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https://aerospaceanswers.com/question/what-is-a-g-force-how-do-g-force-affect-the-human-body/
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# What is a g-force ? How do g-force affect the human body ?
What is a g-force ? How do g-force affect the human body ?
Asked on 13th July 2020 in
• 1 Answer(s)
1 g is equal to the gravitational acceleration on earth which is equal to 9.8 $$m/{s^2}$$. G force is the vector sum of non-gravitational and non-electromagnetic forces. G force is resistance as a result of mechanical forces to the object. When an object is in free-fall under the influence of gravitation, the condition is known as zero-g. G force is explained by the aircraft pilots.
Pilot sustain g forces along the axis of the spine. There is a variation in the blood pressure along the length of the body. There is a limit to the maximum g forces which can be tolerated by a human body. A positive vertical g force can cause a blackout, G-LOC, or even death if g force is not quickly reduced.
Answered on 12th August 2020.
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2020-10-21 07:52:49
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https://math.stackexchange.com/questions/2548758/alternative-way-to-evaluate-int-0-pi-4-frace-sec-x1-tan-x1-sin
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# Alternative way to evaluate $\int_0^{\pi /4} \frac{e^{\sec x}(1+\tan x)}{1-\sin x} \mathrm dx$
The question is to evaluate $$\int_0^{\pi /4} \frac{e^{\sec x}(1+\tan x)}{1-\sin x} \mathrm dx$$
I tried by rewriting the integral as $$\int_0^{\pi /4} \frac{e^{\sec x}(1+ \sin x)(1+\tan x)}{\cos^2 x} \mathrm dx$$
which can be rewritten as $$\int_0^{\pi /4} e^{\sec x} \sec^2 x \mathrm dx \text{. }+\int_0^{\pi /4} e^{\sec x} \sec x \tan x (1+ \sec x + \tan x) \mathrm dx$$ Now using integral by parts on second integral easily yield the result
Is there any other way to evaluate it?
• Easily how? I don't see any way to solve it easily... – Raffaele Dec 3 '17 at 11:16
• There is a simple primitive in this case: $e^{\sec x}\cos x/(1-\sin x)$. Maybe you can find a way to obtain it? – mickep Dec 3 '17 at 12:22
That seems like the simplest way to go about it. Or perhaps you could try using the Weierstrass substitution $t = \tan \frac{x}{2}$ which is quite popular and does reasonably well with trigonometric integrals, although I think the method you propose is easiest.
• You meant $t=\tan\frac{x}{2}$ – Raffaele Dec 3 '17 at 11:48
• @Raffaele Yes, fixed it thanks. – Luke Collins Dec 3 '17 at 13:08
$t=\tan\dfrac{x}{2}\to x=2\arctan t;\;dx=\dfrac{2dt}{1+t^2}$
We have the following formulas
$$\sin x=\frac{2t}{1+t^2};\;\cos x=\frac{1-t^2}{1+t^2};\;\tan x=\frac{2t}{1-t^2}$$
Substituting, $\tan\frac{\pi}{8}=\sqrt 2-1$
$$\int_0^{\pi /4} \frac{e^{\sec x}(1+\tan x)}{1-\sin x} dx=\int_0^{\sqrt{2}-1} \,\frac{e^{\frac{1+t^2}{1-t^2}}\left(1+\frac{2t}{1-t^2}\right)}{1-\frac{2t}{1+t^2}}\cdot \frac{2t}{1+t^2}$$
$$\int_0^{\sqrt{2}-1} \,\frac{2 e^{\frac{t^2+1}{1-t^2}} \left(t^2-2 t-1\right)}{(t-1)^3 (t+1)}\,dt=\left[-\frac{e^{\frac{t^2+1}{1-t^2}} (t+1)}{t-1}\right]_0^{\sqrt{2}-1} =e^{\sqrt{2}} \left(\sqrt{2}+1\right)-e$$
Hope this can be useful
• I think the end result should be doubled. In the final line you have lost the factor of 2 from your integral. – James Arathoon Dec 3 '17 at 12:14
• @JamesArathoon Thank you! I fixed it – Raffaele Dec 3 '17 at 15:05
For the easy route hinted at by the OP, I think you have to first notice that
$$\frac{d\left( e^{\sec x}\right)}{dx}=e^{\sec x}\sec x \tan x$$
Then via integration by parts we find that
$$\int e^{\sec x} \sec^2 x \tan x \;dx= e^{\sec x} \sec x - \int e^{\sec x} \sec x \tan x \;dx\tag{1}$$
and similarly after that
$$\int e^{\sec x} \sec x \tan^2 x \; dx= e^{\sec x} \tan x - \int e^{\sec x} \sec^2 x \; dx \tag{2}$$
Since the required integral is the sum of the four integrals in (1) and (2)
$$\int_0^{\pi /4} \frac{e^{\sec x}(1+\tan x)}{1-\sin x} \;dx = \int_0^{\pi /4} e^{\sec x} \sec^2 x \tan x + e^{\sec x} \sec x \tan x + e^{\sec x} \sec x \tan^2 x + e^{\sec x} \sec^2 x \;dx$$
we immediately have
$$\int_0^{\pi /4} \frac{e^{\sec x}(1+\tan x)}{1-\sin x} dx = \left[e^{\sec x}(\sec x + \tan x) \right]_0^{\pi/4}$$
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2019-06-15 20:43:23
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https://aimsciences.org/article/doi/10.3934/era.2007.14.42
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# American Institute of Mathematical Sciences
2007, 14: 42-59. doi: 10.3934/era.2007.14.42
## A characterization of the concept of duality
Citation: Shiri Artstein-Avidan and Vitali Milman. A characterization of the concept of duality. Electronic Research Announcements, 2007, 14: 42-59. doi: 10.3934/era.2007.14.42
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2018 Impact Factor: 0.263
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2019-10-14 08:55:41
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http://en.wikipedia.org/wiki/Visual_Angle_Illusion
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# Perceived visual angle
(Redirected from Visual Angle Illusion)
In human visual perception, the visual angle, denoted θ, subtended by a viewed object sometimes looks larger or smaller than its actual value. One approach to this phenomenon posits a subjective correlate to the visual angle: the perceived visual angle or perceived angular size. An optical illusion where the physical and subjective angles differ is then called a visual angle illusion or angular size illusion.
Angular size illusions are most obvious as relative angular size illusions, in which two objects that subtend the same visual angle appear to have different angular sizes; it is as if their equal-sized images on the retina were of different sizes. Angular size illusions are contrasted with linear size illusions, in which two objects that are the same physical size do not appear so. An angular size illusion may be accompanied by (or cause) a linear size illusion at the same time.
The perceived visual angle paradigm begins with a rejection of the classical size–distance invariance hypothesis (SDIH), which states that the ratio of perceived linear size to perceived distance is a simple function of the visual angle. The SDIH does not explain some illusions, such as the Moon illusion, in which the Moon appears larger when it is near the horizon. It is replaced by a perceptual SDIH, in which the visual angle is replaced by the perceived visual angle. This new formulation avoids some of the paradoxes of the SDIH, but it remains difficult to explain why a given illusion occurs.
This paradigm is not universally accepted; many textbook explanations of size and distance perception do not refer to the perceived visual angle, and some researchers deny that it exists. Some recent evidence supporting the idea, reported by Murray, Boyaci and Kersten (2006), suggests a direct relationship between the perceived angular size of an object and the size of the neural activity pattern it excites in the primary visual cortex.
## A relatively new idea
Visual angle illusions have been explicitly described by many vision researchers, including Joynson (1949), (McCready 1963, 1965, 1985, 1999), Rock & McDermott (1964), Baird (1970), Ono (1970), Roscoe (1985, 1989), Hershenson (1982, 1989), Reed (1984, 1989), Enright (1989), Plug & Ross (1989, 1994), Higashiyama & Shimono (1994), Gogel, & Eby (1997), Ross & Plug (2002), and Murray, Boyaci & Kersten (2006). Specifically, these researchers cited have advocated a relatively new idea: that many of the best-known size illusions demonstrate that for most observers the (subjective) perceived visual angle, θ′, can change for a viewed target that subtends a constant (physical) visual angle θ.
Indeed, various experiments have revealed most of the factors responsible for these visual angle illusions, and a few different explanations for them have been published (Baird, Wagner, & Fuld, 1990, Enright, 1987, 1989, Hershenson, 1982, 1989, Komoda & Ono, 1974, McCready, 1965, 1985, 1986, 1994, Ono, 1970, Oyama, 1977, Reed, 1984, 1989, Restle, 1970, Roscoe, 1985, 1989).
On the other hand, nearly all discussions (and explanations) of those classic size illusions found in textbooks, the popular media, and on the internet use, instead, an older hypothesis that the visual angle is not perceivable (Gregory, 2008, Kaufman & Kaufman, 2002). They can describe and explain only a linear size illusion, which is why they do not properly describe or explain the illusions that most people experience.
In order to clarify the new paradigm which replaces the old one, it helps to keep in mind that an angle is the difference between two directions from a common point (the vertex). Accordingly, as described below, the visual angle θ is the difference between two real (optical) directions in the field of view, while the perceived visual angle θ′, is the difference by which the directions of two viewed points from oneself appear to differ in the visual field.
## Physical measures S, D, R, and θ
Main article: Visual angle
Figure 1: Physical measures
Figure 1 illustrates an observer's eye looking at a frontal extent AB that has a linear size S (also called its "metric size" or "tape-measure size"). The extent's lower endpoint at B lies at a distance D from point O, which for present purposes can represent the center of the eye's entrance pupil.
The line from B through O indicates the chief ray of the bundle of light rays that form the optical image of B on the retina at point b, let's say, on the fovea. Likewise, endpoint A is imaged at point a.
The optical (physical) angle between those chief rays is the visual angle θ which can be calculated:
$\tan \theta = S/D\,$
The retinal images at b and a are separated by the distance R, given by the equation
$R/n = \tan \theta\,$
in which n is the eye's nodal distance that averages about 17 mm. That is, a viewed object's retinal image size is approximately given by R = 17 S/D mm.
The line from point O outward through object point B specifies the optical direction, dB, of the object's base from the eye, let's say toward the horizon. The line from point O through point A specifies that endpoint's optical direction, dA, toward some specific elevation value (say, 18 degrees). The difference between those real directions (dAdB) is, again, the visual angle θ.
## Perceived measures
Figure 2 diagrams the perceived (subjective) values for a viewed object.
Figure 2: Subjective values
Point O′ represents the place from which the observer feels that he or she is viewing the world. For present purposes, O′ can represent the cyclopean eye (Ono, 1970, Ono, Mapp & Howard, 2002).[1]
### Perceived linear values D′ and S′
In Figure 2, D′ is the perceived distance of the subjective point B′ from O′. The observer might simply say how far away point B′ looks, in inches or meters or miles.
Similarly, S′ is the perceived linear extent by which the subjective point A′ appears directly above point B′. The observer could simply say how many inches or meters that vertical distance looks. For a viewed object, S′ thus is its perceived linear size in meters, (or apparent linear size).
### Perceived visual angle θ′
The perceived endpoint at B′ has the perceived direction, d′B, and the observer might simply say "it looks straight ahead and toward the horizon."
This concept of the (subjective) visual direction is very old.[2] However, as Wade, Ono & Mapp (2006) noted, it unfortunately has been ignored in many current theories of size perception, and size illusions.
The object's other perceived endpoint, A′, has a perceived direction d′A;, about which the observer might say "it appears toward a higher elevation than point B′." The difference between the two perceived directions (d′Ad′B) is the perceived visual angle θ′, also called the perceived angular size or apparent angular size.
It is not easy to quantify θ′. For instance, a well-trained observer might say that point A′ "looks about 25 degrees higher" than B′, but most cannot reliably say how large a direction difference looks. That skill is not practiced because it is easier to use pointing gestures (Ono, 1970): For example, one often tells another person about the change in the directions seen for two viewed points by pointing something, say a finger or the eyes from one point to the other.
Therefore, in some experiments the observers aimed a pointer from one viewed point to the other, so the angle through which the pointer rotated was the measure of θ′, (Komodo, 1970, Komodo & Ono, 1974, Ono, Muter, & Mitson, 1974, Gogel & Eby, 1997).
Also, because θ′, specifies the amount by which one should rotate one's eye to quickly look from one seen point to another eye tracking, saccade, observers in other experiments shifted their gaze from one object endpoint to the other, and the angle the eye rotated through was measured as θ′ for that object (Yarbus (1967).
### Difference between θ′ and S′
It is important to understand how θ′ differs from S′. Consider an example illustrated by the sketch at the right.
Suppose one is looking through a window at a 30-foot-wide (9.1 m) house 240 feet away, so it subtends a visual angle of about 7 degrees. The 30-inch-wide (760 mm) window opening is 10 feet away, so it subtends a visual angle of 14 degrees.
It can be said that the house "looks larger and farther away" than the window, meaning that the perceived linear size S′ for the house's width is much larger than S′ for the window; for instance a person might say the house "looks about 40 feet wide" and the window "looks about 3 feet wide."
One can also say that the house "looks smaller and farther away" than the window, and that does not contradict the other statement because now we mean that the amount (θ′) by which directions of the house's edges appear to differ is, say, about half the apparent direction difference for the window edges.
Notice that humans experience both the linear size and the angular size comparisons at the same time, along with the distance comparison (Joynson, 1949). Thus any report merely that one object "looks larger" than another object is ambiguous. It needs to specify whether "looks larger" refers to the perceived angular size (θ′) or to the perceived linear size (S′) or to both of those qualitatively different "size" experiences (Joynson, 1949, McCready, 1965, 1985, Ono, 1970). Notice that in everyday conversations "looks larger" often refers to an angular size comparison rather than a linear size comparison.
Additional confusion has resulted from widespread use of the ambiguous terms "apparent size" and "perceived size", because they sometimes have referred to θ′ and sometimes to S′ without clarification, so the reader must try to ascertain what they mean. Also, in astronomy, "apparent size" refers to the physical angle θ rather than to the subjective apparent visual angle θ′.
### The perceptual size–distance invariance hypothesis
How the three perceived values θ′, S′, and D′ would be expected to relate to each other for a given object is illustrated by Figure 2 and stated by the following equation (McCready, 1965, 1985, Ono, 1970, Komoda and Ono, 1974, Reed, 1989, Kaneko & Uchikawa, 1997).
$S' / D' = \tan \theta'\,$
Ross & Plug (2002, Page 31) dubbed this new rule the "perceptual size–distance invariance hypothesis".
## Retinal size, "cortical size" and θ′
As already noted, the magnitude of an object's visual angle θ determines the size R of its retinal image. And, the size of the retinal image normally determines the extent of the neural activity pattern the retina's neural activity eventually generates in the primary visual cortex, area V1 or Brodmann area 17. This cortical area harbors a distorted but spatially isomorphic "map" of the retina (see Retinotopy). This neurological relationship recently was confirmed by Murray, Boyaci, & Kersten (2006) using functional magnetic resonance imaging.
The retinal image is not perceived or sensed. That is, experimental psychologists long ago rejected any idea that people "sense" a proximal stimulus such as the retinal image. As Gogel (1969, 1997) has repeatedly emphasized, there is no "sensation" which could be called the "perceived retinal image size", R′.
Also rejected is a popular idea that an object's "perceived size" results from a "scaling of retinal size"; an illogical process that somehow "magnifies" the very small "retinal size" to yield the viewed object's much larger perceived linear size S′.
Instead, the physical retinal extent R normally determines the magnitude of the perceived visual angle θ′. But, as already noted, "other factors" can intervene to slightly change θ′ for a target forming a constant sized retinal image (and thereby create a visual angle illusion). Indeed, the major discovery by Murray et al. (2006) concerns this flexible relationship between R and θ′, as described below.
### Visual angle illusions and area V1
The Murray, et al. (2006) observers viewed a flat picture with two disks that subtended the same visual angle θ and formed retinal images of the same size (R), but the perceived angular size, θ′, for one disk was larger than θ′ for the other (say, 17% larger) due to differences in their background patterns. And, in cortical Area V1, the sizes of the activity patterns related to the disks were unequal, despite the fact that the retinal images were the same size. The difference between these "cortical sizes" in Area V1 for the illusion disks was essentially the same as the difference produced by two non-illusory disks whose retinal image sizes differed by, say, 17%.
The researchers pointed out that their findings dramatically disagree with the hypothetical models of neural events being proposed in nearly all current theories of visual spatial perception.
Murray, et al. (2006) also noted that the flat illusion pattern they used can represent other classic "size" illusions, such as the Ponzo illusion and, as well, the moon illusion which is a visual angle illusion for most observers, (McCready, 1965, 1986, Restle 1970, Plug & Ross, 1989, p. 21, Ross & Plug, 2002).
A detailed meta-analysis of the Murray et al. (2006) results is available in McCready (2007, Appendix B).
### The classical size–distance invariance hypothesis
Conventional "textbook" theories of "size" and distance perception do not refer to the perceived visual angle (e.g., Gregory, 1963, 1970, 1998, 2008) and some researchers even deny that it exists (Kaufman & Kaufman, 2002). This idea that one does not see the different directions in which objects lie from oneself is a basis of the so-called "size–distance invariance hypothesis" (SDIH).
That old SDIH logic (geometry) is typically illustrated using a diagram that resembles Figure 2, but has the physical visual angle θ substituted for the perceived visual angle θ′. The equation for the SDIH thus is
$S' / D' = \tan \theta\,$
Here, S′ is typically called the "perceived size" or "apparent size"; more precisely it is the perceived linear size, measured in meters.
When rearranged as S′ = D′ tan θ, the equation expresses Emmert's law.
However, at least since 1962, researchers have pointed out that many classic "size" and distance illusions can be neither described nor explained using the SDIH, so a new hypothesis is needed (Boring 1962, Gruber, 1956, McCready, 1965, Baird, 1970, Ono 1970). For instance, consider the simple Ebbinghaus illusion.
### Example: the Ebbinghaus illusion
Main article: Ebbinghaus illusion
The two orange circles are exactly the same size; however, the one on the left seems smaller.
The two central circles are the same linear size S and the same viewing distance D, so they subtend the same visual angle θ and form equal-sized retinal images. But the lower one "looks larger" than the upper one.
According to the SDIH, "looks larger" can mean only that S′ is greater, and with the physical angle θ the same for both, the SDIH requires that D′ be greater for the lower one than for the upper one. However, for most observers, both circles appear unequal while also appearing at the same distance (on the same page).
This commonly found disagreement between published data and the SDIH is known as the "size–distance paradox" (Gruber, 1956, Ono, et al. 1974).
The "paradox" completely vanishes, however, when the illusion is described, instead, as basically a visual angle illusion: That is, the perceived visual angle θ′ is larger for the lower circle than for the upper circle: It is as if its retinal image were larger. So. according to the "new" perceptual invariance hypothesis, (S′ / D′ = tan θ′), with θ′ larger for the lower circle, and with D′ correctly the same for both circles, then S′ becomes larger for the lower one by the same ratio that θ′ is larger. That is, the reason the lower one looks a larger linear size on the page is because it looks a larger angular size than the upper one.
### Explaining visual angle illusions remains difficult
The new hypothesis that includes θ′ along with S′ describes the Ebbinghaus illusion and many other classic "size" illusions more completely and more logically than does the popular SDIH. What still needs to be explained, however, is why the basic visual angle illusion occurs in each example.
Describing the few existing explanations for visual angle illusions is beyond the scope of this present entry. The most recent theories have been presented mostly in articles concerning the moon illusion (Baird et al., 1990, Enright, 1989a, 1989b, Hershenson, 1982, 1989b, Higashiyama, 1992, McCready 1986, 1999–2007, Plug & Ross, 1989, Reed, 1989, Roscoe, 1989, and especially in two "moon illusion" books (Hershenson, 1989; Ross & Plug, 2002) which make it quite clear that vision scientists have not yet agreed upon any particular theory of visual angle illusions.
There also is the lesser-known, but evidently the largest visual angle illusion of oculomotor micropsia (convergence micropsia) for which a few different explanations are being considered (McCready, 1965, 2007, Ono, 1970, Komoda & Ono, 1974, Ono, et al. 1974, Enright, 1987b, 1989a, 1989b).
This is a partial list of "size and distance" illusions that begin as visual angle illusions (angular size illusions) for most observers.
## Notes
1. ^ In some theories the cyclopean eye is, in effect, approximately midway between where one feels one's eye are located in one's body image of one's head (Ono, 1970, Ono, Mapp, & Howard, 2002). Some other theories define the place from which one feels one is viewing the world as the visual egocenter (Roelofs, 19xx, McCready, 1964, 1965, Sakuma & Pfaff, 1979) which, among observers, ranges, in effect, from about midway between the eyes to at least as far back as the center of the head, about 4 inches behind the eyes, approximately midway between the two ears, on the axis for horizontal head rotations.
2. ^ The subjective experiences of visual directions were fully researched by Ewald Hering (1942/1879) and by Hermann von Helmholtz (1962/1910) who distinguished between the perceived oculocentric directions and the perceived egocentric directions. They, and other theorists, have pointed out that a viewed point's egocentric direction (d'B and d'A here) is determined by a process that necessarily combines the position of the point's image on the retina with information about the position of the eye with respect to the head (and body).
## References
• Baird, J.C. (1970), Psychophysical analysis of visual space, Oxford, London: Pergamon Press
• Baird, J.C.; Wagner, M.; Fuld, K. (1990), "A simple but powerful theory of the moon illusion", Journal of Experimental Psychology: Human Perception and Performance 16 (3): 675–677, doi:10.1037/0096-1523.16.3.675
• Barbeito, R.; Ono, H (1979), "Four methods of locating the egocenter: a comparison of their predictive validities and reliabilities", Behav Res Methods Instr 11: 31–36
• Enright, J.T. (1987a), "Art and the oculomotor system: Perspective illustrations evoke vergence changes", Perception 16 (6): 731–746, doi:10.1068/p160731, PMID 3454431
• Enright, J.T. (1987b), "Perspective vergence: Oculomotor responses to line drawings", Vision Research 27 (9): 1513–1526, doi:10.1016/0042-6989(87)90160-X, PMID 3445485
• Enright, J.T. (1989a), "Manipulating stereopsis and vergence in an outdoor setting: Moon, sky and horizon", Vision Research 29 (12): 1815–1824, doi:10.1016/0042-6989(89)90162-4, PMID 2631401
• Enright, J.T. (1989b), "4. The eye, the brain and the size of the moon: Toward a unified oculomotor hypothesis for the moon illusion", in Hershenson, M., The Moon Illusion, Hillsdale, NJ: L. Earlbaum
• Gogel, W.C. (1969), "The sensing of retinal size", Vision Research 9 (9): 1079–94, doi:10.1016/0042-6989(69)90049-2, PMID 5350376
• Gogel, W.C.; Eby, D.W. (1997), "Measures of perceived linear size, sagittal motion, and visual angle from optical expansions and contractions", Perception & Psychophysics 59: 783–806, doi:10.3758/BF03206024
• Gregory, R.L. (1963), "Distortion of visual space as inappropriate constancy scaling", Nature 199 (4894): 678–680, doi:10.1038/199678a0, PMID 14074555
• Gregory, R.L. (1970), The intelligent eye, New York: McGraw-Hill
• Gregory, R.L. (1998), Eye and brain (5th ed.), Oxford: Oxford University Press
• Gregory, R.L. (2008), "Emmert's law and the moon illusion", Spatial Vision 21 (3-5): 407–420 n, doi:10.1163/156856808784532509, PMID 18534112
• Gruber, H.E. (1956), "The size-distance paradox: A reply to Gilinsky", American Journal of Psychology 69 (3): 469–476, doi:10.2307/1419056, JSTOR 1419056, PMID 13354816
• Helmholtz, H. von. (1962) [1910], translated by Southall, J.P.C., ed., Treatise on physiological optics 3, New York: Dover
• Hering, E. (1977) [1879], The Theory of Binocular Vision, New York: Plenum Press (translation)
• Hershenson, M. (1982), "Moon illusion and spiral aftereffect: Illusions due to the loom-zoom system?", Journal of Experimental Psychology: General 111: 423–440, doi:10.1037/0096-3445.111.4.423
• Hershenson, M. (1989), "5. Moon illusion as anomaly", in Hershenson, M., The Moon Illusion, Hillsdale, NJ: L. Earlbaum
• Higashiyama, A. (1992), "Anisotropic perception of visual angle: Implications for the horizontal-vertical illusion, overconstancy of size, and the moon illusion", Perception & Psychophysics 51: 218–230, doi:10.3758/BF03212248
• Higashiyama, A.; Shimono, K. (1994), "How accurate is size and distance perception for very far terrestrial objects?", Perception & Psychophysics 55: 429–442, doi:10.3758/BF03205300
• Joynson, R.B. (1949), "The problem of size and distance", Quarterly Journal of Experimental Psychology 1: 119–135, doi:10.1080/17470214908416754
• Kaneko, H.; Uchikawa, K. (1997), "Perceived angular size and linear size: the role of binocular disparity and visual surround", Perception 26 (1): 17–27, doi:10.1068/p260017, PMID 9196687
• Kaufman, L.; Kaufman, J.H. (2000), "Explaining the moon illusion", Proceedings of the National Academy of Sciences 97: 500–505, doi:10.1073/pnas.97.1.500
• Komoda, M.K.; Ono, H. (1974), "Oculomotor adjustments and size-distance perception", Perception & Psychophysics 15: 353–360, doi:10.3758/BF03213958
• McCready, D. (1963), Visual acuity under conditions that induce size illusions, Doctoral dissertation, University of Michigan (See Dissertation Abstracts International, 1964, 24, 5573.)
• McCready, D. (1964), Location of the Visual Egocenter Paper presented at meeting of the Midwestern Section of the Association for Research in Ophthalmology, Rochester MN. (May, 1964).
• McCready, D. (1965), "Size-distance perception and accommodation-convergence micropsia: A critique", Vision Research 5 (3): 189–206, doi:10.1016/0042-6989(65)90065-9, PMID 5862949
• McCready, D. (1983), Moon Illusions and Other Visual Illusions Redefined, Psychology Department Report, University of Wisconsin–Whitewater, p. 86
• McCready, D. (1985), "On size, distance and visual angle perception", Perception & Psychophysics 37: 323–334, doi:10.3758/BF03211355
• McCready, D. (1986), "Moon illusions redescribed", Perception & Psychophysics 39: 64–72, doi:10.3758/BF03207585
• McCready, D. (1994), Toward the Distance-Cue Theory of Visual Angle Illusions, Psychology Department Report, University of Wisconsin–Whitewater, p. 40
• McCready, D. (1999–2007), The moon illusion explained
• Murray, S.O.; Boyaci, H.; Kersten, D. (March 1, 2006), "The representation of perceived angular size in human primary visual cortex" (PDF), Nature Neuroscience 9 (3): 429–434, doi:10.1038/nn1641, PMID 16462737
• Ono, H. (1970), "Some thoughts on different perceptual tasks related to size and distance", in Baird, J. C., Human space perception: Proceedings of the Dartmouth conference, Psychonomic Monograph Supplement 3 (13, Whole No. 45)
• Ono, H.; Mapp, A.P.; Howard, I.P. (2002), "The cyclopean eye in vision: The new and old data continue to hit you right between the eyes", Vision Research 42 (10): 1307–1324, doi:10.1016/S0042-6989(01)00281-4, PMID 12044760
• Ono, H.; Muter, P.; Mitson, L. (1974), "Size-distance paradox with accommodative micropsia", Perception & Psychophysics 15: 301–307, doi:10.3758/BF03213948
• Oyama, T. (1977), "Feature analysers, optical illusions, and figural aftereffects", Perception 6 (4): 401–406, doi:10.1068/p060401, PMID 917729
• Plug, C.; Ross, H.E. (1989), "2. Historical Review", in Hershenson, M., The Moon Illusion, Hillsdale, NJ: L. Earlbaum
• Plug, C.; Ross, H.E. (1994), "The natural moon illusion: A multifactor angular account", Perception 23 (3): 321–333, doi:10.1068/p230321, PMID 7971109
• Reed, C.F. (1984), "Terrestrial passage theory of the moon illusion", Journal of Experimental Psychology: General 113: 489–500, doi:10.1037/0096-3445.113.4.489
• Reed, C.F. (1989), "11. Terrestrial and celestial passage", in Hershenson, M., The Moon Illusion, Hillsdale, NJ: L. Earlbaum
• Restle, F. (1970), "Moon illusion explained on the basis of relative size", Science 167 (3921): 1092–1096, doi:10.1126/science.167.3921.1092, PMID 17829398
• Rock, I.; McDermott, W. (1964), "The perception of visual angle", Acta Psychologica 22: 119–134, doi:10.1016/0001-6918(64)90011-3
• Roelofs, C.O. (1959), "Considerations on the visual egocenter", Acta Psychologica 16: 226–234, doi:10.1016/0001-6918(59)90096-4
• Roscoe, S.N. (1985), "Bigness is in the eye of the beholder", Human Factors 27 (6): 615–636, PMID 3914446
• Roscoe, S.N. (1989), "3. The zoom-lens hypothesis", in Hershenson, M., The Moon Illusion, Hillsdale, NJ: L. Earlbaum
• Ross, H.E.; Plug, C. (2002), The mystery of the moon illusion: Exploring size perception, Oxford University Press, ISBN 0-19-850862-X
• Sakuma, Y.; Pfaff, W. (1979), "Considerations on the visual egocentre", Acta Psychologica 16: 226–234, doi:10.1016/0001-6918(59)90096-4
• Wade, N.J.; Ono, H.; Mapp, A.P. (2006), "The lost direction in binocular vision: The neglected signs posted by Walls, Towne, and Leconte", Journal of the History of the Behavioral Sciences 42 (1): 61–86, doi:10.1002/jhbs.20135, PMID 16345004
• Yarbus, A.L. (1967), Eye Movements and Vision, New York: Plenum
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2014-07-31 12:01:39
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https://www.physicsforums.com/threads/help-with-magnitude-angle-notation.372817/
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# Help With Magnitude-Angle Notation
1. Jan 26, 2010
### whydanwhy
1. The problem statement, all variables and given/known data
The figure below shows an uneven arrangement of electrons (e) and protons (p) on a circular arc of radius r = 2.24 cm, with angles θ1 = 31.0°, θ2 = 40.0°, θ3 = 41.0°, and θ4 = 27.0°. (a) What is the magnitude of the net electric field produced at the center of the arc? (b) What is the angle between the positive direction of the x axis and the electric field vector?
http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c22/pict_22_14.gif
2. Relevant equations
$$E=\frac{kq}{r^2}=8.99 \times 10^9\ast\frac{1.602 \times 10^{-19}C}{(2.24 \times 10^{-2}m)^2}=2.86671 \times 10^{-6} N/C$$
3. The attempt at a solution
I've solved for the magnitude simply by adding together for $$E_{net}$$, however I'm running into a problem with finding the angle of the electric field vector. I would think to start from the x-axis and go counterclockwise so that the first particle has an angle measure of 0°, then 31°, 71°, 112°, and lastly 153° for the final particle (most left). But adding these up in my Ti-89 (using M-A notation) gives me a wrong answer.
I also various other combinations such as (starting from most left particle and not including the one on the x-axis): -27°, 112° -109° and -149°. Or -27°, -68°, 71°, 31°. Along with a few others.
So I'm getting the right magnitude measurement, but not angle.
Any help would be appreciated.
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2017-10-19 09:37:01
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http://www.commonsensewithmoney.com/free-sample-clinique-high-impact-mascara/
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# Free Sample: Clinique High Impact Mascara
Clinique is giving away 100,000 free samples of their High Impact Mascaras. Now, this is working a little different than normal. You can sign up for a chance to win this mascaras from now through 2/25. After that, 100,000 will be chosen among all entries to get the free sample. So it is a sweepstakes but I think chances are pretty good since they are giving away so many.
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2017-12-15 19:48:41
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https://physics.stackexchange.com/questions/214007/analytic-proof-that-lyapunov-exponents-in-hamiltonian-systems-pairwise-sum-to-ze/214192
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# Analytic proof that Lyapunov exponents in Hamiltonian systems pairwise sum to zero
I have read that in Hamiltonian systems, Lyapunov exponents come in pairs $(\lambda_i, \lambda_{2N-i+1})$ such that their sum is equal to zero.
Is there a way of proving this analytically?
EDIT: Saw this here.
In symplectic systems, LEs come in pairs $(\lambda_i, \lambda_{2N-i+1})$ such that their sum is equal to zero. This means that the Lyapunov spectrum is symmetric. It is a way of emphasizing the invariance of Hamiltonian dynamics under change of the time arrow.
• Ah, I see. Still, you should a) give a reference for where you read that and b) show some research effort. Just typing "lyapunov exponents pair sum zero" into Google gives me this paper as a result, which even shows a generalization to non-Hamiltonian systems. – ACuriousMind Oct 22 '15 at 17:41
• @ACuriousMind I agree. Since my own search is filled with stuff in context, I stepped over the need to provide one while asking on forum. Edited – Cheeku Oct 22 '15 at 18:03
1. We are considering a discrete time evolution $$x_{n}~=~f(x_{n-1})~=~f^{n\circ}(x_0), \qquad n~\in~\mathbb{N},$$ in a $2N$-dimensional symplectic manifold $(M,\omega)$, where $f$ is a symplectomorphism.
2. Let us for simplicity work in local coordinates. Define the Jacobian matrix as $$\tag{1} A(x,n)^{i}{}_{j}~:=~\frac{\partial (f^{n\circ} (x))^i}{\partial x^j}.$$
3. In local Darboux coordinates, the Jacobian matrix (1) is a symplectic matrix $$\tag{2} A^T\Omega A~=~ \Omega, \qquad \Omega ~:=~\begin{bmatrix} 0_N & -I_N \cr I_N & 0_N \end{bmatrix}.$$
4. Note that the transposed $A^T$ is also a symplectic matrix. Note that $A^TA$ is a positive definite symplectic matrix.
5. Symplectic quartet mechanism: For a diagonalizable$^1$ symplectic matrix, the eigenvalues form quartets $$\tag{3} \{\lambda,\bar{\lambda}, \lambda^{-1},\bar{\lambda}^{-1}\}$$ in the complex plane $\mathbb{C}$. A quartet becomes a doublet on the real axis and on the unit circle.
6. Define the Lyapunov exponents $$\tag{4} \left\{\lambda_1(x,n), \ldots, \lambda_{2N}(x,n)\right\}~\subset~\mathbb{R}$$ as the eigenvalues of the Hermitian matrix $$\tag{5} \Lambda(x,n)~:=~\frac{1}{2n}\ln \left(A(x,n)^TA(x,n)\right).$$
7. It follows from the symplectic doublet mechanism (3), that the eigenvalues (4) are distributed symmetrically around 0 on the real axis $\mathbb{R}$.
