves the manifold incomplete; there has to be something to the past of that initial spacelike slice. I think it's reasonable to ask what the possibilities are for that past region of the complete manifold; and I also think it's reasonable, physically, to say that a white hole is not one of those possibilities. 17. Nov 12, 2012 Staff: Mentor Actually, since the O-S dust has zero pressure, it can't be static. The most straightforward extension into the past of the O-S model, keeping the assumption of zero pressure, would be the time reverse of the extension into the future; i.e., an expanding FRW region with zero pressure and starting from an initial singularity, joined to a portion of regions IV and I of the maximally extended Schwarzschild spacetime (including the white hole spacelike singularity, which could be thought of as the past endpoint of the spacetime). However, I don't see that as physically reasonable for the same reasons that a white hole in general is not physically reasonable. I suspect that what O-S had in mind was something like a static star with positive pressure, in equilibrium, in which the pressure suddenly declines to zero (or at least to some negligible value compared to its previous one) over a very short time (due to, say, the stoppage of nuclear reactions in its core and a consequent sharp decline in temperature). 18. Nov 12, 2012 TrickyDicky Certainly but I was referring precisely to the static star you depict below in my answer to DS. 19. Nov 12, 2012 Staff: Mentor All the OS model insists is that at some moment there is a spherical dust cloud which is momentarily at rest. There is certainly no implication in the OS model that the "static star" has been in such a state since before the big bang. Plausibility is definitely subjective, so you can choose to disagree. However, to me it is clear that a model which begins from a momentarily stationary sphere of dust is more plausible than a model which begins from a singularity. We have direct experience with things that approximate a momentarily stationary sphere of dust, but not with singularities. So the opposite stance seems tenuous. 20. Nov 12, 2012 TrickyDicky Well, it's not for me to defend singularities' plausibility, or GR's for that matter, you are free to have whatever opinion, I can only refer you to the Hawking-Penrose singularity theorems. 21. Nov 12, 2012 Staff: Mentor Hmm, I thought that was exactly what you were attempting to do in your OP. Are you not claiming that Schwarzschild is more plausible than OS? 22. Nov 12, 2012 Staff: Mentor That's fine, but AFAIK it's not part of the "O-S model" as standardly understood, which is why your use of the phrase "OS collapsing star is a static object" confused me. 23. Nov 13, 2012 TrickyDicky As I said this is a subjective issue in great part, and I have not presented in such simple terms, on the contrary, I was trying to show how the opposite claim by PeterDonis needed some qualifications to have any meaning other than the subjective preference. One of this qualifications I tried to explain was that even if not in the OS model the logical causal future of the collapsing model is a BH with a singularity, and for the non-charged, non-rotating case the only mathematical model we have of that is an exact solution of the EFE is the extended Schwarzschild spacetime. 24. Nov 13, 2012 pervect Staff Emeritus We also have the BKL solutions. Kip Thorne, for one, believes that they are likely to represent actual physical collapse. (This was metioned in his semi-popular book, "Black Holes & Timewarps"). http://en.wikipedia.org/w/index.php?title=BKL_singularity&oldid=490892346 has a brief discussion. I'm not terribly familiar with the details of the BKL solution other than it's very chaotic, Wiki gives the references. Wiki talks about BKL in the context of the early universe, I'd assume time-reversing that gives the solution Kip Thorne is fond of. 25. Nov 13, 2012 TrickyDicky As you mention BKL is a model of evolution of the Universe near the initial singularity, usually applied to cosmological models and time singularities rather than to the spacelike singularities of BHs. I have not read Thorne's semipopular book, so I don't know how or in what context he applied the BKL model in the BH setting.	6,044	27,341	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.75	3	CC-MAIN-2018-30	latest	en	0.939326
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-2-real-numbers-2-2-addition-and-subtraction-of-rational-numbers-problem-set-2-2-page-57/51	1,534,682,524,000,000,000	text/html	crawl-data/CC-MAIN-2018-34/segments/1534221215077.71/warc/CC-MAIN-20180819110157-20180819130157-00108.warc.gz	846,150,160	13,631	## Elementary Algebra $\frac{-17}{24n}$. In order to add or subtract fractions, we start by making a common denominator (bottom of the fraction). Thus, we multiply 1/6n by 4/4 and 7/8n by 3/3 to obtain: $\frac{4}{24n}$-($\frac{21}{24n}$)=$\frac{-17}{24n}$.	92	257	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.125	4	CC-MAIN-2018-34	longest	en	0.782506
https://tellmeanumber.hostcoder.com/231196	1,723,330,513,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640826253.62/warc/CC-MAIN-20240810221853-20240811011853-00562.warc.gz	445,638,030	4,775	# Question Is 231,196 a prime number? The number 231,196 is NOT a PRIME number. #### How to check if the number 231,196 is a prime number A prime number can be divided, without a remainder, only by itself and by 1. For example, 13 can be divided only by 13 and by 1. In this case, the number 231,196 that you looked for, is NOT a PRIME number, so it devides by 1,2, 4, 7, 14, 23, and of course 231,196. # Question Where is the number 231,196 located in π (PI) decimals? The number 231,196 is at position 752144 in π decimals. Search was acomplished in the first 100 milions decimals of PI. # Question What is the roman representation of number 231,196? The roman representation of number 231,196 is CCXXXMCXCVI. #### Large numbers to roman numbers 3,999 is the largest number you can write in Roman numerals. There is a convencion that you can represent numbers larger than 3,999 in Roman numerals using an overline. Matematically speaking, this means means you are multiplying that Roman numeral by 1,000. For example if you would like to write 70,000 in Roman numerals you would use the Roman numeral LXX. This moves the limit to write roman numerals to 3,999,999. # Question How many digits are in the number 231,196? The number 231,196 has 6 digits. #### How to get the lenght of the number 231,196 To find out the lenght of 231,196 we simply count the digits inside it. # Question What is the sum of all digits of the number 231,196? The sum of all digits of number 231,196 is 22. #### How to calculate the sum of all digits of number 231,196 To calculate the sum of all digits of number 231,196 you will have to sum them all like fallows: # Question What is the hash of number 231,196? There is not one, but many hash function. some of the most popular are md5 and sha-1 #### Here are some of the most common cryptographic hashes for the number 231,196 Criptographic function Hash for number 231,196 md5 4b56e3dba00a2bc9cafe7e4a4c3362d0 sha1 9126cb8c7ee5e407b6b40949fdf85f85c3093032 sha256 c4780e8bf0a7e17b2f16be5b72d777cc7b6a40339c07f48a0d852c1e2dc62dd9 sha512 40854d3eae588b0d51c657aa5c065c2ee888120c88fcb0eba909959bbf7012a8064a11a6fbcc5f5beb0e8a9df21763ec11e9c9b766d822f8765b0bf4ed19f74b # Question How to write number 231,196 in English text? In English the number 231,196 is writed as two hundred thirty-one thousand, one hundred ninety-six. #### How to write numbers in words While writing short numbers using words makes your writing look clean, writing longer numbers as words isn't as useful. On the other hand writing big numbers it's a good practice while you're learning. Here are some simple tips about when to wright numbers using letters. Numbers less than ten should always be written in text. On the other hand numbers that are less then 100 and multiple of 10, should also be written using letters not numbers. Example: Number 231,196 should NOT be writed as two hundred thirty-one thousand, one hundred ninety-six, in a sentence Big numbers should be written as the numeral followed by the word thousands, million, billions, trillions, etc. If the number is that big it might be a good idea to round up some digits so that your rider remembers it. Example: Number 231,196 could also be writed as 231.1 thousands, in a sentence, since it is considered to be a big number #### What numbers are before and after 231,196 Previous number is: 231,195 Next number is: 231,197	968	3,425	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.921875	4	CC-MAIN-2024-33	latest	en	0.870331
https://casadimenotti.com/zero-product-property-worksheet/property-of-addition-worksheetsentity-4th-grade-commutative/	1,547,790,030,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547583659890.6/warc/CC-MAIN-20190118045835-20190118071835-00485.warc.gz	455,649,991	13,870	# Zerooductoperty Equations Worksheet Gcf Factoring Solving Quadratic Applying The By Lea Burger at December 27 2018 12:15:13 owever, since the release of Excel 2007 users can now create as many worksheets within one workbook as the memory of the computer can handle. Even if the user does not have access to one of the newest versions of Excel such as Excel 2007 or Excel 2010, they can still make as many worksheets as they would like, but earlier versions of Excel will require more workbooks. How can I learn more about Worksheets and find Tutorials? I have created a website to teach as much about Excel as I can possibly learn. I will be offering valuable advice, knowledge and tutorials about many different features of Excel Worksheets as well as many other aspects of Excel. But are you also aware that math can be fun if you put some thrill and excitement to it? It can be achieved if you incorporate math in fun activity like a game. Summarizing - Summarizing is essential in processing and categorizing all of the information obtained. Students must be able to identify main ideas, discriminate essential and nonessential information, and build this new information into their current schema. So which reading for comprehension worksheets are best? Any activity or worksheet that reinforces one or more of these six general reading for comprehension strategies would be an appropriate use of instructional time in any classroom or homework assignment. ## Gallery of Zero Product Property Worksheet It is widely understood that math has a global use and acceptance. People are also aware of the rate at which math is advancing today at various fields of research and study. Many mathematicians will talk about the pattern and structure of math worksheets which are helpful for people in working fields. Math has helped science and technology reach a higher level of advancement. Letter Books: These are books that frequently use the same phonemes over and over so students can understand them (the link between a letter and the sound it makes). For instance, "Baby bear bounced balls". These books are really good, especially if you have the book as a colouring book that you can fill out together. Here's a good activity: say the sound like "b says...buh buh, ball" and then students race to colour in their balls in their workbook. You can hang these up after and everyone will have fun. What are the Features of a Worksheet? Worksheets provide Excel users with many features. The primary feature provided by worksheets is the ability to store edit and manipulate data in one central location. Different Types of Cell Input Data : Cells are where the magic happen, as they are the individual compartments that hold your data. Cells can contain many different types of data, such as numerical, text or formulaic. Numerical data is just what it sounds like - numbers that can be manipulated using arithmetic or statistical operations.	587	2,951	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.515625	3	CC-MAIN-2019-04	latest	en	0.963692
http://www.learnersplanet.com/class-5-math-printables-multiple-and-factors	1,548,072,512,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547583792338.50/warc/CC-MAIN-20190121111139-20190121133139-00404.warc.gz	319,119,017	9,432	# Multiple & Factors Worksheet-9 Multiple & Factors Worksheet-9 1. L.C.M. of 36 and 72 is: A. 36 B. 72 C. 108 D. 2 1. L.C.M. of 17 and 5 is: A. 105 B. 95 C. 85 D. 5 1. H.C.F. of 36 and 144 is: A. 36 B. 144 C. 4 D. 2 1. HCF of two numbers = _____ A. Product of numbers ÷ their LCM B. Product of numbers × their HCF C. Product of numbers D. none of these 1. HCF of 120, 144 and 216 is: A. 38 B. 24 C. 120 D. 144 1. Three common multiples of 18 and 6 are? A. 18, 6, 9 B. 18, 36, 6 C. 36, 54, 72 D. none of these 1. Every number is a ____ and a ____ of itself. A. factor, multiple B. prime, composite C. even, odd D. none of these 1. A number which is a factor of every number is: A. 0 B. 1 C. 2 D. none of these 1. Difference between a prime (19) and a composite (25) number is: A. 6 B. 8 C. 9 D. 3 1. Number of prime numbers between 50 to 60 is: A. 1 B. 2 C. 3 D. 4	409	1,340	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.78125	4	CC-MAIN-2019-04	latest	en	0.66182
http://blogs.ubc.ca/infiniteseriesmodule/units/unit-1/infinite-series/koch-snowflake-example/?replytocom=12918	1,600,430,228,000,000,000	text/html	crawl-data/CC-MAIN-2020-40/segments/1600400187390.18/warc/CC-MAIN-20200918092913-20200918122913-00042.warc.gz	32,951,661	12,079	# Koch Snowflake Example Previous: A Geometric Series Problem with Shifting Indicies Next: Videos on the Introduction to Infinite Series ## Problem Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below). Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007). We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows: • divide the line segment into three segments of equal length • draw an equilateral triangle that has the middle segment from step 1 as its base and points outward • remove the line segment that is the base of the triangle from step 2 The Koch Snowflake is the limit approached as the number of iterations goes to infinity. ## Solution The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons). Now, to derive an expression for the area of our construction at the iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours. Blue and Green Triangles Assume that the one blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of Blue, Green, and Yellow Triangles Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or the area of a blue triangle. The area of the blue, green, and yellow triangles is Blue, Green, Yellow, and Red Triangles Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or the area of a blue triangle. The area of the blue, green, yellow, and red triangles is Total Area The total area of the snowflake uses the infinite sequence . We will add all the terms of the series together, and add 1, to produce the following sum Seeing that this is a geometric series with a = 1/3 and r = 4/9, we immediately conclude that this series converges and is equal to Previous: A Geometric Series Problem with Shifting Indicies Next: Videos on the Introduction to Infinite Series	574	2,553	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.875	5	CC-MAIN-2020-40	latest	en	0.907494
https://codereview.stackexchange.com/questions/9433/linear-regression-and-data-manipulation	1,618,593,128,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618038088245.37/warc/CC-MAIN-20210416161217-20210416191217-00286.warc.gz	281,971,849	38,425	Linear Regression and data manipulation How could I improve the following code that runs a simple linear regression using matrix algebra? I import a .csv file (link here) called 'cdd.ny.csv', and perform the matrix calculations that solve for the coefficients (intercept and regressor) of Y = XB (i.e., $(X'X)^{-1}X'Y$): import numpy from numpy import * import csv tmp = list(df1) b = numpy.array(tmp).astype('string') b1 = b[1:,3:5] b2 = numpy.array(b1).astype('float') nrow = b1.shape[0] intercept = ones( (nrow,1), dtype=int16 ) b3 = empty( (nrow,1), dtype = float ) i = 0 while i < nrow: b3[i,0] = b2[i,0] i = i + 1 X = numpy.concatenate((intercept, b3), axis=1) X = matrix(X) Y = b2[:,1] Y = matrix(Y).T m1 = dot(X.T,X).I m2 = dot(X.T,Y) beta = m1m2 print beta #[[-7.62101913] # [ 0.5937734 ]] numpy.linalg.lstsq(X,Y) import numpy from numpy import import csv tmp = list(df1) b = numpy.array(tmp).astype('string') b1 = b[1:,3:5] b2 = numpy.array(b1).astype('float') Firstly, I'd avoid all these abbreviated variables. It makes it hard to follow your code. You can also combine the lines a lot more b2 = numpy.array(list(df1))[1:,3:5].astype('float') That way we avoid creating so many variables. nrow = b1.shape[0] intercept = ones( (nrow,1), dtype=int16 ) b3 = empty( (nrow,1), dtype = float ) i = 0 while i < nrow: b3[i,0] = b2[i,0] i = i + 1 This whole can be replaced by b3 = b2[:,0] X = numpy.concatenate((intercept, b3), axis=1) X = matrix(X) If you really want to use matrix, combine these two lines. But really, its probably better to use just array not matrix. Y = b2[:,1] Y = matrix(Y).T m1 = dot(X.T,X).I m2 = dot(X.T,Y) beta = m1*m2 print beta • Thanks! However, the line X = numpy.concatenate((intercept, b3), axis=1) now gives the error "ValueError: arrays must have same number of dimensions" -- this is the reason I added the while loop. Any way around this? – baha-kev Feb 26 '12 at 17:50 • @baha-kev, use b3 = b2[:,0].reshape(-1, 1) – Winston Ewert Feb 26 '12 at 18:39 • Thanks; you mention it's probably better to use arrays - how do you invert an array? The .I command only works on matrix objects. – baha-kev Feb 26 '12 at 19:16 • @baha-kev, use the numpy.lingalg.inv function. – Winston Ewert Feb 26 '12 at 19:27	756	2,277	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.53125	4	CC-MAIN-2021-17	latest	en	0.648317
https://kidsworksheetfun.com/molar-mass-worksheet-pdf/	1,660,205,417,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00533.warc.gz	326,669,125	21,940	# Molar Mass Worksheet Pdf Molar mass practice worksheet author. 1 mole 6 02 x 1023 particles 1 mole molar mass could be atomic mass from periodic table or molecular mass. Chemistry Notes Chemical Equations The Mole And Stoichiometry In 2020 Chemistry Notes Chemistry Worksheets Chemistry Lecture ### Chemistry computing formula mass worksheet problem set up example. Molar mass worksheet pdf. 1 cl2 2 koh 3 becl2 4 fecl3 5 bf3 6 ccl2f2 7 mg oh 2 8 uf6 9 so2 10 h3po4 11 nh4 2so4 12 ch3cooh 13 pb no3 2 14 ga2 so3 3. Mole worksheet use two decimal places for the molar masses and report your answer to the correct. 1 cl 2 71 g mol 2 koh 56 1 g mol 3 becl 2 80 g mol 4 fecl 3 162 3 g mol 5 bf. 2 x 14 0 28 0 o. Three times the molar mass of b d 4 8 times the formula mass of b 9 the atoms of element a are one third as heavy as the atoms of c 12. Mole to grams grams to moles conversions worksheet what are the molecular weights of the following compounds. 2141 grams 5 how many moles are in 2 3 grams of phosphorus. 77 0 grams 3 how many moles are in 22 grams of argon. 1 x 40 1 40 1 n. 3 17 2011 3 42 40 pm. The molar mass of a is. Molar mass worksheet answer key calculate the molar masses of the following chemicals. 0 55 moles 4 how many grams are in 88 1 moles of magnesium. Mole calculation worksheet answer key 1 how many moles are in 15 grams of lithium. 0 46 moles 2 how many grams are in 2 4 moles of sulfur. Molar mass worksheet key water h 2o element number mm h 2 1 01 g mole o 1 16 00 g mole total mm 18 02 g mole carbon dioxide 2co 2 element number mm o 2 16 00 g mole c 1 12 01 g mole total mm 44 01 g mole sodium chloride nacl element number mm cl 1 35 45 g mole. 1 mole molar mass could be atomic mass from periodic table or molecular mass 1 mole 22 4 l of a gas at stp you do not need to worry about this yet each definition can be written as a set of two conversion factors. Place your final answer in the formula mass column. 6 x 16 0 96 0. Find the formula mass of ca no3 2 ca. Round atomic masses to the tenth of a decimal place. 1 naoh 2 h 3po 4 3 h 2o 4 mn 2se 7 5 mgcl 2 6 nh 4 2so 4 there are three definitions equalities of mole. Molar mass worksheet calculate the molar mass of the following chemicals. 1 mole molar mass g can be written as 1 mole or molar mass g. Find the formula mass of the following compounds. Image Result For Igcse Ionic Compounds Worksheet Naming Compounds Worksheet Chemistry Worksheets Practices Worksheets Use This Molar Mass Diagram Worksheet To Help Students Learn How To Find Atomic Mass Molar Mass Is Vi Teaching Chemistry Chemistry Lessons Chemistry Education Pin By Deejay Bobbsters On Knowledge Molar Mass How To Calculate Moles Chemistry Lessons Molar Mass Worksheet Chemistry Worksheets Molar Mass Chemistry Education Molar Mass Worksheet Molar Mass Worksheets Molars Mole Conversions Worksheet Crhs Amazing Mole Conversions Worksheet Crhs Mole Conversion Worksh Mole Conversion Worksheet Mole Conversion Worksheet Template Use This Diagramed Molar Mass Worksheet To Help Students Learn Step By Step How To Find Atomic Mass Chemistry Education Teaching Chemistry Chemistry Classroom Determining Percent Composition From Molecular Formula Worksheet Help Teaching Molecular High School Chemistry Gram Formula Mass Worksheet Chemistry Notes Worksheets Mcat Study Molar Mass Doodle Notes For Teaching Chemistry In 2020 Teaching Chemistry Doodle Notes Molar Mass Molar Mass Worksheet Stem Sheets Molar Mass Teaching Chemistry Worksheets This Set Of Stoichiometry Worksheets Covers Empirical Formulas Molecular Formulas Percent Composi Chemistry Worksheets Chemistry Education Teaching Chemistry Use This Diagrammed Molar Mass Worksheet To Help Students Learn Step By Step How To Find Atomic Mass Teaching Chemistry Chemistry Lessons High School Science Mole Ratio Worksheet Together With Mole Ratio Worksheet Answer Key Moreover Bined Gas Law Worksheet Ans Worksheets Molar Mass Dna Transcription And Translation 9 Moles And Mass Worksheet Answers Key With Work Molar Mass Study Site Mole Conversion Worksheet Molar Mass Maze Worksheet For Review Or Assessment Molar Mass Maze Worksheet Molars Molar Mass Puzzle For Review Or Assessment Molar Mass Molars Teaching Chemistry Mole Conversion Practice Worksheet The Best Worksheets Image In 2020 Practices Worksheets Worksheets Chemistry Worksheets Molar Mass Worksheet Practice Problems And Answer Key General Chemistry Study Materials Study Guide Molar Mass Study Materials Study Guide	1,122	4,525	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.59375	3	CC-MAIN-2022-33	latest	en	0.75703
http://html.rhhz.net/jckxjsgw/html/62884.htm	1,723,781,055,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00073.warc.gz	13,836,205	11,760	法兰式无键液压联轴器安装及运行稳定性验证分析舰船科学技术 2020, Vol. 42 Issue (11): 59-62 DOI: 10.3404/j.issn.1672-7649.2020.11.012 PDF Analysis on installation and operation stability of flange type hydraulic coupling with non key WANG Rui, FAN Hua-tao, SONG Qiang State Key Laboratory of Deep-sea Manned Vehicles, China Ship Scientific Research Center, Wuxi 214082, China Abstract: Aiming at the problem of hydraulic control of hydraulic installation of flange non key coupling of ship shafting and the anti impact performance of hydraulic coupling. A new type flange type hydraulic coupling as the researchobject, buildingthe three-dimensional model of flange type keyless hydraulic couplings, finite element simulation calculating of contact equivalent stress、radial hydraulic pressure、axial thrust and coupling impact resistancebased on the requires of the theory calculation of steel ships of the classification, and the results of simulation analysis and theoretical analysis results are verified. The results show that when the Length of installation reaches a certain position, the axial thrust simulation value and the theoretical calculation value deviation increase due to the increase of the friction force at the contact edge position; In the case of stable operation of hydraulic coupling, the stress of the inner sleeve of the hydraulic coupling is less than that of the outer sleeve, and the weak link of the inner sleeve is in the two ends, After the installation of the hydraulic coupling to run the test, found that the stability of the stability of the coupling is very good. Key words: the coupling hydraulic installation stability length of installation radial oil pressure 0 引　言 1 计算原理 1.1 推入量的计算公式 $\begin{split} {S_1} = & \frac{1}{K}\left[ {47750 \times {{10}^4}\frac{{{N_{\rm{e}}}}}{{A{n_e}}}\left( {\frac{{{C_1}}}{{{E_1}}} + \frac{{{C_2}}}{{{E_2}}}} \right) + } \right.\\ & \left. {({a_2} - {a_1})(35 - t){d_1} + 0.03} \right]\text{，} \end{split}$ (1) ${S_2} = \frac{1}{K}\left[ {0.7{\sigma _s}d{}_1\frac{{K_2^2 - 1}}{{\sqrt {3K_2^4 + 1} }}(\frac{{{C_1}}}{{{E_1}}} + \frac{{{C_2}}}{{E{}_2}}) - ({a_2} - {a_1}){d_1}t} \right]\text{。}$ (2) 1.2 液压安装的基础力 $P = \frac{{{S_F}T}}{{A{B_1}}}\left( { - \frac{{{S_F}K}}{2} + \sqrt {\mu _1^2 + {B_1}{{\left( {\frac{{{F_V}}}{T}} \right)}^2}} } \right)\text{，}$ (3) ${F_0} = A\left( {0.002 + \frac{K}{{20}}} \right) \cdot \left[ {{P_{35}} + \frac{{18}}{{{B_2}}}({\alpha _2} - {\alpha _1})} \right]\text{。}$ (4) 1.3 轴向推力和径向油压 ${F_1} = Ap\left( {{\mu _2} + K/2} \right)\text{，}$ (5) $P = 1.1SK/({d_1}{B_2})\text{。}$ (6) 1.4 应力的计算 $\begin{split} \sigma = & \sqrt {\frac{1}{2}\left[ {{{\left( {{\sigma _1} - {\sigma _2}} \right)}^2} + {{\left( {{\sigma _2} - {\sigma _3}} \right)}^2} + {{\left( {{\sigma _3} - {\sigma _2}} \right)}^2}} \right]} = \\ & 0.9p\left[ {\sqrt {3K_2^4 + 1} /\left( {K_2^4 - 1} \right)} \right]\text{。} \end{split}$ (7) 2 实例计算分析 2.1 模型的网格划分和约束的施加图 1 液压联轴器安装和抗冲击三维模型图 Fig. 1 Three dimensional model of hydraulic coupling installation and shock resistance 2.2 等效应力图 2 液压联轴器应力 Fig. 2 Stress of hydraulic coupling 2.3 径向油压图 3 径向油压的仿真值和理论值对比 Fig. 3 Comparison of simulation value and theoretical value of radial oil pressure 2.4 轴向推力图 4 轴向推力理论和仿真值的对比 Fig. 4 Comparison of theoretical and simulation values of axial thrust 2.5 无键联轴器液压安装方案 3 联轴器的稳定性分析图 5 最大扭矩下联轴器外套和内套的应力分布云图 Fig. 5 Stress distribution nephogram of coupling outer sleeve and inner sleeve under maximum torque 4 结　语 1）由于液压联轴器实际安装过程中边缘等效应力的奇异性，接触的边缘位置具有应力集中的现象，接触面的摩擦力异常增大。当推入量达到一定的值以后，轴向推力仿真的计算结果大于理论计算的值。为了保证液压联轴器正常的工作，安装后的极大等效应力值应该小于联轴器的材料屈服强度。 2）联轴器液压安装过程中径向油压和轴向推力的仿真值和理论值基本都是呈线性增大，当推入量达到某位置以后，仿真值和理论计算值偏差开始增大，并且径向油压的理论计算值几乎都小于有限元的仿真值。为了保证液压联轴器正常的工作，径向油压和轴向推力理论值与仿真值的最大误差要低于船级社钢制海船入级规范中的7%的要求。 3）在液压联轴器运行稳定的情况下，液压联轴器内套的应力小于外套，由于受到边缘等效应力集中现象的影响，内套的最大应力出现在边缘位置，联轴器承内套的薄弱环节出现在两端。 [1] 陈琦, 刘志刚. 液压联轴器外套自紧身有限元分析[J]. 哈尔滨工程大学学报, 2004(3): 319-325. [2] 邵勇, 周建辉, 孙俊洋. 基于接触理论的法兰式液压联轴器结构强度分析[J]. 船海工程, 2014(8): 150-155. [3] 但家俊. 无键螺旋桨液压安装过程测控系统的研发[D]. 武汉: 武汉理工大学, 2012. [4] 胡旭晟, 范世东, 朱汉华. 无键螺旋桨液压安装方案分析[J]. 船舶工程, 2015, 37(7): 30-35. [5] TRUMAN C. E., BOOKER J. D.. Analysis of a shrink-fit failure on a gear hub/shaft assembly[J]. Engineering FailureAnalysis, 2007, 14: 557-572. [6] FREDERIC L, AURELIAN V, BERNARD S. Finite element analysis and contact modelling considerations of interference fits for fretting fatigue strength calculations[J]. Simulation Modelling Practice and Theory, 2009, 17: 1587-1602. DOI:10.1016/j.simpat.2009.06.017 [7] EYERCIOGLU O., KUTUK M. A., YILMAZ N. F.. Shrink fit design for precision gear forgingdies[J]. Journal of Materials Processing Technology, 2009, 209: 2186-2194. DOI:10.1016/j.jmatprotec.2008.05.016 [8] 中国船级社. 钢质海船入级规范(第3分册)[S]. 北京: 人民交通出版社, 2012.	1,987	4,766	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.984375	3	CC-MAIN-2024-33	latest	en	0.74451
https://www.routledge.com/Artificial-Neural-Networks-for-Engineers-and-Scientists-Solving-Ordinary-Differential-Equations/Chakraverty-Mall/p/book/9781498781381	1,713,411,052,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296817187.10/warc/CC-MAIN-20240418030928-20240418060928-00789.warc.gz	901,159,746	34,257	1st Edition Artificial Neural Networks for Engineers and Scientists Solving Ordinary Differential Equations By S. Chakraverty, Susmita Mall Copyright 2017 168 Pages 80 B/W Illustrations by CRC Press 168 Pages 80 B/W Illustrations by CRC Press Also available as eBook on: Differential equations play a vital role in the fields of engineering and science. Problems in engineering and science can be modeled using ordinary or partial differential equations. Analytical solutions of differential equations may not be obtained easily, so numerical methods have been developed to handle them. Machine intelligence methods, such as Artificial Neural Networks (ANN), are being used to solve differential equations, and these methods are presented in Artificial Neural Networks for Engineers and Scientists: Solving Ordinary Differential Equations. This book shows how computation of differential equation becomes faster once the ANN model is properly developed and applied. 1. Preliminaries of Artificial Neural Network 1.1 Introduction 1.2 Architecture of ANN 1.2.1 Feed-Forward Neural Network 1.2.2 Feedback Neural Network 1.3.1 Supervised Learning or Associative Learning 1.3.2 Unsupervised or Self-Organization Learning 1.4 Learning Rules or Learning Processes 1.4.1 Error Back-Propagation Learning Algorithm or Delta Learning Rule 1.5 Activation Functions 1.5.1 Sigmoid Function 1.5.1.1 Unipolar Sigmoid Function 1.5.1.2 Bipolar Sigmoid Function 1.5.2 Tangent Hyperbolic Function References 2. Preliminaries of Ordinary Differential Equations 2.1 Definitions 2.1.1 Order and Degree of DEs 2.1.2 Ordinary Differential Equation 2.1.3 Partial Differential Equation 2.1.4 Linear and Nonlinear Differential Equations 2.1.5 Initial Value Problem 2.1.6 Boundary Value Problem References 3. Multilayer Artificial Neural Network 3.1 Structure of Multilayer ANN Model 3.2 Formulations and Learning Algorithm of Multilayer ANN Model 3.2.1 General Formulation of ODEs Based on ANN Model 3.2.2 Formulation of nth-Order IVPs 3.2.2.1 Formulation of First-Order IVPs 3.2.2.2 Formulation of Second-Order IVPs 3.2.3 Formulation of BVPs 3.2.3.1 Formulation of Second-Order BVPs 3.2.3.2 Formulation of Fourth-Order BVPs 3.2.4 Formulation of a System of First-Order ODEs 3.2.5 Computation of Gradient of ODEs for Multilayer ANN Model 3.3 First-Order Linear ODEs 3.4 Higher-Order ODEs 3.5 System of ODEs References 4. Regression-Based ANN 4.1 Algorithm of RBNN Model 4.2 Structure of RBNN Model 4.3 Formulation and Learning Algorithm of RBNN Model 4.4 Computation of Gradient for RBNN Model 4.5 First-Order Linear ODEs 4.6 Higher-Order Linear ODEs References 5. Single-Layer Functional Link Artificial Neural Network 5.1 Single-Layer FLANN Models 5.1.1 ChNN Model 5.1.1.1 Structure of the ChNN Model 5.1.1.2 Formulation of the ChNN Model 5.1.1.3 Gradient Computation of the ChNN Model 5.1.2 LeNN Model 5.1.2.1 Structure of the LeNN Model 5.1.2.2 Formulation of the LeNN Model 5.1.2.3 Gradient Computation of the LeNN Model 5.1.3 HeNN Model 5.1.3.1 Architecture of the HeNN Model 5.1.3.2 Formulation of the HeNN Model 5.1.4 Simple Orthogonal Polynomial–Based Neural Network (SOPNN) Model 5.1.4.1 Structure of the SOPNN Model 5.1.4.2 Formulation of the SOPNN Model 5.1.4.3 Gradient Computation of the SOPNN Model 5.2 First-Order Linear ODEs 5.3 Higher-Order ODEs 5.4 System of ODEs References 6. Single-Layer Functional Link Artificial Neural Network with Regression-Based Weights 6.1 ChNN Model with Regression-Based Weights 6.1.1 Structure of the ChNN Model of the ChNN Model 6.2 First-Order Linear ODEs 6.3 Higher-Order ODEs References 7. Lane–Emden Equations 7.1 Multilayer ANN-Based Solution of Lane–Emden Equations 7.2 FLANN-Based Solution of Lane–Emden Equations 7.2.1 Homogeneous Lane–Emden Equations 7.2.2 Nonhomogeneous Lane–Emden Equation References 8. Emden–Fowler Equations 8.1 Multilayer ANN-Based Solution of Emden–Fowler Equations 8.2 FLANN-Based Solution of Emden–Fowler Equations References 9. Duffing Oscillator Equations 9.1 Governing Equation 9.2 Unforced Duffing Oscillator Equations 9.3 Forced Duffing Oscillator Equations References 10. Van der Pol–Duffing Oscillator Equation 10.1 Model Equation 10.2 Unforced Van der Pol–Duffing Oscillator Equation 10.3 Forced Van der Pol–Duffing Oscillator Equation References Biography Dr. S. Chakraverty has over 25 years of experience as a researcher and teacher. Currently, he is working at the National Institute of Technology, Rourkela, Odisha as a full Professor and Head of the Department of Mathematics. Prior to this, he was with CSIRCentral Building Research Institute, Roorkee, India. After graduating from St. Columba’s College (Ranchi University), he obtained his M. Sc in Mathematics and M. Phil in Computer Applications from the University of Roorkee (now the Indian Institute of Technology Roorkee), earning First Position in the University honors. Dr. Chakraverty received his Ph. D. from IIT Roorkee in 1992. Afterwards, he did his post-doctoral research at Institute of Sound and Vibration Research (ISVR), University of Southampton, U.K. and at the Faculty of Engineering and Computer Science, Concordia University, Canada. He was also a visiting professor at Concordia and McGill Universities, Canada, during 1997-1999 and visiting professor of University of Johannesburg, South Africa during 2011-2014. Mrs. Susmita Mall received her M. Sc. degree in Mathematics from Ravenshaw University, Cuttack, Odisha, India in 2003. Currently she is a Senior Research Fellow in National Institute of Technology, Rourkela - 769 008, Odisha, India. She has been awarded Women Scientist Scheme-A (WOS-A) fellowship, under Department of Science and Technology (DST), Government of India to undertake her Ph. D. studies. Her current research interest includes Mathematical Modeling, Artificial Neural Network, Differential equations and Numerical analysis. To date, she has published seven research papers in international refereed journals and five in conferences.	1,631	6,116	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2024-18	latest	en	0.806722
https://es.mathworks.com/matlabcentral/answers/1744725-i-have-to-store-3-sets-of-data-with-for-statement	1,719,071,192,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00566.warc.gz	203,456,681	26,624	# I have to store 3 sets of data, with for statement 1 visualización (últimos 30 días) Devesh Kumar el 21 de Jun. de 2022 Comentada: Devesh Kumar el 28 de Jun. de 2022 for i = 1:cf_n x1(i,1) = [2 Vr_on_cf Vr_R1_cf(i) Vr_R2_cf(i) Vr_cf_end]; % Az/D for cross - flow y1(i,1) = [0 0.15 Az1_D(i) Az2_D(i) 0]; % reduced velocity end Here Vr_on_cf,Vr_cf_end is constant = 2.5 and 16 respectivetly , Vr_R1_cf & Vr_R2_cf are vector which contains 3 values say [ a b c] and [d e f], in this particular case cf_n is 3 now I want my output like this x = [2 2.5 a d 16; 2 2.5 b e 16; 2 2.5 c f 16]; basically 5*3 matrix how should I run the for loop ##### 2 comentariosMostrar NingunoOcultar Ninguno Walter Roberson el 21 de Jun. de 2022 for i = 1:cf_n x1(i,:) = [2 Vr_on_cf Vr_R1_cf(i) Vr_R2_cf(i) Vr_cf_end]; % Az/D for cross - flow y1(i,:) = [0 0.15 Az1_D(i) Az2_D(i) 0]; % reduced velocity end Devesh Kumar el 21 de Jun. de 2022 Thanks allot @Walter Roberson Iniciar sesión para comentar. Pooja Kumari el 28 de Jun. de 2022 Dear Devesh, It is my understanding that you want to store three sets of data using for loop. Given that Vr_on_cf, Vr_cf_end is constant = 2.5 and 16 respectively, Vr_R1_cf & Vr_R2_cf are vector which contains 3 values say [ a b c] and [d e f], in this particular case cf_n is 3. You can get the provided output using the following code: Vr_on_cf = 2.5; Vr_cf_end = 16; Vr_R1_cf = [ "a" "b" "c"]; Vr_R2_cf = [ "d" "e" "f"]; % x = [2 2.5 a d 16; 2 2.5 b e 16; 2 2.5 c f 16]; %Required Output for i = 1:3 x1(i,:) = [2 Vr_on_cf Vr_R1_cf(i) Vr_R2_cf(i) Vr_cf_end] % instead of your provided code, you can use this changed to get the required output end Sincerely, Pooja Kumari ##### 1 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos Devesh Kumar el 28 de Jun. de 2022 Iniciar sesión para comentar. ### Categorías Más información sobre Loops and Conditional Statements en Help Center y File Exchange. ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting! Translated by	762	2,093	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.84375	3	CC-MAIN-2024-26	latest	en	0.487095
https://www.eevblog.com/forum/metrology/design-considerations-for-8-5-digit-front-end/	1,708,503,539,000,000,000	text/html	crawl-data/CC-MAIN-2024-10/segments/1707947473401.5/warc/CC-MAIN-20240221070402-20240221100402-00114.warc.gz	775,038,765	21,178	### Author Topic: Design considerations for 8.5 digit front end (Read 6521 times) 0 Members and 1 Guest are viewing this topic. #### sahko123 • Frequent Contributor • Posts: 318 • Country: ##### Design considerations for 8.5 digit front end « on: January 27, 2023, 12:44:18 am » Context: I am designing an 8.5 digit ADC with surrounding voltmeter with a novel self-calibration technique as part of my final year project. (about which i will share but not until the project is finished). As part of the ADC I need an input buffer and believe an ADA4625-1 unity gain buffer should be sufficient. Mostly because this is more about the ADC and getting a +-10V voltage digitized as accurately and precisely as possible. The following project would be about getting other functions such as I and R along with various voltage ranges but for now the ADC is the priority. does anyone have any suggestions on the input buffer or should it be more than enough for the project at the moment? Is there anything Im missing? or should this suffice? Asking for a friend #### coppice • Super Contributor • Posts: 8493 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #1 on: January 27, 2023, 01:06:07 am » Lets says 8.5 digit means a maximum count of +-200,000,000. The step size for 10V would be 50nV. An ADA4625-1 seems to have a offset voltage spec of 80uV typical. Does that seem suitable? Even if you are frequently self calibrating away that offset, just how well temperature controlled would the op-amp need to be tame its offset voltage drift to 50nV between calibrations? #### sahko123 • Frequent Contributor • Posts: 318 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #2 on: January 27, 2023, 01:13:51 am » The offset voltage of 80uV I should be perfectly fine. but the tempco can possibly be better. The typical is actually +-0.2uV/C with a maximum of 1.2uV/c Asking for a friend #### iMo • Super Contributor • Posts: 4562 • Country: • It's important to try new things.. ##### Re: Design considerations for 8.5 digit front end « Reply #3 on: January 27, 2023, 11:48:35 am » FYI - not targeting 8.5digits, but an attempt to find a simplest way to create an AFE for the high-end single chip ADCs. #### Kleinstein • Super Contributor • Posts: 13897 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #4 on: January 27, 2023, 12:01:07 pm » The offset (and drift) is one of the lesser problems if the front end does some kind of auto zero by switching between different inputs, including a 0 V (or similar). The CMRR could be a liniting factor for the linearity. With typ. 130 dB this not enough to guarantee better than 0.3 ppm INL. However chances are that much of the CMRR is still linear - so from this side it can be just acceptable, though not great, when hunting for possible sources of INL. Another important parameter is theg gain, as this effects how good the OP-amp can compensate for internal nonlinearity, especially the output cross over. Here the ADA4625 is quite good - though the cross over error could still be a point to whatch for. With the OPA145 I was able to see that type off error. Adding a constant current load can avoid the cross over error by operating the output in a class A range. For a buffer it is relatively simple to use a bootstrapped supply and this way essentially eliminate the effect of the CMRR. An alternative buffer would be a Zero drift OP-amp like LTC2057 or OPA189. These usually have very good CMRR and gain, at lest for DC and low frequencies. The following users thanked this post: sahko123 #### miro123 • Regular Contributor • Posts: 186 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #5 on: January 27, 2023, 03:56:26 pm » 1. What are the input parameter/requirements - e.g. BW, input R and Z 2. What are parameters of of ADC input - e.g. does ADC uses SC circuit? Does the ADC input have constant/or linear R/C/L/Z? 3. what are the requirement for linearity offsets and drifts? 4. ADC input type - single ended or differential? 4. Do you really need such fast opamp? - the higher power consumption and associated thermals could create headache even at the 6 1/2 digit AFE « Last Edit: January 27, 2023, 04:01:00 pm by miro123 » #### K-Zoltan • Newbie • Posts: 3 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #6 on: September 21, 2023, 06:40:46 am » Here is a drawing with 8.5 digit voltmeter I made. The following users thanked this post: iMo #### Echo88 • Frequent Contributor • Posts: 820 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #7 on: September 21, 2023, 07:50:31 am » Did you read up on the HPM7177? The only real OSHW project that i know off that actually shows a complete project with measurements coming close to a true 8.5 digit voltmeter. You will have to spend plenty of time/money already to design a good DCV analog front end for your ADC, so i assume there wont be time for I/R-measurements. Do you have the means to measure the resulting specs like INL accurately enough, like multiple 3458A or JJA-access? If a high impedance analog frontend is needed you could copy the 3458A-frontend like i intend to do in a coming revision of my HPM7177-implementation, as suggested (not tested, work in progress) here: https://www.eevblog.com/forum/metrology/analog-frontends-for-dmms-approaching-8-5-digits-discussions/msg5034982/#msg5034982 #### K-Zoltan • Newbie • Posts: 3 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #8 on: September 21, 2023, 09:06:07 am » Here is the nstrument I made High impedance frontend I used opa1289 and is in a metal case heat to 40 deg. C In place of relays I used optoflash TLP240. 3458 is an old 35 years (or more ?) old instrument full with discrete components. Now there are much better IC than ampifiers with discrete components. #### dietert1 • Super Contributor • Posts: 1983 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #9 on: September 21, 2023, 09:17:17 am » For our Prema 6048 inspired design there is a voltmeter front end with a +/- 6V bootstrapped OPA189 or similar as single ended buffer, as the integrator is single-ended, too. There are three input relays for the two input polarities and null. The four ranges may become 24 V, 12 V, 2.4 V and 240 mV. While upper ranges are 1x buffered, for the lowest range the bootstrap gets turned off and there is 10x gain. I think the combination of a chopper-stabilized input amplifier with a relay to recalibrate null once the oven has reached stable temperature should be a good solution. The circuit preserves the "continuous integration" concept of the Prema. Don't have measurements yet. Boards just arrived while i am writing. Regards, Dieter #### Echo88 • Frequent Contributor • Posts: 820 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #10 on: September 21, 2023, 11:11:12 am » Very interesting K-Zoltan. Can you share more of your design? The following users thanked this post: Mickle T., Ole #### Mickle T. • Frequent Contributor • Posts: 463 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #11 on: September 21, 2023, 11:43:56 am » Here is the nstrument I made "Video unavailable. This video contains content from SME, who has blocked it in your country on copyright grounds" #### 2N3055 • Super Contributor • Posts: 6264 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #12 on: September 21, 2023, 11:52:03 am » Here is the nstrument I made High impedance frontend I used opa1289 and is in a metal case heat to 40 deg. C In place of relays I used optoflash TLP240. 3458 is an old 35 years (or more ?) old instrument full with discrete components. Now there are much better IC than ampifiers with discrete components. Am I confused, or what? There are 8 digits on display... That would make it 7.5 digit meter, or am I wrong.? #### Ole • Regular Contributor • Posts: 67 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #13 on: September 21, 2023, 12:03:28 pm » Am I confused, or what? There are 8 digits on display... That would make it 7.5 digit meter, or am I wrong.? The Range shown on the thumbnail is the 10V range, which is resolved with 7 digits behind the period. Assuming the range goes only to 9.9999999V it would still be a 8 digit display (as there are 8 fully variable digits). Assuming the range can go up to 12V it would mean that there are 8 fully variable digits and one that can either be a 0 or a 1. The first digit, the one that can only be a 0 or a 1, is hidden because it is not needed. record scratch noise Hey, you. Yes, you. Have an awesome day! #### David Hess • Super Contributor • Posts: 16386 • Country: • DavidH ##### Re: Design considerations for 8.5 digit front end « Reply #14 on: September 21, 2023, 01:34:13 pm » The CMRR could be a liniting factor for the linearity. With typ. 130 dB this not enough to guarantee better than 0.3 ppm INL. However chances are that much of the CMRR is still linear - so from this side it can be just acceptable, though not great, when hunting for possible sources of INL. It is not really practical because of all of the analog switches needed, however Intersil solved the CMRR problem by executing the automatic zero function at the common mode input voltage, so that the automatic zero also corrected the common mode rejection. Siliconix did not and their designs suffered for it. Quote For a buffer it is relatively simple to use a bootstrapped supply and this way essentially eliminate the effect of the CMRR. I would consider bootstrapping the input buffer for another reason also; it would allow for a gigaohm+ input resistance exceeding 15 volts. Most designs are limited to 2 volts or maybe 10 volts, but 15+ volts is very handy and something I am looking for in my next multimeter purchase. Quote An alternative buffer would be a Zero drift OP-amp like LTC2057 or OPA189. These usually have very good CMRR and gain, at lest for DC and low frequencies. I think using a zero drift operational amplifier as a buffer would be a mistake because the input bias current and input current noise will interact with the series input protection (and source impedance) to add offset and noise. The offset voltage of 80uV I should be perfectly fine. but the tempco can possibly be better. The typical is actually +-0.2uV/C with a maximum of 1.2uV/c The input offset voltage drift specification is tough. Even my new favorite low input bias current precision part, the OPA140, has a maximum input offset drift of 1 uV/C, which is great for a JFET input but an order of magnitude worse than the best bipolar inputs. Chopper parts solve this but have worse problems with high impedance inputs. That means grading the input buffer for low drift, correcting the drift somehow, or correcting the drift with automatic zero which is how most designs handle it. One place I worked did the first one with a test chamber and marked the top of the parts with the drift. #### dietert1 • Super Contributor • Posts: 1983 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #15 on: September 21, 2023, 03:54:23 pm » One needs to do something about low frequency noise. If one does it with autozero e.g. 1 PLC alternating between unknown and null input, half of the input signal gets lost. If one does it using a chopper stabilized opamp, it supports continuous integration, yet the chopper frequency will probably be higher with more noise generated at the input. I agree with K-Zoltan that we should try and use integrated circuits where possible. There is a large choice of chopper opamps, some with less input current than the OPA189. In the bootstrap scheme one can also try and compensate the input current to maybe reduce it from 100 pA to 10 pA. As i plan to run the meter inside an oven at constant temperature, there is a good chance to make the compensation work. Anyway, 100 pA at 10 V gives 100 GOhm and with compensation one can expect 1 TOhm. Regards, Dieter #### Kleinstein • Super Contributor • Posts: 13897 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #16 on: September 21, 2023, 04:55:07 pm » There is usually no need to go for the super low noise AZ amplifiers. The input noise of the HP3458 is more like 50 nV/sqtz(Hz) assuming 100% integration. So one can get away with lower input current AZ amplifiers like AD8628, LMP2011 or MCP6V76 and still get a comparable or slightly better noise. These have less input bias and less current noise. However the lower bias types often only comes with a limited supply voltage (e.g. 6 V max) and thus kind of need bootstrapping the supply already for a 10 V range. With suitable filtering at the input one can isolate it from variations in the input impedance reasonably well. The very low bias types (max4238, ICL7650, LTC2055) are a bit on the high side with the noise. Chances are one needs some luck with the amplifier bias and switch leakage to get at least partial compensation or good individual units. The input resistance is more like a differential thing. So one has some input bias and than an input resistance describing how the input current changes with the voltage. So I would not call voltage range divided by input bias the input resistance. One usually needs some filtering anyway to keep EMI out and as part of the ESD protection. In my DMM circuit I have an AD8628 with a BS supply for the main input that works OK. In my case I got some 6 pA bias and some 300-400 Gohm of differential input resistance with a large part of this likely from PCB leakage (changed with cleaning). For bias I consider this a lucky pick of the amplifiers and switches. Actual values can scatter even with the same types used. I have an MCP6V76 with a high voltage divider and this also works OK. The choice of Az amplifier (e.g. Keithley 2000 / 2002, Datron 1271/1281 like) vs AZ switching (e.g. HP 3458 and most other HP) is between frequent chopping with small spikes in the AZ amplifier versus relatively low frequency (e.g. 2.5-25 Hz) but usually larger switching peaks. Both have there pros and cons. #### Echo88 • Frequent Contributor • Posts: 820 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #17 on: September 21, 2023, 05:40:17 pm » I assume the AZ-cycle time can be reduced pretty much when the first amplifier/buffer stage in the AFE (discret or not) can be kept at a stable temperature, like in this paper: https://arxiv.org/pdf/1708.06311.pdf After the buffer youre free to throw AZ-OPs at it. The following users thanked this post: dietert1 #### Kleinstein • Super Contributor • Posts: 13897 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #18 on: September 21, 2023, 06:17:40 pm » The range for the auto zero cycle is usually pretty small: less than 1 PLC is often problematic as mains hum can cause problems. Also many switching spikes are not that desirable. Without Az amplifier one usually wants to suppress much of the 1/f noise and much more than 1 slow reading (e.g. 100 PLC) is not practical. Often 10 PLC is used as a upper limit. So the main choices are some 1 to 10 PLC for the AZ cycle length, rarely more. When there is a low drift (in generalle some kind of AZ amplifier) input buffer the AZ cycle behind it can often be 1 PLC, unless the ADC wants more (especially some older multi-slope ADCs). The JFET amplifier with stablilized temperature is more like an oddity and may still have 1/f noise or slow drift (e.g. from stress or aging) and may thus still want some AZ switching in front - maybe just with a rather slow cycle. The main configurations to look at are: 1) AZ switching close to the input and than non AZ amplifiers (or just 1 amplifier) all the way to the ADC (e.g. HP DMMs). This often needs some pre-charge cicuit to limit the swiching spike. 2) an AZ buffer at the input and than AZ switching after that followed by a non AZ amplifier (e.g. Keithley 2000 / 2002) 3) an AZ amplifier at the front and than AZ switching and a buffer or 2nd amplifier stage at the ADC input (e.g. my DMM circuit, Keithey 2182) 4) an AZ amplifier at the front and than a low drift ADC (e.g. Datron 1281, likely SDM3065, Solartron 7081) with only rare Az switching - if at all Many of the SD-ADC chips have pretty low dirft and may allow this configuration One may mix the cases, and not all input paths must look the same - though with most meters it is. The case with an non AZ buffer at the input is more a thing for electrometers, not so much for a high resolution DMM. The following users thanked this post: splin #### macaba • Regular Contributor • Posts: 199 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #19 on: September 22, 2023, 09:34:00 am » If hyper-focused on a small part of the design, an integrated zero-drift amplifier would appear to make sense, however when considering the whole design, integrated zero-drift amplifiers are not suitable for the input amplifier on a 8.5d voltmeter design (and HP agreed). Designs that use zero-drift amplifier on the input are more like "low source impedance digitiser" rather than "voltmeter" (HPM7177 being a good example where it is continuously monitoring LHC magnets in a highly stable temperature environment). (I'm going to use the term "chopper frequency" going forward, but this could easily refer to the frequency of non-chopper zero-drift mechanisms like autozero, correlated double sampling, and higher order nested zero-drift schemes, as true chopper topology amplifiers are rarely used anymore) 1. IC AZ has no control over the charge injection spikes. With a discrete design, it's easier to have precharge phases. Furthermore, because the chopper frequency of a discrete implementation tends to be lower (1NPLC for example), it's easier to observe the charge injection spikes during development. 2. IC AZ has high input current noise. Good luck measuring high value resistors. Current noise is usually proportional to the chopper frequency. Lower frequency = lower current noise. (therefore you can see the advantage of 1NPLC vs >100kHz...) 3. So the drift of one amplifier is solved... what about the whole signal path? ADCs have 1/f noise too. Even the AD4630 which claims "1/f noise is canceled internally" on the datasheet. So now there are multiple points of the signal chain doing their own zero-drift mechanisms which is a recipe for problems. IMO it's better to do chopping on the whole signal path as one, at a frequency you control. 4. There was mention of "half the input signal gets lost" - there are topologies that avoid this, though I don't think it's a big concern for DMM inputs to have a auto-zero scheme that halves the sample rate. For a modern DMM that only has an IC ADC (no integrating ADC), I think it's sensible to have 2 distinct modes of operation - "NPLC mode" where all the auto-zero/auto-cal mechanisms are enabled, and "Digitiser mode" where they are disabled (perhaps with a single correction at the point where you change modes, and automatically thereafter at any point where the internal DMM temperature has changed 0.1 degC, with an optional user override of this behavior when you absolutely must have a contiguous stream of samples, as in the case of HPM7177). 5. Once you have a discrete zero-drift mechanism in place, other advantages come in too - that analog switch that does the switching between "Input" and "0V" can have another input to a mux that allows the selection of other voltages (like REF+), ideal for self-calibration. 6. JFET front end (whether discrete or OPA140) already has single-digit pA input bias current. Far less compensation required, if any. « Last Edit: September 22, 2023, 09:39:15 am by macaba » The following users thanked this post: Mickle T., NNNI #### tszaboo • Super Contributor • Posts: 7177 • Country: • Current job: ATEX product design ##### Re: Design considerations for 8.5 digit front end « Reply #20 on: September 22, 2023, 09:47:51 am » Context: I am designing an 8.5 digit ADC with surrounding voltmeter with a novel self-calibration technique as part of my final year project. (about which i will share but not until the project is finished). As part of the ADC I need an input buffer and believe an ADA4625-1 unity gain buffer should be sufficient. Mostly because this is more about the ADC and getting a +-10V voltage digitized as accurately and precisely as possible. The following project would be about getting other functions such as I and R along with various voltage ranges but for now the ADC is the priority. does anyone have any suggestions on the input buffer or should it be more than enough for the project at the moment? Is there anything Im missing? or should this suffice? My suggestion is to select another final year project. You will only have time to run 1 maybe 2 PCB iterations. There are a lot of pitfalls in designing a 8.5 digit AFE. University teachers don't grade or appreciate the difficulty of the problem, they grade the paper that you submitted. You need to worry about latex, sources, reading publications, and wrestle with MS Word. Not chasing nanovolts. You simply don't have time to do it right anyway, and you are expected to spend most of your effort on the paper, not the circuit It's an interesting project, but not if you are graded, and not with deadlines. Maybe drop it to 6.5 digit, if you made this for yourself. #### iMo • Super Contributor • Posts: 4562 • Country: • It's important to try new things.. ##### Re: Design considerations for 8.5 digit front end « Reply #21 on: September 22, 2023, 10:21:02 am » As I wrote earlier - would be great to develop a simple AFE with +/- 12V input range only, as a Proof of Concept. Targeting complex designs with U/I/R capabilities in multiple ranges means man-years of work with many iterations, with a result nobody would be able to reproduce (for many reasons) or characterize. Thus a bootstrapped OPA140 (for example), with some switch for calibration/zero, then a 1:10 divider with the final buffer to feed the ADC chip low impedance with a differential +/-1.2V. I think having such a simple AFE with rock solid performance (which would be fully in pair with any single chip ADC we have got handy, like the ADS1263, AD7177, LTC2500-32, AD4630, etc) would be a great practical achievement here. Also for sahko's final year paper, imho.. « Last Edit: September 22, 2023, 10:37:37 am by iMo » The following users thanked this post: MegaVolt #### Kleinstein • Super Contributor • Posts: 13897 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #22 on: September 22, 2023, 11:05:24 am » The integrated AZ OP amps are not that bad: They can use quite some extra effort (more elaborated than the simple precharge) and on chip matching can be quite good with little parasitic capacitance. Another point is that the AZ amplifier are usually chopper stabilized and thus switching with very low voltage at the switches. So the individual switching spike from the AZ amplifier are considerably smaller than the spikes from switching with the full swing like in the HP meters AZ switching cycle. Another point is that AZ switching usually uses a short break to let the spike decay - this can be good, if the decay is fast and also bad if the signal source is high impedance / capacitive and stretches the spike to extend beyound the dead time. With the dead time filtering the spike can be tricky as this is just the kind of source that stretches the pulse. There are meters with discrete build chopper stabilized amplifiers too (datron 1281, Solartron 7081), that can use a moderately low chopper frequency. The reduction in the current noise is for the most part only with the square root of the chopper frequency. It is the input bias that about scales like the frequency. #### dietert1 • Super Contributor • Posts: 1983 • Country: ##### Re: Design considerations for 8.5 digit front end « Reply #23 on: September 22, 2023, 11:29:18 am » If hyper-focused on a small part of the design, an integrated zero-drift amplifier would appear to make sense, however when considering the whole design, integrated zero-drift amplifiers are not suitable for the input amplifier on a 8.5d voltmeter design... Back to reality: The Prema 6048 was an 8.5 meter design using a chopper input amplifier instead of autozero and it is on par with the best HPAK meters, once you have the multiplexer, so one can recalibrate null without touching cables. Look for recent statements of RAX. My own experience is similar, except i want an ovenized/humidity controlled meter. At the same time one can try other improvements. Regards, Dieter #### David Hess • Super Contributor • Posts: 16386 • Country: • DavidH ##### Re: Design considerations for 8.5 digit front end « Reply #24 on: September 22, 2023, 11:57:21 am » 3. So the drift of one amplifier is solved... what about the whole signal path? ADCs have 1/f noise too. Even the AD4630 which claims "1/f noise is canceled internally" on the datasheet. So now there are multiple points of the signal chain doing their own zero-drift mechanisms which is a recipe for problems. IMO it's better to do chopping on the whole signal path as one, at a frequency you control. Does anybody apply chopper stabilization to the whole signal path? The high impedance buffer at the input is the largest contributor of noise before the ADC, and delta-sigma ADCs, at least the instrumentation ones, do cancel their own 1/f noise. They even have the same input offset and input offset drift as chopper stabilized amplifiers built on the same process. (1) Hmm, since the input divider effectively raises the input referred noise of the high impedance buffer, do electrometer style inputs where the input buffer is bootstrapped display lower noise because their divider is after the high impedance buffer? This could be another reason to bootstrap the high impedance buffer. I have not noticed that electrometers have lower noise than multimeters, but I have never had both to compare at the same time. (1) Some delta-sigma converters intended for transducer measurement have a mode where they chop the excitation output with their inputs so they do chop the entire signal conditioning chain, and some multimeters do the same thing in 4-wire ohms mode. Smf	6,605	26,599	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.578125	3	CC-MAIN-2024-10	latest	en	0.907231