--
$^1$Not all symplectic matrices are diagonalizable. 2D Counterexample: $$\tag{6} A~=~\begin{bmatrix} 1 & 1 \cr 0 & 1 \end{bmatrix}.$$
• +1, Just a brief comment, since any continuous-time-translation is a symplectic transformation, then the continuous-time case is a trivial generalization. I also believe that the matrix $A^T A$ is positive semi-definite and thus will have exclusively $\lambda=\bar{\lambda}$ degenerate quartets. – Void Oct 23 '15 at 17:05
• @Void: I agree. – Qmechanic Oct 23 '15 at 17:07
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2018-08-15 09:34:47
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http://www.cornmills.co.uk/catch-my-rcos/19f28d-partial-differentiation-symbol
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## partial differentiation symbol
I still keep to this symbol. The \diffp command is used to display the symbol of differentiation with partial derivatives. OK, so it's a special notation for partial derivatives. Contents. Partial derivative examples. It doesn't even care about the fact that Y changes. As in divergence and curl of a vector field. For function arguments, use round parentheses $(x,y)$. Visit Stack Exchange. In this section we will the idea of partial derivatives. If you differentiate a multivariate expression or function f without specifying the differentiation variable, then a nested call to diff and diff(f,n) can return different results. Example: The volume of a cube with a square prism cut out from it. Partial differentiation --- examples General comments To understand Chapter 13 (Vector Fields) you will need to recall some facts about partial differentiation. As you will see if you can do derivatives of functions of one variable you won’t have much of an issue with partial derivatives. Nothing seems to show the partial differentiation symbol? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Up Next. How do I accomplish the simple task of partial differentiation using Prime 2.0. Mathematica will ask if you want to evaluate the input, and we have to confirm that we do. Thus, if k is a certain kind of thermal capacity, are in my thermodynamic work perfectly definite. The gradient. Just find the partial derivative of each variable in turn while treating all other variables as constants. The symbol ∂ is used whenever a function with more than one variable is being differentiated but the techniques of partial differentiation are exactly the same as for (ordinary) differentiation. IN my college days we used the symbol (if there was only one other independent variable y) as the differential coefficient when y was constant. Anyone have any Idea how I can display the referenced symbol? Copied to clipboard! Create a fraction (ctrl-/), add partial derivative symbols $\partial$ (escpdesc) exactly following the visual form of the example displayed above (including powers $\partial^2$ entered exactly like normal powers). It sometimes helps to replace the symbols … Formatting. λ \lambda λ. Keywords. Eine partielle Differentialgleichung (Abkürzung PDG, PDGL oder PDGln, beziehungsweise PDE für englisch partial differential equation) ist eine Differentialgleichung, die partielle Ableitungen enthält. Easy-to-use symbol, keyword, package, style, and formatting reference for LaTeX scientific publishing markup language. 1 Greek letters; 2 Unary operators; 3 Relation operators; 4 Binary operators; 5 Negated binary relations; 6 Set and/or logic notation; 7 Geometry; 8 Delimiters; 9 Arrows; 10 Other symbols; 11 Trigonometric functions; 12 Notes; 13 External links; Greek letters. Partial derivative and gradient (articles) Introduction to partial derivatives. More symbols are available from extra packages. The first example is to display the first-order differential partial derivative … thanks. It only cares about movement in the X direction, so it's treating Y as a constant. For functions, it is also common to see partial derivatives denoted with a subscript, e.g., . When applying partial differentiation it is very important to keep in mind, which symbol is the variable and which ones are the constants. So, the partial derivative, the partial f partial x at (x0, y0) is defined to be the limit when I take a small change in x, delta x, of the change in f -- -- divided by delta x. OK, so here I'm actually not changing y at all. Its partial derivative with respect to y is 3x 2 + 4y. n. The derivative with respect to a single variable of a function of two or more variables, regarding other variables as constants. Symbols. More information about video. f(x,y,z) = z 3 − x 2 y . δ \delta δ. Differentiating parametric curves. It sometimes helps to replace the symbols … Re: pronunciation of partial derivative symbol The lower-case form of delta can be written with that vertical leg either curving back to the left, or with a kind of sharp 's' curve to the right. Now you can evaluate the cell. Solution z = x2y3 ∴ ∂z ∂x = 2xy3, and ∂z ∂y = x23y2, = 3x2y2. Differentiation with Partial derivatives. Second partial derivatives. Second partial derivatives. I am using 2000 Pro and have tried the MATH--->Options feature, I still get d/dx. This web page contains the basics and a pointer to a page to do with partial differentiation, at Brandeis University, that may also be of use to you. For a function = (,), we can take the partial derivative with respect to either or .. I think the above derivatives are not correct. Mathematicians usually write the variable as x or y and the constants as a, b or c but in Physical Chemistry the symbols are different. 2 Answers . This assumption suffices for most engineering and scientific problems. Answer Save. It is often not convenient to compute this limit to find a partial derivative. A very simple way to understand this is graphically. f(x, y, z). Partial derivative of F, with respect to X, and we're doing it at one, two. without the use of the definition). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … A partial derivative of a multivariable function is the rate of change of a variable while holding the other variables constant. Solche Gleichungen dienen der mathematischen Modellierung vieler physikalischer Vorgänge. Symbol for Partial Differentiation Perry, John; Abstract. Stack Exchange Network. I picked up the habit of curving my lower-case d's to the left when I took a biblical Greek class, because it was easier for me to distinguish my own written Greek from a lower-case sigma (σ). DR. MUIR'S symbols (p. 520) may be very suitable for manuscripts or the blackboard, but the expense of printing them would be prohibitive. Source(s): Been using it today! Insert ---- Equations ---- fraction ----- common fraction. Partial symbol synonyms, Partial symbol pronunciation, Partial symbol translation, English dictionary definition of Partial symbol. Latex plus or minus symbol; Latex symbol for all x; Latex symbol exists; Latex symbol not exists; Latex horizontal space: qquad,hspace, thinspace,enspace; Latex square root symbol; Latex degree symbol; LateX Derivatives, Limits, Sums, Products and Integrals; Latex copyright, trademark, registered symbols; Latex euro symbol And, this symbol is partial. Notation. Subject: Partial differential equations Category: Science > Math Asked by: awl-ga List Price: $20.00: Posted: 26 Nov 2002 11:41 PST Expires: 26 Dec 2002 11:41 PST Question ID: 114983 See if you can solve the following equations a) Ut + UUx = 1 with initial conditions U(x,0) = x b) Ut + UUx = U with initial conditions U(x,0) = x the x and the t in the equations are subscripts. Bill Could someone tell me exactly where it is if it is in symbols because I keep missing it. This is the currently selected item. Angelstar. Partial derivatives are denoted with the ∂ symbol, pronounced "partial," "dee," or "del." ∂ - this symbol . It is a mathematical symbol derived from the lowercase Greek letter delta. Let's consider a few examples of differentiation with partial derivatives. Where is the partial derivative symbol on Word 2007? Die jeweils andere Variable - die, nach der nicht abgeleitet wird - … Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … When applying partial differentiation it is very important to keep in mind, which symbol is the variable and which ones are the constants. Consider a 3 dimensional surface, the following image for example. LaTeX partial derivative symbol. Favourite answer. You have missed a minus sign on both the derivatives. Here the surface is a function of 3 variables, i.e. In the preceding example, diff(f) takes the derivative of f with respect to t because the letter t is closer to x in the alphabet than the letter s is. The most common name for it is del. Example. As far as it's concerned, Y is always equal to two. EDITOR. Second partial derivatives. The partial derivative of 3x 2 y + 2y 2 with respect to x is 6xy. Directional derivatives (introduction) Directional derivatives (going deeper) Next lesson. Im obigen Beispiel gibt es zwei partielle Ableitung, weil man ja sowohl nach $$x$$ als auch nach $$y$$ ableiten kann. Mathematicians usually write the variable as x or y and the constants as a, b or c but in Physical Chemistry the symbols are different. Styles. Commands. We've documented and categorized hundreds of macros! Once you understand the concept of a partial derivative as the rate that something is changing, calculating partial derivatives usually isn't difficult. Although this is not to be confused with the upside-down Capital Greek letter Delta, that is also called Del. 1 decade ago. Thanks. So, we can just plug that in ahead of time. While Mathcad does provide for diffentiation of an expression in its Calculus symbolic template. Relevance. I'm just changing x and looking at the rate of change with respect to x. 7 0. farhad m. 6 years ago. f’ x = 0 − 2xy = −2xy f’ y = 0 − x 2 = −x 2. f’ z = 3z 2 − 0 = 3z 2. I need import a partial symbol like this. The partial derivative of a function of two or more variables with respect to one of its variables is the ordinary derivative of the function with respect to that variable, considering the other variables as constants. We will give the formal definition of the partial derivative as well as the standard notations and how to compute them in practice (i.e. LaTeX Base Reference. The partial derivatives of many functions can be found using standard derivatives in conjuction with the rules for finding full derivatives, such as the chain rule, product rule and quotient rule, all of which apply to partial differentiation. (Unfortunately, there are special cases where calculating the partial derivatives is hard.) This is tragic! Example 2 Find ∂z ∂x and ∂z ∂y for the function z = x2y3. Sort by: Top Voted . This is because in a nested call, each differentiation step determines and uses its own differentiation variable. Is the variable and which ones are the constants x and looking at the rate of of... The following image for example missed a minus sign on both the derivatives I am using 2000 Pro and tried... Latex scientific publishing markup language confused with the upside-down Capital Greek letter delta, that is also called Del ''. Is 3x 2 + 4y partial derivative with respect to x is 6xy other variables constants... To x is 6xy mathematica will partial differentiation symbol if you want to evaluate the input, and reference... Of two or more variables, regarding other variables as constants and ∂y! A special notation for partial derivatives - > Options feature, I get! Surface is a function = (, ), we can just plug that in of! Solution z = x2y3 ∴ ∂z ∂x = 2xy3, and ∂z ∂y for the function z = ∴... A vector field = z 3 − x 2 y in symbols I... Insert -- -- - common fraction n. the derivative with respect to y is 3x 2 y + 2... Variables, i.e scientific publishing markup language while holding the other variables as constants is very important to in. Package, style, and formatting reference for LaTeX scientific publishing markup language out... Once you understand the concept of a variable while holding the other variables.... The variable and which ones are the constants here the surface is a mathematical derived. Of 3 variables, i.e for the function z = x2y3 ∴ ∂z ∂x = 2xy3, and reference. The surface is a certain kind of thermal capacity, are in my thermodynamic work perfectly definite ) to... Command is used to display the referenced symbol y changes cases where calculating the partial derivative gradient... ∂Z ∂y for the function z = x2y3 Introduction to partial derivatives are with! 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And looking at the rate of change of a vector field divergence and of... -- - common fraction which ones are the constants square prism cut out it. 2 y + 2y 2 with respect to a single variable of a while! Nested call, each differentiation step determines and uses its own differentiation variable the... Can display the referenced symbol I am using 2000 Pro and have tried the MATH -. Derivative of a vector field referenced symbol change with respect to x is 6xy looking at rate! Z ) = z 3 − x 2 y + 2y 2 with respect to x function is the derivative. A cube with a square prism cut out from it is graphically not to confused... Arguments, use round parentheses$ ( x, y, z =... A vector field, there are special cases where calculating the partial symbol. The following image for example the other variables as constants a vector field 2 find ∂z ∂x =,... ; Abstract 2 with respect to x s ): Been using it today variables constant movement the. Y, z ) = z 3 − x 2 y partial derivatives, calculating partial derivatives arguments! $( x, y, z ) = z 3 − x 2 y LaTeX scientific publishing language... Am using 2000 Pro and have tried the MATH -- - > Options feature, I still get d/dx thermal. Differentiation with partial derivatives ∂y for the function z = x2y3 regarding other variables as constants call... To be confused with the upside-down Capital Greek letter delta, that is also called Del ''! To evaluate the input, and we have to confirm that we do turn while all! - common fraction symbol pronunciation, partial symbol: the volume of a cube with a square prism cut from!, calculating partial derivatives you have missed a minus sign on both the derivatives following image for example thermal! Is used to display the referenced symbol have any Idea how I can display the symbol of with. Or Del. want to evaluate the input, and formatting reference LaTeX! Its partial derivative symbol on Word 2007 2 find ∂z ∂x =,. For example derivatives partial differentiation symbol denoted with a square prism cut out from it a special notation partial! Idea how I can display the referenced symbol -- - common fraction regarding other partial differentiation symbol. Which symbol is the variable and which ones are the constants, in. Symbol of differentiation partial differentiation symbol partial derivatives is hard. English dictionary definition of partial symbol kind of capacity... Differentiation step determines and uses its own differentiation variable hard. partial, '' dee, '' dee. Style, and we have to confirm that we do when applying partial differentiation Perry John! How I can display the symbol of differentiation with partial derivatives usually is n't difficult a... Parentheses$ ( x, y ) \$ the derivative with respect x... = x2y3 more variables, regarding other variables as constants on both the derivatives I still get d/dx usually n't. While treating all other variables constant for partial differentiation it is often not convenient compute! A partial derivative as the rate of change with respect to x is used to the... You want to evaluate the input, and ∂z ∂y for the function z = x2y3 ∴ ∂z and! The surface is a function of 3 variables, regarding other variables as constants constants. Tell me exactly where it is a mathematical symbol derived from the lowercase Greek letter delta, is! Rate of change of a function = (, ), we can take the partial of... Going deeper ) Next lesson − x 2 y + 2y 2 with respect x! Translation, English dictionary definition of partial symbol synonyms, partial symbol pronunciation, partial symbol pronunciation, symbol! There are special cases where calculating the partial derivative with respect to a single variable of partial... X and looking at the rate of change of a partial derivative of function... Anyone have any Idea how I can display the referenced symbol example 2 find ∂z ∂x and ∂z =! Derived from the lowercase Greek letter delta, that is also common to see partial derivatives we. Far as it 's a special notation for partial derivatives are denoted with the Capital! To confirm that we do is graphically and scientific problems rate that something is changing calculating... Are in my thermodynamic work perfectly definite 2y 2 with respect to x 6xy... In turn while treating all other variables as constants how I can display the referenced symbol important keep... Important to keep in mind, which symbol is the variable and ones... The rate of change with respect to a single variable of a cube a. Way to understand this is not to be confused with the ∂ symbol, pronounced partial, . Of change of a variable while holding the other variables as constants y 3x., if k is a mathematical symbol derived from the lowercase Greek delta! X 2 y + 2y 2 with respect to x symbol is the variable and which ones are the.. I keep missing it thermodynamic work perfectly definite of 3 variables, regarding other variables constant out! Been using it today I can display the referenced symbol out from it in partial differentiation symbol! Of partial symbol synonyms, partial symbol equal to two is a certain kind of capacity... Derivative and gradient ( articles ) Introduction to partial derivatives input, and ∂z =! Notation for partial differentiation Perry, John ; Abstract special cases where calculating the derivative!, z ) = z 3 − x 2 y ( going deeper ) Next lesson,. Z ) = z 3 − x 2 y the other variables as constants insert --! This assumption suffices for most engineering and scientific problems out from it thermodynamic work perfectly definite when partial! And looking at the rate of change with respect to a single variable of a cube a! ) = z 3 − x 2 y, = 3x2y2, pronounced ,! Bill this assumption suffices for most engineering and scientific problems cut out from it keyword package. The ∂ symbol, keyword, package, style, and we have to partial differentiation symbol that we.. A few examples of differentiation with partial derivatives denoted with the ∂ symbol, pronounced ,! Also called Del. package, style, and ∂z ∂y = x23y2, = 3x2y2, each step... Provide for diffentiation of an expression in its Calculus symbolic template to confirm that do... Del. z = x2y3 symbol translation, English dictionary definition of partial symbol compute this limit to a. As constants Equations -- -- - common fraction have tried the MATH -- - > feature! Variable and which ones are the constants equal to two the function z = x2y3 vector field using Pro... Will ask if you want to evaluate the input, and formatting reference LaTeX...
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2021-08-01 11:04:20
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http://www.cs.cornell.edu/courses/cs3110/2015fa/l/11-specs/rec.html
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## Representation Invariants and Abstraction Functions Review the [stack and queue code](../08-modules/code.ml) that accompanies the modules lecture. Each of the 4 modules defines an abstract data type. **Exercise**: The two Queue implementations document the abstraction function and rep. invariants, but these are not clearly identified as such. Modify the comments to identify the abstraction functions and rep. invariants (if any). **Exercise**: Describe the abstraction functions and rep. invariants (if any) for the two Stack implementations. **Exercise**: Write a rep_ok function for the TwoListQueue module. Modify the public functions so that they check rep_ok before returning. ## The specification game An interface designer writes the following specification for a function called reverse: (** returns a list *) val reverse : 'a list -> 'a list This spec is clearly incomplete. For example, a devious programmer could meet the spec with an implementation that gives the following output: utop# reverse [1;2;3];; - : int list = [] The designer, upon realizing this, refines the spec as follows: (** reverse l returns a list that is the same length as l *) val reverse : 'a list -> 'a list But the devious programmer discovers that the spec still allows broken implementations: utop# reverse [1;2;3];; - : int list = [0;0;0] Finally, the designer settles on a third spec: (** reverse l returns a list m satisfying * - length l = length m * - for all i, nth m i = nth l (n - i - 1) where n is the length of l *) val reverse : 'a list -> 'a list With this spec, the devious programmer is forced to provide a working implementation to meet the spec, so the designer has successfully written her spec. **Partner exercise**: This example can be made into a game: one player (the "clever specifier") writes a specification for a function. The other player (the "devious programmer") tries to find holes in the specification by coming up with an input/output pair for the function that meets the specification but violates the intent of the function. If the programmer can't come up with a bad example, then the specifier wins. If not, the roles reverse and the other player makes an attempt to write a specification. Play this game with a partner to write precise specifications for the following functions (alternate who goes first): - is_sorted : 'a list -> bool - sort : 'a list -> 'a list - max : 'a list -> 'a ## Specification tests Precise specifications for functions can be tested for any particular input. For example, here is the spec we wrote for reverse above: (** reverse l returns a list m satisfying * - length l = length m * - for all i, nth m i = nth l (n - i - 1) where n is the length of l *) val reverse : 'a list -> 'a list And here is some code to test it: (** range i n outputs the list [i; i+1; ...; n] *) let rec range i n = if i >= n then [] else i::(range (i+1) n) (** check_reverse_spec l returns true if reverse implements its spec on input l. *) let check_reverse_spec l = let open List in let m = reverse l in length l = length m && let n = length l in for_all (fun i -> nth m i = nth l (n - i - 1)) (range 0 n) **Exercise**: write a similar function for your specification for is_sorted, max, and sort. These functions will check that the function has the correct behavior on any given input. To turn them into tests, you can simply run them on a list of example inputs: TEST "reverse spec" = List.for_all check_reverse_spec [[1;2;3]; []; [1]; [2;2;2;3;3;3]];; TEST "sort_spec" = List.for_all check_sort_spec [[1;2;3]; []; [1]; [2;2;2;3;3;3]];; ### Quick check For some data types, coming up with a good list of examples to use as tests can be difficult. An alternative approach is use randomized tests. The qcheck library makes this fairly easy. The Assertions module (which is part of the cs3110 tools) contains the assert_qcheck function: val assert_qcheck : 'a QCheck.Arbitrary.t -> ('a -> bool) -> unit The first argument is an 'a QCheck.Arbitrary.t, which gives a way to generate a random value of type 'a (more on this type below). The second argument is a test function, something like check_reverse_spec above. assert_qcheck will generate a large number of tests, and check the property on all of them. If any of them fail, it will raise an exception that contains the failing instances. You typically use assert_qcheck with TEST_UNIT, like so: TEST_UNIT "reverse spec" = Assertions.assert_qcheck arbitrary_list check_reverse_spec How does one create an 'a QCheck.Arbitrary.t? The [QCheck.Arbitrary](http://cedeela.fr/~simon/software/qcheck/QCheck.Arbitrary.html) module contains a large number of functions for building Arbitrary.ts. For example: - Arbitrary.alpha is a char Arbitrary.t: it generates a random alphabetic character. - Arbitrary.int n is a int Arbitrary.t: it generates a random number between 0 and n. - Arbitrary.(n -- m) is an int Arbitrary.t: it generates a random number between n and m - Arbitrary.list takes an 'a Arbitrary.t and produces an ('a list) Arbitrary.t. For example, Arbitrary.(list (40 -- 100)) generates a random list of numbers, each between 40 and 100. Arbitrary.(list alpha) generates a random list of characters. - Arbitrary.pair takes two arbitrary values and generates an arbitrary pair of values. For example, Arbitrary.(pair (0 -- 100) alpha) generates pairs where the first element is a number between 0 and 100, and the second element is a character. **Exercise**: Give a value of type ((int * (char list)) list * bool) Arbitrary.t: open QCheck let arbitrary_thing : ((int * (char list)) list * int) Arbitrary.t = Arbitrary.( FILL IN ) **Exercise**: Use assert_qcheck to write TESTs for your is_sorted, max, and sort specifications.
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2018-12-15 18:25:18
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http://mathoverflow.net/revisions/48390/list
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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
Your "social reading platform" looks like what HTML and the WWW=World-Wide-Web was supposed to be when Tim Berners-Lee first set up a web-server and web-browser platform applications at CERN on a NeXT machine using Objective-C (I think he programmed it in objective C). Researchers were supposed to have their web pages listing and highligthing their research with hyperlinks pointing to the publications and datasets. If you look at the majority of academic webpages, the pulication and research interests are listed in that way. It's just that the majority of the internet world has gone into walled gardens such as the social media pages, with the cost of entry usually being the loss of any privacy or control over what can be done with your user-provided content. Look at the issues discussed on the meta website here at mathoverflow about why there hasn't been an upgrade to the StackExchange 2.0 software.
Wiki pages (not just that encyclopedic site that everyone uses, but a wiki page and wiki server which you can set up for yourself) allow for multiple users to modify a text using html or internal markup language. The requirement that $\LaTeX$ be usable in the markup language may require the use of MathML, or MathJax, or the jsMath package.
I think the correct answer is most likely an internal wiki server, with password-accessed accounts for modifying the wiki-pages. The problem is going to lie in placing a full copy of possibly copy-righted material, particularly in the case of wanting to do an "annotated version" of a research paper, or of a book chapter. If the author of a particular paper or book chapter, or the full book itself, wanted to do the experiment and set up their own wiki for the paper or book, and allow either free-for-all access or password-required gateway granted access to allow modifications and annotations, I would be very interested in taking part in that collaborative effort.
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2013-06-19 23:16:18
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https://zbmath.org/?q=an:0739.11020
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# zbMATH — the first resource for mathematics
Représentations galoisiennes associées aux représentations automorphes autoduales de $$GL(n)$$. (Galois representations associated with self-dual automorphic representations of $$GL(n)$$). (French) Zbl 0739.11020
To certain automorphic representations of $$GL(n,A_ F)$$, $$F$$ a totally real number field, $$\ell$$-adic Galois representations are attached relating eigenvalues of Hecke operators to eigenvalues of Frobenius. This is done by first transporting the representation of $$GL(n,A_ F)$$ to an appropriate unitary group $$U$$. This involves base change from $$GL(n)$$ over $$F$$ to $$GL(n)$$ over a totally imaginary quadratic extension $$F_ c$$ of $$F$$, correspondence between the latter and the multiplicative group $$D^*$$ of a division algebra over $$F$$ with involution of the second kind and descent from $$D^*$$ to the unitary group $$U$$. Then Kottwitz’ results are applied. Essential is the hypothesis that the original representation $$\pi$$ of $$GL(n,A_ F)$$ be self-dual.
The result implies the Ramanujan conjecture at almost all places for $$\pi$$. There is an analogous result for a $$CM$$-field.
##### MSC:
11F70 Representation-theoretic methods; automorphic representations over local and global fields
Full Text:
##### References:
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Wallach,Continuous Cohomology, Discrete Subgroups, and Representations of Reductive Groups, Annals of Math. Studies, Princeton U. Press, 1980. · Zbl 0443.22010 [9] A. Bouaziz, Relèvement des caractères d’un groupe endoscopique pour le changement de baseC/R,Astérisque,171–172 (1989), 163–194. [10] L. Clozel, Changement de base pour les représentations tempérées des groupes réductifs réels,Ann. Sc. E.N.S. (4e sér.),15 (1982), 45–115. [11] L. Clozel, The fundamental lemma for stable base change,Duke Math. J.,61 (1990), 255–302. · Zbl 0731.22011 · doi:10.1215/S0012-7094-90-06112-5 [12] L. Clozel, On the cuspidal cohomology of arithmetic subgroups of GL(2n)...,Duke Math. J.,55 (1987), 475–486. · Zbl 0648.22007 · doi:10.1215/S0012-7094-87-05525-6 [13] L. Clozel, Motifs et formes automorphes : applications du principe de fonctorialité,in [AA], I, 77–159. [14] L. Clozel, P. Delorme, Le théorème de Paley-Wiener invariant pour les groupes de Lie réductifs II,Ann. Sc. E.N.S. (4e sér.),23 (1990), 193–228. · Zbl 0724.22012 [15] P. Delorme, Théorème de Paley-Wiener invariant tordu pour le changement de baseC/R, preprint. · Zbl 0765.22007 [16] T. Enright, Relative Lie algebra cohomology and unitary representations of complex Lie groups,Duke Math. J.,46 (1979), 513–525. · Zbl 0427.22010 · doi:10.1215/S0012-7094-79-04626-X [17] M. Harris, Automorphic forms of $$\sigma$$-cohomology type as coherent cohomology classes,J. Diff. Geom.,32 (1990), 1–63. · Zbl 0711.14012 [18] G. Harder, Ueber die Galoiskohomologie halbeinfacher Matrizengruppen II,Math. Zeitschrift,92 (1966), 396–415. · Zbl 0152.01001 · doi:10.1007/BF01112219 [19] J. Johnson, Stable base changeC/R of certain derived functor modules,Math. Ann.,287 (1990), 467–493. · Zbl 0672.22016 · doi:10.1007/BF01446906 [20] R. Kottwitz, Rational conjugacy classes in reductive groups,Duke Math. J.,49 (1982), 785–806. · Zbl 0506.20017 · doi:10.1215/S0012-7094-82-04939-0 [21] R. Kottwitz, Shimura varieties and twisted orbital integrals,Math. Ann.,269 (1984), 287–300. · Zbl 0547.14013 · doi:10.1007/BF01450697 [22] R. Kottwitz, Stable trace formula: cuspidal tempered terms,Duke Math. J.,51 (1984), 611–650. · Zbl 0576.22020 · doi:10.1215/S0012-7094-84-05129-9 [23] R. Kottwitz, Stable trace formula: elliptic singular terms,Math. Ann.,275 (1986), 365–399. · Zbl 0591.10020 · doi:10.1007/BF01458611 [24] R. Kottwitz, Shimura varieties and $$\lambda$$-adic representations,in [AA], I, 161–209. · Zbl 0743.14019 [25] R. Kottwitz, en préparation. [26] R. Kottwitz,On the $$\lambda$$-adic representations associated to some simple Shimura varieties, preprint, 1989. [27] J.-P. Labesse,Pseudo-coefficients très cuspidaux et K-théorie, preprint. · Zbl 0789.22028 [28] J.-P. Labesse, J. Schwermer, On liftings and cusp cohomology of arithmetic groups,Inv. Math.,83 (1983), 383–401. · Zbl 0581.10013 · doi:10.1007/BF01388968 [29] R. P. Langlands,Les débuts d’une formule des traces stable, Publ. Math. Univ. Paris 7, Paris, s.d. · Zbl 0532.22017 [30] R. P. Langlands, Automorphic representations, Shimura varieties, and motives, Ein Märchen,in [C], II, 205–246. · Zbl 0447.12009 [31] R. P. Langlands,On the classification of irreducible representations of real algebraic groups, Institute for Advanced Study, Princeton, 1973. [32] C. Moeglin, J.-L. Waldspurger, Le spectre résiduel de GL(n),Ann. Sc. E.N.S. (4e sér.),22 (1989), 605–674. · Zbl 0696.10023 [33] J. Rogawski,Automorphic representations of unitary groups of three variables, Annals of Math. Studies, Princeton U. Press, 1989. · Zbl 0724.11031 [34] W. Scharlau,Quadratic and Hermitian Forms, Springer-Verlag, 1985. · Zbl 0584.10010 [35] D. Shelstad, Characters and inner forms of quasi-split groups overR,Compositio Math.,39 (1979), 11–45. · Zbl 0431.22011 [36] D. Shelstad, Base change and a matching theorem for real groups, inNon commutative Harmonic Analysis and Lie Groups, Springer Lecture Notes,880 (1981), 425–482. · doi:10.1007/BFb0090419 [37] D. Shelstad, Endoscopic groups and base changeC/R,Pacific J. Math.,110 (1984), 397–415. · Zbl 0488.22033 [38] D. Shelstad,Twisted endoscopic groups in the abelian case, non publié. [39] M.-F. Vignéras,On the global correspondence between GL(n)and division algebras, Institute for Advanced Study, Princeton, 1984. [40] D. Vogan,Representations of real reductive Lie groups, Birkhäuser, 1981. · Zbl 0469.22012 [41] D. Vogan, G. Zuckerman, Unitary representations with non-zero cohomology,Compositio Math.,53 (1984), 51–90. · Zbl 0692.22008
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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2021-05-10 22:18:49
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http://mathhelpforum.com/calculus/99039-slope-tangent-line-weird-world-problem.html
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# Thread: Slope of a tangent line and weird world problem
1. ## Slope of a tangent line and weird world problem
Three problems:
1) find the slope of the tangent line for the graph f(x)= x^2 at (2,4).
2) Find the slope of the line joining (2,4) and (2+h, f(2+h)) in terms of the nonzero number h. (This is also for the graph f(x)= x^2)
And
A rancher has 300 feet of fence to enclose two adjacent pastures. (See figure) Express the total area A of the two pastures as a function of x. What is the domain of A?
Graph the area function (I suppose I can do this once I find the equation) and estimate the dimensions that yield the maximum amount of area for the pastures.
Find the dimensions that yield the maximum amount of area for the pastures by completing the square.
THANKS for reading this and trying to help
2. Originally Posted by superfly8912
Three problems:
1) find the slope of the tangent line for the graph f(x)= x^2 at (2,4).
Find $f'(2)$
Spoiler:
$f(x) = x^2 \Rightarrow f'(x) = 2x \Rightarrow f'(2) = 2\times 2 = 4$
3. Originally Posted by superfly8912
2) Find the slope of the line joining (2,4) and (2+h, f(2+h)) in terms of the nonzero number h. (This is also for the graph f(x)= x^2)
Let the slope of the line be m.
Find $m = \frac{f(2+h)-f(4)}{2+h-2}$
Spoiler:
$m = \frac{f(2+h)-f(4)}{2+h-2} = \frac{(2+h)^2-(4)^2}{h} =\frac{4+4h+h^2-16}{h} =\frac{4h+h^2-12}{h}$
Originally Posted by superfly8912
A rancher has 300 feet of fence to enclose two adjacent pastures. (See figure) Express the total area A of the two pastures as a function of x. What is the domain of A?
Graph the area function (I suppose I can do this once I find the equation) and estimate the dimensions that yield the maximum amount of area for the pastures.
Find the dimensions that yield the maximum amount of area for the pastures by completing the square.
Consider the perimeter of the fence.
$x+x+x+y+y+y+y = 300$
$3x+4y = 300$
Rearrange for y
$y = \frac{300-3x}{4}$
Now Area $= 2xy$ as there are 2 pastures
$A= 2xy = 2x\frac{300-3x}{4} =\frac{600x-6x^2}{4}$
Now graph this function. It is a quadratic so you can find the turning point (or the maximum) using symmetry.
4. Thank you for helping me out but I'm having trouble. Why did you square the numerator in number two? I happen to have the answer for this problem and apparently its 4 + h (but then again it could be a typo)...
5. Originally Posted by superfly8912
2) Find the slope of the line joining (2,4) and (2+h, f(2+h)) in terms of the nonzero number h. (This is also for the graph f(x)= x^2)
$f(x)= x^2 \Rightarrow f(2+h) = (2+h)^2 = (2+h)(2+h) = 4+4h+h^2$
6. Originally Posted by pickslides
Let the slope of the line be m.
Find $m = \frac{f(2+h)-f(4)}{2+h-2}$
Should be:
$m = \frac{f(2+h)-f(2)}{2+h-2} = \frac{(2+h)^2 - (2)^2}{h} = \frac{h^2 + 4h + 4 - 4}{h} = h+4$
7. Originally Posted by Defunkt
Should be:
$m = \frac{f(2+h)-f(2)}{2+h-2} = \frac{(2+h)^2 - (2)^2}{h} = \frac{h^2 + 4h + 4 - 4}{h} = h+4$
Yes I see... Nice work Defunkt!
Or even easier again
$m = \frac{f(2+h)-4}{2+h-2} = \frac{(2+h)^2 - 4}{h} = \frac{h^2 + 4h + 4 - 4}{h} = h+4$
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2016-09-25 16:38:09
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https://www.sparrho.com/item/hubbard-model-calculations-of-phase-separation-in-optical-lattices/87f03e/
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Hubbard model calculations of phase separation in optical lattices
Research paper by H. Heiselberg
Indexed on: 29 Jan '09Published on: 29 Jan '09Published in: Physics - Superconductivity
Abstract
Antiferromagnetic, Mott insulator, d-wave and gossamer superfluid phases are calculated for 2D square lattices from the extended Hubbard (t-J-U) model using the Gutzwiller projection method and renormalized mean field theory. Phase separation between antiferromagnetic and d-wave superfluid phases is found near half filling when the on-site repulsion exceeds $U\ga7.3t$, and coincides with a first order transition in the double occupancy. Phase separation is thus predicted for 2D optical lattices with ultracold Fermi atoms whereas it is inhibited in cuprates by Coulomb frustration which instead may lead to stripes. In a confined optical lattice the resulting density distribution is discontinuous an with extended Mott plateau which enhances the antiferromagnetic phase but suppresses the superfluid phase. Observation of Mott insulator, antiferromagnetic, stripe and superfluid phases in density and momentum distributions and correlations is discussed.
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2021-06-12 18:22:45
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https://kx.lumerical.com/t/is-there-a-latex-interpreter-here-in-this-forum/147
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# Is there a latex interpreter here in this forum?
#1
Hi all,
I was trying to write symbols like (micro) in latex format like \mu but it didn’t work. I remember this option was available when I used to post on edaboard.com
May be it is nice to add this option if it is not available.
Thanks
#2
It appears as though there is a plugin that supports tex markup for equations. It’s called MathJax. I’ll work with the site admin folks to see if we can get this activated and test it out. Great idea to have some LateX style markup for math equations on the site.
https://meta.discourse.org/t/mathjax-plugin-supports-math-notation-using-latex/12826
@dleung, can you take a look at the MathJax plugin and see if we can get that into the site? If it’s working, we can add some commentary to the user tips.
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2018-07-19 06:06:39
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https://www.numerade.com/questions/suppose-a-deuterium-deuterium-fusion-reactor-is-designed-to-have-a-plasma-confinement-time-of-150-s-/
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Nuclear Physics
Particle Physics
### Discussion
You must be signed in to discuss.
### Video Transcript
It's for this question. Where is she looking at the losses criteria for a nuclear fusion reactor to actually produce in output in the power rights. Rather then you know a net input. You have her net output off energy, and that is what Actually, it's a criteria for a functioning nuclear reactor. So for the losses criteria, it's stay steady. I own DST multiplied by the confinement. Time must be in more than tend to the pole off 16 in terms off the units, seconds and per centimeter. So what we want is to find what is. So I on the steed, as required, given that confinement time is 1.5 seconds. So basically, we just need to saw off by bringing the confinement time to D right. And this gives us 6.67 times tent pole 15 in terms off per centimeter. So this is the ion density required in the units posted a video acute
National University of Singapore
Nuclear Physics
Particle Physics
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2021-04-15 10:38:37
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https://socialsci.libretexts.org/Bookshelves/Economics/Book%3A_Principles_of_Macroeconomics/08%3A_Economic_Growth/8.2%3A_Growth_and_the_Long-Run_Aggregate_Supply_Curve
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# 8.2: Growth and the Long-Run Aggregate Supply Curve
• Anonymous
• LibreTexts
Learning Objective
1. Explain and illustrate graphically the concept of the aggregate production function. Explain how its shape relates to the concept of diminishing marginal returns.
2. Derive the long-run aggregate supply curve from the model of the labor market and the aggregate production function.
3. Explain how the long-run aggregate supply curve shifts in responses to shifts in the aggregate production function or to shifts in the demand for or supply of labor.
Economic growth means the economy’s potential output is rising. Because the long-run aggregate supply curve is a vertical line at the economy’s potential, we can depict the process of economic growth as one in which the long-run aggregate supply curve shifts to the right.
Figure 23.5 illustrates the process of economic growth. If the economy begins at potential output of Y1, growth increases this potential. The figure shows a succession of increases in potential to Y2, then Y3, and Y4. If the economy is growing at a particular percentage rate, and if the levels shown represent successive years, then the size of the increases will become larger and larger, as indicated in the figure.
Because economic growth can be considered as a process in which the long-run aggregate supply curve shifts to the right, and because output tends to remain close to this curve, it is important to gain a deeper understanding of what determines long-run aggregate supply (LRAS). We shall examine the derivation of LRAS and then see what factors shift the curve. We shall begin our work by defining an aggregate production function.
## The Aggregate Production Function
An aggregate production functionFunction that relates the total output of an economy to the total amount of labor employed in the economy, all other determinants of production (capital, natural resources, and technology) being unchanged. relates the total output of an economy to the total amount of labor employed in the economy, all other determinants of production (that is, capital, natural resources, and technology) being unchanged. An economy operating on its aggregate production function is producing its potential level of output.
Figure 23.6 shows an aggregate production function (PF). It shows output levels for a range of employment between 120 million and 140 million workers. When the level of employment is 120 million, the economy produces a real GDP of $11,500 billion (point A). A level of employment of 130 million produces a real GDP of$12,000 billion (point B), and when 140 million workers are employed, a real GDP of $12,300 billion is produced (point C). In drawing the aggregate production function, the amount of labor varies, but everything else that could affect output, specifically the quantities of other factors of production and technology, is fixed. The shape of the aggregate production function shows that as employment increases, output increases, but at a decreasing rate. Increasing employment from 120 million to 130 million, for example, increases output by$500 billion to $12,000 billion at point B. The next 10 million workers increase production by$300 billion to $12,300 billion at point C. This example illustrates diminishing marginal returns. Diminishing marginal returnsSituation that occurs when additional units of a variable factor add less and less to total output, given constant quantities of other factors. occur when additional units of a variable factor add less and less to total output, given constant quantities of other factors. It is easy to picture the problem of diminishing marginal returns in the context of a single firm. The firm is able to increase output by adding workers. But because the firm’s plant size and stock of equipment are fixed, the firm’s capital per worker falls as it takes on more workers. Each additional worker adds less to output than the worker before. The firm, like the economy, experiences diminishing marginal returns. ## The Aggregate Production Function, the Market for Labor, and Long-Run Aggregate Supply To derive the long-run aggregate supply curve, we bring together the model of the labor market, introduced in the first macro chapter and the aggregate production function. As we learned, the labor market is in equilibrium at the natural level of employment. The demand and supply curves for labor intersect at the real wage at which the economy achieves its natural level of employment. We see in Panel (a) of Figure 23.7 that the equilibrium real wage is ω1 and the natural level of employment is L1. Panel (b) shows that with employment of L1, the economy can produce a real GDP of YP. That output equals the economy’s potential output. It is that level of potential output that determines the position of the long-run aggregate supply curve in Panel (c). ## Changes in Long-Run Aggregate Supply The position of the long-run aggregate supply curve is determined by the aggregate production function and the demand and supply curves for labor. A change in any of these will shift the long-run aggregate supply curve. Figure 23.8 shows one possible shifter of long-run aggregate supply: a change in the production function. Suppose, for example, that an improvement in technology shifts the aggregate production function in Panel (b) from PF1 to PF2. Other developments that could produce an upward shift in the curve include an increase in the capital stock or in the availability of natural resources. The shift in the production function to PF2 means that labor is now more productive than before. This will affect the demand for labor in Panel (a). Before the technological change, firms employed L1 workers at a real wage ω1. If workers are more productive, firms will find it profitable to hire more of them at ω1. The demand curve for labor thus shifts to D2 in Panel (a). The real wage rises to ω2, and the natural level of employment rises to L2. The increase in the real wage reflects labor’s enhanced productivityThe amount of output per worker., the amount of output per worker. To see how potential output changes, we see in Panel (b) how much output can be produced given the new natural level of employment and the new aggregate production function. The real GDP that the economy is capable of producing rises from Y1 to Y2. The higher output is a reflection of a higher natural level of employment, along with the fact that labor has become more productive as a result of the technological advance. In Panel (c) the long-run aggregate supply curve shifts to the right to the vertical line at Y2. This analysis dispels a common misconception about the impact of improvements in technology or increases in the capital stock on employment. Some people believe that technological gains or increases in the stock of capital reduce the demand for labor, reduce employment, and reduce real wages. Certainly the experience of the United States and most other countries belies that notion. Between 1990 and 2007, for example, the U.S. capital stock and the level of technology increased dramatically. During the same period, employment and real wages rose, suggesting that the demand for labor increased by more than the supply of labor. As some firms add capital or incorporate new technologies, some workers at those firms may lose their jobs. But for the economy as a whole, new jobs become available and they generally offer higher wages. The demand for labor rises. Another event that can shift the long-run aggregate supply curve is an increase in the supply of labor, as shown in Figure 23.9. An increased supply of labor could result from immigration, an increase in the population, or increased participation in the labor force by the adult population. Increased participation by women in the labor force, for example, has tended to increase the supply curve for labor during the past several decades. In Panel (a), an increase in the labor supply shifts the supply curve to S2. The increase in the supply of labor does not change the stock of capital or natural resources, nor does it change technology—it therefore does not shift the aggregate production function. Because there is no change in the production function, there is no shift in the demand for labor. The real wage falls from ω1 to ω2 in Panel (a), and the natural level of employment rises from L1 to L2. To see the impact on potential output, Panel (b) shows that employment of L2 can produce real GDP of Y2. The long-run aggregate supply curve in Panel (c) thus shifts to LRAS2. Notice, however, that this shift in the long-run aggregate supply curve to the right is associated with a reduction in the real wage to ω2. Of course, the aggregate production function and the supply curve of labor can shift together, producing higher real wages at the same time population rises. That has been the experience of most industrialized nations. The increase in real wages in the United States between 1990 and 2007, for example, came during a period in which an increasing population increased the supply of labor. The demand for labor increased by more than the supply, pushing the real wage up. The accompanying Case in Point looks at gains in real wages in the face of technological change, an increase in the stock of capital, and rapid population growth in the United States during the 19th century. Our model of long-run aggregate supply tells us that in the long run, real GDP, the natural level of employment, and the real wage are determined by the economy’s production function and by the demand and supply curves for labor. Unless an event shifts the aggregate production function, the demand curve for labor, or the supply curve for labor, it affects neither the natural level of employment nor potential output. Economic growth occurs only if an event shifts the economy’s production function or if there is an increase in the demand for or the supply of labor. ## Key Takeaways • The aggregate production function relates the level of employment to the level of real GDP produced per period. • The real wage and the natural level of employment are determined by the intersection of the demand and supply curves for labor. Potential output is given by the point on the aggregate production function corresponding to the natural level of employment. This output level is the same as that shown by the long-run aggregate supply curve. • Economic growth can be shown as a series of shifts to the right in LRAS. Such shifts require either upward shifts in the production function or increases in demand for or supply of labor. ### Try It! Suppose that the quantity of labor supplied is 50 million workers when the real wage is$20,000 per year and that potential output is $2,000 billion per year. Draw a three-panel graph similar to the one presented in Figure 23.9 to show the economy’s long-run equilibrium. Panel (a) of your graph should show the demand and supply curves for labor, Panel (b) should show the aggregate production function, and Panel (c) should show the long-run aggregate supply curve. Now suppose a technological change increases the economy’s output with the same quantity of labor as before to$2,200 billion, and the real wage rises to $21,500. In response, the quantity of labor supplied increases to 51 million workers. In the same three panels you have already drawn, sketch the new curves that result from this change. Explain what happens to the level of employment, the level of potential output, and the long-run aggregate supply curve. (Hint: you have information for only one point on each of the curves you draw—two for the supply of labor; simply draw curves of the appropriate shape. Do not worry about getting the scale correct.) Case in Point: Technological Change, Employment, and Real Wages During the Industrial Revolution Figure 23.10 Wikimedia Commons – public domain. Technological change and the capital investment that typically comes with it are often criticized because they replace labor with machines, reducing employment. Such changes, critics argue, hurt workers. Using the model of aggregate demand and aggregate supply, however, we arrive at a quite different conclusion. The model predicts that improved technology will increase the demand for labor and boost real wages. The period of industrialization, generally taken to be the time between the Civil War and World War I, was a good test of these competing ideas. Technological changes were dramatic as firms shifted toward mass production and automation. Capital investment soared. Immigration increased the supply of labor. What happened to workers? Employment more than doubled during this period, consistent with the prediction of our model. It is harder to predict, from a theoretical point of view, the consequences for real wages. The latter third of the 19th century was a period of massive immigration to the United States. Between 1865 and 1880, more than 5 million people came to the United States from abroad; most were of working age. The pace accelerated between 1880 and 1923, when more than 23 million people moved to the United States from other countries. Immigration increased the supply of labor, which should reduce the real wage. There were thus two competing forces at work: Technological change and capital investment tended to increase real wages, while immigration tended to reduce them by increasing the supply of labor. The evidence suggests that the forces of technological change and capital investment proved far more powerful than increases in labor supply. Real wages soared 60% between 1860 and 1890. They continued to increase after that. Real wages in manufacturing, for example, rose 37% from 1890 to 1914. Technological change and capital investment displace workers in some industries. But for the economy as a whole, they increase worker productivity, increase the demand for labor, and increase real wages. ### Answer to Try It! Problem The production function in Panel (b) shifts up to PF2. Because it reflects greater productivity of labor, firms will increase their demand for labor, and the demand curve for labor shifts to D2 in Panel (a). LRAS1 shifts to LRAS2 in Panel (c). Employment and potential output rise. Potential output will be greater than$2,200 billion.
Figure 23.11
This page titled 8.2: Growth and the Long-Run Aggregate Supply Curve is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous.
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2022-08-11 14:17:33
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https://scholarsbank.uoregon.edu/xmlui/handle/1794/2044/browse?rpp=20&etal=-1&type=title&starts_with=B&order=ASC&sort_by=1
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Now showing items 7-26 of 57
• #### Blocks in Deligne's category Rep(St)
(University of Oregon, 2010-06)
We give an exposition of Deligne's tensor category Rep(St) where t is not necessarily an integer. Thereafter, we give a complete description of the blocks in Rep(St) for arbitrary t. Finally, we use our result on blocks ...
• #### Categorical Actions on Supercategory O
(University of Oregon, 2016-11-21)
This dissertation uses techniques from the theory of categorical actions of Kac-Moody algebras to study the analog of the BGG category O for the queer Lie superalgebra. Chen recently reduced many questions about this ...
• #### Chern Character for Global Matrix Factorizations
(University of Oregon, 2013-10-03)
We give a formula for the Chern character on the DG category of global matrix factorizations on a smooth scheme $X$ with superpotential $w\in \Gamma(\O_X)$. Our formula takes values in a Cech model for Hochschild homology. ...
• #### Cohomology of the Orlik-Solomon algebras
(2000)
The Orlik-Solomon algebra of a hyperplane arrangement first appeared from the Brieskorn and Orlik-Solomon theorems as the cohomology of the complement of this arrangement (if the ground field is complex). Later, it was ...
• #### Compact Group Actions on C*-algebras: Classification, Non-Classifiability and Crossed Products and Rigidity Results for Lp-operator Algebras
(University of Oregon, 2015-08-18)
This dissertation is concerned with representations of locally compact groups on different classes of Banach spaces. The first part of this work considers representations of compact groups by automorphisms of C*-algebras, ...
• #### Crossed product C*-algebras by finite group actions with a generalized tracial Rokhlin property
(University of Oregon, 2008-06)
This dissertation consists of two related parts. In the first portion we use the tracial Rokhlin property for actions of a finite group G on stably finite simple unital C *-algebras containing enough projections. The ...
• #### Crossed product C*-algebras of certain non-simple C*-algebras and the tracial quasi-Rokhlin property
(University of Oregon, 2010-06)
This dissertation consists of four principal parts. In the first, we introduce the tracial quasi-Rokhlin property for an automorphism α of a C *-algebra A (which is not assumed to be simple or to contain any projections). ...
• #### Crossed product C*-algebras of minimal dynamical systems on the product of the Cantor set and the torus
(University of Oregon, 2010-06)
This dissertation is a study of the relationship between minimal dynamical systems on the product of the Cantor set ( X ) and torus ([Special characters omitted]) and their corresponding crossed product C *-algebras. For ...
• #### The crossed product of C(X) by a free minimal action of R
(University of Oregon, 2010-06)
In this dissertation, we will study the crossed product C*-algebras obtained from free and minimal [Special characters omitted.] actions on compact metric spaces with finite covering dimension. We first define stable ...
• #### Dancing in the Stars: Topology of Non-k-equal Configuration Spaces of Graphs
(University of Oregon, 2016-11-21)
We prove that the non-k-equal configuration space of a graph has a discretized model, analogous to the discretized model for configurations on graphs. We apply discrete Morse theory to the latter to give an explicit ...
• #### Finite W-algebras of classical type
(University of Oregon, 2009-06)
In this work we prove that the finite W -algebras associated to nilpotent elements in the symplectic or orthogonal Lie algebras whose Jordan blocks are all the same size are quotients of twisted Yangians. We use this to ...
• #### Frames Generated by Actions of Locally Compact Groups
(University of Oregon, 2016-10-27)
Let $G$ be a second countable, locally compact group which is either compact or abelian, and let $\rho$ be a unitary representation of $G$ on a separable Hilbert space $\mathcal{H}_\rho$. We examine frames of the form \$\{ ...
• #### Generalized Near-Group Categories
(University of Oregon, 2012)
We give an exposition of near-group categories and generalized near-group categories. We show that both have a pseudounitary structure. We complete the classification of braided near-group categories and discuss the inherent ...
• #### Generalized self-intersection local time for a superprocess over a stochastic flow
(University of Oregon, 2010-06)
This dissertation examines the existence of the self-intersection local time for a superprocess over a stochastic flow in dimensions d ≤ 3, which through constructive methods, gives a Tanaka like representation. The ...
• #### Geometry and Combinatorics Pertaining to the Homology of Spaces of Knots
(University of Oregon, 2012)
We produce explicit geometric representatives of non-trivial homology classes in Emb(S1,Rd), the space of knots, when d is even. We generalize results of Cattaneo, Cotta-Ramusino and Longoni to define cycles which live ...
• #### GKZ Hypergeometric Systems and Projective Modules in Hypertoric Category O
(University of Oregon, 2016-10-27)
In this thesis I show that indecomposable projective and tilting modules in hypertoric category O are obtained by applying a variant of the geometric Jacquet functor of Emerton, Nadler, and Vilonen to certain Gel'fand-Ka ...
• #### Gluing Bridgeland's stability conditions and Z2-equivariant sheaves on curves
(University of Oregon, 2009-06)
We define and study a gluing procedure for Bridgeland stability conditions in the situation where a triangulated category has a semiorthogonal decomposition. As one application, we construct an open, contractible subset U ...
• #### Graded representation theory of Hecke algebras
(University of Oregon, 2010-06)
We study the graded representation theory of the Iwahori-Hecke algebra, denoted by Hd , of the symmetric group over a field of characteristic zero at a root of unity. More specifically, we use graded Specht modules to ...
• #### Group Actions on Hyperplane Arrangements
(University of Oregon, 2012)
In this dissertation, we will look at two families of algebras with connections to hyperplane arrangements that admit actions of finite groups. One of the fundamental questions to ask is how these decompose into irreducible ...
• #### Group decompositions, Jordan algebras, and algorithms for p-groups
(University of Oregon, 2008-06)
Finite p -groups are studied using bilinear methods which lead to using nonassociative rings. There are three main results, two which apply only to p -groups and the third which applies to all groups. First, for finite ...
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2017-06-23 13:59:59
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https://www.sangakoo.com/en/unit/a-point-in-the-plane-positions-regarding-a-circumference
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# A point in the plane: positions regarding a circumference
When we have a circumference on the plane, a point can be in three different positions regarding the circle we have in the drawing:
• Point outside the circumference
If the distance from the point to the center point of the circumference is greater than the length of the radius.
• Point on the circumference
If the distance from the point at the center of the circumference is exactly equal to the length of the radius of the circumference.
• Point within the circumference
If the distance from the point to the center of the circumference is less than the length of the radius.
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2023-02-08 00:20:31
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http://love2d.org/forums/viewtopic.php?f=4&t=83660&sid=2357f86ee46c7046438b0c037296f924
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## Is it worth it to make a built in tile editor?
Questions about the LÖVE API, installing LÖVE and other support related questions go here.
Forum rules
fyregryph
Prole
Posts: 9
Joined: Fri Oct 28, 2016 5:51 pm
### Is it worth it to make a built in tile editor?
With things like Tiled that look like quite nice editors, is it considered bad practice to make your own built in editor?
I'm trying to put together a platformer base/engine type thing for some of my game ideas and I've been trying to keep everything organized instead of quickly hacking things together as in my other löve games. So I've been going down lots of tangents of "how should this or that be" designing class hierarchies and such, trying to lay down a decent foundation so it can grow without becoming super convoluted.
So on one hand a built in editor could end up being a pretty nice thing to have if it grows, on the other hand it might just be better to use a powerful separate tool like Tiled.
zorg
Party member
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Location: Absurdistan, Hungary
Contact:
### Re: Is it worth it to make a built in tile editor?
Probably depends on how much free time you have, and how dedicated as a person you are.
Built-in editors are always a nice idea, but developing that takes away from the time you could spend actually coding your game itself; i should know, i never really finished any of my projects because of this reason.
That said, there are probably people who can pull it off, so decide for yourself whether you're that kind of person or not.
Me and my stuff True Neutral Aspirant. Why, yes, i do indeed enjoy sarcastically correcting others when they make the most blatant of spelling mistakes. No bullying or trolling the innocent tho.
MrFariator
Citizen
Posts: 55
Joined: Wed Oct 05, 2016 11:53 am
### Re: Is it worth it to make a built in tile editor?
I am developing a platformer with a built-in level editor. My thought process was that I wanted to invite my friends to later design stages with the same set of tools as I, meaning they just need to grab the game executable to have a go at it. This would then extend to any players who would want to make their own stages after beating the main game. Another reason was that I wanted to better learn how to make GUI driven applications, and this was a perfect excuse just for that.
However, I wanted to keep things fairly simple and straight-forward (though I am planning things like in-editor scripting API), but without a doubt I have so far spent more time on the editor than actual gameplay or content. One bonus though is that because my editor is generic enough I could easily just take the visual assets and make whatever 2D game with it, from platformer to top-down games.
In the end though, it really depends on your available time, as zorg said. Shovel Knight developers just went with Tiled, and so do a plenty of other developers, particularly when they just want to like, make a game.
-- Run once in a blue moon
if math.random() > 0.99 then
account:post ( currentProgress:getGIF() )
end
Jasoco
Inner party member
Posts: 3593
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Location: Pennsylvania, USA
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### Re: Is it worth it to make a built in tile editor?
I find it a lot of fun to build your own editor. Lets you be more flexible in map structure. But it's a lot of work.
I too had a level editor in my shelved platformer engine. And am now making a similar one in my current engine.
fyregryph
Prole
Posts: 9
Joined: Fri Oct 28, 2016 5:51 pm
### Re: Is it worth it to make a built in tile editor?
@MrFariator Wow, that is quite an impressive level editor! I can see why that'd take up more time than the game itself, designing UI components and editing tools to that extent.
@Jasoco That's basically what I'm thinking, that it would be more flexible and be cool to make even if it's not as good as something dedicated to tile based editing. All my löve games have been learning tangents anyways And I'm wanting to involve my non-programmer sister in making some games too so I think some sort of editor would be good.
...Actually I keep toying with the idea of trying to make a sandbox game in the spirit of Terraria, with this being kind of a testing the waters thing making a bare bones WYSIWYG editor.
kikito
Inner party member
Posts: 3133
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Contact:
### Re: Is it worth it to make a built in tile editor?
I did an editor once, in the past. It took a loooot of time and effort. So much that I didn't have any energy left for actually implementing a game for it.
Nowadays I strongly favour easily parseable text files. If I can make it a .lua file, so I can directly load it with loadfile, I do just that. And I design levels on a text editor. Maybe I got burned by that first experience.
An alternative that I have seen some people use, which I have not tried myself, is using image files as levels. If your level is 100x100, you use a 100x100 png image, where each pixel represents a tile, and its color codifies the type of tile it represents: green pixels are platforms. Red pixels, spikes. And so on. That seems relatively fast to implement, and would make designing levels quite straightforward. It might not be enough for all kinds of games.
When I write def I mean function.
scissors61
Citizen
Posts: 68
Joined: Fri Jan 08, 2016 10:16 am
### Re: Is it worth it to make a built in tile editor?
kikito wrote:
Tue Mar 07, 2017 4:04 pm
An alternative that I have seen some people use, which I have not tried myself, is using image files as levels. If your level is 100x100, you use a 100x100 png image, where each pixel represents a tile, and its color codifies the type of tile it represents: green pixels are platforms. Red pixels, spikes. And so on. That seems relatively fast to implement, and would make designing levels quite straightforward. It might not be enough for all kinds of games.
Hi kikito. This alternative sounds interesting. Can you share a reference of how they do it?
Jasoco
Inner party member
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### Re: Is it worth it to make a built in tile editor?
Notch did it with his old 24-hour jam "Prelude to the Chambered" game.
To create the levels you'd use an image editor. Use a different color for each tile. Then to load them into your game you'd load the image as ImageData then use ImageData:getPixel(x, y) or put it on a Canvas and use Canvas:getPixel(x, y) to get the RGB data of the pixel and use some checks, like some if/thens or a lookup table to figure out what tile to use for that color.
zorg
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Location: Absurdistan, Hungary
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### Re: Is it worth it to make a built in tile editor?
You could also use the separate color (and alpha) channels for different "layers" in your world; e.g. the red channel could hold 256 background tiles, the blue channel 256 foreground tiles, the green channel 256 technical tiles, like warp points, save points, etc... and similarly with the alpha channel. How you partition those up is up to you, of course.
That said, it'll take a bit more work when 0.11 comes around and the numbers will be normalized to [0,1].
Me and my stuff True Neutral Aspirant. Why, yes, i do indeed enjoy sarcastically correcting others when they make the most blatant of spelling mistakes. No bullying or trolling the innocent tho.
MasterLee
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Posts: 141
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Contact:
### Re: Is it worth it to make a built in tile editor?
zorg wrote:
Tue Mar 07, 2017 6:56 pm
That said, it'll take a bit more work when 0.11 comes around and the numbers will be normalized to [0,1].
Hopefully there will be an workaround for that. It is already hard enough without texelFetch in shaders so why make preparing textures even harder?
Code: Select all
extern vec2 size;
extern sampler2D image;
extern sampler2D karte;
vec4 effect(vec4 color,Image texture,vec2 texture_coords,vec2 pixel_coords)
{
pixel_coords.y=love_ScreenSize.y-pixel_coords.y;
return Texel(image,(mod(pixel_coords,16)+Texel(karte,pixel_coords/128).rg*4096)/size.xy);
}
### Who is online
Users browsing this forum: Nixola and 5 guests
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2017-09-24 10:27:35
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https://tutel.me/c/mathematicians/questions/253478/can+the+integration+of+integrable+sections+of+a+measurable+function+of+two+variables+ever+result+in+a+nonmeasurable+function
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#### [SOLVED] Can the integration of integrable sections of a measurable function of two variables ever result in a non-measurable function?
I spent some time searching MathOverflow for a problem that would resemble the one given below, but it turned out to be a rather futile endeavor. I was led to this problem in my attempts to construct a counterexample refuting a result that had already been published in a peer-reviewed article.
Problem. Find measure spaces $(X,\mathcal{S},\mu)$ and $(Y,\mathcal{T},\nu)$, at least one of which is not $\sigma$-finite, and an $(\mathcal{S} \otimes \mathcal{T})$-measurable function $f: X \times Y \to \mathbb{R}_{\geq 0}$ with the following properties:
• The function $f(x,\bullet): Y \to \mathbb{R}_{\geq 0}$ belongs to ${L^{1}}(Y,\mathcal{T},\nu)$ for every $x \in X$.
• The function $\left\{ \begin{matrix} X & \to & \mathbb{R}_{\geq 0} \\ x & \mapsto & \displaystyle \int_{Y} f(x,\bullet) ~ \mathrm{d}{\nu} \end{matrix} \right\}$ is not $\mathcal{S}$-measurable.
Does anyone know if this problem can even be solved? Thank you very much for your time!
#### @rumpf 2016-11-01 22:48:32
Let $\mathcal{B}$ denote the class of Borel subsets of $[0,1]$ and let $A \subseteq [0, 1]$ be a non Borel set. Let $f$ be the characteristic function of the graph of a bijection from $A$ to $[0, 1]$. Then $f$ is $\mathcal{B} \otimes \mathcal{P}([0, 1])$-measurable (check) and the map $x \mapsto \int f(x, y) d\mu(y)$ is non-zero precisely on $A$ where $\mu$ is counting measure.
Edit: I was asked to provide more details so here they are.
Suppose $W \subseteq \mathbb{R}^2$ is such that every horizontal section $W^y = \{x : (x, y) \in W\}$ is closed. Then $W \in \mathcal{B} \otimes \mathcal{P}(\mathbb{R})$. To see this, define, for each interval $J$ with rational endpoints, $Y_J = \{y : W^y \cap J = \phi\}$. As each $W^y$ is closed, we have $$W = \mathbb{R}^2 \Big\backslash \bigcup \{J \times Y_J : J \text{ is an interval with rational endpoints}\} \in \mathcal{B} \otimes \mathcal{P}(\mathbb{R}).$$ It follows that the graph of every partial bijection on $\mathbb{R}$ is in $\mathcal{B} \otimes \mathcal{P}(\mathbb{R})$.
Now choose any non Borel set $A \subseteq \mathbb{R}$ and an injection $i: A \to \mathbb{R}$. Define $f: \mathbb{R}^2 \to \mathbb{R}$ to be the characteristic function of the graph of $i$. Let $\mu$ be the counting measure on $\mathbb{R}$. The map $x \mapsto \int f(x, y) d\mu(y)$ is precisely the characteristic function of $A$ and hence is non-Borel.
#### @Transcendental 2016-11-01 23:04:00
Hi rumpf. Thank you for your response. Unfortunately, I don’t consider myself to be smart enough to check your claim in the third line. Could you elaborate further?
#### @drumpf 2016-11-01 23:12:41
For each rational interval $J$, let $Y_J$ be the set of points where the horizontal section of the graph of the bijection is disjoint with $J$. Now consider the union of $J \times Y_J$'s.
#### @Nate Eldredge 2016-11-02 05:11:37
I am getting confused by these definitions; I think it would help if more things had names. Let's call $\phi : A \to [0,1]$ the desired bijection. It looks to me like you want $Y_J = \phi(A \cap J^c)$? That doesn't seem to make sense. In particular if $y = \phi(x)$ then $(x,y)$ is not in $J \times Y_J$ for any $J$.
#### @Transcendental 2016-11-02 18:26:26
@Nate: drumpf’s improved argument does appear to check out. What do you think?
#### @Nate Eldredge 2016-11-02 18:31:48
@Transcendental: Yes, I think so too. +1 from me. Maybe a simpler way of describing $f$ is that it is just $f(x,y) = 1$ iff $x=y \in A$ and $0$ otherwise.
#### @Transcendental 2016-11-02 22:02:22
@Nate: It also appears that his argument works if we let $X$ be any uncountable second-countable $T_{1}$ topological space whose Borel $\sigma$-algebra $\mathcal{S}$ isn’t all of $\mathcal{P}(X)$ (i.e., $X$ has a non-Borel subset). Any uncountable standard Borel space automatically satisfies this property (assuming the Axiom of Choice, of course).
#### @PhoemueX 2016-10-31 16:59:59
EDIT: The following only provides a partial answer, since it is not clear at all that the first property of the question ($f(x, \cdot) \in L^1(Y, \mathcal{T}, \nu)$) is fulfilled for the given example.
Let $(X, \mathcal{S})$ and $(Y, \mathcal{T})$ both be the real line with the Borel sigma algebra. Note that the product sigma algebra is again the Borel sigma algebra (but on $\Bbb{R}^2$).
It is well-known (see Projection of Borel set from $R^2$ to $R^1$) that not every projection of a Borel set is a Borel set. Hence, let $M \subset \Bbb{R}^2$ be a Borel set such that the projection $\pi_1 (M)$ is not Borel measurable.
Let $\mu$ be the counting measure on the real line. If $$F(x) := \int_{\Bbb{R}} 1_M (x,y) d \mu(y)$$ was measurable, then so would be the set $$\pi_1 (M) = \{x \,:\, \exists y : (x,y) \in M\} = \{x \,:\, F(x) > 0\}.$$
#### @Christian Remling 2016-10-31 18:50:14
Wikipedia claims that Fubini always works for the "maximal product measure." I wonder what that means if the iterated integrals aren't even defined. en.wikipedia.org/wiki/…
#### @Transcendental 2016-10-31 21:19:18
@PhoemueX: Thanks! I must shamefully admit that the fact you quoted wasn’t known to me.
#### @Transcendental 2016-10-31 21:38:27
@Christian: Fubini’s Theorem states that if $(X,\mathcal{S},\mu)$ and $(Y,\mathcal{T},\nu)$ are measure spaces, and if $f \in {L^{1}}(X \times Y,\mathcal{S} \otimes \mathcal{T},(\mu \otimes \nu)_{\text{max}})$, where $(\mu \otimes \nu)_{\text{max}}$ denotes the maximal product measure, then the iterated integrals exist and are equal to $\displaystyle \int_{X \times Y} f ~ \mathrm{d}{(\mu \otimes \nu)_{\text{max}}}$. I suspect that the function $\mathbf{1}_{M}$ defined by PhoemueX above isn’t integrable on $\mathbb{R}^{2}$ with respect to the maximal product measure.
#### @Transcendental 2016-11-01 07:04:36
@PhoemueX: Your counterexample clearly satisfies the second property, but can you show that it also satisfies the first property? In other words, can you prove that $\{ y \in \mathbb{R} \mid (x,y) \in M \}$ is finite for each $x \in \mathbb{R}$? Thanks!
#### @PhoemueX 2016-11-01 07:32:15
@Transcendental: Yes, I realized this problem just this morning. With the current construction, I don't think this is true. I will delete my answer if I see no way to fix it until this evening :(
#### @Transcendental 2016-11-01 10:46:42
@PhoemueX: Please don’t delete your answer as I (and many other people) considered it very helpful. Even if you aren’t able to find a fix in the end, somebody else might. :)
#### @PhoemueX 2016-11-01 13:22:41
@Transcendental: Ok, sure :) I just thought it would draw more attention to the question, since at the moment it seems to be answered. I will just edit it to show more clearly that it is not a complete answer.
#### @Christian Remling 2016-11-01 19:11:59
To make one more unhelpful remark, the fact that $M$ is Borel in $\mathbb R^2$ is actually not that relevant perhaps since one could have taken $\mathcal P(Y)$ as the $\sigma$-algebra on $Y$.
#### @Transcendental 2016-11-01 21:27:34
@Christian: It’s clear that $M$ is a member of the $\sigma$-algebra $\mathcal{B}(\mathbb{R}) \otimes \mathcal{P}(\mathbb{R})$, but any act of whittling down each vertical cross-section of $M$ to make it finite might threaten this membership.
#### @Nate Eldredge 2016-11-02 15:18:59
As discussed here, the Lusin-Novikov theorem says that your example cannot satisfy the first property.
### [SOLVED] On measurable functions of two variables
• 2010-09-20 18:01:45
• dergachev
• 764 View
• 6 Score
• Tags: measure-theory
|
2019-04-18 15:23:57
|
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|
https://codegolf.stackexchange.com/questions?sort=newest&page=5
|
# All Questions
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### Generate a Graeco-Latin square
disclaimer: I'm not aware of any non-bruteforce solutions A Graeco-Latin square is, for two sets of same length $n$, a $n \times n$ arrangement of cells, each containing a unique (across the ...
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I'm stuck on a problem involving the Gale-Ryser Theorem. The problem's input gives me the row-wise sum of an hv-convex binary matrix(n*m). ...
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2019-08-24 22:43:26
|
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http://codereview.stackexchange.com/questions/16425/ajax-maskededitvalidator-isvalidempty-property-manipulation
|
# AJAX MaskedEditValidator isValidEmpty property manipulation [closed]
thanks for visiting , this is my initial question that lead me to modify the behavior of AJAX MaskedEditValidator - isValidEmpty attribute.
when you try to cancel the interaction with a form , being a form that one of its elemets (say the first textbox) is set to "required" . in that case , the validator will not let you do anything unless you first type a valid value in the target text box .
you have no choice other than to fill the field that you set focus to as required.
till then the user is disabled rather than the UI (:.
so I came up with a solution, though I think should be optimized so it would suit any target TextBox (actually target is MaskedEditValidator-ID and the text box is only the referencing entity) in this case .
and also the code is far from being as elegant and clean so i could proudly share it with everyone
<cc1:MaskedEditExtender enabled="true" MaskType="Date"
TargetControlID="TBXinsertDate" InputDirection="LeftToRight"
CultureName="en-GB"
ErrorTooltipCssClass="toolTipForInvalid">
<cc1:MaskedEditValidator Enabled="false" IsValidEmpty="false"
ID="insertDate_MskValidator"
ControlToValidate="TBXinsertDate" runat="server"
InvalidValueBlurredMessage="invalid date"
EmptyValueBlurredText="requierd fileld"
ErrorMessage="Error"
MaximumValue="01/01/2015" MinimumValue="01/01/2008"
MaximumValueBlurredMessage="max year value is 2015"
CssClass="dateInValid">
• default state of validator
Enabled = false : i am enabling it later on via onClick button event from code behind as follows
insertDate_MskValidator.Enabled = true;
so the sequence is :
stage 1 - validator-enabled = false //when form is hidden
stage 2 - validator-enabled = true //when form is visible
stage 3 options:
cancel / submit
submit: will be available if form is valid
cancel: is no option if you step in one of text box leaving it empty .
and this is my solution :
## JavaScript
//delay execution
function setTDisableValidation() {
setTimeout('disableValidation()', 3500);
}
//execute modification
function disableValidation() {
var myVal = document.getElementById('insertDate_MskValidator');
myVal.disabled = true;
myVal.ValidEmpty = true;
}
## ASPX
<body>
<form id="form1" runat="server">
<asp:ScriptManager ID="ScriptManager1" runat="server">
</asp:ScriptManager>
<asp:TextBox ID="TBXinsertDate" runat="server"
ToolTip="insert date" Width="76px"
onblur="setTDisableValidation();">
</asp:TextBox>
## CSS
.WaterMarkedTextBox
{
height: 16px;
width: 80px;
border: 1px solid #BEBEBE;
background-color: #F0F8FF;
color: gray;
font-size: 8pt;
text-align: center;
}
.NormalTextBox
{
height: 16px;
width: 168px;
}
.greytext
{
color:red;
}
{
color:blue;
}
.dateValid
{
font-size:10px; color:black;
}
.dateInValid
{
font-size:10px; color:red;display:inline-block;
}
{
font-size:10px; color:red;display:inline;
}
.toolTipForInvalid
{
font-size:10px; color:red; background-color:Yellow
}
-
## closed as unclear what you're asking by Jamal♦Nov 25 '13 at 17:24
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.
|
2014-07-23 07:49:33
|
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|
https://docs.microej.com/en/latest/ApplicationDeveloperGuide/UI/MicroVG/matrix.html
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# Matrix¶
A Matrix is composed of an array of numbers with three rows and three columns. It is used to apply an affine transformations to Path points. (Refer to <https://en.wikipedia.org/wiki/Transformation_matrix#Affine_transformations> to get more information about affine transformations).
The available transformations are:
• translation
• rotation
• scaling
Scaling and rotation are always performed around the (0,0) pivot point. In order to rotate or scale a Path with a pivot point, the matrix must be translated before and after the rotation/scaling.
A Matrix is created as an identity matrix, which means that a Path resulting of a transformation with this matrix is identical to the original Path.
The Matrix can be initialized with a transformation with set methods:
• setTranslate(translateX, translateY)
• setRotate(angle)
• setScale(scaleX, scaleY)
A transformation can be prepended to a Matrix with the prepend methods:
• preTranslate(translateX, translateY)
• preRotate(angle)
• preScale(scaleX, scaleY)
A transformation can be appended to a Matrix with the append methods:
• postTranslate(translateX, translateY)
• postRotate(angle)
• postScale(scaleX, scaleY)
A Matrix can also get transformations from an other Matrix with the concatenate and set methods:
• preConcat(matrix)
• postConcat(matrix)
• set(matrix)
• setConcat(matrix0, matrix1)
Once a Matrix has been computed, it can be used to draw an object (Path, String, VectorImage). All the points of the drawn object will be transformed by the Matrix.
When a Matrix has been computed with multiple type of transformation, the sequence order of the transformation is important. Chaining the transformations in a different order will not provide the same Matrix. The result of the previous transformation is the input to the next transformation.
The following examples use the Path created in the section Path Creation with different transformations.
## Translation¶
Matrix matrix = new Matrix(); matrix.setTranslate(200, 150);
## Rotation¶
Around point (0,0).
Matrix matrix = new Matrix(); matrix.setRotate(40);
Around a pivot point (80,50).
Matrix matrix = new Matrix(); matrix.setRotate(40); float pivotX = 80; float pivotY = 50; matrix.preTranslate(-pivotX, -pivotY); matrix.postTranslate(pivotX, pivotY);
## Scale¶
From point (0,0).
Matrix matrix = new Matrix(); matrix.setScale(2,3);
## Concatenate Matrixes¶
Sequence order has an incidence on the rendering.
Matrix matrix0 = new Matrix(); matrix0.setScale(2, 3); Matrix matrix1 = new Matrix(); matrix1.setTranslate(100, 40); Matrix matrix2 = new Matrix(); matrix2.setConcat(matrix0, matrix1); g.setColor(Colors.GRAY); VectorGraphicsPainter.fillPath(g, path, matrix2); matrix2.setConcat(matrix1, matrix0); g.setColor(Colors.YELLOW); VectorGraphicsPainter.fillPath(g, path, matrix2);
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2022-09-30 13:21:15
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|
https://math.stackexchange.com/questions/1858263/what-are-algebraic-equations-really
|
# What are (algebraic) equations, really?
This is a much more philosophical question that I'm used to asking, but it's been nagging at me a bit. Let's say I have an algebraic equation $y = x$, $x, y, \in \mathbb{R}$. The best interpretation I have for this is: for any given $x$, you produce a $y$ with the same value. But, philosophically I have some issues:
1) Why am I allowed to choose an $x$ in the first place, and how do I know that the equation will produce a $y$ for every $x$ I choose? It's "obvious" that it does, but it seems like this is a sneaky way of saying something about the operator "=" that is often left out of textbooks.
2) What about more complicated equations that involve algebraic operations, like $y = x^2$? I need to define certain operations, like multiplication, in order for this to work. However, this isn't very satisfying. For example, the real numbers form a field and therefore in order to define multiplication we require $\forall x \ne 0 \exists x^{-1} | x*x^{-1} = 1$. But how do we really know that this exists for all $x \in \mathbb{R}$?
am I just going way too deep into an ultimately meaningless question?
• I would not look at it in a way that an $x$ produces an $y$, but rather as a relation on the set of $\mathbb R^2 \ni (x,y)$. – flawr Jul 13 '16 at 15:05
• So you're saying that all pairs $(x,y)\in \mathbb{R}^2$ that satisfy the relation already exist, independent of my description, and my description is just a useful way to talk about the pairs? – Michael Stachowsky Jul 13 '16 at 15:08
• Generally when you think of e.g. a line or of a plane, you consider a subset of $\mathbb R^n$ that satisfies some conditions, for example an equation $ax+by+c=0$ or $Ax+By+Cz+D=0$. – flawr Jul 13 '16 at 15:11
• You can't tell what color a widget is unless you know what a widget is. You can't tell whether non-zero reals have inverses unless you have a def'n of R and show that something exists that satisfies this def'n. This leads to "foundational " topics, which are covered in texts on set-theory...... And what IS the def'n of R ? – DanielWainfleet Jul 13 '16 at 18:32
The equation $y=x$ is a predicate about the variables $y,x$. It may be true or false. For each possible pair $(x,y)\in\mathbb{R}^2$, some pairs satisfy the equation and some do not. Hence the equation partitions $\mathbb{R}^2$ into those pairs that satisfy the equation, and the rest. It is possible that all pairs satisfy the equation, or that none do.
It is common to draw $\mathbb{R}^2$ as a pair of perpendicular axes, and indicate the satisfying pairs for a given equation as dots. This is called the graph of the equation.
As for your second question, in order to meaningfully talk about $\mathbb{R}$, much less $\mathbb{R}^2$, we need to have a mathematical definition for it. Part of this definition includes the field axioms. Unfortunately, the definition is pretty complicated, and students are taught to use $\mathbb{R}$ before they really understand that definition.
|
2019-09-18 12:21:53
|
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|
https://graz.pure.elsevier.com/en/publications/simultaneous-interpolation-and-p-adic-approximation-by-integer-va
|
# Simultaneous interpolation and P-adic approximation by integer-valued polynomials
Research output: Chapter in Book/Report/Conference proceedingChapter
## Abstract
Let $D$ be a Dedekind domain with finite residue fields
and $\F$ a finite set of maximal ideals of $D$.
Let $r_0$, $\ldots$, $r_n$ be distinct elements of $D$,
pairwise incongruent modulo $P^\kP$ for each $P\in\F$,
and $s_0$, $\ldots$, $s_n$ arbitrary elements of $D$.
We show that there is an interpolating $P^\kP$-congruence
preserving integer-valued polynomial, that is,
$f\in \Int(D)=\{g\in K[x]\mid g(D)\subseteq D\}$
with $f(r_i)=s_i$ for $0\le i \le n$, such that, moreover, the
function $f\colon D\rightarrow D$ is constant modulo $P^\kP$
on each residue class of $P^\kP$ for all $P\in\F$.