https://kr.mathworks.com/matlabcentral/cody/problems/68-kaprekar-steps/solutions/1863684	1,591,147,332,000,000,000	text/html	crawl-data/CC-MAIN-2020-24/segments/1590347426956.82/warc/CC-MAIN-20200602224517-20200603014517-00265.warc.gz	399,816,755	15,880	Cody # Problem 68. Kaprekar Steps Solution 1863684 Submitted on 3 Jul 2019 by Ingrid Odlén This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1 Pass x = 3276; y_correct = 5; assert(isequal(KaprekarSteps(x),y_correct)) x = 5265 x = 3996 x = 6264 x = 4176 x = 6174 2 Pass x = 3; y_correct = 6; assert(isequal(KaprekarSteps(x),y_correct)) x = 2997 x = 7173 x = 6354 x = 3087 x = 8352 x = 6174 3 Pass x = 691; y_correct = 7; assert(isequal(KaprekarSteps(x),y_correct)) x = 9441 x = 7992 x = 7173 x = 6354 x = 3087 x = 8352 x = 6174 4 Pass x = 3333; y_correct = Inf; assert(isequal(KaprekarSteps(x),y_correct)) x = 0 5 Pass x = 1; y_correct = 5; assert(isequal(KaprekarSteps(x),y_correct)) x = 999 x = 8991 x = 8082 x = 8532 x = 6174 6 Pass x = 6174; y_correct = 0; assert(isequal(KaprekarSteps(x),y_correct)) 7 Pass x = 1234; y_correct = 3; assert(isequal(KaprekarSteps(x),y_correct)) x = 3087 x = 8352 x = 6174 8 Pass x = 3141; y_correct = 5; assert(isequal(KaprekarSteps(x),y_correct)) x = 3177 x = 6354 x = 3087 x = 8352 x = 6174 9 Pass x = 8080; y_correct = 6; assert(isequal(KaprekarSteps(x),y_correct)) x = 8712 x = 7443 x = 3996 x = 6264 x = 4176 x = 6174	530	1,303	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.8125	4	CC-MAIN-2020-24	latest	en	0.433444
http://metamath.tirix.org/mpests/cbvrexvw	1,721,359,043,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514866.33/warc/CC-MAIN-20240719012903-20240719042903-00593.warc.gz	22,303,091	2,301	# Metamath Proof Explorer ## Theorem cbvrexvw Description: Change the bound variable of a restricted existential quantifier using implicit substitution. Version of cbvrexv with a disjoint variable condition, which does not require ax-10 , ax-11 , ax-12 , ax-13 . (Contributed by NM, 2-Jun-1998) (Revised by Gino Giotto, 10-Jan-2024) Ref Expression Hypothesis cbvralvw.1 ${⊢}{x}={y}\to \left({\phi }↔{\psi }\right)$ Assertion cbvrexvw ${⊢}\exists {x}\in {A}\phantom{\rule{.4em}{0ex}}{\phi }↔\exists {y}\in {A}\phantom{\rule{.4em}{0ex}}{\psi }$ ### Proof Step Hyp Ref Expression 1 cbvralvw.1 ${⊢}{x}={y}\to \left({\phi }↔{\psi }\right)$ 2 eleq1w ${⊢}{x}={y}\to \left({x}\in {A}↔{y}\in {A}\right)$ 3 2 1 anbi12d ${⊢}{x}={y}\to \left(\left({x}\in {A}\wedge {\phi }\right)↔\left({y}\in {A}\wedge {\psi }\right)\right)$ 4 3 cbvexvw ${⊢}\exists {x}\phantom{\rule{.4em}{0ex}}\left({x}\in {A}\wedge {\phi }\right)↔\exists {y}\phantom{\rule{.4em}{0ex}}\left({y}\in {A}\wedge {\psi }\right)$ 5 df-rex ${⊢}\exists {x}\in {A}\phantom{\rule{.4em}{0ex}}{\phi }↔\exists {x}\phantom{\rule{.4em}{0ex}}\left({x}\in {A}\wedge {\phi }\right)$ 6 df-rex ${⊢}\exists {y}\in {A}\phantom{\rule{.4em}{0ex}}{\psi }↔\exists {y}\phantom{\rule{.4em}{0ex}}\left({y}\in {A}\wedge {\psi }\right)$ 7 4 5 6 3bitr4i ${⊢}\exists {x}\in {A}\phantom{\rule{.4em}{0ex}}{\phi }↔\exists {y}\in {A}\phantom{\rule{.4em}{0ex}}{\psi }$	602	1,392	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.640625	4	CC-MAIN-2024-30	latest	en	0.525414
https://quizizz.com/en-in/perimeter-of-a-rectangle-worksheets-class-1	1,719,355,334,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00290.warc.gz	413,974,494	27,537	Area and Perimeter 16 Q 1st - 4th Perimeter & Area 11 Q 1st - 4th Bab 10 : Perimeter dan Luas (Tingkatan 1) 20 Q 1st - 10th THE STARS IN ARCHITECTURE 20 Q 1st - PD 17 Q 1st - 4th Perimeter and Area 10 Q 7th - 1st PERIMETER DAN LUAS 20 Q 1st - 5th lukisan berskala tingkatan 3 12 Q 1st - 3rd Perimeter 10 Q 1st - 4th Perimeter 15 Q 1st - 12th Geometry- a basic 10 Q 1st - 3rd KSSM MATHS FORM 1 CHAPTER 6-10 (TASK 1) 13 Q 1st - 4th Area and Perimeter Vocabulary Words 10 Q 1st - 8th perimeter 4 Q 1st - 5th Perimeter and Area 18 Q 1st - 6th Area and Perimeter 10 Q 1st - 5th Lukisan berskala 20 Q 1st - 12th Maths Quiz 9 Q KG - 1st ATS MATHS GCSE 8F 17 Q 1st - 3rd PERIMETER AND AREA 15 Q 1st - 3rd Area & Perimeter 16 Q 1st Area and Perimeter 20 Q 1st - 2nd Y6 Area and Perimeter 11 Q 1st - 9th area and perimeter 35 Q 1st - PD ## Explore printable Perimeter of a Rectangle worksheets for 1st Class Perimeter of a Rectangle worksheets for Class 1 are essential tools for teachers looking to introduce their students to the world of Math and Geometry. These worksheets provide a fun and engaging way for young learners to grasp the concept of Perimeter and apply it to real-life situations. With a variety of exercises and activities, these worksheets help Class 1 students develop their problem-solving skills and build a strong foundation in Math. Teachers can easily incorporate these worksheets into their lesson plans, ensuring that their students have a thorough understanding of the Perimeter of a Rectangle. By using these worksheets, teachers can create a dynamic and interactive learning environment that fosters a love for Math and Geometry in their Class 1 students. Quizizz is an excellent platform for teachers to find a wide range of resources, including Perimeter of a Rectangle worksheets for Class 1, to enhance their students' learning experience. This platform offers a variety of interactive quizzes, games, and activities that cater to different learning styles and help students grasp complex concepts with ease. Teachers can easily customize the content to suit their lesson plans and track their students' progress through detailed reports. In addition to Math and Geometry resources, Quizizz also offers materials for other subjects, making it a one-stop solution for all educational needs. By incorporating Quizizz into their teaching strategies, teachers can ensure that their Class 1 students receive a well-rounded education that prepares them for future success.	687	2,564	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.9375	3	CC-MAIN-2024-26	latest	en	0.932716
http://functions.wolfram.com/ElementaryFunctions/Sqrt/27/01/0004/	1,527,351,044,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794867559.54/warc/CC-MAIN-20180526151207-20180526171207-00524.warc.gz	123,245,117	7,256	html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; } Sqrt http://functions.wolfram.com/01.01.27.0003.01 Input Form Sqrt[z^2] == -z /; Inequality[Pi/2, Less, Arg[z], LessEqual, Pi] \|\| Inequality[-Pi, Less, Arg[z], LessEqual, Pi/2] Standard Form Cell[BoxData[RowBox[List[RowBox[List[SqrtBox[SuperscriptBox["z", "2"]], "\[Equal]", RowBox[List["-", "z"]]]], "/;", RowBox[List[RowBox[List[FractionBox["\[Pi]", "2"], "<", RowBox[List["Arg", "[", "z", "]"]], "\[LessEqual]", "\[Pi]"]], "\[Or]", RowBox[List[RowBox[List["-", "\[Pi]"]], "<", RowBox[List["Arg", "[", "z", "]"]], "\[LessEqual]", FractionBox["\[Pi]", "2"]]]]]]]]] MathML Form z 2 - z /; π 2 < arg ( z ) π - π < arg ( z ) π 2 Condition z 2 1 2 -1 z Inequality 2 -1 z Inequality -1 z 2 -1 [/itex] Rule Form Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", SqrtBox[SuperscriptBox["z_", "2"]], "]"]], "\[RuleDelayed]", RowBox[List[RowBox[List["-", "z"]], "/;", RowBox[List[RowBox[List[FractionBox["\[Pi]", "2"], "<", RowBox[List["Arg", "[", "z", "]"]], "\[LessEqual]", "\[Pi]"]], "\|\|", RowBox[List[RowBox[List["-", "\[Pi]"]], "<", RowBox[List["Arg", "[", "z", "]"]], "\[LessEqual]", FractionBox["\[Pi]", "2"]]]]]]]]]]] Date Added to functions.wolfram.com (modification date) 2001-10-29	589	1,660	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.796875	3	CC-MAIN-2018-22	latest	en	0.250852
https://www.heyiamindians.com/is-tangential-speed-a-vector/	1,719,139,895,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198862466.81/warc/CC-MAIN-20240623100101-20240623130101-00179.warc.gz	708,670,372	9,920	## Is tangential speed a vector? From physics, we define a vector as a quantity having both magnitude and direction. For example, velocity is a vector where the magnitude is the speed. For tangential velocity, we are describing the motion along the edge of a circle and the direction at any given point on the circle is always along the tangent line. ## Is tangential velocity the same as tangential speed? Tangential velocity is the linear component of the speed of any object which is moving along a circular path. This is termed as tangential velocity. Also, we may say that the linear velocity is its tangential velocity at any instant. What’s another name for tangential speed? Linear speed is the distance travelled per unit of time, while tangential speed (or tangential velocity) is the linear speed of something moving along a circular path. A point on the outside edge of a merry-go-round or turntable travels a greater distance in one complete rotation than a point nearer the center. ### What is an example of tangential speed? Divide the circumference by the amount of time it takes to complete one rotation to find the tangential speed. For example, if it takes 12 seconds to complete one rotation, divide 18.84 by 12 to find the tangential velocity equals 1.57 feet per second. ### Is tangential speed a scalar? Thus “tangental speed” should be a scalar describing how fast the object is moving in the tangental direction, and “tangental vector” should be a vector which is in the tangental direction and has a magnitude equal to the tangental speed. Is tangential velocity constant? The tangential speed is constant, but the direction of the tangential velocity vector changes as the object rotates. Note: The direction of the centripital acceleration is always inwards along the radius vector of the circular motion. #### How are tangential speed and rotational speed related? Tangential speed and rotational speed are related. Tangential speed is directly proportional to the rotational speed and the radial distance from the axis of rotation. where v is tangential speed and w (pronounced oh MAY guh) is rotational speed. You move faster if the rate of rotation increases (bigger w). #### Do all points have the same tangential speed? Yes – if the wheel is circular all points on the wheel have the same tangential speed. Moving in a circular track means that the direction of motion is changing, thus the velocity of the vehicle changing – it has acceleration. What does tangential mean in physics? Tangential force is defined as the force acting on a moving body in the direction of the tangent to the curved path of the body. If the velocity of the object is positive, the acceleration will be negative. This is called a tangential force. ## What direction is her tangential velocity? Tangential velocity vector is always parallel to the radius of the circular path along which the object moves. Tangential velocity vector is always perpendicular to the radius of the circular path along which the object moves. ## Where is tangential speed the greatest? Hawaii has the greatest tangential speed of the 50 states. It is the state that is closest to the Equator, and therefore has the largest radius of revolution. What’s the difference between speed and tangential velocity? As far as I know, Speed is scalar quantity defined only with its magnitude while Velocity is a vector defined with both magnitude and direction. The direction of tangential velocity is always tangent to the circle so that it’s always changed while its magnitude is constant (in case of uniform circular motion) ### Is the speed of a velocity a scalar or a vector? Speed is a scalar. It is the magnitude of velocity, which is a vector whose direction must be specified. In there dimensional space, speed can be given as the length of a velocity 3-vector. Thus speed is scalar. ### Which is an example of a scalar quantity? SCALAR QUANTITIES or SCALARS – Those physical quantities which can be described by magnitude ( or numerical value) only and requires no direction for its specification are called scalars. for e. g., speed, distance, temperature, mass, electric current etc. “Speed” of course a Scalar Quantity indeed. Which is the magnitude of the null vector? The magnitude of the null vector is 0. Therefore, the average velocity had magnitude 0, but the average speed was 20 m/s, which does not match the magnitude of the average velocity. The magnitude of the instantaneous velocity vector equals the instantaneous speed.	932	4,556	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.34375	4	CC-MAIN-2024-26	latest	en	0.898761
https://www.nuclear-power.com/Nobelium-specific-heat-latent-heat-vaporization-fusion/	1,725,709,520,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00111.warc.gz	873,992,432	27,294	# Nobelium – Specific Heat, Latent Heat of Fusion, Latent Heat of Vaporization ## Nobelium – Specific Heat, Latent Heat of Fusion, Latent Heat of Vaporization Specific heat of Nobelium is — J/g K. Latent Heat of Fusion of Nobelium is — kJ/mol. Latent Heat of Vaporization of Nobelium is — kJ/mol. Specific Heat Specific heat, or specific heat capacity, is a property related to internal energy that is very important in thermodynamics. The intensive properties cv and cp are defined for pure, simple compressible substances as partial derivatives of the internal energy u(T, v) and enthalpy h(T, p), respectively: where the subscripts v and p denote the variables held fixed during differentiation. The properties cv and cp are referred to as specific heats(or heat capacities) because under certain special conditions, they relate the temperature change of a system to the amount of energy added by heat transfer. Their SI units are J/kg.K or J/mol K. Different substances are affected to different magnitudes by the addition of heat. When a given amount of heat is added to different substances, their temperatures increase by different amounts. Heat capacity is an extensive property of matter, meaning it is proportional to the size of the system. Heat capacity C has the unit of energy per degree or energy per kelvin. When expressing the same phenomenon as an intensive property, the heat capacity is divided by the amount of substance, mass, or volume. Thus the quantity is independent of the size or extent of the sample. Latent Heat of Vaporization In general, when a material changes phase from solid to liquid or from liquid to gas, a certain amount of energy is involved in this change of phase. In the case of liquid to gas phase change, this amount of energy is known as the enthalpy of vaporization (symbol ∆Hvap; unit: J), also known as the (latent) heat of vaporization or heat of evaporation. As an example, see the figure, which describes the phase transitions of water. Latent heat is the amount of heat added to or removed from a substance to produce a phase change. This energy breaks down the attractive intermolecular forces and must provide the energy necessary to expand the gas (the pΔV work). When latent heat is added, no temperature change occurs. The enthalpy of vaporization is a function of the pressure at which that transformation takes place. Latent Heat of Fusion In the case of solid to liquid phase change, the change in enthalpy required to change its state is known as the enthalpy of fusion (symbol ∆Hfus; unit: J), also known as the (latent) heat of fusion. Latent heat is the amount of heat added to or removed from a substance to produce a phase change. This energy breaks down the attractive intermolecular forces and must provide the energy necessary to expand the system (the pΔV work). The liquid phase has higher internal energy than the solid phase. This means energy must be supplied to a solid to melt it. Energy is released from a liquid when it freezes because the molecules in the liquid experience weaker intermolecular forces and have higher potential energy (a kind of bond-dissociation energy for intermolecular forces). The temperature at which the phase transition occurs is the melting point. When latent heat is added, no temperature change occurs. The enthalpy of fusion is a function of the pressure at which that transformation takes place. By convention, the pressure is assumed to be 1 atm (101.325 kPa) unless otherwise specified. ## Nobelium – Properties Element Nobelium Atomic Number 102 Symbol No Element Category Rare Earth Metal Phase at STP Synthetic Atomic Mass [amu] 259 Density at STP [g/cm3] Electron Configuration [Rn] 5f14 7s2 Possible Oxidation States +2,3 Electron Affinity [kJ/mol] Electronegativity [Pauling scale] 1.3 1st Ionization Energy [eV] 6.65 Year of Discovery 1958 Discoverer Albert Ghiorso, Glenn T. Seaborg, Torbørn Sikkeland, John R. Walton Thermal properties Melting Point [Celsius scale] 827 Boiling Point [Celsius scale] Thermal Conductivity [W/m K] Specific Heat [J/g K] Heat of Fusion [kJ/mol] Heat of Vaporization [kJ/mol]	987	4,146	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.578125	3	CC-MAIN-2024-38	latest	en	0.780614
https://www.kli.org/tlhIngan-Hol/2015/January/msg00036.html	1,680,322,649,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296949701.0/warc/CC-MAIN-20230401032604-20230401062604-00213.warc.gz	926,008,093	3,690	# Re: [Tlhingan-hol] mathematics terminology ### Bellerophon, modeler (bellerophon.modeler@gmail.com) ```<div dir="ltr"><div class="gmail_extra"><div class="gmail_quote">On Wed, Jan 7, 2015 at 4:20 PM, Brad Wilson <span dir="ltr"><<a href="mailto:bmacliam@aol.com"; target="_blank">bmacliam@aol.com</a>></span> wrote:<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><font color="black" face="arial"><div style="font-size:10pt;color:black"><div style="margin:0px;font-size:12px;color:rgb(0,0,0);background-color:rgb(255,255,255)"><pre style="font-size:9pt"><tt><font face="Tahoma, Verdana, Arial, sans-serif">As for {juvmeH mI'}, things are rarely "measured" in integers, so using this for real numbers seems logical. Rational vs irrational numbers would be subsets of the real numbers. I like your idea of using "precision of measurement" to differentiate these. I could see {mI' pup} used for rational numbers, ie. those that can be measured with precision.</font></tt></pre><pre style="font-size:9pt"><tt><font face="Tahoma, Verdana, Arial, sans-serif"></font></tt></pre></div></div></font></blockquote><div>I would expect {juvmeH mI'} to correspond to rational numbers, since the result of a measurement is always a rational number. I wonder what the words would be for precision, error, and tolerance. One word used in this context might be {DIch}, since the point is to make certain the measurement matches the actual quantity within a known margin of error. {Qagh} might not be the word for this type of error: why would being able to determine the distance to the target within 0.1 kellicams have any relationship to forgetting to seal the airlock?</div></div><div>~'eD</div>-- <br><div class="gmail_signature">My modeling blog: <a href="http://bellerophon-modeler.blogspot.com/"; target="_blank">http://bellerophon-modeler.blogspot.com/</a><br>My other modeling blog: <a href="http://bellerophon.blog.com/"; target="_blank">http://bellerophon.blog.com/</a><br></div> </div></div> ``` ```_______________________________________________ Tlhingan-hol mailing list Tlhingan-hol@kli.org http://mail.kli.org/mailman/listinfo/tlhingan-hol ```	629	2,250	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3	3	CC-MAIN-2023-14	latest	en	0.614439
micasamontessori.ie	1,725,796,575,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00474.warc.gz	377,186,883	11,105	Mortensen [ssba_hide] What is Mortensen Maths?. It is named for Jerry Mortensen, who designed the program after working for years in Montessori schools. The blocks are an adjunct to the method. This is a unique approach to teaching Mathematics that also employs manipulatives. Principles 1. All you need to know is how to count to 9 and to build a rectangle 2. One of the key principles is “to see all numbers as rectangles.” What does this mean? Traditionally when a speaker says “five” the listener has a mental image of the numeral 5. Yet, when a speaker says “Peter!” the listener sees Peters’s face in his mind’s eye. Similarly with other nouns we form a mental image of the object as it is mentioned by name. We do not have a mental image of the name, as in block capital letters, say. We don’t “see” A-P-P-L-L-E in our mind’s eye, we see what that object looks like.A rectangle has area and dimension, two of them, Mortensen calls them the “over” and “up.” Dimensions. 3. All we do in math is count is another key principle. What do we do with numbers? Even the youngest child knows that: we count with them. To count you first must know what one is. See one as a rectangle that is “one over” and “one up”. Now, all operations are just “building rectangles” and therefore all a child needs to know to do all the math there is, is to count to nine and to build a rectangle. As in all skill-learning, knowledge of results is important. When relying on memory to do math in the traditional way, there is no certainty, no way of knowing that you have done it right. With Mortensen Math there is a very good visual check. If the rectangle has no holes in it, and no bits sticking out, then it is correct. The learner can see that it is right. This builds confidence.	419	1,777	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.75	4	CC-MAIN-2024-38	latest	en	0.961169
https://www.flexiprep.com/NCERT-Exemplar-Solutions/Physics/Class-12/NCERT-Physics-Class-12-Exemplar-Ch-11-Dual-Nature-of-Radiation-and-Matter-Part-1.html	1,579,554,744,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579250599789.45/warc/CC-MAIN-20200120195035-20200120224035-00269.warc.gz	886,265,861	12,941	# NCERT Physics Class 12 Exemplar Ch 11 Dual Nature of Radiation and Matter Part 1 14 A proton and anα-particle are accelerated, using the same potential difference. How are the deBroglie wavelengths and related to each other? Ans: 15 (i) In the explanation of photo electric effect, we assume one photon of frequency ν collides with an electron and transfers its energy. This leads to the equation for the maximum energy of the emitted electron as Where is the work function of the metal. If an electron absorbs 2 photons (each of frequency ν) what will be the maximum energy for the emitted electron? (ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential? Ans: (ii) The probability of absorbing 2 photons by the same electron is very low. Hence such emissions will be negligible. 16 There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength. Ans: • In the first case energy given out is less than the energy supplied. • In the second case, the material has to supply the energy as the emitted photon has more energy. • This cannot happen for stable substances. 17 Do all the electrons that absorb a photon come out as photoelectrons? Ans: No, most electrons get scattered into the metal. Only a few come out of the surface of the metal. 18 There are two sources of light, each emitting with a power of W. One emits X-rays of wavelength 1nm and the other visible light at nm. Find the ratio of number of photons of X-rays to the photons of visible light of the given wavelength? Ans: • Total E is constant • Let and be the number of photons of X-rays and visible region Doorsteptutor material for NCO is prepared by worlds top subject experts- fully solved questions with step-by-step exaplanation- practice your way to success. Developed by:	419	1,983	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3	3	CC-MAIN-2020-05	longest	en	0.919028
http://www.enotes.com/homework-help/simplify-following-expressions-solve-equations-444714	1,386,361,465,000,000,000	text/html	crawl-data/CC-MAIN-2013-48/segments/1386163052469/warc/CC-MAIN-20131204131732-00055-ip-10-33-133-15.ec2.internal.warc.gz	315,032,281	7,286	Homework Help # Simplify the following expressions and solve the equations. Student Honors • Up • 1 • Down Simplify the following expressions and solve the equations. Images: This image has been Flagged as inappropriate Click to unflag Image (1 of 1) Posted by anya4one on July 24, 2013 at 6:17 PM via web and tagged with algebra, equations, help, homework, math College Teacher (Level 1) Educator Emeritus • Up • 1 • Down I will only do first part as it is instructed to avoid any multiple questions. ` 1/(x-2)+(2x)/((x-2)(x-8)) = x/(2(x-8))` `(x-8)/((x-2)(x-8))+(2x)/((x-2)(x-8)) = x/(2(x-8))` `(x-8)+2x = (x(x-2))/2` `2x-16+4x = x^2-2x` `x^2-8x+16 = 0` `(x-4)^2 = 0` `x = 4` So the answer is x = 4 Posted by jeew-m on July 24, 2013 at 6:32 PM (Answer #1)	295	777	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.796875	4	CC-MAIN-2013-48	latest	en	0.898508
https://fabianfuchsml.github.io/equivariance1of2/	1,721,471,713,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763515079.90/warc/CC-MAIN-20240720083242-20240720113242-00595.warc.gz	211,126,074	10,172	# CNNs and Equivariance - Part 1/2 CNNs are famously equivariant with respect to translation. This means that translating the input to a convolutional layer will result in translating the output. Arguably, this property played a pivotal role in the advent of deep learning, reducing the number of trainable parameters by orders of magnitude. So, given the importance of equivariance in deep learning, wouldn’t it be nice to design a convolutional layer which is equivariant not only with respect to translation but also, for instance, rotation and mirroring? Taco Cohen and Max Welling’s 2016 paper Group Equivariant Convolutional Networks pioneered a group theoretical approach to extending translation equivariance of CNNs to further symmetry classes. The paper has an important role in the literature as it arguably kicked off an entire stream of work. However, it is, as attested by many, not easy to understand, especially if you aren’t familiar with the relevant mathematical background. In this series of two blog posts, we aim to make this topic a bit more approachable. In this post, we start by introducing the concept of equivariance, from both a practical and a mathematical point of view. We explain why it plays such a crucial role in deep learning, how it is enforced in the case of translation equivariance, and how it can be enforced more generally. In the second part, we dive into Taco Cohen and Max Welling’s paper, which proposes a concrete approach for extending equivariance in CNNs to rotations and mirroring. ## Intro In machine learning, we’re often concerned with the flexibility of our models. We’d like to know that the model we choose is actually capable of the task that we want it to perform. For instance, the universal approximation theorem for neural networks gives us confidence that neural networks can approximate a very broad class of functions to any desired degree of accuracy1. There is a downside to complete flexibility. While we know that we can learn our target function, there’s also a whole universe of incorrect functions that look exactly the same on our training data. If we’re totally flexible, our model could learn any one of these functions, and once we move away from the training data, we might fail to generalise. For this reason, it’s often desirable to restrict our flexibility a little. If we can identify a smaller class of functions which still contains our target, and build an architecture which only learns functions in this class, we rule out many wrong answers while still allowing our model enough flexibility to learn the right answer. This might make a big difference to the necessary amount of training data or, given the same training data, make the difference between a highly successful model and one that performs very poorly. A famous success story for this reduction in flexibility is the CNN. The convolutional structure of the CNN encodes the translation symmetries of image data. We’ve given up total flexibility and can no longer learn functions which lack translation symmetry, but we don’t lose any useful flexibility, because we know that our target function does have translation symmetry. In return, we have a much smaller universe of functions to explore, and we’ve reduced the number of parameters that we need to train. Of course, images are not the only kind of data with useful structure, and translation symmetries are not the only kind of symmetry. The CNN is extremely successful on the right kind of data, but can we do something similar to encode other useful structure in other kinds of data? Cohen and Welling’s work gives us a general framework for doing just this. We’ve been waving our hands a lot in this introduction talking about “structure” and “symmetries”, and the first thing we’ll need to do to understand Cohen and Welling is to say precisely what we mean here. The crucial concept we need to make this precise is equivariance. ## What is equivariance? There’s more than one way, of course, of taking the fuzzy idea of “encoding structure in a model” and making it precise. Here’s one way of starting with the fuzzy intuition and getting gradually more precise – this particular train of thought is going to lead us to equivariance. 1. One aspect of the structure in our data is that it satisfies some symmetries. We should build a model which incorporates our knowledge of these symmetries. 2. Our model should preserve the symmetries of our data. That is, the outputs of the model should retain the symmetries of the inputs. 3. For any symmetry operation $\sigma$, applying $\sigma$ to our input and then passing it through the model should be the same as applying $\sigma$ to the output of the model. If we write $f$ for the function learned by our model, this final statement is pretty clearly something that we can straightforwardly write as an equation: $f\big(\sigma (x)\big) = \sigma \big(f(x)\big)$ If this equation holds, we say that $f$ is equivariant with respect to $\sigma$. If it holds for every $\sigma$ in some collection $S$ of symmetry operations, we say that $f$ is equivariant with respect to $S$. Making the above fully precise requires some mathematical machinery (specifically, group theory2), but for now, we’ll take this as our definition of equivariance. In particular, the full definition of equivariance doesn’t require the function $\sigma$ applied to the output to be identical to the one applied to the input. However, to keep things from getting too complex, it is helpful to assume that they are identical. Moreover, we will later see that Group Equivariant Convolutional Networks fulfil this stricter definition of equivariance3. Our definition of equivariance looks a little like part of the definition of linearity. With linearity, we can do scalar multiplication either inside or outside the function. With equivariance, we can do symmetry operations either inside or outside the function. This property of linear functions can actually be stated as “linear functions are equivariant with respect to scalar multiplication”. So even though equivariance may be an unfamiliar concept, you’ve already met it in disguise if you’re familiar with linear functions! To get a more concrete grasp on equivariance, let’s think about CNNs again. The following video shows the equivariance of CNNs with respect to translation. A shift to the input image directly corresponds to a shift of the output features. (The full video is available here). It is also helpful to relate the concept of equivariance to the perhaps more familiar concept of invariance. $f$ is invariant if its output doesn’t change upon applying $\sigma$ to the input. Hence, the above equation reduces to: $f\big(\sigma(x)\big) = f(x)$ If a layer is an equivariant embedding of the input with respect to a certain symmetry, it can always be turned into an invariant embedding in a later layer. Whether this can be done in a straightforward and meaningful manner depends on how the equivariance is implemented. Famous examples where this works are networks where multiple convolutional layers are followed by a global average pooling layer (GAP). In this case, everything until the GAP layer is translation equivariant, but the output of the GAP layer (and of the entire network) is invariant with respect to translations of the input. We will later see that this principle can also be applied to group equivariant convolutional layers. If we think about convolutional layers in a neural network trained for object classification, it becomes clear why this is a very useful property. We typically want the final output to be invariant with respect to translations (within reason – obviously shifting the entire object out of the frame should change the output). However, the first few layers should be equivariant, not invariant. Let’s look at the following image: Assume that after a few layers, the perceptive field and the complexity of the operations are both large enough to detect features like noses and eyes but not yet large enough to detect faces. Dropping spatial information at this point (e.g. becoming translation invariant) would make it impossible for the network to see that both eyes of the person are located on the same cheek: Picasso’s surrealistic painting would look like a regular face to the network (credit for this analogy goes to Daniel Worrall). Hence there is an interesting interplay between perceptive field size, network depth, invariance and equivariance. ## How CNNs enforce equivariance We’ve stated above that CNNs are equivariant with respect to translation, but now let’s dig into why this is so. It’s worth being a little careful here – the familiar intuitive picture of how CNNs work is most easily framed in terms of motion, but of course nothing here is really in motion. In reality there are many stationary copies of a convolutional filter, each of which is centred over a different pixel. The intuition of motion is very helpful though, so we’ll keep talking in those terms. Let’s think of a single filter in a convolutional layer. We’ll pass an image through this, and get an image-shaped map of activations. Recalling briefly how a convolutional layer works, we’re going to slide our filter over the input image. The activation at each point in the output is computed when the convolutional filter sits directly above that point in the input image. Now let’s change our perspective a little. Usually, we think of holding the image fixed while the convolutional filter moves above it. But what if we imagine holding the filter fixed, and moving the image? The two perspectives are visualised in the video below: (Pixel art from here.) On the left hand side we have a fixed image and a moving filter. On the right hand side we have a fixed filter and a moving image. Imagining that these are physical objects in motion, we’re looking at the same physical system from two different viewpoints – one viewpoint follows the image, and one follows the filter. This is actually how we created the video – it’s one scene rendered from two different cameras. But remember that motion isn’t really the right way to think about what’s happening here. In reality, our old perspective had a single image and many translated copies of the filter, and our new perspective has a single filter and many translated copies of the image. We can think about the behaviour of the CNN from either point of view. From the single-filter perspective, here’s how the convolutional layer operates: 1. We have a function on images (convolution with our filter). 2. We feed many translated copies of our input image to this function. 3. For each translation, we store the computed function value. Now, what happens if we have a new input image, which is just a translated version of our old one? Do we need to recompute anything? Of course not! We’ve already looked at every translated version of our old input image in step 2. The results of the required computations are already there in our output – we just need to look them up. The only problem is that they’re not going to live in exactly the right place. Instead, they’re translated over by the same amount as our new image is. In other words, if we translate the input, we just translate the output by the same amount. This is exactly translation equivariance! We’ll need to fill in some details here. It’s clear that the original output really does contain all the information we need to construct our new output. We might already have an intuition that the way to recover this information is with a translation in the output space, and it turns out that this intuition is correct, but we’ll need to do some maths to prove it. We’ll come back to this in part 2. ## Enforcing equivariance more generally Now suppose that we have some other group of symmetry transformations, and we want to be equivariant with respect to these. It’s just a case of saying “transform” instead of “translate” in our 3-step process from above: 1. We have a function on images (convolution with our filter). 2. We feed many transformed copies of our input image to this function. 3. For each transformation, we store the computed function value. Once again, it’s clear that if we see a transformed version of an old image, we don’t need to do any new computation. We’ve already seen the transformed version in step 2, and we just need to look up the relevant outputs. How? By applying the same transformation in the output space! And just like that, we can get equivariance with respect to any group of transformations we like. ## Conclusion This was an introduction to the concept of equivariance. If you don’t want to go deeper into the maths and just wanted to get a broad sense of what this research area is about, feel free to stop here. Otherwise, stick around to read the second part of this two-part series, where we explain how the paradigm from the last section can be put into practice. It should be ready next week. It is ready! 🙂 Credit: Ed Wagstaff is funded by the EPSRC Centre for Doctoral Training in Autonomous Intelligent Machines and Systems. I am funded by the Bosch Center for AI where I am currently undertaking a research sabbatical. A big thank you also to Daniel Worrall for a number of conversations on this topic, and Adam Golinski and Mike Osborne for proofreading and feedback. 1. In theory, at least! In practice the degree of accuracy may be limited by the finite amount of computation we can do. 2. Even more specifically, the main group-theoretic concept at play here is that of an action. We can (and will!) safely ignore the distinction between groups and actions for the purpose of these blog posts, but if you want to dig deeper into the maths in the paper, a basic understanding of group actions would be helpful. 3. There’s a caveat to this: the strict definition does hold for all the intermediate layers, but the first layer of the network looks a bit different, and only has the less strict property.	2,883	14,045	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2024-30	latest	en	0.927237
http://www.gurufocus.com/term/NCAV/CR/Net%2BCurrent%2BAsset%2BValue/Crane%2BCompany	1,480,789,837,000,000,000	text/html	crawl-data/CC-MAIN-2016-50/segments/1480698541066.79/warc/CC-MAIN-20161202170901-00473-ip-10-31-129-80.ec2.internal.warc.gz	484,022,839	30,831	Switch to: Crane Co (NYSE:CR) Net-Net Working Capital Per Share \$-19.77 (As of Sep. 2016) In calculating the Net-Net Working Capital (NNWC), Benjamin Graham assumed that a companys accounts receivable is only worth 75% its value, its inventory is only worth 50% of its value, but its liabilities have to be paid in full. In addition, Graham believed that preferred stock belongs on the liability side of the balance sheet, not as part of capital and surplus. This is a conservative way of estimating the companys value. Crane Co's net-net working capital per share for the quarter that ended in Sep. 2016 was \$-19.77. Definition Crane Co's Net-Net Working Capital (NNWC) per share for the fiscal year that ended in Dec. 2015 is calculated as Net-Net Working Capital Per Share (A: Dec. 2015 ) = (Cash And Cash Equivalents + 0.75 * Acct. Receivable + 0.5 * Inventory - Total Liabilities - Preferred Stock) / Shares Outstanding = (363.5 + 0.75 * 397.6 + 0.5 * 376.9 - 2197.5 - 0) / 58.11 = -23.19 Crane Co's Net-Net Working Capital (NNWC) per share for the quarter that ended in Sep. 2016 is calculated as Net-Net Working Capital Per Share (Q: Sep. 2016 ) = (Cash And Cash Equivalents + 0.75 * Acct. Receivable + 0.5 * Inventory - Total Liabilities - Preferred Stock) / Shares Outstanding = (436.3 + 0.75 * 426.7 + 0.5 * 372.3 - 2097.1 - 0) / 58.40 = -19.77 * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. In calculating the Net-Net Working Capital (NNWC), Benjamin Graham assumed that a companys accounts receivable is only worth 75% its value, its inventory is only worth 50% of its value, but its liabilities have to be paid in full. In addition, Graham believed that preferred stock belongs on the liability side of the balance sheet, not as part of capital and surplus. In "Security Analysis", preferred stock is dubbed "an imperfect creditorship position" that is best placed on the balance sheet alongside funded debt. This is a conservative way of estimating the companys value. Explanation One research study, covering the years 1970 through 1983 showed that portfolios picked at the beginning of each year, and held for one year, returned 29.4 percent, on average, over the 13-year period, compared to 11.5 percent for the S&P 500 Index. Other studies of Grahams strategy produced similar results. Benjamin Graham looked for companies whose market values were less than two-thirds of their net-net value. They are collected under our Net-Net screener. GuruFocus also publishes a monthly Net-Net newsletter. Related Terms Historical Data * All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency. Crane Co Annual Data Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 NNWC -16.10 -21.38 -23.05 -18.80 -18.27 -23.31 -19.63 -27.02 -26.70 -23.19 Crane Co Quarterly Data Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 NNWC -24.64 -24.67 -26.70 -26.95 -25.90 -24.89 -23.19 -22.37 -21.31 -19.77 Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names \| Earn affiliate commissions by embedding GuruFocus Charts GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More)	890	3,292	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.671875	3	CC-MAIN-2016-50	longest	en	0.895962
http://www.slideshare.net/xforbidden20/informe-analisis-de-algoritmos-mitad-de-cuadrado	1,462,477,021,000,000,000	text/html	crawl-data/CC-MAIN-2016-18/segments/1461860127878.90/warc/CC-MAIN-20160428161527-00001-ip-10-239-7-51.ec2.internal.warc.gz	799,164,192	37,050	Upcoming SlideShare × 653 views 526 views Published on Informe Análisis de algoritos Published in: Education 0 Likes Statistics Notes • Full Name Comment goes here. Are you sure you want to Yes No • Be the first to comment • Be the first to like this Views Total views 653 On SlideShare 0 From Embeds 0 Number of Embeds 2 Actions Shares 0 10 0 Likes 0 Embeds 0 No embeds No notes for slide 1. 1. INGENIERIA EN INFORMATICA Análisis de algoritmos Mitad del cuadrado Asignatura: Análisis de algoritmos Integrantes: Eduardo Leiva Jonathan García Sergio Ormeño Docente: Pilar Pardo H. Fecha: 16/04/2014 2. 2. 1 Contenido Introducción................................................................................................................................. 2 Algoritmos de Búsqueda ........................................................................................................... 3 Búsqueda Binaria:...................................................................................................................... 3 Búsqueda Lineal:........................................................................................................................ 3 Hashing........................................................................................................................................ 3 Truncamiento: ............................................................................................................................. 4 Plegamiento: ............................................................................................................................... 5 Aritmética Modular: .................................................................................................................... 5 Mitad del Cuadrado:................................................................................................................... 6 Caso Promedio ........................................................................................................................... 7 Peor Caso.................................................................................................................................... 7 Tratamiento de colisiones: ........................................................................................................ 8 Sondeo lineal............................................................................................................................... 8 Doble hashing ............................................................................................................................. 9 Encadenamiento de sinónimos ................................................................................................ 9 Direccionamiento por cubetas................................................................................................ 10 Cuadro comparativo de los métodos de búsqueda de un algoritmo................................ 10 Ventajas y Desventajas del Hashing..................................................................................... 11 Comportamiento de un Algoritmo .......................................................................................... 11 Ventajas y desventajas de los algoritmos ............................................................................ 11 Conclusión................................................................................................................................. 13	436	3,452	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.890625	3	CC-MAIN-2016-18	latest	en	0.18434
https://www.teacherspayteachers.com/Product/Fraction-Snack-Shop-389308	1,544,975,689,000,000,000	text/html	crawl-data/CC-MAIN-2018-51/segments/1544376827769.75/warc/CC-MAIN-20181216143418-20181216165418-00335.warc.gz	1,065,279,534	17,783	# Fraction Snack Shop Subject Resource Type Product Rating File Type PDF (Acrobat) Document File 575 KB\|13 pages Share Product Description The focus of this fraction lesson is the differences between one half, one third, one fourth as equal shares. Materials Included: * Teacher Directions * Rectangle Fraction Shapes Circle Fraction Shapes Dividing Doughnuts assignment * Dividing Candy Bars assignment * Fraction Practice assignment * Show What You Know assessment Standard 1: Make sense of problems and persevere in solving them. Standard 2: Reason abstractly and quantitatively. Standard 3: Construct viable arguments and critique reasoning of others. Standard 4: Model with mathematics. Standard 5: Use appropriate tools strategically. Standard 6: Attend to precision. CCSS: 2.G.3 Partition circles and rectangles into two, three, or four equal shares, describe the shares using the words, halves, thirds, half of, a third of, etc., and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of identical wholes need not have the same shape. Realistic-teacher.blogspot.com Total Pages 13 pages N/A Teaching Duration 90 minutes Report this Resource \$3.50	270	1,205	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.078125	3	CC-MAIN-2018-51	latest	en	0.844282
https://wannagofast.com/gbual/duration-and-convexity-formula-1f2f2b	1,621,324,911,000,000,000	text/html	crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00346.warc.gz	618,267,412	8,920	Woolacombe Bay Camping, Canyon Independent School District, How To Connect Grand Videoke To Phone, Avian Control Bird Repellent 1 Gallon, Sydney Rainfall 2019, Sea Cliff Isle Of Man, Spider-man Edge Of Time Ps4, Family Guy Brian Returns Episode, Crash Team Racing Nitro-fueled Gameplay, Ar-15 Trigger Spring Kit, Viper Vx Plate Carrier, Anglesey Weather Bbc, " /> Woolacombe Bay Camping, Canyon Independent School District, How To Connect Grand Videoke To Phone, Avian Control Bird Repellent 1 Gallon, Sydney Rainfall 2019, Sea Cliff Isle Of Man, Spider-man Edge Of Time Ps4, Family Guy Brian Returns Episode, Crash Team Racing Nitro-fueled Gameplay, Ar-15 Trigger Spring Kit, Viper Vx Plate Carrier, Anglesey Weather Bbc, " /> # duration and convexity formula Duration & Convexity Calculation Example: Working with Convexity and Sensitivity Interest Rate Risk: Convexity Duration, Convexity and Asset Liability Management – Calculation reference For a more advanced understanding of Duration & Convexity, please review the Asset Liability Management – The ALM Crash course and survival guide . It is calculated as Macaulay Duration divided by 1 + yield to maturity. By including convexity in our price change formula. Its Macaulay duration is 2 and its modified duration is 1.8110 (= 2/1.10433927). Also known as the Modified Duration. Effective Duration Formula = (51 – 48) / (2 * 50 * 0.0005) = 60 Years Example #2 Suppose a bond, which is valued at $100 now, will be priced at 102 when the index curve is lowered by 50 bps and at 97 when the index curve goes up by 50 bps. The convexity adjustment is the annual convexity statistic, AnnConvexity, times one-half, multiplied by the change in the yield-to-maturity squared. A:Pays$610 at the end of year 1 and $1,000 at the end of year 3 B:Pays$450 at the end of year 1, $600 at the end of year 2 and$500 at the end of year 3. Duration & Convexity: The Price/Yield Relationship Investors who own fixed income securities should be aware of the relationship between interest rates and a bond’s price. To get the curve duration and convexity, first shift the underlying yield curve, … Duration and Convexity 443 That duration is a measure of interest rate risk is demonstrated as fol-lows. Most textbooks give the following formula using modified duration to approximate the change in the present value of a cash flow series due to a change in interest rate: The column "(PV(t^2+t))" is used for calculating the Convexity of the Bond. Step 3: Next, determine the yield to maturity of the bond based on the ongoing market rate for bonds with similar risk profiles. Chapter 11 - Duration, Convexity and Immunization Section 11.2 - Duration Consider two opportunities for an investment of \$1,000. The formula for calculating bond convexity is shown below. We can get a better approximation of the new price as follows: Price Change = (- Duration x Price Yield) + (0.5 x Convexity x (Yield Change)^2)) Using our previous example, if the 8% 10-year note has a 0.60 convexity, the new estimated price change is calculated as follows: The formula for convexity can be computed by using the following steps: Step 1: Firstly, determine the price of the bond which is denoted by P. Step 2: Next, determine the frequency of the coupon payment or the number of payments made during a year. Explanation. more accurate than the usual second-order approximation using modified duration and convexity. Those are the yield duration and convexity statistics. These Macaulay approximations are found in formulas (4.2) and (6.2) below. This amount adds to the linear estimate provided by the duration alone, which brings the adjusted estimate … Of course, there are formulas that you can type in (see below), but they aren’t easy for most people to remember and are tedious to enter. On the other hand, using our formula above gives: $$\Delta D \approx (7.52^2 - 72.17)(0.25/100) = -0.04$$ share \| improve this answer \| follow \| edited Nov 20 at 20:45 Its convexity is 4.9198 [= (2*3)/(1.10433927)2]. As a general rule, the price of a bond moves inversely to changes in interest rates: a bond’s price will increase as rates decline and will decrease as rates move up. Convexity - The degree to which the duration changes when the yield to maturity changes. Duration and convexity are important numbers in bond portfolio management, but it is far from obvious how to calculate them on the HP 12C. It's the reason why bond price changes do not exactly match changes in interest rates times duration. Both have a yield rate of i = :25because (1:25) 1 = :8,	1,107	4,587	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.171875	3	CC-MAIN-2021-21	longest	en	0.86165