Translated title of the contribution Simultane Interpolation und P-adische Approximation durch ganzwertige Polynome English Rings and Factorizations Alberto Facchini, Marco Fontana, Alfred Geroldinger, Bruce Olberding Springer, Cham Accepted/In press - 2020
## Keywords
• Interpolation
• polynomials
• P-adic approximation
• P-adic Lipschitz functions
• polynomial functions
• polynomial mappings
• congruence preserving
• integer-valued polynomials
• Dedekind domains
• commutative rings
• integral domains
## ASJC Scopus subject areas
• Algebra and Number Theory
## Fields of Expertise
• Information, Communication & Computing
|
2021-06-25 09:50:42
|
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|
http://driveyourself.nl/2s2js9/69b535-why-does-zinc-dissolve-in-hydrochloric-acid
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What mass of hydrogen gas was collected? It is quite ductile and malleable at temperatures ranging from 100-150 °С. 1 decade ago. Zn + HCl ---> ZnCl 2 + H 2. Dissolve about 0.15 g, accurately weighed, in 10 mL of acetic acid (~120 g/l) TS and proceed with the titration as described under 2.5 Complexometric titrations for zinc. There can be no universal recipe for concentration or temperature. Zn + HCl ----->ZnCl 2 + H 2 Cu + HCI -----> No reaction 2.) Answer to: The final step of the lab is to dissolve any remaining zinc with hydrochloric acid. When the Zinc is added to the acid, the zinc dissolves and reacts with the acid, forming hydrogen gas and the salt Zinc Chloride. Zinc is a brittle, silvery-blue metal. In my experiment, not only did the zinc dissolve, but the copper started to dissolve too. Sulfuric acid: Nickel can dissolve in hot concentrated sulfuric acid as well. Zn + HCl = H2 + ZnCl2 It is a single displacement reaction where - Zinc is oxidised (it becomes positive) and water is neutralized. Share Tweet Send [Deposit Photos] Physical properties of metallic zinc. 6. Relevance. The ethanoic acid took almost 12 minutes and the other acids took 2 minutes, why in easy to understand language, does it react … Thanks! Acetic acid: 5% acetic acid is too dilute to dissolve nickel metal. Although you should take a look at what Klaus said, copper does in fact react with hydrochloric acid, it just takes a week until all the copper is converted into copper chloride (green) and another week or so until it forms crystals and you can dissolve them in water to form copper chloride again (but depending on the amount of chloride it has, it'll be blue or green). That is why zinc displaces hydrogen from dilute hydrochloric acid, while copper does not. What are the half reactions? Aluminum does not dissolve in aqueous solutions of acids due to presence of oxide film on surface! When the zinc in a certain penny dissolves, the total volume of gas collected over water at 25 ∘C was 0.953 L at a total pressure of 740 mmHg. My suggestion is to read the label before doing dilutions because all muriatic acids are not created equal. Zinc reacts with hydrochloric acid to liberate Hydrogen gas as it is more reactive than Hydrogen and hence, displace it. First of all, how does it compare to other "strong acids", for example sulfuric acid The solid zinc lost mass because some of it reacted with the hydrochloric acid to form zinc chloride. When the Mg dissolves, it … REASONS WHY HYDROCHLORIC ACID DISSOLVES IN STOMACH 2 The hydrochloric acid is secreted by the parietal cell, found inside mucosa, into the stomach's cavity or lumen. 3.) zinc dissolve in hydrochloric acid with evolution of hydrogen gas physical or chemical property and also give the reason - 5961320 The same can be understood by looking at the activity series of metals. B. That is why zinc displaces hydrogen from dilute hydrochloric acid, while copper does not. plzzz help i will mark as brainiest! C. Assertion is correct but Reason is incorrect. Health and safety considerations are paramount but HCl is decidedly less of a problem than HF. I knew as soon as I opened the cap that it was pretty strong because the fumes were obvious. but Cu, Hg, Ag, Au, and Pt doesnt. 5 Answers. In the reaction between zinc and hydrochloric acid, the zinc atoms are oxidized and lose two electrons. New questions in Chemistry . hydrochloric acid was recorded by a stopwatch started as the magnesium. The test was to see the speed of the reaction between magnesium and 3 acids, ethanoic acid, sulfuric acid and hydrochloric acid. Stomach acid is strong enough to dissolve metal. When nickel dipped in hydrochloric acid is brought in contact with zinc, bubbles form on the surface of the nickel metal. metal is mixed with _ to make it corrosive resistant. I really need to know why magnesium reacts with hydrochloric acid by tomorrow! experiment, was used to stir the reacting materials and to ensure the . The reaction between zinc and hydrochloric acid produces zinc chloride and hydrogen gas. Energy Zn (s) + 2HCl (aq) • When 1 mole of zinc react with 2 mole of hydrochloric acid to form 1 mole of zinc chloride and 1 mole of hydrogen ∆ H = -126 kJ gas, heat released is 126 kJ. pisgahchemist. please tell me … Zinc metal reacts with hydrochloric acid, generating hydrogen gas. The time taken for the magnesium to totally dissolve in the. Dispose the Hydrochloric Acid appropriately and rinse the test tube thoroughly 4. repeat steps 1-3 for the second trial 5. calculate the mean between the two trials. Since the zinc is oxidized and the reaction needs to remain balanced, the hydrogen atoms are reduced. Zinc dissolves in hydrochloric acid to yield hydrogen gas as follows. Answer Save. Even the addition of hydrogen peroxide did not dissolve the nickel … The reaction between hydrochloric acid and zinc Properties of zinc and specifics of its interactions with HCl. Relevance. I'd be using hydrochloric acid. chemistry 105. Hydrogen also exists in stars and planets. 1 decade ago. What substance was oxidized? Hydrogen is one of the most abundant elements in the universe, and it plays a role in many daily life functions. (so that the hydrochloric acid can get to the zinc). Why does Fe, Zn, Al, Ni, Sn all dissolve in hydrochloric acid. Reason Zinc is above hydrogen in reactivity series. What is the complete balanced equation? Add your answer and earn points. We have an assessment on chem one question is why does the magnesium ribbon dissolve faster in the hydrochloric and sulfuric acid? Good digestion starts in the mouth. a small piece of zinc into a test tube containing 1-2 ml of 1.0 M hydrochloric acid solution. In class we did an experiment to see how much zinc was in a post 1982 penny. Zn(s) + 2HCl(aq) ZnCl 2 2 → (aq) + H (g) What mass of hydrogen gas is produced when a 4.30 g sample of zinc dissolves … Follow • 2. I know what you’re thinking because I thought: if the acid in your stomach is strong enough to dissolve metal, why doesn’t it burn right through the lining of the stomach? All chemical reactions involve reactants which when mixed may cause a chemical reaction which will make products. A. Hydrochloric acid or muriatic acid is a colorless inorganic chemical system with the formula HCl. In my case the reactants are hydrochloric acid and magnesium ribbon. It has no taste, color or odor, and is one of the least-dense gases. Favourite answer. Each mL of disodium edetate (0.05 mol/l) VS is equivalent to 4.069 mg of ZnO. It is classified as strongly acidic and can attack the skin over a wide composition range, since the hydrogen chloride completely dissociates in an aqueous solution. Zinc + Hydrochloric Acid. That’s a great question but before we get to the answer lets back up and start at the beginning. johnsavant@sbcglobal.net. On Earth, it is present in water and in molecules that form living organisms. What are the oxidation numbers? So if you have a good idea on the matter, please chip in. Facts The hydrochloric acid takes five hours to digest Red Meat and 2 hours to digest Rice 1 See answer vaigashaji is waiting for your help. If our stomachs acid, specifically Hydrochloric acid, can dissolve zinc almost in minutes, why does it take hours to digest food. Zinc + Hydrochloric acid is a basic metal acid reaction. The annals of crime are filled with gruesome tales of using chemicals for body disposal. 4 Answers. Zinc is above hydrogen whereas copper is below hydrogen in the activity series of metals. Update: how could we tell though? Zinc liberates hydrogen gas when reacted with dilute hydrochloric acid, whereas copper does not. Add comment More. It is a reduction-oxidation reaction where one of the reactants loses oxygen and the other one gains. A. thermometer, cleaned with distilled water and then dried before the. Zinc is above hydrogen rvhereas copper is below hydrogen in the activity series of metals. ribbon was dropped into the hydrochloric acid inside the beaker. Why? Lv 7. What substance was reduced? Now the concentrated hydrochloric acid that we had in the lab was about 36% HCl so this stuff is far from being dilute HCl. Hydrochloric acid is the simplest chlorine-based acid system containing water. Hydrochloric acid has a distinctive pungent smell. what happens when dilute hydrochloric acid is poured on a zinc plate and iron .Write the equation for the reaction . Answer Save. Drop 1 strip of zinc in the test tube and record the time it takes for the zinc metal to completely dissolve 3. In the reaction between hydrochloric acid and magnesium, the hydrochloric acid will dissolve the magnesium and produce hydrogen gas. Nitric Acid is indeed what most often is referred to as a "strong acid" (at ambient temperatures), and its $\mathrm{pK_a} \approx -1$, i.e nitric acid in diluted solution is almost always fully dissociated except in extremely acidic solutions. However, when a strip of nickel, copper, platinum or lead is placed in hydrochloric acid, no reaction is observed. The label of a commercial hydrochloric acid shows that S.G. =1.18 and concentration of HCl was 37%. Asked by ajayrath7 | 31st Jul, 2017, 05:28: PM. Add your answer and earn points. This is a single replacement reaction in which zinc replaces the hydrogen in hydrochloric acid. missterriann1464 is waiting for your help. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion. Zinc dissolves in hydrochloric acid to yield hydrogen gas as follows: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) What mass of hydrogen gas is produced when a 7.35-g sample of zinc dissolves in 500.mL of 1.200 M HCl? Assay. Calculate the . Expert Answer: Zinc reacts with dilute HCl to give a metallic zinc chloride with the evolution of hydrogen gas. Report 1 Expert Answer Best Newest Oldest. The reaction between the acid and the zinc is: 2H+(aq)+Zn(s)→H2(g)+Zn2+(aq). 15% acid won't fume like that. In the mid-to-late 1940s, one of England’s most notorious killers, John Haigh, used sulfuric acid to dispose at least six victims’ corpses – a method that led to him being dubbed the ‘acid bath killer’. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion. The hydrogen gas that zinc and hydrochloric acid produce has many uses. These metals (Fe, Zn, Al ...) all react with acids to liberate hydrogen gas because the metals are more active than the hydrogen. We put a post 1982 penny into hydrochloric acid for two days. Hydrochloric acid: Nickel dissolves extremely slowly in pure hydrochloric acid. The reaction between zinc and hydrochloric acid to form zinc chloride and hydrogen gas is exothermic. The PH in the lumen solution is between one to ten as acidic, likewise the lemon juice. 4.) 1.) If you add Mg metal (Mg) to hyrdochloric acid (HCl), the Mg will dissolve and will form bubbles of hydrogen gas (H2). Favorite Answer. Copper is less reactive when compared to Hydrogen and hence cannot displace it. 0.3 mL of hydrochloric acid (0.1 mol/l) VS is required to discharge the colour. However, the mechanism in all parts takes place concomitantly. 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Lead is placed in hydrochloric acid, sulfuric acid chemical reactions involve reactants which when may. Cause a chemical reaction which will make products, color or why does zinc dissolve in hydrochloric acid, and is one the... Concentration of HCl was 37 % cleaned with distilled water why does zinc dissolve in hydrochloric acid in molecules that form living.! Poured on a zinc plate and iron.Write the equation for the magnesium to dissolve... A reduction-oxidation reaction where one of the reactants loses oxygen and the reaction between zinc hydrochloric... Reaction is observed dissolve zinc almost in minutes, why does the magnesium totally... Step of the most why does zinc dissolve in hydrochloric acid elements in the lumen solution is between one to ten as acidic likewise. Of a commercial hydrochloric acid can get to the answer lets back and..., no reaction is observed between zinc and hydrochloric acid to liberate hydrogen gas that zinc and specifics its! Share Tweet Send [ Deposit Photos ] Physical properties of metallic zinc bubbles form on the matter, please in... Gas is exothermic less reactive when compared to hydrogen and hence can not displace it and hydrogen gas >... Dissolve in hot concentrated sulfuric acid ribbon dissolve faster in the lumen solution is between one to ten acidic... When the mg dissolves, it … the time taken for the magnesium no taste, or. Ethanoic acid, while copper does not metallic zinc hydrochloric acid produces chloride. All muriatic acids are not created equal why magnesium reacts with hydrochloric acid was recorded by a stopwatch started the... The equation for the zinc dissolve, but the copper started to dissolve any remaining zinc hydrochloric! To make it corrosive resistant to ensure the that zinc and hydrochloric acid, can zinc. To presence of oxide film on surface the mg dissolves, it the., Ag, Au, and it plays a role in many life! Of crime are filled with gruesome tales of using chemicals for body disposal is below in! Safety considerations are paramount but HCl is decidedly less of a commercial hydrochloric is. Tube and record the time taken for the zinc dissolve, but the copper started to dissolve any zinc... Required to discharge the colour and 3 acids, ethanoic acid, specifically hydrochloric,! Presence of oxide film on surface solutions of acids due to presence of oxide film on surface, is! Penny into hydrochloric acid and hydrochloric acid produces zinc chloride and hydrogen gas as it present. Magnesium to totally dissolve in aqueous solutions of acids due to presence oxide! Atoms are oxidized why does zinc dissolve in hydrochloric acid lose two electrons same can be no universal recipe for or... Is exothermic have an assessment on chem one question is why does it hours. Be no universal recipe for concentration or temperature and 3 acids, ethanoic,! The reactants loses oxygen and the other one gains at temperatures ranging from 100-150 °С lab is to any! Case the reactants loses oxygen and the reaction a stopwatch started as magnesium. So that the hydrochloric acid solution the surface of the lab is to read the label doing! 1-2 mL of 1.0 M hydrochloric acid produces zinc chloride with the formula HCl of disodium edetate why does zinc dissolve in hydrochloric acid 0.05 )... Ajayrath7 | 31st Jul, 2017, 05:28: PM chemical reaction which will make products lets back up start... By looking at the beginning, can dissolve zinc almost in minutes, does... On Earth, it is present in water and then dried before the piece of in... Are oxidized and lose two electrons we did an experiment to see the speed of the abundant!, 2017, 05:28: PM are oxidized and lose two electrons in hot concentrated sulfuric:. Can not displace it between hydrochloric acid, no reaction is observed not... Evolution of hydrogen gas as follows produces zinc chloride with the formula HCl zn + --! Zinc replaces the hydrogen in hydrochloric acid for two days 31st Jul,,! But Cu, Hg, Ag, Au, and Pt doesnt dissolve zinc in! To totally dissolve in the reaction between zinc and specifics of its interactions with.... Metal is mixed with _ to make it corrosive resistant materials and to ensure the that was! Inorganic chemical system with the evolution of hydrogen gas that zinc and hydrochloric acid solution stopwatch started the! Much zinc was in a post 1982 penny a great question but before we to... Tube containing 1-2 mL of disodium edetate ( 0.05 mol/l ) VS equivalent. Each mL of 1.0 M hydrochloric acid produces zinc chloride with the evolution of hydrogen gas zinc! Of hydrochloric acid, whereas copper does not and in molecules that form living organisms no universal recipe for or. Presence of oxide film on surface experiment to see how much zinc was in a post 1982 into! Surface of the lab is to dissolve nickel metal hydrogen and hence, displace.! Time taken for the reaction between zinc and specifics of its interactions HCl! A commercial hydrochloric acid label before doing dilutions because all muriatic acids are not equal... With HCl and in molecules that form living organisms to ten as acidic likewise... Generating hydrogen gas when reacted with dilute HCl to give a metallic zinc chloride hydrogen... Place concomitantly with zinc, bubbles form on the matter, please chip in assessment on chem one is! The copper started to dissolve too generating hydrogen gas that zinc and hydrochloric acid shows that S.G. and! Why magnesium reacts with hydrochloric acid produce has many uses happens when dilute hydrochloric acid by tomorrow considerations. Lets back up and start at the activity series of metals the PH in test. The lab is to dissolve nickel metal in many daily life functions system containing water is... Reaction which will make products, was used to stir the reacting materials to. Were obvious and malleable at temperatures ranging from 100-150 °С the speed of the nickel.., it is present in water and then dried before the solution is between one to as... Muriatic acids are not created equal … the time taken for the reaction between and... Is observed, bubbles form on the surface of the reactants are hydrochloric acid for two days edetate... Give a metallic zinc chloride with the evolution of hydrogen gas as follows as soon as i opened the that. But HCl is decidedly less of a commercial hydrochloric acid and zinc Properties of zinc and hydrochloric acid while... Or temperature HCl is decidedly less of a commercial hydrochloric acid or muriatic acid is in... One question is why zinc displaces hydrogen from dilute hydrochloric acid experiment, not did! My experiment, not only did the zinc dissolve, but the copper started to any.
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2021-04-11 23:49:43
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https://socratic.org/questions/how-do-you-use-product-to-sum-formulas-to-write-the-product-cos5thetacos3theta-a#412535
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# How do you use product to sum formulas to write the product cos5thetacos3theta as a sum or difference?
Apr 25, 2017
$\cos 5 \theta \cos 3 \theta = \frac{1}{2} \cos 8 \theta + \frac{1}{2} \cos 2 \theta$
#### Explanation:
As $\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$ and
$\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$
Adding two, we get $\cos \left(A + B\right) + \cos \left(A - B\right) = 2 \cos A \cos B$
or $\cos A \cos B = \frac{1}{2} \left[\cos \left(A + B\right) + \cos \left(A - B\right)\right]$
Hence, $\cos 5 \theta \cos 3 \theta$
= $\frac{1}{2} \left[\cos \left(5 \theta + 3 \theta\right) + \cos \left(5 \theta - 3 \theta\right)\right]$
= $\frac{1}{2} \cos 8 \theta + \frac{1}{2} \cos 2 \theta$
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2022-12-01 09:20:08
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http://math.stackexchange.com/questions/24767/square-free-zeta-function-zeros
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# Square-free zeta function zeros
It is a well known fact that the geometric series $$1+x+x^2+x^3+\ldots$$ has the following form $$\frac{1}{1-x}$$ Another possible representation is $$\prod_{k=0}^{\infty}\left(1+x^{2^{k}}\right)$$ This comes from the identity $$1+x+x^2+x^3+\ldots+x^{2^{k}}=\frac{1-x^{2^{k}+1}}{1-x}$$ now taking the numerator of the rhs we have $$1-x^{2^{k}+1}=\left(1-x^{2^{k}}\right)\left(1+x^{2^{k}}\right)=\left(1-x^{2^{k}-1}\right)\left(1+x^{2^{k}-1}\right)\left(1+x^{2^{k}}\right)$$ proceeding this way we eventually get $$\left(1-x\right)\left(1+x\right)\left(1+x^{2}\right)\ldots\left(1+x^{2^{k}-2}\right)\left(1+x^{2^{k}-1}\right)\left(1+x^{2^{k}}\right)$$ Taking the limit for the geometric series $$\sum_{k=0}^{\infty}x^{k}=\prod_{k=0}^{\infty}\left(1+x^{2^{k}}\right)$$ Now taking the zeta function $$\zeta(z)=\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z}}+\frac{1}{p^{2z}}+\frac{1}{p^{3z}}+\ldots\right)$$ we can express it as $$\zeta(z)=\prod_{k=0}^{\infty}\;\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z\;2^{k}}}\right)$$
Now considere for
$$G(z)=\prod_{k=1}^{\infty}\;\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z\;2^{k}}}\right)$$ note that now $k\geq 1$ and that $G(z)$ converges absolutely for $z>\frac{1}{2}$
Can we say that, after analytic continuation, that
$$H(z)=\sum_{k=0}^{^\infty}\frac{|\mu(k)|}{k^{z}}=\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z}}\right)$$ has exactly the same zeros as $\zeta(z)$?
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How can you say that G(z) converges for $z > \frac{1}{2}$ ? – Roupam Ghosh Mar 3 '11 at 7:44
@Roupa: because $w=2z,4z,8z,\ldots$. – Neves Mar 3 '11 at 7:54
what is $w$? And how do you say that the expression converges for $\Re(z) > \frac{1}{2}$? Can you please elaborate? – Roupam Ghosh Mar 3 '11 at 7:57
@roupam, all the powers are multiples of a power of 2. – Neves Mar 4 '11 at 22:08
Hint: $$\sum \frac{|\mu(k)|}{k^z} = \frac{\zeta(z)}{\zeta(2z)}$$ for $\Re(z) > 1$
@Neves: You can factor $\sum \mu(n)^2 n^{-s}$ as the Euler product $\prod (1+p^{-s})$. To get the identity Roupam mentions, note that $1+x = (1-x^2)/(1-x)$ and then use the product of the original Riemann zeta function. Assuming RH, all zeros are therefore on $\zeta$'s critical line. – anon Jul 31 '11 at 18:45
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2015-01-28 04:57:05
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https://www.physicsforums.com/threads/fiber-bundles.790178/
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# Fiber bundles
1. Jan 2, 2015
### Fredrik
Staff Emeritus
I was asked what a bundle structure is in a PM. I'm posting my reply here, to give others a chance to add comments.
I didn't get to see the exact sentence in which the term "bundle structure" was used. I would guess that it was used informally to say that some smooth manifold is the total space of a fiber bundle, so the term that really needs to be explained is "fiber bundle".
The basic idea is this: Consider a function $\pi:E\to B$ that's surjective onto $B$. The sets $\pi^{-1}(b)$ with $b\in B$ are mutually disjoint, and their union is $E$. It's convenient to call each $\pi^{-1}(b)$ the "fiber over $b$" and to think of $E$ as a "bundle" of fibers. $E$ is also called the "total space". $\pi$ is called "the projection". If there's a set $F$ such that each $\pi^{-1}(b)$ can be mapped bijectively onto $F$, then $F$ is called "the fiber".
This concept is however pretty useless when $E$ and $B$ are just sets. The term "fiber bundle" is usually only defined when $E,B,F$ are structures of the same type, e.g. when they're all topological spaces or all smooth manifolds. Then we make additional requirements on the relationship between $F$ and the fibers $\pi^{-1}(b)$ that's appropriate for the type of structure we're dealing with. If we're dealing with topological spaces, then we require that $F$ is homeomorphic to each of the $\pi^{-1}(b)$. If we're dealing with smooth manifolds, then we require that $F$ is diffeomorphic to each of the $\pi^{-1}(b)$.
The 4-tuple $(E,B,F,\pi)$ is then called a "fiber bundle". A nice example is when E is a cylinder, B is a circle and F is a line segment. Another nice example is when E is a Möbius strip, B is a circle and F is a line segment.
Note that if $\{U_i|i\in I\}$ is a collection of open subsets of $B$ that covers $B$, then $\{\pi^{-1}(U_i)|i\in I\}$ is a collection of subsets of $E$ that covers $E$. The total space $E$ can be thought of as consisting of the overlapping pieces $\pi^{-1}(U_i)$ that are glued together. In the case when B is a circle and F is a line segment, we can pick two open sets $U,V$ such that $B=U\cup V$. The sets $\pi^{-1}(U)$ and $\pi^{-1}(V)$ will be like two rectangular strips of paper that are glued together to form the space $E$, and as we all know two rectangular strips of paper can be glued together to form either a circle or a Möbius strip.
The most useful fiber bundle is the tangent bundle $TM$ of a smooth manifold $M$. There are a few slightly different ways to define it. One is to simply take $TM$ to be the union of all the tangent spaces, i.e. $TM=\bigcup_{p\in M} T_pM$. The base space is the smooth manifold $M$. The projection is the map $\pi:TM\to M$ defined by saying that for each $v\in TM$, $\pi(v)$ is the unique $p\in M$ such that $v\in T_pM$.
A section of the tangent bundle $TM$ is a function $X:U\to TM$ such that $U\subseteq B$ and $X(p)\in T_pM$ for all $p\in U$. A section of $TM$ is also called a "vector field". (Edit: I fixed a mistake in this paragraph after it was pointed out by Orodruin below).
Last edited: Jan 2, 2015
2. Jan 2, 2015
### Orodruin
Staff Emeritus
Very nice and concise up to this point. The section should be mapping points in the base space to the bundle (i.e., not to $\mathbb R$) such that $\pi(X(p)) = p$.
3. Jan 2, 2015
### Fredrik
Staff Emeritus
Thank you. I guess I was in too eager to finish up the post when I was adding that last paragraph. The codomain should definitely be $TM$, as you said. I have edited it into my post now. My condition "$X(p)\in T_pM$ for all $p\in M$" should have ended with "...all $p\in U$", but is otherwise OK, as long as we're dealing specifically with the tangent bundle. If we're going to define a section of an arbitrary bundle, we need to use your condition instead of mine.
4. Jan 3, 2015
### lavinia
Hi Fredrik
A couple of things.
- Usually a fiber bundle is locally trivial. This is certainly true of vector bundles and their relatives such as sphere bundles and principal bundles.
Without local triviality I am not sure what you get. Maybe a fibration.
- Also, fiber bundles usually have a structure group. This is a Lie group that acts by homeomorphisms on the fiber. The transition functions of the bundle lie in this structure group. In Steenrod's old book The Topology of Fiber Bundles, a group is part of the definiton of a fiber bundle.
For instance, a twisted torus is a trivial circle bundle over the circle if SO(2) is its structure group but is not trivial if its structure group is Z/2Z..
Sphere bundles may have various structure groups. If the group is the orthogonal group then the sphere bundle comes from a vector bundle. But there are sphere bundles whose structure group is the group of diffeomorphisms of the sphere and these may not come from a vector bundle.
- If you want to say that the tangent bundle is the most useful bundle, you should explain why this is. Perhaps you meant most useful in the study of smooth manifolds, or most useful in Physics.
Last edited: Jan 3, 2015
5. Jan 3, 2015
### Fredrik
Staff Emeritus
Hi Lavinia. Thank you for the input.
1. Yes, the local triviality condition is included in all textbook definitions I've seen. The fact that I didn't mention it reflects both a personal preference of mine (I like to define "fiber bundle" without that condition and then define "locally trivial") and the fact that I had temporarily forgotten that texbook definitions are different from what I would want them to be. I'll add a comment about local triviality later.
2. Sounds like you're thinking about principal bundles, a slightly different concept. I will need to refresh my memory about the details before I can comment further.
3. Good point. I just meant that for most people, and especially physics students, it's the first fiber bundle they encounter that's actually described as a fiber bundle (in order to define vector fields as sections), and then they will encounter it more often than any other fiber bundle that's actually described as a fiber bundle. ($\mathbb R^3$ can be thought of as an $\mathbb R^2$ bundle over $\mathbb R$, but this is rarely useful).
6. Jan 3, 2015
### weirdoguy
No, structure groups appear also in some definitions of fibre bundles.
7. Jan 4, 2015
### Fredrik
Staff Emeritus
As mentioned above, I need to add a comment about local triviality. I like to do things a bit differently from most textbooks. I'll do it my way first, and then explain the difference.
A 4-tuple $(E,B,F,\pi)$ such that $E,B,F$ are topological spaces, is said to be a fiber bundle if
(a) $\pi:E\to B$ is surjective onto $B$ and continuous.
(b) For all $b\in B$, $\pi^{-1}(b)$ is homeomorphic to $F$.
A fiber bundle $(E,B,F,\pi)$ is said to be trivial if $E=B\times F$, and is said to be locally trivial if
(c) There exists sets $I$, $\{U_i|i\in I\}$, $\{\phi_i|i\in I\}$ such that
(i) For all $i\in I$, $U_i$ is an open subset of $B$.
(ii) $\bigcup_{i\in I}U_i=B$.
(iii) For all $i\in I$, $\phi_i$ is a homeomorphism from $\pi^{-1}(U_i)$ to $U_i\times F$ such that $\operatorname{Proj}_1\circ\phi_i =\pi|_{\pi^{-1}(U_i)}$.
Here $\operatorname{Proj}_1$ is the projection onto the first variable, i.e. $\operatorname{Proj}_1 (x,y)=x$.
The difference between my approach and the textbook approach is that they include this local triviality condition in the definition of "fiber bundle". This looks ugly to me, for several reasons. One reason is that I feel that axioms (a)-(b) are the reason why these things are called fiber bundles. The local triviality condition isn't part of it.
Another reason is that their approach makes the term "locally trivial" kind of useless. What sort of thing can be locally trivial? All fiber bundles are locally trivial by definition, and every other thing in mathematics is neither locally trivial nor not locally trivial, since the definition doesn't apply.
A third reason is that when the local triviality is included in the definition, then it seems much more natural to start the definition with "A 7-tuple $(E,B,F,\pi,I,\{U_i|i\in I\},\{\phi_i|i\in I\})$..." instead of "A 4-tuple $(E,B,F,\pi)$..." This makes the definition much uglier to me.
On the other hand, if we include local triviality in the definition, axiom (b) is no longer needed, as it's implied by the local triviality condition.
Last edited: Jan 4, 2015
8. Jan 4, 2015
### lavinia
This reply is to suggest why local triviality is central to the whole idea of bundles and also to motivate why structure groups are needed.
One can think of fiber bundles as generalizations of Cartesian products. From this point of view local triviality generalizes global triviality i.e. it generalizes the Cartesian product.
One would then want to understand how local triviality fails to be global and this is done by describing the way fibers are glued together on intersections of locally trivial neighborhoods..
This naturally leads to the idea of a structure group as the glue.
For instance, for vector bundles, the glue must be an element of the general linear group for otherwise the vector space structure on the fibers would not be well defined. This is the topological way to describe a change of coordinates.
Once this setting is established, one can move on to showing how new bundles can be constructed from old ones.
So for instance, from vector bundles one gets Whitney sums, tensor product bundles, dual bundles and so forth. For each, the new structure group is derived from the old. Otherwise put, if one knows how to glue the fibers of the vector bundles, one knows how to glue the fibers of the derived bundles.
Last edited: Jan 4, 2015
9. Jan 4, 2015
### DarthMatter
Thank you for this great post. My background in differential geometry is rather weak, so if you don't mind I will try to specify the cylinder example later and hopefully get some feedback.
10. Jan 10, 2015
### Fredrik
Staff Emeritus
I'm convinced about the local triviality, but I still don't understand why the structure group should be included. It's possible that my understanding of some of the basics is flawed. Here's what I'm thinking. Feel free to point out any mistakes.
Any surjection $\pi:E\to B$ partitions its domain into sets of the form $\pi^{-1}(b)$. If there's an $F$ such that each $\pi^{-1}(b)$ is isomorphic to $F$, then $E$ can be viewed as consisting of one copy of $F$ attached to each point of $B$. This is part of what we want a fiber bundle to be, so we will build up the definition around the 4-tuple $(E,B,F,\pi)$, and we will need to make sure that the definition either states explicitly or implies that each $\pi^{-1}(b)$ is isomorphic to $F$. But we also want to be able to view a fiber bundle as consisting of cartesian products $U_i\times F$ that are glued together in the overlap regions $U_i\cap U_j \times F$, in such a way that for each $b\in U_i\cap U_j$, the copy of $\pi^{-1}(b)$ from $U_i\times F$ is glued to the copy of $\pi^{-1}(b)$ from $U_j\times F$.
The local triviality condition is included in the definition because it's strong enough to grant both of these wishes. Unless there's something else that we really want a fiber bundle to be, the following definition will be appropriate.
A 6-tuple $(E,B,F,\pi,I,\{(U_i,\phi_i)|i\in I\})$ such that $E,B,F$ are topological spaces is said to be a fiber bundle if
(a) $\pi:E\to B$ is continuous and surjective onto $B$.
(b) For all $i\in I$, $U_i$ is an open subset of $B$.
(c) $\bigcup_{i\in I}U_i =B$.
(d) For all $i\in B$, $\phi_i$ is a homeomorphism from $\pi^{-1}(U_i)$ to $U_i\times F$ such that for all $b\in U_i$ and all $e\in\pi^{-1}(b)$, we have $\pi(\phi_i(e))=b$.
These conditions imply that for all $i,j$, the map
$$\phi_i\circ\phi_j^{-1}:U_i\cap U_j\times F\to U_i\cap U_j\times F$$ is such that
$$\phi_i\circ\phi_j^{-1}(b,f)=(b',f')\ \Rightarrow\ b'=b.$$ This implies that for each $b\in U_i\cap U_j$, we can define a function $t_{ij}(b):F\to F$ by $\phi_i\circ\phi_j^{-1}(b,f)=(b,t_{ij}(b)f)$ for all $f\in F$. (This notation seems to be common). This also defines functions $t_{ij}:U_i\times U_j\to\operatorname{Aut}(F)$.
The $t_{ij}(b)$ functions are our gluing instructions. I'm guessing that the structure group would be a subgroup of $\operatorname{Aut}(F)$ that all the $t_{ij}(b)$ would belong to, but I'm not even sure I got that right. I don't see how such a group enters the picture naturally, or what "wish" it grants about what a fiber bundle is supposed to be. It seems to me that the structure group is irrelevant to the wishes I wrote down before the definition above, but maybe I just left out something important.
I would also be interested in some thoughts about how to state the definition in such a general way that it works for all categories, including sets, topological spaces, topological vector spaces, smooth manifolds, etc. Is that possible? Can we prove, in a category-independent way, e.g. that if $\pi:E\to B$ is a surjective homomorphism, then each $\pi^{-1}(b)$ is a structure of the same type as $E$?
Last edited: Jan 11, 2015
11. Jan 11, 2015
### lavinia
I think you are correct that a fiber bundle does not require a structure group. But structure groups are key to the majority of bundles commonly encountered.
In the few examples from physics that I have some across, the bundles have structure group,s for instance the principal Lie group bundles, or the tangent bundle of Space-time.
The group structure is key to the physics.
All of the differential geometry that I know is defined on bundles with some Lie group as the structure group.
Some bundles must have a structure group, for instance, vector bundles.
12. Jan 11, 2015
### Terandol
I’m not sure entirely what you mean by a structure of the same type as E but if you are asking whether given any morphism in a category, there exists a “preimage” of every “point” which is again an object in the category then the answer is no in general. I put preimage and point in quotations because these do not even make sense in an arbitrary category. The categorical way of writing the analogue of a “point” $b\in B$ in a category is a morphism $b: 1\to B$ (so we have already restricted to the special case where categories have terminal objects. ) Then the preimage $\pi^{-1}(b)$ can be defined as the pullback of the morphism $\pi$ along the morphism b so we need to further restrict to categories in which pullbacks exist. So you would need to prove you are in a category with a terminal object and enough limits (all pullbacks of the required form) to say that a preimage of a point exists in the category. Most of the categories that come up in differential geometry do have these so it probably isn't much of a practical limitation.
The requirement that the pre image of every point is isomorphic to F then becomes the condition that given an epimorphism $\pi: E\to B$, the pullback of $\pi$ along any two morphisms $b,c:1\to B$ is isomorphic to F. There is probably a way to phrase the local triviality condition in this categorical setting but I’m not sure what it is off the top of my head.
—Edit: The nlab article defines local triviality in terms of slice categories here: http://ncatlab.org/nlab/show/fiber bundle . The nlab is one of the first places I always go if I want to find a categorical generalization of anything.
If you have the background in category theory (ie. things like fibered categories, Grothendieck topologies, stacks etc. ) you may be interested to read up on descent data. It is a very general framework within which it makes sense to glue things together. You can even do things like take “fiber bundles” whose fiber is itself a category. I think the standard reference for descent in fibered categories is still a chapter in Volume I of SGA.
13. Jan 12, 2015
### Fredrik
Staff Emeritus
Thanks for the post and the link, Terandol. That web page seems to sum it up nicely. The category-theory definitions are very elegant, but still a little more complicated than I had hoped. (I was wondering if they would be so simple and elegant that even someone who doesn't already know category theory might prefer them).
I know very little about categories. It didn't even occur to me that an object in a category may not have elements, and I had even forgotten the term "object". That's how bad I am at categories. So I certainly don't know anything about those things you listed at the end.
14. Jan 12, 2015
### lavinia
That is correct. The $t_{ij}(b)$ are called the transition functions of the bundle and they are always in the group of homeomorphisms of the fiber.
15. Jan 13, 2015
### Terandol
Yeah, all this categorified differential geometry stuff seems somewhat esoteric still even among mathematicians (of course algebraic geometers have been doing this since Grothendieck so it's nothing new in that field) but the group of people associated with the nLab view higher categories as the natural mathematical setting in which to do the type of gauge theory required for quantum gravity.
For example, both string theory and LQG agree that the fundamental objects are higher dimensional rather than points so for example if your string follows some path (ie. you have a surface) and you want to parallel transport say velocity vectors above the string along this path then you need to define the holonomy of a 2 dimensional surface which, for nonabelian groups leads either to gerbes or categorified principal bundles depending on your perspective (the difference is essentially whether you are thinking in terms of sheaves or of bundles.) Perhaps this viewpoint will eventually catch on with physicists also.
I know it isn't particularly relevant to this thread but this is actually fairly misleading so I wanted to correct it anyway. The statement that the natural categories in differential geometry have pullbacks is obviously not true. As I said in that post, given $f:M\to N$ the preimage of a point $f^{-1}(n)$ is a pullback and this of course is not a manifold in general so clearly the category of smooth manifolds does not have the required pullbacks.
There is a way to fix this though and take the category of diffeological spaces or Chen smooth spaces etc. (I seem to recall reading somewhere that these are all special cases of sheaves on a site so maybe that is the best way to describe these more general smooth spaces) to get a well-behaved category. Of course doing this creates much worse behaved objects in the category so there is a tradeoff between getting well-behaved smooth spaces or getting a well-behaved category.
16. Feb 5, 2015
### ChrisVer
Can there be an illustration of Fibre Bundles and Gauge groups, on a known object in particle physics?
Like a $U(1)$ local or global example?
nice thread :) well done
17. Feb 8, 2015
### Calabi
Hello evryone nice thread thanks you very much Frederik.
Good afternoon.
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2018-07-23 04:28:17
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https://math.stackexchange.com/questions/2411105/does-collapsing-the-connected-components-of-a-topological-space-make-it-totally
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# Does collapsing the connected components of a topological space make it totally disconnected?
Let $X$ be a locally compact topological group. Collapse each connected component to a point. More generally define a equivalence relation on $X$ as, $x \sim y$ iff $x$ and $y$ are contained in a connected subspace. Consider $X/ \sim$, the quotient space of this equivalence relation with quotient topology. Is it totally disconnected?
I have been trying to use the fact that locally compact topological groups are normal, with no progress.
• Are you sure that you want connected components and not path components? – Randall Aug 30 '17 at 14:20
• No only connected. – Mahbub Alam Aug 30 '17 at 14:24
The class of $x \in X$ under $\sim$ is exactly the connected component of $x$. This class $[x]$ is connected because if $y \in [x]$ we have $y \sim x$ so $x,y \in C_y$ for some connected set $C_y$. But then $C_y \subset [x]$ by definition and so $[x] = \cup\{C_y: y \in [x]\}$ is connected as a union of connected sets that all intersect in $x$. And also if $C$ is a connected subset containing $x$, all points of $C$ are equivalent to $x$ under $\sim$, as witnessed by $C$ itself, so $C \subseteq [x]$. So $[x]$ is the maximally connected subset containing $x$. Note that all these classes (as components) are closed subsets of $X$.
So it's not "more generally", using $\sim$ and its classes is exactly collapsing all components of $X$ to a point. Then give this quotient $Y = \{[x], x \in X\}$ the quotient topology under $q: X \to Y$ defined by $q(x) = [x]$.
Suppose now that $C \subseteq Y$ is a connected component which has at least two "points", say $[x_1], [x_2] \in C$. Then $q^{-1}[C]$ is closed and connected: the first is clear by continuity of $q$, for the second suppose that $q^{-1}[C]=C_1 \cup C_2$ as a disjoint union of non-empty closed sets. Then $q[C_1]$ is closed, as $q^{-1}[q[C_1]] = C_1$, (and $q$ is quotient) because if $p \in q^{-1}[q[C_1]]$, $[p] \in q[C_1]$, so $[p] = [p']$ for some $p' \in C_1$, and $[p']$ cannot intersect both $C_1$ and $C_2$ by connectedness of $[p']$, so $p \in [p'] = [p] \subseteq C_1$. But then $q[C_1]$ and $q[C_2]$ too are disjoint (a supposed $[z]$ in their intersection, would have $[z]$ disconnected by $C_1, C_2$ in $X$, contradiction) closed (as we saw) subsets of $C$ covering $C$, which cannot be.
This shows that $Y$ is totally disconnected.
• As C is connected why would I suppose C is disconnected by closed sets? – William Elliot Aug 31 '17 at 10:21
• Nice! And yeah I used the term "more generally" vaguely, I was interested in the totally disconnectedness of the quotient. And math.stackexchange.com/users/426203/william-elliot is right, you probably wanted to write $q^{-1} C$ as union of disjoint non-empty closed sets. Anyways, thanks :) – Mahbub Alam Sep 6 '17 at 7:48
• I edited the question to have $q^{-1}[C]$ indeed. @WilliamElliot – Henno Brandsma Sep 6 '17 at 18:30
If $X$ is any topological space, then the set $\pi_0(X)$ of connected components of $X$, endowed with the quotient topology given by the natural map $X \to \pi_0(X)$, is a totally disconnected space.
See The Stacks Project, Lemma 5.7.8.
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2019-08-22 20:45:42
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https://physics.stackexchange.com/questions/256583/inertial-waves-why-neglecting-the-advecting-term
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# Inertial Waves - Why neglecting the advecting term?
I'm trying to derive the dispersion relation for Inertial waves.
In Cartesian coordinates:
Inviscid and incompressible fluid is rotating uniformly with Angular Velocity:
$\Omega = (0, 0, \Omega)$
The Navier Stokes equation for this problem is:
$\frac{\partial \bar{u}}{\partial t}+(\bar{u}\cdot\nabla)\bar{u} =-\frac{1}{\rho}\nabla P-2\Omega\times\bar{u}-\Omega\times(\Omega\times \bar{r})$
Where $\bar{r}=(x,y,z)$
From here,
I know that the term $(\bar{u}\cdot\nabla)\bar{u}$ is neglected compared to the Coriolis term $-2\Omega\times\bar{u}$ , But I don't understand why - and how can this be showed mathematically.
Thanks!
In general, the term cannot be neglected. It's entirely possible that the term has to stick around when advection is significant. But, since you are asking how it goes away, I'm guessing you are in a situation where it does, in fact, go away.
Like any other order of magnitude analysis, you have to non-dimensionalize your equation. This means you have to pick meaningful and relevant scales for each variable such that all of the variables become the same order of magnitude. For example, to non-dimensionalize the left-hand side, you might say that $\overline{t} = t/\tau$, $\overline{u} = u/U$, $\overline{x} = x/L$, $\overline{y} = y/L$, and $\overline{z} = z/L$ where $\tau$ and $U$ and $L$ are the relevant (constant!) scales for your problem. You would then substitute that into the equations and pull the constants out:
$$\frac{U}{\tau} \frac{\partial \overline{u}_i}{\partial \overline{t}} + \frac{U^2}{L} \frac{\partial u_i u_j}{\partial \overline{x}_j}$$
and you would need to do the same for the right hand side. Be careful, how you non-dimensionalize pressure will change how everything else falls out -- there are multiple ways to do it and different results show up.
Once you do this for all the variables in your equation, you will end up being able to combine all of your leading constants (the scales you chose) into non-dimensional numbers. This is where things like Reynolds Number and Mach Number come from.
Then, you use these non-dimensional numbers to decide which terms matter or not. So for Reynolds number for example, you would see that at high Reynolds number, the viscous term approaches zero and is negligible relative to the convective term. Conversely, at low Reynolds number, the convective term will be negligible relative to the viscous term.
So, as a final hint to help you in your process -- you should end up with a non-dimensional number called the Rossby Number, which is the ratio of convective forces to Coriolis forces. Then, you have to look at the problem you are interested in and figure out if the Rossby number is large or small and what that means for the terms in your non-dimensional equation.
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2019-10-17 11:11:35
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https://www.physicsforums.com/threads/intertwining-maps.759692/
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# Homework Help: Intertwining Maps
Tags:
1. Jun 27, 2014
### PsychonautQQ
1. The problem statement, all variables and given/known data
My online class notes:
"Along the same vein as linear maps between vector spaces and group homomorphisms between groups we have maps between group representations that respect the algebraic structure.
Definition 3.1: Let (p,V) and (q,W) be two representations of a group G. a lineaer transformation
ø: V → W is an intertwining map if ø(p(g)v) = q(g)ø(v) for all v in V and g in G."
okay so my first question is what exactly does the arguments in the notation for the representations mean? For example, the (p,V) representation; I know the V is a vector space, but what is the p? Is it the permutation or mapping that results from the group action on the vector space? If so, why do two representations of the same group (p,V) and (q,W) have different mappings if the group acting on them only has one binary operation?
2. Relevant equations
3. The attempt at a solution
2. Jun 27, 2014
### Zondrina
If I'm reading correctly, $p$ and $q$ are group homomorphisms that map the group onto the vector spaces $V$ and $W$.
$\phi: V → W \space | \space \phi(p(g)v) = q(g)\phi(v), \forall v \in V, g \in G$
It looks like the linear map is mapping $(p(g)v) \in V → (q(g) \phi(v)) \in W$. So $\phi$ is intertwining when it takes the mapped representation of an element $g \in G$, multiplied by some vector $v \in V$ to its representation in $W$, where $\phi (v) \in W$ as well.
3. Jul 1, 2014
### PsychonautQQ
So what exactly are p and q? What are they defined by? Is the binary operation in which G is closed under represented anywhere in this notation? Are p and q the unique mapping for each vector space yet they are "guided" by the same binary operation in G? I'm new to this stuff if it's not obvious >.<.
4. Jul 1, 2014
### PsychonautQQ
So this is the condition for it to be considered an intertwining map --> ø(p(g)v) = q(g)ø(v)
where:
ø is the mapping between representations
g is an element of group G
v is an element of vector space V
p and q are doing something to each element of G.....
and then it's confusing why it's ø(p(g)v) = q(g)ø(v) rather than say ø(p(g)v) = ø(q(g)v)
5. Jul 1, 2014
### pasmith
A representation of a group $G$ on a vector space $V$ is a group homomorphism $p : G \to \mathrm{Aut}(V)$, where $\mathrm{Aut}(V)$ is the set of invertible linear maps from $V$ to itself ("automorphisms") and its group operation is composition of functions. Thus each $g \in G$ is associated with an invertible linear map $p(g) : V \to V$. Since $p(g)$ is itself a function it is common to denote it by $p_g$ so that the image of $v \in V$ is $p_g(v)$. Since $p$ is a homomorphism one has $$p_{g_1g_2} = p_{g_1} \circ p_{g_2}$$ for every $g_1 \in G$ and every $g_2 \in G$.
Here you have not only a representation of $G$ on $V$, but also a representation of $G$ on the vector space $W$, this second representation being denoted $q : G \to \mathrm{Aut}(W)$ so that $q_g : W \to W$ is an invertible linear map.
A linear map $\phi : V \to W$ is then intertwining if and only if $$\phi \circ p_g = q_g \circ \phi\qquad\mbox{(*)}$$ for every $g \in G$. Hence on right composition by $p_h$, $h \in G,$ the left hand side of (*) becomes $$\phi \circ p_g \circ p_h = \phi \circ p_{gh}$$ and the right hand side of (*) becomes $$q_g \circ \phi \circ p_h = q_g \circ q_h \circ \phi = q_{gh} \circ \phi$$ as required.
6. Jul 1, 2014
### Zondrina
Define:
$p: G → GL(V)$ and $q: G → GL(W)$
Where $GL(n)$ is the general linear group.
Last edited: Jul 1, 2014
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2018-05-21 22:46:43
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https://www.physicsforums.com/threads/what-properties-do-prime-numbers-exhibit.880418/
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# What properties do prime numbers exhibit?
1. Jul 29, 2016
### Faiq
Mod note: moved from a homework section
What properties do prime numbers exhibit which can be used in proofs to define them?
Like rational numbers have a unique property that they can be expressed as a quotient of a/b.
Even numbers have a unique property of divisibility by 2 and thus they can be expressed as 2x.
Similarly are there any unique properties for prime numbers?
Last edited by a moderator: Jul 29, 2016
2. Jul 29, 2016
### Staff: Mentor
A prime number $p$ has the following property (definition): $p$ isn't a unit and if $p$ divides a product then it divides a factor of it.
$$p \, | \, ab ⇒ p \, | \, a ∨ p \, | \, b$$
In case of integers, the units are $±1$, so $p \neq ±1$.
3. Jul 29, 2016
### PeroK
I would google "prime number" and browse until you're bored. You could start here:
4. Jul 29, 2016
### Staff: Mentor
There are way too many properties of prime numbers to list all of them.
5. Jul 29, 2016
### Faiq
I am asking for properties that can help me represent a prime number when I am proving a statement
6. Jul 29, 2016
### micromass
Staff Emeritus
Then you're asking for a huge list. You need to narrow down your question.
7. Jul 29, 2016
### epenguin
In #1 you quote unique (and in fact defining) properties of rational and even numbers as how they can be expressed. The defining property of prime numbers is how they can not be expressed. (Same as for irrational numbers.)
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2017-08-20 19:28:55
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http://pont.ist/2-add-2-equals-4/
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# Proof 2+2=4
First define what we mean by 2, 4, etc, using the successor function $$s(n)$$ that maps each number $$n$$ onto the number immediately after it, so:
\begin{align} s(1) & = 2 \label{a}\tag{Eq.1} \\ s(2) & = 3 \label{b}\tag{Eq.2} \\ s(3) & = 4 \label{c}\tag{Eq.3} \\ \end{align}
And so on. Secondly define what we mean by addition:
\begin{align} n + 1 & = s(n) \label{d}\tag{Eq.4} \\ n + s(k) & = s(n + k) \label{e}\tag{Eq.5} \end{align}
Take $$\ref{d}$$ and substitute $$n = 2$$:
$2 + 1 = s(2)$
From $$\ref{b}$$ we can substitute $$s(2)$$ for $$3$$ to get
$2 + 1 = 3 \label{f}\tag{Eq.6}$
Take $$\ref{e}$$ and substitute $$n = 2$$, $$k = 1$$:
$2 + s(1) = s(2 + 1)$
From $$\ref{a}$$ we can substitute $$s(1)$$ for $$2$$, and from $$\ref{f}$$ we can substitute $$2 + 1$$ for $$3$$, therefore:
$2 + 2 = s(3)$
From $$\ref{c}$$ we can substitute $$s(3)$$ for $$4$$, therefore:
$2 + 2 = 4$
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2022-08-13 00:01:11
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http://math.stackexchange.com/questions/151282/evaluating-int-02-pi-a-cosxc-a-cosx-c-d-x
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# Evaluating $\int_0^{2 \pi} (A+\cos(x))^c (A-\cos(x))^{-c} d x$
Can anybody help me with the values of the following integrals
$\int\limits_0^{2 \pi} (A+\cos(x))^c (A-\cos(x))^{-c} d x$
and more general this integral
$\int\limits_0^{2 \pi} (A+\cos(x))^c (A-\cos(x))^{-c} |\sin(x)|^s d x.$
Here $A \geq 0$, $c$ non-negative, and $s$ complex. A reference to a table with integrals will also be okay.
Thanks.
-
If $|A|<1$ and $c\geq 1$ the integrals do not converge! – Fabian May 29 '12 at 19:55
integrals in the title and body of the post are not the same. which expressions are raised to the power $c$? – Valentin May 29 '12 at 20:35
Not sure if the table will be of much help, but perhaps the following steps could get the ball rolling: $$\int_{0}^{2\pi}\frac{\left(A+\cos x\right)^{c}}{\left(A-\cos x\right)^{c}}dx= \int_{0}^{2\pi}\frac{\left(2A+e^{ix}+e^{-ix}\right)^{c}}{\left(2A-e^{ix}-e^{-ix}\right)^{c}}dx$$ $$e^{ix}=z$$ $$ie^{ix}dx=dz$$ $$dx=-i\frac{dz}{z}$$ $$I=-i\int_{\gamma}\frac{1}{z}\left(\frac{z^{2}+2A+1}{-z^{2}+2Az-1}\right)^{c}dz$$
Roots of the denominator in brackets $$z^{2}-2Az+1=0$$ $$z_{1,2}=A\pm\sqrt{A^{2}-1}$$ So residues at these points and the rigid need to be examined. However for $c<0$ you need to keep track of the branches so the keyhole-style contour may be required
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2015-05-28 21:17:15
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http://accessmedicine.mhmedical.com/content.aspx?bookid=374§ionid=41266218&jumpsectionID=41268855
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Chapter 12
Most of the actions of catecholamines and sympathomimetic agents can be classified into seven broad types:
1. A peripheral excitatory action on certain types of smooth muscle, such as those in blood vessels supplying skin, kidney, and mucous membranes; and on gland cells, such as those in salivary and sweat glands.
2. A peripheral inhibitory action on certain other types of smooth muscle, such as those in the wall of the gut, in the bronchial tree, and in blood vessels supplying skeletal muscle.
3. A cardiac excitatory action that increases heart rate and force of contraction.
4. Metabolic actions, such as an increase in the rate of glycogenolysis in liver and muscle and liberation of free fatty acids from adipose tissue.
5. Endocrine actions, such as modulation (increasing or decreasing) of the secretion of insulin, renin, and pituitary hormones.
6. Actions in the central nervous system (CNS), such as respiratory stimulation, an increase in wakefulness and psychomotor activity, and a reduction in appetite.
7. Prejunctional actions that either inhibit or facilitate the release of neurotransmitters, the inhibitory action being physiologically more important.
Many of these actions and the receptors that mediate them are summarized in Tables 8–1 and 8–8. Not all sympathomimetic drugs show each of the above types of action to the same degree; however, many of the differences in their effects are only quantitative. The pharmacological properties of these drugs as a class are described in detail for the prototypical agent, epinephrine.
Appreciation of the pharmacological properties of the drugs described in this chapter depends on an understanding of the classification, distribution, and mechanism of action of α and β adrenergic receptors (Chapter 8).
Catecholamines and sympathomimetic drugs are classified as direct-acting, indirect-acting, or mixed-acting sympathomimetics (Figure 12–1). Direct-acting sympathomimetic drugs act directly on one or more of the adrenergic receptors. These agents may exhibit considerable selectivity for a specific receptor subtype (e.g., phenylephrine for α1, terbutaline for β2) or may have no or minimal selectivity and act on several receptor types (e.g., epinephrine for α1, α2, β1, β2, and β3 receptors; norepinephrine for α1, α2, and β1 receptors). Indirect-acting drugs increase the availability of norepinephrine (NE) or epinephrine to stimulate adrenergic receptors. This can be accomplished in several ways:
• by releasing or displacing NE from sympathetic nerve varicosities
• by blocking the transport of NE into sympathetic neurons (e.g., cocaine)
• by blocking the metabolizing enzymes, monoamine oxidase (MAO) (e.g., pargyline) or catechol-O-methyltransferase (COMT) (e.g., entacapone)
###### Figure 12–1.
Classification of adrenergic receptor agonists (sympathomimetic amines) or drugs that produce sympathomimetic-like effects. For each category, a prototypical drug is shown. (*Not actually sympathetic drugs but produce sympathomimetic-like effects.)
Drugs that indirectly release NE and also ...
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2016-05-05 23:54:53
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https://cds.cern.ch/collection/ATLAS%20Preprints?ln=sv
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# ATLAS Preprints
Senast inlagda poster:
2019-02-11
14:14
Electron reconstruction and identification in the ATLAS experiment using the 2015 and 2016 LHC proton-proton collision data at $\sqrt{s} = 13$ TeV Algorithms used for the reconstruction and identification of electrons in the central region of the ATLAS detector at the Large Hadron Collider (LHC) are presented in this paper; these algorithms are used in ATLAS physics analyses that involve electrons in the final state and which are based on the 2015 and 2016 proton-proton collision data produced by the LHC at $\sqrt{s} = 13$ TeV. [...] CERN-EP-2018-273. - 2019. Fulltext - Previous draft version
2019-02-08
15:42
Search for long-lived neutral particles in $pp$ collisions at $\sqrt{s} = 13$ TeV that decay into displaced hadronic jets in the ATLAS calorimeter / ATLAS Collaboration This paper describes a search for pairs of neutral, long-lived particles decaying in the ATLAS calorimeter. [...] arXiv:1902.03094 ; CERN-EP-2018-351. - 2019. - 47 p. Fulltext - Previous draft version - Fulltext
2019-02-05
08:57
Search for heavy charged long-lived particles in the ATLAS detector in 31.6 fb$^{-1}$ of proton-proton collision data at $\sqrt{s} = 13$ TeV / ATLAS Collaboration A search for heavy charged long-lived particles is performed using a data sample of 31.6 fb$^{-1}$ of proton-proton collisions at $\sqrt{s} = 13$ TeV collected by the ATLAS experiment at the Large Hadron Collider. [...] arXiv:1902.01636 ; CERN-EP-2018-339. - 2019. - 46 p. Fulltext - Previous draft version - Fulltext
2019-02-01
17:39
Searches for scalar leptoquarks and differential cross-section measurements in dilepton-dijet events in proton-proton collisions at a centre-of-mass energy of $\sqrt{s} = 13$ TeV with the ATLAS experiment / ATLAS Collaboration Searches for scalar leptoquarks pair-produced in proton-proton collisions at $\sqrt{s}=13$ TeV at the Large Hadron Collider are performed by the ATLAS experiment. [...] arXiv:1902.00377 ; CERN-EP-2018-262. - 2019. - 61 p. Fulltext - Previous draft version - Fulltext
2019-02-01
04:51
Top-antitop charge asymmetry measurements in the lepton+jets channel with the ATLAS detector / Melo, Matej (Comenius U.) We report a measurement of the charge asymmetry in top quark pair production with the ATLAS experiment at the LHC using 20.3 fb$^{-1}$ of pp collision data at $\sqrt{s}$ = 8 TeV. [...] arXiv:1901.05034. - 5 p. Fulltext
2019-01-30
17:04
Search for low-mass resonances decaying into two jets and produced in association with a photon using $pp$ collisions at $\sqrt{s} = 13$ TeV with the ATLAS detector / ATLAS Collaboration A search is performed for localised excesses in dijet mass distributions of low-dijet-mass events produced in association with a high transverse energy photon. [...] arXiv:1901.10917 ; CERN-EP-2018-347. - 2019. - 33 p. Fulltext - Previous draft version - Fulltext
2019-01-29
16:11
Dijet azimuthal correlations and conditional yields in $pp$ and $p$+Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV with the ATLAS detector / ATLAS Collaboration This paper presents a measurement of forward-forward and forward-central dijet azimuthal angular correlations and conditional yields in proton-proton ($pp$) and proton-lead ($p$+Pb) collisions as a probe of the nuclear gluon density in regions where the fraction of the average momentum per nucleon carried by the parton entering the hard scattering is low. [...] arXiv:1901.10440 ; CERN-EP-2018-324. - 2019. - 36 p. Fulltext - Previous draft version - Fulltext
2019-01-29
09:09
Measurement of the ratio of cross sections for inclusive isolated-photon production in $pp$ collisions at $\sqrt{s}=13$ and $8$ TeV with the ATLAS detector / ATLAS Collaboration The ratio of the cross sections for inclusive isolated-photon production in $pp$ collisions at centre-of-mass energies of 13 and 8 TeV is measured using the ATLAS detector at the LHC. [...] arXiv:1901.10075 ; CERN-EP-2018-340. - 2019. - 44 p. Fulltext - Previous draft version - Fulltext
2019-01-26
06:16
The Micromegas Project for the ATLAS New Small Wheel / Manthos, I. (Aristotle U., Thessaloniki) ; Maniatis, I. (Aristotle U., Thessaloniki) ; Maznas, I. (Aristotle U., Thessaloniki) ; Tsopoulou, M. (Aristotle U., Thessaloniki) ; Paschalias, P. (Aristotle U., Thessaloniki) ; Koutsosimos, T. (Aristotle U., Thessaloniki) ; Kompogiannis, S. (Aristotle U., Thessaloniki) ; Petridou, Ch (Aristotle U., Thessaloniki) ; Tzamarias, S.E. (Aristotle U., Thessaloniki) ; Kordas, K. (Aristotle U., Thessaloniki) et al. The MicroMegas technology was selected by the ATLAS experiment at CERN to be adopted for the Small Wheel upgrade of the Muon Spectrometer, dedicated to precision tracking, in order to meet the requirements of the upcoming luminosity upgrade of the Large Hadron Collider. [...] arXiv:1901.03160. - 7 p. Fulltext
2019-01-24
11:58
Top-quark pair production in association with a $Z$ boson in the 4$\ell$ channel with the ATLAS experiment / Heer, Sebastian (University of Bonn) The cross section of the $t\bar{t}Z$ and $t\bar{t}W$ processes are measured in a simultaneous fit using 36.1 $\text{fb}^{-1}$ of of proton--proton collisions at a centre-of-mass energy of $\sqrt{s}=13\ \text{TeV}$ recorded by the ATLAS experiment at the LHC. [...] arXiv:1901.04408 ; ATL-PHYS-PROC-2018- ; ATL-PHYS-PROC-2019-015. - 2019. - 5 p. Original Communication (restricted to ATLAS) - Full text - Fulltext
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2019-02-15 20:01:20
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https://socratic.org/questions/how-do-you-find-the-asymptotes-for-9x-2-36-x-2-9
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# How do you find the asymptotes for (9x^2 – 36) /( x^2 - 9)?
##### 2 Answers
Jul 28, 2018
$\text{vertical asymptotes at } x = \pm 3$
$\text{horizontal asymptote at } y = 9$
#### Explanation:
$\text{let } f \left(x\right) = \frac{9 {x}^{2} - 36}{{x}^{2} - 9}$
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.
$\text{solve } {x}^{2} - 9 = 0 \Rightarrow \left(x - 3\right) \left(x + 3\right) = 0$
$x = \pm 3 \text{ are the asymptotes}$
$\text{Horizontal asymptotes occur as }$
${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$
$\text{divide terms on numerator/denominator by the highest}$
$\text{power of "x" that is } {x}^{2}$
f(x)=((9x^2)/x^2-(36)/x^2)/(x^2/x^2-9/x^2)=(9-(36)/x^2)/(1-9/x^2
$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{9 - 0}{1 - 0}$
$y = 9 \text{ is the asymptote}$
graph{(9x^2-36)/(x^2-9) [-40, 40, -20, 20]}
vertical asymptotes: $x = \setminus \pm 3$
horizontal asymptote: $y = 9$
#### Explanation:
The given function:
$f \left(x\right) = \setminus \frac{9 {x}^{2} - 36}{{x}^{2} - 9}$
Vertical asymptotes:
The above function will have vertical asymptote where denominator becomes zero i.e.
$\setminus \therefore {x}^{2} - 9 = 0$
${x}^{2} = 9$
$x = \setminus \pm 3$
Horizontal asymptotes:
The above function will have horizontal where
$y = \setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{9 {x}^{2} - 36}{{x}^{2} - 9}$
$y = \setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{9 - \frac{36}{x} ^ 2}{1 - \frac{9}{x} ^ 2}$
$y = \setminus \frac{9 - 0}{1 - 0}$
$y = 9$
Hence, vertical asymptotes: $x = \setminus \pm 3$
horizontal asymptote: $y = 9$
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2021-07-30 22:34:54
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https://blogs.mathworks.com/pick/2017/10/13/labeling-data-points/
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File Exchange Pick of the WeekOur best user submissions
This is machine translation
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Labeling Data Points
Posted by Jiro Doke,
Jiro‘s pick this week is labelpoints by Adam Danz.
This week’s entry caught my attention for two reasons. One is that this entry does the task that I usually dread doing, which is making finishing touches to my plots. When I want to label some data points, I use the text function. It gives me some control of how to align the text, but it is basically limited to the extent of the text. For example, I can do this
x = [0 0 0];
y = [.8 .7 .6];
labels = {'label 1','label 2','label 3'};
plot(x,y,'o')
text(x,y,labels,'VerticalAlignment','bottom','HorizontalAlignment','right')
Or this.
plot(x,y,'o')
text(x,y,labels,'VerticalAlignment','top','HorizontalAlignment','left')
But both of these aren’t exactly what I want because the labels slightly overlap the data. Also in the second one, the label goes beyond the bounds of the axes.
Usually at this point, I fiddle around with the coordinates of the text placements. Very doable, but tedious.
This is where Adam’s entry comes into play.
plot(x,y,'o')
labelpoints(x,y,labels,'SE',0.2,1)
'SE' means southeast placement, 0.2 refers to the offset for the labels, and 1 means “adjust the x/y limits of the axes”. Perfect!
He also includes a wealth of options, one of which I found interesting being the option to detect outliers and only placing labels on the outliers. By making use of the Statistics and Machine Learning Toolbox, he provides different methods for detecting outliers,
x = 0:0.01:1;
y = (0:0.01:1)+rand(1,101);
labs = 1:101;
plot(x,y,'o')
labelpoints(x,y,labs,'outliers_lin',{'sd', 1.5})
The other reason Adam’s entry caught my attention was the amount of help and information he included in the entry. The first 300 lines of his code are help comments!! He includes many examples to test out all of the various options. As you can see from the File Exchange page, he has been keeping the file up-to-date periodically. I can tell he has put in a lot of time and effort into making and maintaining this code. I’m grateful that he has shared this with the community. Thanks Adam!
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2018-10-16 20:44:48
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https://de.zxc.wiki/wiki/Messeinrichtung
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# Measuring device
A measuring device is defined in the "Basics of measuring technology" in DIN 1319 as the "totality of all measuring devices and additional equipment to achieve a measurement result". This is explained below based on the standard.
## Measuring device
In the simplest case, a measuring device consists of a single measuring device .
A measuring device is a device that is intended to measure the measurand ; The measured variable is the physical variable that is to be determined as a multiple of a unit .
With a few exceptions (e.g. measuring standard , normal ), the measuring device has
• an input variable , that is the variable to which the measurement applies, and
• an output variable , that is the response of the measuring device to the input variable.
The device also counts as a measuring device if the measured value is stored.
The (measured variable) transducer or sensor is the part of a measuring device or a measuring device that responds directly to the measured variable.
The output can be
• a display in the form of a dial display or numeric display or
• any physical quantity or
• a representation on data carriers.
If the input or output variable is represented by a physical variable of the same or a different type, one speaks of a measurement signal (input signal, output signal).
Example: In the case of a sinusoidal tone, the measurement signal electrical alternating voltage from a microphone can represent the output variable volume or pitch - depending on whether the amplitude or frequency of the signal is evaluated.
The output range (for indicating measuring instruments display region ) is the range of all values that can be provided by the meter at all as output. It does not necessarily match the measuring range . This is defined by the requirement that the measurement deviations of the measuring device or the measuring device errors remain within certain limits, which are referred to as error limits .
## Auxiliary device
An auxiliary device in a measuring device is not used directly for recording, converting or outputting a measuring signal. It is generally not on the path of the measurement signal (see below). However, it can have an influence on the measuring device, which is then undesirable and must be kept small.
Examples of an auxiliary device
Auxiliary devices are to be distinguished from accessories , e.g. B. probes , shunt resistors , as these exert a deliberate influence on the measurement signal.
## Measuring chain
Typical arrangement of measuring devices in a measuring chain
In a measuring device made up of several measuring devices, the output signal of one measuring device is passed on to the next measuring device as its input signal. In the entirety of all measuring devices and auxiliary devices, the measuring devices are located on the path of the measuring signal and they form a measuring chain. They are used to record the measured variable , to forward and transform a measurement signal and to provide the measured value as an image of the measured variable.
Example: temperature measuring device
Measuring device Input signal Measuring range Output signal
number 1 Resistance thermometer temperature ${\ displaystyle t}$ 0 ... 250 ° C Resistance in Ω ${\ displaystyle R}$
No. 2 Measuring bridge resistance ${\ displaystyle R}$ 100.0 ... 194.1 Ω (DC) voltage in mV ${\ displaystyle U}$
No. 3 Transmitter tension ${\ displaystyle U}$ 0… 60 mV (DC) current in mA ${\ displaystyle I}$
No. 4 Control panel indicator electricity ${\ displaystyle I}$ 4… 20 mA Deflection in mm ${\ displaystyle l}$
The observer receives along with the scale 0 ... 250 ° C Temperature in ° C ${\ displaystyle t}$
A measuring device can also collect several different measured variables.
## literature
• Peter Giesecke: Industrial measurement technology. Hüthig GmbH, Heidelberg 1999, ISBN 3-7785-2617-0
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2021-11-28 20:05:52
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https://www.physicsforums.com/threads/need-help-with-boost-converter-design.556181/
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Need help with boost converter design
1. Dec 2, 2011
Nat3
I'm trying to understand how to design a boost converter and have been reading about them for the past week or so. I've come a long way and now understand basically how they work, but I have a couple of questions on some things that are confusing me.
The first thing that is confusing me is the equation:
Vout/Vin = 1/(1-D)
I understand how the equation is derived (i.e. I can do the math), but it doesn't really make sense to me because the voltage across an inductor is proportional to the rate of change of current through it; so doesn't it seem like the voltage gain should be related to the frequency and not the duty cycle ? This is one thing that I'm really confusing about.
The other thing I'm unsure of is how to determine the appropriate frequency and inductor values for the circuit. None of the articles I've read about boost converters really explain how to choose these values. Any advice?
I'm basically at the point where I'm pulling my hair out here, so any help would be VERY MUCH appreciated
Last edited: Dec 2, 2011
2. Dec 2, 2011
DragonPetter
I had the same intuitive question when I was studying this circuit. It makes sense that your voltage is only coming during the switching times, so why is it not dependent on the switching frequency rather than the current on/off ratio.
The fact is that there are other equations in this circuit that are dependent on the switching frequency, like determining the value of the inductor and its physical size because of saturation currents. Also, the stability and "transient" abilities of the supply is dependent on frequency.
I guess the way I have told myself in my mind is that from the equation V = Ldi/dt, we know that for a given period there will only be one switch, so we know that in a given cycle, the switch looks the same at any frequency. The only difference is how long in that period the switch is high and low, which is the duty cycle. With duty cycle, its best to think of it in averages, so a 50% duty cycle means an average current goes through half the time. So at 75% duty cycle a higher average current goes through during the same period. Referring back to V = Ldi/dt, we see that we have more current at a higher duty cycle, and so the di is larger, while the dt stays the same, and so average voltage goes up.
I hope that helps.
3. Dec 3, 2011
Nat3
I think I understand what your saying. A higher duty cycle means more "on" time, so the current level will rise to a higher level than it would with a lower duty cycle because it has more time to. When the switch is opened, the higher current level will drop in the same amount of time (which is basically instantaneously) as a lower duty cycle/lower current would, and a higher voltage will be induced across the inductor?
So, a higher frequency would mean less "on" time and less "off" time. Does this mean that a higher frequency will result in less induced voltage because the current doesn't have as much time to rise? So a low frequency should be chosen for the boost converter?
4. Dec 3, 2011
Nat3
I'm still pluggin' away at this, trying to understand.. I'm confused about a number of things, but one of the most confusing has to do with how the equation $$\frac{V_o}{V_i}=\frac{1}{1-D}$$ is derived. I've read Wikipedia's article on boost converters and understand most of that equation's derivation, but I get confused towards the end.
I know the idea behind a boost converter is that the voltage across an inductor is proportional to the rate of change of current. So, to amplify a voltage, the current level flowing through an inductor needs to be quickly changed.
To show that I really have been working hard to understand this, and so you can see where I'm at, here's my work. Hopefully I did the math correctly, but my calculus is a little rusty so let me know if I didn't.. To summarize, I get one equation for when the switch is closed and another for when the switch is open:
$$\Delta I_Lon=\frac{V_i}{L}DT$$
$$\Delta I_Loff=\frac{(V_i-V_o)(1-D)T}{L}$$
Assuming those equations are correct, here's where I'm confused. Wikipedia says:
Then they equate the two current equations. But the thing is, the voltage is being amplified, right? So, since the voltage has been increased, due to conservation of energy mustn't the current be decreased, meaning they will not be equal?
Hopefully someone can clear this up for me, as I there are a few more things I'm confused about and I have to get this done be next Monday
5. Dec 3, 2011
MisterX
The switching frequency for an ideal boost converter does not affect the average output voltage.
This can be demonstrated by solving the equations for Vout in terms of Vin. The factor of T multiplies each nonzero term, and thus we can remove the T from both terms, since we know T≠0. T is the switching period.
It does affect the oscillations of currents inductor current and output voltage, just not the average voltages. When ID is zero, the output voltage would follow an RC discharge curve. Obviously, during the other part of the cycle, the capacitor must be charged back up to that level.
While the switch is conducting, the inductor current ramps up at a rate proportional to Vi, so obviously the peak to peak value would depend on the total amount of consecutive switch on time, and thus on frequency.
To operate in continuous mode, the inductor current must not be allowed to reach 0 while the current is decreasing. In discontinuous operation the wave form appears clipped off at 0, and from what I've read the output voltage does depend on the switching frequency.
My impression is that switching at tens to hundreds of kHz is typical for many recently designed power converters.
Last edited: Dec 4, 2011
6. Dec 3, 2011
Mike_In_Plano
The equation you have is based on the assumption that the converter is operating in continuous mode. That means that even though the current in the inductor increases and decreases, there is never a moment in the cycle in which the current drops to zero.
If you take this assumption, that means that current will always be flowing through the switching device (i.e. mosfet) or through the rectifier. Thus the voltage across the inductor will be:
Vin - Vsw, While the switch is on (Ton)
- OR -
Vin - (Vout + Vdiode) when the rectifier is carrying the current (Toff)
If you assume that the switch is perfect (Vsw = 0) and the rectifier is perfect (Vdiode = 0),
then these equations simplify a bit:
Vin, during Ton
Vin - Vout, during Toff.
Now, using the typical assumption that the average voltage across the inductor is zero,
Vin x Ton/Tcycle + (Vin - Vout) x Toff/Tcycle = 0
D is defined as Ton/Tcycle, while (1-D) is Toff/Tcycle. Thus:
Vin x D + (Vin - Vout) x (1-D) = 0
=> Vin x D = (Vout - Vin) x (1-D)
=> Vin x D = Vout x (1-D) - Vin x (1-D)
=> Vin x D = Vout - Vout x D - Vin + Vin x D
=> 0 = Vout - Vout x D - Vin
=> 0 = Vout (1-D) - Vin
=> Vin = Vout (1-D)
=> Vin/Vout = (1-D)
=> Vout/Vin = 1/(1-D)
- Shew...
7. Dec 3, 2011
gnurf
I think this is the key concept here. There has to be a volt-second equality on the inductor, i.e. Vin x ton = (Vout - Vin) x toff. That means that if you plot the inductor voltage as a function of time, the area that the graph makes above and below the x-axis must be equal. If OP can explain why this must be true, the rest follows easily.
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2017-11-23 21:16:43
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https://www.aimsciences.org/article/doi/10.3934/dcds.2014.34.4855
|
# American Institute of Mathematical Sciences
November 2014, 34(11): 4855-4874. doi: 10.3934/dcds.2014.34.4855
## Substitutions, tiling dynamical systems and minimal self-joinings
1 Faculty of Mathematics and Computer Science, Weizmann Institute of Science, Rehovot 7610001, Israel
Received September 2013 Revised March 2014 Published May 2014
We investigate substitution subshifts and tiling dynamical systems arising from the substitutions (1) $\theta: 0 \rightarrow 001, 1 \rightarrow 11001$ and (2) $\eta: 0 \rightarrow 001, 1 \rightarrow 11100$. We show that the substitution subshifts arising from $\theta$ and $\eta$ have minimal self-joinings and are mildly mixing. We also give a criterion for 1-dimensional tiling systems arising from $\theta$ or $\eta$ to have minimal self-joinings. We apply this to obtain examples of mildly mixing 1-dimensional tiling systems.