https://www.fxsolver.com/browse/?like=2113&p=31	1,653,782,134,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00052.warc.gz	885,067,231	44,081	' Search results Found 952 matches Thermal energy of an ideal gas Thermal energy is a term sometimes used to refer to the internal energy present in a system in a state of thermodynamic equilibrium by virtue of its ... more Degree of Combined Leverage In finance, leverage is a general term for any technique to multiply gains and losses.Most often it involves buying more of an asset by using borrowed ... more Logarithm of the odds ratio The logit function is the inverse of the sigmoidal “logistic” function or logistic transform used in mathematics, especially in statistics. ... more Darcy friction factor - Goudarâ€“Sonnad equation (Variable q) In fluid dynamics, the Darcy friction factor formulae are equations that allow the calculation of the Darcy friction factor, a dimensionless quantity used ... more Darcy friction factor - Serghides's solution (Variable A) In fluid dynamics, the Darcy friction factor formulae are equations that allow the calculation of the Darcy friction factor, a dimensionless quantity used ... more Darcy friction factor - Goudarâ€“Sonnad equation (Variable d) In fluid dynamics, the Darcy friction factor formulae are equations that allow the calculation of the Darcy friction factor, a dimensionless quantity used ... more Dirichlet kernel (In mathematical analysis) The Dirichlet kernel is a collection of functions and its importance comes from its relation to Fourier series, that decomposes any periodic function or ... more Probability density function of a log-normal distribution In probability theory, a probability distribution assigns a probability to each measurable subset of the possible outcomes of a random experiment, survey, ... more In classical geometry, a radius of a circle or sphere is any of the line segments from its center to its perimeter, and in more modern usage, it is also ... more Drift velocity in a current-carrying metallic ohmic conductor The drift velocity is the average velocity that a particle, such as an electron, attains due to an electric field. In terms of the basic ... more ...can't find what you're looking for? Create a new formula Search criteria: Similar to formula Category	454	2,190	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.953125	3	CC-MAIN-2022-21	latest	en	0.890186
http://www.nelson.com/mathfocus/grade2/teacher/surf/ch3_2.html	1,553,619,721,000,000,000	text/html	crawl-data/CC-MAIN-2019-13/segments/1552912205597.84/warc/CC-MAIN-20190326160044-20190326182044-00070.warc.gz	317,574,646	3,191	Math Focus 2 Teacher Centre • Professional Resources • Surf for More Math # Surf for More Math ## CHAPTER 3: ADDITION AND SUBTRACTION STRATEGIES ### Lesson 2: Equal or Not Equal? Use these interactive activities to encourage students to have fun on the Web while learning about strategies for addition and subtraction. Students can try these activities on their own or in pairs. Goal Demonstrate, explain, and record equalities and inequalities. [C, CN, PS, R, V] Instructions for Use To use Balance Addition Equations, enter the number in the space provided that will make the addition equation balanced. Click “Submit” to check your answer. When you have completed a certain number of questions, click “Submit & finish” to view a summary of your score. Balance Subtraction Equations has students balancing subtraction equations. To use Balance Subtraction Equations, enter the number in the space provided that will make the subtraction equation balanced. Click “Submit” to check your answer. When you have completed a certain number of questions, click “Submit & finish” to view a summary of your score. Definition: Does Not Equal gives students the definition of does not equal and the symbol used. The inequality symbols are also presented. .	258	1,260	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.703125	3	CC-MAIN-2019-13	latest	en	0.900647
https://www.jiskha.com/display.cgi?id=1335132187	1,500,693,519,000,000,000	text/html	crawl-data/CC-MAIN-2017-30/segments/1500549423842.79/warc/CC-MAIN-20170722022441-20170722042441-00251.warc.gz	775,852,607	3,736	# physics posted by . An athlete executing a long jump leaves the ground at a 28.0 angle and lands 7.60m away. part a)What was the takeoff speed? part b) If this speed were increased by just 8.0 % , how much longer would the jump be? i got for part a)9.48 m/s but i need help with part b • physics - L = v^2•sin2α/g , v = sqrt(L•g/sin2α) = 9.48 m/s. New velocity V =1.08•v = 1.08•9.48= 10.24 m/sþ New range is L1 = V^2•sin2α/g = 8.87 m. ΔL = 8.87 – 7.6 = 1.27 m	186	467	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2017-30	longest	en	0.94551
https://www.jiskha.com/similar?question=1%29Find+the+sixth+term+of+the+geometric+sequence+for+which+a1%3D5+and+r%3D3+A%291215+B%293645+C%299375+D%2923+I+chose+A+2%29Write+an+equation+for+the+nth+term+of+the+geometric+sequence+-12%2C4%2C-4%2F3...+A%29aN%3D-12%281%2F3%29n-1+B%29aN%3D12%28-1%2F3%29n-1+C%29aN%3D-12%28-1%2F3%29-n%2B1+D%29aN%3D-12%28-1%2F3%29n-1+I+chose+C+3%29Find&page=648	1,571,231,363,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570986668569.22/warc/CC-MAIN-20191016113040-20191016140540-00173.warc.gz	954,240,741	28,364	# 1)Find the sixth term of the geometric sequence for which a1=5 and r=3 A)1215 B)3645 C)9375 D)23 I chose A 2)Write an equation for the nth term of the geometric sequence -12,4,-4/3... A)aN=-12(1/3)n-1 B)aN=12(-1/3)n-1 C)aN=-12(-1/3)-n+1 D)aN=-12(-1/3)n-1 I chose C 3)Find 93,362 questions, page 648 1. ## calculus helpppp! a 20m ladder rests vertically against the side of a barn. a pig that has been hitched to the ladder starts to pull the base of the ladder away from the wall at a constant rate of 40 cm per second. find the rate of change of the height of the top of the asked by chris on December 8, 2008 2. ## physics A plane progressive wave on a water surface is given by the equation y=2sin2pi(100t-x/30) where x-is the distance covered in a time t, x, y and t are cm and second respectively. Find (i) the wavelength and frequency of the wave motion. (ii) the phase asked by Edwin on March 23, 2010 3. ## Math Joel has \$20. He needs a calculator that costs \$12 and some notebooks that cost \$2 each. 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Find a formula for d, your distance from Pittsburgh, as a function of t, the number of hours of asked by Lucy Ripply on October 5, 2013 35. ## chem h/w i did a lab and i need to find out the melting point for lauric acid and lauric acid with another substance. the teacher told us to figure it out using excel. this is my graph. tinypic. com/view.php?pic=1605yev&s=7 can someone help me in figuring out the asked by RiChy on September 18, 2010 36. ## physics Hi! could anyone help me to solve this problem please. Three point charges are located at the corners of an equilateral triangle as in the figure below. Find the magnitude and direction of the net electric force on the 1.60 µC charge. (A = 1.60 µC, B = asked by Chan on May 15, 2010 37. ## Statistics A recent survey found that 70% of all adults over 50 wear glasses for driving. You randomly select 30 adults over 50, and ask if he or she wears glasses. Decide whether you can use the normal distribution to approximate the binomial distribution. If so, asked by Clifford on February 16, 2013 38. ## Physics A football player runs the pattern given in the drawing by the three displacement vectors A, B, and C. The magnitudes of these vectors are A = 5 m, B = 13.0 m, and C = 16.0 m. Using the component method, find the magnitude and direction è of the resultant asked by yoyoma on August 18, 2010 39. ## Middle Chinese Which has become an art form in the Mandarin speaking world? A.Drinking tea B.Making congee C.Making sticky rice in a bamboo or reed leaf. I think it is drinking tea. I can't find it in my lesson, I searched this site, I even goggled it, nothing. asked by Hi!!! on December 20, 2017 40. ## Physics a batman deflects a boy by an angle 90 without change in acc. speed of 54km/h if mass of ball is 0.15kg. what is the impulse exerted on the ball also find force exerted on ball if ball remain for 0.1sec In connected with wall? asked by Mani on September 7, 2016 41. ## Chemistry Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 4.0010^2 mL of solution and then titrate the solution with 0.138 M NaOH. C6H5CO2H(aq) + OH-(aq) C6H5CO2-(aq) + H2O(l) Find concentration of Na+ at equivalence asked by john on April 21, 2013 42. ## Chemistry Hello, we recently did a lab in our chemistry class to analyze an Unknown amount Calcium Carbonate which was titrated with EDTA. Our titration results were: mass of the unknown CaCO3 was 0.172 g and the mL of EDTA added was 30.7. How do I find the percent asked by Anna on July 1, 2010 43. ## Molecular Formula How do you work out the molecular formula? Generally you convert grams to mols, then find the ratio of the atoms to reach other. But this is all generality. Why don't you post the entire problem, tell us what you know how to do and what you don't asked by L0bster_quadrille on February 3, 2007 44. ## Physics In a cathode-ray tube, electrons are projected horizontally at 6.4 x 106 m/s and travel a horizontal distance of 4.7 cm across the tube. Find the vertical distance the electrons fall under the influence of gravity during their flight. Give your answer in asked by Angelina on February 6, 2019 45. ## Math The amount of an investment of P dollars for t years at simple interest rate r is given by A= P+Prt a) Rewrite this formula by factoring out the greatest common factor on the right-hand side. b) Find A if \$8300 is invested for 3 years at a simple interest asked by Roger on March 4, 2009 46. ## math In 1 through 6, find each sum. Simplify, if possible. Estimate for reasonableness. 1. 7 __2 3 + 8 __5 6 2. 4 __3 4 + 2 __2 5 3. 11 ___9 10 + 3 ___1 20 4. 7 __6 7 + 5 __2 7 5. 5 __8 9 + 3 __1 2 6. 21 ___ 11 12 + 17 __2 3 7. Number Sense Write two mixed asked by lizzy on December 20, 2011 47. ## math a loan for \$2000 with a simple interest rate of 15% was made on September 29th and was due November 30th I have to find the ordinary interest rate and don't know how I know there are 62 days between those two dates with just two months and one day I don't asked by a on May 20, 2015 48. ## Math: Algebra II You start 50 miles east of Pittsburgh and drive east at a constant speed of 65 miles per hour. (Assume that the road is straight and permits you to do this.) Find a formula for d, your distance from Pittsburgh, as a function of t, the number of hours of asked by Lucy Ripply on October 5, 2013 49. ## Mathematics The price for an adult carnival ticket is 6.5 more than a carnival ticket for a child. Bob takes his son to the carnival. He buys cotton candy for \$10.25, and spends \$55. Write and solve a linear equation to find the prices for each of their carnival asked by Austin on January 23, 2018 50. ## Math 7th Mrs. Jones bought 3 1/2 pounds of ajita meat at \$2.50 per pound, 1 3/4 pounds of pinto beans and a package of tortillas for \$2.23. She gave the cashier 20.00. What other information is needed to find the correct change Mrs.Jones will receive? asked by Danny on December 7, 2010 51. ## Pre Cal---Fix to the question below Let P be the point on the unit circle U that corresponds to t. Find the coordinates of P and the exact values of the trigonometric functions of t, whenever possible. (If there is no solution, enter NO SOLUTION.) 4pi = ( , ) sin(4pi) = cos(4pi) = tan(pi) = asked by Karmen on February 9, 2010 52. ## History Why did the United States become interested in central and South American countries such as Cuba and Mexico during the gilded age?was the gilded age a Period of continuity or change in American foreign policy? * need help even a website would help to find asked by Lula on February 27, 2019 53. ## Calculus Find the limits: lim {[1/(2+x)]-(1/2)}/x as x->0 lim (e^-x)cos x as x->infinity ...First off, I'm having trouble getting rid of complex fractions. I tried multiplying the whole numerator by [(2+x)/2], but you'd still get 0/undefined answer (the answer in asked by Momo on February 14, 2010 54. ## Physics A racecar driver in an 825-kg car accelerates from rest to a velocity of 21.5 m/s in 0.55 s. What is the net force acting on the car? I found out using A=v/t that the horizontal acceleration is 39.1 m/s^2. My problem now is this... Is the net force just asked by Hannah on June 20, 2009 55. ## Chemistry The equilibrium constant for the reaction 2NO(g) + O2(g) 2NO2(g) is Kp=1.48x10^4 at 184C. Calculate Kp for 2NO2(g) 2NO(g) + O2(g). I know the equation is Kp=Kc(RT)x Delta n Kp is what we're solving for Kc= I can't seem to find. R=.0821 T=184+273 Delta n= 1 asked by Anonymous on March 9, 2010 56. ## physics a 70 kg box is pulled by a rope that exerts a 400 N force at angle of 30 degrees to the horizontal the leading coefficient of kinetic friction id 0.50. draw a diagram. draw a free body diagram . find the acceleration of the box. asked by tionne on November 13, 2012 57. ## Physics,chemistry,maths On turning a corner a car drIver driving at 36 km per hour find a child on the road 55 metre ahead.he immediately applies brakes so as stop within 5 metre. Calculate the retardation produced and the time taken by the car to stop. asked by Anamika roy on August 3, 2017 58. ## Calculus In the diagram below, the y-coordinate of point P is increasing at a rate of 4 units per minute while point Q remains in its fixed position. Find the rate of change of angle, in radians per minute, when y = 20. y ^ \| P (0, y) \| \| \| \| \|---\| \|___\|________ Q asked by Katie on February 6, 2011 59. ## math A company has sales (measured in millions of dollars) of 50, 60, and 75 during the first three consecutive years. Find a quadratic function that fits these data, and use the result to predict the sales during the fourth year. Assume that the quadratic asked by bryant on May 31, 2010 60. ## physics A satellite is in a circular earth orbit that has a radius of 6.63E+6 m. A model airplane is flying on a 16.4 m guideline in a horizontal circle. The guideline is nearly parallel to the ground. Find the speed of the plane such that the plane and the asked by Anonymous on February 12, 2012 61. ## Physics II Please can anyone explain. Thank you. A rocket is launched vertically from ground with a constant acceleration. If the rocket emits a burst of sound every five seconds after launching, find the difference of intensity levels observed at the launching site asked by ANA on January 31, 2015 62. ## Physics A pitcher throws horizontally a fast ball at 156 km/h toward the home plate, which is 18.2 m away. Neglecting air resistance (not a good idea if you are the batter), find how far the ball drops because of gravity by the time it reaches the home plate. asked by Neal parikh on June 4, 2016 63. ## Physics 2 The wire in the figure below carries a current of 12 A. Suppose that a second long, straight wire is placed right next to this wire. The current in the second wire is 30 A. Use Ampère's law to find the magnitude of the magnetic field at a distance of r = asked by Anne on February 13, 2012 64. ## math a rectangular prism has a width of 92ft. and a volume of 240ft^3. Find the volume of a similar prism with a width of 23ft. Round to the nearest tenth if necessary. a)3.8ft^3 b)60ft.^3 c)15ft.^3 d)10.4ft.^3 I am totally cinfused by this. Can somebody please asked by Shannon on August 30, 2008 65. ## Ag Science I have a HUGE project due for ag. It's on California and Hawaii. There are questions like "how many of miles of hiking trail is in the state?" and "how many big game species are in the state?" I am desperately searching all over google for answers to these asked by Taylor on March 2, 2008 66. ## Chemistry When the power supplying a 30.0 gallon water heater was turned off, the temperature dropped from 75.0 degrees celsius to 22.5 degrees. How much heat was lost to the surrounding? I know I need to find the energy and or heat lost, but how will i get the mass asked by Un-Known on April 24, 2016 67. ## physic/math What happens when ocean waves with (a) wavelength=10m and (b) wavelength=200m encounters a small island of width 100m. Discuss the amount of shelter you will find in each case by taking your small boat to the lee side of the island. asked by AAAA on February 9, 2012 68. ## probability A company ships computer components in boxes that contain 20 items. Assume that the probability of a defective computer component is 0.2. Find the probability that the first defect is found in the seventh component tested. Round your answer to four decimal asked by GIBSON on November 27, 2013 69. ## physics A 380 turn solenoid of length 32.0 cm and radius 3.10 cm carries a current of 4.80 A. Find the following. (a) the magnetic field strength inside the coil at its midpoint mT (b) the magnetic flux through a circular cross-sectional area of the solenoid at asked by christiam on March 9, 2010 70. ## Physics The wire in the drawing carries a current of 13 A. Suppose that a second long, straight wire is placed right next to this wire. The current in the second wire is 28 A. Use Ampere's law to find the magnitude of the magnetic field at a distance of r = 0.76 m asked by Anonymous on February 25, 2013 71. ## Algebra 2 Use technology or a z-distribution table to find the indicated area. The weights of grapefruits in a bin are normally distributed with a mean of 261 grams and a standard deviation of 9.4 grams. Approximately 20% of the grapefruits weigh less than which asked by student on March 20, 2019 72. ## statistical reasoning One hundred people are selected at random and tested for colorblindness to determine whether gender and colorblindness are independent. The following counts were observed. Colorblind Not Colorblind Total Male 8 52 60 Female 2 38 40 Total 10 90 100 Find the asked by Anonymous on April 14, 2017 73. ## science A 3.5-cm tall object is placed 10.0 cm in front of a concave mirror having a focal length of 14.5 cm. Find the location position and height by drawing a ray diagram to scale. Verify your answer using the mirror and magnification equations. cm (location) cm asked by Maggie on March 26, 2014 74. ## physics In a water balloon toss a contestant throws her balloon with a speed of 4.84 m/s at an angle of 16.5 degrees above the horizontal. Find the maximum height, the time of flight and the range of the balloon assuming her partner catches the balloon at the same asked by becky on February 6, 2012 75. ## statistics a pizza shop owner wishes to find the 95% confidence interval of the true mean cost of a large plain pizza. how large should the sample be if she wishes to be accurate to within \$0.15? a previous study showed that the standard deviation of the price was asked by paula on March 17, 2013 76. ## Calculus Find all points on the graph of the function f(x) = 2 cos(x) + (cos(x))2 at which the tangent line is horizontal. Consider the domain x = [0,2π). I have pi/2 and 3pi/2 for x values. But when I plug them I get zero. Is this correct as y values or am I asked by Calculus on January 11, 2016 77. ## English What does the aunt in "The StoryTeller" find improper about the bachelor's story? A. It makes fun of Bertha's goodness. B.It has too many fantasy elements. C.It unrealistically suggests that virtue is always rewarded. D. The bachelor tells it without asked by Caitlyn on September 28, 2015 78. ## Physics I (Classical) A kayaker needs to paddle north across a 85-m-wide harbor. The tide is going out, creating a tidal current that flows to the east at 2.0 m/s. The kayaker can paddle with a speed of 3.0 m/s. Find the angle West of North he needs to travel and the time it asked by Victoria on June 7, 2016 79. ## social studies when whalers first began hunting, they used small rowboats and were able to find many whalers near the shore. Later, the whalers needed bigger ships, and their whaling trip lasted much longer. Sometimes, whalers were gone for many moths or even years. asked by Celest on December 17, 2012 80. ## physics Two horse pull horizontally on ropes attached to a stump. The two forces f1 and f2 that they apply to the stump are such that the net(RESULTANT) force R has a magnitude equal to that of F and makes an angle of 90 with F1. Let F1=1300 N and R =1300 N also. asked by evelyn on March 4, 2012 81. ## Math A man traveled a distance of 22km in one hour,partly in a car at a speed of 30km/hr and partly on a motorcycle at 18km/hr.find the distance travelled by him in the car. Ans is 10km plz gve me fst sltn asked by Simran kaur on November 18, 2012 82. ## Physics 2 The wire in the figure below carries a current of 13 A. Suppose that a second long, straight wire is placed right next to this wire. The current in the second wire is 26 A. Use Ampère's law to find the magnitude of the magnetic field at a distance of r = asked by Kellsey on March 2, 2011 83. ## Physics 2 The wire in the figure below carries a current of 13 A. Suppose that a second long, straight wire is placed right next to this wire. The current in the second wire is 26 A. Use Ampère's law to find the magnitude of the magnetic field at a distance of r = asked by Kellsey on March 2, 2011 84. ## Physics The wire in the drawing carries a current of 14 A. Suppose that a second long, straight wire is placed right next to this wire. The current in the second wire is 33 A. Use Ampere's law to find the magnitude of the magnetic field at a distance of r = 1.0 m asked by Mareena on February 12, 2012 85. ## Algebra The first three equations I have solved, I would just appreciate someone checking them over to make sure I'm doing them right. Find the GCF: 16x^2z ,40xz^2 , 72z^3 = 3^3z Factor our GCF: a(a+1) - 3(a+1) = a+1 Factor Polynomial: 9a^2 - 64b^2 = asked by KB on January 22, 2009 86. ## Biology Im doing an experiment in Biology on a drug, but im not sure where I can find information on the drug Chantix. i also have a few general questions I was hoping you could help me with... 1. Why is it improtant to have as many subjects as possible in an asked by rainy on January 14, 2011 87. ## math help The price for an adult carnival ticket is 6.5 more than a carnival ticket for a child. Bob takes his son to the carnival. He buys cotton candy for \$10.25, and spends \$55. Write and solve a linear equation to find the prices for each of their carnival asked by kora on January 23, 2018 88. ## Math Okay, really I just need to know if this is going to be division, multiplication, subtraction or addition. Damon's town received 12.3 inches of rain during a 24-hour period. First, estimate the amount of rain that fell per hour. Then, find the exact amount asked by Barry on January 10, 2019 89. ## English I am trying to look for 20 lines of blank verse in shakespeare's play Measure for Measure. I found this: we CANnot WEIGH our BROther WITH ourSELF. great MEN may JEST with SAINTS; tis WIT in THEM. (2.2.16) Did anyone find a bigger chunk of blank verse? asked by Jasmin on November 13, 2011 90. ## physics A 1.00-kg beaker containing 2.29 kg of oil (density = 916 kg/m3) rests on a scale. A 2.33-kg block of iron is suspended from a spring scale and is completely submerged in the oil (see figure below). Find the equilibrium readings of both scales. top scale N asked by anna on April 11, 2016 91. ## Business maths An amount of k20.000 is invested for 5 years in a credit union which pays an annual untreated rate of 4.5 percent. Find accumulated value after 4 years if intrest rate is a: annually b:semi annually c:quarterly e:monthly asked by Margaret on August 17, 2016 92. ## math A round bar 40mm diameter is subjected to an axial pull of 80kN and reduction in diameter was found to be 0.00775mm. Find Poisson's ratio and Young's modules for the material of the bar. Take value of shear modules as 40kN/mm0…5. asked by aswath on July 25, 2013 93. ## math Land area of US states Stem-Leaf asked by nfKBJadk on April 21, 2019 94. ## Math Mr. Perry took his family out to eat. They ordered three meals costing \$8.99 each, two sodas at \$1.50 each and one cup of coffee for \$1.25 . Write an expression to find the total amount the Perry family spent on dinner before taxes and tip. Afterwards, asked by Vei on October 20, 2017 95. ## phusics We did not find any results for 'a ball is projected vertically up with an initial speed of 20m/sec and acceleration due to gravity is 10m/s^2then how long will it take for the ball to reach a point 10m above the point of projection second line' asked by dipu on September 11, 2014 96. ## physics Find the coefficient of kinetic friction between a 4.3-kg block and the horizontal surface on which it rests if an 80 Newton/meter spring must be stretched by 2.7 cm to pull the block with constant speed. Assume the spring pulls in a direction 13 degree asked by emily on January 19, 2011 97. ## Chemistry When 1.560 g of liquid hexane (C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.76 ∘C to 38.31 ∘C. Find ΔErxn for the reaction in kJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate asked by just need help on April 19, 2019 98. ## MTH 157 A company will need \$50,000 in five years for a new addition.To meet this goal the company deposits the money in an account today that pays 4% annual interest compounded quarterly. Find the amount that should be invested to total \$50,000 in 5 years. Thanks asked by Kim on July 31, 2015 99. ## physics the coefficients of static and kinetic friction for stone on wood are, respectively, 0.5 and 0.4. if a 150 kg stone statue is pushed with just enough force to start moving across a wooden floor and the same force continues to act afterward. find the asked by tom on November 7, 2012 100. ## Chemistry Calculate the molarity of the solutions made by dissolving the following amounts of solute in water and making the volume up to the stated value. 5.0mol nitric acid in 2.0L Do I just find the molarity for nitric acid only, because when I write out the asked by Jiskhaa on August 28, 2012	8,476	30,494	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.1875	3	CC-MAIN-2019-43	latest	en	0.913429
https://www.oddscoach.com/betonline-NBA-odds-03-19-2018/	1,521,510,912,000,000,000	text/html	crawl-data/CC-MAIN-2018-13/segments/1521257647251.74/warc/CC-MAIN-20180320013620-20180320033620-00412.warc.gz	821,867,035	21,316	# betonline NBA Odds Monday March 19, 2018 Published on March 19, 2018 Updated on March 19th, 2018 BetOnline Basketball NBA Odds Updated 3/19/2018 8:39 PM away / homepointsoddsedgeoddsedgepointsoddsedge Golden State Warriors8-1054.71%+2903.42%201.5-1074.74% San Antonio Spurs-8-115-350-113 Detroit Pistons-4-1084.75%-1802.75%205-1104.76% Sacramento Kings4-112+160-110 Toronto Raptors-10-1104.76%214-1104.76% Orlando Magic10-110-110 Dallas Mavericks9-1104.76%219-1104.76% New Orleans Pelicans-9-110-110 Los Angeles Clippers3-1054.71%227-1104.76% Minnesota Timberwolves-3-115-110 Atlanta Hawks13-1054.71%209-1104.76% Utah Jazz-13-115-110 Houston Rockets-4.5-1164.68%214.5-1104.76% Portland Trail Blazers4.5-104-110 Rotation Number - Rotation Numbers are a shorthand used by sportsbooks to refer to a side or team in a game. They are standardized throughout North American sportsbooks and casinos. Also, the rotation number allows each book to list the games in similar order and format. It keeps all of the games that are posted each day and throughout the week organized. That makes it easy for the bettor and the book, particularly in noisy casinos, to efficiently place a bet. Spread/Puckline/Runline - Spreads are the number of points/goals/runs to add or subtract from the final score to determine if the bet is won. Spreads help give the bettor a different look at the odds and encourage betting on longshot teams or lower the chalk on big favorites. here are videos/articles explaining Spreads, Runline and Puckline. Implied Probability - Implied Probability is best explained as odds of any type converted to percent likelihood. It can give the bettor a general idea the chance of the outcome occurring. There is an important caveat: Implied Probability always contains the sportsbooks edge, sometimes called their profit. Implied Probabilities for a single match always add up to more than 100%. The extra percent is the edge, as explained below. Here is a good article on Implied Probability and a useful Implied Probability Calculator. Sportsbook Edge - Edge represents the percent advantage that the sportsbook has over the bettor. A lower Sportsbook Edge is always better for gamblers. The best way to think about it, is that if you flip coins but the person you are playing against has a 2.5% advantage you are going to lose to them over time. If that percentage is 4.75% you are going to lose much sooner. Totals or Over/Under - Totals betting is not specific to a particular team. The sportsbook sets the total number of points it expects to be scored in any given match and bettors choose to bet that the score is going to be over or under this mark. The most common over/under market is total combined points for both teams, but sometimes can be a total points scored for a single team or side. Here is a video/article that explains Over Under or Totals Betting	735	2,883	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.703125	3	CC-MAIN-2018-13	latest	en	0.871982
https://myhomeworkhelp.com/know-how-to-solve-assignment-problem-in-excel-for-quick-solutions/	1,539,706,513,000,000,000	text/html	crawl-data/CC-MAIN-2018-43/segments/1539583510853.25/warc/CC-MAIN-20181016155643-20181016181143-00249.warc.gz	729,060,880	117,328	## Enter your keyword ### Know How to Solve Assignment Problem in Excel for Quick Solutions Assignment without proper help is like breaking bricks with a head! So, professionals around the world came up with different solutions for people like me and you to use certain techniques and accomplish tasks efficiently. Excel happens to be one of the approaches which helps people in solving tasks efficiently without going through much trouble. You can now also manage your work easily like the rest of us. How to solve assignment problem in excel? This is a good question. Excel aids in optimizing work. Take a look at excel in detail! What is the excel solver exactly? It refers to a Microsoft Excel that is a tool to solve constrained optimization issues that is minimize or maximize available constraints’ objective function. General assignment issues Optimization of a general assignment is daily trouble you might be facing. The standard is number of assignments or tasks at hand (n) and labors or machines available to complete it (m). Each of these has a particular cost when performing and also time for each unit to complete work. If you can optimize adequately, then you can minimize production cost and thereby, maximizing profit. Time constraints and gains are what excel helps in solving. However, take a look at installing the ad-ins before starting to solve task issues. It is not a default function in excel; hence, you need to add it before understanding How to solve assignment problem in excel. To add you require following steps which are below. Clicking on option make a window pop up on your computer or laptop display. From there you can go to Excel’s top left corner for opening file menu. Next, choose add-ins which will pop up another window. Then click on after checking solver will install it in your excel. Note: If you use 2007 Ms Office then press office button on left upper corner and click Excel option. Now that you are aware of how to install the solver, you are ready to learn in-depth about how to solve your assignments problem step by step using excel. Excel solver for solutions This portion will show you how to solve problems using this tool. Through different formulas, you can find out the constraints and objective function. By implementing the data, one receives through the formula one can incorporate it into excel. This shows How to solve assignment problem in excel. Data entry: You should enter the available data first in table form in excel. Each worker’s cost data, as well as time spent, is available. Also, total time along with D and W sets’ conditions are present. Hence, you should first place all available data in excel for a solution to the optimization problem. Let’s see an example; there are 15 tasks and 5 workers. There is a particular cost for each worker’s task. Hence, there should be a 5×15 table for each worker’s cost. This is your cost table. Also, time is a factor when workers are completing a job. Hence another 5×15 table which indicates each worker’s required time for each specific task. This is one answer of How to solve assignment problem in excel! Variables defining: You must also find which worker did what tasks. Therefore, if there are 5 different workers and 15 jobs, then variables should be 75. This is also updated in table. If the element’s value in table is 1, then it gives a worker’s accomplished task. However, if it offers a zero value, then that particular worker hasn’t done that task. The primary objective here for you is to minimize production costing. If it is zero, then that cost is not considered, but if the numerical is one, then it will add to cost. So, when you know How to solve assignment problem in excel you get variable too. Objective functions: Minimizing cost is the chief objective. In problem formulation, we learn that selected products’ sum should be minimum. Hence the table of cost which offers the data for workers when working as well as not working. Conditions: There are different conditions which you need to go through such as binary variables, single task for each worker, availability of total time, and large group tasks. Now you should place all the conditions mentioned above in excel solver. Proceeding to data in the menu bar you can find solver. Clicking it will display a pop up solver parameter window. All these offer guidance on How to solve assignment problem in excel. Also, if you require adding constraints, all you need is to click on excel solver window’s right-hand side. After putting all the available data all one requires is to press solve button at the bottom of the screen and receive a solution in the objective function form. After getting this, adding other constraints in solver which are in the problem is the way to go. Examples of further constraints include novice workers having less than three or four tasks, group one of worker must have at least one or more jobs, etc. Hence, all these complicated calculations are quickly done by the excel solver, and a person gets an adequate final result. You get the solution you require quickly without having to waste much time. So, knowing How to solve assignment problem in excel is exceptionally efficient for all people. Things this tool is used for Since this program is utilized for finding workforce schedules to minimize costing. Some concepts on which this is used are financial solver, capital budgeting, pricing model, network flow and inventory model, and demand sensitivity when price changes. Apart from this, this solver is also used in fantasy sports, negotiation solver, and many more. This application solves issues of various concept which is why you should know about it in detail. So, now that you are aware of the all the functions and have quite a precise knowledge about How to solve assignment problem in excel, no more assignment will be an issue for you from now on. Just put in all the variables in the excel solver and get the solution you need instantly. Try it today!	1,202	6,043	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2018-43	longest	en	0.933518
http://www.topperlearning.com/forums/ask-experts-19/solve-cos-sin-1-x-1-9-mathematics-inverse-trigonometric-functions-61372/reply	1,490,892,695,000,000,000	text/html	crawl-data/CC-MAIN-2017-13/segments/1490218195419.89/warc/CC-MAIN-20170322212955-00420-ip-10-233-31-227.ec2.internal.warc.gz	738,019,130	35,295	Question Mon December 17, 2012 By: Amee Solve: cos(sin-1 x)=1/9 Solve: cos(sin-1 x)=1/9 Thu December 20, 2012 LET sin-1 x=a therefor sin a=X given cos a=1/9 therefor X^2=(sin a)^2=1-1/81=80/81 therefor x=(80/81)^(1/2)=(4root 5)/9 Related Questions Fri March 24, 2017 Wed May 11, 2016	133	288	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.546875	4	CC-MAIN-2017-13	latest	en	0.823461
https://mathspace.co/textbooks/syllabuses/Syllabus-440/topics/Topic-8231/subtopics/Subtopic-107890/?activeTab=interactive	1,642,741,045,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320302723.60/warc/CC-MAIN-20220121040956-20220121070956-00520.warc.gz	406,762,916	40,004	# Multiplying fractions ## Interactive practice questions Evaluate $\frac{3}{8}\times\frac{5}{7}$38×57. Easy Less than a minute Evaluate $\frac{6}{7}\times\frac{3}{11}$67×311. Evaluate $\frac{4}{11}\times\frac{1}{10}$411×110. Evaluate $\frac{7}{10}\times\frac{2}{9}$710×29. ### Outcomes #### 9P.NA2.01 Simplify numerical expressions involving integers and rational numbers, with and without the use of technology	129	421	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.921875	3	CC-MAIN-2022-05	longest	en	0.581064
http://oeis.org/A144221	1,571,341,161,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570986675598.53/warc/CC-MAIN-20191017172920-20191017200420-00329.warc.gz	138,970,353	4,013	This site is supported by donations to The OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A144221 Triangle read by rows, A118433 * A007318^(-1) * A000012 0 1, -1, -1, 1, 1, -1, -1, 5, -1, 1, 5, -11, 1, 1, 1, 9, -11, -11, 9, -1, -1, -9, -9, 51, -29, 1, 1, -1, -1, -57, 111, -29, -29, 13, -1, 1, -15, 113, -111, -111, 169, -55, 1, 1, 1, -31, 113, 113, -559, 449, -55, -55, 17, -1 (list; table; graph; refs; listen; history; text; internal format) OFFSET 0,8 COMMENTS Row sums = (1, 0, 1, 2, -3, -4, 5, 6, -7, -8,...). LINKS FORMULA Triangle read by rows, A118433 * A007318^(-1) * A000012. A118433 = the self-inverse triangle. A007318^(-1) = the inverse of Pascal's triangle and A000012 = an infinite lower triangular matrix with all 1's. EXAMPLE First few rows of the triangle = 1; 1, -1; -1, 1, 1; -1, -1, 5, -1; 1, 5, -11, 1, 1; 1, 9, -11, -11, 9, -1; -1, -9, -9, 51, -29, 1, 1 -1, -1, -57, 111, -29, -29, 13, -1; 1, -15, 113, -111, -111, 169, -55, 1, 1; ... CROSSREFS Sequence in context: A152717 A071856 A242133 * A209575 A159570 A280374 Adjacent sequences: A144218 A144219 A144220 * A144222 A144223 A144224 KEYWORD tabl,sign AUTHOR Gary W. Adamson, Sep 14 2008 STATUS approved Lookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam Contribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent The OEIS Community \| Maintained by The OEIS Foundation Inc. Last modified October 17 15:32 EDT 2019. Contains 328116 sequences. (Running on oeis4.)	640	1,558	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.640625	4	CC-MAIN-2019-43	latest	en	0.580838
https://nubtrek.com/maths/algebra-basics/algebra-expressions-polynomials/degree-of-polynomials	1,566,329,258,000,000,000	text/html	crawl-data/CC-MAIN-2019-35/segments/1566027315558.25/warc/CC-MAIN-20190820180442-20190820202442-00559.warc.gz	564,075,259	4,548	Server Error Server Not Reachable. This may be due to your internet connection or the nubtrek server is offline. Thought-Process to Discover Knowledge Welcome to nubtrek. Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge. In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators. Read in the blogs more about the unique learning experience at nubtrek. mathsIntroduction to Algebra, Polynomials, and IdentitiesPolynomials - Basics ### Degree of Polynomials Based on the highest value of sum of exponents of terms, polynomials are classified as linear, quadratic, cubic, etc. click on the content to continue.. What is the similarity between the given polynomials of single variable? x^2+3 sqrt(5)x^2-8x-2 4x^2+8x -2x^2 • no similarity is found • the highest power or exponent of the variable is 2 • the highest power or exponent of the variable is 2 The answer is 'the highest power or exponent of the variable is 2' When the highest power is equal, a generic form can be derived. For example, ax^2+bx+c is the generic form for all the following polynomials. ax^2+bx+c (given generic form) x^2+3 (a=1; b=0; c=3) sqrt(5)x^2-8x-2 (a=sqrt(5); b=-8; c=-2 ) 4x^2+8x (a=4; b=8; c=0) -2x^2 (a=-2; b=0; c=0) What is the highest power of x^3y^2+xy^2+x^4+3? • 3 • 4 • 5 • 5 • maximum power is not defined for polynomial of two variables The answer is '5'. The term x^3y^2 is the one with the highest power of variables. The highest power is 3+2 = 5. The highest sum of powers of variables in a term is called degree of the polynomial. Which of the following is a meaning for the word 'degree'? • amount or level at which something is present • amount or level at which something is present • rotation in clockwise direction The answer is 'amount or level at which something is present'. The degree of a polynomial is the highest power of the variables in the terms. What is the word or phrase used to refer the highest power of the variables in a polynomial? • Pronunciation : Say the answer once Spelling: Write the answer once The answer is 'degree of the polynomial'. Degree of a polynomial is the highest exponent value with all variables combined. Degree of a Polynomial : In a polynomial of multiple variables, for each of the terms, the sum of exponents of all variables is compared and the highest sum is the degree of the polynomial. What is the degree of the polynomial 23 x+2? • 1 • 1 • 2 • 23 The answer is '1'. This is an example of polynomial of degree 1, also called "linear polynomial". what is the degree of the polynomial 3x^2+3x+3? • 1 • 2 • 2 • 3 The answer is '2'. This is an example of polynomial of degree 2, also called "quadratic polynomial". what is the degree of the polynomial xy^2+2xyz+yz? • 1 • 2 • 3 • 3 The answer is '3'. This is an example of polynomial of degree 3, also called "cubic polynomial". What is the degree of the polynomial 23? • 1 • 0 • 0 • 23 The answer is '0'. This can be called polynomial of degree 0, which is a "constant". Which of the following is a meaning for the word 'linear'? • vertically downwards • along a straight line • along a straight line The answer is 'along a straight line'. The graph of a linear polynomial is a straight line. Which of the following is a meaning for the word 'quadratic'? • a four sided figure • related to square • related to square The answer is 'related to square'. The square is in two dimensions and degree of quadratic polynomials is 2. The word "quadri" is used to denote 4 as a square is of four sides. Which of the following is a meaning for the word 'cubic'? • related to 3D shape cube • related to 3D shape cube • a personal cell or room The answer is 'related to 3D shape cube'. The cube is in 3 dimensions and degree of cubic polynomial is 3. What is the word or phrase used to refer a polynomial of degree 1? • Pronunciation : Say the answer once Spelling: Write the answer once The answer is 'linear polynomial'. What is the word or phrase used to refer a polynomial of degree 2? • Pronunciation : Say the answer once Spelling: Write the answer once What is the word or phrase used to refer a polynomial of degree 3? • Pronunciation : Say the answer once Spelling: Write the answer once The answer is 'cubic polynomial'. What is the word or phrase used to refer a polynomial of degree 0? • Pronunciation : Say the answer once Spelling: Write the answer once The answer is 'constant'. Based on the degree of a polynomial, they are classified as constant, linear, quadratic, cubic, and polynomial of degree n. Classification of Polynomial based on degree of polynomial : • constant : polynomial of degree 0. • linear : polynomial of degree 1. • quadratic : polynomial of degree 2. • cubic : polynomial of degree 3. • polynomial of degree n. Solved Exercise Problem: what is the degree of the polynomial 3x^2y+3xy^2+3z^2? • 6 • 5 • 3 • 3 The answer is '3'. slide-show version coming soon	1,355	5,097	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.125	4	CC-MAIN-2019-35	latest	en	0.894451
http://mathhelpforum.com/discrete-math/33606-one-one-onto.html	1,529,598,330,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267864191.74/warc/CC-MAIN-20180621153153-20180621173153-00056.warc.gz	192,865,151	8,741	1. ## one-to-one and onto a) {(x,y): x,y e Z, x + y = 0} b) {(x,y): x,y e Z, xy = 0} c) {(x,y): x,y e Z, x^2 + y^2 = 1} For each of these, I have to answer: 1) is this a function with range Z? 2) If it is a function, what is its domain and image? 3) If its a function, is it 1-1? onto? a bijection? 4) If its a bijection, what is its inverse function? x,y e Z means x and y can be any integer (pos or neg). Sorry for the notation.	157	434	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.6875	3	CC-MAIN-2018-26	latest	en	0.892221
https://mcqsolve.com/if-every-seconds-saturday-and-all-sundays-are-holidays-in-a-30-days-month-beginning-on-saturday-then-how-many-working-days-are-there-in-that-month-month-starts-from-saturday/	1,701,495,774,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100327.70/warc/CC-MAIN-20231202042052-20231202072052-00240.warc.gz	450,462,915	46,200	# If every seconds Saturday and all Sundays are holidays in a 30 days month beginning on Saturday, then how many working days are there in that month ? (Month starts from Saturday) A) 25 B) 22 C) 24 D) 23 Option – D. • ### If the sum and diference of two numbers are 20 and 8 respectively, then the difference of their squares is : error: Content is protected !!	95	369	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.921875	3	CC-MAIN-2023-50	longest	en	0.87415
http://www.slideserve.com/eshe/the-network-layer	1,490,221,850,000,000,000	text/html	crawl-data/CC-MAIN-2017-13/segments/1490218186353.38/warc/CC-MAIN-20170322212946-00010-ip-10-233-31-227.ec2.internal.warc.gz	706,763,507	26,558	This presentation is the property of its rightful owner. 1 / 83 # The Network Layer PowerPoint PPT Presentation The Network Layer. Chapter 5. Network Layer Design Isues. Store-and-Forward Packet Switching Services Provided to the Transport Layer Implementation of Connectionless Service Implementation of Connection-Oriented Service Comparison of Virtual-Circuit and Datagram Subnets. The Network Layer Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - ## The Network Layer Chapter 5 ### Network Layer Design Isues • Store-and-Forward Packet Switching • Services Provided to the Transport Layer • Implementation of Connectionless Service • Implementation of Connection-Oriented Service • Comparison of Virtual-Circuit and Datagram Subnets ### Store-and-Forward Packet Switching fig 5-1 The environment of the network layer protocols. ### Implementation of Connectionless Service Routing within a diagram subnet. ### Implementation of Connection-Oriented Service Routing within a virtual-circuit subnet. 5-4 ### Routing Algorithms • The Optimality Principle • Shortest Path Routing • Flooding • Distance Vector Routing • Hierarchical Routing • Multicast Routing • Routing for Mobile Hosts • Routing in Ad Hoc Networks ### Routing Algorithms (2) Conflict between fairness and optimality. ### The Optimality Principle (a) A subnet. (b) A sink tree for router B. ### Shortest Path Routing The first 5 steps used in computing the shortest path from A to D. The arrows indicate the working node. ### Flooding 5-8 top Dijkstra's algorithm to compute the shortest path through a graph. ### Flooding (2) 5-8 bottom Dijkstra's algorithm to compute the shortest path through a graph. ### Distance Vector Routing (a) A subnet. (b) Input from A, I, H, K, and the new routing table for J. ### Distance Vector Routing (2) The count-to-infinity problem. Each router must do the following: • Discover its neighbors, learn their network address. • Measure the delay or cost to each of its neighbors. • Construct a packet telling all it has just learned. • Send this packet to all other routers. • Compute the shortest path to every other router. (a) Nine routers and a LAN. (b) A graph model of (a). ### Measuring Line Cost A subnet in which the East and West parts are connected by two lines. (a) A subnet. (b) The link state packets for this subnet. ### Distributing the Link State Packets The packet buffer for router B in the previous slide (Fig. 5-13). ### Hierarchical Routing Hierarchical routing. Reverse path forwarding. (a) A subnet. (b) a Sink tree. (c) The tree built by reverse path forwarding. ### Multicast Routing (a) A network. (b) A spanning tree for the leftmost router. (c) A multicast tree for group 1. (d) A multicast tree for group 2. ### Routing for Mobile Hosts A WAN to which LANs, MANs, and wireless cells are attached. ### Routing for Mobile Hosts (2) Packet routing for mobile users. ### Routing in Ad Hoc Networks Possibilities when the routers are mobile: • Military vehicles on battlefield. • No infrastructure. • A fleet of ships at sea. • All moving all the time • Emergency works at earthquake . • The infrastructure destroyed. • A gathering of people with notebook computers. • In an area lacking 802.11. ### Route Discovery • (a) Range of A's broadcast. Shaded nodes are new recipients. Arrows show possible reverse routes. ### Route Discovery (2) Format of a ROUTE REQUEST packet. ### Route Discovery (3) Format of a ROUTE REPLY packet. ### Route Maintenance (a) D's routing table before G goes down. (b) The graph after G has gone down. ### Node Lookup in Peer-to-Peer Networks (a) A set of 32 node identifiers arranged in a circle. The shaded ones correspond to actual machines. The arcs show the fingers from nodes 1, 4, and 12. The labels on the arcs are the table indices. (b) Examples of the finger tables. ### Congestion Control Algorithms • General Principles of Congestion Control • Congestion Prevention Policies • Congestion Control in Virtual-Circuit Subnets • Congestion Control in Datagram Subnets • Jitter Control ### Congestion When too much traffic is offered, congestion sets in and performance degrades sharply. ### General Principles of Congestion Control • Monitor the system . • detect when and where congestion occurs. • Pass information to where action can be taken. • Adjust system operation to correct the problem. ### Congestion Prevention Policies 5-26 Policies that affect congestion. ### Congestion Control in Virtual-Circuit Subnets (a) A congested subnet. (b) A redrawn subnet, eliminates congestion and a virtual circuit from A to B. ### Hop-by-Hop Choke Packets (a) A choke packet that affects only the source. (b) A choke packet that affects each hop it passes through. ### Jitter Control (a) High jitter. (b) Low jitter. ### Quality of Service • Requirements • Techniques for Achieving Good Quality of Service • Integrated Services • Differentiated Services • Label Switching and MPLS ### Requirements 5-30 How stringent the quality-of-service requirements are. ### Buffering Smoothing the output stream by buffering packets. ### The Leaky Bucket Algorithm (a) A leaky bucket with water. (b) a leaky bucket with packets. ### The Leaky Bucket Algorithm (a) Input to a leaky bucket. (b) Output from a leaky bucket. Output from a token bucket with capacities of (c) 250 KB, (d) 500 KB, (e) 750 KB, (f) Output from a 500KB token bucket feeding a 10-MB/sec leaky bucket. ### The Token Bucket Algorithm 5-34 (a) Before. (b) After. 5-34 An example of flow specification. ### Packet Scheduling (a) A router with five packets queued for line O. (b) Finishing times for the five packets. ### RSVP-The ReSerVation Protocol (a) A network, (b) The multicast spanning tree for host 1. (c) The multicast spanning tree for host 2. ### RSVP-The ReSerVation Protocol (2) (a) Host 3 requests a channel to host 1. (b) Host 3 then requests a second channel, to host 2. (c) Host 5 requests a channel to host 1. ### Expedited Forwarding Expedited packets experience a traffic-free network. ### Assured Forwarding A possible implementation of the data flow for assured forwarding. ### Label Switching and MPLS Transmitting a TCP segment using IP, MPLS, and PPP. ### Internetworking • How Networks Differ • How Networks Can Be Connected • Concatenated Virtual Circuits • Connectionless Internetworking • Tunneling • Internetwork Routing • Fragmentation ### Connecting Networks A collection of interconnected networks. ### How Networks Differ 5-43 Some of the many ways networks can differ. ### How Networks Can Be Connected (a) Two Ethernets connected by a switch. (b) Two Ethernets connected by routers. ### Concatenated Virtual Circuits Internetworking using concatenated virtual circuits. ### Connectionless Internetworking A connectionless internet. ### Tunneling Tunneling a packet from Paris to London. ### Tunneling (2) Tunneling a car from France to England. ### Internetwork Routing (a) An internetwork. (b) A graph of the internetwork. ### Fragmentation (a) Transparent fragmentation. (b) Nontransparent fragmentation. ### Fragmentation (2) Fragmentation when the elementary data size is 1 byte. (a) Original packet, containing 10 data bytes. (b) Fragments after passing through a network with maximum packet size of 8 payload bytes plus header. (c) Fragments after passing through a size 5 gateway. ### The Network Layer in the Internet • The IP Protocol • Internet Control Protocols • OSPF – The Interior Gateway Routing Protocol • BGP – The Exterior Gateway Routing Protocol • Internet Multicasting • Mobile IP • IPv6 ### Design Principles for Internet • Make sure it works. • Keep it simple. • Make clear choices. • Exploit modularity. • Expect heterogeneity. • Avoid static options and parameters. • Look for a good design; it need not be perfect. • Be strict when sending and tolerant when receiving. • Consider performance and cost. ### Collection of Subnetworks The Internet is an interconnected collection of many networks. ### The IP Protocol (2) 5-54 Some of the IP options. ### Subnets A campus network consisting of LANs for various departments. ### Subnets (2) A class B network subnetted into 64 subnets. ### CDR – Classless InterDomain Routing 5-59 A set of IP address assignments. ### NAT – Network Address Translation Placement and operation of a NAT box. ### Internet Control Message Protocol 5-61 The principal ICMP message types. ### ARP– The Address Resolution Protocol Three interconnected /24 networks: two Ethernets and an FDDI ring. ### Dynamic Host Configuration Protocol Operation of DHCP. ### OSPF – The Interior Gateway Routing Protocol (a) An autonomous system. (b) A graph representation of (a). ### OSPF (2) The relation between ASes, backbones, and areas in OSPF. ### OSPF (3) 5-66 The five types of OSPF messeges. ### BGP – The Exterior Gateway Routing Protocol (a) A set of BGP routers. (b) Information sent to F. 5-69	2,213	9,527	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.703125	3	CC-MAIN-2017-13	longest	en	0.779627