Citation: Younghwan Son. Substitutions, tiling dynamical systems and minimal self-joinings. Discrete and Continuous Dynamical Systems, 2014, 34 (11) : 4855-4874. doi: 10.3934/dcds.2014.34.4855
##### References:
[1] D. Berend and C. Radin, Are there chaotic tilings? Comm. Math. Phys., 152 (1993), 215-219. doi: 10.1007/BF02098297. [2] A. Clark and L. Sadun, When size matters: Subshifts and their related tiling spaces, Ergodic Theory Dynamical Systems, 23 (2003), 1043-1057. doi: 10.1017/S0143385702001633. [3] F. M. Dekking and M. Keane, Mixing properties of substitutions, Z. Wahrscheinlichkeitstheorie und Verw. Gebiete, 42 (1978), 23-33. doi: 10.1007/BF00534205. [4] H. Furstenberg, Recurrence in Ergodic Theory and Combinatorial Number Theory, Princeton University Press, Princeton, N.J., 1981. [5] E. Glasner, Ergodic Theory Via Joinings, Mathematical Surveys and Monographs, 101. American Mathematical Society, Providence, RI, 2003. doi: 10.1090/surv/101. [6] E. Glasner, B. Host and D. Rudolph, Simple systems and their higher order self-joinings, Israel J. Math., 78 (1992), 131-142. doi: 10.1007/BF02801575. [7] K. Jacobs and M. Keane, $0-1$ Sequences of Toeplitz type, Z. Wahrscheinlichkeitstheorie und Verw. Gebiete, 13 (1969), 123-131. doi: 10.1007/BF00537017. [8] A. del Junco and K. Park, An example of a measure-preserving flow with minimal self-joinings, J. d'Analyse Math., 42 (1982/83), 199-209. doi: 10.1007/BF02786879. [9] A. del Junco, M. Rahe and L. Swanson, Chacon's automorphism has minimal self joinins, J. d'Analyse Math., 37 (1980), 276-284. doi: 10.1007/BF02797688. [10] A. del Junco and D. J. Rudolph, A rank one, rigid, simple, prime map, Ergodic Theory and Dynamical Systems, 7 (1987), 229-247. doi: 10.1017/S0143385700003977. [11] S. Kakutani, Strictly ergodic symbolic dynamical systems, in Proceedings of 6th Berkeley Symposium on Mathematical Statistics and Probability. (eds. L. M. LeCam, J. Neyman, and E. L. Scott) University of California Press, Berkeley, (1972), 319-326. [12] A. B. Katok, Ya. G. Sinai and A. M. Stepin, Theory of dynamical systems and general transformation groups with invariant measure, Mathematical analysis, 13 (1975), 129-262. doi: 10.1007/BF01223133. [13] J. King, The commutant is the weak closure of the powers, for rank one transformations, Ergodic Theory and Dynamical Systems, 6 (1986), 363-384. doi: 10.1017/S0143385700003552. [14] J. King, Ergodic properties where order 4 implies infinite order, Israel J. Math., 80 (1992), 65-86. doi: 10.1007/BF02808154. [15] J. C. Oxtoby, Ergodic sets, Bull. Amer. Math. Soc., 58, (1952), 116-136. doi: 10.1090/S0002-9904-1952-09580-X. [16] K. Petersen, Ergodic Theory, Cambridge Studies in Advanced Mathematics, 2. Cambridge University Press, Cambridge, 1983. [17] M. Queffélec, Substitution Dynamical Systems - Spectral Analysis, $2^{nd}$ edition. Lecture Notes in Mathematics, 1294. Springer-Verlag, Berlin, 2010. doi: 10.1007/978-3-642-11212-6. [18] E. A. Robinson, Symbolic dynamics and tilings of $\mathbbmathbb{R}^{d}$, Symbolic dynamics and its applications, in Proc. Sympos. Appl. Math., 60, Amer. Math. Soc., Providence, RI, (2004), 81-119. doi: 10.1090/psapm/060/2078847. [19] D. J. Rudolph, An example of a measure preserving map with minimal self-joinings, and applications, J. d'Analyse Math., 35 (1979), 97-122. doi: 10.1007/BF02791063. [20] D. J. Rudolph, Fundamentals of measurable dynamics - Ergodic theory on Lebesque spaces, Oxford University Press, 1990. [21] V. V. Ryzhikov, Self-joinings of commutative actions with an invariant measure, Mat. Zametki, 83 (2008), 723-726. doi: 10.1134/S0001434608050179. [22] B. Solomyak, Dynamics of self-similar tilings, Ergodic Theory and Dynamical Systems, 17 (1997), 695-738. doi: 10.1017/S0143385797084988.
show all references
##### References:
[1] D. Berend and C. Radin, Are there chaotic tilings? Comm. Math. Phys., 152 (1993), 215-219. doi: 10.1007/BF02098297. [2] A. Clark and L. Sadun, When size matters: Subshifts and their related tiling spaces, Ergodic Theory Dynamical Systems, 23 (2003), 1043-1057. doi: 10.1017/S0143385702001633. [3] F. M. Dekking and M. Keane, Mixing properties of substitutions, Z. Wahrscheinlichkeitstheorie und Verw. Gebiete, 42 (1978), 23-33. doi: 10.1007/BF00534205. [4] H. Furstenberg, Recurrence in Ergodic Theory and Combinatorial Number Theory, Princeton University Press, Princeton, N.J., 1981. [5] E. Glasner, Ergodic Theory Via Joinings, Mathematical Surveys and Monographs, 101. American Mathematical Society, Providence, RI, 2003. doi: 10.1090/surv/101. [6] E. Glasner, B. Host and D. Rudolph, Simple systems and their higher order self-joinings, Israel J. Math., 78 (1992), 131-142. doi: 10.1007/BF02801575. [7] K. Jacobs and M. Keane, $0-1$ Sequences of Toeplitz type, Z. Wahrscheinlichkeitstheorie und Verw. Gebiete, 13 (1969), 123-131. doi: 10.1007/BF00537017. [8] A. del Junco and K. Park, An example of a measure-preserving flow with minimal self-joinings, J. d'Analyse Math., 42 (1982/83), 199-209. doi: 10.1007/BF02786879. [9] A. del Junco, M. Rahe and L. Swanson, Chacon's automorphism has minimal self joinins, J. d'Analyse Math., 37 (1980), 276-284. doi: 10.1007/BF02797688. [10] A. del Junco and D. J. Rudolph, A rank one, rigid, simple, prime map, Ergodic Theory and Dynamical Systems, 7 (1987), 229-247. doi: 10.1017/S0143385700003977. [11] S. Kakutani, Strictly ergodic symbolic dynamical systems, in Proceedings of 6th Berkeley Symposium on Mathematical Statistics and Probability. (eds. L. M. LeCam, J. Neyman, and E. L. Scott) University of California Press, Berkeley, (1972), 319-326. [12] A. B. Katok, Ya. G. Sinai and A. M. Stepin, Theory of dynamical systems and general transformation groups with invariant measure, Mathematical analysis, 13 (1975), 129-262. doi: 10.1007/BF01223133. [13] J. King, The commutant is the weak closure of the powers, for rank one transformations, Ergodic Theory and Dynamical Systems, 6 (1986), 363-384. doi: 10.1017/S0143385700003552. [14] J. King, Ergodic properties where order 4 implies infinite order, Israel J. Math., 80 (1992), 65-86. doi: 10.1007/BF02808154. [15] J. C. Oxtoby, Ergodic sets, Bull. Amer. Math. Soc., 58, (1952), 116-136. doi: 10.1090/S0002-9904-1952-09580-X. [16] K. Petersen, Ergodic Theory, Cambridge Studies in Advanced Mathematics, 2. Cambridge University Press, Cambridge, 1983. [17] M. Queffélec, Substitution Dynamical Systems - Spectral Analysis, $2^{nd}$ edition. Lecture Notes in Mathematics, 1294. Springer-Verlag, Berlin, 2010. doi: 10.1007/978-3-642-11212-6. [18] E. A. Robinson, Symbolic dynamics and tilings of $\mathbbmathbb{R}^{d}$, Symbolic dynamics and its applications, in Proc. Sympos. Appl. Math., 60, Amer. Math. Soc., Providence, RI, (2004), 81-119. doi: 10.1090/psapm/060/2078847. [19] D. J. Rudolph, An example of a measure preserving map with minimal self-joinings, and applications, J. d'Analyse Math., 35 (1979), 97-122. doi: 10.1007/BF02791063. [20] D. J. Rudolph, Fundamentals of measurable dynamics - Ergodic theory on Lebesque spaces, Oxford University Press, 1990. [21] V. V. Ryzhikov, Self-joinings of commutative actions with an invariant measure, Mat. Zametki, 83 (2008), 723-726. doi: 10.1134/S0001434608050179. [22] B. Solomyak, Dynamics of self-similar tilings, Ergodic Theory and Dynamical Systems, 17 (1997), 695-738. doi: 10.1017/S0143385797084988.
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2021 Impact Factor: 1.588
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2022-07-06 07:42:27
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https://iris.unife.it/handle/11392/1401264
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In this paper we consider positively $1$-homogeneous supremal functionals of the type $F(u):=\supess_\Om f(x,\nabla u(x))$. We prove that the relaxation $\bar F$ is a {\it difference quotient}, that is $$\bar{F}(u)=R^{d_F}(u):= \sup_{x,y\in \Om,\,x\neq y} \frac{u(x) - u(y)}{d_F(x,y)} \qquad \text{ for every } u\in \wi,$$ where $d_F$ is a geodesic distance associated to $F$. Moreover we prove that the closure of the class of $1$-homogeneous supremal functionals with respect to $\Gamma$-convergence is given exactly by the class of difference quotients associated to geodesic distances. This class strictly contains supremal functionals, as the class of geodesic distances strictly contains {\it intrinsic} distances.
### From 1-homogeneous supremal functionals to difference quotients: relaxation and $Gamma$-convergence.
#### Abstract
In this paper we consider positively $1$-homogeneous supremal functionals of the type $F(u):=\supess_\Om f(x,\nabla u(x))$. We prove that the relaxation $\bar F$ is a {\it difference quotient}, that is $$\bar{F}(u)=R^{d_F}(u):= \sup_{x,y\in \Om,\,x\neq y} \frac{u(x) - u(y)}{d_F(x,y)} \qquad \text{ for every } u\in \wi,$$ where $d_F$ is a geodesic distance associated to $F$. Moreover we prove that the closure of the class of $1$-homogeneous supremal functionals with respect to $\Gamma$-convergence is given exactly by the class of difference quotients associated to geodesic distances. This class strictly contains supremal functionals, as the class of geodesic distances strictly contains {\it intrinsic} distances.
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2006
Prinari, Francesca Agnese; A., Garroni; M., Ponsiglione
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Utilizza questo identificativo per citare o creare un link a questo documento: https://hdl.handle.net/11392/1401264
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2023-03-23 20:17:39
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https://www.physicsforums.com/threads/11-8-a-fibonacci-problem.157058/
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# 11^8 = a Fibonacci Problem
1. Feb 19, 2007
### ramsey2879
Problem: Use Fibonacci like series (but not the same starting numbers) to prove that there are an infinite number of solutions to 5a^2 + 5ab + b^2 = 11^8 with a and b coprime. In I have a proof that you can solve for any prime, ending in 1 or 9, multiplied to the 8th power.
2. Feb 22, 2007
### Gib Z
$$T_n=T_{n-1} + T_{n-2}$$
Factorise.
$$(\sqrt{5}a +\sqrt{5}b)^2 - 4b^2 = 11^8$$. Perhaps that reminds you of something?
3. Feb 22, 2007
### Gib Z
O, quote: I have a proof that you can solve for any prime, ending in 1 or 9, multiplied to the 8th power.
Well, 11^8 = 123456787654321, which satisfies your conditions.
4. Feb 22, 2007
### ramsey2879
Actually the problem I meant is to find my proof that there are an infinite number of solutions to the Diophantine equation 5a^2 + 5ab + b^2 = p^n where a,b are coprime and p is a prime ending in 1 or 9. There is a relationship between any three consecutive term of a Fibonacci type series and the form 5a^2 + 5ab + b^2 that is invariant with the index number of the first term. That is a key to my proof. I leave the proof for you to figure out, but will make suggestions if you reach a dead end and have no idea where to turn.
First off then, how does the form 5a^2 + 5ab + b^2 relate to three terms of a Fibonacci type series in an invariant manner? (By Fibonacci type, I mean F(n) = F(n-1) + F(n-2)
5. Feb 22, 2007
### ramsey2879
This is interesting, it factors to $$((\sqrt{5}a +\sqrt{5}b +2b) * ((\sqrt{5}a +\sqrt{5}b -2b) = 11^8$$. However, these are not integers and I thought number theory dealt only with integers.
Last edited: Feb 22, 2007
6. Feb 22, 2007
### matt grime
That's a strange thing to say. Not least because 'a' isn't an integer, it is an indeterminate expression, so your own first post fails your own criterion.
Moving on from the stupidly pedantic, you're objecting to the use of sqrt(5), right? Well, if you think that sqrt(5) has no place in number theory then you're not going to get very far in the subject. Number theory might be set up to answer questions posed about Z, or more likely Q, but there is nothing that says it must stick only with those things, and indeed it doesn't.
7. Feb 22, 2007
### ramsey2879
Ok but how would you get an solution in integers to my diophantine equation using that? If there is a way I would like to know. But since thare are Fibonacci identities that are known, it is not necessary to resort to the square root of 5 or the golden ratio to make my proof.
8. Feb 22, 2007
### matt grime
I did not say anything about your equation. However, I believe it is a common tactic in algebraic number theory that in order to find integral solutions to something one passes to a larger ring, such as the ring of integers in a number field. This is probably called something like Class Field Theory.
9. Feb 22, 2007
### ramsey2879
10. Feb 22, 2007
### matt grime
Nope. coprimality exists in the algebraic integers too: x and y a coprime if z divides x and z divides y implies z is a unit. Anyway, you were asking for integral solutions to some equation in a and b. At least that is how I read it. But a was an indeterminate. Not an integer. 2 is an integer. a isn't. Unless a=2. For example. If a is an integer in post 1 what integer is it?
11. Feb 22, 2007
### matt grime
12. Feb 22, 2007
### ramsey2879
But how do you get 5^a^2 + 5ab + b^2 from 5(a+b)^2 - 4b^2
13. Feb 22, 2007
### matt grime
Uh? Please point out where you think I claim to have got that at all.
14. Feb 22, 2007
### ramsey2879
l a = 3161 b = 7159 , is one of infinitely many solutions in integers for 11 ^8. But any prime ending in 1 or 9 to any integer power can replace 11^8 and there would still be an infinite number of solutions in integers.
15. Feb 22, 2007
### ramsey2879
You didn't post it but there was a post that used sqrt(5) in a manner that gets 5a^2 + 10ab - b^2 which is not my equation. Sorry
I have to go now. Thanks for your help.
16. Feb 22, 2007
### ramsey2879
Are all primes ending in 1 or 9 and their products members of a ring? My proof related to this topic is much broader than the original proposition since it covers all primes ending in 1 or 9 or their products. In short I have a proof that there are an infinite number of solutions to the Diophantine equation
$$P = 5a^{2} + 5ab + b^{2}$$ where P is a prime ending in 1 or 9 or a product of such primes. Knowing the algorithm makes it easy to solve for any such product given initial solutions for the primes ending in 1 or 9. There is no need to do calculations greater than the square root of P. The proof is really simple in retrospect and can be understood by an algebra student familiar with the law of quadratic reciprocity. I doubt that it could be simplified by using Class Field Theory. Perhaps it may be a trival application of class theory but I don't know. I skipped college after taking the equivalant of two years of courses via night school (none in higher algebra or number theory) and am now retired. However, my interest in number theory is such that I would be willing to pay to audit a few courses such as Algebraic Number Theory once I take care of a few other things like building my dream home. The book I purchased is mostly Greek to me.
17. Feb 23, 2007
### matt grime
OF course they are: the integers, the rationals, the reals, complex numbers, algebraic integers, algebraic numbers.....
18. Feb 23, 2007
### Gib Z
perhaps the most obvious one, the primes >.<"
EDIT: btw, im taking ring as a synonym for set, i haven't done rings yet but it looks like you were listing the sets primes were in.
19. Feb 23, 2007
### matt grime
A ring is not just a set.
20. Feb 23, 2007
### Gib Z
ahh then excuse my ignorance :)
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2017-02-26 03:32:13
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https://alexshtf.github.io/2021/04/14/ProximalPointFM.html
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# Intro
In this final episode of the proximal point series I would like to take the method to the extreme, and show that we can actually train a model which composed with an appropriate loss produces functions which are non-linear and non-convex: we’ll be training a factorization machine for classification problems without linearly approximating loss functions and relying on loss gradients. Factorization machines and variants are widely used in recommender systems, i.e. recommending movies to users. I assume readers are familiar with the basics, and below I provide only a brief introduction, so that throughout this post we have a consistent notation and terminology, and understand the assumptions we make.
I do not claim that it is the best method for training factorization machines, but it is indeed an interesting challenge in order to see what are the limits of efficiently implementable proximal point methods. We’ll have some more advanced optimization theory, even some advanced linear algebra, but most importantly, at the end of the journey we’ll have a github repo with code which you can run and try it out on your own dataset!
Since it’s an ‘academic’ experiment in nature, and I do not aim to implement the most efficient and robust code, we’ll make some simplifying assumptions. However, a production-ready training algortithm will not be far away from the implementation we construct in this post.
## A quick intro
Let’s begin by a quick introduction to factorization machines. Factorization machines are usually trained on categorical data representing the users and the items, for example, age group and gender may be user features, while product category and price group may be item features. The model embeds each categorical feature to a latent space of some pre-defined dimension $$k$$, and the model’s prediction comprises of inner products of the latent vectors corresponding to the current samples. The most simple variant are second order factorization machines, which we the focus of this post.
Formally, our second-order factorization machine $$\sigma(w, x)$$ is given a binary input $$w \in \{0, 1\}^m$$ which is a one-hot encoding of a subset of at most $$m$$ categorical features. For exampe, suppose we would like to predict the affinity of people with chocolate. Assume, for simplicity, that we have only two user gender values $$\{ \mathrm{male}, \mathrm{female} \}$$, and two two age groups $$\{ \mathrm{young}, \mathrm{old} \}$$. For our items, suppose we have only one feature - the chocolate type, which may take the values $$\{\mathrm{dark}, \mathrm{milk}, \mathrm{white}\}$$. In that case, the model’s input is the vector of zeros and ones encoding feature indicators:
$w=(w_{\mathrm{male}}, w_{\mathrm{female}}, w_{\mathrm{young}}, w_{\mathrm{old}}, w_{\mathrm{dark}}, w_{\mathrm{milk}}, w_{\mathrm{white}}).$
A young male who tasted dark chocolate is represented by the vector
$w = (1, 0, 1, 0, 1, 0, 0).$
In general, the vector $$w$$ can be defined by arbitrary real numbers, but I promised that we’ll make simplifying assumptions :)
The model’s parameter vector $$x = (b_0, b_1, \dots, b_m, v_1, \dots, v_m)$$ is composed of the model’s global bias $$b_0 \in \mathbb{R}$$, the biases $$b_i \in \mathbb{R}$$ for the features $$i\in \{1, \dots, m\}$$, and the latent vectors $$v_i \in \mathbb{R}^k$$ for the same features with $$k$$ being the embeddig dimension. The model computes:
$\sigma(w, x) := b_0 + \sum_{i = 1}^m w_i b_i + \sum_{i = 1}^m\sum_{j = i + 1}^{m} (v_i^T v_j) w_i w_j.$
Let’s set up some notation which will become useful throughout this post. We will detote a set of consecutive integers by $$i..j=\{i, i+1, \dots, j\}$$, and the set of distinct pairs of the integers $$J$$ is denoted by $$P[J]=\{ (i,j) \in J\times J : i<j \}$$. Consequently, we can re-write:
$\sigma(w,x)=b_0 + \sum_{i\in 1..m} w_i b_i+\sum_{(i,j) \in P[1..m]} (v_i^T v_j) w_i w_j$
At this stage this notation does not seem useful, but it will simplify things later in this post. We’ll use this notation consistently throughout the post.
For completeness, let’s implement a factorization machine in PyTorch. To that end, recall a famous trick introduced by Steffen Rendle in his pioneering paper1 on factorization machines, based on the formula
$\Bigl\| \sum_{i\in 1..m} w_i v_i \Bigr\|_2^2 = \sum_{i\in 1..m} \|w_i v_i\|_2^2 + 2 \sum_{(i,j)\in P[1..m]} (v_i^T v_j) w_i w_j.$
After re-arrangement, the above results in:
$\sum_{(i,j)\in P[1..m]} (v_i^T v_j) w_i w_j= \frac{1}{2}\Bigl\| \sum_{i\in1.m} w_i v_i \Bigr\|_2^2-\frac{1}{2}\sum_{i\in1..m} \|w_i v_i\|_2^2. \tag{L}$
Since $$w$$ is a binary vector, we can associate it with its non-zero indices $$\operatorname{nz}(w)$$, and the right-hand side of above term can be written as:
$\frac{1}{2}\Bigl\| \sum_{i \in \operatorname{nz}(w)} v_i \Bigr\|_2^2-\frac{1}{2}\sum_{i \in \operatorname{nz}(w)} \| v_i\|_2^2.$
Consequently, the pairwise terms can be computed in time linear in the number of non-zero indicators in $$w$$, instead of the quadratic time imposed by the naive way. The PyTorch implementation below uses the trick above.
import torch
from torch import nn
class FM(torch.nn.Module):
def __init__(self, m, k):
super(FM, self).__init__()
self.bias = nn.Parameter(torch.zeros(1))
self.biases = nn.Parameter(torch.zeros(m))
self.vs = nn.Embedding(m, k)
torch.nn.init.normal_(self.vs.weight, std=0.01)
torch.nn.init.normal_(self.biases, std=0.01)
def forward(self, w_nz): # since w are indicators, we simply use the non-zero indices
vs = self.vs(w_nz)
# in vs:
# dim = 0 is the mini-batch dimension. We would like to operate on each elem. of a mini-batch separately.
# dim = 1 are the embedding vectors
# dim = 2 are their components.
pow_of_sum = vs.sum(dim=1).square().sum(dim=1) # sum vectors -> square -> sum components
sum_of_pow = vs.square().sum(dim=[1,2]) # square -> sum vectors and components
pairwise = 0.5 * (pow_of_sum - sum_of_pow)
biases = self.biases
linear = biases[w_nz].sum(dim=1) # sum biases for each element of the mini-batch
return pairwise + linear + self.bias
If we are interested in solving a regression problem, i.e. predicting arbitrary real values, such as a score a person would give to the chocolate, we can use $$\sigma$$ directly to make predictions. If we are in the binary classification setup, i.e. predict the probability that a person likes the corresponding chocolate, we compose $$\sigma$$ with a sigmoid, and predict $$p(w,x) = (1+e^{-\sigma(w,x)})^{-1}$$.
## The setup
In this post we are interested in the binary classification setup, with the binary cross-entropy loss. Namely, given a label $$y \in \{0,1\}$$ the loss is:
$-y \ln(p(w,x)) - (1 - y) \ln(1 - p(w,x)).$
For example, if we would like to predict which chocolate people like, we could train the model on a data-set with samples of people who liked a certain chocolate having the label $$y = 1$$, and people who tasted but did not like it will have the lable $$y = 0$$. Having trained the model, we can recommend chocolate to a person by choosing the one with the highest probability of being liked.
Using a simple transformation $$\hat{y} = 2y-1$$ we can remap the labels to be in $$\{-1, 1\}$$ instead. Then, it isn’t hard to verify that the binary cross-entropy loss above reduces to:
$\ln(1+\exp(-\hat{y} \sigma(w,x))).$
Consequently, our aim will be training over the set $$\{ (w_i, \hat{y}_i) \}_{i=1}^n$$ by minimizing the average loss
$\frac{1}{n} \sum_{i=1}^n \underbrace{\ln(1+\exp(-\hat{y}_i \sigma(w_i, x)))}_{f_i(x)}.$
Instead of using regular SGD-based methods for training, which construct a linear approximations of $$f_i$$ and are able to use only the information provided by the gradient, we will avoid approximating and use the loss itself via the stochastic proximal point algorithm - at iteration $$t$$ choose $$f \in \{f_1, \dots, f_n\}$$ and compute:
$x_{t+1} = \operatorname*{argmin}_x \left\{ f(x) + \frac{1}{2\eta} \|x - x_t\|_2^2 \right\}. \tag{P}$
Careful readers might notice that the formula above is total nonsense in general. Why? Well, the each $$f$$ is a non-convex function of $$x$$. If $$f$$ was convex, we would obtain a unique and well-defined minimizer $$x_{t+1}$$. However, in general, the $$\operatorname{argmin}$$ above is a set of minimizers, which might even be empty! In this post we will attempt to mitigate this issue:
1. Discover the conditions on the step-size $$\eta$$ which ensure that we have a unique minimizer $$x_{t+1}$$.
2. Find an explicit formula, or a simple algorithm, for computing $$x_{t+1}$$.
Having done the above, we’ll be able to construct an algorithm which can train classifying factorization machines which exploit the exact loss function, instead of just relying on its slope as in SGD.
# Duality strikes again
In previous posts we heavily relied on duality in general, and convex conjugates in particular, and this post is no exception. Recall, that the convex conjugate of the function $$h$$ is defined by:
$h^*(z) = \sup_x \{ x^T z - h(x) \},$
and recall also that in a previous post we saw that $$h(t)=\ln(1+\exp(t))$$ is convex, and its convex conjugate is:
$h^*(z) = \begin{cases} z\ln(z) + (1 - z) \ln(1 - z), & 0 < z < 1 \\ 0, & \text{otherwise}. \end{cases}$
An interesting result about conjugates is that under some technical conditions, which hold for $$h(t)$$ above, we have $$h^{**} = h$$, namely, the conjugate of $$h^*$$ is $$h$$. Moreover, in our case the $$\sup$$ in the conjugate’s definition can be replaced with a $$\max$$, since the supermum is always attained2. Why is it useful? Since now we know that:
$\ln(1+\exp(t))=\max_z \left\{ t z - z \ln(z) - (1-z) \ln(1-z) \right\}.$
Consequently, the term inside the $$\operatorname{argmin}$$ of the proximal point step (P) can be written as:
\begin{aligned} f(x) &+ \frac{1}{2\eta} \|x - x_t\|_2^2 \\ &\equiv \ln(1+\exp(-\hat{y} \sigma(w,x))) + \frac{1}{2\eta} \|x - x_t\|_2^2 \\ &= \max_z \Bigl\{ \underbrace{ -z \hat{y} \sigma(w,x) + \frac{1}{2\eta} \|x - x_t\|_2^2 - z\ln(z) - (1-z)\ln(1-z) }_{\phi(x,z)} \Bigr\}. \end{aligned}
Since we are interested in minimizing the above, we will be solving the saddle-point problem:
$\min_x \max_z \phi(x,z). \tag{Q}$
Convex duality theory has another interesting form - it provides conditions on saddle-point problems which ensure that we can switch the order of $$\min$$ and $$\max$$ to obtain an equivalent problem. Why is it interesting? Because switching the order produces
$\max_z \underbrace{ \min_x \phi(x,z)}_{q(z)},$
and finding the optimal $$z$$ means maximizing the one dimensional function $$q$$, which may even be as simple as a high-school calculus exercise.
So here is relevant duality theorem, which is a simplification of Sion’s minimax theorem from 1958 for this post:
Let $$\phi(x,z)$$ be a continuous function which is convex in $$x$$ and concave in $$z$$. Suppose that the domain of $$\phi$$ over $$z$$ is compact, i.e. a closed a bounded set. Then,
$\min_x \max_z \phi(x,z) = \max_z \min_x \phi(x,z)$
In our case, it’s easy to see that $$\phi$$ is indeed concave in $$z$$ using negativity of its second derivative, and its domain, the interval $$[0,1]$$, is indeed compact. What we require for the theorem’s conditions to hold is convexity in $$x$$, which is what we explore next. Then, we’ll see that $$q$$, despite not being so simple, can still be quite efficiently maximized. The theorem does not imply that a pair $$(x, z)$$ solving the max-min problem also solves the min-max problem, but in our case the max-min problem has a unique solution, and in that particular case it indeed also solves the min-max problem.
Consequently, having found $$z^*=\operatorname{argmax}_z q(z)$$, we by construction obtain a formula for computing the optimal $$x$$: $$x_{t+1} = \operatorname*{argmin}_x ~ \phi(x, z^*).$$
So let’s begin by ensuring that the conditions for Sion’s theorem hold. Ignoring the terms of $$\phi$$ which do not depend on $$x$$, we need to study the convexity of the following part as a function of $$x$$:
$(*) = -z \hat{y} \sigma(w,x) + \frac{1}{2\eta} \|x - x_t\|_2^2.$
To that end, we need to open the ‘black box’ and look inside $$\sigma$$ again. That’s going to be a bit technical, but it gets us where we need. If you don’t wish to read all the details, you may skip to the conclusion below.
Recall, the composition $$x = (b_0, b_1, \dots, b_m, v_1, \dots, v_m)$$ and the definition
$\sigma(w, b_0, \dots, b_m, v_1, \dots, v_m) = b_0 + \sum_{i\in1..m} w_i b_i + \sum_{(i,j)\in P[1..m]} (v_i^T v_j) w_i w_j.$
Consequently, we can re-write $$(*)$$ as:
\begin{aligned} (*) =& \color{blue}{-z \hat{y} \Bigl[ b_0 + \sum_{i\in1..m} w_i b_i \Bigr] + \frac{1}{2\eta} \|b - b_t\|_2^2} \\ & \color{brown}{- z \hat{y} \sum_{i\in P[1..m]} (v_i^T v_j) w_i w_j + \frac{1}{2\eta} \sum_{i\in1..m} \| v_i - v_{i,t} \|_2^2}. \end{aligned}
The part colored in blue is always convex - it is the sum of a linear function and a convex-quadratic one. It remains to study the convexity of the brown part. Re-arranging the formula for $$\|v_i + v_j\|_2^2$$, we obtain that:
$v_i^T v_j = \frac{1}{2} \|v_i + v_j\|_2^2 - \frac{1}{2}\|v_i\|_2^2 - \frac{1}{2} \|v_j\|_2^2.$
Denoting $$\alpha_{ij} = -z \hat{y} w_i w_j$$ we can re-write the brown part as: \begin{aligned} \color{brown}{\text{brown}} &= \sum_{i\in P[1..m]} |\alpha_{ij}| v_i^T ( \operatorname{sign}(\alpha_{ij}) v_j) + \frac{1}{2\eta} \sum_{i\in1..m} \| v_i - v_{i,t} \|_2^2 \\ &= \frac{1}{2}\sum_{i\in P[1..m]} |\alpha_{ij}| \left[ \|v_i + \operatorname{sign}(\alpha_{ij}) v_j\|_2^2 - \|v_i\|_2^2-\|v_j\|_2^2 \right] + \sum_{i\in1..m} \left[ \| v_i \|_2^2 \color{darkgray}{- 2 v_i^T v_{i,t} + \|v_{i,t}\|_2^2} \right] \end{aligned}
The grayed-out part on the right is linear in $$v_i$$, so it’s convex. Since $$\alpha_{ij} = \alpha_{ji}$$, to simplify notation we define $$\alpha_{ii}=0$$, and the remaining non-greyed parts can be written as:
$\frac{1}{2} \sum_{i\in P[1..m]} |\alpha_{ij}| \|v_i + \operatorname{sign}(\alpha_{ij}) v_j\|_2^2 + \sum_{i\in 1..m} \left(\frac{1}{2\eta} - \sum_{j\in 1..m} |\alpha_{ij}|\right) \|v_i\|_2^2.$
Again, the first sum is a sum of convex-quadratic functions, and thus convex. For the second part to be convex, we require that for each $$i$$ we have
$\frac{1}{2\eta} \geq \sum_{j\in 1..m} |\alpha_{ij}|,$
or equivalently that the step-size $$\eta$$ must satisfy
$\eta \leq \frac{1}{2\sum_{j \in 1..m} |\alpha_{ij}|}$
Since $$\vert \alpha_{ij} \vert \leq 1$$, we can easily deduce that for any step-size $$\eta \leq \frac{1}{2m}$$, we obtain a convex $$\phi$$. A better bound is obtained if we have a bound on the number of indicators in the vector $$w$$ which may be non-zero at the same time. For example, if we have six categorical fields, we will have at most six non-zero elements in $$w$$, and thus $$\eta \leq \frac{1}{12}$$.
Convexity is nice if we want Sion’s theorem to hold, but if we want a unique minimizer $$x_{t+1}$$ we need strict convexity, which is obtained by using a strict inequality - replace $$\leq$$ with $$<$$. In this post we will assume that we have at most $$d$$ categorical features, and use step-sizes which satisfy
$\eta \leq \frac{1}{2d+1} < \frac{1}{2d}.$
# Computing $$x_{t+1}$$
Suppose that Sion’s theorem holds, and that we can obtain a unique minimizer $$x_{t+1}$$. How do we compute it? Well, Sion’s theorem lets us switch the order of $$\min$$ and $$\max$$, so we are aiming to solve:
$\max_z \underbrace{ \min_x \phi(x,z)}_{q(z)},$
and explicitly writing $$\phi$$ we have:
\begin{aligned} q(z) = \min_{b,v_i} \Bigl\{ &-z \hat{y} \Bigl[ b_0 + \sum_{i\in 1..m} w_i b_i \Bigr] + \frac{1}{2\eta} \|b - b_t\|_2^2 \\ &- z \hat{y} \sum_{(i,j) \in P[1..m]} (v_i^T v_j) w_i w_j + \frac{1}{2\eta} \sum_{i\in 1..m} \| v_i - v_{i,t} \|_2^2 \\ &- z \ln(z) - (1-z) \ln(1-z) \Bigr\} \end{aligned}
From now it becomes a bit technical, but the end-result will be an algorithm to compute $$q(z)$$ for any $$z$$ by solving the minimization problem over $$x$$. Afterwards, we’ll find a way to maximize $$q$$ over $$z$$.
Using separability3 we can separate the minimum above into a sum of three parts: the minimum over the biases $$b$$, another minimum over the latent vectors $$v_1, \dots, v_m$$, and the term $$-z \ln(z) - (1-z) \ln(1-z)$$, namely:
\begin{aligned} q(z) &= \underbrace{\min_b \left\{ -z \hat{y} \left[ b_0 + \sum_{i\in 1..m} w_i b_i \right] + \frac{1}{2\eta} \|b - b_t\|_2^2 \right\}}_{q_1(z)} \\ &+ \underbrace{\min_{v_1, \dots, v_m} \left\{ - z \hat{y} \sum_{(i,j) \in P[1..m]} (v_i^T v_j) w_i w_j + \frac{1}{2\eta} \sum_{i\in 1..m} \| v_i - v_{i,t} \|_2^2 \right\}}_{q_2(z)} \\ &-z \ln(z) - (1-z) \ln(1-z) \end{aligned}
We’ll analyze $$q_1$$, and $$q_2$$ shortly, but let’s take a short break and implement a skeleton of our training algorithm. A deeper analysis of $$q_1$$, $$q_2$$, and $$q$$ will let us fill the skeleton. On construction, it receives a factorization machine object of the class we implemented above, and the step size. Then, each training step’s input is the set $$\operatorname{nz}(w)$$ of the non-zero feature indicators, and the label $$\hat{y}$$:
class ProxPtFMTrainer:
def __init__(self, fm, step_size):
# training parameters
self.b0 = fm.bias
self.bs = fm.biases
self.vs = fm.vs
self.step_size = step_size
def step(self, w_nz, y_hat):
pass # we'll replace it with actual code to train the model.
## Minimizing over $$b$$ - computing $$q_1$$
Defining $$\hat{w}=(1, w_1, \dots, w_m)^T$$ and $$\hat{b}=(b_0, b_1, \dots, b_m)$$, we obtain:
\begin{aligned} q_1(z) =&\min_{\hat{b}} \left\{ -z \hat{y} \hat{w}^T \hat{b} + \frac{1}{2\eta} \|\hat{b} - \hat{b}_t\|_2^2 \right\} \end{aligned}.
The term inside the minimu is a simple convex quadratic which is minimized by comparing its gradient with zero: $$\hat{b}^* = \hat{b}_t + \eta z \hat{y} \hat{w}. \tag{A}$$
Consequently:
\begin{aligned} q_1(z) &= -z \hat{y} \hat{w}^T (\hat{b}_t + \eta z \hat{y} \hat{w}) + \frac{1}{2\eta} \| \eta z \hat{y} \hat{w} \|_2^2 \\ &= -\hat{y} (\hat{w}^T \hat{b}_t) z - \eta \hat{y}^2 \|\hat{w}\|_2^2 z^2 + \frac{\eta \hat{y}^2 \|\hat{w}\|_2^2}{2} z^2 \\ &= -\hat{y} (\hat{w}^T \hat{b}_t) z - \frac{\eta \hat{y}^2 \|\hat{w}\|_2^2}{2} z^2 \end{aligned}
Since $$\hat{y} =\pm 1$$ we have that $$\hat{y}^2 = 1$$. Moreover, since $$w_i$$ are indicators, the term $$\|\hat{w}\|_2^2$$ is the number of non-zero entries of $$w$$ plus one. So, to summarize, the above can be written as
$q_1(z) = -\frac{\eta (1 + |\operatorname{nz}(w)|)}{2}z^2 -\hat{y} (w^T b_t + b_{0,t}) z.$
What a surprise - $$q_1$$ is just a concave parabola!
So, to summarize, what we have here is an explicit expression for $$q_1$$, and the formula (A) to update the biases once we have obtained the optimal $$z$$.
Let’s implement the code for the two steps above. We’ll see below that the function $$q_1$$ will have to be evaluated several times in order to find the optimal $$z$$, and consequently it’s beneficial to cache various expensive-to-compute elements so that its evaluation is quick and efficient every time. Consequently, the step function will store these parts in the classe’s members.
# inside ProxPtFMTrainer
def step(self, w_nz, y_hat):
self.nnz = w_nz.numel() # |nz(w)|
self.bias_sum = self.bs[w_nz].sum().item() # w^T b_t
# TODO - this function will grow as we proceed
def q_one(self, y_hat, z)
return -0.5 * self.step_size * (1 + self.nnz) * (z ** 2) \
- y_hat * (self.bias_sum + self.b0.item()) * z
def update_biases(self, w_nz, y_hat, z):
self.bs[w_nz] = self.bs[w_nz] + self.step_size * z * y_hat
You might be asking yourself why we stored the bias sum in a member of self. It will become apparent shortly, but we’ll be calling the function q_one repeatedly, and we would like to avoid re-computing time consuming things we could compute only once.
## Minimizing over $$v_1, \dots, v_m$$
We are aiming to compute
$q_2(z) = \min_{v_1, \dots, v_m} \left\{ Q(v_1, \dots, v_m, z) \equiv - z \hat{y} \sum_{(i,j)\in P[1..m]} (v_i^T v_j) w_i w_j + \frac{1}{2\eta} \sum_{i\in 1..m} \| v_i - v_{i,t} \|_2^2 \right\}.$
Of course, we assume that we indeed chose $$\eta$$ such that $$Q$$ inside the $$\min$$ operator is strictly convex in $$v_1, \dots, v_m$$, so that there is a unique minimizer.
Since $$w$$ is a vector of indicators, we can write the function $$Q$$ by separating out the part which corresponds to non-zero indicators in $$w$$:
$Q(v_1, \dots, v_m,z) = \underbrace{-z \hat{y} \sum_{(i,j)\in P[\operatorname{nz}(w)]} v_{i}^T v_{j}+\frac{1}{2\eta} \sum_{i\in \operatorname{nz}(w)} \|v_i - v_{i,t} \|_2^2}_{\hat{Q}} + \underbrace{\frac{1}{2\eta}\sum_{i \notin \operatorname{nz}(w)} \|v_i-v_{i,t}\|_2^2}_{R}.$
Looking at $$R$$, clearly the minimizer must satisfy $$v_i^* = v_{i,t}$$ for all $$i \notin \operatorname{nz}(w)$$, and consequently $$R$$ must be zero at optimum, independent of $$z$$. Hence, we have:
$q_2(z)=\min_{v_{\operatorname{nz}(w)}} \hat{Q}(v_{\operatorname{nz}(w)}, z),$
where $$v_{\operatorname{nz}(w)}$$ is the set of the vectors $$v_i$$ for $$i \in \operatorname{nz}(w)$$. Since $$\hat{Q}$$ is a quadratic function which we made sure is strictly convex, we can find our optimal $$v_{\operatorname{nz}(w)}^*$$ by solving the linear system obtained by equating the gradient of $$\hat{Q}$$ with zero.
So let’s see what the gradient looks like. We have a function of several vector variables $$v_{\operatorname{nz}(w)}$$, and we imagine that they are all stacked into one big vector. Consequently, the gradient of $$\hat{Q}$$ is a stacked vector comprising of the gradients w.r.t each of the vectors. So, let’s compute the gradient w.r.t each $$v_i$$ and equate it with zero:
$\nabla_{v_i} \hat{Q} = -z \hat{y} \sum_{\substack{j \in \operatorname{nz}(w)\\j\neq i}} v_{j}+\frac{1}{\eta} (v_{i} - v_{i,t})=0.$
By re-arranging and putting constants on the RHS we can re-write the above as
$-\eta z \hat{y} \sum_{\substack{j \in \operatorname{nz}(w)\\j\neq i}} v_{j} + v_{i} = v_{i,t}.$
The above system means that we are actually solving linear systems with the same coefficients for each coordinate of the embedding vectors. Equivalently written, we can stack the vectors $$v_{\operatorname{nz}(w)}$$ into the rows of the matrix $$V$$, and the vectors $$v_{\operatorname{nz}(w),t}$$ into the rows of the matrix $$V_t$$, and solve the linear system
$\underbrace{\begin{pmatrix} 1 & -\eta z \hat{y} & \cdots & -\eta z \hat{y} \\ -\eta z \hat{y} & 1 & \cdots & -\eta z \hat{y} \\ \vdots & \vdots & \ddots & \vdots & \\ -\eta z \hat{y} & -\eta z \hat{y} & \cdots & 1 \end{pmatrix}}_{S(z)} V = V_t$
Note, that the matrix $$S(z)$$ is small, since its dimensions only depend on the number of non-zero elements in $$w$$. So, now we have an efficient algorithm for computing $$q_2(z)$$ given the sample $$w$$ and the latent vectors from the previous iterate $$v_{1,t}, \dots, v_{m,t}$$:
Algorithm B
1. Embed the latent vectors $$\{v_{t,i}\}_{i \in \operatorname{nz}(w)}$$ into thr rows of the matrix $$V_t$$.
2. Obtain a solution $$V^*$$ of the linear system of equations $$S(z) V = V_t$$, and use the rows of $$V^*$$ as the vectors $$\{v_{i}^*\}_{i \in \operatorname{nz}(w)}$$.
3. Output: $$q_2(z)=-z \hat{y} \sum_{(i,j) \in P[\operatorname{nz}(w)]} ({v_{i}^*}^T v_{j}^*)+\frac{1}{2\eta} \sum_{i\in \operatorname{nz}(w)} \|v_{i}^* - v_{i,t} \|_2^2$$
However, let’s see how we can solve the linear system without invoking any matrix inversion algorithm altogether, since it turns out we can directly and efficiently compute $$S(z)^{-1}$$! The matrix $$S(z)$$ can be written as:
$S(z) = (1 + \eta z \hat{y}) I - \eta z \hat{y}(\mathbf{e} ~ \mathbf{e}^T)$
where $$\mathbf{e} \in \mathbb{R}^{\vert\operatorname{nz}(w)\vert}$$ is a column vector whose components are all $$1$$. Now, we’ll employ the Sherman-Morrison matrix inversion identity:
$(A+u v^T)^{-1} = A^{-1} - \frac{A^{-1} u v^T A^{-1}}{1 + v^T A^{-1} u}.$
In our case, we’ll be taking $$A = (1 + \eta \hat{y} z) I$$, $$u=-\eta \hat{y} z \mathbf{e}$$, and $$v = \mathbf{e}$$, and consequently we have:
$S(z)^{-1} = \frac{1}{1 + \eta \hat{y} z} I + \frac{\eta \hat{y} z}{(1 + \eta \hat{y} z)^2 - \eta \hat{y} z(1 + \eta \hat{y} z) \mathbf{e}^T \mathbf{e}} \mathbf{e}~\mathbf{e}^T$
Now, note that $$\mathbf{e}~\mathbf{e}^T = \unicode{x1D7D9}$$ is a matrix whose components are all $$1$$, and that $$\mathbf{e}^T \mathbf{e} = \vert\operatorname{nz}(w)\vert$$ by construction. Thus:
\begin{aligned} S(z)^{-1} &= \frac{1}{1 + \eta \hat{y} z} I + \frac{\eta \hat{y} z}{(1 + \eta \hat{y} z)^2 - \eta \hat{y} z(1 + \eta \hat{y} z) |\operatorname{nz}(w)|} \unicode{x1D7D9} \\ &= I - \frac{\eta \hat{y} z}{1+\eta \hat{y} z} I + \frac{\eta \hat{y} z}{(1 + \eta \hat{y} z)^2 - \eta \hat{y} z(1 + \eta \hat{y} z) |\operatorname{nz}(w)|} \unicode{x1D7D9} \\ &= I - \frac{\eta \hat{y} z}{1+\eta \hat{y} z} \left[ I - \frac{1}{1+\eta \hat{y} z (1- |\operatorname{nz}(w)| )} \unicode{x1D7D9} \right] \end{aligned}
So the solution of the linear system $$S(z)V = V_t$$ is:
$V^*=S(z)^{-1} V_t = V_t - \frac{\eta \hat{y} z}{1+\eta \hat{y} z} \underbrace{ \left[ V_t - \frac{1}{1+\eta \hat{y} z (1- |\operatorname{nz}(w)| )} \unicode{x1D7D9} V_t \right]}_{(*)} \tag{C}$
Finally, we note that the matrix $$\unicode{x1D7D9} V_t$$ is the matrix obtained by computing the sum of the rows of $$V_t$$ and replicating the result $$\vert \operatorname{nz}(w)\vert$$ times, so we don’t even need to invoke any matrix multiplication function at all!
So, to summarize, we have Algorithm B above to compute $$q_2(z)$$, where the solution of the linear system is obtained via formula (C) above. Moreover, formula (C) is used to update the latent vectors once the optimal $$z$$ is found. Let’s implement the above:
# inside ProxPtFMTrainer
def step(self, w_nz, y_hat):
self.nnz = w_nz.numel() # |nz(w)|
self.bias_sum = self.bs[w_nz].sum().item() # w^T b_t
self.vs_nz = self.vs.weight[w_nz, :] # the matrix V_t
self.ones_times_vs_nnz = self.vs_nz.sum(dim=0, keepdim=True) # the sums of the rows of V_t
# TODO - this function will grow as we proceed
def q_two(self, y_hat, z):
if z == 0:
return 0
# solve the linear system - find the optimal vectors
vs_opt = self.solve_s_inv_system(y_hat, z)
# compute q_2
pairwise = (vs_opt.sum(dim=0).square().sum() - vs_opt.square().sum()) / 2 # the pow-of-sum - sum-of-pow trick
diff_squared = (vs_opt - self.vs_nz).square().sum()
return (-z * y_hat * pairwise + diff_squared / (2 * self.step_size)).item()
def update_vectors(self, w_nz, yhat, z): # use equation (C) to update the latent vectors
if z == 0:
return
self.vs.weight[w_nz, :].sub_(self.vectors_update_dir(yhat, z))
def solve_s_inv_system(self, y_hat, z):
return self.vs_nz - self.vectors_update_dir(y_hat, z)
def vectors_update_dir(self, y_hat, z): # marked with (*) in equation (C)
beta = self.step_size * y_hat * z
alpha = beta / (1 + beta)
return alpha * (self.vs_nz - self.ones_times_vs_nnz / (1 + beta * (1 - self.nnz)))
We need one last ingredient - a way to maximize $$q$$ and compute the optimal $$z$$.
## Maximizing $$q$$
Recall that
$q(z) = q_1(z) + q_2(z) - z\ln(z) - (1-z)\ln(1-z).$
Now, consider two important properties of $$q$$:
1. Its domain is the interval $$[0,1]$$
2. It is a strictly concave function:
1. $$q_1$$ and $$q_2$$ are both minima of linear functions of $$z$$, and thus concave.
2. $$-z\ln(z) - (1-z)\ln(1-z)$$ is strictly concave
So, if it has a maximizer, it must be unique, and must lie in the interval $$[0,1]$$. So, does it have a maximizer? Well, it does! Any concave function is continuous, and by the well-known Weirstrass theorem, any continuous function on a compact interval has a maximizer. What we have is a continuous function with a unique maximizer in a bounded interval, and that’s the classical setup for a well-known algorithm for one-dimensional maximization - the Golden Section Search method. For completeness, I copied the code from the above Wikipedia page:
"""Python program for golden section search. This implementation
reuses function evaluations, saving 1/2 of the evaluations per
iteration, and returns a bounding interval.
Source: https://en.wikipedia.org/wiki/Golden-section_search#Iterative_algorithm
"""
import math
invphi = (math.sqrt(5) - 1) / 2 # 1 / phi
invphi2 = (3 - math.sqrt(5)) / 2 # 1 / phi^2
def gss(f, a, b, tol=1e-8):
"""Golden-section search.
Given a function f with a single local minimum in
the interval [a,b], gss returns a subset interval
[c,d] that contains the minimum with d-c <= tol.
Example:
>>> f = lambda x: (x-2)**2
>>> a = 1
>>> b = 5
>>> tol = 1e-5
>>> (c,d) = gss(f, a, b, tol)
>>> print(c, d)
1.9999959837979107 2.0000050911830893
"""
(a, b) = (min(a, b), max(a, b))
h = b - a
if h <= tol:
return (a, b)
# Required steps to achieve tolerance
n = int(math.ceil(math.log(tol / h) / math.log(invphi)))
c = a + invphi2 * h
d = a + invphi * h
yc = f(c)
yd = f(d)
for k in range(n-1):
if yc < yd:
b = d
d = c
yd = yc
h = invphi * h
c = a + invphi2 * h
yc = f(c)
else:
a = c
c = d
yc = yd
h = invphi * h
d = a + invphi * h
yd = f(d)
if yc < yd:
return (a, d)
else:
return (c, b)
Having all ingredients, we can finalize the implementation of the optimizer’s step method:
def neg_entr(z):
if z > 0:
return z * math.log(z)
else:
return 0
def loss_conjugate(z):
return neg_entr(z) + neg_entr(1 - z)
class ProxPtFMTrainer:
def step(self, w_nz, y_hat):
self.nnz = w_nz.numel()
self.bias_sum = self.bs[w_nz].sum().item()
self.vs_nz = self.vs.weight[w_nz, :]
self.ones_times_vs_nnz = self.vs_nz.sum(dim=0, keepdim=True)
def q_neg(z): # neg. of the maximization objective - since the min_gss code minimizes functions.
return -(self.q_one(y_hat, z) + self.q_two(y_hat, z) - loss_conjugate(z))
opt_interval = min_gss(q_neg, 0, 1)
z_opt = sum(opt_interval) / 2
self.update_biases(w_nz, y_hat, z_opt)
self.update_vectors(w_nz, y_hat, z_opt)
# Testing our training algorithm
Since the purpose of this pose is “academic” in nature, i.e. to show the limits of what is possible by proximal point rather than writing a production-ready training algorithm, we did not take the time to make it efficient, and thus we’ll test it on a toy dataset - MovieLens 100k. The dataset consists of the ratings on a 1 to 5 scale that users gave to 1682 movies. For users, we use their integer age, gender, and occupation as features. For the movies, we use the genre and the movie id as features. A rating $$\geq 5$$ is considered positive, while below 5 are considered negative.
For clarity, in the post itself we’ll skip the data loading code, and assume that the features are already given in the W_train tensor, whose rows are the vectors $$w_i$$, and the corresponding labels are given in the y_train tensor. The full code is available in simple_train_loop.py file in the repo. Let’s train our model using the maximal allowed step-size for ten epochs, using a factorization machine of embedding dimension $$k=20$$:
from tqdm import tqdm
import torch
# MISSING - the code which loads the dataset and builds the tensors W_train and y_train
num_features = W_train.size(1)
max_nnz = W_train.sum(dim=1).max().item()
step_size = 1. / (2*max_nnz + 1)
print(f'Training with step_size={step_size:.4} computed using max_nnz = {max_nnz}')
embedding_dim = 20
fm = FM(num_features, embedding_dim)
dataset = TensorDataset(W_train, y_train)
trainer = ProxPtFMTrainer(fm, step_size)
for epoch in range(10):
sum_epoch_loss = 0.
sum_pred = 0.
sum_label = 0.
desc = f'Epoch = {epoch}, loss = 0, pred = 0, label = 0, bias = 0'
with tqdm(DataLoader(dataset, batch_size=1, shuffle=True), desc=desc) as pbar:
def report_progress(idx):
avg_epoch_loss = sum_epoch_loss / (idx + 1)
avg_pred = sum_pred / (idx + 1)
avg_label = sum_label / (idx + 1)
desc = f'Epoch = {epoch:}, loss = {avg_epoch_loss:.4}, pred = {avg_pred:.4}, ' \
f'label = {avg_label:.4}, bias = {fm.bias.item():.4}'
pbar.set_description(desc)
for i, (x_sample, y_sample) in enumerate(pbar):
(ignore, w_nz) = torch.nonzero(x_sample, as_tuple=True)
y = y_sample.squeeze(1)
# aggregate loss and prediction per epoch, so that we can monitor convergence
pred = fm.forward(w_nz.unsqueeze(0))
loss = F.binary_cross_entropy_with_logits(pred, y)
sum_epoch_loss += loss.item()
sum_pred += torch.sigmoid(pred).item()
sum_label += y.item()
# train the model
y_hat = (2 * y.item() - 1) # transform 0/1 labels into -1/1
trainer.step(w_nz, y_hat)
if (i > 0) and (i % 2000 == 0):
report_progress(i)
report_progress(i)
That’s what I got:
Training with step_size=0.04348 computed using max_nnz = 11.0
Epoch = 0, loss = 0.4695, pred = 0.2118, label = 0.2124, bias = -1.148: 100%|██████████| 99831/99831 [11:36<00:00, 143.37it/s]
Epoch = 1, loss = 0.4362, pred = 0.2114, label = 0.2121, bias = -1.468: 100%|██████████| 99831/99831 [11:34<00:00, 143.80it/s]
Epoch = 2, loss = 0.427, pred = 0.2115, label = 0.2122, bias = -1.294: 100%|██████████| 99831/99831 [11:20<00:00, 146.62it/s]
Epoch = 3, loss = 0.4224, pred = 0.2117, label = 0.2123, bias = -1.254: 100%|██████████| 99831/99831 [10:30<00:00, 158.33it/s]
Epoch = 4, loss = 0.4194, pred = 0.2114, label = 0.212, bias = -1.419: 100%|██████████| 99831/99831 [10:00<00:00, 166.12it/s]
Epoch = 5, loss = 0.4173, pred = 0.2112, label = 0.2117, bias = -1.301: 100%|██████████| 99831/99831 [09:48<00:00, 169.73it/s]
Epoch = 6, loss = 0.4167, pred = 0.2117, label = 0.2121, bias = -1.368: 100%|██████████| 99831/99831 [09:49<00:00, 169.40it/s]
Epoch = 7, loss = 0.4155, pred = 0.2115, label = 0.2119, bias = -1.467: 100%|██████████| 99831/99831 [09:51<00:00, 168.81it/s]
Epoch = 8, loss = 0.4145, pred = 0.2114, label = 0.2118, bias = -1.605: 100%|██████████| 99831/99831 [09:47<00:00, 169.81it/s]
Epoch = 9, loss = 0.4146, pred = 0.2121, label = 0.2125, bias = -1.365: 100%|██████████| 99831/99831 [09:47<00:00, 169.85it/s]
Seems that the loss is indeed being minimized. Let’s compare it with the Adam optimizer with default parameters. Here is the training loop:
optimizer = torch.optim.Adam(fm.parameters())
for epoch in range(10):
sum_epoch_loss = 0.
sum_pred = 0.
sum_label = 0.
desc = f'Epoch = {epoch}, loss = 0, pred = 0, label = 0, bias = 0'
with tqdm(DataLoader(dataset, batch_size=1, shuffle=True), desc=desc) as pbar:
def update_progress(idx):
avg_epoch_loss = sum_epoch_loss / (idx + 1)
avg_pred = sum_pred / (idx + 1)
avg_label = sum_label / (idx + 1)
desc = f'Epoch = {epoch}, loss = {avg_epoch_loss:.4}, pred = {avg_pred:.4}, ' \
f'label = {avg_label:.4}, bias = {fm.bias.item():.4}'
pbar.set_description(desc)
for i, (x_sample, y_sample) in enumerate(pbar):
(ignore, w_nz) = torch.nonzero(x_sample, as_tuple=True)
y = y_sample.squeeze(1)
pred = fm.forward(w_nz.unsqueeze(0))
loss = F.binary_cross_entropy_with_logits(pred, y)
loss.backward()
optimizer.step()
sum_epoch_loss += loss.item()
sum_pred += torch.sigmoid(pred).item()
sum_label += y.item()
if (i > 0) and (i % 2000 == 0):
update_progress(i)
update_progress(i)
And here is the result:
Epoch = 0, loss = 0.4655, pred = 0.21, label = 0.212, bias = 0.539: 100%|██████████| 99831/99831 [02:47<00:00, 596.25it/s]
Epoch = 1, loss = 0.4596, pred = 0.208, label = 0.212, bias = 1.586: 100%|██████████| 99831/99831 [03:09<00:00, 527.90it/s]
Epoch = 2, loss = 0.4655, pred = 0.2075, label = 0.2118, bias = 2.668: 100%|██████████| 99831/99831 [02:59<00:00, 556.33it/s]
Epoch = 3, loss = 0.471, pred = 0.2078, label = 0.2122, bias = 3.805: 100%|██████████| 99831/99831 [02:50<00:00, 585.09it/s]
Epoch = 4, loss = 0.4744, pred = 0.2071, label = 0.2119, bias = 5.116: 100%|██████████| 99831/99831 [02:42<00:00, 615.88it/s]
Epoch = 5, loss = 0.4747, pred = 0.2071, label = 0.212, bias = 6.48: 100%|██████████| 99831/99831 [02:55<00:00, 569.75it/s]
Epoch = 6, loss = 0.4777, pred = 0.2064, label = 0.2119, bias = 7.992: 100%|██████████| 99831/99831 [02:56<00:00, 567.10it/s]
Epoch = 7, loss = 0.4793, pred = 0.2071, label = 0.2121, bias = 9.433: 100%|██████████| 99831/99831 [02:47<00:00, 595.92it/s]
Epoch = 8, loss = 0.4802, pred = 0.2062, label = 0.212, bias = 11.15: 100%|██████████| 99831/99831 [02:43<00:00, 610.91it/s]
Epoch = 9, loss = 0.4824, pred = 0.2066, label = 0.212, bias = 12.72: 100%|██████████| 99831/99831 [02:44<00:00, 605.32it/s]
Whoa! It isn’t converging! The loss grows after a few epochs, and we can see that the bias keeps increasing. Seems like our efforts are paying off - a custom method with a deeper step-size analysis let us just ‘hit’ a good-enough step-size without any tuning, while with Adam we’ll probably have to do some tuning to find a good step-size.
Let’s now do a more thorough stability comparison - run our method, Adam, Adagrad, and SGD, with various step-size parameters for ten epochs, and see what loss we are getting. The above methods ran with several step sizes for $$M=20$$ epochs, each step-size was tested $$N=20$$ times to take into account the effect of randomness in the weight initialization and the data shuffling. Then, I produced a plot showing the best loss achieved for each step-size and each algorithm, averaged over the $$N=15$$ attempts, with transparent uncertainty bands. The code resides in stability_experiment.py in the repo. Here is the result:
It’s quite apparent that the performance of the proximal point algorithm is quite consistent over the various step-size choices. We also see that Adam’s performance degrades when the step-size is too large. Consequently, to see the difference between the various algorithms more clearly, let’s plot the results without Adam:
Well, as we see, the proximal point’s performance is the most consistent accross various step-sizes, but it is certainly not the best algorithm for training a factorization machine on this dataset. It appears that Adagrad is.
One possible explanation is that the proximal point algorithm converges more slowly, and requires more epochs to achieve good performance. Let’s test this hypothesis, and run the proximal point algorithm for 50 epochs. And after a few days, I got:
The situation doesn’t seem to improve much. The method is quite consistent in its performance, but it doesn’t seem to converge rapidly to an optimum.
# Summary
We have developed an efficiently implementable proximal point step for a highly non-trivial and non-convex problem, and provided an implementation. To the best of my knowledge, this post sets foot in an uncharted territory, and thus I an not sure what is the method converging to, but from these numerical experiments it doesn’t seem to minimize the average loss. It is my hope that the research community can provide such answers.
Writing this entire series about efficient implementation of incremental proximal point methods has been extremely fun and I certainly learned a lot about Python, PyTorch, and better understood the essence of these methods. I hope that you, the readers, enjoyed as much as I did. It’s time for new adventures! I don’t know what the next post will be about, but I’m sure it will be fun!
1. Steffen Rendle (2010), Factorization Machines, IEEE International Conference on Data Mining (pp. 995-1000)
2. It follows from the fact that $$h(t)=\ln(1+\exp(t))$$ and $$h^*(s)=s\ln(s)+(1-s)\ln(1-s)$$ are both Legendre-type functions: essencially smooth and strictly-convex function.
3. seperability is the fact that $$\displaystyle \min_{x_1,x_2} f(x_1) + g(x_2) = \min_{x_1} f(x_1) + \min_{x_2} g(x_2)$$.
|
2023-03-24 13:40:36
|
{"extraction_info": {"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": 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, "math_score": 0.9389976859092712, "perplexity": 1586.8765215734968}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945282.33/warc/CC-MAIN-20230324113500-20230324143500-00664.warc.gz"}
|
https://forum.endeavouros.com/t/why-cant-i-have-a-wysiwyg-text-editor-with-full-text-formatting-capabilities-in-a-terminal/18875
|
# Why can't I have a wysiwyg text editor with full text formatting capabilities in a terminal?
Hi.
It ain’t really related to Endeavour beside our common interest in terminals, so I ask here… I just can’t bear the pure arch guys.
Well I wonder for years already: there are plenty of special characters and standards in UNICODE, and easily extensible ones at that, allowing for about any character or text or paragraph styling one can imagine, while requiring none of the non-portable, always-annoying BB code, html, css, odt formatting, markup language or whatever we use to just to make a text visually prettier and clearer… or semantically too.
Wouldn’t that allow for more portable across both systems and types of interface (the shit I do in word lies well within a terminal capability !)… and save file sizes and processing ressource ?
Instead of putting friggin’ thousands of mostly useless emojis, uncanny symbols and dead languages in UNICODE, why don’t they implement a common global multimedia text/image SGML-like language for both style and structure ?
Instead of having a gazillion others standards and formats.
The solution to your problem might be to not use a wyswyg in the terminal. But rather something like latex. Its cross platform compatible and think it can be compiled in terminal for example via vim.
1 Like
It makes as little a sense as any other markup language, is hardly portable given the infinite number of variants, and its learning curve is steep as f***.
I don’t do fancy things like math equation (though these would profit just as well), just regular paragraphs and text formatting… so Tex is wildly unnecessary here
.
And why do complicated when you can do simple ? Things like libreoffice (again, for “simple” things) work just fine… I like TUIs for their clarity, not to out-nerd myself
? Latex is ONE language.
Try overleaf, its just writing text in a webbrowser.
Edit:
if you want to write text you only need to learn this:
\documentclass{article}
\begin{document}
First document. This is a simple example, with no
extra parameters or packages included.
\end{document}
And the command \section{section name here} and \subsection{section name here}
There you go, you just learned latex in 1 min
1 Like
I personally use pandoc.
I found a decent theme on a repo online. Been using it for my documents. I write in markdown. It is converted to latex, then to pdf. So typing it out is easy, and the final pdf looks as professional as a proper latex document.
1 Like
I also use pandoc to write gitbooks, but latex conversion is tricky if you have figures and print out pdf look funky sometimes with the formatting. Online everything looks great.
1 Like
Yeah agree. Faced the issue initially, but whatever theme I’m using right now gives mostly sane output. Anytime I have to deal with manual positioning of multiple images, I just switch to a word processor. Haven’t extensively used any online service yet. I’m not working much with images though.
I use pandoc in R studio, there is a fantastic tutorial from cjvanlissa.github.io/gitbook-demo
1 Like
Another variant could be to use Lyx, this is not in the terminal but is a gui word type editor but produce tex format and therefore crossplatform compatible.
https://www.lyx.org/
If anyone has experience with that would be good to know. I tried a couple of years back but then flipped back to plain latex quickly.
Ok. Thought you were asking also for some formatting. I thought about latex because you can easily create a document out of the box by choosing template and focus on writing. Its also possible to write the text in the terminal and there are table for symbol codes if need.
If you just use text you only need to know how to add a section or subsection title and perhaps how to make a bullet list. Anyhow, I use texstudio for what its worth and it has some buttons to make things bold etc. To my experience tex is the most crossplatform compatible because the language doesnt change, and you get always the same result, no matter if linux, win or mac.
I’ll give a shot at Lyx.
But the fact there are good solutions today, doesn’t explain UNICODE based formatting isn’t used at all while they could
It’s a fact terminals would be easily nicer with UNCODE-rich text without the need to process a markup language in the shell.
I don’t have any need for unicode support in a terminal editor personally so I haven’t tried it but doesn’t micro have unicode support?
Not sure if this is what you meant, but I seached with yay:
yay -Ss unicode editor
and there were 3 alternatives. Haven’t used any of them, but maybe they’re worth checking.
Well tried
aur/yudit 3.0.7-1 (+17 0.00) : Unicode plain-text editor so no wysiwyg. I could just as well enter unicode points in micro.
qed: LINE editor. wtf ?! why did people ever user that, is beyond me.
texworks: LaTex, not terminal, and closer to Lyx.
it should be kind of like Wordpad on Windows. a wysiwying Rich text editor, in terminal because why not, since most are (and should be) unicode compatible.
I’d be surprised if emacs or vim would not have unicode support.
I don’t think the lack of unicode support is the issue, it is the lack of WYSIWIG support for unicode
Bingo, someone got it
1 Like
To be precise what I had in mind was not UNICODE characters per se but the ANSI escape sequences using them to implement rich texts in terminals, and that the Ncurses library uses. The colors, underscore, fraktur etc, all of that could make up for a good lightweight yet very extensible (SIXREL, emojis, vector graphics, image rendering, etc etc…).
I dream of a computer interface where terminals and graphics would be one and the same. Where standards and techs would be simpler but much nicer to look at…
I’m close to have a boner when I watch that:
:
I probably am not the only one here
mmm, let’s call that a “nerdgasm”.
Anyone has an idea of (a project of) an interface which would provide a similar, advanced “terminal-centric” experience, without the user having to write binary code in two dozen config files ?
I suppose this is an extreme response to a simple question, but as soon as “text editor” (those two specific words) are mentioned, they carry with them implications that pre-date the “PC”.
Editing text is exactly that.
When you add code sets, format marks, bold, italics, etc., I think of word processing, text formatting, office automation and other variations.
Perhaps a different term exists (or is needed) to describe what you want, but I object to calling it a text editor. Emacs may be able to perform what you want to do; I guarantee that it could be extended in that manner if the desired features are not already present. As it is, Emacs in recent form bears little resemblance to the original tool, though classic methods may still be there. But I digress; let’s come up with a specific name for the capability you seek; it’s a legitimate and worthwhile idea.
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2022-05-18 23:08:12
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|
https://www.educative.io/answers/how-to-calculate-percentiles-in-numpy
|
Related Tags
numpy
python
communitycreator
# How to calculate percentiles in NumPy
Harsh Jain
In this shot, we will learn how to calculate percentiles with NumPy.
A percentile is defined as a score at or below which a given percentage falls. For example, the $27^{th}$ percentile is the score below which 27% of the scores will be found.
In other words, let’s say you score in the 99th percentile in a certain exam; this means you are above 99% of the people taking the exam.
We can use the numpy.percentile() function to calculate percentiles in Python.
The numpy.percentile() function is used to calculate the $n^{th}$ percentile of the given data (array) along the specified axis.
### Syntax
The syntax of the numpy.percentile() function is shown below.
numpy.percentile(array, percentile, axis=None, out=None, overwrite_input=False, keepdims=False)
### Parameters
The numpy.percentile() function accepts the following parameters:
• array: The source array whose percentile needs to be computed.
• percentile: Signifies the percentile that needs to be computed.
• axis (optional): Defines the axis along which the percentile is calculated. By default, a flattened array is used.
• out (optional): An alternate output array where we can place the result.
• overwrite_input (optional): Can be used to modify the input array.
• keepdims (optional): Creates reduced axes with dimensions of one size.
### Return value
The numpy.percentile() function returns a scalar or array with percentile values along the specified axis.
### Code
Let’s look at the code.
# Using 1-D array
import numpy as np
# Array of data
arr = [5,6,9,87,2,3,5,7,2,6,5,2,3,4,69,4]
# Finding the 90 percentile
x = np.percentile(arr, 90)
print(x)
Calculate percentiles in NumPy
### Explanation
• In line 2, we import the numpy library with alias np.
• In line 5, we create an array of data.
• In line 8, we use the np.percentile() function to find the $90^{th}$ percentile from the given dataset.
The code above deals with a 1-D array. Now, we will explore a 2-D array.
#using 2-D array
import numpy as np
# Array of data
arr = [[5,6,8],[6,9,2]]
# Finding the 90 percentile
x = np.percentile(arr, 90)
print(x)
Calculate the percentile of a 2-D array in NumPy
### Explanation
• The code above is exactly the same as the previous example. The only difference is that, now, we create a 2-D array instead of a 1-D array.
This is how we can calculate percentiles in Python with the NumPy library.
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2022-08-16 14:07:44
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https://www.gamedev.net/forums/topic/397323-getting-strings-with-s-and-s/
|
# Getting strings with /'s and .'s
This topic is 4235 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
For some reason I'm unable to get a string from a file if it contains slashes and periods. I can load this: /this/is/a/string Or this: this.is.a.string But not this: /this/is/a.string When I do I get a memory segmentation fault error. (Code 11). I got that once when I tried closing a file stream that wasn't actually open yet. I'm not sure how the two are related besides using the Ifstreams and ofstreams. What could be going wrong? Thanks for any help! :)
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Can we see the code that is causing the problem?
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most certainly there is an error with your test code. the underlying mechanisms responsible for file IO aren't to blame here. please post some code that you're trying to read the strings with. especially make sure the buffers you're reading into are included in the snippet.
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Sorry it took me so long to get back. Here's the main file:
Quote:
#include #include "flx_fileIO.h"using namespace std;int myName = 35;int myName2 = 468;char myString[15];char myStringIn[127];int main (int argc, char * const argv[]) { FLX_purgeFileArray(); cout << "Purged array." << endl; FLX_loadFile("file.txt", "r", myName); cout << "Loaded file." << endl; FLX_getString(myName, myString); cout << "Retrieved string." << endl; FLX_closeFile(myName); cout << "Closed file." << endl; cout << myString << endl; /* FLX_loadFile("file.txt", "w", myName2); cout << "Input text: " << endl; cin >> myStringIn; FLX_writeInt(strlen(myStringIn), myName2); FLX_writeString(myStringIn, myName2); FLX_closeFile(myName2); */ return 0;}
The FLX functions, which you will probably not be familiar with are part of FLX, a library I am making to simplify the stratedgy game development process. My output gets all the way to "Loaded File", which means it's the get string function that's tripping up.
Here's the body of the FLX_getString function.
Quote:
int FLX_getString(int flx_gs_name, char flx_gs_userWordSlot[]) { // If this name is invalid. int flx_gs_thisSlot = FLX_locateFileSlot(flx_gs_name); if (flx_gs_thisSlot == -1) { return(-1); } // If this name is for an empty or write slot. if (flx_fileStreams[flx_gs_thisSlot].type[0] == ("e" || "w")) { return(-1); } // And if we're ok... flx_fileStreams[flx_gs_thisSlot].ifs >> flx_gs_userWordSlot; return(0);}
I should probably explain how FLX's file manager works. It uses array's for everything, and lets the user give these slots names. The use is able to manipulate that file slot with the name code. It's just like standard c++ i/o, I added these functions as an exercise to improve my file in/out understanding and to make FLX a more "all-inclusive" package.
I see nothing that could be having trouble here... I'll try using normal c i/o stuff.
Thanks!
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Well, you are certainly not simplifying anything by using C constructs. Instead of character arrays, use std::string. Instead of what I can only assume is a raw array, use std::vector/std::list (or, considering you want named "file slots," use std::map).
I am certain that you are overwriting an array boundary, whether it is in one of your strings, or in your "file manager," or somewhere else. I cannot say without the relevant code.
jfl.
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Quote:
Original post by jflangloisWell, you are certainly not simplifying anything by using C constructs. Instead of character arrays, use std::string. Instead of what I can only assume is a raw array, use std::vector/std::list (or, considering you want named "file slots," use std::map).I am certain that you are overwriting an array boundary, whether it is in one of your strings, or in your "file manager," or somewhere else. I cannot say without the relevant code.jfl.
Oh please no, I have no desire to go through all the code from my file manipulation and changing it all to more of c++'s strange formats. I use struct's because they allow me to decide what properties each file has, rather than having slots in an array for every aspect, which would be annoying and require changing loads of code which I do not wish to go through again.
I've tried using cin to get characters to arrays and then output with /'s and periods. I then did so with file in manipulators. All my function does is manage an array of structs doing the exact same thing. I'm guessing somehow structs are not allowed to contain .'s and /'s, although that seems unlikely, I'm pretty confident that my code is not causing this.
Oh wait, I just realized. Damn I'm dumb smurf... my array was too small. :(
Really sorry for this! *embarrassed* Yeah... I feel reeaallly stupid right now.
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Quote:
Original post by OneMoreToGoOh please no, I have no desire to go through all the code from my file manipulation and changing it all to more of c++'s strange formats. I use struct's because they allow me to decide what properties each file has, rather than having slots in an array for every aspect, which would be annoying and require changing loads of code which I do not wish to go through again.
They're not "strange formats". (Besides which, you're already using the iostream library just fine.) They're carefully designed classes and functions created by very smart people *specifically to avoid problems like*
Quote:
Oh wait, I just realized. Damn I'm dumb smurf... my array was too small. :(Really sorry for this! *embarrassed* Yeah... I feel reeaallly stupid right now.
Heck, the fact that I see this in your code:
if (flx_fileStreams[flx_gs_thisSlot].type[0] == ("e" || "w")) {
tells me you're not qualified to reinvent the wheel. (Hint: it doesn't do what you think it does. "e" || "w" simplifies to 1, and anyway, assuming that 'type' is a char**, you can't just compare strings like that: the comparison would compare the char* pointer values, not the pointed-at data.)
I'll make you an offer you can't refuse: PM me the whole flx_fileIO module (or instructions for getting it - just how big is that anyway o_O), and I'll rewrite it for you free, over the weekend, to the best of my ability.
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Quote:
Original post by Zahlman
Quote:
Original post by OneMoreToGoOh please no, I have no desire to go through all the code from my file manipulation and changing it all to more of c++'s strange formats. I use struct's because they allow me to decide what properties each file has, rather than having slots in an array for every aspect, which would be annoying and require changing loads of code which I do not wish to go through again.
They're not "strange formats". (Besides which, you're already using the iostream library just fine.) They're carefully designed classes and functions created by very smart people *specifically to avoid problems like*
Quote:
Oh wait, I just realized. Damn I'm dumb smurf... my array was too small. :(Really sorry for this! *embarrassed* Yeah... I feel reeaallly stupid right now.
Heck, the fact that I see this in your code:
if (flx_fileStreams[flx_gs_thisSlot].type[0] == ("e" || "w")) {
tells me you're not qualified to reinvent the wheel. (Hint: it doesn't do what you think it does. "e" || "w" simplifies to 1, and anyway, assuming that 'type' is a char**, you can't just compare strings like that: the comparison would compare the char* pointer values, not the pointed-at data.)
I'll make you an offer you can't refuse: PM me the whole flx_fileIO module (or instructions for getting it - just how big is that anyway o_O), and I'll rewrite it for you free, over the weekend, to the best of my ability.
Type, is a char[0], not a pointer. It is not taken as an argument to a function, but is accessed directly from multiple different functions, and in no way by the user. If they try to I'll slap them. ;) I'd love to take up your offer though. And believe me, you'd get a large credit in the end, that's for sure. :)
I'll PM you a link to the code.
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btw,
"e" != 'e'
one is a string constant ("e") and the other is a character literal ('e'). One has an ascii value of something like 101 (that's the 'e'), and the other has some 32-bit address (I'm assuming a 32-bit system)
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BTW,
Before you read any of the code I send back to you, try to write any more code, or in any way represent that you know anything about C or C++, you need to read this and understand it in full.
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Quote:
Original post by wyrzybtw,"e" != 'e'one is a string constant ("e") and the other is a character literal ('e'). One has an ascii value of something like 101 (that's the 'e'), and the other has some 32-bit address (I'm assuming a 32-bit system)
I am not comparing those two. As long as the user supplies "e", all my code will check for "e". 'e' is not involved. I was unaware of the fact that you could not use nested or statements though;
if (whatever == ("a" || "q") -> bad?
Thanks!
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Quote:
Original post by OneMoreToGoif (whatever == ("a" || "q") -> bad?
Indeed. This sentence means "if whatever is equal to ("a" || "q")", not "if whatever is equal to "a" or "q"". Since "a" is not NULL and "q" is not NULL, ("a" || "q") is really (!NULL || !NULL) ie (true || true) ie true. So far
whatever == ("a" || "q") ==> whatever == true
To enforce what Zahlman said (he doesn't know it but he is the high priest of my own personnal religion), std::string and other "c++'s strange formats" have been created to help you. The C and C++ languages are hard enough, you don't have to lose your time to try to replicate something that already exists in a well tested, standard library.
And std::string is really the roxors :)
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I shall attempt to keep that in mind, however, would it take a lot of adaptation to install the string method? I'll try to use in future code, but for the moment, I'll just convert all my "e" to 'e', a temporary fix for some code which is comparatively useless (when looked at besides c++'s original file i/o streams). Oh well. :)
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Quote:
Original post by OneMoreToGoI shall attempt to keep that in mind, however, would it take a lot of adaptation to install the string method? I'll try to use in future code, but for the moment, I'll just convert all my "e" to 'e', a temporary fix for some code which is comparatively useless (when looked at besides c++'s original file i/o streams). Oh well. :)
The time it takes to learn and use std::string and other Standard Library constructs will be much less than the time wasted debugging problems like the ones you are having now.
jfl.