https://www.justanswer.com/multiple-problems/89yxi-for-finley-this-set-questions-attachment.html	1,624,126,743,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623487649688.44/warc/CC-MAIN-20210619172612-20210619202612-00515.warc.gz	758,122,258	52,240	Multiple Problems For Help with Multiple Homework Problems, Ask an Expert Related Multiple Problems Questions Do NOT use the dollar sign. Do NOT use the percent symbol. Do NOT use the dollar sign. Do NOT use the percent symbol. Do NOT use words in your answers. Round your answers to nearest hundredths or hundredths percent, as needed.***What will a deposit of \$4,500 … read more Manal Elkhoshkhany Bachelor's Degree 1,417 satisfied customers I have many questions... I am completing a quiz and am I have many questions... I am completing a quiz and am completely stuck. Quiz is 24 questions, how much will it cost me?… read more Johnmark1900 Finance Manager Chartered Accountant 205 satisfied customers 1. find the interezt paid on a loan of \$2400 years at a 1. find the interezt paid on a loan of \$2400 for two years at a simple interest rate of 12% per year. The interest on a loan is \$ 2. Find the maturity value of a loan of \$2400.00 after three yearz. Th… read more linda_us Master's Degree Master's Degree 1,402 satisfied customers There are only 7 questions and would need by Wednesday. There are only 7 questions and would need by Wednesday. Question 1 Your finance text book sold 53,250 copies in its first year. The publishing company expects the sales to grow at a rate of 20 percent… read more linda_us Master's Degree Master's Degree 1,402 satisfied customers 1) Find the simple interest on the loan. \$1400 at 8% 1) Find the simple interest on the loan. \$1400 at 8% for 10 years. \$____ 2) Find the total amount due for the simple interest loan. \$1300 at 8% for 10 years. \$____ 3) Find the interest rate on a loan … read more akch2002 Master's Degree 3,433 satisfied customers I need these answered please 1. Which of the following would result in a decrease in cash flow and a use of cash? A. A decrease in notes payable B. An increase in long-term debt C. A decrease in inven… read more linda_us Master's Degree Master's Degree 1,402 satisfied customers QUESTION 1 Assume interest rate of 12%. A company receives QUESTION 1 Assume interest rate of 12%. A company receives cash flows of \$596 at the end of year 5, \$238 at the end of year 7, and \$542 at the end of year 10. Compute the future value of this cash flo… read more linda_us Master's Degree Master's Degree 1,402 satisfied customers 1. find the interest paid on a loan of \$2800 for two years 1. find the interest paid on a loan of \$2800 for two years at a simple interest rate of 11% per year. The interest on a loan is \$ 2. Find the maturity value of a loan of \$2400.00 after three years. Th… read more linda_us Master's Degree Master's Degree 1,402 satisfied customers Use the formula for the present value of an ordinary annuity Use the formula for the present value of an ordinary annuity or the amortization formula to solve the following problem. PV=\$7000; i=0.025; PMT=\$450; n = Round up to the nearest integer Solve the foll… read more getsetgo Master's Degree 2,385 satisfied customers there are text questions please respond soonest Using there are text questions please respond soonest Using the formula for simple interest and the given values find i P= \$300, r=5%, t=4 years; I=\$ Use the formula for simple interest, I=prt to find the i… read more linda_us Master's Degree Master's Degree 1,402 satisfied customers 1. Find the interest paid on a loan of \$2500 for one year at a simple 1. Find the interest paid on a loan of \$2500 for one year at a simple interest rate of 10% per year. Interest rate is ____ 2. Find the maturity value of a loan of \$2400.00 after three years. The loan … read more F. Naz B.Com CA Finalist & Completed B.com 1,145 satisfied customers The liabilities and owners equity for Campbell Industries The liabilities and owners' equity for Campbell Industries is found here: Accounts payable \$502,000 Notes payable \$244,000 Current liabilities \$746,000 Long-term debt \$1,239,000 Common equity \$5,266,0… read more linda_us Master's Degree Master's Degree 1,402 satisfied customers 1. (TCO 1) Which of the these activities is a capital budgeting 1. (TCO 1) Which of the these activities is a capital budgeting task? (Points : 4) determining the amount of cash needed on a daily basis to operate a firm . identifying assets that produce value in e… read more linda_us Master's Degree Master's Degree 1,402 satisfied customers I need help with abt 3 practice and 3 problem questions in I need help with abt 3 practice and 3 problem questions in my MyFinance Lab. Can anyone help me out?… read more PDtax Owner Master's Degree 15,084 satisfied customers 1. Ben Woolmer has an investment that will pay him the following 1. Ben Woolmer has an investment that will pay him the following cash flows over the next five years: \$8,061, \$4,266, \$9,547, \$4,399, and \$7,327. If his investments typically earn 9.72 percent, what i… read more linda_us Master's Degree Master's Degree 1,402 satisfied customers a. Calculate the future value of \$1000, given that it will a. Calculate the future value of \$1000, given that it will be held in the bank for 9 yrs and earn an annual interest rate of 5%. b. Recalculate part (a) using a compounding period that is (1) semiannu… read more linda_us Master's Degree Master's Degree 1,402 satisfied customers 2. Calculate the present value of \$5,000 received five years 2. Calculate the present value of \$5,000 received five years from today if your investments pay a. 6 percent compounded annually b. 8 percent compounded annually c. 10 percent compounded annually d. 1… read more linda_us Master's Degree Master's Degree 1,402 satisfied customers Disclaimer: Information in questions, answers, and other posts on this site ("Posts") comes from individual users, not JustAnswer; JustAnswer is not responsible for Posts. Posts are for general information, are not intended to substitute for informed professional advice (medical, legal, veterinary, financial, etc.), or to establish a professional-client relationship. The site and services are provided "as is" with no warranty or representations by JustAnswer regarding the qualifications of Experts. To see what credentials have been verified by a third-party service, please click on the "Verified" symbol in some Experts' profiles. 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Show MoreShow Less	2,362	9,825	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.90625	3	CC-MAIN-2021-25	latest	en	0.915431
https://testbook.com/question-answer/in-an-excel-spreadsheet-the-contents-of-cells-are--60961d170f039e2c3e4d9878	1,638,539,973,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964362879.45/warc/CC-MAIN-20211203121459-20211203151459-00375.warc.gz	623,147,287	29,003	# In an Excel spreadsheet, the contents of cells are as follows:a) A1 contains the value 2b) A2 contains the value 3c) A3 contains the formula: =$A$1+A2Contents of cell A3 are copied to Cell A4. The value displayed in Cell A4 will be: This question was previously asked in PSPCL LDC Previous Paper 9 (Held On : 30 Dec 2019 Shift 1) View all PSPCL LDC Papers > 1. 3 2. 8 3. 5 4. 7 Option 4 : 7 ## Detailed Solution • And if A3 contains the formula: =$A$1+A2 • But if contents of cell A3 are copied to cell A4 =$A$1+A3	176	521	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.171875	3	CC-MAIN-2021-49	latest	en	0.844171
http://anothermysqldba.blogspot.com/2015_06_01_archive.html	1,490,708,606,000,000,000	text/html	crawl-data/CC-MAIN-2017-13/segments/1490218189771.94/warc/CC-MAIN-20170322212949-00228-ip-10-233-31-227.ec2.internal.warc.gz	18,697,409	16,950	Sunday, June 28, 2015 Fibonacci Sequence Procedure Well it has been awhile since I posted, sorry about that. Today I was watching reruns of the TV show Fringe and when Walter referenced the Fibonacci Sequence I got side tracked with MySQL options for this. So I took that post and expanded on it a little, the result is a procedure that you can call and return the range within the Fibonacci Sequence that you are after. The procedure is below: ``` delimiter // CREATE PROCEDURE `Fibonacci`(IN POS INT, IN RANG INT, IN LIMTED INT) BEGIN select FORMAT(Fibonacci,0) AS Fibonacci from ( select @f0 Fibonacci, @fn:=@f1+@f0, @f0:=@f1, @f1:=@fn from (select @f0:=0, @f1:=1, @fn:=1) x, information_schema.STATISTICS p limit LIMTED) y LIMIT POS, RANG; END// delimiter ; ``` You can call this and pass whatever values and ranges you are after. So if you want the 5th value (starting from 0) in the sequence and the next value ``` > CALL Fibonacci(5,2,100); +-----------+ \| Fibonacci \| +-----------+ \| 5 \| \| 8 \| +-----------+ ``` So if you want the 30th value (starting from 0) in the sequence and the next value ``` > CALL Fibonacci(30,2,100); +-----------+ \| Fibonacci \| +-----------+ \| 832,040 \| \| 1,346,269 \| +-----------+ ``` So if you want the 150th value (starting from 0) in the sequence and the next value ``` > CALL Fibonacci(150,2,1000); +--------------------------------------------+ \| Fibonacci \| +--------------------------------------------+ \| 9,969,216,677,189,305,000,000,000,000,000 \| \| 16,130,531,424,904,583,000,000,000,000,000 \| +--------------------------------------------+ ``` So you get the idea. Now you can also expand the range of results if you want more than 2 just change the 2nd value in the procedure call. ``` > CALL Fibonacci(0,10,100); +-----------+ \| Fibonacci \| +-----------+ \| 0 \| \| 1 \| \| 1 \| \| 2 \| \| 3 \| \| 5 \| \| 8 \| \| 13 \| \| 21 \| \| 34 \| +-----------+ > CALL Fibonacci(30,5,100); +-----------+ \| Fibonacci \| +-----------+ \| 832,040 \| \| 1,346,269 \| \| 2,178,309 \| \| 3,524,578 \| \| 5,702,887 \| +-----------+ ``` Anyway, hope someone finds it helpful and credit for the base of query does go to the original post.	652	2,308	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.109375	3	CC-MAIN-2017-13	longest	en	0.735748
https://countspeed.com/speed-of-rhino/	1,685,453,832,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224645810.57/warc/CC-MAIN-20230530131531-20230530161531-00087.warc.gz	224,521,056	15,240	Have you ever been chased by a Rhino? And you didn’t have an idea about the speed of Rhino. Here, we go to know how much the speed of a Rhino is. You might find it very interesting to know how such a big creature can reach such speeds. Because, when you see them, you can not guess their speed, as they are so heavy on weight. Despite being heavy on weight or bulky like tanks, their powerful hind legs propel them forward, even on uneven surfaces. The interesting thing about Rhinos is that, when they reach top speeds, they will actually run on their toes. In this article, you may get to learn everything about Rhino’s speed and its calculations. Let’s get started and know the speed of Rhino. ## What Is The Speed Of Rhino? The speed of rhino is between 50 to 55 km/h, which depends upon the type of rhino. Among all the species of Rhinos, the Black rhino is the fastest rhino. There are a total of five species of rhino. Let’s know the speed of all these rhinos. 1. The speed of an Indian rhino, whose scientific name is Rhinoceros Unicornis, runs at a speed of 40 km/h. But, some estimated that the Indian Rhino can run at 55 km/h. 2. The speed of Javan Rhinos or Rhinoceros sondaicus is 48 km/h. 3. The speed of White rhinos or Ceratotherium simum is between 40 to 50 km/h. 4. The speed of Black rhinos or Diceros bicornis is 55 km/h and this is their top speed. 5. The speed of Sumatran Rhinos or Dicerorhinus sumatrensis is 55 km/h. Now, you can figure out what is the top speed of a rhino and it is 55 km/h. The average running speed of rhino is 50 km/h. Many people get confused between rhinos and hippos. When you see a rhino vs hippo, then you find that the hippo has an immense body and a large head, a short neck, and a broad chest. While a rhino has a long face and a pronounced hump on its neck. Now, the question is which is faster hippo or rhino. Rhino is much faster than hippos and they can run for 2,750 kilometers. Now, you know how fast a rhino run and how long. Let’s know how to find the speed of a rhino. Click here – What Is The Speed Of Sound In fps? ## Calculate The Speed Of Rhino Calculating the speed of a rhino is very interesting yet easy. So, for that, you need to know a formula called the speed formula. The seed formula is, Speed = Distance/Time In symbolic form, S = D/T Where, S is the speed of rhino D is the total distance covered by a rhino T is the time taken to cover the distance by rhino Let me explain it to you by using an example. For example, a group of rhinos is running continuously for 1 hour after hearing the roar of a lion. Without any break, they covered a distance of 52 kilometers and reached a safe place. What will be the speed of the rhino? By using a basic formula of speed, Speed of rhino = Distance/Time Speed of rhino = 52/1 = 52 km/h So, the speed of a group of rhinos is 52 km/h. Rhino is a member of any 5 extant species of odd-toed ungulates in the family of Rhinocerotidae. The five species of rhinos are, white, black, greater one-horned, Javan, and Sumatran. Additionally, a number of other animals have rhinoceros as part of their names, including the rhinoceros auklet, rhinoceros beetle, rhinoceros chameleon, etc. Rhinos are especially known for their giant body, and horns that grow from their snouts. Rhino grows tall up to 1.8 m and weighs a massive 2,500 kg, sometimes less or more. Now, you can imagine how strong is a rhino Despite their bulky body, Rhinos do not eat other animals meaning they are herbivores, and instead like to munch on lots of grass and plants at night, dawn, and dusk. Assemble more facts on different topics like these on Countspeed ## FAQ ### Can You Outrun A Rhino? Can a human outrun a rhino? No, not even if you’re an elite athlete. Rhinos run faster than any human on record. For example, in 2009, Usain Bolt got to almost 27.8 mph (44.7 km/h) in a 100-meter race during the World Championships in Berlin. ### Which Is Faster A Rhino Or A Hippo? Hippos can reach speeds of 30mph! And in a race with a rhino, it would depend on the rhino, a couch potato rhino would probably lose to hippo, but a well-trained athlete rhino would win. Rhinos have been recorded at speeds of 34mph, so just a tad faster than hippos. ### Can A Tiger Take Down A Rhino? “It’s not rare that tigers kill and eat rhino. Rhino comes as an easy hunt for a tiger who can not chase a deer,” he said. Assam’s Kaziranga National Park, which shelters the biggest population of rhinos, has about 15 to 20 rhino cubs getting killed in tiger attacks every year. ### Why Can’t Rhino Take His Suit Off? His first suit, more crude in overall design, was originally bonded to his skin and he was unable to remove it. ### How Fast Is A Rhino In Miles? Despite their weight and their bulk, rhinos move fast! They can run up to 30 – 40 miles per hour. To put that in context, Usain bolt can run 28 miles per hour. ### What Is The Fastest Rhino In The World? Black rhinos Black rhinos are the fastest rhino species and can run 34 miles per hour. Click here – What Is The Speed Of Sound In Space? ## Conclusion After reading this article you come to know about the top speed of rhino, rhino weight, and also how long can a rhino run. Rhinos can not bear a lot of temperatures, so to escape from the hot temperature they used to sleep in the shade or wallow in muddy pools to cool off their body. They are black in color, so their skin attracts more sun rays. Rhino are solitary animals and like to avoid each other but white rhino lives in a group and the group is known as ‘Crash”. Well, now you know everything about the speed of rhino	1,442	5,647	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2023-23	longest	en	0.942634
https://www.youtobia.com/blog/pages/artificial-neural-networks-activation-functio-4081122198	1,582,453,296,000,000,000	text/html	crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00280.warc.gz	957,260,427	15,366	# AI News, Artificial Neural Networks/Activation Functions ## Artificial Neural Networks/Activation Functions There are a number of common activation functions in use with neural networks. The output is a certain value, A1, if the input sum is above a certain threshold and A0 if the input sum is below a certain threshold. These kinds of step activation functions are useful for binary classification schemes. In other words, when we want to classify an input pattern into one of two groups, we can use a binary classifier with a step activation function. Each identifier would be a small network that would output a 1 if a particular input feature is present, and a 0 otherwise. Combining multiple feature detectors into a single network would allow a very complicated clustering or classification problem to be solved. linear combination is where the weighted sum input of the neuron plus a linearly dependent bias becomes the system output. In these cases, the sign of the output is considered to be equivalent to the 1 or 0 of the step function systems, which enables the two methods be to equivalent if This is called the log-sigmoid because a sigmoid can also be constructed using the hyperbolic tangent function instead of this relation, in which case it would be called a tan-sigmoid. Sigmoid functions in this respect are very similar to the input-output relationships of biological neurons, although not exactly the same. Sigmoid functions are also prized because their derivatives are easy to calculate, which is helpful for calculating the weight updates in certain training algorithms. The softmax activation function is useful predominantly in the output layer of a clustering system. ## Understanding Activation Functions in Neural Networks Recently, a colleague of mine asked me a few questions like “why do we have so many activation functions?”, “why is that one works better than the other?”, ”how do we know which one to use?”, “is it hardcore maths?” and so on. So I thought, why not write an article on it for those who are familiar with neural network only at a basic level and is therefore, wondering about activation functions and their “why-how-mathematics!”. Simply put, it calculates a “weighted sum” of its input, adds a bias and then decides whether it should be “fired” or not ( yeah right, an activation function does this, but let’s go with the flow for a moment ). Because we learnt it from biology that’s the way brain works and brain is a working testimony of an awesome and intelligent system ). To check the Y value produced by a neuron and decide whether outside connections should consider this neuron as “fired” or not. You would want the network to activate only 1 neuron and others should be 0 ( only then would you be able to say it classified properly/identified the class ). And then if more than 1 neuron activates, you could find which neuron has the “highest activation” and so on ( better than max, a softmax, but let’s leave that for now ). But..since there are intermediate activation values for the output, learning can be smoother and easier ( less wiggly ) and chances of more than 1 neuron being 100% activated is lesser when compared to step function while training ( also depending on what you are training and the data ). Ok, so we want something to give us intermediate ( analog ) activation values rather than saying “activated” or not ( binary ). straight line function where activation is proportional to input ( which is the weighted sum from neuron ). We can definitely connect a few neurons together and if more than 1 fires, we could take the max ( or softmax) and decide based on that. If there is an error in prediction, the changes made by back propagation is constant and not depending on the change in input delta(x) !!! That activation in turn goes into the next level as input and the second layer calculates weighted sum on that input and it in turn, fires based on another linear activation function. No matter how many layers we have, if all are linear in nature, the final activation function of last layer is nothing but just a linear function of the input of first layer! No matter how we stack, the whole network is still equivalent to a single layer with linear activation ( a combination of linear functions in a linear manner is still another linear function ). It tends to bring the activations to either side of the curve ( above x = 2 and below x = -2 for example). Another advantage of this activation function is, unlike linear function, the output of the activation function is always going to be in range (0,1) compared to (-inf, inf) of linear function. The network refuses to learn further or is drastically slow ( depending on use case and until gradient /computation gets hit by floating point value limits ). Imagine a network with random initialized weights ( or normalised ) and almost 50% of the network yields 0 activation because of the characteristic of ReLu ( output 0 for negative values of x ). That means, those neurons which go into that state will stop responding to variations in error/ input ( simply because gradient is 0, nothing changes ). When you know the function you are trying to approximate has certain characteristics, you can choose an activation function which will approximate the function faster leading to faster training process. For example, a sigmoid works well for a classifier ( see the graph of sigmoid, doesn’t it show the properties of an ideal classifier? ## Activation function In artificial neural networks, the activation function of a node defines the output of that node given an input or set of inputs. A standard computer chip circuit can be seen as a digital network of activation functions that can be 'ON' (1) or 'OFF' (0), depending on input. This is similar to the behavior of the linear perceptron in neural networks. However, only nonlinear activation functions allow such networks to compute nontrivial problems using only a small number of nodes[1]. In artificial neural networks this function is also called the transfer function. In biologically inspired neural networks, the activation function is usually an abstraction representing the rate of action potential firing in the cell.[2] In its simplest form, this function is binary—that is, either the neuron is firing or not. The function looks like ( v i ) {\displaystyle \phi (v_{i})=U(v_{i})} is the Heaviside step function. In this case many neurons must be used in computation beyond linear separation of categories. line of positive slope may be used to reflect the increase in firing rate that occurs as input current increases. Such a function would be of the form {\displaystyle \phi (v_{i})=\mu v_{i}} {\displaystyle \mu } This activation function is linear, and therefore has the same problems as the binary function. In addition, networks constructed using this model have unstable convergence because neuron inputs along favored paths tend to increase without bound, as this function is not normalizable. All problems mentioned above can be handled by using a normalizable sigmoid activation function. One realistic model stays at zero until input current is received, at which point the firing frequency increases quickly at first, but gradually approaches an asymptote at 100% firing rate. Mathematically, this looks like {\displaystyle \phi (v_{i})=U(v_{i})\tanh(v_{i})} where the hyperbolic tangent function can be replaced by any sigmoid function. This behavior is realistically reflected in the neuron, as neurons cannot physically fire faster than a certain rate. This model runs into problems, however, in computational networks as it is not differentiable, a requirement to calculate backpropagation. The final model, then, that is used in multilayer perceptrons is a sigmoidal activation function in the form of a hyperbolic tangent. Two forms of this function are commonly used: {\displaystyle \phi (v_{i})=\tanh(v_{i})} whose range is normalized from -1 to 1, and {\displaystyle \phi (v_{i})=(1+\exp(-v_{i}))^{-1}} is vertically translated to normalize from 0 to 1. The latter model is often considered more biologically realistic, but it runs into theoretical and experimental difficulties with certain types of computational problems. special class of activation functions known as radial basis functions (RBFs) are used in RBF networks, which are extremely efficient as universal function approximators. These activation functions can take many forms, but they are usually found as one of three functions: {\displaystyle c_{i}} is the vector representing the function center and {\displaystyle a} {\displaystyle \sigma } are parameters affecting the spread of the radius. Support vector machines (SVMs) can effectively utilize a class of activation functions that includes both sigmoids and RBFs. In this case, the input is transformed to reflect a decision boundary hyperplane based on a few training inputs called support vectors {\displaystyle x} The activation function for the hidden layer of these machines is referred to as the inner product kernel, {\displaystyle K(v_{i},x)=\phi (v_{i})} The support vectors are represented as the centers in RBFs with the kernel equal to the activation function, but they take a unique form in the perceptron as {\displaystyle \beta _{0}} {\displaystyle \beta _{1}} must satisfy certain conditions for convergence. These machines can also accept arbitrary-order polynomial activation functions where Activation function having types: Some desirable properties in an activation function include: The following table compares the properties of several activation functions that are functions of one fold x from the previous layer or layers: The following table lists activation functions that are not functions of a single fold x from the previous layer or layers: {\displaystyle \delta _{ij}} ## Fundamentals of Deep Learning – Activation Functions and When to Use Them? Internet provides access to plethora of information today. When our brain is fed with a lot of information simultaneously, it tries hard to understand and classify the information between useful and not-so-useful information. Let us go through these activation functions, how they work and figure out which activation functions fits well into what kind of problem statement. Before I delve into the details of activation functions, let’s do a little review of what are neural networks and how they function. A neural network is a very powerful machine learning mechanism which basically mimics how a human brain learns. The brain receives the stimulus from the outside world, does the processing on the input, and then generates the output. As the task gets complicated multiple neurons form a complex network, passing information among themselves. The black circles in the picture above are neurons. Each neuron is characterized by its weight, bias and activation function. A linear equation is simple to solve but is limited in its capacity to solve complex problems. A neural network without an activation function is essentially just a linear regression model. The activation function does the non-linear transformation to the input making it capable to learn and perform more complex tasks. We would want our neural networks to work on complicated tasks like language translations and image classifications. Activation functions make the back-propagation possible since the gradients are supplied along with the error to update the weights and biases. If the value Y is above a given threshold value then activate the neuron else leave it deactivated. When we simply need to say yes or no for a single class, step function would be the best choice, as it would either activate the neuron or leave it to zero. The function is more theoretical than practical since in most cases we would be classifying the data into multiple classes than just a single class. This makes the step function not so useful since during back-propagation when the gradients of the activation functions are sent for error calculations to improve and optimize the results. The gradient of the step function reduces it all to zero and improvement of the models doesn’t really happen. We saw the problem with the step function, the gradient being zero, it was impossible to update gradient during the backpropagation. Now if each layer has a linear transformation, no matter how many layers we have the final output is nothing but a linear transformation of the input. Our choice of using sigmoid or tanh would basically depend on the requirement of gradient in the problem statement. First things first, the ReLU function is non linear, which means we can easily backpropagate the errors and have multiple layers of neurons being activated by the ReLU function. But ReLU also falls a prey to the gradients moving towards zero. If you look at the negative side of the graph, the gradient is zero, which means for activations in that region, the gradient is zero and the weights are not updated during back propagation. So in this case the gradient of the left side of the graph is non zero and so we would no longer encounter dead neurons in that region. The parametrised ReLU function is used when the leaky ReLU function still fails to solve the problem of dead neurons and the relevant information is not successfully passed to the next layer. The softmax function is also a type of sigmoid function but is handy when we are trying to handle classification problems. The softmax function would squeeze the outputs for each class between 0 and 1 and would also divide by the sum of the outputs. The softmax function is ideally used in the output layer of the classifier where we are actually trying to attain the probabilities to define the class of each input. Now that we have seen so many activation functions, we need some logic / heuristics to know which activation function should be used in which situation. However depending upon the properties of the problem we might be able to make a better choice for easy and quicker convergence of the network. In this article I have discussed the various types of activation functions and what are the types of problems one might encounter while using each of them. ## Artificial Neural Networks/Activation Functions There are a number of common activation functions in use with neural networks. The output is a certain value, A1, if the input sum is above a certain threshold and A0 if the input sum is below a certain threshold. These kinds of step activation functions are useful for binary classification schemes. In other words, when we want to classify an input pattern into one of two groups, we can use a binary classifier with a step activation function. Each identifier would be a small network that would output a 1 if a particular input feature is present, and a 0 otherwise. Combining multiple feature detectors into a single network would allow a very complicated clustering or classification problem to be solved. linear combination is where the weighted sum input of the neuron plus a linearly dependent bias becomes the system output. In these cases, the sign of the output is considered to be equivalent to the 1 or 0 of the step function systems, which enables the two methods be to equivalent if This is called the log-sigmoid because a sigmoid can also be constructed using the hyperbolic tangent function instead of this relation, in which case it would be called a tan-sigmoid. Sigmoid functions in this respect are very similar to the input-output relationships of biological neurons, although not exactly the same. Sigmoid functions are also prized because their derivatives are easy to calculate, which is helpful for calculating the weight updates in certain training algorithms. The softmax activation function is useful predominantly in the output layer of a clustering system. ## Activation Functions: Neural Networks As you can see the function is a line or linear.Therefore, the output of the functions will not be confined between any range. Equation : f(x) = x Range : (-infinity to infinity) It doesn’t help with the complexity or various parameters of usual data that is fed to the neural networks. Nonlinearity helps to makes the graph look something like this It makes it easy for the model to generalize or adapt with variety of data and to differentiate between the output. Therefore, it is especially used for models where we have to predict the probability as an output.Since probability of anything exists only between the range of 0 and 1, sigmoid is the right choice. But the issue is that all the negative values become zero immediately which decreases the ability of the model to fit or train from the data properly. That means any negative input given to the ReLU activation function turns the value into zero immediately in the graph, which in turns affects the resulting graph by not mapping the negative values appropriately. Activation Functions in Neural Networks (Sigmoid, ReLU, tanh, softmax) ActivationFunctions #ReLU #Sigmoid #Softmax #MachineLearning Activation Functions in Neural Networks are used to contain the output between fixed values ... Neural Network Calculation (Part 2): Activation Functions & Basic Calculation From In this part we see how to calculate one section of a neural network. This calculation will be repeated many times to .. Deep Learning with Tensorflow - Activation Functions Enroll in the course for free at: Deep Learning with TensorFlow Introduction The majority of data .. Activation Functions In a neural network, the output value of a neuron is almost always transformed in some way using a function. A trivial choice would be a linear transformation ... Derivative of the sigmoid activation function, 9/2/2015 Activation Function using Sigmoid & ReLU using TensorFlow Mod-08 Lec-26 Multilayer Feedforward Neural networks with Sigmoidal activation functions; Pattern Recognition by Prof. P.S. Sastry, Department of Electronics & Communication Engineering, IISc Bangalore. For more details on NPTEL visit ... Training a two input perceptron to build an AND gate, 4/2/2015 The University of Jordan; Lutfi Al-Sharif; Mechatronics Engineering. Neural networks [1.2] : Feedforward neural network - activation function OptimizersLossesAndMetrics - Keras Here I go over the nitty-gritty parts of models, including the optimizers, the losses and the metrics. I first go over the usage of optimizers. Optimizers are ...	3,701	18,673	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.59375	3	CC-MAIN-2020-10	longest	en	0.909099
https://www.physicsforums.com/threads/so-i-have-another-matlab-problem.164034/	1,708,866,783,000,000,000	text/html	crawl-data/CC-MAIN-2024-10/segments/1707947474595.59/warc/CC-MAIN-20240225103506-20240225133506-00268.warc.gz	952,820,152	18,493	# So I have another matlab problem • MATLAB • fasterthanjoao In summary, the conversation discusses a problem with comparing two sets of data with different time intervals. The solution proposed is to plot both data sets and compare them, or to use the theoretical model to calculate values for the experimental time intervals and then compare. It is suggested to use Matlab's ODE solver and interpolation functions to interpolate the solution to the coarser time grid. It is also mentioned that the solver can be forced to take specific time values without affecting accuracy. There is also a discussion about the potential danger of interpolating the experimental data and the suggestion to average out the theoretical data according to the time scale of the device used to measure the experimental data. #### fasterthanjoao Should be a simple one, just can't get it to work. Basically, I have two sets of data that I want to compare. One set of data was observed at set intervals of time, and the second set of data is calculated - the problem is that the theoretical data has entries for non-integer values of t - when the observed is in single integer steps. This means that for the period of a minute in the theoretical data, I'll have several values but only a single observed value. Unfortunately, the number of theoretical values in a given range differs throughout so I feel that the best way to sort this out would be to lose a bit of accuracy and plot the observed data then somehow automatically calculate a point for each bit in time which would correspond to my theoretical. Is this possible? Plot both data sets and compare them or alternatively using the theoretical model, calculate the values given the experimental t's then plot and compare. I would calculate the experimental t's, but the theoretical data is calculated in an ODE solver with a defined tolerance (as far as I know, there's no way to define a definite step in ODE solvers, only an initial suggestion?). Leaves me with plotting the two results, is there any way to subtract two plots? Just to give a nicer comparison. Take the experimental output and the theoretical output and do a term by term search for the coincident times and subtract them. I'd just plot both then make the comparision, it'd look better to me and more professional. If you're using Matlab's ODE solver you can give it the time values at which you want the solution. Also, you can use the interp functions to interpolate your solution down to the coarser time grid. If you're using Matlab's ODE solver you can give it the time values at which you want the solution. Also, you can use the interp functions to interpolate your solution down to the coarser time grid. Very true...Forgot about those functions. Dr Transport said: Take the experimental output and the theoretical output and do a term by term search for the coincident times and subtract them. I'd just plot both then make the comparision, it'd look better to me and more professional. If you're using Matlab's ODE solver you can give it the time values at which you want the solution. Also, you can use the interp functions to interpolate your solution down to the coarser time grid. How do I force the ODE solver to take specific time values? Will that affect the accuracy of the solutions it finds at those points? As far as I knew (this is the first time I've used multiple ODE solvers in matlab) all I can do is give the solver an initial step suggestion and let it work through relative/absolute error tolerances to find the rest. thanks also, many a-far-more knowledgeable than I. Last edited: It seems to me a bit dangerous interpolating the experimental data -- eg. the theoretical data could show much higher frequency dynamics than the experimental data. In which case, I would average out the theoretical data according to the time scale of the device used to measure the experimental stuff. fasterthanjoao said: How do I force the ODE solver to take specific time values? Will that affect the accuracy of the solutions it finds at those points? As far as I knew (this is the first time I've used multiple ODE solvers in matlab) all I can do is give the solver an initial step suggestion and let it work through relative/absolute error tolerances to find the rest. thanks also, many a-far-more knowledgeable than I. If tspan is the time input argument to the ode solver, before you had tspan = [t0, tf] where t0 is the starting time and tf is the final time, just change it to tspan = [t0, t1, t2, t3, t4, tf] where t0,t1,t2,... are the points you want the solution at. And no, it does not affect the accuracy of the integration scheme, the ode solver solves it at many more points but only returns values for the specified points. J77 said: It seems to me a bit dangerous interpolating the experimental data -- eg. the theoretical data could show much higher frequency dynamics than the experimental data. In which case, I would average out the theoretical data according to the time scale of the device used to measure the experimental stuff. This is what I meant by interpolate down to the coarser time scale. excellent. Very much appreciated LeBrad. ## 1. What is Matlab and how is it used in scientific research? Matlab is a high-level programming language and interactive environment commonly used in scientific research to manipulate, analyze, and visualize data. It allows scientists to perform complex calculations and create sophisticated algorithms, making it a valuable tool in fields such as mathematics, engineering, and physics. ## 2. What are the common challenges when solving a problem using Matlab? Some common challenges when using Matlab include understanding the syntax and structure of the language, debugging errors, and optimizing code for efficiency. It is important to have a strong understanding of mathematical concepts and algorithms in order to effectively use Matlab for problem-solving. ## 3. What are the steps to solving a problem using Matlab? The first step is to clearly define the problem and identify the necessary inputs and desired outputs. Next, develop an algorithm or plan for solving the problem. Then, code the algorithm in Matlab, test it, and make any necessary revisions. Finally, run the code and analyze the results to ensure they align with the desired outputs. ## 4. How can I improve my skills in using Matlab? Practice and experimentation are key to improving skills in using Matlab. There are also many online resources and tutorials available, as well as courses and workshops offered by Matlab itself. Additionally, collaborating with other researchers and discussing different approaches to problem-solving can also help improve skills in using Matlab. ## 5. Are there any common mistakes to avoid when using Matlab? Some common mistakes to avoid when using Matlab include not properly defining variables, using incorrect syntax, and not debugging code for errors. It is also important to regularly save work and backup code in case of any unexpected errors or crashes.	1,433	7,092	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.171875	3	CC-MAIN-2024-10	latest	en	0.942362
http://waeconline.org.ng/e-learning/Agriculture/Agric118mq2.html	1,516,210,945,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084886952.14/warc/CC-MAIN-20180117173312-20180117193312-00634.warc.gz	377,759,209	5,986	Home Technical Mathematics Languages Science Social Science Art Literature Arabic Islamic Studies C.R.KHistory MusicVisual Art Clothing/Textile Home Management Shorthand Agricultural Science Paper 1, May/June 2009 Questions: 1 2 3 4 Main Weakness/Remedies Strength Question 2 (a) For each of specimens H, I and J name two farm tools that can perform functions similar to those of the specimens. [3 marks] (b) Describe how each of specimens I and J is used. [4 marks] (c) Describe each of specimens H and J. [4 marks] (d) Enumerate four advantages of using farm tools for farm work [4 marks] _____________________________________________________________________________________________________ OBSERVATION This question was not popular with the candidates. In question 2(a), majority of the candidates could not name farm tools that can perform functions similar to specimens H (Spade), I (Secateurs) and J (Open-ended spanner). Also in question 2(b) many candidates were unable to describe how specimens I and J are used. In question 2(c), most of the candidates could not describe specimens H and J. Also, in question 2(d), majority of the candidates were unable to enumerate advantages of using farm tools for farm work. Tools that can perform similar functions as specimens H, I and J - Hoe, shovel and hand trowel Specimen I (Secateurs) - Saw, sickle, cutlass and shears Specimen J (open-ended spanner) - Wrench, ring spanner and pliers How Specimen I (Secateurs) is used • Open the handle to widen the blades • Fix the opened blades against the material to be cut • Press back the handles to close the blades and cut the materials • Release the handle How specimen J (open-ended spanner) is used • Fix opened and directly on top of the bolt (appropriate size) • Turn the handle clockwise (to tighten) or anticlockwise (to loosen) the bolt. • Continue to turn until the bolt is firmly fixed or loosened before releasing the handle. Description of each of specimens H and J • Has a long wooden/metallic handle • Blade is rectangular, flat and sharp at one end for easy penetration into the soil • Handle widens at the posterior end to form a triangular block or “D” shape for firm handling Specimen J (open-ended spanner) • It is metallic • Both sides end in a hexagonal shape of various sizes that fit the heads of bolts • Flattened metallic handle Advantages of using farm tools for farm work - Minimize injury to the farmer - Make work easier/minimize drudgery - Enhance faster completion of operation/saves time - Enhance certain operations, e.g. budding - Not expensive to acquire - Easy to maintain - Increases farmers’ productivity - They are sometimes the only way to get certain operations done on the farm - Easy to operate and requires no high technical know-how - Do not require any specialised sources of power to operate	690	2,975	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.203125	3	CC-MAIN-2018-05	latest	en	0.809802
https://www.tutorialspoint.com/questions/Pythonista	1,547,723,561,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547583658928.22/warc/CC-MAIN-20190117102635-20190117124635-00239.warc.gz	949,895,976	9,068	Drag to reposition Contributed Questions # How to perform square root without using math module in Python? How to perform square root without using math module in Python? # How to calculate square root of a number in Python? How to calculate square root of a number in Python? # How do you round up a float number in Python? How do you round up a float number in Python? # How to round down to 2 decimals a float using Python? How to round down to 2 decimals a float using Python? # How to round off a floating number using Python? How to round off a floating number using Python? # How to Find the Power of a Number Using Recursion in Python? How to Find the Power of a Number Using Recursion in Python? # How to find power of a number in Python? How to find power of a number in Python? # How to find the fractional and integer parts of x in a two-item tuple in Python? How to find the fractional and integer parts of x in a two-item tuple in Python? # How to find the value closest to negative infinity in Python? How to find the value closest to negative infinity in Python? # How to find the value closest to positive infinity in Python? How to find the value closest to positive infinity in Python?	266	1,224	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.640625	3	CC-MAIN-2019-04	latest	en	0.814561
http://www.vias.org/kimberlyee/ee_09_01.html	1,701,424,319,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100286.10/warc/CC-MAIN-20231201084429-20231201114429-00275.warc.gz	88,070,894	4,069	Electrical Engineering is a free introductory textbook to the basics of electrical engineering. See the editorial for more information.... # Magnetic Circuit Author: E.E. Kimberly Magnetism that depends on a flow of electric current is called electromagnetism. Fig. 7-1 shows a typical magnetic pattern produced by a loosely wound coil or solenoid with air core. Fig. 7-1. Solenoid With Air Core While most of the magnetic flux links all the turns of the solenoid, some of it leaks through, as at a, and does not link the end turns. This is called flux leakage with respect to the end turns. It may be shown by calculation1 that, if a solenoid whose length is many times its diameter be wound with N turns uniformly distributed over its length l, the field intensity at its center will be and where I = current flowing, in amperes; l = length of solenoid, in centimeters; N = number of turns in solenoid; H = field intensity, in gausses. Fig. 7-2. Closed-Loop Solenoid If the solenoid were bent so that its axis formed a closed loop, as in Fig. 7-2, and there were no flux leakage, the amperes required to produce a field intensity H with N turns would be The flux leakage from such a solenoid or coil may be reduced to an amount negligible for most purposes by substituting for the air core a core of some other material such as iron in which a magnetic field may be established more easily than in air. Thus, for a magnetic circuit all or most of which is in iron, equation (7-1) may be used except as modified in equation (7-5). See the solution of Example 7-1. It is much easier to magnetize iron than a vacuum, or air. The ease with which a material may be made to carry magnetic lines of induction, compared to the ease with which they may be established in a vacuum, is called its permeability. The following comparison may be made between the electric circuit and the magnetic circuit. Electric Circuit Magnetic Circuit Electromotive Force, emf or EProduced by any one of several means. Unit is the volt. Magnetomotive Force, mmf or MProduced only by ampere-turns, Unit is the gilbert, which equals 0.4πNI. Conductivity, KVaries for different materials, from practically zero to high values. Has its highest values for a few pure metals, notably silver and copper. Is independent of current density, but varies with temperature, the change being different for different metals. Permeability, μIs unity for a vacuum, air, and many other materials. Varies from very slightly below 1 for a few diamagnetic materials to more than 1000 for iron and some alloys. Varies greatly with flux density B, but is not materially affected by moderate temperature. The amount of change with B varies greatly for different alloys. Resistivity, ΡReciprocal of conductivity. ReluctivityReciprocal of permeability. If equation (7-1) were used to find the flux density in the core of a solenoid like that in Fig. 7-1, which is provided with an iron core of uniform-cross-section and of permeability μ, then Also, or where B = flux density in the iron, in lines per square centimeter; ϕ = total flux in the iron; A = cross-sectional area of the iron, in square centimeters; I = length of the iron, in centimeters. The term is called reluctance, and its symbol is Ρ (rho). Reluctance in a magnetic circuit corresponds to resistance in an electric circuit. Thus, [7-6] or [7-7] in which M = magnetomotive force. Equation (7-6) or equation (7-7) is called Ohm's Law of the magnetic circuit because of its similarity to the equation of the electric circuit. The resistance of air to low electrical potentials is infinite, and so it is possible to "open" an electric circuit. The reluctivity of air, however, is 1; so it is not possible to have a magnetic path of infinite reluctance and hence, it may be said that a magnetic circuit cannot be opened. 1 Principles of Direct-Current Machines, by A, S. Langsdorf, McGraw-Hill Book Co. Last Update: 2010-11-22	947	3,975	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.046875	3	CC-MAIN-2023-50	latest	en	0.957437
https://stoplearn.com/question/if-the-mass-of-the-earth-is-5-781024kg-and-g-equals-6-6710-11-calculate-the-gravitational-force-intensity-due-to-the-earth-if-the-radius-of-the-earth-is-6400km-what-will-be-the-escape-velocity/	1,709,210,744,000,000,000	text/html	crawl-data/CC-MAIN-2024-10/segments/1707947474808.39/warc/CC-MAIN-20240229103115-20240229133115-00387.warc.gz	527,352,433	15,382	# if the mass of the earth is 5.7810^24kg and g equals 6.6710^-11. calculate the gravitational force intensity due to the earth. if the radius of the earth is 6400km. what will be the escape velocity All QuestionsCategory: Secondary Schoolif the mass of the earth is 5.7810^24kg and g equals 6.6710^-11. calculate the gravitational force intensity due to the earth. if the radius of the earth is 6400km. what will be the escape velocity Get paid for every topic you create in: Shoutam.com Forum! StopLearn Team Staff answered 9 months ago To calculate the gravitational force intensity due to the Earth, we can use the formula: F = (G * m1 * m2) / r^2 where F is the gravitational force, G is the gravitational constant (6.67 x 10^-11 N m^2/kg^2), m1 is the mass of the Earth, m2 is the mass of the object, and r is the distance between the centers of the Earth and the object. Given: Mass of the Earth (m1) = 5.78 x 10^24 kg Gravitational constant (G) = 6.67 x 10^-11 N m^2/kg^2 Radius of the Earth (r) = 6400 km = 6,400,000 m Now, let’s calculate the gravitational force intensity due to the Earth: F = (G * m1) / r^2 F = (6.67 x 10^-11 N m^2/kg^2 * 5.78 x 10^24 kg) / (6,400,000 m)^2 F = (3.8766 x 10^14 N m^2/kg^2) / (40,960,000,000 m^2) F ≈ 9.48 x 10^3 N/kg The gravitational force intensity due to the Earth is approximately 9.48 x 10^3 N/kg. To calculate the escape velocity, we can use the formula: v = √(2 * G * m1 / r) where v is the escape velocity. Now, let’s calculate the escape velocity: v = √(2 * G * m1 / r) v = √(2 * 6.67 x 10^-11 N m^2/kg^2 * 5.78 x 10^24 kg / 6,400,000 m) v ≈ √(9.17 x 10^14 N m^2/kg^2 / 6,400,000 m) v ≈ √(1.432 x 10^8 N m^2/kg^2) v ≈ 1.195 x 10^4 m/s The escape velocity from the Earth is approximately 1.195 x 10^4 m/s. Get paid for every topic you create in: Shoutam.com Forum! Noah Ilemona David answered 9 months ago To calculate the gravitational force intensity due to the Earth, you can use the formula: F = (G * m1 * m2) / r^2 where: F is the gravitational force intensity, G is the gravitational constant (6.67 * 10^-11 N(m/kg)^2), m1 is the mass of the Earth (5.78 * 10^24 kg), m2 is the mass of an object (assuming it is much smaller than the Earth’s mass, we can consider it negligible), and r is the radius of the Earth (6400 km or 6.4 * 10^6 m). Substituting the values into the formula: F = (6.67 * 10^-11 N(m/kg)^2 * 5.78 * 10^24 kg * m2) / (6.4 * 10^6 m)^2 F = (4.44906 * 10^14 N * m2) / 4.096 * 10^13 m^2 F ≈ 10.88 N * m2 / m^2 The gravitational force intensity due to the Earth is approximately 10.88 N. Now, to calculate the escape velocity, we can use the formula: v = √(2 * G * M / r) where: v is the escape velocity, G is the gravitational constant (6.67 * 10^-11 N(m/kg)^2), M is the mass of the Earth (5.78 * 10^24 kg), and r is the radius of the Earth (6400 km or 6.4 * 10^6 m). Substituting the values into the formula: v = √(2 * 6.67 * 10^-11 N(m/kg)^2 * 5.78 * 10^24 kg / 6.4 * 10^6 m) v = √(7.53676 * 10^13 N * m / 4.096 * 10^6 m) v ≈ √(18.3789 * 10^6 m^2/s^2) v ≈ 4285.5 m/s Therefore, the escape velocity from the Earth is approximately 4285.5 m/s.	1,122	3,146	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.09375	4	CC-MAIN-2024-10	latest	en	0.742017
https://plainmath.org/algebra-ii/93199-use-synthetic-division-to-divide-the-pol	1,718,940,614,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00838.warc.gz	394,874,416	20,053	sengihantq 2022-10-08 Use synthetic division to divide the polynomial. $\left(5{x}^{2}-17x-12\right)÷\left(x-4\right)$ Sarahi Gallegos Divide the polynomial using synthetic division. polyhorner scheme $\left[x=4\right]\left\{5{x}^{2}-17x-12\right\}$ The quotient for the given is 5x+3 Result: 5x+3 Do you have a similar question?	118	334	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 19, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.828125	4	CC-MAIN-2024-26	latest	en	0.722987
https://www.jiskha.com/display.cgi?id=1220806312	1,516,567,812,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084890874.84/warc/CC-MAIN-20180121195145-20180121215145-00428.warc.gz	919,826,358	4,551	posted by . Wal-Mart, a discount store chain, is planning to build a new store in Rock Springs, Maryland. The parcel of land the company owns is large enough to accommodate a store with 140,000 square feet of floor space. Based on marketing and demographic surveys of the area and historical data from its other stores, Wal-Mart estimates its annual profit contribution per square foot for each of the store's departments to be as shown in the following table. Department Profit contribution per ft2 Men's clothing \$4.25 Women's clothing \$5.10 Children's clothing \$4.50 Toys \$5.20 Housewares \$4.10 Electronics \$4.90 Auto supplies \$3.80 Each department must have at least 15,000 ft2 of floor space and no department can have more than 20% of the total retail floor space. Men's women's and children's clothing plus housewares keep all their stock on the retail floor; however, toys, electronics, and auto supplies keep some items (bicycles, televisions, tires, etc.) in inventory. Thus, 10% of the total retail floor space devoted to these three departments must be set aside outside the retail area for stocking inventory. Formulate a linear programming model that can be used to determine the floor space that should be devoted to each department in order to maximize profit contribution. ## Similar Questions Wal-Mart, a discount store chain, is planning to build a new store in Rock Springs, Maryland. The parcel of land the company owns is large enough to accommodate a store with 140,000 square feet of floor space. Based on marketing and … 2. ### finance Home Builder Supply a retailer in the home improvement industry, currently operate seven retail outlets in GA and SC. Management is contemplating building an eighth retail store across town from its most successful retail outlet. The … 3. ### Maths Bill Ding plans to build a new hardware store. He buys a rectangular lot that is 50 ft by 200 ft, the 50-foot dimension being along the street. The store is to have an area of 4000 square feet. Construction costs \$100 per linear foot … 4. ### Operations Management Wal-Mart is the largest corporation in the world, and it has obviously enjoyed tremendous success. But while many welcome its location in their communities, others do not. Some complain that its presence has too many negative effects … 5. ### linear programming A company is planning to purchase and store two items, gadgets and widgets. Each gadget costs \$2.00 and occupies 2 square meters of floor space; each widget costs \$3.00 and occupies 1 square meter of floor space. \$1,200 is available … 6. ### math A company is planning to purchase and store two items, gadgets and widgets. Each gadget costs \$2.00 and occupies 2 square meters of floor space; each widget costs \$3.00 and occupies 1 square meter of floor space. \$1,200 is available … 7. ### economics Suppose that the company Mama's Pies adds another store to sell its pastries. Supposed that it costs \$400,000 to build the new store and assume that the new store will generate revenues of \$450,000. What is the rate of return on this … 8. ### math A large bicycle retailer collects data on the number of bicycles in each store compared to floor space of each store. The data is given in the table below. Number of Bicycles 60 56 208 52 70 55 Floor Space (sq ft) 1400 1140 3250 1100 … 9. ### math A store is located at (-1, -2). the owner of the store plans to build another store that is 5 blocks east and 4 blocks north of the original store. plot both stores on the coordinate plane. Give directions from the original store to …	828	3,597	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.671875	3	CC-MAIN-2018-05	latest	en	0.941169
https://docs.trifacta.com/aac/en/trifacta-classic/wrangle-language/language-appendices/transformation-examples/example---pow-and-sqrt-functions.html	1,726,659,438,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00698.warc.gz	196,073,843	5,355	# EXAMPLE - POW and SQRT Functions In this example, you learn how to compute exponentials and square roots on your numeric data. Functions: Item Description POW Function Computes the value of the first argument raised to the value of the second argument. SQRT Function Computes the square root of the input parameter. Input value can be a Decimal or Integer literal or a reference to a column containing numeric values. All generated values are non-negative. Source: The dataset below contains values for x and y: X Y 3 4 4 9 8 10 30 40 Transformation: You can use the following transformation to generate values for z2. Note Transformation Name New formula Single row formula (POW(x,2) + POW(y,2)) 'Z' You can see how column Z is generated as the sum of squares of the other two columns, which yields z2. Now, edit the transformation to wrap the value computation in a SQRT function. This step is done to compute the value for z, which is the distance between the two points based on the Pythagorean theorem. Transformation Name New formula Single row formula SQRT((POW(x,2) + POW(y,2))) 'Z' Results: X Y Z 3 4 5 4 9 9.848857801796104 8 10 12.806248474865697 30 40 50	303	1,211	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.015625	3	CC-MAIN-2024-38	latest	en	0.728931
https://yugashakthi.org/chicken-cacciatore-rolok/pascal%27s-triangle-explained-a07810	1,624,149,314,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623487653461.74/warc/CC-MAIN-20210619233720-20210620023720-00383.warc.gz	1,004,346,870	11,633	The triangle is constructed using a simple additive principle, explained in the following figure. Omissions? Because of this connection, the entries in Pascal's Triangle are called the _binomial_coefficients_. For example, the numbers in row 4 are 1, 4, 6, 4, and 1 and 11^4 is equal to 14,641. The first diagonal is, of course, just "1"s. The next diagonal has the Counting Numbers (1,2,3, etc). If you have any doubts then you can ask it in comment section. The numbers on the left side have identical matching numbers on the right side, like a mirror image. We may already be familiar with the need to expand brackets when squaring such quantities. Ring in the new year with a Britannica Membership, https://www.britannica.com/science/Pascals-triangle. Example Of a Pascal Triangle For example, drawing parallel “shallow diagonals” and adding the numbers on each line together produces the Fibonacci numbers (1, 1, 2, 3, 5, 8, 13, 21,…,), which were first noted by the medieval Italian mathematician Leonardo Pisano (“Fibonacci”) in his Liber abaci (1202; “Book of the Abacus”). The third row has 3 numbers, which is 1, 2, 1 and so on. This can be very useful ... you can now work out any value in Pascal's Triangle directly (without calculating the whole triangle above it). It is one of the classic and basic examples taught in any programming language. His triangle was further studied and popularized by Chinese mathematician Yang Hui in the 13th century, for which reason in China it is often called the Yanghui triangle. (The Fibonacci Sequence starts "0, 1" and then continues by adding the two previous numbers, for example 3+5=8, then 5+8=13, etc), If you color the Odd and Even numbers, you end up with a pattern the same as the Sierpinski Triangle. The number on each peg shows us how many different paths can be taken to get to that peg. The triangle also shows you how many Combinations of objects are possible. It contains all binomial coefficients, as well as many other number sequences and patterns., named after the French mathematician Blaise Pascal Blaise Pascal (1623 – 1662) was a French mathematician, physicist and philosopher. So the probability is 6/16, or 37.5%. It is named after Blaise Pascal. (x + 3) 2 = (x + 3) (x + 3) (x + 3) 2 = x 2 + 3x + 3x + 9. Pascal's Triangle can show you how many ways heads and tails can combine. For … We take an input n from the user and print n lines of the pascal triangle. note: the Pascal number is coming from row 3 of Pascal’s Triangle. Be on the lookout for your Britannica newsletter to get trusted stories delivered right to your inbox. Blaise Pascal was a French mathematician, and he gets the credit for making this triangle famous. (Hint: 42=6+10, 6=3+2+1, and 10=4+3+2+1), Try this: make a pattern by going up and then along, then add up the values (as illustrated) ... you will get the Fibonacci Sequence. He used a technique called recursion, in which he derived the next numbers in a pattern by adding up the previous numbers. We will know, for example, that. ), and in the book it says the triangle was known about more than two centuries before that. Each number is the numbers directly above it added together. Corrections? It’s known as Pascal’s triangle in the Western world, but centuries before that, it was the Staircase of Mount Meru in India, the Khayyam Triangle in Iran, and Yang Hui’s Triangle in China. An interesting property of Pascal's triangle is that the rows are the powers of 11. In the twelfth century, both Persian and Chinese mathematicians were working on a so-called arithmetic triangle that is relatively easily constructed and that gives the coefficients of the expansion of the algebraic expression (a + b) n for different integer values of n (Boyer, 1991, pp. We can use Pascal's Triangle. The midpoints of the sides of the resulting three internal triangles can be connected to form three new triangles that can be removed to form nine smaller internal triangles. He discovered many patterns in this triangle, and it can be used to prove this identity. Each number is the sum of the two directly above it. The process of cutting away triangular pieces continues indefinitely, producing a region with a Hausdorff dimension of a bit more than 1.5 (indicating that it is more than a one-dimensional figure but less than a two-dimensional figure). The triangle can be constructed by first placing a 1 (Chinese “—”) along the left and right edges. The four steps explained above have been summarized in the diagram shown below. Pascal also did extensive other work on combinatorics, including work on Pascal's triangle, which bears his name. Our editors will review what you’ve submitted and determine whether to revise the article. The principle was … Basically Pascal’s triangle is a triangular array of binomial coefficients. The natural Number sequence can be found in Pascal's Triangle. For example, if you toss a coin three times, there is only one combination that will give you three heads (HHH), but there are three that will give two heads and one tail (HHT, HTH, THH), also three that give one head and two tails (HTT, THT, TTH) and one for all Tails (TTT). Pascal's triangle is made up of the coefficients of the Binomial Theorem which we learned that the sum of a row n is equal to 2 n. So any probability problem that has two equally possible outcomes can be solved using Pascal's Triangle. Pascal's Triangle is a mathematical triangular array.It is named after French mathematician Blaise Pascal, but it was used in China 3 centuries before his time.. Pascal's triangle can be made as follows. The sum of all the elements of a row is twice the sum of all the elements of its preceding row. One of the most interesting Number Patterns is Pascal's Triangle (named after Blaise Pascal, a famous French Mathematician and Philosopher). In the … One of the most interesting Number Patterns is Pascal's Triangle. Adding the numbers along each “shallow diagonal” of Pascal's triangle produces the Fibonacci sequence: 1, 1, 2, 3, 5,…. The triangle displays many interesting patterns. (x + 3) 2 = x 2 + 6x + 9. Examples: So Pascal's Triangle could also be Then the triangle can be filled out from the top by adding together the two numbers just above to the left and right of each position in the triangle. Balls are dropped onto the first peg and then bounce down to the bottom of the triangle where they collect in little bins. 1 2 1. Donate The Pascal’s triangle is a graphical device used to predict the ratio of heights of lines in a split NMR peak. The "!" Each number equals to the sum of two numbers at its shoulder. View Full Image. In Pascal's words (and with a reference to his arrangement), In every arithmetical triangle each cell is equal to the sum of all the cells of the preceding row from its column to the first, inclusive(Corollary 2). Polish mathematician Wacław Sierpiński described the fractal that bears his name in 1915, although the design as an art motif dates at least to 13th-century Italy. Try another value for yourself. Pascal's Triangle is probably the easiest way to expand binomials. Natural Number Sequence. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. Chinese mathematician Jia Xian devised a triangular representation for the coefficients in the 11th century. Each line is also the powers (exponents) of 11: But what happens with 115 ? Another interesting property of the triangle is that if all the positions containing odd numbers are shaded black and all the positions containing even numbers are shaded white, a fractal known as the Sierpinski gadget, after 20th-century Polish mathematician Wacław Sierpiński, will be formed. There are 1+4+6+4+1 = 16 (or 24=16) possible results, and 6 of them give exactly two heads. In fact, if Pascal's triangle was expanded further past Row 15, you would see that the sum of the numbers of any nth row would equal to 2^n. Each number is the numbers directly above it added together. An example for how pascal triangle is generated is illustrated in below image. On the first row, write only the number 1. Named after the French mathematician, Blaise Pascal, the Pascal’s Triangle is a triangular structure of numbers. It is very easy to construct his triangle, and when you do, amazin… There is a good reason, too ... can you think of it? Step 1: Draw a short, vertical line and write number one next to it. Pascal's identity was probably first derived by Blaise Pascal, a 17th century French mathematician, whom the theorem is named after. His triangle was further studied and popularized by Chinese mathematician Yang Hui in the 13th century, for which reason in China it is often called the Yanghui triangle. Fibonacci history how things work math numbers patterns shapes TED Ed triangle. The numbers at edges of triangle will be 1. To construct the Pascal’s triangle, use the following procedure. At first it looks completely random (and it is), but then you find the balls pile up in a nice pattern: the Normal Distribution. This is the pattern "1,3,3,1" in Pascal's Triangle. Display the Pascal's triangle: ----- Input number of rows: 8 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 Flowchart: C# Sharp Code Editor: Contribute your code and comments through Disqus. It can look complicated at first, but when you start to spend time with some of the incredible patterns hidden within this infinite … The rows of Pascal's triangle are conventionally enumerated starting with row n = 0 at the top. Updates? William L. Hosch was an editor at Encyclopædia Britannica. Let us know if you have suggestions to improve this article (requires login). 1 3 3 1. Pascal’s triangle and the binomial theorem mc-TY-pascal-2009-1.1 A binomial expression is the sum, or diﬀerence, of two terms. Get a Britannica Premium subscription and gain access to exclusive content. A Pascal Triangle consists of binomial coefficients stored in a triangular array. The third diagonal has the triangular numbers, (The fourth diagonal, not highlighted, has the tetrahedral numbers.). (Note how the top row is row zero Or we can use this formula from the subject of Combinations: This is commonly called "n choose k" and is also written C(n,k). In fact there is a formula from Combinations for working out the value at any place in Pascal's triangle: It is commonly called "n choose k" and written like this: Notation: "n choose k" can also be written C(n,k), nCk or even nCk. The triangle is also symmetrical. and also the leftmost column is zero). Pascal’s triangle, in algebra, a triangular arrangement of numbers that gives the coefficients in the expansion of any binomial expression, such as (x + y) n. It is named for the 17th-century French mathematician Blaise Pascal, but it is far older. Hence, the expansion of (3x + 4y) 4 is (3x + 4y) 4 = 81 x 4 + 432x 3 y + 864x 2 y 2 + 768 xy 3 + 256y 4 It was included as an illustration in Zhu Shijie's. The formula for Pascal's Triangle comes from a relationship that you yourself might be able to see in the coefficients below. Simple! The digits just overlap, like this: For the second diagonal, the square of a number is equal to the sum of the numbers next to it and below both of those. 204 and 242).Here's how it works: Start with a row with just one entry, a 1. It is called The Quincunx. This sounds very complicated, but it can be explained more clearly by the example in the diagram below: 1 1. Notation: "n choose k" can also be written C (n,k), nCk or … Pascal's Triangle can also show you the coefficients in binomial expansion: For reference, I have included row 0 to 14 of Pascal's Triangle, This drawing is entitled "The Old Method Chart of the Seven Multiplying Squares". An amazing little machine created by Sir Francis Galton is a Pascal's Triangle made out of pegs. Magic 11's. is "factorial" and means to multiply a series of descending natural numbers. The triangle that we associate with Pascal was actually discovered several times and represents one of the most interesting patterns in all of mathematics. Pascal’s triangle is a number pyramid in which every cell is the sum of the two cells directly above. In fact, the Quincunx is just like Pascal's Triangle, with pegs instead of numbers. For example, x + 2, 2x + 3y, p - q. an "n choose k" triangle like this one. PASCAL'S TRIANGLE AND THE BINOMIAL THEOREM. To build the triangle, start with "1" at the top, then continue placing numbers below it in a triangular pattern. It's much simpler to use than the Binomial Theorem , which provides a formula for expanding binomials. Begin with a solid equilateral triangle, and remove the triangle formed by connecting the midpoints of each side. This can then show you the probability of any combination. Amazing but true. A binomial expression is the sum, or difference, of two terms. A Formula for Any Entry in The Triangle. Yes, it works! Pascal Triangle is a triangle made of numbers. The first row, or just 1, gives the coefficient for the expansion of (x + y)0 = 1; the second row, or 1 1, gives the coefficients for (x + y)1 = x + y; the third row, or 1 2 1, gives the coefficients for (x + y)2 = x2 + 2xy + y2; and so forth. Pascal’s triangle, in algebra, a triangular arrangement of numbers that gives the coefficients in the expansion of any binomial expression, such as (x + y)n. It is named for the 17th-century French mathematician Blaise Pascal, but it is far older. It is from the front of Chu Shi-Chieh's book "Ssu Yuan Yü Chien" (Precious Mirror of the Four Elements), written in AD 1303 (over 700 years ago, and more than 300 years before Pascal! In mathematics, Pascal's triangle is a triangular array of the binomial coefficients that arises in probability theory, combinatorics, and algebra. Pascal's triangle is a triangular array constructed by summing adjacent elements in preceding rows. If there were 4 children then t would come from row 4 etc… By making this table you can see the ordered ratios next to the corresponding row for Pascal’s Triangle for every possible combination.The only thing left is to find the part of the table you will need to solve this particular problem( 2 boys and 1 girl): It was included as an illustration in Chinese mathematician Zhu Shijie’s Siyuan yujian (1303; “Precious Mirror of Four Elements”), where it was already called the “Old Method.” The remarkable pattern of coefficients was also studied in the 11th century by Persian poet and astronomer Omar Khayyam. What do you notice about the horizontal sums? It is named after the 17^\text {th} 17th century French mathematician, Blaise Pascal (1623 - 1662). Chinese mathematician Jia Xian devised a triangular representation for the coefficients in an expansion of binomial expressions in the 11th century. Just a few fun properties of Pascal's Triangle - discussed by Casandra Monroe, undergraduate math major at Princeton University. To build the triangle, start with "1" at the top, then continue placing numbers below it in a triangular pattern. at each level you're really counting the different ways that you can get to the different nodes. The method of proof using that is called block walking. To build the triangle, always start with "1" at the top, then continue placing numbers below it in a triangular pattern.. Each number is the two numbers above it added … Thus, the third row, in Hindu-Arabic numerals, is 1 2 1, the fourth row is 1 4 6 4 1, the fifth row is 1 5 10 10 5 1, and so forth. Principle of Pascal’s Triangle Each entry, except the boundary of ones, is formed by adding the above adjacent elements. Pascal's triangle contains the values of the binomial coefficient. I have explained exactly where the powers of 11 can be found, including how to interpret rows with two digit numbers. Pascal's Triangle! Answer: go down to the start of row 16 (the top row is 0), and then along 3 places (the first place is 0) and the value there is your answer, 560. The entries in each row are numbered from the left beginning Pascal’s principle, also called Pascal’s law, in fluid (gas or liquid) mechanics, statement that, in a fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the fluid and to the walls of the container. …of what is now called Pascal’s triangle and the same place-value representation (, …in the array often called Pascal’s triangle…. In much of the Western world, it is named after the French mathematician Blaise Pascal, although other mathematicians studied it centuries before him in India, Persia, China, Germany, and Italy. 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Sir Francis Galton is a good reason, too... can you think of it of 11 can be,!	5,689	24,714	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.46875	4	CC-MAIN-2021-25	latest	en	0.938763
https://www.multicharts.com/discussion/viewtopic.php?t=48195	1,544,989,999,000,000,000	text/html	crawl-data/CC-MAIN-2018-51/segments/1544376827992.73/warc/CC-MAIN-20181216191351-20181216213351-00131.warc.gz	979,048,566	8,397	# Help for Create function from indicator code Studies that have been contributed to the community by other users. If you’ve got something useful to share, that’s great! valabhi Posts: 45 Joined: 25 Jun 2007 ### Help for Create function from indicator code I have Ehlers Trend Indicator code. I would like to create function so that can be used with other indicators. Can some one help. code for indicator are ; Inputs: Price((H+L)/2); Vars: InPhase(0), Phase(0), DeltaPhase(0), count(0), InstPeriod(0), Period(0), Trendline(0); If CurrentBar > 5 then begin Value1 = Price - Price[6]; Value2 =Value1[3]; Value3 =.75(Value1 - Value1[6]) + .25(Value1[2] - Value1[4]); InPhase = .33Value2 + .67InPhase[1]; {Use ArcTangent to compute the current phase} If AbsValue(InPhase +InPhase[1]) > 0 then Phase = {Resolve the ArcTangent ambiguity} If InPhase < 0 and Quadrature > 0 then Phase = 180 - Phase; If InPhase < 0 and Quadrature < 0 then Phase = 180 + Phase; If InPhase > 0 and Quadrature < 0 then Phase = 360- Phase; {Compute a differential phase, resolve phase wraparound, and limit delta phase errors} DeltaPhase = Phase[1] - Phase; If Phase[1] < 90 and Phase > 270 then DeltaPhase = 360 + Phase[1] - Phase; If DeltaPhase < 1 then DeltaPhase = 1; If DeltaPhase > 60 then Deltaphase = 60; {Sum DeltaPhases to reach 360 degrees. The sum is the instantaneous period.} InstPeriod = 0; Value4 = 0; For count = 0 to 40 begin Value4 = Value4 + DeltaPhase[count]; If Value4 > 360 and InstPeriod = 0 then begin InstPeriod = count; end; end; {Resolve Instantaneous Period errors and smooth} If InstPeriod = 0 then InstPeriod = InstPeriod[1]; Value5 = .25(InstPeriod) + .75Value5[1]; {Compute Trendline as simple average over the measured dominant cycle period} Period = IntPortion(Value5); Trendline = 0; For count = 0 to Period + 1 begin Trendline = Trendline + Price[count]; end; If Period > 0 then Trendline = Trendline / (Period + 2); Value11 = .33(Price + .5(Price - Price[3])) + .67*Value11[1]; if CurrentBar < 26 then begin Trendline = Price; Value11 = Price; end; Plot1(Trendline, "TR"); Plot2(Value11, "ZL"); end;	645	2,133	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.375	3	CC-MAIN-2018-51	latest	en	0.663464
http://www.controlglobal.com/articles/2007/097.html	1,369,168,340,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368700497024/warc/CC-MAIN-20130516103457-00091-ip-10-60-113-184.ec2.internal.warc.gz	393,979,480	13,781	# Envelope optimization ## Using envelope optimization as a tool, the next generation of process control engineers will be able to convert today's technology into a force that will turn a profit and improve the quality of our lives. 01/15/1998 1 vote Text size: - + To most of us, the term "optimization" implies complexity. It suggests partial differential equations and multidimensional peak searching. It implies the stuff that is for egghead theoreticians and not for the down-to-earth engineers in real-world processing plants. In our plants, where pipes leak, sensors plug, and pumps cavitate, the role of the plant engineer is more like that of a fireman than that of a scientist. Yet optimization is a friend of the plant engineer! It does not need to be esoteric or theoretical-it can provide practical solutions to real problems, it can maximize productivity while minimizing costs. And most importantly-because of its fully automated nature-it can actually free the plant engineer to fix those leaking pipes and pump seals. Optimization is nature's way of control: A tree, for example, is simultaneously reacting to all the variables that affect it. Similarly, in a fully optimized plant, levels, temperatures, pressures, or flows should only be constraints. They should all be allowed to float within their predetermined safe limits while the efficiency or productivity of the operation is continuously maximized. This is different from the earlier design philosophy, where levels and temperatures were held at constant values (if you keep a level constant, what is the tank for?) and unit productivity was an accidental and uncontrolled consequence of these rigidly held values. Control Evelope The key tool of optimization is the multivariable control envelope. This is a polygon with its sides representing the various constraints of the particular process. In case of a boiler, for example, one might simultaneously monitor excess oxygen, carbon monoxide, unburned hydrocarbons, stack temperature, and opacity, and assign the sides of the polygon to represent the allowable limits on each. FIGURE 1: THE CONTROL ENVELOPE The multivariable control envelope, a polygon with sides representing the constraints of the process, is the key tool of optimization. Whenever a limit is approached, for example, the CO limit on a boiler, control is switched to that controller, but as soon as the process approaches another limit, control is transferred to that other variable. Whenever a limit is approached (for example, the CO limit in Figure 1 above), control is switched to that controller, but as soon as the process approaches another limit, control is transferred from the CO controller to that other variable. Through this multivariable control strategy, the boiler is "herded" to stay within the envelope and thereby a much faster and more sensitive control is provided (see Figure 2 below). FIGURE 2: MORE SENSITIVE CONTROL In a boiler, for example, the envelope control strategy is faster and more sensitive than excess oxygen control. Another advantage of envelope control is that it provides an extra degree of freedom. This means that an additional controller can be activated when all constraints are within their limits (when the process is inside the control envelope). During these periods, we are free to manipulate the process by a new controller, which can maximize efficiency. The efficiency of combustion would be at a maximum when every carbon atom found two oxygens to unite into a CO2 molecule. Under these ideal conditions, no unreacted excess oxygen would remain to leave with the flue gases. In real combustion, the mixing of fuel with air is never perfect. Therefore some air (including its 79% nitrogen) will always enter the combustion process at ambient temperature and will travel through the boiler "just for the ride," picking up valuable heat and then wasting that heat as it leaves with the flue gases. In a combustion process, the losses are the sum of the radiation losses, the flue gas heat losses (through the stack), and the unburned fuel losses caused by incomplete combustion. For each boiler, the sum of these three loss curves (the total loss curve) has a minimum (see Figure 3 below), which identifies the point where efficiency is the maximum. When the boiler is operating inside its constraint envelope, the optimizer controller will automatically shift it towards this maximum-efficiency point. FIGURE 3: CUTTING LOSES The minimum of the sum of the loss curves identifies the air-fuel ratio point where efficiency is the maximum. The efficiency target for coal-fired boilers is 88-89%; for oil-fired, 85-87.5% ; and for gas-fired, 82-82.5%. An added benefit of such optimization strategies is that whenever the actual boiler efficiency drops below the above listed targets, the operator knows that it is time for maintenance. Finding the Optimum Just as the total loss curve of a combustion process has a minimum, so do most other processes. For example, the cost of meeting the cooling water needs of a plant can also be represented by a curve that has a minimum. Such a minimum point results whenever the total cost curve is the sum of two curves of opposite slopes. In the case of the cooling water supply system, these are the pumping and fan cost curves. If the temperature of the plant cooling water supply is reduced (approach to wet-bulb lowered), the cost of cooling water pumping also drops because less of this colder water needs to be pumped, but the cost of cooling-tower fan operation is increased because more air flow is needed to generate the cooling water supply at this lower temperature. The total cost will optimized when the sum of the fan and pump station costs is minimized. The task of the optimizer controls (see Figure 4 below) is to bring the levels of fan and pump station operations to this optimum point. This is achieved by setting the approach (Tctws-Twb) setpoint on TDIC-1 to match the minimum point on the cost curve (see Figure 5 below). FIGURE 4: OPTIMIZER CONTROLS Optimizer controls on the fan and pump station can be used to bring operations to the desired point. FIGURE 5: MINIMIZING COST Cost can be optimized by setting the approach (Tctws-Twb) setpoint on TDIC-1 to match the minimum point on the cost curve.	1,291	6,336	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.828125	3	CC-MAIN-2013-20	longest	en	0.940851
http://math.ivyglobal.com/questions/2/16/250	1,656,166,506,000,000,000	text/html	crawl-data/CC-MAIN-2022-27/segments/1656103035636.10/warc/CC-MAIN-20220625125944-20220625155944-00708.warc.gz	37,500,638	3,641	1 EASY Round 24 to the nearest 10. 2 EASY Round 40 to the nearest 10. 3 EASY Round 148 to the nearest 10. 4 EASY Round 72 to the nearest 10. 5 MEDIUM Add 16 to 32, then round your answer to the nearest 10. 6 MEDIUM Round 35 to the nearest 10. 7 MEDIUM What digit must every number that is rounded to the nearest 10 end with? 8 HARD Ms. Steinar assigned Jack a two-digit even number. Jack rounded his number to the nearest 10, added the result to 12, and rounded the sum again to the nearest 10. If the final number is 50, which of the following CANNOT be Jack's original number? 3436384042 9 HARD John rounds 48 to the nearest ten, then rounds 54 to the nearest ten, then adds his two numbers together. Jean adds 48 and 54, then rounds the result. Who's final number is bigger, John's or Jean's? JohnJeanTheir numbers are equalImpossible to determineNone of the above 10 HARD Jack rounds 27 to the nearest 10, then rounds 35 to the nearest 10, then adds the two numbers. Jacques adds 27 and 35, then rounds the sum to the nearest 10. Who's final number is bigger, Jack's or Jacques's? JackJacquesTheir numbers are equalImpossible to determineNone of the above	317	1,177	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.578125	3	CC-MAIN-2022-27	latest	en	0.885315
http://biosig-consulting.com/matlab/help/freetb4matlab/statistics/tests/z_test_2.html	1,369,039,077,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368698646863/warc/CC-MAIN-20130516100406-00080-ip-10-60-113-184.ec2.internal.warc.gz	28,206,108	1,888	Home > freetb4matlab > statistics > tests > z_test_2.m # z_test_2 ## PURPOSE % For two samples @var{x} and @var{y} from normal distributions with ## SYNOPSIS function [pval, z] = z_test_2 (x, y, v_x, v_y, alt) ## DESCRIPTION ```% -- texinfo -- % @deftypefn {Function File} {[@var{pval}, @var{z}] =} z_test_2 (@var{x}, @var{y}, @var{v_x}, @var{v_y}, @var{alt}) % For two samples @var{x} and @var{y} from normal distributions with % unknown means and known variances @var{v_x} and @var{v_y}, perform a % Z-test of the hypothesis of equal means. Under the null, the test % statistic @var{z} follows a standard normal distribution. % % With the optional argument string @var{alt}, the alternative of % interest can be selected. If @var{alt} is @code{'~='} or % @code{'<>'}, the null is tested against the two-sided alternative % @code{mean (@var{x}) ~= mean (@var{y})}. If alt is @code{'>'}, the % one-sided alternative @code{mean (@var{x}) > mean (@var{y})} is used. % Similarly for @code{'<'}, the one-sided alternative @code{mean % (@var{x}) < mean (@var{y})} is used. The default is the two-sided % case. % % The p-value of the test is returned in @var{pval}. % % If no output argument is given, the p-value of the test is displayed % along with some information. % @end deftypefn``` ## CROSS-REFERENCE INFORMATION This function calls: This function is called by: Generated on Sat 16-May-2009 00:04:49 by m2html © 2003	450	1,434	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.765625	3	CC-MAIN-2013-20	longest	en	0.702117
http://www.numbersaplenty.com/322243	1,586,149,208,000,000,000	text/html	crawl-data/CC-MAIN-2020-16/segments/1585371618784.58/warc/CC-MAIN-20200406035448-20200406065948-00433.warc.gz	262,769,994	3,259	Search a number 322243 is a prime number BaseRepresentation bin1001110101011000011 3121101000221 41032223003 540302433 610523511 72511325 oct1165303 9541027 10322243 11200119 12136597 13b389c 1485615 156572d hex4eac3 322243 has 2 divisors, whose sum is σ = 322244. Its totient is φ = 322242. The previous prime is 322237. The next prime is 322247. The reversal of 322243 is 342223. Adding to 322243 its reverse (342223), we get a palindrome (664466). It can be divided in two parts, 322 and 243, that added together give a palindrome (565). It is a strong prime. It is a cyclic number. It is a de Polignac number, because none of the positive numbers 2k-322243 is a prime. It is not a weakly prime, because it can be changed into another prime (322247) by changing a digit. It is a polite number, since it can be written as a sum of consecutive naturals, namely, 161121 + 161122. It is an arithmetic number, because the mean of its divisors is an integer number (161122). 2322243 is an apocalyptic number. 322243 is a deficient number, since it is larger than the sum of its proper divisors (1). 322243 is an equidigital number, since it uses as much as digits as its factorization. 322243 is an evil number, because the sum of its binary digits is even. The product of its digits is 288, while the sum is 16. The square root of 322243 is about 567.6645135994. The cubic root of 322243 is about 68.5584774234. The spelling of 322243 in words is "three hundred twenty-two thousand, two hundred forty-three", and thus it is an iban number.	444	1,554	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.09375	3	CC-MAIN-2020-16	latest	en	0.888941
http://ecovirtual.ib.usp.br/doku.php?id=en:ecovirt:roteiro:math:zerosumrcmdr	1,725,967,693,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651241.17/warc/CC-MAIN-20240910093422-20240910123422-00325.warc.gz	10,851,237	7,889	This translation is older than the original page and might be outdated. ## Tutorials Population Structure Dynamics Density-independent Density-dependent Strutured Population Metapopulation Single Species Two Species Community Estrutura Dynamics and Disturbances Dinâmicas Neutras Mathematics & Statistics Differential and Integral Calculus Stochastic Processes ## Visitors en:ecovirt:roteiro:math:zerosumrcmdr # Zero-sum dynamics - Tutorial for EcoVirtual The concept of a zero-sum game comes from game theory, and describes a division of a fixed amount of resource between each participant, in such a way that one player can only win what the other players lose. If we model the profits and subsequent losses as happening with a certain probability, the game acquires a stochastic dynamic, similar to the one present in the neutral theory of biodiversity. Its author, Stephen Hubbell, assumes that communities are saturated with individuals, such that a new individua may only become established if another one dies. The random succession of deaths, births, and arrival of migrants would then create a zero-sum dynamic, explaining the patterns found in natural communities. In this tutorial, we will simulate a very simple stochastic zero-sum dynamic. Then, you can study its application on our tutorial about the neutral theory of biodiversity. ## A silly game Let's imagine a simple gambling game between two players, with no ties. Every round, the player that lost the gamble pays a fixed amount to the winner. Both players have the same probability of winning at each round. This is a zero-sum game, as the total amount of money in possession of both players never changes. The only thing that changes is the fraction of that amount that each player possesses. If we allowed for different winning probabilities for winning, or for the transferred amount to change in any way, it would still be a zero-sum game. In our simulation, "the game only ends when it's over", what means that we only stop the simulation when one player has lost all its money1). To proceed, you must have the R environment with the Rcmdr and Ecovirtual packages installed and loaded. If you do not have and do not know how to have them, see the Installation page. Let's simulate this situation with the function Zero Sum Game. Open the EcoVirtual menu in Rcmdr: EcoVirtual>Biogeograph models> Zero Sum Game. The following window will open: In this function we have 3 parameters: Option Effect Total amount the total amount of money in the game\\In the start of the game, it is evenly divided between the players Bet size the amount that the loser pays to the winner Maximum game time maximum game time (in real world minutes) The argument Maximum game time is not part of the games rules - it's just a precaution against the simulation taking too long. Fix it around 10 minutes, but most simulations should end way before this time. ## What determines the game length? The simulation will run until the game is over. Vary the total amount of the game and assess what difference does it make on the game length. Some suggestions: • Total amount = 20, bet size = 1 • Total amount = 20, bet size = 5 • Total amount = 100, bet size = 1 • Total amount = 100, bet size = 5 In stochastic dynamics, the result may change every time the simulation is run, even if the parameters are exactly the same. Thus, it's important that you repeat each simulation a couple of times to be sure of the results main tendency. ### Questions 1. What is the effect of the total amount and of the bet size over the game length? 2. Is this game a process of one-dimensional random walk? Explain your answer. 1) if the simulation takes to long, you can also interrupt it en/ecovirt/roteiro/math/zerosumrcmdr.txt · Last modified: 2017/11/21 10:07 by melina.leite	844	3,835	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2024-38	latest	en	0.919965
https://gmatclub.com/forum/ds-finding-angles-74189.html	1,534,824,709,000,000,000	text/html	crawl-data/CC-MAIN-2018-34/segments/1534221217951.76/warc/CC-MAIN-20180821034002-20180821054002-00202.warc.gz	696,543,688	48,408	GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 20 Aug 2018, 21:11 ### 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 # DS: Finding Angles Author Message Manager Joined: 05 Aug 2008 Posts: 88 Schools: McCombs Class of 2012 ### Show Tags 24 Dec 2008, 16:23 00:00 Difficulty: (N/A) Question Stats: 0% (00:00) correct 0% (00:00) wrong based on 0 sessions ### HideShow timer Statistics Attachment: Q1.JPG [ 45.95 KiB \| Viewed 985 times ] --== Message from GMAT Club Team ==-- This is not a quality discussion. It has been retired. If you would like to discuss this question please re-post it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative \| Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you. SVP Joined: 07 Nov 2007 Posts: 1728 Location: New York ### Show Tags 25 Dec 2008, 00:04 smarinov wrote: Attachment: Q1.JPG assume Angle PQR=x Angle QPR=z Angle PRQ=y statement 1) z=30 x+y=150 Question Angle PRS - Angle PQR = (180-y)-x = 180- (x+y) = 180-150 =30 Sufficient Statment 2) same as above suffcient D _________________ Smiling wins more friends than frowning Manager Joined: 10 Aug 2008 Posts: 72 ### Show Tags 25 Dec 2008, 10:40 Statement 1 ------------- PRS = QPR + PQR PRS - PQR = QPR = 30 Sufficient Statement 2 ------------- PQR + PRQ = 150 => QPR = 30, because sum of angle of a triangel is 180 Hence again SUFFICIENT ! Ans is D. --== Message from GMAT Club Team ==-- This is not a quality discussion. It has been retired. If you would like to discuss this question please re-post it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative \| Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you. Re: DS: Finding Angles &nbs [#permalink] 25 Dec 2008, 10:40 Display posts from previous: Sort by # Events & Promotions Powered by phpBB © phpBB Group \| Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.	735	2,701	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.0625	4	CC-MAIN-2018-34	latest	en	0.856726
http://www.thestreet.com/story/12226463/1/markets-will-tilt-downward-in-early-2014.html?cm_ven=RSSFeed	1,455,438,914,000,000,000	text/html	crawl-data/CC-MAIN-2016-07/segments/1454701171770.2/warc/CC-MAIN-20160205193931-00337-ip-10-236-182-209.ec2.internal.warc.gz	688,194,241	29,239	This Day On The Street Continue to site This account is pending registration confirmation. Please click on the link within the confirmation email previously sent you to complete registration. # Markets Will Tilt Downward in Early 2014 NEW YORK (TheStreet) -- So far in 2014, the Dow Jones Industrial Average (^DJI) has dropped for five of the first seven trading days. Conversely, this index has risen for two of the first seven trading days, for a simple 28% probability of an up day. If markets are simply continuing last year's upward march, how do these patterns fit in so far? How do these early year-to-date market statistics match up versus prior years? And what do they imply about the Bayesian tilt with which 2014 has begun? First we look at the theoretical distribution provided by a probability tree, in which each successive day has an equal 50% chance of either branching up or branching down. Then, for a specific tree path over the first seven days, a red value was given to indicate a down day, while a green value was given for an up day. For more on combinatorics, see this note We see that the first trading day for 2014 was down. And this had a 50% theoretical chance of happening. Then the second trading day was up, and so we end that day with a score of one down day out of two days. This score too has a 50% chance as shown (e.g., there is another 25% chance we'd instead have two up days in a row, and a 25% chance to instead have two down days in a row). Through this morning, Jan. 13, we've collected five down days out of seven trading days. We see on the tree above, that this outcome had a 16% chance of occurring. We also see that the probability of having five or more down days, out of seven trading days, is slightly larger at 23% (or ~16%+5%+1%). A different interpretation of this 23% chance is that over the 14 years prior to this one, three of them (23%*14) should have also had at least 5 down days out of the first 7 trading days. And empirically this is true: 2001, 2005, 2009. That's some company. But these probabilities were again derived assuming a fair 50% chance for either an up day or a down day. That indicates a close to trendless market, which lacks general direction, even though we technically understand that, over the long run, markets offer a slight upward trend. What would have been the probability of seeing five drops out of seven trading days, if we instead had assumed different up-day probabilities, ranging from 25%, to 75%? We see in the chart below, that as our assumption for the daily chance for an up day increases, the probability of having five drops (out of seven trading days) falls well below our 16% baseline. Put differently, having five drops out of seven is less probable (from 16% to 6%) if we switch from assuming a fair 50% up-day probability to a slight up streak characterized by 60% up-day probabilities. Conversely, as the daily chance for an up day decreases, the probability of having five drops (out of seven trading days) rises well above our 16% baseline. Let's then use our assumed probability for an up day to better explore our likelihood of outcome. We know that the chance that we are in a general phase that is far from "fair" (i.e., normal 50% probability of an up day) decreases the farther away the up-day probability is from 50%. Being in a strong up streak (e.g., 75% up-day probability) or in a strong down streak (e.g., 25% probability of an up day) for extended periods is simply not as likely versus being in a phase closer to a 50% up-day probability. 1 of 2 ### Check Out Our Best Services for Investors Portfolio Manager Jim Cramer and Director of Research Jack Mohr reveal their investment tactics while giving advanced notice before every trade. ###### Product Features: • \$2.5+ million portfolio • Large-cap and dividend focus ##### Quant Ratings Access the tool that DOMINATES the Russell 2000 and the S&P 500. ###### Product Features: • Buy, hold, or sell recommendations for over 4,300 stocks • Unlimited research reports on your favorite stocks • A custom stock screener ##### Stocks Under \$10 David Peltier uncovers low dollar stocks with serious upside potential that are flying under Wall Street's radar. ###### Product Features: • Model portfolio 14-Days Free Only \$9.95 14-Days Free David Peltier identifies the best of breed dividend stocks that will pay a reliable AND significant income stream. ###### Product Features: • Diversified model portfolio of dividend stocks ##### Trifecta Stocks Every recommendation goes through 3 layers of intense scrutiny—quantitative, fundamental and technical analysis—to maximize profit potential and minimize risk. ###### Product Features: • Model Portfolio ##### Real Money More than 30 investing pros with skin in the game give you actionable insight and investment ideas. Only \$49.95 14-Days Free 14-Days Free	1,114	4,899	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.546875	3	CC-MAIN-2016-07	latest	en	0.961199
http://math.stackexchange.com/questions/tagged/probability-distributions+functional-equations	1,405,314,638,000,000,000	text/html	crawl-data/CC-MAIN-2014-23/segments/1404776439950.90/warc/CC-MAIN-20140707234039-00064-ip-10-180-212-248.ec2.internal.warc.gz	74,485,907	12,293	# Tagged Questions 88 views ### Tricky probability problem I am having trouble with proving the following assertion: $X,Y$ are i.i.d. with mean $0$ and variance $1$. If $X+Y$ and $X-Y$ are independent then $X,Y$ are normally distributed. Should I be ... 51 views ### Probability distribution satisfying constraints? Continuing from this question. Given two random variables $X$ and $Y$ where $X \sim \operatorname{Beta}(a, b)$ and $Y \sim \operatorname{Beta}(c, d)$, I'm looking for a random variable $Z$ with a ... 45 views ### Differential Equation for CDF Consider the following differential equation $$F(cx) = F(x) + x F'(x)$$ for $c>1$. Does this differential equation belong to a some well known class? Is there a way to find all the solutions ... 73 views ### How to solve a functional equation of the form $1-g(f(s))=m(1-g(s))$? I have arrived to this equation in several contexts within branching processes. It arises from textbook exercises, so it must be solvable somehow. Here $f$ is a probability generating function which ... Maxwell's theorem (after James Clerk Maxwell) says that if a function $f(x_1,\ldots,x_n)$ of $n$ real variables is a product $f_1(x_1)\cdots f_n(x_n)$ and is rotation-invariant in the sense that the ...	325	1,251	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.953125	3	CC-MAIN-2014-23	longest	en	0.878618
https://howmanyzeros.com/zeros-in-4-lakhs	1,709,483,226,000,000,000	text/html	crawl-data/CC-MAIN-2024-10/segments/1707947476396.49/warc/CC-MAIN-20240303142747-20240303172747-00390.warc.gz	301,665,521	21,522	Home » Number of Zeros » How many Zeros in 4 Lakhs? # How many Zeros in 4 Lakhs? If you have been wondering about how many zeros in 4 lakhs?, then you have come to the right post. Apart from the answer to this question, we provide you with related information regarding the number of zeros in 4 lakhs. Read on to learn all about the amount of 0s in 4L, and make sure to check out our app right below. Reset ## How many Zeros in Four Lakhs? 4 lakhs in figures equals 400000, or 4,00,000 when written in the Indian numbering system, which makes it easier to count the occurrences of 0. Thus, we get: Zeros in 4 Lakhs = 5 5 is the answer to 4 lakhs has how many 0s? Similar questions on our website include, for example: If you like to learn how many zeros for a numeral different from 4 lakhs, fill in our application at the beginning of this post. After inserting your number, our tool displays the amount of 0’s automatically. In the next part of how many zeros in four lakhs, we show you how many 100 and 1000 there are in 4 lakhs, and other related information. ## 4 Lakhs has how many Zeros? You already know the answer to this question, but we are left with telling you how many 10, 100, 1000 et cetera there are in four lakhs: • How many tens in 4 lakhs? Answer: 40,000 tens. • How many hundreds in 4 lakhs? Answer: 4,000 hundreds. • How many thousands in 4 lakhs? Answer: 400 thousands. • How many ten thousands in 4 lakhs? Answer: 40 ten thousands. • How many hundred thousands in 4 lakhs? Answer: 4 hundred thousands. • How many millions in 4 lakhs? Answer: 0.4 million. In the concluding section ahead we wrap how many 0 in four lakhs up, and explain the use of our search form located in the sidebar and our menu. ## Summary The image sums how many zeros in 4 lakhs? up: Comments, questions and everything else you might have about how many 0 in 4 lakhs are truly appreciated, and can be left in the designated form at the bottom. As an alternative means, you can send us an email with a meaningful title, and we will get back to you as soon as possible. Observe that by means of our search form in the menu you can learn about the number of zeros for lots of numerals including 4L. Insert, for example, number of zeros 4 lakhs. If our information about the number of 0’s in 4L has been useful to, don’t forget to bookmark our site and to hit some of the share buttons at your disposal. Thanks for visiting us. – Article written by Mark	638	2,472	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.65625	4	CC-MAIN-2024-10	latest	en	0.941559
http://www.chegg.com/homework-help/questions-and-answers/fig-4-38-ball-launched-velocity-850-m-s-angle-50-horizontal-launch-point-base-ramp-horizon-q950605	1,469,751,596,000,000,000	text/html	crawl-data/CC-MAIN-2016-30/segments/1469257829320.91/warc/CC-MAIN-20160723071029-00002-ip-10-185-27-174.ec2.internal.warc.gz	369,192,803	13,317	In Fig. 4-38, a ball is launched with a velocity of 8.50 m/s, at an angle of 50° to the horizontal. The launch point is at the base of a ramp of horizontal length d1 = 3.70 m and height d2 = 3.60 m. A plateau is located at the top of the ramp. (a) Does the ball land on the ramp or the plateau? - ramp or plateau (b) What is the magnitude of its displacement from the launch point when it lands? --------m (c) What is the angle of its displacement from the launch point when it lands? (counterclockwise from the horizontal axis directed to the right)	146	553	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.765625	3	CC-MAIN-2016-30	latest	en	0.914568
https://bookiesoft.com/gamble/you-asked-what-does-perfect-pair-pay-in-blackjack.html	1,642,700,437,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320302355.97/warc/CC-MAIN-20220120160411-20220120190411-00227.warc.gz	187,599,713	18,949	# You asked: What does perfect pair pay in blackjack? Contents Perfect Pair (e.g. Jack hearts – Jack hearts): This is a pair of cards of the same rank AND suit. It pays 30/1. Perfect Pairs strategy and tips. It’s important to understand how many cards are available when you play Perfect Pairs European Blackjack. A standard online Perfect Pairs game may use two decks of cards. ## What does perfect pair mean in blackjack? Perfect Pairs Blackjack is a variation of the standard Blackjack game that allows the players to place an additional bet to cover the possibility of the first two cards dealt, being a pair. A Perfect Pairs bet may only be placed on a Perfect Pairs Blackjack table. ## What is a 3 2 payout in blackjack? Understand what “Blackjack 3 to 2” means. Somewhere on the blackjack table there will be a sign that says, “blackjack pays 3 to 2”. All this means is that you, the player, will get \$3 for every \$2 you wager. This is standard, and gives the house slightly elevated odds. ## Does blackjack always pay 3 to 2? If you have a winning blackjack hand, you get paid 3 dollars for every 2 that you bet, or 1.5:1 odds. In 6:5 you get paid \$6 for every \$5 you bet, which is 1.2:1 odds. ## What is PP in Black Jack? Perfect Pairs is an optional side bet placed at the same time as the opening wagers in the basic game. Perfect Pairs is a wager that the first two cards dealt to a hand will be a pair of the same value (for example, a pair of twos, threes, four etc). ## Is perfect pairs a good bet? Perfect Pairs strategy and tips Therefore, the odds of you hitting two cards the same (a “perfect pair”) are better than the 30/1 offered in the payouts. Even if a shoe has 6-8 decks, the chances of being dealt two cards the same are in the thousands. The side bets in Perfect Pairs add a different element to the game. ## What are the odds of getting any pair in blackjack? What are the real odds of getting a perfect pair? Taking a game using eight decks of cards for an example, there are seven cards out of the 415 in the dealer’s shoe that can make the player a perfect pair. This equates to just over a one-in-59 chance of the card you need coming up. ## What does 1 to 9 odds pay? Standard Win Bets and Payouts Odds \$ Payout Odds 1/9 \$2.20 7/1 1/5 \$2.40 8/1 2/5 \$2.80 9/1 1/2 \$3.00 10/1 ## Which blackjack has the best odds? Single-deck Blackjack offers the best odds of any online casino game in the US. The house edge stands at just 0.13%. Using just one deck instead of six or eight makes the biggest difference. The house edge is also brought down by the dealer standing on soft 17. IMPORTANT: Quick Answer: What is casino laundering? ## What do the odds 3 2 mean? 3 to 2 means you will get paid: 3 units if you bet 2. 6 units if you bet 4. 12 units if you bet 8. So, on the 6/5 you have to bet 10 to get 12 while on the 3/2 you have to bet 8 to get 12. ## What happens if dealer and player both have blackjack? If you get Blackjack, the dealer pays you 3 to 2. If you and the dealer both get Blackjack, it is a push and no chips are given or taken away. If you have a higher total than the dealer (or the dealer busts), the dealer matches the amount of your chips.	843	3,226	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.71875	3	CC-MAIN-2022-05	latest	en	0.934633
https://www.coursehero.com/file/5989635/C14-activities/	1,529,669,534,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267864391.61/warc/CC-MAIN-20180622104200-20180622124200-00076.warc.gz	782,603,411	417,530	C14-activities # C14-activities - ACTIVITY 14-1 3 k ACTIVITY 14-1 i 5 F 1 k... This preview shows pages 1–2. Sign up to view the full content. 1 ACTIVITY 14-1 C v + i C v + 40 V 50 V 3 k Ω 1 k Ω 4 k Ω 5 F μ Switch closed for a long time Opens at t 0 = i C Find i(t) and v (t) for t 0 ACTIVITY 14-1 Write Down v c (t), i(t) for t > 0: t/ C CSS C0 CSS v (t) v (v v ) e τ = + t/ SS 0 SS i(t) i (i i ) e τ = + C0 0 CSS SS Need to find: v , i , v , i , τ Follow a 4 Step Approach 0 t 0 = ACTIVITY 14-1 Step 1 40 V 50 V 3 k Ω 1 k Ω 4 k Ω Draw Circuit at t 0 = C Find i(0 ), v (0 ) i(0 ) C v (0 ) + 0 50 40 V source and 3 k resistor do not affect circuit Ω Open Circuit Switch Closed ACTIVITY 14-1 50 V 1 k Ω 4 k Ω i(0 ) 50 i(0 ) 10 mA 1 4 = = − + C v (0 ) 10 x 4 40 V = − = − C v (0 ) + DC Steady State i(0 ) Step 1 t 0 = C v (0 ) + = ACTIVITY 14-1 Step 2 Draw Circuit at t 0 + = 0 C0 Find i , v 40 V C0 C v v (0 ) 40 V = = − 40 50 V 3 k Ω 1 k Ω 4 k Ω 5 F μ C0 v + 0 i 0 40 Open 0 40 ( 40) i 20 mA 3 1 − − = = + ACTIVITY 14-1 v c Cannot Change Instantaneously: => v c (0 + ) = v co = v c (0 - ) = - 40 V i Can Change Instantaneously: Changed from -10 mA at t = 0 - to + 20 mA at t = 0 + i is Not the Current through an Inductor This preview has intentionally blurred sections. Sign up to view the full version. View Full Document 2 ACTIVITY 14-1 Step 3 40 V 50 V 3 k Ω 1 k Ω 4 k Ω Open Ckt This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]} ### What students are saying • As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students. Kiran Temple University Fox School of Business ‘17, Course Hero Intern • I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero. Dana University of Pennsylvania ‘17, Course Hero Intern • The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time. Jill Tulane University ‘16, Course Hero Intern	806	2,425	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2018-26	latest	en	0.802115
https://grantland.com/the-triangle/red-sox-pitching-stats-sample-size-regression-mean/	1,569,155,314,000,000,000	text/html	crawl-data/CC-MAIN-2019-39/segments/1568514575513.97/warc/CC-MAIN-20190922114839-20190922140839-00125.warc.gz	502,940,369	20,161	## Is It Time to Give Up on the Red Sox Pitching Staff? What Statistical Thinking Can Tell Us Rich Gagnon/Getty Images Here are some troubling numbers: Clay Buchholz’s ERA is 5.73; Wade Miley’s is 5.60. Collectively, the Boston Red Sox’s staff ERA is 4.90, second worst in the majors. By the time Justin Masterson walked off the mound on Tuesday night, his ERA had risen to 6.37. The staff WHIP is 1.43, third worst in baseball. Sox pitchers have allowed 38 home runs, fourth most in the league. Maintaining calm in the face of numbers like these is difficult, especially on the heels of a losing season and an offseason in which the coveted pitchers — James Shields, Cole Hamels, and most especially the beloved Jon Lester — got away. As Sox starters continue to get pounded, the fans are left to wonder and argue about what to believe. Baseball’s statistical revolution arrived long ago, but a lot of people still don’t know what kind of confidence to place in numbers. Even while sophisticated baseball statistics like OPS, BABIP, and UZR1 have grown in popularity, statistical thinking remains relatively obscure to most fans. To put this more plainly: It’s early May, and Buchholz’s ERA (or his FIP!)2 can still come down. It’s a small sample size, as someone on a radio show is possibly saying right this minute. But what exactly does that mean? ## Size Matters, But Small Samples Aren’t Meaningless A sample is just a set of individual figures from a population. As Aubrey Clayton, an insurance risk analyst with a PhD in mathematics (and a lifelong attachment to the Rangers), explains it, the popular phrase “small sample size” does not have a specific meaning. The size matters, he says, not because smaller samples are worthless, but because they change what kinds of questions we can ask and answer about an event. “The amount of information you can extract from [a sample] depends on how big it is,” Clayton says. In other words, it would be a mistake to go to a ballgame, watch Wade Miley get shelled, and chalk it up as meaningless. But the information you gather is cumulative — you get a little bit more each time Miley takes the hill. More than that, Clayton explained, you can’t properly draw conclusions from statistical data unless you start with a hypothesis. Rather than looking at Miley’s bad outing and trying broadly to decide how good he is, a statistician would come to the park with an expectation, then use what he sees to evaluate its accuracy. This may sound like a pedantic point, but it’s actually clarifying, maybe even comforting, because we do have an expectation for how good Wade Miley is, based on what statisticians call prior information, i.e., Miley’s career so far. And in fact, our expectation is based on far better data than what we have for 2015; Miley has pitched fewer than 40 innings in a Red Sox uniform, but more than 600 innings with the Diamondbacks, during which time he had an ERA of 3.79. This means we have a strong hypothesis for him, namely that he is a pitcher who gives up fewer than four runs every nine innings. It also means that he has stretches when it might seem like he gets battered every fifth night. The same goes for Buchholz, Masterson, and Joe Kelly, all of whom are currently sporting ERAs well above their career averages, and none of whom has pitched even 40 innings in this humbling stretch. ## Data Samples Have Properties Other Than Size Is it a coincidence that three of the five Sox starters are also new to the team and the AL East? There are many senses in which the data we’re collecting right now — the outcomes of the first 34 games of the 2015 season — may not be representative of the starting rotation’s collective performance over the long term. Generalizing from an unrepresentative sample is another classic statistics mistake. As Clayton explained, if you wanted to test a hypothesis that 2 percent of Americans had red hair, it would be a mistake to sample only people with freckles. Accounting for such problems gets arcane pretty quickly. You’ve got to mathematically determine whether your sample is representative of the overall population and then mathematically correct for it if you want to make a useful generalization from your data. Changes like pitching in a new park, with a new set of teammates, or in a new league need to be taken into consideration. They can affect performance significantly enough to change future expectations. Remember, for instance, that John Lackey allowed the most earned runs in baseball in 2011 before turning right back into his old self after elbow surgery. Health is surely the most significant factor that affects a sample’s representativeness, but it is not the only one. All this is why good statistics sites like FanGraphs try to look for statistics that are subject to fewer variables, such as FIP. Buchholz’s FIP is notably far better than his ERA, which is a good reason for optimism in his case. His most recent turn, a 6.1-inning, three-run effort in Toronto, may well be more predictive than some of the early-inning flameouts he’s had in other games. In fact, it’s more or less in line with his career averages. Which brings us to one last point: ## Regression to the Mean Is Not Regression Below the Mean Confusion over the concept of regression to the mean is widespread, and not just in baseball circles. It is common in casinos, where it has acquired the name the gambler’s fallacy. The general (mistaken) idea here is that when an abnormal pattern occurs, it must be offset or corrected by some opposite abnormal pattern in the future. For instance, if Joe Kelly normally gives up four home runs in a month, and for two months he gives up six each, he’ll be likely to give up two in each of the two months following. In plain English, this idea sounds ridiculous, because it is ridiculous. Or at least, it asks a lot from our notions of cause and effect. How could two bad months of pitching create two good months of pitching? Regression to the mean exists, of course, and is a valuable statistical concept. But it looks different from what some eager gamblers may hope for. Imagine a roulette wheel, Clayton said. “If you have a sample where you’ve got an abnormally high percentage of red spins on a roulette wheel, it is now likely that over the next period of time the number of reds you get will be less than it was in that sample … but not necessarily less than it always has been on average.” If you have 100 roulette spins that turn up 75 percent red, the next 100 spins are likely to regress to 50 percent red. The converse is just as true, too: If your first 100 spins are 25 percent red, the next hundred spins are likely to regress to 50 percent. That’s still regression to the mean, even though we tend not to talk about it that way. Underperforming players are just as likely to improve as overperforming players are to revert. If they weren’t, the mean wouldn’t be the mean. So while we can’t assume that an underperforming pitcher like Kelly is “due” for a few months of his best baseball, we can reasonably expect that he’ll play like his old self. Whether that’s enough to win the AL East is another question, especially given that the Sox have put themselves in a hole. But it’s still the most useful predictor of what we’ll get from Kelly from here on out. And the same goes for his fellow aces, Buchholz, Miley, Masterson, and Rick Porcello. Statistically, then, it would seem that patience is the right approach, which is perhaps why Sox GM Ben Cherington suggested recently that the team isn’t likely to shake up its rotation right now. Then again, the team seems to think change will help, since pitching coach Juan Nieves was recently given his walking papers. Boston’s logic seems obvious enough — as noted above, four of the five starters have ERAs much higher than their career numbers would predict. But they have reasons to maintain confidence in their hypotheses about those players’ capabilities. To really test this thinking, consider Masterson, whose career ERA, established over more than 1,100 lifetime innings, is tied with Porcello for highest of the group at 4.31. He’s got the second-highest ERA on the team, up to 6.37 after a two-inning performance Tuesday night against the A’s. He’s over 30 (just barely), and his pitch velocity is down (as per an appropriately headlined article at SB Nation). It’s tempting to say that Masterson is finished — and the early word Wednesday is that he is headed for the disabled list. But this likely has more to do with those velocity numbers than Masterson’s ERA, because pitch velocity isn’t affected by as many variables. Thus it’s much easier to establish that a small sample of this number is representative, at least in terms of thinking about who Masterson is now. A likely scenario is that he’s got some kind of injury. If Masterson gets his arm strength back, don’t be distracted by the current outcomes; the prior information on Masterson’s career is far more substantial than any six-week stretch of games. From April 9 to May 13 (i.e., his season so far), Masterson has surrendered 25 earned runs in 35.1 innings; remarkably, from April 17 to May 13, 2012, he also gave up 25 earned runs in 35.1 innings. This didn’t stop him from having a (marginally) acceptable 4.78 ERA the rest of that year, or from having one of the best seasons of his career, including his only All-Star appearance, in 2013. In fact, that 2012 season illustrates that even an entire year of below-average pitching may only incrementally change your expectations for a pitcher with a substantial track record. If you want to establish a timeline for giving up on a player’s season, you have to take into account other factors: the quality of available replacements, say, or balancing the team’s other needs. What the numbers can tell you is not what you might expect: be patient. Regression can be a good thing. Ben Adams (@bendadams) is a book editor at PublicAffairs. His writing has appeared at Vice Sports, SB Nation, and Sports on Earth, among other places. See all from See all MLB ## More The Triangle • ### NHL Grab Bag: Let’s Get Spooky October 30, 2015 See all The Triangle	2,285	10,232	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.78125	3	CC-MAIN-2019-39	longest	en	0.946837
https://www.nafems.org/blog/posts/nbc07-solution/	1,561,029,835,000,000,000	text/html	crawl-data/CC-MAIN-2019-26/segments/1560627999210.22/warc/CC-MAIN-20190620105329-20190620131329-00400.warc.gz	835,738,917	9,529	Solution - The NAFEMS Benchmark Challenge 07 February 21st, 2017 Often, when faced with a commercial FE system, the engineer will find that it contains a veritable plethora of beam elements each purporting to be suitable for different type of beam. Different beams will typically be based on different beam theories and different FE formulations of the particular theory. For a given beam problem, the different elements will often produce quite different results and often mesh refinement is required in order to home in on the theoretical solution. This challenge presented two problems, the first of which was used as a software verification problem. Based on the mesh independence of the p-type refinement study conducted on the first problem, the same mesh was used for a design problem in which the load was moved from the center to the three-quarter point. However, the results for the design problem created using the software function for tabulating element data showed moments different to those reported at the nodes. The difference observed in the results led to the challenge of determining the quantities of engineering interest, maximum deflection and moment, for the design problem. Two Beam Theories The primary actions seen in beams are shear and bending. For ‘thin’ beams, where the span to thickness ratio is large, the deformations are predominantly due to bending. However, as the span to thickness ratio decreases then deformations due to shear become significant and need to be accounted for both in terms of deflections. The Euler-Bernoulli (EB) theory ignores shear deformation and is thus appropriate to ‘thin’ beams whereas the Timoshenko theory includes a representation of shear deformation and is thus appropriate for thicker beams. The kinematics of the two beam formulations differ. Both formulations assume that plane sections remain plane but whereas the EB formulation assumes that sections normal to the neutral axis in the undeformed state remain so, in the deformed state, this condition is relaxed for the Timoshenko (T) formulation as illustrated in Figure 2. Figure 2: Euler-Bernoulli and Timoshenko kinematic assumptions - nafe.ms/2jwa57f Theoretical Solutions to the Design Problem The EB solution for the design problem is available in many structural engineering texts and is presented in Figure 3. Figure 3: Euler-Bernoulli solution to the Design Problem - nafe.ms/2jw1Vvu The theoretical solutions for the design problem using the two-beam theories are shown in Table 1. In addition to the significant increase in the displacement under the load with shear deformable theory, the statics, in terms of the shear forces and moments, also change. The theoretical displaced shapes (exaggerated by scale) for the two theories are shown in Figure 4. Figure 4: Theoretical Displaced shapes for the Design Problem Finite Element Solutions to the Design Problem Many finite elements have been formulated based on the two beam theories already discussed. Perhaps the most common beam element based on the EB theory is the two-noded, Hermitian element which interpolates displacement as a cubic polynomial, i.e., it is capable of modelling the linear bending moments of the design problem exactly. In terms of the T theory then there is more variation in the formulation of available elements. The most widely available element is, perhaps, that based on the Mindlin formulation. This element is often of variable degree and interpolates both the displacement and rotations in the same manner, e.g., if the element is a linear element (p=1) then both displacements and rotations are interpolated linearly. The FE code used to generate the results for this response was of the Mindlin formulation with linear, quadratic and cubic degrees. It is worth noting that a finite element system should only require a single beam element if the same element could be used reliably for both ‘thick’ and ‘thin’ beams. There is, however, a numerical issue with conforming (displacement) finite elements (CFE) formulated on T theory when the beam becomes ‘thin’. This issue is known as shear-locking and it can pollute the results of ‘thin’ beams. For this reason, many FE systems offer elements based on both beam theories and this at least allows the engineer to compare the results produced for both theories and confirm whether or not shear-locking is influencing the results. FE results, using a two element mesh of variable degree Mindlin elements, for both the problems considered in the challenge are presented in Tables 1 and 2. It is seen (Table 2) that whilst the shear forces and moments do not change with p-type refinement, the displacements under the load do. There is a significant change between p=1 and p=2 and only a small change, in the fifth significant digit, for p=3. The assumption, made in the challenge, that mesh independence was obtained for p=1, is clearly erroneous. For the design problem, with the load at the three-quarter position, both shear forces, moments and displacements change as the element degree is increased between linear and quadratic. However, mesh independence does appear to be observed as the results for the cubic element are identical to those for the quadratic element. Discussion In the design problem the beam is moderately thick with a span to thickness ratio of 10 and the maximum deflections are 0.32 and 0.47mm respectively for the EB and T theories. Thus, if the EB displacement had been taken, the maximum deflection would have been underestimated by some 30% and this could make the difference between the beam passing and failing an SLS check on maximum deflection. Theoretical solutions for the design problem were obtained using both beam theories. For both theories, the maximum deflection was seen to occur away from the point load, at 0.60m and 0.65m respectively for the EB and T theories. Many commercial FE systems only report displacements at nodes and unless the maximum displacement occurs at the node then it will not be available to the engineer. The theoretical EB solution for the design problem can, as already noted, be recovered exactly using two cubic Hermitian beam elements. However, in order to recover the maximum displacement the engineer would have had to perform mesh refinement to ‘home in’ on the maximum displacement. This is an example of how poorly implemented post-processing in commercial software can frustrate the engineer’s task. Had the engineer (erroneously) used an element based on EB theory and taken the maximum displacement from a two-element model then he would have obtained 0.25mm which is almost 50% below the correct value! Of course, if one is adopting an inappropriate mathematical model in addition to not picking up the maximum displacement then simulation governance, the matching of numerical simulation with measured results, will be impossible. The design problem is hyperstatic (statically indeterminate) thus the different kinematic assumptions of the two beam theories, which result in different beam stiffnesses, also, in addition to the displacements, lead to the different forces and moments. Both sets are, however, in equilibrium with the applied load. The maximum moment for both theories is at the right-hand support and it is seen that the EB theory predicts a moment of 14.06kNm whereas the T theory is about 3% less at 13.64kNm. In an Allowable Stress Design approach, this difference would lead to different factors of safety but in a ULS calculation of the plastic limit load, the collapse load would be unchanged. For this response, a variable degree Mindlin element was used to model the Timoshenko beam theory. The advice offered by the vendor for this element is that the quadratic (p=2) element is capable of representing linearly varying bending moments exactly. The results almost bear this out except that there is a small difference in the displacement under the load for the Software Verification Problem – see Table 2 – as the degree is increased from quadratic to cubic. The quality of the result for the p=2 Mindlin element is though somewhat surprising since we know, from the theoretical solution shown in Figure 4, that the displacement field is more or less cubic. In investigating this apparent anomaly further, the displacement for a three-noded, quadratic Mindlin element was compared with the exact Timoshenko solution – see Figure 5. Figure 5: Comparison of theoretical displacements with quadratic Mindlin element The nodal displacements are, more or less, exact but clearly since they are quadratic then between nodes there is significant discrepancy. The moments inside the element are examined to see whether or not they agree with those reported at the nodes - Figure 6. In the case of the linear Mindlin element, a single integration point is used and so the variation of the internal moment field is assumed to be constant. This leads to significant differences between the internally generated moments and the nodal moments. Whilst the nodal moments are in equilibrium with the applied load, the internal moments, extrapolated from integration points, are clearly not. For the quadratic element, two integration points are used and it is seen that these appear to be exact as a linear extrapolation to the nodes leads to the same values as the nodal moments. Thus, in responding to the challenge, it might be noted that with a two-element mesh of quadratic Mindlin elements, a very close approximation to the theoretically exact solution is obtained. It is noted, however, that the maximum displacement is not available from this mesh and it has already been noted how this inadequacy might stymy the engineer’s task. The same is, of course, true for moments. It is rather easy to construct a problem where the maximum moment occurs somewhere between nodes. Thus both serviceability and ‘strength’ calculations, based on maximum moment, might be compromised for the engineer by the inadequacies of commercial software. In closing this response, it is important to recognise a potential cause of finite element malpractice when using Mindlin-type elements. When using a linear element the internal moments (extrapolated from integration points) would not have been the same as the nodal values. In fact, in this example, they would have been significantly less than the true values - see Figure 6(a). The erroneously underestimated moments would lead, in an Allowable Stress Design, to an overestimation of the factor of safety and this is clearly of concern to the engineer. The reason that this point is mentioned is that in some industries the standard technique for assessing structural members is based on internal stress resultants extrapolated from integration points. As demonstrated in this response, if the degree of the element is not appropriate for the loading seen by the beam then this approach can lead to significantly erroneous stress resultants that would, under code assessment, give an erroneous view of the safety of the structure. The solution to this potentially significant issue, is to always use stress resultants calculated directly at the nodes from the basic equilibrium equations. These are guaranteed to be in equilibrium with the applied loads even if the degree of the element is inappropriate to the applied loading, and provided sufficient mesh refinement (either/both p-type and h-type) has been undertaken then these resultants will form an appropriate set on which the structure can safely be assessed. Download all the Benchmark Challenges and their solutions available so far in the Education and Training Working group section of the NAFEMS website. Empty List Sorry nothing to list	2,347	11,758	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2019-26	latest	en	0.950028
http://mathhelpforum.com/advanced-statistics/72254-unbiasedness-print.html	1,526,957,522,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794864622.33/warc/CC-MAIN-20180522014949-20180522034949-00388.warc.gz	185,251,328	2,650	# Unbiasedness • Feb 6th 2009, 09:05 PM eigenvector11 Unbiasedness A sample of size 1 is drawn from the uniform pdf defined over the interval [0, theta]. Find an unbiased estimator for theta^2. Hint: is theta hat = Y^2 unbiased? • Feb 7th 2009, 02:40 AM mr fantastic Quote: Originally Posted by eigenvector11 A sample of size 1 is drawn from the uniform pdf defined over the interval [0, theta]. Find an unbiased estimator for theta^2. Hint: is theta hat = Y^2 unbiased? Use the hint! $\displaystyle E(Y^2) = k \theta^2$ therefore $\displaystyle E\left( \frac{Y^2}{k}\right) = \theta^2$ therefore $\displaystyle \frac{Y^2}{k}$ is an unbiased estimator for $\displaystyle \theta^2$. I leave it to you to get the value of k by solving an appropriate integral.	229	762	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3	3	CC-MAIN-2018-22	latest	en	0.771011
https://edustrings.com/mathematics/1700557.html	1,660,720,209,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882572870.85/warc/CC-MAIN-20220817062258-20220817092258-00001.warc.gz	232,465,179	7,372	1 March, 18:17 # Best Deal 1Quantity of Flour Price3 pounds \$5.255 pounds \$9.757 pounds \$12.6010 pounds \$14.2012 pounds \$18.24A baker purchases flour each week from a wholesale warehouse. The chart shows the quantities and prices available this week. If the best deal last week for flour was \$1.60 per pound. What is the difference between the unit price for last week's best deal and this week's best deal?A) 8 centsB) 15 centsC) 18 centsD) 20 cents 0 1. 1 March, 19:30 0 C) 18 cents Step-by-step explanation: Given, This weeks deal - Quantity of Flour Price (\$) Price per pound (\$) 3 pounds 5.25 (5.25/3) = 1.75 5 pounds 9.75 (9.75/3) = 1.95 7 pounds 12.60 (12.60/3) = 1.80 10 pounds 14.20 (14.20/3) = 1.42 12 pounds 18.24 (18.24/3) = 1.52 The lowest price per pound is \$1.42. Therefore, this is the current week's best deal. The last week's best deal was = \$'1.60/pound The difference between the unit price for previous week's best deal and this week's best deal = \$1.60 - \$1.42 = \$0.18 Therefore, 18 cents.	344	1,039	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.84375	4	CC-MAIN-2022-33	latest	en	0.86745
https://www.homeworklib.com/question/1429303/problem-4-linear-time-invariant-systems-a-linear	1,657,027,878,000,000,000	text/html	crawl-data/CC-MAIN-2022-27/segments/1656104576719.83/warc/CC-MAIN-20220705113756-20220705143756-00580.warc.gz	852,447,388	10,712	Question # Problem 4. Linear Time-Invariant System.s A linear system has the block diagram y(t) z(t) →\| Delay... Problem 4. Linear Time-Invariant System.s A linear system has the block diagram y(t) z(t) →\| Delay by 1 dt h(t) where g(t) sinc(t Since this is a linear time invariant system, we can represent it as a convolution with a single impulse response h(t) a) Find the impulse response h(t). You don't need to explicitly differentiate. b) Find the frequency response H(j for this system. We need at least 10 more requests to produce the answer. 0 / 10 have requested this problem solution The more requests, the faster the answer. All students who have requested the answer will be notified once they are available. #### Earn Coins Coins can be redeemed for fabulous gifts. Similar Homework Help Questions • ### 4. A linear time invariant system has the following impulse response: h(t) =2e-at u(t) Use convolution... 4. A linear time invariant system has the following impulse response: h(t) =2e-at u(t) Use convolution to find the response y(t) to the following input: x(t) = u(t)-u(t-4) Sketch y(t) for the case when a = 1 • ### For a continuous time linear time-invariant system, the input-output relation is the following (x(t) the input, y(t) the... For a continuous time linear time-invariant system, the input-output relation is the following (x(t) the input, y(t) the output): , where h(t) is the impulse response function of the system. Please explain why a signal like e/“ is always an eigenvector of this linear map for any w. Also, if ¥(w),X(w),and H(w) are the Fourier transforms of y(t),x(t),and h(t), respectively. Please derive in detail the relation between Y(w),X(w),and H(w), which means to reproduce the proof of the basic convolution property... • ### 2. Let y(t)(e')u(t) represent the output of a causal, linear and time-invariant continuous-time system with unit... 2. Let y(t)(e')u(t) represent the output of a causal, linear and time-invariant continuous-time system with unit impulse response h[nu(t) for some input signal z(t). Find r(t) Hint: Use the Laplace transform of y(t) and h(t) to first find the Laplace transform of r(t), and then find r(t) using inverse Laplace transform. 25 points • ### From homework section of CD: Continuous-time convolution Consider the linear time-invariant system shown below. 7ylt) (t)... From homework section of CD: Continuous-time convolution Consider the linear time-invariant system shown below. 7ylt) (t) The input alt) and the impulse response h(t) are shown in the figures below. ht) Time, sec Time, sec Calculate (using convolution) the output of this system, yo). • ### 2. Consider a linear time-invariant system with transfer function H(s)Find the (s + α)(s + β) impulse response, h(t), of the system 2. Consider a linear time-invariant system with transfer f... 2. Consider a linear time-invariant system with transfer function H(s)Find the (s + α)(s + β) impulse response, h(t), of the system 2. Consider a linear time-invariant system with transfer function H(s)Find the (s + α)(s + β) impulse response, h(t), of the system • ### solve all 22. The input-output relationship for a linear, time-invariant system is described by differential equation... solve all 22. The input-output relationship for a linear, time-invariant system is described by differential equation y") +5y'()+6y(1)=2x'()+x(1) This system is excited from rest by a unit-strength impulse, i.e., X(t) = 8(t). Find the corresponding response y(t) using Fourier transform methods. 23. A signal x(1) = 2 + cos (215001)+cos (210001)+cos (2.15001). a) Sketch the Fourier transform X b) Signal x() is input to a filter with impulse response (1) given below. In each case, sketch the associated frequency response... • ### A linear time invariant system has an impulse response given by h[n] = 2(-0.5)" u[n] –... A linear time invariant system has an impulse response given by h[n] = 2(-0.5)" u[n] – 3(0.5)2º u[n] where u[n] is the unit step function. a) Find the z-domain transfer function H(2). b) Draw pole-zero plot of the system and indicate the region of convergence. c) is the system stable? Explain. d) is the system causal? Explain. e) Find the unit step response s[n] of the system, that is, the response to the unit step input. f) Provide a linear... • ### 4- Let the step response of a linear, time-invariant, causal system be (-1).uln] ylnl.ynl-ler uln... 4- Let the step response of a linear, time-invariant, causal system be (-1).uln] ylnl.ynl-ler uln].. 15 3 3 12 a) Find the transfer function H(Z) of this system b) Find the impulse response of the system. Is this system stable? c) Find the difference equation representation of this system. 4- Let the step response of a linear, time-invariant, causal system be (-1).uln] ylnl.ynl-ler uln].. 15 3 3 12 a) Find the transfer function H(Z) of this system b) Find the... • ### The impulse response h(t) of a linear time-invariant system is 2pi[(t-2)/2]. Find and plot the output... The impulse response h(t) of a linear time-invariant system is 2pi[(t-2)/2]. Find and plot the output when the system is driven by an input signal that is identical to the impulse response. • ### Question 1: (2 marks) Find the zero-input response yz(t) for a linear time-invariant (LTI) system described... Question 1: (2 marks) Find the zero-input response yz(t) for a linear time-invariant (LTI) system described by the following differential equation: j(t) + 5y(t) + 6y(t) = f(t) + 2x(t) with the initial conditions yz (0) = 0 and jz (0) = 10. Question 2: (4 marks) The impulse response of an LTI system is given by: h(t) = 3e?'u(t) Find the zero-state response yzs (t) of the system for each the following input signals using convolution with direct integration....	1,528	5,792	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.59375	4	CC-MAIN-2022-27	latest	en	0.849093
https://brilliant.org/problems/complexity-6-2/	1,501,239,152,000,000,000	text/html	crawl-data/CC-MAIN-2017-30/segments/1500549448198.69/warc/CC-MAIN-20170728103510-20170728123510-00266.warc.gz	630,622,696	18,352	# Complexity! (6) If $$\displaystyle \alpha = e^{\frac{i2\pi}{7}}$$ and $$f(x) = 1 + \displaystyle \sum_{k=1}^{6}{a_{k}x^{k}} + \sum_{k=8}^{13}{a_{k}x^{k}} + \sum_{k=15}^{20}{a_{k}x^{k}}$$, then find the value of the expression below: $f(x) + f(\alpha x) + f({\alpha}^{2} x) + f({\alpha}^{3} x) + f({\alpha}^{4} x) + f({\alpha}^{5} x) + f({\alpha}^{6} x)$ Notations:	174	369	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.125	3	CC-MAIN-2017-30	longest	en	0.390667
https://brainly.in/question/319789	1,485,153,037,000,000,000	text/html	crawl-data/CC-MAIN-2017-04/segments/1484560282140.72/warc/CC-MAIN-20170116095122-00185-ip-10-171-10-70.ec2.internal.warc.gz	792,054,156	10,484	# Momu has got 3 cards They are red,white and black.. Chomu wants to select a red card ...he takes away the red card... Noe, momu has white and blue card left.... Rindi tells momu that if he gives her some candies, she will triple his cards and remove the fraction number of cards having probability of getting a blue card initially. Now, Momu gives her a sweet candy. She does the same as she said. Now, how may cards does momu have ? Out of the left cards, he keeps 2 cards with him and gives the rest to chinmay....How may cards does chinmay have ? 2 by sjajvah 2016-04-07T17:31:55+05:30 Total cards momu has initially = 3 Cards left after chomu takes his card = 3-1 = 2 Probability of geetiing a blue card = 1/2 Triple the number of cards = 2(3) = 6 Number of cards left with momu after the deal with Rindi = 6 - 1/2(6) = 3 Number of cards Chinmay got = 3-2 = 1	259	871	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.09375	4	CC-MAIN-2017-04	latest	en	0.964948
http://serc.carleton.edu/sp/library/earthsystem/nutshell/courses.html?results_start=81	1,455,109,527,000,000,000	text/html	crawl-data/CC-MAIN-2016-07/segments/1454701159376.39/warc/CC-MAIN-20160205193919-00079-ip-10-236-182-209.ec2.internal.warc.gz	208,633,721	9,255	Pedagogy in Action > Library > Using an Earth System Approach > Earth System Science in a Nutshell > Earth System Science Courses # Earth System Science Courses Help Results 81 - 90 of 1284 matches Human Demographics part of Pedagogy in Action:Library:Mathematical and Statistical Models:Examples In this biology simulation students explore factors that change human population growth including age at which women begin to bear children, fertility rate and death rate. Genetic Drift part of Pedagogy in Action:Library:Mathematical and Statistical Models:Examples In this biology simulation, students use a mathematical simulation of genetic drift to answer questions about the factors that influence this evolutionary process. Students run a series of simulations varying allele frequency and population size and then analyze their data and propose a model to explain their results. Earthquake Demonstration part of Pedagogy in Action:Library:Interactive Lecture Demonstrations:Examples This demonstration uses an "earthquake machine" constructed from bricks, sand paper, and a winch, to simulate the buildup of elastic strain energy prior to a seismic event and the release of that energy during an earthquake. Crystallization from Melt Demonstration part of Pedagogy in Action:Library:Interactive Lecture Demonstrations:Examples This demonstration uses melted phenyl salicylate to show how crystals nucleate and grow as the temperature of the liquid melt decreases. Copper Extraction Demonstration Tutorial part of Pedagogy in Action:Library:Interactive Lecture Demonstrations:Examples Summary This demonstration uses sulfuric acid and crushed copper ore (malachite) to produce a solution of copper sulfate and carbonic acid in a beaker. When a freshly sanded nail is dropped into the copper sulfate ... Fog Chamber part of Pedagogy in Action:Library:Interactive Lecture Demonstrations:Examples Show how clouds and fog are created with a very simple physical model. Materials needed are: A large 1 gallon jar, latex glove, a little water, and matches. Phases of the Moon part of Pedagogy in Action:Library:Interactive Lecture Demonstrations:Examples This exercise has students use a simple physical model of the Earth, sun, and moon to understand why the moon changes phases from the perspective of Earthly observers. Slinky and Waves part of Pedagogy in Action:Library:Interactive Lecture Demonstrations:Examples Use a Slinky to show:P and S waves, Wave reflection, and Standing waves in interactive lecture demonstration. Magma Viscosity Demos part of Pedagogy in Action:Library:Interactive Lectures:Examples This is an interactive lecture where students answer questions about demonstrations shown in several movie files. They learn to connect what they have learned about molecules, phases of matter, silicate crystal structures, and igneous rock classification with magma viscosity, and to connect magma viscosity with volcano explosiveness and morphology. Using Mass Balance to Understand Atmospheric CFCs part of Pedagogy in Action:Library:Teaching with Data:Examples Students use an interactive online mass balance model help understand the observed levels of chlorofluorocarbon CFC-12 over the recent past.	644	3,230	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.796875	3	CC-MAIN-2016-07	latest	en	0.857693
https://www.bartleby.com/essay/Biology-Design-Lab-PKNB6Z8K86TS	1,596,885,324,000,000,000	text/html	crawl-data/CC-MAIN-2020-34/segments/1596439737645.2/warc/CC-MAIN-20200808110257-20200808140257-00548.warc.gz	599,547,205	11,244	# Biology Design Lab 4465 Words18 Pages Standard Level Biology Design Lab: How do Smarties and M&Ms compare when examined in 5 areas: mass, shell solubility, volume, density, and nutrition facts? , Erin MacNeil SL Biology Kathy DeGrasse Halifax Grammar School March 3rd, 2013 Section 1.1 Planning (a) * Introduction * Research Question * Hypothesis * Explanation of Hypothesis * Variables Introduction: In this lab, M&Ms and Smarties will be compared in five different ways: mass, shell solubility, volume, density, and nutrition facts. For each comparison 10 Smarties will be used and 10 M&Ms will be used. This is to ensure that the results are more accurate, as each piece of candy is not the same size, weight etc… A total of 50…show more content… Observe both candies until the candy coating is completely dissolved. When there is only the chocolate left in the water, record the time on to an Excel spreadsheet. Repeat these steps four more times to complete 10 trails for Smarties. When all of the trials are completed with Smarties, dump out all of the water and chocolate, rinse the beakers thoroughly and dry them with paper towel. Repeat the same steps used for Smarties using M&Ms. After all of the data has been collected and entered into the Excel spreadsheet; use the “Insert Function” feature to calculate the average time of solubility and the standard deviation. Nutrition Facts: Look on the back of the boxes of Smarties and M&Ms for the Nutrition Facts. Calculate the amount of sugar in a single piece of a Smartie. Divide the grams of sugar, as stated on the package, by the total grams of the package. Example: one package of Smarties contains approximately 30g (30 Smartie candies), and 22g of sugar. 22g of sugar divided by the total of 30g equals 0.73g of sugar for every 1g (1 Smartie). Repeat these steps using the following nutrition facts: fat, protein, calories and carbohydrates. Also repeat all of the steps above using M&Ms. Enter all data into an Excel spreadsheet and use the “Insert Function” feature to calculate the mean and standard deviation for the sugar, fat, protein, calories and carbohydrates in both Smarties and	499	2,183	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.78125	4	CC-MAIN-2020-34	latest	en	0.862534
www.studygems.com	1,716,570,643,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00777.warc.gz	54,214,210	12,240	# Simplification tricks, rule, methods with example BODMAS Rule (Mathematics)- Simplification BODMAS rule, rule, methods, solution with the help of various examples and short methods to solve simplification questions in exam ## Rule for Simplification BODMAS rule, tricks, methods, solution with the help of various examples and short methods to solve simplification questions in exam Name of Rule: BODMAS Rule B: Bracket O: of D: Division M: Multiplication S: Subtraction Thus, simplifying an expression, first of all brackets must be removed, strictly in the Order: 1. () 2. {} 3. [] After removing the bracket, we must use the following operations strictly in order 1. Of 2. Division 3. Multiplication 5. Subtraction Question 1) 5005 – 5000 ÷ 10 Steps 1) Apply BODMAS Step 2) Change Sign of Divide into Multiply i.e ( ÷ 10 ) convert with ( × ) Solution: = 5005 – 5000 × = 5005 – 500 Question 2) Question regarding – Simplification tricks, rule, methods, solution 4 + 3 + ? + 2 = 13 Solution) 4 + 3 + ? + 2 = 13 Step 1) Convert mixed fraction into simple fraction i.e. mixed fraction (4 ) convert into ( ) simple fraction ————————— Method of Conversion mixed fraction into Simple fraction = 4 = 4 + = ——————————– + + x + = Firstly find the value of x Shift these values to the right hand side ( Note: while shifting, if the values are positive before shifting, then the values become negative after shifting ) x = – ( + + ) x = – ( ) x = – x = – 10 x = x = or It can also be written as 3 Question 3 : 1 ÷ [1 + 1 ÷ { 1 + 1 ÷ ( 1 + 1 ÷ 2 )}] Solution: Apply BODMAS Rule Step 1) Change Sign of Divide into Multiply i.e ( ÷ 2 ) convert with ( × ) = 1 ÷ [1 + 1 ÷ { 1 + 1 ÷ ( 1 + 1 × )}] Step 2) Start solving from the inner most bracket = 1 ÷ [1 + 1 ÷ { 1 + 1 ÷ ( 1 + )}] = 1 ÷ [1 + 1 ÷ { 1 + 1 ÷ }] = 1 ÷ [1 + 1 ÷ { 1 + 1 ÷ }] = 1 ÷ [1 + 1 ÷ { 1 + 1 × }] = 1 ÷ [1 + 1 ÷ { 1 + }] = 1 ÷ [1 + 1 ÷ ] = 1 ÷ [1 + 1 × ] = 1 ÷ [1 + ] = 1 ÷ = 1 × = Question 4) Question regarding – Simplification tricks, rule, methods, solution for the same. A man divides Rs. 8600 among 5 sons, 4 daughters and 2 nephews. If each daughter receives four times as much as each nephew and each son receives five times as much as each nephew, how much does each daughter receive? Solution) In the above statement the man is sharing money between Sons, Daughters and nephew and everybody is receiving more money that nephew. So, Let’s take the share for nephew = x Share for each daughter = Rs. 4x (i.e. 4 times more that nephew) Share for each son = Rs. 5x (i.e. 5 times more that nephew) Therefore, Total amount 8600 = 5 Sons (5x) + 4 Daughters (4x) +2 nephew (x) So, 8600 = 5 (5x) + 4 (4x) + 2x 8600 = 25x + 16x + 2x 8600 = 43x = x 200 = x Share of Daughter’s = 4x = 4 × 200 = 800 Check / Verify: Total amount Rs. 8600 = 5 Sons × (5 × 200) + 4 Daughters × (4× 200) +2 nephew × 200 = 5000 + 3200 + 400 = 8600	1,075	3,183	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.78125	5	CC-MAIN-2024-22	latest	en	0.673099
http://www.bendwavy.org/klitzing/incmats/ditatha.htm	1,369,179,195,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368700871976/warc/CC-MAIN-20130516104111-00056-ip-10-60-113-184.ec2.internal.warc.gz	353,398,927	1,092	Acronym ditatha Name ditrigonary trigonal hemiapeirogonal tesselation Vertex figure [(3,∞)3]/2 = [(3/2,∞)3] Incidence matrix (N→∞) NN \| 6 \| 3 3 -----+-----+----- 2 \| 3NN \| 1 1 -----+-----+----- 3 \| 3 \| NN * 3N \| 3N \| * N	105	231	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.84375	3	CC-MAIN-2013-20	latest	en	0.631043
http://conversion.org/area/cord/barn	1,721,844,927,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763518427.68/warc/CC-MAIN-20240724171328-20240724201328-00704.warc.gz	6,659,434	6,721	# cord to barn conversion Conversion number between cord and barn [b] is 1.48644864 × 10+28. This means, that cord is bigger unit than barn. ### Contents [show][hide] Switch to reverse conversion: from barn to cord conversion ### Enter the number in cord: Decimal Fraction Exponential Expression cord eg.: 10.12345 or 1.123e5 Result in barn ? precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential: ### Calculation process of conversion value • 1 cord = (exactly) (1.48644864) / (110^-28) = 1.48644864 × 10+28 barn • 1 barn = (exactly) (110^-28) / (1.48644864) = 6.7274440104436 × 10-29 cord • ? cord × (1.48644864 ("m²"/"cord")) / (1*10^-28 ("m²"/"barn")) = ? barn ### High precision conversion If conversion between cord to square-metre and square-metre to barn is exactly definied, high precision conversion from cord to barn is enabled. Decimal places: (0-800) cord Result in barn: ? ### cord to barn conversion chart Start value: [cord] Step size [cord] How many lines? (max 100) visual: cordbarn 00 101.48644864 × 10+29 202.97289728 × 10+29 304.45934592 × 10+29 405.94579456 × 10+29 507.4322432 × 10+29 608.91869184 × 10+29 701.040514048 × 10+30 801.189158912 × 10+30 901.337803776 × 10+30 1001.48644864 × 10+30 1101.635093504 × 10+30 Copy to Excel ## Multiple conversion Enter numbers in cord and click convert button. One number per line. Converted numbers in barn: Click to select all ## Details about cord and barn units: Convert Cord to other unit: ### cord Definition of cord unit: ≡ 192 bd. = 1.48644864 m² Convert Barn to other unit: ### barn Definition of barn unit: ≡ 10−28 m². Originally used in nuclear physics for expressing the cross sectional area of nuclei and nuclear reactions. ← Back to Area units	568	1,759	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.4375	3	CC-MAIN-2024-30	latest	en	0.603475
http://www.hindawi.com/journals/jam/2014/703178/	1,472,330,306,000,000,000	application/xhtml+xml	crawl-data/CC-MAIN-2016-36/segments/1471982925602.40/warc/CC-MAIN-20160823200845-00215-ip-10-153-172-175.ec2.internal.warc.gz	486,928,544	100,776	• Views 465 • Citations 0 • ePub 17 • PDF 353 `Journal of Applied MathematicsVolume 2014 (2014), Article ID 703178, 9 pageshttp://dx.doi.org/10.1155/2014/703178` Research Article ## Solutions of a Quadratic Inverse Eigenvalue Problem for Damped Gyroscopic Second-Order Systems Department of Mathematics, East China Normal University, Shanghai 200241, China Received 2 September 2013; Accepted 15 December 2013; Published 21 January 2014 Copyright © 2014 Hong-Xiu Zhong et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. #### Abstract Given pairs of complex numbers and vectors (closed under conjugation), we consider the inverse quadratic eigenvalue problem of constructing real matrices , , , and , where , and are symmetric, and is skew-symmetric, so that the quadratic pencil has the given pairs as eigenpairs. First, we construct a general solution to this problem with . Then, with the special properties and , we construct a particular solution. Numerical results illustrate these solutions. #### 1. Introduction Vibrating structures such as buildings, bridges, highways, and airplanes are distributed parameter systems [1]. Very often a distributed parameter system is first discretized to a matrix second-order using techniques of finite element or finite difference, and then an approximate solution is obtained for the discretized model. Associated with the matrix second-order model is the eigenvalue problem of the quadratic pencil, where , , , and are, respectively, mass, damping, gyroscopic and stiffness matrices. The system represented by (1) is called damped gyroscopic system. In general, the gyroscopic matrix is always skew-symmetric, the damping matrix and the stiffness matrix are symmetric, the mass matrix is symmetric positive definite, and they are all real matrices. If , the system is called damped nongyroscopic system, and if , the system is called undamped gyroscopic system. The damped gyroscopic system has been widely studied in two aspects: the quadratic eigenvalue problem (QEP) and the quadratic inverse eigenvalue problem (QIEP). The QEP involves finding scalars and nonzero vectors , called the eigenvalues and eigenvectors of the system, to satisfy the algebraic equation , when the coefficient matrices are given. Many authors have been devoted to this kind of problems and a series of good results have been made (see, e.g., [28]). The QIEP determines or estimates the parameters of the system from observed or expected eigeninformation of . Our main interest in this paper is the corresponding inverse problem: given partially measured information about eigenvalues and eigenvectors, we reconstruct matrices , , , and , satisfied with several conditions, so that has the given eigenpairs. The problem we considered is stated as follows. Problem 1. Given an eigeninformation pair , where with find real matrices , , , and , with being symmetric definite, and being symmetric, and being skew-symmetric, so that In [9], Gohberg et al. developed a powerful GLR theory to solve the QIEP of the undamped gyroscopic system. In [10], Chu and Xu developed an elegant procedure to obtain a real-valued spectral decomposition of the damped nongyroscopic system. And then, Jia and Wei [11] derived a real-valued spectral decomposition of the undamped gyroscopic system. However, [10, 11] both need all the eigen-information of to obtain the parameters, and it is often impractical or impossible to obtain complete spectral information. Thus, it becomes very interesting to consider a QIEP with only a subset of eigenpairs known. In [12], Kuo et al. constructed the solutions of the QIEP of the damped nongyroscopic system with given eigenpairs. And for the same system, Cai et al. [13] solved the QIEP with given eigenpairs. Meanwhile, for the damped gyroscopic system, Yuan [14] solved the QIEP with given eigenpairs. In [14], Yuan constructed symmetric positive semidefinite matrix and skew-symmetric matrix for , with and as given matrices. So, it becomes challenging to construct for damped gyroscopic system (1) with given eigenpairs, and this is the goal of this paper. This paper is organized as follows. In Section 2, we establish the solubility theory of the Problem 1. In Section 3, we develop a simple method to compute a particular solution to the Problem 1 with . Moreover, for and , a simple algorithm is developed to compute a solution in Section 4. Some numerical results are presented in Section 5 to illustrate our main results. In the last section, some conclusions and acknowledgements are given. Throughout this paper, we use capital letters to denote matrices, and lowercase (bold) letters to denote scalars (vectors). denotes the transpose of the matrix , denotes the identity matrix, and denotes the Moore-Penrose generalized inverse of . We write if is real symmetric positive (semidefinite). The spectrum of is denoted by . For simplicity, we make the following assumptions.(A1)The eigenvector matrix in Problem 1 has full column rank, that is, rank .(A2)The eigenvalue matrix in Problem 1 has only simple eigenvalues, that is, . Remark 2. For the case that , using the assumption of simple eigenvalues, we can partition , where has no zero eigenvalue, and then do discussion with . So, in this paper, we only consider the case that has no zero eigenvalue. #### 2. General Solution of the Problem In this section, we will give a general solution to the Problem 1 for given matrix pair as in (2) and (3). At the beginning, we will introduce some lemmas. Lemma 3 (see [15]). Let and ; then has a solution if and only if where is the Moore-Penrose generalized inverse of . When condition (6) is satisfied, the general solution of (5) is where is arbitrary and is constrained only by the symmetry requirement that Lemma 3 directly results in the following lemma. Lemma 4. Let be a nonsingular matrix and ; then has a solution if and only if in which case the general solution is , where is an arbitrary matrix. Given matrix pair as in Problem 1: let be the -factorization of , where is orthogonal and is upper triangular. We may require that has positive diagonal entries, since is of full column rank. Let , so that finding , , , and which satisfy (4) is equivalent to finding , , and which satisfy and the relations of , , and are Let ; we can see exists by using . Denoting where , , , , , , , , , , we will obtain the following main theorem. Theorem 5. Let , , and be defined as in (14)–(16); then there are real matrices , and satisfy (4) if and only if(i) is arbitrarily symmetric positive definite,(ii), , are arbitrary,(iii) is arbitrary symmetric,(iv), where is arbitrary symmetric,(v),(vi).Furthermore, and can be expressed as in (13). Proof. Necessity. From (14)–(16), we know , , and ; substituting them and (11) into (12), we have Thus, finding , , and which satisfy (12) is equivalent to finding the submatrices , , , , , and which satisfy (17) and (18). Clearly, it follows from (18) that is determined by where and are arbitrary. As and are required to be symmetric positive definite and symmetric, respectively, so are and in (14) and (16). From (17) it follows that Let be an arbitrary symmetric positive definite matrix. We need to find such that is symmetric; that is, it satisfies After rearrangement, (21) becomes Because and is nonsingular, we can get from Lemma 4 that where is arbitrary symmetric. Substituting (23) into (20) yields (v). Furthermore, and can be expressed as in (13). Sufficiency. From the description of (i)–(vi), we can obtain that (12) holds; thus (4) holds with Remark 6. The general solution to the Problem 1 with and prescribed eigenpairs has been given in [12], here, we generalize its solution to the case of . Remark 7. It is complicated for the more general case , and we will discuss it in our next work. However, here we provide a simple solution. We can select the linear independent columns and the relevant eigenvalues to construct a new and , then do discussion with them. Remark 8. When , by Theorem 5, the general solution of the Problem 1 is given by where with and which can be arbitrarily chosen and is arbitrarily symmetric. Using Theorem 5, we can construct a solution to the Problem 1 as follows. Algorithm 9. An algorithm for solving Problem 1 is proposed as follows.(1)Input and , compute the decomposition of according to (11), and compute .(2) Choose a symmetric positive definite matrix and a symmetric matrix , arbitrarily. Compute and by (iv) and (v) in Theorem 5, respectively.(3) Choose arbitrary and , and compute by (vi) in Theorem 5, and .(4)Choose a symmetric positive definite matrix ; compute .(5)Choose arbitrary matrices and and a symmetric matrix , and form where is given by (11). Compute and by (13). #### 3. Particular Solutions with As we all know, the applications of the undamped gyroscopic system (i.e., ) exist in many fields; for details, see [5]. In this section, we will discuss the particular solutions of Problem 1 with and prescribed eigenpairs. And in this case, in (1) becomes It is well known that the eigenvalues of (28) have a Hamiltonian structure; that is, they occur in quadruples , possibly collapsing to real or imaginary pairs or single-zero eigenvalues. In [11], Jia and Wei discussed the eigenvalues of in (28) and separated them into four categories. From the assumption (A2), we can know is even. Here we rewrite the given eigeninformation pair of Problem 1 as with where are eigenpairs, and . In this section, the Problem 1 becomes the following problem. Given an eigeninformation pair with (29)–(31), find real matrices , , and , with being symmetric definite, being symmetric, and being skew-symmetric, so that, As well as in Section 2, let where is partitioned conforming with that of in (14), and we can easily calculate that also satisfies the action of in Section 2, except that has an additional property, that is, . In the following theorem, we will discuss the solubility of Problem 1 with and . Theorem 10. Let , , and be defined as in (14), (16) and (34); then there are real matrices , , and which satisfy (33) if and only if(i) is arbitrarily symmetric positive definite,(ii) and are arbitrary,(iii) is arbitrary symmetric,(iv),(v),(vi),in which with and being arbitrary real numbers. Proof. Necessity. Same as the proof of Theorem 5, we can get where and are arbitrary. We also have and satisfies After rearrangement, (38) becomes It is easily seen that (39) has a particular solution Next, we consider the homogeneous equation Substituting into (41), we get Write , where is partitioned conforming with that of in (29). Then we observe that When , (43) can be rewritten as in which stands for the Kronecker product and vec stands for the column vectorization of a matrix. Because , and assumption (A2), we know that is nonsingular; therefore , so . Now we discuss the structures of matrices , , which are skew-symmetric. For simplicity, we denote by . Then we need to solve Since has the form in (30), we can easily compute that the general solution of (45) has the form Thus, the general solution of the homogeneous equation (41) has the form with defined in (35). This, together with (40), gives rise to the general solution of (39) Substituting (48) into (37) yields (v). Sufficiency. From the description of (i)–(vi), we can obtain that (33) holds. Remark 11. When , by Theorem 10, the solution of the Problem 1 with is given by where with and which can be arbitrarily chosen. #### 4. Particular Solution with and In practice, the matrix in the Problem 1 with is sometimes required to be symmetric negative definite [5]. In this section, we will apply Theorem 10 to construct such a solution. We first prove the following lemma. Lemma 12. For any given matrix defined in (35), we can construct a symmetric positive definite matrix so that defined in Theorem 10 is symmetric negative definite. Proof. Since , it is easy to see that in Theorem 10 is symmetric negative definite if and only if the matrix is symmetric negative definite. By the assumption (A2), we can first construct a symmetric positive definite matrix so that . Then we use to construct the desired . From (35) and (30), we denote with Here and are arbitrary real numbers. Take with Using (53), if we choose , , , , , and such that then and . Obviously, such real numbers , , , , , and can be easily chosen. Once is determined, the required can be chosen by Furthermore, Using Lemma 12, we can construct a particular solution to the Problem 1 with , as follows. Algorithm 13. An algorithm for solving the Problem 1 with , is proposed as follows.(1)Input and , compute the decomposition of according to (11), and compute .(2)Choose as in (35) arbitrarily and compute by (52) and (53).(3)Construct a symmetric positive definite matrix by (54)–(57), compute by (iv), and compute by (v) in Theorem 10 or by (58).(4)Choose arbitrary and , and compute by (vi) in Theorem 10, and .(5)Choose a symmetric positive definite matrix and a symmetric negative definite matrix ; compute , .(6)Choose an arbitrary skew-symmetric matrix , and form where is given by (11). Remark 14. When , we only need to choose by (54)–(56), and compute by (57), that is, is the whole ; then use the same method described in Remark 8, we can obtain the particular solution of the Problem 1 with and . #### 5. Numerical Examples In this section, we present two numerical examples to illustrate the solutions constructed in Sections 2 and 4, respectively. For presentation, we report all numbers in 5 significant digits only, though all calculations are carried out in full precision. Example 1. In this example, we use Algorithm 9 to construct the general solution of the Problem 1. The partially prescribed eigeninformation as in (2)-(3) is given by the following eigenvalues and eigenvectors, which are from [12]: It is easy to check that the matrix pair satisfy the assumptions (A1) and (A2). According to Algorithm 9, by randomly choosing we can figure out It is easy to check that is symmetric positive definite, and are symmetric, and is skew-symmetric. We define the residual as and the numerical results are shown in Table 1. This shows that Algorithm 9 to construct the general solution of the Problem 1 is effective. Table 1: Example 1. Example 2. In this example, we use Algorithm 13 to construct the general solution of the Problem 1 with and . The partially prescribed eigeninformation as in (29)–(31) is given by randomly generated eigenvalues and eigenvectors It is easy to check that the matrix pair satisfy the assumptions (A1) and (A2). According to Algorithm 13, by randomly choosing and choosing we can figure out It is easy to check that is symmetric positive definite, is symmetric negative definite, and is skew-symmetric. We define the residual as and the numerical results are shown in Table 2. This shows that Algorithm 13 to construct the particular solution of the Problem 1 with and is effective. Table 2: Example 2. #### 6. Conclusions In this paper, we first use techniques involving matrix decompositions to derive an expression of the general solution to the question, for a set of given pairs of complex numbers and vectors (closed under conjugation), under assumptions (A1) and (A2). Then, with the special properties and , we construct a particular solution. Numerical results illustrate these solutions. For another case of , it is rather complex, and the proof method in Theorem 5 seems not to be used directly to find a solution of Problem 1. Fortunately, for the damped nongyroscopic system, Cai et al. [13] solved the QIEP with given eigenpairs. However, case has never been discussed in the literature for damped gyroscopic system. It might be an interesting research and needs further investigation. #### Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper. #### Acknowledgment Guo-Liang Chen is supported by the National Natural Science Foundation of China (no. 11071079). #### References 1. M. J. Balas, “Trends in large space structure control theory: fondest hopes, wildest dreams,” IEEE Transactions on Automatic Control, vol. 27, no. 3, pp. 522–535, 1982. 2. Z. Bai and Y. Su, “Soar: a second-order arnoldi method for the solution of the quadratic eigenvalue problem,” SIAM Journal on Matrix Analysis and Applications, vol. 26, no. 3, pp. 640–659, 2005. 3. C. Guo, “Numerical solution of a quadratic eigenvalue problem,” Linear Algebra and Its Applications, vol. 385, no. 1–3, pp. 391–406, 2004. 4. Z. Jia and Y. Sun, “A refined Jacobi-Davidson method for the quadratic eigenvalue problem,” in Proceedings of the 10th WSEAS International Confenrence on APPLIED MATHEMATICS, pp. 1150–3155, Dallas, Tex, USA, November 2006. 5. K. Meerbergen, “The quadratic arnoldi method for the solution of the quadratic eigenvalue problem,” SIAM Journal on Matrix Analysis and Applications, vol. 30, no. 4, pp. 1463–1482, 2008. 6. J. Qian and W. Lin, “A numerical method for quadratic eigenvalue problems of gyroscopic systems,” Journal of Sound and Vibration, vol. 306, no. 1-2, pp. 284–296, 2007. 7. F. Tisseur and K. Meerbergen, “The quadratic eigenvalue problem,” SIAM Review, vol. 43, no. 2, pp. 235–286, 2001. 8. L. Zhou, L. Bao, Y. Lin, Y. Wei, and Q. Wu, “Restarted generalized second-order krylov subspace methods for solving quadratic eigenvalue problems,” World Academy of Science, Engineering and Technology, vol. 67, pp. 429–436, 2010. 9. I. Gohberg, P. Lancaster, and L. Rodman, “On selfadjoint matrix polynomials,” Integral Equations and Operator Theory, vol. 2, no. 3, pp. 434–439, 1979. 10. M. Chu and S. Xu, “Spectral decomposition of real symmetric quadratic λ-matrices and its applications,” Mathematics of Computation, vol. 78, no. 265, pp. 293–313, 2009. 11. Z. Jia and M. Wei, “A real-valued spectral decomposition of the undamped gyroscopic system with applications,” SIAM Journal on Matrix Analysis and Applications, vol. 32, no. 2, pp. 584–604, 2011. 12. Y. Kuo, W. Lin, and S. Xu, “Solutions of the partially described inverse quadratic eigenvalue problem,” SIAM Journal on Matrix Analysis and Applications, vol. 29, no. 1, pp. 33–53, 2006. 13. Y. Cai, Y. Kuo, W. Lin, and S. Xu, “Solutions to a quadratic inverse eigenvalue problem,” Linear Algebra and Its Applications, vol. 430, no. 5-6, pp. 1590–1606, 2009. 14. Y. Yuan, “An inverse eigenvalue problem for damped gyroscopic second-order systems,” Mathematical Problems in Engineering, vol. 2009, Article ID 725616, 10 pages, 2009. 15. H. W. Braden, “The equations ${A}^{T}X±{X}^{T}A=B$,” Journal on Matrix Analysis and Applications, vol. 20, no. 2, pp. 295–302, 1999.	4,565	18,904	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.859375	3	CC-MAIN-2016-36	latest	en	0.850162
https://www.teacherspayteachers.com/Product/Open-Ended-Math-Questions-Money-2234804	1,529,930,671,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267867666.97/warc/CC-MAIN-20180625111632-20180625131632-00196.warc.gz	915,654,620	17,836	# Open-Ended Math Questions - Money Subject Resource Type Product Rating 3.8 6 Ratings File Type Compressed Zip File 24 MB\|56 pages Share Product Description Open-Ended Math – Money (With American and Canadian coins) I created this resource because I wanted a bank of open-ended questions. I like giving open-ended questions because it reveals students’ thought processes. They can approach problems in their own way. You can differentiate for all of the needs in your class and many students will have success. I also love that I can challenge my fast finishers to come up with several solutions while the rest work at their slower pace and provide one or a couple examples. This package includes: • 16 questions • Colored graphics • 4 different cards on a page • Print on cardstock and laminating I use these for small groups. I provide blank paper and students choose manipulatives, if required, to work on the problem. MATH JOURNAL PAGES • 30 questions • Black and white graphics (Ink friendly!) • 2 to a page • There is a math notebook page for each task card • Most of the math notebook pages have a customizable version For example: - Regular version: I have 5 coins. How much money could I have? - Customizable version: I have _____ coins. How much money could I have? I use these when I want to students to work on problems individually. I like using the customizable versions so that they aren’t tempted to copy each other! COIN PAGES • American penny, nickel, dime, quarter • Canadian penny, nickel, dime, quarter, loonie, toonie RUBRIC • Generic • Editable • Choose from 1 or 2 per page REFERENCE CHART • I Can statements • Choose from 1, 2, or 3 per page This product is part of a Bundle of Open-Ended Math Questions. Don’t forget to check out the preview! Thanks for looking! Nicole Brown Brownie Points Total Pages 56 pages N/A Teaching Duration N/A Report this Resource Get this as part of a bundle: 1. Open-Ended Math Questions I created these resources because I wanted a bank of open-ended questions. I like giving open-ended questions because it reveals students’ thought processes. They can approach problems in their own way. You can differentiate for all of the needs in your class and many s \$5.00	530	2,237	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.515625	3	CC-MAIN-2018-26	latest	en	0.908377
https://danmackinlay.name/notebook/kernel_nn	1,709,518,072,000,000,000	text/html	crawl-data/CC-MAIN-2024-10/segments/1707947476409.38/warc/CC-MAIN-20240304002142-20240304032142-00658.warc.gz	185,929,735	12,380	# Neural net kernels September 16, 2019 — May 24, 2021 Hilbert space kernel tricks machine learning metrics probabilistic algorithms signal processing spheres statistics stochastic processes Random infinite-width NN induce covariances which are nearly dot product kernels in the input parameters. Say we wish to compare the outputs given two input examples $$.$$ They depend on the several dot products, $$\mathbf{x}^{\top} \mathbf{x}$$, $$\mathbf{x}^{\top} \mathbf{y}$$ and $$\mathbf{y}^{\top} \mathbf{y}$$. Often it is convenient to discuss the angle $$\theta$$ between the inputs: $\theta=\cos ^{-1}\left(\frac{\mathbf{x} ^{\top} \mathbf{y}}{\\|\mathbf{x}\\|\\|\mathbf{y}\\|}\right)$ The classic result is that in a single layer wide-neural net, \begin{aligned} K(\mathbf{x}, \mathbf{y}) &= \mathbb{E}\big[ \psi(Z_x) \psi(Z_y) \big], \quad \text{ where} \\ \begin{pmatrix} Z_x \\ Z_y \end{pmatrix} &\sim \mathcal{N} \Bigg( \mathbf{0}, \underbrace{\begin{pmatrix} \mathbf{x}^\top \mathbf{x} & \mathbf{x}^\top \mathbf{y} \\ \mathbf{y}^\top \mathbf{x} & \mathbf{y}^\top \mathbf{y} \end{pmatrix}}_{:=\Sigma} \Bigg). \end{aligned} It is sometimes useful to note that $$\begin{pmatrix} Z_x \\ Z_y \end{pmatrix}\overset{d}{=} \operatorname{Chol}(\Sigma)\boldsymbol{Z}_1,$$ where $$\boldsymbol{Z}_1\sim \mathcal{N} \Bigg( \mathbf{0}, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \Bigg)$$ and $$\operatorname{Chol}(\Sigma)= \begin{pmatrix} \\|\mathbf{x}\\| & \\|\mathbf{y}\\|\cos \theta \\ 0 & \\|\mathbf{y}\\|\sqrt{1-\cos^2 \theta} \end{pmatrix}.$$ These $$Z_{x}$$ terms arise from the (appropriately scaled limit of) the random weight matrix \begin{aligned} Z_x &= \mathbf{W}^\top\mathbf{x} \\ Z_y &= \mathbf{W}^\top \mathbf{y}. \end{aligned} Now, define \begin{aligned} Z_{xi} :&= W_{i} x_{i}, \\ Z_{yj} :&= W_{j} y_{j}, \\ Z'_{xi} :&= W_i, \\ Z'_{yj} :&= W_j. \end{aligned} We have that \begin{aligned} \kappa &= \mathbb{E} \big[ \psi\big(Z_x\big) \psi\big(Z_y \big) \big] \\ \frac{\partial \kappa}{\partial x_{i}} x_{i} &= \mathbb{E} \big[ \psi'\big(Z_x\big) \psi\big(Z_y \big) Z_{xi}\big] \\ \frac{\partial^2 \kappa}{\partial x_{i} \partial y_{j}} x_{i} y_{j} &= \mathbb{E} \big[ \psi'\big(Z_x\big) \psi'\big(Z_y \big) Z_{xi} Z_{yj} \big] \\ \frac{\partial^2 \kappa}{\partial x_{i} \partial x_{j}} x_{i}x_{j} &= \mathbb{E} \big[ \psi''\big(Z_x\big) \psi\big(Z_y \big) Z_{xi} Z_{xj} \big]\end{aligned} and thus \begin{align} \frac{\partial \kappa}{\partial x_{i}} &= \mathbb{E} \big[ \psi'\big(Z_x\big) \psi\big(Z_y \big) Z_{xi}'\big] \\ \frac{\partial^2 \kappa}{\partial x_{i} \partial y_{j}} &= \mathbb{E} \big[ \psi'\big(Z_x\big) \psi'\big(Z_y \big) Z_{xi}' Z_{yj}' \big] \\ \frac{\partial^2 \kappa}{\partial x_{i} \partial x_{j}} &= \mathbb{E} \big[ \psi''\big(Z_x\big) \psi\big(Z_y \big) Z_{xi}' Z_{xj}'\big] . \end{align} ## 1 Erf kernel Williams (1996) recover a kernel that corresponds to the Erf sigmoidal activation in the infinite width limit. Let $$\tilde{\mathbf{x}}=\left(1, x_{1}, \ldots, x_{d}\right)$$ be an augmented copy of the inputs with a 1 prepended so that it includes the bias, and let $$\Sigma$$ be the covariance matrix of the weights (which are usually isotropic, $$\Sigma=\mathrm{I}$$ ). Then $$K_{\mathrm{erf}}\left(\mathbf{x}, \mathbf{y}\right)$$ can be written as $K_{\mathrm{erf}}\left(\mathbf{x}, \mathbf{y}\right)=\frac{1}{(2 \pi)^{\frac{d+1}{2}}\|\Sigma\|^{1 / 2}} \int \Phi\left(\mathbf{w}^{\top} \tilde{\mathbf{x}}\right) \Phi\left(\mathbf{w}^{\top} \tilde{\mathbf{y}}\right) \exp \left(-\frac{1}{2} \mathbf{w}^{\top} \Sigma^{-1} \mathbf{w}\right) \mathrm{d}\mathbf{w}.$ This integral can be evaluated analytically to give $K_{\mathrm{erf}}(\mathbf{x}, \mathbf{y}) =\frac{2}{\pi} \sin^{-1} \frac{ 2 \tilde{\mathbf{x}}^{\top} \Sigma \tilde{\mathbf{y}} }{ \sqrt{\left( 1+2 \tilde{\mathbf{x}}^{\top} \Sigma \tilde{\mathbf{x}} \right)\left( 1+2 \tilde{\mathbf{y}}^{\top} \Sigma \tilde{\mathbf{y}} \right)}}.$ If there is no bias term, you can lop those tildes off and a factor of $$\sqrt{2\pi}$$ and the result should still hold. If the weights are isotropic, the s vanish also. ## 2 Arc-cosine kernel An interesting dot-product kernel is the arc-cosine kernel : $K_{n}(\mathbf{x}, \mathbf{y})= \frac{2}{(2 \pi)^{\frac{d}{2}}} \int \Theta(\mathbf{w} ^{\top} \mathbf{x}) \Theta(\mathbf{w} ^{\top} \mathbf{y})(\mathbf{w} ^{\top} \mathbf{x})^{n}(\mathbf{w} ^{\top} \mathbf{y})^{n} \exp\left(-\frac{1}{2}\mathbf{w}^{\top}\mathbf{w}\right) \mathrm{d}\mathbf{w}$ Specifically, $K_{n}(\mathbf{x}, \mathbf{y})=\frac{1}{\pi}\\|\mathbf{x}\\|^{n}\\|\mathbf{y}\\|^{n} J_{n}(\theta)$ where $J_{n}()$ is given by: $J_{n}(\theta)=(-1)^{n}(\sin \theta)^{2 n+1}\left(\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\right)^{n}\left(\frac{\pi-\theta}{\sin \theta}\right)$ The first few $$J_{n}$$ are $\begin{array}{l} J_{0}(\theta)=\pi-\theta \\ J_{1}(\theta)=\sin \theta+(\pi-\theta) \cos \theta. \end{array}$ $$J_{1}$$ recovers the ReLU activation in the infinite width limit. i.e. The arc-cosine kernel of order $$1$$ corresponding to the case where $$\psi$$ is the ReLU is \begin{aligned} k(\mathbf{x}, \mathbf{y}) &= \frac{\sigma_w^2 \Vert \mathbf{x} \Vert \Vert \mathbf{y} \Vert }{2\pi} \Big( \sin \|\theta\| + \big(\pi - \|\theta\| \big) \cos\theta \Big) \end{aligned} Observation: This appears related to Grothendieck’s identity, that any fixed vectors $$u, v \in \mathbb{S}^{n-1},$$ we have $\mathbb{E} \operatorname{sign}X_{u} \operatorname{sign}X_{v}=\frac{2}{\pi} \arcsin u^{\top} v.$ I don’t have any use for that, it is just a cool identity I wanted to note down. In an aside Djalil Chafaï observes that the Rademacher RV is the distribution over the 1 dimensional sphere, $$\in \mathbb{S}^{0}.$$ Is that what makes this go? ## 3 Absolutely homogenous Activation functions which are absolutely homogeneous of degree $$r$$ satisfying $$\psi(\|a\|z)=\|a\|^r\psi(z)$$ have additional structure. This class includes the ReLU and leaky ReLU activations (which are also included as the first order arc-cosine kernel above.) It follows from the definition that functions $$f$$ drawn from an NN with such an activation a.s. satisfy $$f(\|a\|\mathbf{x}) = \|a\|^r f(\mathbf{x})$$. For absolutely homogeneous activation we can sum the derivatives over the coordinate indices \begin{aligned} \sum_{i,j=1}^d \frac{\partial^2 \kappa}{\partial x_{i} \partial x_{j}} x_{i} x_{j} &= \mathbb{E} \big[ \psi''\big(Z_x\big) \psi\big(Z_y \big) (Z_x)^2 \big] = 0 \\ \sum_{i,j=1}^d \frac{\partial^2 \kappa}{\partial y_{i} \partial y_{j}} y_{i} y_{j} &= \mathbb{E} \big[ \psi''\big(Z_y\big) \psi\big(Z_x \big) (Z_y)^2 \big] = 0 \\ \sum_{i,j=1}^d \frac{\partial^2 \kappa}{\partial x_{i} \partial y_{j}} x_{i} y_{j}&= \kappa. \end{aligned} i.e. \begin{aligned} \mathbf{x}\frac{\partial^2 \kappa}{ \partial \mathbf{x}_{p} \partial \mathbf{x}_{q}^\top} \mathbf{y}^{\top} &=\kappa\\ \mathbf{x}\frac{\partial^2 \kappa}{ \partial \mathbf{x}_{p} \partial \mathbf{x}_{p}^\top} \mathbf{x}^{\top} &=0\\ \mathbf{y}\frac{\partial^2 \kappa}{ \partial \mathbf{x}_{q} \partial \mathbf{x}_{q}^\top} \mathbf{y}^{\top} &=0. \end{aligned} ## 4 References Adlam, Lee, Xiao, et al. 2020. arXiv:2010.07355 [Cs, Stat]. Arora, Du, Hu, et al. 2019. “On Exact Computation with an Infinitely Wide Neural Net.” In Advances in Neural Information Processing Systems. Belkin, Ma, and Mandal. 2018. In International Conference on Machine Learning. Chen, and Xu. 2020. arXiv:2009.10683 [Cs, Math, Stat]. Cho, and Saul. 2009. In Proceedings of the 22nd International Conference on Neural Information Processing Systems. NIPS’09. Domingos. 2020. arXiv:2012.00152 [Cs, Stat]. Fan, and Wang. 2020. In Advances in Neural Information Processing Systems. Fort, Dziugaite, Paul, et al. 2020. In Advances in Neural Information Processing Systems. Geifman, Yadav, Kasten, et al. 2020. In arXiv:2007.01580 [Cs, Stat]. He, Lakshminarayanan, and Teh. 2020. In Advances in Neural Information Processing Systems. Jacot, Gabriel, and Hongler. 2018. In Advances in Neural Information Processing Systems. NIPS’18. Neal. 1996. In Bayesian Learning for Neural Networks. Lecture Notes in Statistics. Pearce, Tsuchida, Zaki, et al. 2019. “Expressive Priors in Bayesian Neural Networks: Kernel Combinations and Periodic Functions.” In Uncertainty in Artificial Intelligence. Simon, Anand, and DeWeese. 2022. Tsuchida, Roosta, and Gallagher. 2018. In International Conference on Machine Learning. Williams. 1996. In Proceedings of the 9th International Conference on Neural Information Processing Systems. NIPS’96.	3,152	8,560	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.796875	3	CC-MAIN-2024-10	latest	en	0.415648
http://www.stata.com/statalist/archive/2013-02/msg01224.html	1,495,955,087,000,000,000	text/html	crawl-data/CC-MAIN-2017-22/segments/1495463609605.31/warc/CC-MAIN-20170528062748-20170528082748-00351.warc.gz	812,774,880	3,716	Notice: On March 31, it was announced that Statalist is moving from an email list to a forum. The old list will shut down on April 23, and its replacement, statalist.org is already up and running. Re: st: Ksmirnov one-sided test interpretation From Nick Cox To statalist@hsphsun2.harvard.edu Subject Re: st: Ksmirnov one-sided test interpretation Date Thu, 28 Feb 2013 18:06:28 +0000 ```Why not plot the data to show what is going on? Nick On Thu, Feb 28, 2013 at 5:23 PM, Tsankova, Teodora <TsankovT@ebrd.com> wrote: > I have a question related to a previous post: > > http://www.stata.com/statalist/archive/2009-01/msg00525.html > > The Stata output from this message is as follows: > > Two-sample Kolmogorov-Smirnov test for equality of distribution functions: > > Smaller group D P-value Corrected > ---------------------------------------------- > male: 0.2468 0.002 > female: 0.0000 1.000 > Combined K-S: 0.2468 0.005 0.003 > > > From the one sided tests (first two lines) on can say which distribution tends to be lower - for males or for females. However, I am not sure how to interpret it. > > Given that the pvalue from the first line is low and that D in the second line is 0, can we say that this is a proof that the distribution of male is lower than that of female? To rephrase it - can we claim that the distribution of male stochastically dominates the one of female which would imply that the values of the underlying variable tend to be larger for male than for female? Or, do we interpret it in the exactly opposite way - that the values for male tend to be lower than the values for female? * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/faqs/resources/statalist-faq/ * http://www.ats.ucla.edu/stat/stata/ ```	502	1,859	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.6875	3	CC-MAIN-2017-22	latest	en	0.894704
https://www.popularmechanics.com/science/math/a28569610/viral-math-problem-2019-solved/	1,680,413,855,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296950383.8/warc/CC-MAIN-20230402043600-20230402073600-00235.warc.gz	1,013,347,841	87,847	There it appeared on Twitter, like a cruel taunt: This content is imported from twitter. You may be able to find the same content in another format, or you may be able to find more information, at their web site. Every few months, the Internet eats itself over some kind of viral riddle or illusion, each more infuriating than the last. And so, like clockwork, this maddening math problem has gone viral, following in the grand tradition of such traumatic events as The Dress and Yanny/Laurel. These kinds of conundrums are purposely meant to divide and conquer, and predictably, the seemingly simple problem posed in the offending tweet—8÷2(2+2)—practically caused a civil war in the Popular Mechanics office, which we also share with our (former) friends at Runner’s World and Bicycling magazines. ###### More From Popular Mechanics You love challenging math problems. So do we. Let's solve them together. Naturally, we took to Slack to hash out our differences. Here’s a heated chat between the editors who stopped doing any semblance of actual work for the day to solve an equation designed to flummox fourth graders—and make many enemies in the process—followed by insight from real mathematicians and physicists who begrudgingly responded to our request for comment to solve the enraging math debate, once and for all. ## The Slack War, Part I Derek Call, video producer: 8 divided by 8 is 1. Jeff Dengate, Runner’s World runner-in-chief: PEMDAS. 16. Bobby Lea, test editor (and three-time Olympic cyclist): i ride bikes Pat Heine, video producer: ...she writes out PEMDAS and then does PEDMAS Matt: you clearly didn't listen Pat: i didn't...i was busy correcting her math. Derek: When they reinvent math? Matt: ok, Derek, the video's for you Pat: if you get 16 it's because you don't know the difference between brackets and parentheses. Morgan Petruny, test editor: I agree with Derek and disagree with YouTube. What if you want to do it the long way and use the distributive property and distribute the 2 first? You would do: 8 / (4+4) = 1. Or does the distributive property suddenly no longer apply? That's what I would say proves that 1 is correct. Derek: I trust Morgan because she's had a math class this decade. Pat: Wikipedia says you hate America if you get 16. Dan Roe, test editor: Right but it's multiplication/division, not multiplication then division Morgan: BUT multiplication with parentheses trumps division. So you still have 8 / 2(4). So you have to do the 2x4 first. At least, that is what I was taught. Dan: smart Berkley people say it's too ambiguous to say; PEMDAS isn't a mathematical convention as much as a teaching method Pat: multiplication/division::right/wrong Taylor Rojek, associate features editor: Biggest takeaway isn't that anyone sucks at doing math, but that this person sucks at writing out clear equations Bill Strickland, editorial director: MAKE IT CONTENT! instagram story with our staff debating? can we call a famous mathematician? Bobby: This sounds like a conversation the belongs on the Not My Job segment of Wait Wait Don't Tell Me Katie Fogel, social media editor: polling our IG audience on this now... Pat: The equation is not written according to ISO standards, leaving ambiguity of interpretation and the real answer is we need to teach better math writing. Ambiguous PEMDAS Ambiguous problems, order of operations, PEMDAS, BEMDAS, BEDMAS aka...what Taylor said, but from Harvard Morgan: aka...teach the distributive property instead of random acronyms Pat: When written according to ISO standard, the answer is 1. Andrew Daniels, how-to editor: honestly, we could post this slack thread word for word and then get a scholar to chime in and school us Katie Fogel: From our IG audience... Kit Fox, special projects editor: Isn’t the question and ambiguity here on when the parentheses disappear? Like, do the parens stick around after you do 2 + 2? Or do they go away once you solve the mini equation inside the parens first. I say they do not go away. I am on team 1 I also have not taken a math class in over 10 years Trevor Raab, photographer: My question is to what real world scenario would this apply to Brad Ford, test editor: Math class? Trevor: ahh the classic learn to do math to learn to do more math Bobby: school ain't real world Morgan: Generating heated and polarizing office discussion Brad: Bobby, tell that to a 6th grader. Bobby: i'll work on preparing my argument now Taylor: You've got, what, 11 years to perfect it Bobby: time is on my side which is code for: I can put this off for a reeeaaaalllly long time Pat: which is code for "ask your mother" Bobby: she likes to claim she's good at math. She may come to rue the day she bragged about that Pat: "This won't help me win millions of dollars playing Fortnite tho" ## A Brief Statement from Mike Breen, the Public Awareness Officer for the American Mathematical Society, Whose Job Is to “Try to Tell People How Great Math Is” According to order of operations, you solve whatever is in the parentheses first. That gives you 4. Then, in PEMDAS, multiplication and division take equal precedence, so you’d do the first that occurs from left to right. So you’d do 8 divided by 2 first, which is 4. Thus, it’s 16 according to classic order of operations. But the way it’s written, it’s ambiguous. In math, a lot of times there are ambiguities. Mathematicians try to make rules as precise as possible. According to strict order of operations, you’d get 16, but I wouldn’t hit someone on the wrist with a ruler if they said 1. ## The Slack War, Part II Andrew: hooooo boy i just got off the phone with the american mathematical society what a rollercoaster this is turning out to be my man mike with the AMS, whose job it is to explicitly answer questions like this one, says the answer is ... Tyler Daswick, associate features editor: secretly the best answer here Andrew: SIXTEEN Andrew, minutes later: why is no one reacting appropriately to this news Brad: Because he's wrong. Trevor: but doesn’t that go against PEMDAS? Andrew: he says (and i'll have to go back to the transcript) that using traditional order of operations, the answer is 16 Matt Phillips, senior test editor: Andrew, my brother has a PhD in theoretical physics and writes papers with titles like… “Angular Dependence for ν‘, j’-Resolved States in F + H2 → HF(ν‘, j’) + H Reactive Scattering Using a New Atomic Fluorine Beam Source” I can see if he wants to weigh in… Andrew: yes! please do [Editor's note: Matt's brother hasn't responded.] Taylor: is there a way that 1 is also a valid answer for this? Trevor: PEMDAS Andrew: i'll also fire off the request to my go-to physicist who also just answered the POP question of how to jump from a moving train Taylor: tbh it would be awesome if we could find experts who disagree Trevor: wait revisited my interpretation of PEMDAS back to 16 this is why I went to art school Taylor: I asked my friend [REDACTED], who is about to graduate with her phd in statistics from [REDACTED] and has three or four math masters degrees and i am so pleased to report she's on my side Derek: [REDACTED] wins Andrew: but what did [REDACTED] say was the answer??! Taylor: there is no answer, fake question designed to stoke outrage Bill: maybe our smart take is: math is not subjective, nobody writes math like this, here is what's wrong Taylor: she's just getting started Kit: Sounds like [REDACTED] needs to write the sweaty math take Andrew: daaaaang [REDACTED] go off Bobby: no we're onto something! A Parting Shot from Rhett Allain, Ph.D., Associate Professor of Physics at Southeastern Louisiana University, Who Delivered the Final Verdict and Decisively Shut Us All Up This is the math version of, ‘What color is this dress? Blue and black or gold and white?’ My answer is that you do parentheses first, so that becomes: 8/24 Next, you go from left to right. 8/2 is 4, so it is 44 Now you get 16. Of course this isn't math. This is convention. We have conventions on how to write these things just like we have conventions on how to spell stuff. But still, there are different conventions. Some people spell it as ‘gray’ and others as ‘grey.’ We still understand what's going on. For me, I would write this more explicitly so that there is no confusion. Like this: 8/(2*(2+2)), if that's what you are trying to do. That way no one will get it wrong. ##### Popular Mechanics T-Shirts on Amazon ###### Popular Mechanics June 1926 Cover T-Shirt Andrew Daniels Director of News Andrew Daniels is the Director of News for Popular Mechanics, Runner's World, Bicycling, Best Products, and Biography. In a past life, he was a senior editor at Men’s Health and wrote for Playboy, among lots of other publications that have since deleted his work. He’s also the author of The Barstool Book of Sports: Stats, Stories, and Other Stuff for Drunken Debate, which one Amazon reviewer called “the perfect book for the crapper,” and another called “moronic.” He lives in Allentown, Pennsylvania with his wife and dog, Draper.	2,219	9,138	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.84375	3	CC-MAIN-2023-14	latest	en	0.934156
https://workingrede.com/10568334-how-to-calculate-the-standard-hour	1,669,681,328,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710684.84/warc/CC-MAIN-20221128235805-20221129025805-00278.warc.gz	660,633,658	11,329	# How To Calculate The Standard Hour ## Video: How To Calculate The Standard Hour The efficient operation of any enterprise is impossible without planning. When drawing up a plan for the production of products or service, you need to know such a value as the standard hour. At its core, the standard hour is a temporary standard for the performance of a particular production operation and reflects its labor intensity and, ultimately, has a direct impact on the cost of products or services provided. You can calculate it yourself. ## Instructions ### Step 1 The gross number of working hours is equal to the number of employees of the enterprise employed in production, multiplied by the time that was spent on the manufacture of a product by the joint efforts of these workers. It will not be equal to the actually spent hours, which could serve as a standard. This is due to the fact that during the production process, every minute of working time was not used with an equal degree of intensity. ### Step 2 Please note that some of the time was used for rest breaks. Suppose you are calculating standard hours for a production unit that employs 10 people for 1 workweek for a total of 40 hours. During the day, they take two breaks of 10 minutes each. Thus, the total time that 10 workers spent on breaks during a five-day work week would be: (10 minutes * 2 * 5 days) * 10 people = 1000 minutes or 16, 7 hours. Therefore, taking into account the time spent on the break, the total time for manufacturing products was: 10 * 40 hours - 16, 7 = 383 hours. ### Step 3 To make your calculations more accurate, they should take into account the days of temporary disability and absenteeism. This figure can fluctuate depending on the season and holidays falling on different periods. As practice shows, on average for the year it is 4%. Refine the calculated values taking into account this parameter, the number of spent man-hours will be equal to: 383 - (383 * 0.04) = 367.7 man-hours. ### Step 4 This indicator is also theoretical and needs to be clarified, since labor productivity during one working day is also different. At the beginning of the day, workers need time to get ready for work, and at the end - to get ready for home. In addition, some of the time can be lost due to lack of necessary materials, tool breakage. Such losses usually do not amount to more than 7% of working time. With this in mind, the potential number of man-hours will be: 367, 7 - (0, 07 * 367, 7) = 367, 7 - 27, 7 = 342 man-hours practically available. ### Step 5 Now calculate your normal hours. If the labor efficiency of this working group does not exceed the norm and is equal to 100%, then the number of standard hours will be 342, if the labor efficiency in this group is higher and equal to 110%, then you will have 342 * 1, 10 = 376, 2 standards -hour. ### Step 6 From these calculations, you can see that if this group is entrusted with a work order, the estimated execution time of which is 400 hours, then the workers will not have time to complete it in a week. Consider this and solve the problem by increasing the number of workers or transferring part of the order to another department.	729	3,213	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2022-49	latest	en	0.958587
https://www.reference.com/math/cube-polygon-e22dba8c1501e6a3	1,488,216,205,000,000,000	text/html	crawl-data/CC-MAIN-2017-09/segments/1487501172902.42/warc/CC-MAIN-20170219104612-00369-ip-10-171-10-108.ec2.internal.warc.gz	882,137,859	20,508	Q: # Is a cube a polygon? A: A cube is a three-dimensional solid, more properly called a polyhedron. It is made up of six congruent square faces all set at perpendicular angles. It is one of the five convex polyhedrons known as the Platonic solids. ## Keep Learning The other Platonic solids are the tetrahedron, the octahedron, the dodecahedron and the icosahedron. These are the only convex polyhedrons made up entirely of congruent, regular polygonal faces. The faces of the dodecahedron are all regular pentagons, while the faces of the tetrahedron, octahedron and icosahedron are equilateral triangles. The Platonic solids are sometimes used as dice because their symmetry assures that there is an equal chance of the die landing on any face. Sources: ## Related Questions • A: A tetrahedron is a four-faced polyhedron, or three-dimensional shape, created with triangles. The triangles in a tetrahedron are usually equilateral. To ma... Full Answer > Filed Under: • A: The number of edges of any polyhedron can be calculated using the formula: E = V + F - 2, where "E'" denotes the number of edges, "V" indicates the number ... Full Answer > Filed Under: • A: A pyramid refers to a polyhedron created by joining a polygonal base to triangular lateral faces. There are several types of pyramids, and each is named af... Full Answer > Filed Under: • A: A cube is a six-faced, three-dimensional figure composed of square-shaped faces of the same size that meet at 90-degree angles, whereas a cuboid is a box-s... Full Answer > Filed Under: PEOPLE SEARCH FOR	381	1,567	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.84375	3	CC-MAIN-2017-09	longest	en	0.930124
https://de.zxc.wiki/wiki/Ikosaedergruppe	1,709,282,689,000,000,000	text/html	crawl-data/CC-MAIN-2024-10/segments/1707947475203.41/warc/CC-MAIN-20240301062009-20240301092009-00812.warc.gz	187,999,939	6,877	# Icosahedral group Icosahedron Division of the spherical surface into fundamental areas according to icosahedral or dodecahedral symmetry The icosahedron group is the point group of the homogeneous icosahedron (and the homogeneous dodecahedron , which is dual to the icosahedron). It consists of the rotations and reflections that the icosahedron transforms into itself and is of order 120. It is too isomorphic, whereby the alternating group is of order 5 (group of even permutations of 5 objects) and the cyclic group is of order 2 (consisting of the identity and the space reflection at the center of the icosahedron). ${\ displaystyle A_ {5} \ times C_ {2}}$${\ displaystyle A_ {5}}$${\ displaystyle C_ {2}}$ The subgroup that is too isomorphic, the icosahedron rotation group, consists of the orientation-maintaining symmetries of motion of the icosahedron (rotations). You can use this rotating group z. B. can be realized as a group of even permutations of the five cubes inscribed in a regular dodecahedron. is the smallest simple non-commutative group and has order 60. ${\ displaystyle A_ {5}}$${\ displaystyle A_ {5}}$ The icosahedral group contains five-fold rotations and is therefore incompatible with long-range crystalline order ( see space group ). Quasicrystals , on the other hand, often have icosahedral symmetry. The character table of the icosahedron group contains the golden ratio and related numbers, which is a direct consequence of the fivefold rotational symmetry. Since the soccer ball is derived from a truncated icosahedron , it also has the icosahedron group as a symmetry group, as does the “soccer molecule” C 60 (buckyball). The icosahedron group has diverse applications in mathematics, which are presented in Felix Klein's classical work, lectures on the icosahedron and the solution of the equations of the fifth degree . According to Galois theory, the general equation of the fifth degree has no solution in radicals, since it is not solvable (it is a finite simple group ). ${\ displaystyle A_ {5}}$ In crystallography , the rotation group of the icosahedron is designated with the Schoenfliess symbol and the complete symmetry group with . ${\ displaystyle I}$${\ displaystyle I_ {h}}$	555	2,236	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.84375	3	CC-MAIN-2024-10	latest	en	0.927
http://functions.wolfram.com/ElementaryFunctions/ArcTan/16/01/01/02/01/0005/	1,526,810,263,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794863277.18/warc/CC-MAIN-20180520092830-20180520112830-00331.warc.gz	116,120,907	7,650	html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; } ArcTan http://functions.wolfram.com/01.14.16.0020.01 Input Form ArcTan[1/z] == ((Pi z)/2) Sqrt[1/z^2] Sqrt[1/(1 + z^2)] Sqrt[1 + z^2] - ArcTan[z] Standard Form Cell[BoxData[RowBox[List[RowBox[List["ArcTan", "[", FractionBox["1", "z"], "]"]], "\[Equal]", RowBox[List[RowBox[List[FractionBox[RowBox[List["\[Pi]", " ", "z"]], "2"], SqrtBox[FractionBox["1", SuperscriptBox["z", "2"]]], SqrtBox[FractionBox["1", RowBox[List["1", "+", SuperscriptBox["z", "2"]]]]], " ", SqrtBox[RowBox[List["1", "+", SuperscriptBox["z", "2"]]]]]], "-", RowBox[List["ArcTan", "[", "z", "]"]]]]]]]] MathML Form tan - 1 ( 1 z ) π z 2 1 z 2 1 z 2 + 1 z 2 + 1 - tan - 1 ( z ) 1 z -1 z 2 -1 1 z 2 -1 1 2 1 z 2 1 -1 1 2 z 2 1 1 2 -1 z [/itex] Rule Form Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["ArcTan", "[", FractionBox["1", "z_"], "]"]], "]"]], "\[RuleDelayed]", RowBox[List[RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List["\[Pi]", " ", "z"]], ")"]], " ", SqrtBox[FractionBox["1", SuperscriptBox["z", "2"]]], " ", SqrtBox[FractionBox["1", RowBox[List["1", "+", SuperscriptBox["z", "2"]]]]], " ", SqrtBox[RowBox[List["1", "+", SuperscriptBox["z", "2"]]]]]], "-", RowBox[List["ArcTan", "[", "z", "]"]]]]]]]] Date Added to functions.wolfram.com (modification date) 2003-08-21	670	1,762	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.71875	3	CC-MAIN-2018-22	latest	en	0.238424
http://slideplayer.com/slide/4347554/	1,508,493,496,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187823997.21/warc/CC-MAIN-20171020082720-20171020102720-00226.warc.gz	314,173,186	26,152	# Reading Quiz A light bulb is connected to a battery so that current flows through the bulb, which gives off light. Choose the correct statement: The amount. ## Presentation on theme: "Reading Quiz A light bulb is connected to a battery so that current flows through the bulb, which gives off light. Choose the correct statement: The amount."— Presentation transcript: Reading Quiz A light bulb is connected to a battery so that current flows through the bulb, which gives off light. Choose the correct statement: The amount of current going into the bulb equals the current leaving the bulb. The amount of current entering the bulb is greater than the amount of current leaving the bulb. The bulb is converting electric charge into light. Physics Help Center NEXT EXAM Wednesday April 7th @7pm to 9pm Room 237 Physics Building: 8am to 5:30pm Ask for help from graduate students on homework and exams Can enter solutions on the computers in the room to check your solution. NEXT EXAM Wednesday April to 9pm Chapters 10,11,12,13,14,15 Summary of electric circuits: The amount of current is the same at every place in a series circuit; I=q/t. The power provided by the battery (P=IDV) is exactly equal to the power dissipated in the resistors (P=I2R). Ohm’s Law applies to resistors: DV=IR Series circuit: effective R = R1 + R2 + R3 Parallel circuit: effective R is Fig. 13.3 Fig. 13.3 Fig. 13.1 Fig. 13.1 Fig. 13.2 Fig. 13.2 Fig. 13.4 Fig. 13.4 Summary 2 Fig. 13.p270b Ohm’s Law applies to resistors: DV=IR Battery increases energy of charges (DPE); voltage on battery is called “EMF”. This amount of energy is expended in the resistor. For a given battery (V) and resistor (R), the current is given by Ohm’s Law: DV=IR EMF volts I V R The unit of resistance is an Ohm. Fig. 13.6 Fig. 13.6 Fig. 13.7 Fig. 13.7 Fig. 13.5 Fig. 13.5 A simple battery does which of the following: Lecture Quiz 26 - Question 1: A simple battery does which of the following: A. It creates charges. B. It does work on charges. C. It creates energy. Is the voltage drop the same across all three resistors? Series Resistance EMF= DV1 + DV2 + DV3 OR V = IR1 + IR2 + IR3 = I(R1 + R2 + R3) = I Rseries Where Rseries = R1 + R2 + R3 EMF volts R1 V R2 R3 Is the voltage drop the same across all three resistors? Is the current the same through all three resistors? Fig. 13.9 Fig. 13.9 What is the total resistance of this circuit? Exercise 9 What is the total resistance of this circuit? 1.) 66 ohms 2.) 60 ohms 3.) 54 ohms 4.) 25 ohms 5.) 15 ohms Fig. 13.p274a What is the current that flows in this circuit? Exercise 9 What is the current that flows in this circuit? 1.)10 amps 2.) 1 amp 3.) 0.1 amp 4.) 0.3 amp 5.) 0.03 amps Fig. 13.p274a Fig Fig V = I Reffective Parallel Resistors Effective resistance 1 2 3 V Where EMF volts 1 2 3 V Where V = I Reffective How much current flows through the three resistors? Is the voltage drop the same for all three resistors? Exercise 12 What is the total resistance of this circuit? 1.) 72 ohms Fig. 13.p274b A . What is the current that flows in this circuit at point A? 1.)288 amps 2.) 1.5 amp 3.) 0.1 amp 4.) 0.5 amp 5.) 0.03 amps Fig. 13.p274b B. What is the current that flows in this circuit at point B? 1.)288 amps 2.) 1.5 amp 3.) 0.1 amp 4.) 0.5 amp 5.) 0.03 amps Fig. 13.p274b By placing electrical items in series they receive the same current By placing all electric items in parallel they receive the same voltage Fig By placing electrical items in series they receive the same current A volt meter has a high resistance so take very little current Fig A volt meter has a high resistance so take very little current Fig Fig An Amp meter has very little resistance so take very little voltage Summary of electric circuits: The amount of current is the same at every place in a circuit; I=q/t. The power provided by the battery (P=IDV) is exactly equal to the power dissipated in the resistors (P=I2R). Ohm’s Law applies to resistors: DV=IR Series circuit: effective R = R1 + R2 + R3 Parallel circuit: effective R is Question 14 Fig. 13.p272g For an “ohmic” resistor, V=IR, so P = I2R The power provided by the battery (P=IDV) is exactly equal to the power dissipated in the resistors (P=I2R). Power P= work/time Power=DPE/t =qDV/t =IDV =IV For an “ohmic” resistor, V=IR, so P = I2R Lecture Quiz #26 Question 2 Consider a standard flashlight which is turned on. A. The batteries create electrons, which get used up in the light bulb. B. Energy is created in the batteries, which is destroyed in the light bulb. C. The power used up by the light bulb originates in the batteries. D. The light bulb converts electrons into light. Lecture Quiz: Question 3 How much current I runs through this circuit: A Amps B A C A D A E A 3 V 4 W 4 W Lecture Quiz: Question 4 For three resistors in parallel, choose the correct statement: A. The current and the voltage across all three are the same. B. The voltage drop and the power dissipated across all three are the same. C. The energy given up per electron is the same for all three. D. The current through one resistor is the same as the current through the battery. EMF volts 1 2 3 What does the battery do? Where does the light energy come from? The amount of current is the same at every place in a (single-loop) circuit; I=q/t. I Battery Light Bulb I I What does the battery do? Where does the light energy come from? Why do batteries burn out? I Download ppt "Reading Quiz A light bulb is connected to a battery so that current flows through the bulb, which gives off light. Choose the correct statement: The amount." Similar presentations	1,561	5,641	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.71875	4	CC-MAIN-2017-43	longest	en	0.910337
https://www.jiskha.com/display.cgi?id=1320937097	1,503,040,935,000,000,000	text/html	crawl-data/CC-MAIN-2017-34/segments/1502886104612.83/warc/CC-MAIN-20170818063421-20170818083421-00040.warc.gz	938,094,983	3,743	ALGEBRA posted by . write the expression using exponents mm • ALGEBRA - m2 Respond to this Question First Name School Subject Your Answer Similar Questions 1. Algebra Simplify each expression. Assume that no denominators are zero. Write each answer without using negative exponents. The ^'s mean they're exponents y^-3y^-4y^0 ------------ <- division (2y^-2)^3 numerator: y^-3y^-4y^0 = y^(-3-4)= … 2. algebra write the expression using only positive exponents 1.13^6 2.8^-9 3.xy^0 4.6g^-5 3. pre-algebra write the expression using only positive exponents 1.13^6 2.8^-9 3.xy^0 4.6g^-5 how do you do this? 4. algebra *show work evaluate the expression: 7^-6 (-6)^-9 simplify the expression. write your answer using only positive exponents. 4x^-2-x -2a^0a^2 5. algebra 2 Write an equivalent expression using radical notation 1.(a^2b^2)^1/9 Use the laws of exponents to simplify 2.6^3/5 6^5/10 6. I NEED HELP (ALGEBRA) Simplify the expression. Write your answer using only positive exponents. Explain as well. 1. (8x)^38^-4 2. 2x^-5y^-3 3. (4x^4y^-3)^-2 4. 1/(3x)^-2 7. Pre-Algebra Simplify the expression. Write the result using positive exponents only. (-2x^3y^-4)(5x^-1y) 8. College Algebra I have a few more questions that I'm not understanding. Its multiple choice however I would appreciate a breakdown on how to solve. 4.Simplify the compound rational expression. Use either method. 1+(5/y-10)/y+(25/y-10) a. 1/y-5 b. … 9. Algebra 1 HELP I am not sure how to do these 2 problems can someone show me. Simplify the expression Write answers using exponents. 8^6/8^48^2 4^51/4^2 <---is by the 4 only 10. Algebra Write the simplified form of the expression. {q^-2/3} {q^1/3} --------------- q^4/3 Using positive exponents. ------------------------------------- Solution: {q^-2/3} {q^1/3} = q^-1/3 = 1 --------------- ------ ---- q^4/3 q^4/3 q^5 … More Similar Questions	629	1,897	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.484375	3	CC-MAIN-2017-34	latest	en	0.701934
errotically.com	1,638,216,700,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964358842.4/warc/CC-MAIN-20211129194957-20211129224957-00277.warc.gz	318,837,079	10,745	# How to make the audio shift faster Audio frequency shift is a very simple and easy way to make your audio more consistent and less distracting. You can achieve this effect by using a frequency shift algorithm, and this article will show you how to use audio frequency shift. Frequency shift is not the same as amplitude shift, it can be useful to have both, or both can be used for audio frequency. For example, you can use the audio frequency to create an increase in audio volume. Here is how to apply a frequency to your audio: Frequency Shift Frequency is an integer, it’s defined as the amount of time the audio will have a different frequency. If you have audio at a frequency of 0, and your speakers have a frequency at a certain frequency, they will be slightly different. The difference will be greater than 1.0, or -1.0dB. In this example, we have a audio at frequency 1.025Hz. At this frequency, the sound will be louder than normal, and will sound a bit more “pop” than it normally would. However, it will still sound quite good. Frequency Shift The first thing to do is to calculate the audio amplitude. If we have an audio at 0.0333Hz, the audio volume is 0.0047 dB. If the audio has a frequency from 0.1 to 1.2, the volume will be 1.4dB. Frequency is equal to the ratio of the number of audio samples in the system to the total number of samples in each channel. So, we can use our frequency shift formula to calculate how many samples will have different frequencies. We can then convert that value to a frequency, and use that to make our audio frequency change. Frequency: Audio Amplitude = (0.0 + 1.3) / (0 – 1.1) = 1.15Hz Frequency: Amplitude Shift = (1.15 – 0.5) / 0.2 = 0.65Hz The frequency of our audio is 0, so we need to multiply the amplitude by 1.5 to get the frequency shift, which is -1, so the audio becomes 0.55Hz. FrequencyShift Frequency shift can be applied to your entire audio system, and you can also use it to apply it to your speakers. Here’s how to do it: FrequencyShift = Frequency * (1 – 0) / 1 = 1Hz FrequencyShift is also known as amplitude modulation. When you set up a frequency modulation, you have to first add a frequency in series with the audio. So you have two frequencies, 0.015 and 0.005Hz. Then, you add one more frequency, which will be 0.025 Hz. Finally, you set a new frequency, so now your audio is 1.25 Hz. So now the audio is at 0,1,1.25. Now it is the same sound, but you have the added frequency at the beginning. So if you had two frequencies at 1.125Hz and 1.225Hz, and also had a frequency modulator set at 1Hz, then you’d get 1.525Hz. This is called amplitude modulation, because it’s only when you increase the frequency that your sound will shift. When amplitude modulation is used to make a frequency change, you usually need to change the ratio between the frequencies. So for example, if we have audio from 0 to 1Hz and 0 to 2.625Hz, we will need to add one frequency, then we will have to subtract one frequency from the audio signal. So we will want to use the ratio 0.125. So here’s how we can do it. Frequency = 1 – (0) * (0 / 1.8) = 0Hz Frequency = 0 * (2.625 / 0) = -1Hz Frequency shift: Frequency = ((1 – 1) * 0.05) / ((0.8 – 0)*1.8 = 0) So, when we increase the audio, we get a shift in frequency. Frequency shifts are usually applied to stereo channels. Here, we’ve added a frequency on one side, and subtracted the audio from the other side. This means the audio now has a 0.95% shift in audio amplitude, so it’s louder than the original. When we subtract the audio and change the audio’s frequency, we do the same thing. FrequencyModulation = ( 0.975 / 1) / 3 = 0dB This is why frequency shift can sometimes sound a little bit “pop”. Frequency Shift When frequency shift has been applied to a stereo audio channel, you will see the audio increase in volume. Frequency modulators often also work with audio channels. For instance, a stereo stereo channel can have a high gain and a low gain. A high gain channel will have more audio in the channel, while a low input will be quieter. In audio, the difference between the loudness of the audio in a high input	1,057	4,184	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.84375	4	CC-MAIN-2021-49	latest	en	0.911214
https://ell.stackexchange.com/questions/20958/a-triangular-prism-what-else-its-shape-could-be?answertab=oldest	1,575,700,552,000,000,000	text/html	crawl-data/CC-MAIN-2019-51/segments/1575540496492.8/warc/CC-MAIN-20191207055244-20191207083244-00528.warc.gz	364,996,030	30,961	# A triangular prism? What else its shape could be? Looking at this question, first, I thought for a while about redundancy. A triangular prism. I searched on Wikipedia for the terms prism that says At least two of the flat surfaces must have an angle between them. (means triangle, isn't it? and triangular prism that says a three-sided prism. In addition, I searched Google Images with triangle prism and prism. Most of them were triangle! Of course, the latter one talks about geometry but still, a triangular prism? Why simply not say prism and it includes its shape i.e. triangle! Fun is, I searched for round prism and rectangle prism - both have results! Google is great! • "At least two of the flat surfaces..." – Helix Quar Apr 10 '14 at 5:31 • @helix true but what else shape you can come up for being a prism? – Maulik V Apr 10 '14 at 5:38 • I can conceive of a trapezoidal prism, though I have no idea of its practical application – toandfro Apr 10 '14 at 5:41 • @MaulikV lots – Helix Quar Apr 10 '14 at 5:42 • A cylinder is a prism. So are these. Why say "triangular prism"? Because a prism isn't necessarily triangular. A prism's base can have any shape or number of sides. – J.R. Apr 10 '14 at 9:30 Although there are lots of other kinds, people often use the word prism to refer to a triangular prism. From the same Wikipedia article you linked... The traditional geometrical shape is that of a triangular prism with a triangular base and rectangular sides, and in colloquial use "prism" usually refers to this type. From Wiktionary A transparent block in the shape of a prism (typically with triangular ends), used to split or reflect light. • Yes, that's what I was a bit confused about! – Maulik V Apr 10 '14 at 5:58 In a single lense reflex (SLR) photo camera, you often have a pentaprism to transfer the immage that is seen through the lens, and reflected from the mirror before the shutter to the ocular (eyepiece). If you look at the shape, you will see it fits your definition, and yet, it is not triangular :) • Thank you. True..it has to be specified to mean a triangular prism. +1 :) – Maulik V Apr 10 '14 at 5:59	547	2,153	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.890625	3	CC-MAIN-2019-51	latest	en	0.951095
http://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-1-section-1-1-linear-equations-1-1-assess-your-understanding-page-92/95	1,524,257,941,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524125944682.35/warc/CC-MAIN-20180420194306-20180420214306-00014.warc.gz	417,333,557	12,216	## College Algebra (10th Edition) Published by Pearson # Chapter 1 - Section 1.1 - Linear Equations - 1.1 Assess Your Understanding: 95 #### Answer Width = 11 Length = 8 + 11 = 19 #### Work Step by Step Perimeter = 2(length + width) Perimeter = 60 feet Length = 8 + width (given) Replacing in formula Perimeter = 2((8 + width) + width) 60 = 2(8 + 2width) $\frac{60}{2}$ = (8 + 2width) 30 = 8 + 2width 30 - 8 = 2width 22 = 2 width $\frac{22}{2}$ = 2 width Width = 11 Length = 8 + 11 = 19 After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.	214	653	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.09375	4	CC-MAIN-2018-17	latest	en	0.768358
https://es.mathworks.com/matlabcentral/answers/1589434-need-help-verifying-i-have-the-correct-polyfit-line-for-my-function?s_tid=prof_contriblnk	1,652,913,905,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00280.warc.gz	302,904,662	25,971	# Need help verifying I have the correct polyfit line for my function. 1 view (last 30 days) Kristin Aldridge on 18 Nov 2021 Commented: Walter Roberson on 18 Nov 2021 Question: Ps7.mat contain scores for 100 subjects on two different tests. Use the linear regression function you wrote in class to fit a linear model to these data. Plot the data using a scatter plot and plot the fitted model with a line. Report the estimated parameters a and b in the figure title. Then, use polyfit to find the best-fit line predicting score2 from score1. Your coefficients should be the same as the ones you found using your linear regression function in problem 2. List these coefficients in a comment or show them on the figure. Below I have included my regression line, my a (named alpha) and b (named beta) parameters (that I hope are correct), and what I plugged in for polyfit. I am a bit concerned because my polyfit gives me 0 -0.1606 , and my b is also equal to -0.1606 so I want to make sure that is correct, and if not where I'm going wrong. function [b,a,rsq] = regressline(x,y) n=length(x); xbar=mean(x); ybar=mean(y); sigxsq= sum(x.^2) - (sum(x)^2)/n; sigxy=sum(x.y)-(sum(x)sum(y))/n; beta=sigxy./sigxsq; % = -0.1606 alpha=ybar-betaxbar;% = -0.5841 plot(x,y,'o') hold on title('alpha=0.5841, beta=-0.1606') plot(x,(betax) + alpha, 'r') end p=polyfit(alpha,beta,1) plot(p) %p=0 -0.1606 Thanks! Walter Roberson on 18 Nov 2021 No you should polyfit x and y. ##### 2 CommentsShowHide 1 older comment Walter Roberson on 18 Nov 2021 Do not plot() the output of polyfit. If you want to plot the prediction line then use polyval to create projected locations and plot() those ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting! Translated by	508	1,817	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.484375	3	CC-MAIN-2022-21	latest	en	0.83414
https://www.electronicspoint.com/forums/threads/circuit-protection-ideas.116973/	1,618,541,403,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618038088471.40/warc/CC-MAIN-20210416012946-20210416042946-00085.warc.gz	760,252,065	12,929	# Circuit protection ideas? Discussion in 'Electronic Design' started by [email protected], Jul 2, 2008. 1. ### Guest I am putting together a design for an RGB LED controller using PWM to control color and intensity on RGB LED's. I want the controller to work with various LED's, so I am using a 4 position screw terminal. The controller just supplies 12 volts and switches ground on the 3 other terminals to effect the PWM. Then, on the LED module, I will use the appropriate resistors to work with the 12V power supplied from the controller. I use an NPN transistor with a current rating of 800mA to switch the grounds, and the transistor is controlled by a microcontroller. I am trying to add some sort of circuit protection such that the transistor will not be destroyed, even if the user of the device does something they shouldn't - like shorts the PWM-switched grounds right to power (which would happen if the LED wires are not inserted well into the controller). In my prototype design I tried resettable fuses from Littelfuse, but the transistor was still destroyed when I tried shorting it to ground. I think the fuses are not fast enough, despite being marketed as super fast. Can anyone offer suggestions? I am sure there are feedback circuit designs out there that would turn off the output if the current rose above a given value, or perhaps there is an easier solution out there? Board space is at a premium so I am trying to keep the circuit as physically small as possible. Any input is appreciated! Thanks CJ 2. ### Frank BussGuest Did you use the right one? Tripping current is much higher than hold current. E.g. if you use the 2920L030, which has a hold current of 0.3A, it needs up to 3 seconds @1.5A tripping current. And the min resistance is 1.2 Ohm, which means there could be high current spikes, e.g. 100ms with 3A, see diagram on page 2: http://www.littelfuse.com/data/en/Data_Sheets/Littelfuse_2920L.pdf I would suggest to build a constant current source. If space is a problem, there are some nice LED drivers from Maxim, with integrated output short-circuit protection: http://www.maxim-ic.com/quick_view2.cfm/qv_pk/5755 Digikey doesn't have it, but you can buy it from Maxim: 3. ### JamieGuest The transistor may be just fine how ever, you didn't specify in which manner the device is destroying it self? So allow me to come up with a scenario. Assuming you're using a open-collector (CLC), did you put in the circuit a resistor in series in the emitter for example to limit the current? Selecting a resistor to allow for max current when the transistor is in saturation is normally what I do, others may do it other ways. Also, did you account for wheeling voltages that form from arc and inductive loads? A unidirectional TVS diode works very well for this. It'll protect the transistor for reverse states and for over voltage states. Selecting one that is very close to the normal operating voltage of the transistor is normally a good idea. That is my contribution to this subject be is useable or not. http://webpages.charter.net/jamie_5"	756	3,098	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.515625	3	CC-MAIN-2021-17	longest	en	0.937325
https://m.hanspub.org/journal/paper/26692	1,611,591,886,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703587074.70/warc/CC-MAIN-20210125154534-20210125184534-00485.warc.gz	443,508,948	13,996	基于微穿孔板的通风管道噪声自适应控制算法研究 # 基于微穿孔板的通风管道噪声自适应控制算法研究Study on Adaptive Control Algorithm of Ventilation Duct Noise Based on Micro-Perforated Pane Abstract: Pipeline noise includes low, medium and high frequency noise. In order to realize the active control of broadband pipeline noise, this paper combined with impedance compound muffler structure and active noise reduction method, proposed resonance sound absorption structure weaken noise source of the sound radiation and combined with the impedance characteristics of microperforated panel acoustic absorption mode optimization matching method. This paper combines two adaptive control algorithms to study the sound absorption law of pipelines, proposes different active noise reduction schemes, simulates the iterative waveforms of different algorithms, and summarizes the influence of the parameters of its adaptive control algorithm on the sound absorption effect. The feed forward control system of the pipe sound absorption platform was established to further verify the conclusion of the experiment. 1. 引言 2. 前馈管道噪声控制模型 2.1. 阻抗复合式消声结构 Figure 1. The system model of feed-forward ANC (a) (b) Figure 2. The structure diagram of micro-perforated plate sound absorption 2.2. 主动控制策略 $\omega \left(n+1\right)=\omega \left(n\right)-2\upsilon x\left(n\right)e\left(n\right)$ (1) 1) 确定L、 $\upsilon$$\omega \left(0\right)$ ，其中L为滤波器的长度， $\upsilon$ 为步长， $\omega \left(0\right)$ 为时间n = 0时的滤波器的初始化值； 2) 计算自适应滤波器的输出y(n)； $y\left(n\right)=\underset{l=0}{\overset{L-1}{\sum }}{\omega }_{l}\left(n\right)x\left(n-l\right)$ (2) 3) 计算误差信号e(n); $e\left(n\right)=x\left(n\right)-y\left(n\right)$ (3) 4) 通过FXLMS算法更新自适应滤波器的权重； ${\omega }_{l}\left(n+1\right)={\omega }_{l}\left(n\right)+\upsilon x\left(n-l\right)e\left(n\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}l=0,1,\cdots ,L-1$ (4) h = adaptfilt.algorithm (input 1, input 2, ∙∙∙∙∙∙) (5) Figure 3. The block diagram of adaptive filtering algorithm Figure 4. The flow diagram of FXLMS algorithm program h = dsp.FilteredXLMSFilter (l, 'StepSize', mu, 'LeakageFactor', ∙∙∙∙∙∙ 1, 'SecondaryPathCoefficients', b) (6) 3. 前馈系统最优控制策略 3.1. 主动控制策略仿真 3.2. 前馈控制系统设计 Figure 5. Adaptive filtering of FXLMS algorithm Figure 6. The original noise signal (a)(b) Figure 7. The noise reduction effect of two adaptive algorithms Figure 8. The system diagram of feed-forward adaptive control Figure 9. The software block diagram of control system 3.3. 前馈控制系统实验 4. 总结 Figure 10. Physical drawings of the experimental system (a)(b) Figure 11. The sound pressure level at the measuring point under the primary noise 1) 在基于柔性微穿孔板的被动噪声控制下，加入主动噪声控制的方法，有利于低、中噪声频率的消除，使得整个管道实现宽频带噪声的控制； 2) 在主动控制策略方面，自适应算法的效果与自适应权重(自适应滤波系数)的选取有关，为了得到较好的减噪效果，通常采用编程的方法获取； 3) 通过比较自适应滤波的减噪效果，可以看出基于RLS算法的主被动控制系统具有更好的宽频带消声，但同时计算量也会增大。 [1] 陈克安. 有源噪声控制[M]. 北京: 国防工业出版社, 2003. [2] Sellen, N., Cuesta, M. and Galland, M.A. (2006) Noise Reduction in a Flow Duct: Implementation of a Hybrid Passive/Active Solution. Journal of Sound & Vibration, 297, 492-511. https://doi.org/10.1016/j.jsv.2006.03.049 [3] 赵栋, 李凯翔, 张飞, 庞彦宾. 主动噪声控制系统在C6000DSP上的设计与实现[J]. 测控技术, 2017, 36(1): 63-66. [4] Sen M. Kuo, Bob H. Lee, Wenshun Tian. 实时数字信号处理: 实践方法与应用[M]. 北京: 清华大学出版社, 2012. [5] 孙琳. 基于DSP的窄带主动噪声控制系统实现[J]. 仪器仪表学报, 2011, 32(2): 252-257. [6] 姜占才, 孙燕, 王得芳. 基于谱减和LMS的自适应语音增强[J]. 计算机工程与应用, 2012, 48(7): 42-45. Top	1,308	3,377	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.828125	3	CC-MAIN-2021-04	latest	en	0.588876
https://warwick.ac.uk/fac/sci/statistics/staff/academic-research/brettschneider/st222-resources-2017-18/previousweeks/	1,716,688,990,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058861.60/warc/CC-MAIN-20240526013241-20240526043241-00256.warc.gz	534,421,724	28,577	# Material from previous years (2016/17) - for archive only! #### Week 1: Tuesday/Friday: What are decisions under uncertainty? Some examples: investment, job offer, sport betting, non-emergency medicial treatment (e.g. chronic pain surgery, immunisation), simple game, St Peterburg paradox. In these examples, we have see key concepts that will be picked up later during Part I of this module. What is relevant in taking decisions? Mathematical models based on outcomes, (subjective) probabilities and values of these, utilities and risk attitudes are used to deteremine optimal (in some sense) decisions. It has some, but not full validity for homo sapiens, the species actually populating this planet, and issue we will pick up again in Part III. Friday: What is probability? Normative probability (Kolmogorov's axioms) based on set algebras to model events as subsets of outcomes and probability measures as functions of these. In other words, probability measures are functions on set algebras. Concept of atoms as fundamental events. House price decision model as an example for a very simple finite algebra on an a priori more complex (continuous) outcome space. Examples to illustrate the use axiomatic probability. (i) Two-headed coin: From a box with n-1 normal coins and one two-headed coin you draw a coin at random and toss it three times. It shows three heads. Which box do you think it came from, and how sure are you? Generalisation to k instead of 3 tosses. The example is discussed in Section 2.5.1 of the ST111 Lecture Notes (see r.h.s.). This is decision problem of the type seen in statistical tests, where the task is in "re-engineering" which was the random process that generated an outcome using likelihood ratios. (ii) Nick & Penny: Flipping coins with different probabilities for heads, p1 and p2. Model for infinitely many coin tosses as 0-1 sequences obtained as an (infinite) product (sigma-)algebra. The number of heads in n coin flips distributed following a Binomial distribution with parameters (n, p1) for Nick and (n, p2) for Penny (seen in first year probability). Determine the distribution of random variables N = the first time that both Nick and Penny obtain heads, and of random variable M = first time at least one of them get heads. ###### Study advice for week 1: If you do not feel entirely confident with the axioms of probability and the calculations in the examples, read also the simpler examples in the lecture notes and/or review your first year (discrete) probability notes. And then tackle exercise sheet 1. ###### Questions by students in week 1: • What is a nickel? A 10 pence coin. A silly joke in that example with Nick and Penny... • Why do we need that infinite product of algebras in the Nick and Penny example? We could set an upper limit to the number of coin tosses, and then we would not need that. However, if we want to look at random variables such as M and N above, or even simpler ones such as the waiting time T for the first head, they do not have an a priori upper bound. While it's extremly unlikely that it takes Nick 1 million tosses to obtain the first head, the probability for this event (T=1 million), is larger than zero. Of course, you could obtain arbitrarily good approximations by setting increasingly large upper bounds, and that's a common technique used on proofs of asymptotic results. #### Week 2 Tuesday: Continuation of the Nick & Penny example (ii) shows that the random time until both of them get heads is a waiting time and has a geometric distribution with parameter p1p2. The random time until at least one gets heads is also a waiting time and has a geometric distribution with parameter (1-(1-p1)(1-p2)). (iii) Class size example shows how two different perspectives can lead to different answers to a question if it is not phrased precisely enough. One could estimate the size of a class picked at random from all classes. Or one could estimate the size of a class a randomly picked student sits in. Depending on the class size distribution, this can lead to very different answers, a phenomenon called size bias, which is also behind the bus paradox. What does probability mean? Classical probability refers to the calculus of probabilities. Axiomatic probability is the rigorous mathematical set up for probability as a function from an algebra of events to [0,1]. There is also the frequency interpretation of probability, which can not serves as a stand alone definition, because it requires the law of large numbers to ensure existence of the limit of average numbers of relative frequency of an event in the first n trials as n goes to infinity. However, it serves as an interpretation. Friday noon: Then notion of subjective probability puts an emphasis on subjectiveness and individuality. How can we quantify a person’s degree of belief an event will happen? The key idea is elicitation through bets. The subjective probability for an event A is defined as the amount of money the person would be prepared to pay to enter the bet b(M,A) which pays M if A occurs and nothing otherwise. This is assumed to the same as the minimum the person would ask someone else to pay when offering that bet - technically intuitive, but actually a psychological assumption. After normalisation, and assuming there is not a nonlinear effect of M, this leads to a concept of probability that can be shown to obey the axioms of probability. We introduce the notions of collections of bets and of equivalent bets in the sense of giving equal rewards and study examples. Is the resulting probability mathematically sensible? We proof that under basic rationality assumptions, that people are willing to trade equivalent bets, subjective probability is additive, and hence fullfill the probability axioms. Friday 4pm: What happens if a personal has irrational beliefs about probabilities? We look at four examples. (i) In coin toss, J believes P(heads)=P(tails)=0.6. Students sell her the collection of bets b(1,{H}) & b({T}) for £1.2 which guaranties them a sure win of £0.2. (ii) Similar, but no J believes both probabilities are 0.4. Students get her to sell them for \$0.4 each, which again gives them a £0.2 sure profit. (iii) Let Omega be an outcome space. J believes P(Omega)=a<1. She is hence willing to sell b(£1,Omega) it for £a to the students, who end up making a sure profit of £(1-a)>0. (iv) Similar, but now instead of a it is a’>1. Students get her to buy from them b(£1,Omega) for £b and make a sure profic of £(b-1). A Dutch book is collection of bets for which one party can not loose. As examples we look at a lottery ticket seller and at a bookies strategies in horse betting using an explicit simple example with 4 horses. ###### Study advice for week 2: • If you forgot most of your probability theory, review it (e.g. use your own material from the first year, texbooks, ST111 lecture notes (2012)). • Read additional examples in the ST114 lecture notes not covered in class, such as an example for Dutch book using a race (with creative horse names such as "Go Lightening"). • Find ways to contruct Dutch books in real life. For example, elicit probability believes of your house mates, friends or family and see if you can benefit from these. Questions by students in week 2: • Why does the bookie make a profit? Because he set up the bets this way. He gets all the betting prices summing up to £210. But whichever horse wins, he only has to pay out £200 to the winner. In the languate of probability, he constructed a world where people pay for bets that add up to 1.05 total probability rather than 1, and he benefits from the extra 0.05. • Does this always work? What if unequal numbers of people bet on the hourses? You are spot on, I run out of time and didn't complete the discussion. Will pick that up again next week. Bookies need to adjust their odds depending on how many people buy bets on which horse. Also, they need to adjust them if a horse drops out before the race. #### Weeks 10-3 They will soon be erased to make room for this year's summaries. However, they will stay here until the end of week 2 so you can find more information about ST222 while finalising your module choices. Summary of content ST222 in 2016/17: Short summary sheet ##### Week 10 (scroll down for earlier weeks) Tuesday: Framing of contingencies, introduction to prospect theory. We consider three different preference choices between lotteries. Two of these are mathematically equivalent, but one is framed as a two stage process. This leads to subjects making inconsistent choices. Instead, they make the same choices as they do in a set up that only consists of the second stage, a phenomenon called isolation effect. Behind the preferences itself is the certainty effect. The mathematical basis for prospect theory consists of two ingredients: an S-shaped probability weighting function w capturing the underrating of probabilities close to but not equal to 1, the overestimation of probabilities close to 0, and the unrealistic low increase for average probabilities. While there is a continuum of suitable functions, we introduce two common ones (by KT and by Prelec). The value function v replaces the utility function in EUT; special features include the reference point and the asymmetry between gains and losses. These functions have been validated by lab experiments. Multiplying the values of outcomes with their weighted probabilities leads to the value of a prospect, which corresponds to the expected value of a decision in EUT. Friday noon: Common consequence, common ratio, modelling. After reviewing the prospect theory model, we introduce abstract versions of paradoxes initially formulated by Allais. In many studies, a majority of subjects does not follow EUT in any of these situations. In contrast, PT (prospect theory) can explain their choices. We give an explicit example for w and v that explains the choices in the common ratio situation. A similar construction can be done for the common consequences situation (exercise sheet 5). In these examples, we did not even need to use a value function v other than the identity, but it was crucial to invoke a nonlinear probability weighting function w. These provide two classes of examples, where prospect theory can explain behaviour driven by the certainty effect. In contrast, the value function is crucial to explain the framing effect. How can biases in decision making be reduced? We need to address this in the whole process of decision making. In the editing phase information is collected and prepared. Biases can arise for example from changing or misunderstanding information and from selection of sources or insufficient monitoring of their quality. In the evaluation phase probabilities and values of outcomes are being combined. It is mostly in this step here that mathematical training can improve the process. Emotions have been shown to interfere with both editing and evaluation phases, also known as affect heuristics. For example, sunny weather has been shown to affect the stock market, stress has an impact on preferences, and feelings of insecurity lead to a decreased tolerance of cognitive dissonance hence enhance confirmation bias. Mental balance of the decision maker usually improves the decision process. Friday 4pm: Exercises, modelling. Resources for Week 10 • Which mathematical models do you know? Are they appropriate? Have they been empirically validated? How? • Solve problems from exercise sheet 5 (will be posted by end of this week) Questions/comments by students in week 10: • Summery of results of the questions (corresponding to "Problem 5, 6, 7" in the lecture slides) distributed in Lecture 1. • What if they could entre the lotteries in the questions (distributed in lecture, see above) many times? Would people make choices following EUT more closely? Answer: I suppose yes, but it would have to be tried. • Why is the probability weighting function continuous? Answer: It's a model for something we can only indirectly observe. Building models means we want to have a simple but sufficiently complicated representation of the reality (here people's minds during decicion making under uncertainty), that can describe and predict such observations. A discontinuous probability weighting function would be a different model, but is appealing in that one could treat p=1 entirely differently from p<1. However, sticking to continious - actually, even better, differentiable functions - has the advantage that we can apply techniques from calculus. • Are there any field studies where prospect theory can explained observed behaviour better than expected utility theory? Answer: Here is a collection of study summaries. ##### Week 9 Tuesday: Paradoxes, heuristics and biases. Normative and descriptive theory of decision making exist in parallel coming from different angles. In addition, the prescriptive approach sitting at their intersection aim to train humans to make use of probability in the normative sense. The example of judging sample variation shows that this can work: Warwick ST222 students with training in probability gave much better answers to a question about the dependency of sample variation on sample size initially used by Kahneman and Tversky (K&T). However, a majority of Warwick ST222 students showed the same choice patterns in the Allais paradox as subjects in the literature, that is, not following EUT. In an experiment by Ellsberg with balls in urns subjects also show what looks like an inconsistent preference pattern that can be explained by ambiguity avoidance. In addition to randomly drawn outcomes, this set up features additional ambiguity by concealing the exact composition of colours in a bet. Friday noon: Base rate neglect, reason based choice and more incoherent choices. In many situations, people forget to take into account the initial probabilities (prevalences) of outcomes or swap around the role of events in conditional probabilities. This can lead to very incorrect probability judgements. Further examples for inconherent choices include an availability bias in numbers of path in different graphic structures, resonses to lost bills/tickets and to missing a flight. Another question is the motivation behind decisions. It has been found that people make faster decisions in conditions where one options clearly dominates the other. In conflict situations, people hesitate and will even try to obtain more options. Uncertainty about another outcomes, even if irrelevant, can also delay decision making. Friday 4pm: Conjunction fallacy, representativeness, confirmation bias. The Linda problem and its variations demonstrate how people give a higher priority to matching different parts of a story than to following the axioms of probability theory. More specifically, they chose rankings that suggest that P(B) is smaller than P(B and F), even though it should be the other way aournd.This has consistently been demonstrated under different types of experimental conditions and for subjects of different levels of justification, including ST222 students. It can be explain by representativeness theory. Another important and traditional phenomenon is confirmation bias, a tendency to select, weigh and deform information on order to, unconciously, make it conform to pre-existing beliefs. This is driven by human desire to minimise cognitive dissonance, though there is a lot of variation between humans when it comes to how much internal contradictions they can tolerate. Resources for Week 9: • Use some of the questions considered during the lectures to test how your friends and family answer and try to find out from them what motivated their choices. • Find examples where you delay a decision (D) because you are waiting for another decision (D'). Determine whether D' was actually relevant for D. Can you find examples where D' was irrelevant for D, but you still waited? Questions/comments by students in week 9: Please submit if you have any... ##### Week 8 Tuesday: Overview of Part I, II and III, normative versus descriptive theory, models of human beings. We look at overall structure of the module connecting including Part I, II and III. Part I is based on the assumption that human beings are rational, in the sense of maximising their expected rewards. The validity of these assumptions will be challenged in Part III of this module. One of the learning objectives of this module is mathematical/statistical modelling. A model is not intended to be perfect, but to be a simple representation of a part of the real world; good enough to make explanations and predictions for specified aims. To put the modelling task in the context of our objective to understand human decision making, we first ask a more general question: What are models of human beings? To illustrate on a very concrete example what this could be, we look at the mathematically inspired homme moyen that forms the basis for Le Corbusier's architectural theory. The rationality assumption we used in Part I is also based on a model of human being, nicknamed homo economicus, which is very common in traditional economic theories (that means, before the rise of behavioural economics). Friday noon: Imitating coin tossing. We conduct an experiment to get some experience with the process of generating sequences obtained by repeatedly tossing a fair coin. In one part of the classroom, students generate by themselves sequences that look like sequences of coin tosses (singles condition). In the rest of the classroom, students do that same but work in groups of three (trios condition). Mathematically speaking, we aiming to generate 0-1 sequences from independent identically distributed (i.i.d.) random variables with values 0 and 1, with equal probabilities for the two outcomes. Given a 0-1 sequence, how can you tell whether it was created by this model? You can not tell for sure, but you look at characteristics of the sequence and compare this with what be typical for a sequence created by the model. You can quantify this using probabilistic calculations and use the obtain probabilities to make judgements about the processes underlying the generation of your data (models). In sequences with just two outcomes, the main characteristics is how often they alternate or, in other words, how often they do not alternate. For example, we can look at runs. It is relatively simple to calculate the expected number of runs of a given length using indicator method. A more complex question is the distribution of the longest run. We derive a recursive formula that enables us to explicitly calculate the cumulative distribution function (CDF). Friday 4pm: Fallacies (Gambler's fallacy, hot hand, clustering illusion, perception of random patterns and sequences, anchoring bias, framing effect). The gambler's fallacy describes the belief that after a long run of outcomes of one kine, the other kind of outcomes because more likely. In the case of a sequence of independently created events such as coin tossing or roulette, this is wrong. People typically believe this, because of their intuitive understanding of what the law of large numbers says, that relative frequencies converge to the theoretically expected probabilities. However, it is wrong to apply this thinking on small samples. For independently sampled random variables, there is no memory in the sequence that would change the probabilities of outcomes to balance things out on a finite horizon. Of course, there are also games where independence is not true, such as Black Jack, which needs to be modelled taking into account that cards are not replaced. Practical applications of gambler's fallacy in the 'real world' have been found in situations where people make a lot of similar type of decisions (judges, load officers, referees etc). The hot hand as a form of opposite of the gambler's fallacy has been investigated in sports. Poisson point patterns tend to look not random to viewers who did not yet study this process, because they have more clusters than people think would occur at random. The perception of random sequences can differ depending on whether or not students could see the whole sequence. If being asked to create black & white sequences, 90% of people start with black. This is an example for anchoring bias. Traditional examples of anchoring bias involve estimations of geographical facts (e.g. length of Mississippi river) or numerical calculations. Surprisingly high differences in answers are observed depending on people were exposed to smaller or larger numbers, or even entirely unrelated tasks involving length, before being asked the question. Finally, the framing effect is introduced with the classical example by Tversky & Kahneman about a disease prevention programme. Resources for week 8: Questions/comments by students in week 8: • Are we going to learn about prospect theory? Answer: Yes, in particularly we will introduce the mathematical model of probability deformation. • How do we really know whether a given sequence is really from a fair coin? After many heads in a row, maybe one should conclude that it is actually two-headed coin? Answer: You can never know anything for sure. What statistics can do for you, however, is to tell you how likely your observations are under a certain model. This is the basic principle of Starting from observations (data), extract characteristics or summaries and then calculate how likely these were generated by a model. Compare different models and choose the most likely to explain the data (Maximum Likelihood Principle). We actually did some toy examples for this in Week 1 involving two-headed coins. There are also methods detecting dependency. However, it can be difficult to distinguish, for example, between time inhomogeneous probabilities and dependency just from observations (model unidentifiability) and contextual information may have to be invoked to decide on a most credible model. ##### Week 7 Tuesday: Separability. Our set-up for games is so general. To find out more about properties and strategies, we need to look at special classes of games. The concept of separability describes games where there is no interaction between the players' decisions. Formally, this is captured by a decomposition of the reward function (which depends on both players' moves) into a sum of two reward functions that each only depend on the move of one of the players. Note that these functions are not unique. The prisoner's dilemma is an example for a separable game, as can be seen by an explicit construction of such function. We derive necessary conditions for a generalised version of the prisoner's dilemma to be separable. As an immediate consequence of the additivity of the expected value, we obtain a simplified formula for the expected rewards of a separable game. Friday noon: Domination, purely competitive games. The concept of domination captures that some moves may be universally better or equal for a player than any of the other moves in the sense that regardless of what other player does they lead to a better or equal reward. Dominant moves should be played, because their expected reward is at least as high as it would be for any other move. This is a direct implication of the monotonicity of the expectation. A move is domimated if there is another move that dominates it. Moves that are dominated by others can be eliminated from the game in an iterative process. The use of these techniques can potentially lead to a simplified game matrix or even to a discriminating strategy. Examples show the techniques in action. Note that the assumption underlying all this is rationality of both players in the strict (and debated) definition that rationality can be reduced to the aim of maximising expected reward in each round of a game. Another important class are zero-sum games. They are a normal form (rewards adding up to 0) of what is captured in the concept of purely competitive games, where any amount that one of the players wins is at the other player's loss. A maximin strategy means to choose a move that brings a maximal reward over the worse cases determined as a function of what the other player does. In RSP, for example, that leads to the unfortunate outcome that each of the players expects to loose. Obviously, they can't both loose! What has not been taken into account is the interaction between them. Friday 4pm: Solving zero-sum games, mixed strategies, fundamental theorem. To resolve the paradox occurring with maximin strategies we extend the available options to probability distributions of moves (mixed strategies) instead of only deterministic moves (pure strategies). The expected rewards can be expressed as expectations of the rewards over these distributions. Maximin for Player 1 corresponds to minimax for Player 2. The strategies can be calculated for each of the players and an obvious questions is whether the corresponding expected rewards are the same. This feels intuitive for zero-sum games and is indeed the case, as stated by the fundamental theorem of zero-sum two player games. However, the proof is not simple. One inequality yields in more generality and can be shown easily in a lemma. The other direction is indirect and based on a smart contruction of a version of the game which then leads to a contradiction invoking the separating hyperplane theorem for convex sets. Note that the proof of this theorem is not constructive. It does not actually provide the mixed strategies where the maximin/minimax is attained, but only demonstrates their existence. Also, note that the maximin attained is unique and the same for both players, but the strategies do not have to be unique. Resources for week 7: • Write down some random game matrices and check if they are separable. • Construct game matrices with dominant moves. Find one that has dominant moves for Player 1, but not for Player 2. Find one that initially does not have dominant moves for Player 2, but does so after dominant moves for Player 2 were removed. • Look at shapes in your environment and decide which of them is convex. Start at breakfast: a bun? a croissant? a bagel? a pretzel? (Do this everywhere and long enough with every shape you notice, and you life will have changed forever! You can not revert back to not seeing convexity wherever it is.) • Can you see why the Separating Hyperplane Theorem (see slides of the proof of the fundamental theorem) needs the convexity assumption? Questions by students in week 7: • How do I find these functions in the definition of a separable game? Answer: You can stare at the numbers and guess and this works sometimes... Or you can rewrite the condition as a system of linear equations, which you can than tackle with standard methods you learned in linear algebra. The system may not have a solution, which implies the game is not separable. If it has a solution, no matter whether unique or not, then the game is separable. • In an economics module we learned a definition for strictly competitive similar to the definition of purely competitive in this module, but involving inequalities. What is the connection? Answer: It looks like the class you defined in economics is more general. It is a less quantitative condition. If player 1 gains, then player 2 looses, but unlike in our definition, the loss does not have to be by the same amount as the gain. Taking away this equality covers more examples and is more applicable, but it makes it much harder to develop precise theory as for example today with the fundamental theorem of zero-sum games. Of course, you could use the theory for purely competitive games to derive approximate answers or upper/lower bounds for questions about a strictly competitive game. ##### Week 6 Tuesday: Archimedian axiom, independence, von Neumann-Morgenstein representation, Allais paradox. Introduction of two more properties of binary relations that are needed for a more explicit representation of utilities. The Archimedian property describes a preference order where, given any strongest x, medium y and weakest z, mixing in a little bit of the weakest z to the strongest x does not reverse the preference of the strongest over the medium y. In other words, the weakest can not be incommensurably weaker than the medium. A corresponding property holds on the other end, when mixing in a bit of the strongest x to the weakest z. There is a conceptual connection to the Archimedian property on the real numbers, which says that for any epsilon > 0 and any real x there is an integer n such than nespsilon > x. Examples where the Archimedian property is not true can, for example, be generated by including outcomes that are hard to compare on the same scale such as monetary outcomes and death. The independence property captures the phenomenon that adding an additional option on both sides of a preference relation does not change the preference relation. Examples where the Archemedian property is not true occur, for example, when there is some kind of interaction between the additional option and one of the other ones. For example, providing a can opener is a useful addition to a can of soup but not for a load of dry bread. The von Neymann-Morgenstein representation theorem says that preferences relations that also these two properties can be represented as expected values of a utility function directly defined on the outcomes, with respect to the probability distributions that define the actions involved. In his 1957 seminal paper, the French economist Maurice Allais introduced en example involving lotteries challenging the dogma that humans make their choices based on expected utility theory. He conjectured, that people’s preferences in two pairs of lotteries would not be following expected utility theory. His conjecture has been confirmed in many empirical studies with human subjects in the sense that a large majority of people made their choices according to Allais’ predictions. Demonstrating that this contradicts expected utility theory amounts to showing that there exists no utility function that could represent them by constructing a contradiction between two inequalities derived from the conditions following from people’s preferences. The set-up for this was done in lecture, but you need to complete the calculations as homework, and you will also discuss that this can be connected to the independence property not being generally valid (see exercise sheet 4). Friday noon: Classtest. Results will appear online within 4 weeks. You will receive an email notice. Friday 4pm: Introduction to games. What is a game and how do we describe this mathematically? Games we are thinking of are rock-paper-scissors, board games, card games, prisoners dilemma. From a decision theory point of view, a game is just two (or more) people are taking turns making decisions. We consider games of two in this modules. For each combination of decisions, there are reward functions for each player which leads to a description of a game as a payoff matrix. Through playing both RSP and prisoners dilemma for a few rounds, we discovered and experienced some of the concepts that are relevant to describe and analyse games: sources of uncertainty, hypothesising about the opponents moves, iteration of this strategy and consequences, competetive versus cooperative games, the role of trust, the consideration of long term and short term goals and different implications, the role and the value of information and the gain of information as a potential intermediate gain in repeated games. If Player A has a probability distribution for the move Player B will make, then player A can use this to calculate his expected reward and vica versa. Natural questions are when expected rewards are independent of such distributions and what is the role and the implications of assuming rationality. Resources for week 6: • Continue to solve the problems from Exercise Sheet 3. • Play games with your friends, family, roommates. • Explore playing RPS against a computer that was trained on humans. • Think about what distinguishes games from other games. What are sensible classes for payoff matrices? Questions by students in week 6: • Can the payoffs be negative? Answer: Yes. And note they are not always uniquely defined. They are just a description of a game. For example, in RPS, we can use -1, 0, 1 to code for loss, draw, win, but we might have used other numbers. • What about utility? Answer: Yes, very important point. You could convert the given payoff matrix by replacing each players rewards by their utilities. Since the players may have very different utilities based on their personalities or circumstances, this may fundamentally change the game. A fair game may become unfair, for example, or what was an advantage for one player by become an advantage for the other. ##### Week 5 Tuesday: Certainty money equivalent and utility. Look at certainty money equivalent (CME) as a function of p. Example of a naive’s person’s CME for bets based on rolling dice, which shows a convex shape. Last time we defined utility as the inverse of CME. We can alternatively start with a utility function, and construct CME m as a function of probability through the equation U(m)=E[U(b(p)], where b(p) is the bet that pays £t with probability p and £s otherwise, for fixed parameters s, t. Both CME and utility are subjective context dependent functions. We consider the example of buying fire insurance for a house. The contrast between the utility for the owner of the house the the seller of the insurance explains why there is room for a deal. The owner is risk averse, because relative to the owner’s wealth the house means a lot. The insurer is risk neutral, because this deal is just one of a huge number of similar deals, hence it is justified to approach this with a simple expectation of wealth without further transformations. Risk aversion corresponds to a concave utility function and risk neutral corresponds to a linear one. A different example is lottery. Similar calculations (see resources below) show that playing the lottery can be explained with risk seeking attitude (convex utility). Friday noon: Binary relations and their properties. We define binary relations on action spaces (e.g. spaces of bets) and introduce some of the fundamental properties. Asymmetry (A) states that given a relation, the opposite relation is wrong. Completeness (C) essentially means that for any two actions x and y, people need to make up their minds whether x is preferable to y, or y to x, or whether they consider them being equally preferable. While this sounds banale, it is not always justified in practical examples. For example, not all people can make sense of the alternative “be stupid and satisfied” or “be smart and dissatisfied”. Also, choices must be available at the same time in the same location (e.g. “foie gras” and “hot dog”). Transitivity (T) means that a relation between x and y combined with a relation between y and z implies that same relation for x and z. Rational people should have transitive preferences, because otherwise their preference would have cycles and one could engage them in a series of exchange trades that would cause them to loose money with certainty. However, transitivity may not be always true in real world situation and needs to be checked when preference are used in models. For example, relations like friendship, or multi-attribute based preferences (e.g. ijkphone example), may not be transitive. Also, incremental changes can clash with transitivity of the relation equally preferable. For example, coffee with n grains of sugar is equally preferable to coffee with n+1 grains of sugar for all n, but iterating transitivity would imply that coffee with 1 grain of sugar is equally preferable to coffee with 1 million grains of sugar, which is probably unrealistic for most people. Negative transitivity (NT) is a condition similar to transitivity and can be linked to transitivity when (C) and (A) are also assumed. Friday 4pm: Preference relation, numerical representation, representation theorem, lexicographical order. A binary relation with (A), (C) and (NT) is called preference relation. A numerical representation of a preference relation is a function mapping the actions to the real numbers that preserves the order induced by the preference relation. We proof a theorem saying that preference relations on action spaces that have a countable order dense subset have a numerical representation. The numerical representation is not unique, which can be seen by going through the proof (several choices where made), or by showing that from a given numerical representation we can easily construct more (e.g. by multiplying them with a constant). The lexicographical order on the plane is defined by first ordering by the first coordinate, and if tied the second coordinate. We show that this order does not have a numerical representation. Resources for week 5: • Think about what is your utility function in some real world situations. Compare with those created by your fellow students. Consider situations with gains and situations with losses. • Solve the problems on Sheet 3 Questions by students in week 5: • In the proof that there is no numerical representation for the lexicographical order you construct a map by saying for every r there is a q between u(r,2) and u(r,1). Is the map well defined? Is that unique? Answer: It is well defined, but it is not unique. As the choices are still countable, you do not even need the axiom of choice to justify you can pick one of the choices. • Can you give an explicit construction of this q? Answer: Yes, we can. See these notes (not covered during the lecture and not examinable). • Why does the theorem not apply to the lexicographical order? Which condition(s) of the theorem are not valid in this case? Explain why. Answer: Will appear here by Wednesday, try yourself first. ##### Week 4 Tuesday: EMV decision rule, reward perspective, decision trees, eye disease treatment decision. As decision rule is in an algorithm to determine a decision under uncertainty based on a set D of decision options, as set Chi of outcomes with (subjective) probabilities p_i and a loss function L. One very common decision rule is the expected monetary value strategy (EMV) which defines the optimal decision d* as one that minimises the expected loss E[L(d,X)]. Sometimes, it is more natural to phrase the consequences using a reward function R rather than a loss function. No new model is used for this, but we simply define L=-R and obtain that d* maximises the reward. Decisions under uncertainty can be visualised with decision trees. With more complex and multistage decisions, these can get very big (see next lecture using slides). Example for a medical decision based on uncertain evidence: A test for an eye disease with prevalence p is conducted involving the distinction of n images. People without the disease perform better than those with the disease, but either group performs perfectly. The physician needs to decide whether to perform an inexpensive treatement now (d_1), or do nothing. The latter means potentially having to perform a more costly treatment later, if the patient did have/was developing the disease. Some evidence comes from how many images r the patient recognised correctly in the test. This can be phrased as a decision problem and we can determine which is the optimal decision as a function r, given fixed parameters n and p. We set up everything, but didn't entirely finish the calculation - try this yourself and see below for the solution. Friday noon preview: Complex multilevel decision tree example (Oil drilling). Value of information. We look at an decision about drilling for oil in either field A or field B. The decision is made more complex by the additional options to conduct test drills in A or B which give some evidence but not certainty about the presence of oil in these locations. All relevant probability estimates are given, but it needs Bayes rule to derive all probabilities need to apply the EMV strategy to derive the optimal decision. An interesting concept in this context is the value of information. In fact, there are two alternative definitions. One allows for imperfect information that is typically practically available. The other one is for perfect information that may not be available, but it may still be of interest to calculate its values; for example to serve as an upper bound for the value of any even hypothetical sources of information. Friday 4pm: Pros and cons of EMV strategies and alternatives. Example Farmer. The EMV strategy is a transparent and systematics way of arriving at an optimal decision for quantitative outcomes such as money (gains or losses), time, or anything where quantitative measures are available (education outcome, health status etc). However, it requires that subjective probabilities are available. An alternative are, for example, maximin and maximax strategies that represent an optimist's and a pessimist's approach to decisions making and are very intuitive. A disadvantage is that they are driven by extremes. Another disadvantage of the EMV approach is that it does not take into account the subjective value of money, which has been proven wrong in empirical studies with people in many situations - including in a survey conducted with ST222 students last year! This can be overcome by applying a utility function to the loss function. We introduce the certainty money equivalent and define utility as its inverse. This provides a concept of utility based on elicitation through bets. Resources for week 4: • Draw some decision trees of your own • Complete the missing calculations in the eye disease example • Solve the problems on Sheet 2 Questions by students in week 4: • Nobody really got a chance to ask a question yet, but a good question to ask would be: "Is d* unique?". The answer is no, it is not unique. There could be two (or more) optimal decisions in the sense that they all lead to the same expected monetary value. • After the Friday noon lecture there were some comments on the suitability of EMV. They were spot on, as we are about to consider alternatives to EMV in the Friday 4pm class. A particular issue is the appropriateness of using expectations for decisions that are only taken once. It will depend on the situation. If this is about the one time decision that can lead to ruin or death, expected values can be of limited use for an individual or a small company. Thinks of Keynes quote "In the long run, we are all dead!" questioning the overuse of asumptotic approaches in economics. On the other hand, if this is just one of many decisions taken by a big company, there is a justification to apply expectations. ##### Week 3 Tuesday: Dutch book example, elicitation of probabilities, rational individuals are coherent. In horse betting example for Dutch books from last week it is important to note that bookie may have to close bets or keep adjusting odds (e.g. horse dropped, unequal sales). Elicitation of subjective probabilities through comparison with bets in familiar situations such as simple probabilistic experiments (e.g. spinner, balls in urn). An individual is called coherent if her subjective probability assignments obey the axioms of probability. A person is rational if she would not make any deal that may be disadvantageous to her, in particularly would not bring herself in a situation that a Dutch book could be constructed against her. This can be used to show that a rational agent’s subjective probabilities must be coherent. This is discussed in detail in the lecture notes Theorems 3.1 and 3.2. We show one of the two cases of the proof for Theorem 3.1, the others are very similar. The key idea is to convert the subjective probabilities into bets and then construct Dutch books from suitable bets. Friday noon: Review and examples conditional probability, random variables, expected value. Review of definition of conditional probability, non-symmetry, relationship with independence, law of total probability, Bayes theorem, Bayes rule. Called off bet b(A\|B) is the best that only happens if B occurs. Its value is the same as the value of a simple bet b(C) on an event C with probability P(A\|B). Example: Screening test for a condition (e.g. disease, drug use). Information about the quality of a tests is provided as the probability that the test shows a positive results given the subject has the condition and the probability that the test shows a negative ressults given the subject does not have the condition. From the point of view of a subject who was tested positive is relevant to find out what is the probability that this is actually correct. Bayes rule is used to calculate this. With imperfect tests and low prevalence of the condition (e.g. popultation wide rare disease testing without symptoms, drug test in people without specific evidence or history of use), the probability that given a positive test the subject really has the condition can be suprisingly low (in our example is was lower than 10%). Friday 4pm: Expectation of random variables, prediction, loss function, decision making. Expectation of a discrete random variable as sum. If (countably) infinite the sum may or may not converge. (For continuous random variables they become integrals.) What is the best predictor for the value of a random variable? It depends on how you measure “best”. We introduce three alternative loss functions measuring deviations from the prediction and define the best predictor as the value that minimises the expected value of the loss function L applied to the random variable X. For show that for a simplistic L distinguishing only right and wrong, the best predictor b is the mode of X (follows from definition of the mode). For L measuring the absolute deviation: homework (solution). For L measuring the squared error, b is the expectation of X (proof using calculus). Loss functions also have a central role in decision theory. Model for decision making includes decision options, outcomes, subjective probabilities and a loss function. We model the task of decision making as finding the decision that minimises the expectation of a loss function on the product space of decision space and outcome space. Burglary insurance example. Resources for week 3: • Study the extra examples about Dutch books in the lecture notes not covered in class. • Try to proof Theorems 3.1 and 3.2 yourself. • Construct decision models for examples in your own life. • Solve problems from exercise sheet 2. Questions by students in week 3: There were not really questions suitable to discuss here, but many thanks for comments that helped resolving typos on the board, and also for questions to clarify things - it’s very good for me to know whenever something needs to be explained better. Also, I have some question which are not examinable at all, but may be good for passing your time while waiting for the U1 (or, maybe, for conversation with your housemates or at a party): • Can any rational person pretend to be irrational? • Can any irrational person pretend to be rational?	9,608	47,426	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.453125	3	CC-MAIN-2024-22	latest	en	0.93379
https://www.gradesaver.com/textbooks/math/prealgebra/prealgebra-7th-edition/chapter-3-section-3-1-simplifying-algebraic-expressions-exercise-set-page-169/3	1,537,670,827,000,000,000	text/html	crawl-data/CC-MAIN-2018-39/segments/1537267158958.72/warc/CC-MAIN-20180923020407-20180923040807-00514.warc.gz	734,205,458	12,927	## Prealgebra (7th Edition) $-1n$ $2n-3n=-1n$	25	46	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.5625	3	CC-MAIN-2018-39	longest	en	0.596537
http://www.opengl.org/discussion_boards/showthread.php/139495-Interleaved-arrays	1,369,450,843,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368705352205/warc/CC-MAIN-20130516115552-00025-ip-10-60-113-184.ec2.internal.warc.gz	623,911,507	10,522	# The Industry's Foundation for High Performance Graphics 1. ## Interleaved arrays?? Is it possible to display an indexed interleaved array?? If i have a gigantic interleaved array of verticies, texture coords, and normals, and an array of indexes. Can I use a function that will display it?? So far ive found that index arrays only work with vertices, is that true?? 2. ## Re: Interleaved arrays?? Vertex arrays can contain vertex position, color, texture coordinate, normals, fog coordinare, and perhaps some more stuff. Whether you dereference this array using indices, or draw them as they are stored, is irrelevant. How they are stored physically in memory is also irrelevand (that is, separate or interleaved arrays). As long as you can specify the arrays using a start pointer, what part of a vertex it represent, the type of data stored, how many componets it contains, and the distance between two consecutive entries, you can use any of OpenGL's functions to draw the array. 3. ## Re: Interleaved arrays?? Ok so say I have this setup float vertexinfo[]= {v1, v2, v3, t1, t2, n1, n2, n3, v4, v5, v6, t3, t4, n4, n5, n6} where v is the verticies, t is texture coords n is normals, and so. Then i have int index[]= {1, 2, 6, 3, 6, 4, 2} etc. i call glVertexPointer and point it to vertexinfo(only the parts i need) i call glNormalPoint and do the same thing and so on Then i call glInterleavedArray and point it to index??? 4. ## Re: Interleaved arrays?? I believe you have some things to learn concerning vertex arrays, like what the different commands do. Go read the Red Book , chapter two. Basically, glInterleavedArray() is a wrapper for all the gl[Vertex\|Normal\|Color\|TexCoord]Pointer() commands. You use one, and only one, of them to setup an array, never both. None of them draw anything, that is a job for glDrawElements/glDrawArrays, and similar commands. 5. ## Re: Interleaved arrays?? BRAIN FART, i knew that you only use one at a time, god im so mentaly challenged. Sorry i was at work when i wrote that, inputting database quote AHHHHHHHHH. anywho thanks for the help... #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •	582	2,256	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2013-20	latest	en	0.874041
https://classroomsecrets.co.uk/mixed-age-year-5-and-6-converting-units-step-5-resource-pack/	1,716,750,864,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00116.warc.gz	132,053,537	79,149	Mixed Age Year 5 and 6 Converting Units Step 5 Resource Pack – Classroom Secrets \| Classroom Secrets All › Mixed Age Year 5 and 6 Converting Units Step 5 Resource Pack # Mixed Age Year 5 and 6 Converting Units Step 5 Resource Pack ## Step 5: Year 5 and 6 Mixed Age Money Step 5 Resource Pack Year 5 and 6 Mixed Age Money Step 5 Resource Pack includes a teaching PowerPoint and differentiated varied fluency and reasoning and problem solving resources for this step which covers Year 6 Imperial Measures for Summer Block 1. ### What's included in the Pack? This Year 3 and 4 Mixed Age Money Step 5 pack includes: • Year 5 and 6 Mixed Age Money Step 5 Teaching PowerPoint with examples. • Year 6 Imperial Measures Varied Fluency with answers. • Year 6 Imperial Measures Reasoning and Problem Solving with answers. #### National Curriculum Objectives Differentiation for Year 6 Imperial Measures: Varied Fluency Developing Questions to support comparing and converting imperial measures. Converting using whole numbers, doubling or halving. Expected Questions to support comparing and converting imperial measures. Converting numbers with up to 2 decimal places and using direct fractions of the conversions, e.g. 1/8 when converting ounces to pounds. Greater Depth Questions to support comparing and converting imperial measures. Converting numbers up to 2 decimal places, percentages and using equivalent fractions of the conversions, e.g. 2/16 when converting ounces to pounds. Reasoning and Problem Solving Questions 1, 4 and 7 (Reasoning) Developing Explain whether a statement is true or false. Converting using whole numbers - doubling or halving. Expected Explain whether a statement is true or false. Converting numbers with up to 2 decimal places and using direct fractions of the conversions, e.g. 1/8 when converting ounces to pounds. Greater Depth Explain whether a statement is true or false. Converting numbers with up to 2 decimal places, percentages and using equivalent fractions of the conversions, e.g. 2/16 when converting ounces to pounds. Questions 2, 5 and 8 (Reasoning) Developing Explain an error. Converting using whole numbers - doubling or halving. Expected Explain an error. Converting numbers with up to 2 decimal places and using direct fractions of the conversions, e.g. 1/8 when converting ounces to pounds. Greater Depth Explain an error. Converting numbers with up to 2 decimal places, percentages and using equivalent fractions of the conversions, e.g. 2/16 when converting ounces to pounds. Questions 3, 6 and 9 (Problem Solving) Developing Convert an imperial measure. Converting using whole numbers - doubling or halving. Expected Convert an imperial measure. Converting numbers with up to 2 decimal places and using direct fractions of the conversions, e.g. 1/8 when converting ounces to pounds. Greater Depth Convert an imperial measure. Converting numbers with up to 2 decimal places, percentages and using equivalent fractions of the conversions, e.g. 2/16 when converting ounces to pounds.	664	3,045	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.921875	4	CC-MAIN-2024-22	latest	en	0.769131

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