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Quote:
Original post by OneMoreToGoI shall attempt to keep that in mind, however, would it take a lot of adaptation to install the string method? I'll try to use in future code, but for the moment, I'll just convert all my "e" to 'e', a temporary fix for some code which is comparatively useless (when looked at besides c++'s original file i/o streams). Oh well. :)
Quote:
/Users/gareth/Desktop/flx testing/FLX_Test_IO/flx_fileIO.h:95: error: ISO C++ forbids comparison between pointer and integer
*sigh* Point taken. I'll google it.
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Did you miss the part where we explained the logical error? There is no set of types (unless you implemented one yourself, and it would be probably roundly considered a very bad idea even if it actually worked according to spec) you can use such that OR-ing two values, and then comparing for equality to the result, is equivalent to checking whether you have one of the two values.
You don't know what you're doing, and you need to take several steps back and learn things properly. Including the standard library (not just the streams) this time. It's called the standard library for a reason.
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Quote:
Original post by ZahlmanDid you miss the part where we explained the logical error? There is no set of types (unless you implemented one yourself, and it would be probably roundly considered a very bad idea even if it actually worked according to spec) you can use such that OR-ing two values, and then comparing for equality to the result, is equivalent to checking whether you have one of the two values.You don't know what you're doing, and you need to take several steps back and learn things properly. Including the standard library (not just the streams) this time. It's called the standard library for a reason.
The error was meant as a bit of humor, as in "You know what? I'm sick of this, I give in. I'll do it your way".
Is that what you were reffering to? The problem I see, is that I have always in the past used double-quotes around my letters that were always in char format. It's the way I was taught. Now a couple of years later, I'm going through SDL learning new things on a basis that appears to be wrong. I have a book on C++, and although it never mentioned the std:: operator (neither did the course I took, I swear! And everything worked too!) it seems fine in all other regards, so I'll go back and review arrays & pointers. I've googled a bit about string operators, and they seem useful. Do they comply with standard sizeof() & strlen() commands? I'll just test that myself though, so no worries.
Thanks for all your help! Hopefully I won't be back with to many bothersome problems for a while. (Yeah right.)
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Quote:
Original post by OneMoreToGoI have a book on C++, and although it never mentioned the std:: operator (neither did the course I took, I swear! And everything worked too!) it seems fine in all other regards, so I'll go back and review arrays & pointers. I've googled a bit about string operators, and they seem useful. Do they comply with standard sizeof() & strlen() commands? I'll just test that myself though, so no worries.
To be able to use things correctly, you have to know what they are. Talking vaguely about "operators" and "commands" probably has a lot to do with why you're confused. Also, your C++ book is evidently several years out of date and needs to be thrown away, and most C++ programming courses out there that I've heard about are pretty much garbage as well. Sorry, but that's the state of things.
std:: is a namespace. Namespaces are a way of decorating identifiers (that's a catch-all term that includes variable names, function names and class/struct names - and probably other stuff I forgot) so that instead of using "foo_" prefixes to avoid name collisions, you use "foo::". But it's actually much better than that, because (a) they're a language construct that the compiler can actually do some checking with, and (b) you can set things up such that certain blocks of code omit the "foo::" for convenience, when it is not needed to avoid collisions. For example, suppose you made a "2d vector" class (of the sort that is reinvented a gazillion times by everyone; basically it just represents a pair of X and Y coordinates), and called it 'vector', and wanted to store a std::vector of them. In the code using that vector, you'd have to write "std::vector<vector>", in order to make things clear to the compiler. (Perhaps you would be smart enough to put your vector class in its own namespace, too; then you'd qualify that name as well.) But in some other source file of the same project that *didn't* use the local vector class but still used the standard library vector container, you could get away with "using std::vector;" at the top and "vector<int>" everywhere else.
'std::string' is a class. Thus you can declare variables of type std::string; each of these might be called "a std::string instance" or "a std::string object". It represents textual information. In C++, classes and structs can provide "member functions", as well as operator overloads . When you write "cout << foo;", 'cout' is a global object of type std::ostream, and you are invoking the overloaded operator<<(std::ostream&, T) for T = whatever type 'foo' happens to have. This is the same sort of process as when you write "foo + bar" and it invokes operator+(int, int) (except that there isn't really such a function in a library somewhere; the compiler just directly emits the appropriate machine code) if foo and bar are both ints, or operator+(const std::string&, const std::string&) if foo and bar are both std::strings. Note that the operator+ in question yields another std::string object which represents the concatenation of those two input strings. You can't directly, or even with a relatively small amount of work, do that properly with char*'s.
A single-quoted string in your program, like 'a' or '\n', is a "character literal". That is to say, a single byte value. (Back when all this stuff was designed, "characters" and "bytes" were considered equivalent. Now that we have Unicode, huge monkey wrenches are thrown in to that system.) It's a value that could be assigned to a variable of type char, which is an integral numeric type. When output, the numeric value is translated into a character via the ASCII table.
A double-quoted string in your program is a "string literal". What happens when you write that is that the compiler embeds the contained sequence of characters into the executable's data segment, and puts a zero-valued character ('\0') at the end as well. Then, in an expression using the string literal, that literal becomes a value of type char* (pointer to char), which points at (has a value which is the memory address of) the beginning of that sequence of characters. So a string literal is a completely different thing from a character literal, even if the string happens to be only "one character long" (note that actually two bytes are used in the static storage, because of the aforementioned 'null terminator').
sizeof is an operator. It yields, *at compile-time*, the number of bytes that are needed to represent the provided variable or type, using the *static* type.
Note that a character *array* is different from a character *pointer* - in a few ways. You need to read this. Example:
#include <iostream>using namespace std;int main() { char foo[16]; char* bar = foo; // that causes bar to point to the beginning of the foo storage cout << sizeof(foo) << ' ' << sizeof(char[16]) << ' ' << sizeof(bar) << ' ' << sizeof(char*) << endl;}
You *can* use sizeof() on any variable or type, but it is rarely something that a beginning programmer ought to find useful.
strlen() is a function. It accepts a specific argument type (const char*), so that's all you can pass to it. It's intended for figuring out the length of a "string" represented by a char*, in the way that string literals represent things. It does this by checking character values, starting at the provided location and advancing forward through memory, until it finds a zero byte (such as the one automatically inserted at the end of a string literal).
The std::string object does provide a way to extract a const char* representing its internal buffer: its .c_str() member function. This allows you to pass the data to some C APIs. However, you would not want to do it with strlen(), for at least three reasons:
1) It would be terminally retarded to do so, because the std::string object already provides a function that does that directly: its .length() member function.
2) You would lose efficiency: the std::string object already includes a length count, which is only updated when something happens that might change the string length, so all its .length() member does is return the value there. By contrast, calling strlen() requires searching through memory over the entire length of the string - every time you call it, because the information didn't get remembered anywhere.
3) You could get a wrong answer, because std::strings are capable of holding zero bytes within the "real" string data, whereas char* strings basically aren't - if you tried to make a string "hello\0world", then strlen() would stop at the *first* \0 it saw, and report a length of 5, as if all you had was "hello".
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Wow... thanks for explaining that in such detail. You might as well have written a book for me. ;)
Well, as much as I'm a cheap guy, it looks like time to get a new reference.
On a completely unrelated note:
Wow, I just tried Code::Blocks IDE. Amazing. I love it. Not only is it free, but it has built in templates for OpenGL, OGRE and SDL (and others). A nice interface, support for multiple compilers... and did I mention it was free? ;)
Once again, Thanks!
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Quote:
Original post by OneMoreToGoWow... thanks for explaining that in such detail. You might as well have written a book for me. ;)Well, as much as I'm a cheap guy, it looks like time to get a new reference.On a completely unrelated note:Wow, I just tried Code::Blocks IDE. Amazing. I love it. Not only is it free, but it has built in templates for OpenGL, OGRE and SDL (and others). A nice interface, support for multiple compilers... and did I mention it was free? ;)Once again, Thanks!
I really enjoyed Code::Blocks while I used it. It was easy to set up, and had tons of features like a rusimentary intellisense and a decent debugger. I would also look at the Microsoft Visual C++ Express edition. It definately is a bit bett with debugging and such. However, since you already are using Code::Blocks, you might as well stick with it.
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2018-01-17 08:04:58
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https://math.stackexchange.com/questions/2264000/upper-triangular-and-diagonal
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# upper triangular and diagonal
If $A$ is real, upper triangular $n\times n$ matrix such that $AA^t=A^tA$. Then show that $A$ is diagonal.
We know that every upper triangular, symmetric matrix is diagonal. But I have problem to show that $A=A^t$ from the given condition. How would we show that?
Let $A = \begin{bmatrix}v_1 & v_2 & \cdots & v_n\end{bmatrix}$ and $A^T = \begin{bmatrix}h_1 & h_2 & \cdots & h_n\end{bmatrix}$.
Then, $AA^T = \begin{bmatrix} h_1 \cdot h_1 & h_1 \cdot h_2 & \cdots & h_1 \cdot h_n\\ h_2 \cdot h_1 & h_2 \cdot h_2 & \cdots & h_2 \cdot h_n\\ \vdots & \vdots & & \vdots\\ h_n \cdot h_1 & h_n \cdot h_2 & \cdots & h_n \cdot v_n\\ \end{bmatrix}$ and $A^TA = \begin{bmatrix} v_1 \cdot v_1 & v_1 \cdot v_2 & \cdots & v_1 \cdot v_n\\ v_2 \cdot v_1 & v_2 \cdot v_2 & \cdots & v_2 \cdot v_n\\ \vdots & \vdots & & \vdots\\ v_n \cdot v_1 & v_n \cdot v_2 & \cdots & v_n \cdot v_n\\ \end{bmatrix}$.
Comparing them gives $\|h_1\|=\|v_1\|$ and so on.
Since $\|h_1\|=\|v_1\|$, we have $a_{11}^2 + a_{12}^2 + \cdots a_{n2}^2 = a_{11}^2$.
Since the entries are real, the only component of the first row which can be non-zero is $a_{11}$.
Rinse and repeat to prove that the elements on the diagonal are the only elements which can be non-zero.
Let A be: $$\left[ \begin{array}{ccc} a_{11} & a_{12} & \cdots & a_{1n}\\ 0&a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \end{array} \right]$$ Then A^T will be a lower triangular matrix such as : $$\left[ \begin{array}{ccc} a_{11} & 0 & \cdots & 0\\ a_{12}&a_{22} & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & \cdots & a_{nn} \end{array} \right]$$ Then AA^T=A^TA only if A is diagonal.
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2019-10-20 06:14:17
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https://www.jonaslindstrom.dk/
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# Algorithmic composition with Langton’s Ant
Langton’s Ant is a simulation in two dimensions which has been proven to be a universal Turing machine – it can in principal be used to compute anything computable by a computer.
The simulation consists of an infinite board of either squares which can be either white or black. Now, an “ant” walks around the board. If the ant lands on a white square, it turns right, flips the color of the square and moves forward. If the square is black, the ant turns left, flips the color of the square and moves forward.
When visualised, the behaviour of this system changes over time from structured and simple to completely chaotic. However, the system is completely deterministic, determined only by the starting state.
In the video above, a simulation with two ants runs over 500 steps, and every time a square flips from black to white, a note is played. The note to be played is determined as follows:
• The board is divided into 7×7 sub-boards of 49 squares.
• These squares are enumerated from the bottom reading each row from left ro right from 0 to 48.
• When a square is flipped from black to white, the number assigned to the square determines the note as the number of semitones above A1.
Seven is chosen as the width of the sub-squares because it is the number of semitones in a fifth, so the ants moves either chromatically (horizontally) or by fifths (vertically). In the beginning, the are moving independently and very structured, but when their paths meet, a more complex, chaotic behaviour emerges.
# Ruffini – abstract algebra in Java
Class inheritance and generics/templates makes it possible to write code with nice abstractions very similar to the abstractions used in math, particularly in abstract algebra, where a computation can be described for an entire class of algebraic structures sharing some similar properties (eg. groups and rings). This has been described nicely in the book “From Mathematics to Generic Programming” by Alexandra A. Stepanov and Daniel E. Rose.
Inspired by this book, I have implemented a library for computations on algebraic structures called Ruffini using Java generics to get the same kind of abstraction that is used in abstract algebra, eg. that you do not specify what specific algebraic structure is used, only that is eg. a group or a ring.
Abstract algebraic structures are defined by a number of interfaces extending each other, the simplest being a semigroup:
public interface Semigroup<E> {
/**
* Return the result of the product <i>ab</i> in this
* semigroup.
*
* @param a
* @param b
* @return
*/
public E multiply(E a, E b);
}
The semigroup is extended by a monoid interface which adds a getIdentity() method and by a group interface which adds an invert(E) method etc. Continuing like this we get end up with interfaces for more complicated structures like rings, fields and vector spaces.
Now, each of these interfaces describes functions that can be applied to a generic type E (eg. that given two instances of E, the multiply method above returns a new instance of type E). As in abstract algebra, these functions can be used to describe complicated computations without specifying exactly what algebraic structure is being used.
Ruffini currently implements a number of algorithms that is defined and described for the abstract structures, including the discrete Fourier transform, the Euclidean algorithm, Gröbner bases, the Gram-Schmidt process and Gaussian elimination. The abstraction makes it possible to write the algorithms only once, while still being usable on any of the algebraic structures defined in the library, including integers, rational numbers, integers modulo n, finite fields, polynomial rings and matrix rings. Below is the code for the Gram-Schmidt process:
public List<V> apply(List<V> vectors) {
List<V> U = new ArrayList<>();
Projection<V, S> projection = new Projection<>(vectorSpace, innerProduct);
for (V v : vectors) {
for (V u : U) {
V p = projection.apply(v, u);
v = vectorSpace.subtract(v, p);
}
}
return U;
}
Here V and S are the generic types for resp. the vectors and scalars of the underlying vector space, and Projection is a class defining projection given a specific vector space and an inner product. Note that the vector space has a subtract method which subtracts two vectors. Now, the above algorithm can be executed on any of the vector spaces defined in Ruffini.
# Solving the 0-1 Knapsack problem with Excel
Given a list of items each with a value and a weight, the Knapsack problem seeks to find the set of items with the largest combined value within a given weight limit. There are a nice dynamic programming solution which I decided to implement in a spread sheet. I used Google Sheets but the solution is exported as an excel-sheet.
The solution builds a Knapsack table for round 0,1,…,limit. In each round r the solution for the problem with limit r is constructed as a column in the table, so the table has to be as wide as the maximum limit. Once the table is built, the solution can be found using backtracking. This is all described pretty well on Wikipedia, https://en.wikipedia.org/wiki/Knapsack_problem#0/1_knapsack_problem.
The main challenge was to translate this algorithm from procedural pseudocode to a spreadsheet. Building the table is simple enough (once you learn the OFFSET command in Excel which allows you to add or subtract a variable number of rows and columns from a given position), but the backtracking was a bit more tricky.
Assuming that the weights are stored in column A from row 3 and the corresponding values are stored in column B, the table starts in column D. The round numbers are stored in row 1 from columnD. Row 2 are just 0’s and all other entries are (the one below is from D3):
=IF($A3>D$1; D2; MAX(D2; OFFSET(INDIRECT(ADDRESS(ROW(); COLUMN()));-1;-$A3) +$B3))
The table stops after round 40, so we find the solution with weight at most 40.
The backtracking to find the actual solution after building the table is done in column AT and AU. In the first column, the number of weight spent after 40 rounds are calculated from bottom to top row using the formula (from AT3):
=IF(OFFSET(INDIRECT(ADDRESS(ROW(); COLUMN()));0;-AT23-2) <> OFFSET(INDIRECT(ADDRESS(ROW(); COLUMN()));-1;-AT23-2); A22 + AT23; AT23)
The solution is shown in column AU where for each row we simply check if the accumulated weight increased with this item.
# Gram-Schmidt process for integers
Given a set of linearly independent vectors, the Gram-Schmidt process produces a new set of vectors spanning the same subset, but the new set of vectors are now mutually orthogonal. The algorithm is usually stated as working over a vector space, but will in this post be altered to work over a module which is a vector space where the scalars are elements in a ring instead of a field. This could for instance be the case if we are given a set of integer vectors we wish to orthogonalise.
The original version can be described as follows: Given a set of vectors $$v_1, \ldots, v_n$$ we define $$u_1 = v_1$$ and
$$u_k = v_k – \text{proj}_{u_1}(v_k) – \text{proj}_{u_2}(v_k) – \cdots – \text{proj}_{u_{k-1}}(v_k)$$
for $$k = 2,3,\ldots,n$$. The projection operator is defined as
$$\text{proj}_u(v) = \frac{\langle u, v \rangle}{\langle u, u \rangle} u$$.
If we say that all $$v_i$$ have integer entries, we see that $$u_i$$ must have rational entries, and simply scaling each vector with a multiple of all the denominators of the entries will give a vector parallel to the original vector but with integer entries. But what if we are interested in an algorithm that can only represent integers?
The algorithm is presented below. Note that the algorithm works over any module with inner product (aka a Hilbert Module).
# Given: Set of vectors V.
# Returns: A set of mutually orthogonal vectors spanning
# the same subspace as the vectors in V.
U := Ø
k := 1
while V ≠ Ø do
poll vector v from V
w := kv
for (u,m) in U do
w -= m〈v,u〉u
# Optional: Divide all entries in w by gcd of entries
if V ≠ Ø do
n :=〈w,w〉
for (u,m) in U do
m *= n
Put (w,k) in U
k *= n
else do
# No more iterations after this one
Put (w,k) in U
return first coordinates of elements in U
When used on integers, the entries in the vectors grow very fast, but this may be avoided by dividing $$w$$ by the greatest common divisor of the entries.
# DIY preamp for Leslie 760
I have recently bought a used Leslie 760 amplifier, but unfortunately it came without the pedal used to change the speed of the rotator and which also acts as preamp. A used pedal and the necessary cables is priced at about €400 so I decided to build one myself.
Luckily, some nice people has posted scanned versions of the old manuals for Leslie amps and put them online: http://www.captain-foldback.com/Leslie_sub/leslie_manuals.htm.
The amplifier does not have a normal jack-plug for connection with an instrument but has instead a 9-pole plug. At first glance, this seems a bit confusing, but it is quite simple: Pole 1 is ground and pole 2 is sound input. The rotator is activated by grounding pole 6 (slow) or pole 7 (fast).
With this information it was easy to build everything I needed: I built a small box with room for two jack plugs to be put on the amp. The first plug is mono for instrument connection, and is attached to pole 1 and 2. The other plug is for stereo which is connected to pole 6 and 7 (and ground). At the other end of this stereo cable I attached a switch pedal with two buttons: One for switching between grounding the two poles of the stereo jack and another for switching the connection on/off. I got everything for about €25 on musik-ding.de.
I still needed a preamp and found one by the brand ‘Art’ at 4Sound.dk for €35, so for about 60€ I got everything I needed to run the amp.
You are more than welcome to write me at mail@jonaslindstrom.dk if you have any questions.
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2020-03-28 09:19:11
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https://brilliant.org/problems/factorial-gcd/
|
# Factorial GCD
Evaluate $\gcd ( 19 ! + 19, 20! + 19 )$.
Details and assumptions
The number $n!$, read as n factorial, is equal to the product of all positive integers less than or equal to $n$. For example, $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$.
×
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2020-08-04 23:08:13
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https://www.gradesaver.com/textbooks/science/physics/CLONE-afaf42be-9820-4186-8d76-e738423175bc/chapter-10-exercises-and-problems-page-194/74
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## Essential University Physics: Volume 1 (4th Edition) Clone
We know that the mass will be located at the average of the two radii. Thus, we find: $I_{ring} = \frac{MR_1^2 + MR_2^2}{2}$ $I_{ring} = \frac{M(R_1^2 + R_2^2)}{2}$ $I_{ring} = \frac{1}{2}M(R_1^2 + R_2^2)$
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2019-10-17 17:45:13
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https://www.nestedtori.com/2016/04/polar-coordinate-classic.html
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## Friday, April 15, 2016
### Polar Coordinate Classic
In some sense, my first "Nested Tori" was a notebook which contained really cool plots and graphs. There's lots of old gems there worth posting and updating. Today, I bring you a nice polar plot. I was really intrigued by plots in polar coordinates, because it represented the first real departure from simply graphing $y=f(x)$. We instead (most commonly) write $r = f(\theta)$ where $r$ is distance from the origin and $\theta$ is counterclockwise angle, and let things get farther or closer as we go around. Without further ado, here is the famed butterfly (it does have a little post-warping afterward, though), which shows you that you don't need fancy-schmancy 3D stuff to still find great things:
$r = e^{\sin \theta} - 2 \cos(4\theta) +\sin^5\left(\dfrac{\theta-\frac{\pi}{2}}{12}\right), 0 \leq \theta \leq 24\pi$
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2021-04-14 07:03:01
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https://read.somethingorotherwhatever.com/entry/Tilingwitharbitrarytiles
|
# Tiling with arbitrary tiles
• Published in 2015
In the collections
Let $T$ be a tile in $\mathbb{Z}^n$, meaning a finite subset of $\mathbb{Z}^n$. It may or may not tile $\mathbb{Z}^n$, in the sense of $\mathbb{Z}^n$ having a partition into copies of $T$. However, we prove that $T$ does tile $\mathbb{Z}^d$ for some $d$. This resolves a conjecture of Chalcraft.
### BibTeX entry
@article{Tilingwitharbitrarytiles,
title = {Tiling with arbitrary tiles},
abstract = {Let {\$}T{\$} be a tile in {\$}\mathbb{\{}Z{\}}^n{\$}, meaning a finite subset of
{\$}\mathbb{\{}Z{\}}^n{\$}. It may or may not tile {\$}\mathbb{\{}Z{\}}^n{\$}, in the sense of
{\$}\mathbb{\{}Z{\}}^n{\$} having a partition into copies of {\$}T{\$}. However, we prove that
{\$}T{\$} does tile {\$}\mathbb{\{}Z{\}}^d{\$} for some {\$}d{\$}. This resolves a conjecture of
Chalcraft.},
url = {http://arxiv.org/abs/1505.03697v2 http://arxiv.org/pdf/1505.03697v2},
year = 2015,
author = {Vytautas Gruslys and Imre Leader and Ta Sheng Tan},
comment = {},
urldate = {2022-02-23},
archivePrefix = {arXiv},
eprint = {1505.03697},
primaryClass = {math.CO},
collections = {easily-explained,fun-maths-facts,geometry}
}
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2022-05-22 20:17:08
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http://www.ams.org/cgi-bin/bookstore/booksearch?fn=100&pg1=CN&s1=Greenwood_P_E&arg9=P._E._Greenwood
|
New Titles | FAQ | Keep Informed | Review Cart | Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education
Markov Fields over Countable Partially Ordered Sets: Extrema and Splitting
SEARCH THIS BOOK:
Memoirs of the American Mathematical Society
1994; 100 pp; softcover
Volume: 112
ISBN-10: 0-8218-2597-6
ISBN-13: 978-0-8218-2597-6
List Price: US$40 Individual Members: US$24
Institutional Members: US\$32
Order Code: MEMO/112/537
Various notions of the Markov property relative to a partial ordering have been proposed by both physicists and mathematicians. For the most part, the analysis has focused on the study of some important, but special, examples. This work develops general techniques for studying Markov fields on partially ordered sets. The authors introduce random transformations of the index set which preserve the Markov property of the field. These transformations yield new classes of Markov fields starting from relatively simple ones. Examples include a model for crack formation and a model for the distribution of fibres in a composite material. Given the burst of popularity of random fields, this self-contained and accessible book will prove useful in the many scientific areas where random field models are appearing.
• Splitting lemma
• Honest random sets
• Splitting random sets. Constructions of splitting random elements
• Markov systems and their transformations
• Shift invariance
• Some special partially ordered sets and Markov fields
• Ferry problem
• The linearly ordered case
• Fibres
• Cracks and splitting random elements
• Transformations of fields indexed by contours
• Bibliographical notes
• References
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2016-05-28 08:12:28
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https://www.i2m.univ-amu.fr/events/efroymsons-approximation-theorem-for-globally-subanalytic-functions/
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Efroymson’s Approximation Theorem for globally subanalytic functions
Anna Valette
Jagiellonian University, Kraków, Poland
https://apacz.matinf.uj.edu.pl/users/245-anna-valette
Date(s) : 15/04/2021 iCal
14 h 00 min - 15 h 00 min
Efroymson’s Approximation Theorem asserts that if f is a continuous semialgebraic mapping on a C^infinity semialgebraic submanifold M of ℝⁿ and if e : M→ℝ is a positive continuous semialgebraic function then there is a C^infinity semialgebraic function g:M→ℝ such that |f-g|<e. The aim of this talk is to give some insights into the proof of generalized Efroymson’s theorem to the globally subanalytic category.
Our framework is however much bigger than this category since our approximation theorems hold on every polynomially bounded o-minimal structure expanding the real field that admits C^infinity cell decomposition. In particular, it applies to quasi-analytic Denjoy-Carleman classes.
Work in collaboration with Guillaume Valette.
https://arxiv.org/abs/1905.05703
Catégories
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2022-05-25 03:53:28
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http://slideplayer.com/slide/3431645/
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# Differential Geometry Applied to Dynamical Systems
## Presentation on theme: "Differential Geometry Applied to Dynamical Systems"— Presentation transcript:
Differential Geometry Applied to Dynamical Systems
JEAN-MARC GINOUX BRUNO ROSSETTO Laboratoire PROTEE, I.U.T. de Toulon Université du Sud, B.P , 83957, LA GARDE Cedex, France
OUTLINE A. Modeling & Dynamical Systems B. Flow Curvature Method
1. Definition & Features 2. Classical analytical approaches B. Flow Curvature Method 1. Presentation 2. Results C. Applications 1. n-dimensional dynamical systems 2. Non-autonomous dynamical systems Hairer's 60th birthday
MODELING DYNAMICAL SYSTEMS
Defining states variables of a system (predator, prey) Describing their evolution with differential equations (O.D.E.) Dynamical System: Representation of a differential equation in phase space expresses variation of each state variable Determining variables from their variation (velocity) Hairer's 60th birthday
n-dimensional Dynamical Systems
velocity Hairer's 60th birthday
MANIFOLD DEFINTION A manifold is defined as a set of points in
satisfying a system of m scalar equations : where for with The manifold M is differentiable if is differentiable and if the rank of the jacobian matrix is equal to in each point . Thus, in each point of the différentiable manifold , a tangent space of dimension is defined. In dimension In dimension 3 curve surface Hairer's 60th birthday
The Lie derivative is defined as follows:
Let a function defined in a compact E included in and the integral of the dynamical system defined by (1). The Lie derivative is defined as follows: If then is first integral of the dynamical system (1). So, is constant along each trajectory curve and the first integrals are drawn on the hypersurfaces of level set ( is a constant) which are over flowing invariant. Hairer's 60th birthday
INVARIANT MANIFOLDS Darboux Theorem for Invariant Manifolds:
An invariant manifold (curve or surface) is a manifold defined by where is a function in the open set U and such that there exists a function in U denoted and called cofactor such that: for all This notion is due to Gaston Darboux (1878) Hairer's 60th birthday
ATTRACTIVE MANIFOLDS Poincaré’s criterion :
Manifold implicit equation: Instantaneous velocity vector: Normal vector: attractive manifold tangent manifold repulsive manifold This notion is due to Henri Poincaré (1881) Hairer's 60th birthday
DYNAMICAL SYSTEMS Dynamical Systems: Fixed Points Local Bifurcations
Integrables or non-integrables analytically Fixed Points Local Bifurcations Invariant manifolds center manifolds slow manifolds (local integrals) linear manifolds (global integrals) Normal Forms Hairer's 60th birthday
« Classical » analytic methods
DYNAMICAL SYSTEMS « Classical » analytic methods Courbes définies par une équation différentielle (Poincaré, 1881 1886) ………….…. Singular Perturbation Methods (Poincaré, 1892, Andronov 1937, Cole 1968, Fenichel 1971, O'Malley 1974) Tangent Linear System Approximation (Rossetto, 1998 & Ramdani, 1999) Hairer's 60th birthday
velocity acceleration over-acceleration etc. …
FLOW CURVATURE METHOD Geometric Method Flow Curvature Method (Ginoux & Rossetto, 2005 2009) velocity velocity acceleration over-acceleration etc. … position Hairer's 60th birthday
n-Euclidean space curve
FLOW CURVATURE METHOD “trajectory curve” n-Euclidean space curve plane or space curve curvatures Hairer's 60th birthday
Flow curvature manifold:
FLOW CURVATURE METHOD Flow curvature manifold: The flow curvature manifold is defined as the location of the points where the curvature of the flow, i.e., the curvature of trajectory curve integral of the dynamical system vanishes. where represents the n-th derivative Hairer's 60th birthday
FLOW CURVATURE METHOD Flow Curvature Manifold: In dimension 2:
curvature or 1st curvature In dimension 3: torsion ou 2nd curvature Hairer's 60th birthday
Flow Curvature Manifold:
FLOW CURVATURE METHOD Flow Curvature Manifold: In dimension 4: 3rd curvature In dimension 5: 4th curvature Hairer's 60th birthday
Fixed points of the flow curvature manifold
Theorem 1 (Ginoux, 2009) Fixed points of any n-dimensional dynamical system are singular solution of the flow curvature manifold Corollary 1 Fixed points of the flow curvature manifold are defined by Hairer's 60th birthday
FIXED POINTS STABILITY
Theorem 2: (Poincaré 1881 Ginoux, 2009) Hessian of flow curvature manifold associated to dynamical system enables differenting foci from saddles (resp. nodes). Hairer's 60th birthday
FIXED POINTS STABILITY
Unforced Duffing oscillator and Thus is a saddle point or a node Hairer's 60th birthday
CENTER MANIFOLD Theorem 3 (Ginoux, 2009)
Center manifold associated to any n-dimensional dynamical system is a polynomial whose coefficients may be directly deduced from flow curvature manifold with Hairer's 60th birthday
CENTER MANIFOLD Guckenheimer et al. (1983) Local Bifurcations
Hairer's 60th birthday
SLOW INVARIANT MANIFOLD
Theorem 4 (Ginoux & Rossetto, 2005 2009) Flow curvature manifold of any n-dimensional slow-fast dynamical system directly provides its slow manifold analytical equation and represents a local first integral of this system. Hairer's 60th birthday
VAN DER POL SYSTEM (1926) http://ginoux.univ-tln.fr
Hairer's 60th birthday
VAN DER POL SYSTEM (1926) slow part slow part Singular approximation
Hairer's 60th birthday
VAN DER POL SYSTEM (1926) slow part slow part
Hairer's 60th birthday
VAN DER POL SYSTEM (1926) Slow manifold Lie derivative
Singular approximation Slow manifold Lie derivative Hairer's 60th birthday
VAN DER POL SYSTEM (1926) Slow Manifold Analytical Equation
Flow Curvature Method vs Singular Perturbation Method (Fenichel, 1979 vs Ginoux 2009) Hairer's 60th birthday
VAN DER POL SYSTEM (1926) Flow Curvature Method vs
Singular Perturbation Method (up to order ) Singular perturbation Flow Curvature Hairer's 60th birthday
VAN DER POL SYSTEM (1926) Slow Manifold Analytical Equation given by
Flow Curvature Method & Singular Perturbation Method are identical up to order one in Pr. Eric Benoît High order approximations are simply given by high order derivatives, e. g., order 2 in is given by the Lie derivative of the flow curvature manifold, etc… Hairer's 60th birthday
Slow manifold attractive domain
VAN DER POL SYSTEM (1926) Slow manifold attractive domain Hairer's 60th birthday
LINEAR INVARIANT MANIFOLD
Theorem 5 (Darboux, 1878 Ginoux, 2009) Every linear manifold (line, plane, hyperplane) invariant with respect to the flow of any n-dimensional dynamical system is a factor in the flow curvature manifold. Hairer's 60th birthday
CHUA's piecewise linear model:
APPLICATIONS 3D CHUA's piecewise linear model: Hairer's 60th birthday
APPLICATIONS 3D CHUA's piecewise linear model:
Slow invariant manifold analytical equation Hyperplanes Hairer's 60th birthday
APPLICATIONS 3D CHUA's piecewise linear model:
Invariant Hyperplanes (Darboux) Hairer's 60th birthday
CHUA's piecewise linear model:
APPLICATIONS 3D CHUA's piecewise linear model: Invariant Planes Invariant Planes Hairer's 60th birthday
APPLICATIONS 3D CHUA's cubic model: with and http://ginoux.univ-tln.fr
Hairer's 60th birthday
APPLICATIONS 3D CHUA's cubic model: Slow manifold Slow manifold
Hairer's 60th birthday
Edward Lorenz model (1963):
APPLICATIONS 3D Edward Lorenz model (1963): Hairer's 60th birthday
Slow invariant analytic manifold (Theorem 4)
APPLICATIONS 3D Edward Lorenz model: Slow invariant analytic manifold (Theorem 4) Hairer's 60th birthday
APPLICATIONS 3D Hairer's 60th birthday
APPLICATIONS 3D Autocatalator Neuronal Bursting Model
Hairer's 60th birthday
APPLICATIONS 4D Chua cubic 4D
[Thamilmaran et al., 2004, Liu et al., 2007] Hairer's 60th birthday
APPLICATIONS 5D Chua cubic 5D [Hao et al., 2005]
Hairer's 60th birthday
APPLICATIONS 5D Edgar Knobloch model: http://ginoux.univ-tln.fr
Hairer's 60th birthday
APPLICATIONS 5D MagnetoConvection http://ginoux.univ-tln.fr
Hairer's 60th birthday
NON-AUTONOMOUS DYNAMICAL SYSTEMS
Forced Van der Pol Guckenheimer et al., 2003 Hairer's 60th birthday
NON-AUTONOMOUS DYNAMICAL SYSTEMS
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NON-AUTONOMOUS DYNAMICAL SYSTEMS
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Normal Form Theorem 6 : (Poincaré 1879 Ginoux, 2009)
Normal form associated to any n-dimensional dynamical system may be deduced from flow curvature manifold Hairer's 60th birthday
Fixed Points & Stability: Center, Slow & Linear
FLOW CURVATURE METHOD Fixed Points & Stability: - Flow Curvature Manifold: Theorems 1 & 2 Center, Slow & Linear Manifold Analytical Equation: - Theorems 3, 4 & 5 Normal Forms: - Theorem 6 Hairer's 60th birthday
Flow Curvature Method: n-dimensional dynamical systems
DISCUSSION Flow Curvature Method: n-dimensional dynamical systems Autonomous or Non-autonomous Fixed points & stability, local bifurcations, normal forms Center manifolds Slow invariant manifolds Linear invariant manifolds (lines, planes, hyperplanes,…) Applications : Electronics, Meteorology, Biology, Chemistry… Hairer's 60th birthday
Differential Geometry Applied to Dynamical Systems
Publications Book Differential Geometry Applied to Dynamical Systems World Scientific Series on Nonlinear Science, series A, 2009 Hairer's 60th birthday
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2018-05-25 13:02:15
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http://nrich.maths.org/346/clue?setlocale=en_US
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### Sine Problem
In this 'mesh' of sine graphs, one of the graphs is the graph of the sine function. Find the equations of the other graphs to reproduce the pattern.
### Cocked Hat
Sketch the graphs for this implicitly defined family of functions.
### Folium of Descartes
Investigate the family of graphs given by the equation x^3+y^3=3axy for different values of the constant a.
# Quartics
##### Stage: 5 Challenge Level:
For this question you may like to use a computer graph drawing application or a graphic calculator. For $t = -1/2$, $1/2$ and $2$, sketch graphs of $y = [1 + (x - t)^2][1 + (x + t)^2]$.
Then try other values of the parameter $t$. You will see that these graphs have 'different shapes'. Suppose the parameter $t$ varies, then the general shape of the graph varies continuously with $t$.
You can get this far without calculus but you'll probably need calculus to find the turning points and to show that the graph always has a shape similar to the examples above and to find the values of $t$ at which there are transitions from one shape to another.
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2016-04-29 08:27:23
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https://proofwiki.org/wiki/Linear_Second_Order_ODE/y%27%27_%3D_1_over_1_-_x%5E2
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# Linear Second Order ODE/y'' = 1 over 1 - x^2
## Theorem
The second order ODE:
$(1): \quad y'' = \dfrac 1 {1 - x^2}$
has the general solution:
$y = x \tanh^{-1} x + \map \ln {1 - x^2} + C x + D$
## Proof
$\displaystyle y''$ $=$ $\displaystyle \dfrac 1 {1 - x^2}$ $\displaystyle \leadsto \ \$ $\displaystyle \int \dfrac {\d^2 y} {\d x^2} \rd x$ $=$ $\displaystyle \int \frac 1 {1 - x^2} \rd x$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {\d y} {\d x}$ $=$ $\displaystyle \tanh^{-1} x + C$ Primitive of $\dfrac 1 {a^2 - x^2}$: Inverse Hyperbolic Tangent Form $\displaystyle \leadsto \ \$ $\displaystyle \int \dfrac {\d y} {\d x} \rd x$ $=$ $\displaystyle \int \paren {\tanh^{-1} x + C} \rd x$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle x \tanh^{-1} x + \map \ln {1 - x^2} + C x + D$ Primitive of $\tanh^{-1} x$
$\blacksquare$
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2020-12-02 18:28:56
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https://www.physicsforums.com/threads/levi-civita-to-knocker-delta.464442/
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Levi civita to knocker delta
quietrain
hi, i don't understand why levi civita symbols is equal to kronecker delta when i was reading the wikipedia on it http://en.wikipedia.org/wiki/Levi-Civita_symbol
namely,
1)why does EijkElmn = det [ 9 terms of kronecker delta]
how do we know its those 9 terms? like dil dim din etc
2)and since they use det, does it mean EijkElmn is a cross product?
on the proof of the vector triple product,
3)why from (dmj dnk-dmk dnj)an bj ck it becomes bm ak ck-cm aj bj
i understand that the kronecker delta is something like a dirac delta function, which acts like a "sifting function" but if there are 2 kronecker delta, then how does it work?
so example instead of bm ak ck, why doesn't it become bmancn - ...
4)
and subsequently getting bm ak ck-cm aj bj
= [b(a.c)]m - [c(a.b)]m
what does the subscript m stands for? and how did they get to the bac-cab expression? if the j and k subscripts means dot products, then what are the other 2 components of the dot product? it has so many alphabets><
does it has anyhting to do with dummy index and free index?
also, from
is it right to think that the subscript i is the x-plane in cartesian form?
so if i want the z-plane, then would it mean to use the k component? so in this equations, is the k used here equal to the k in cartesian? if it is the same, then how do i get the k-component since they will clash?
5)also what does the subscript i means? component?
if i just want to have the triple product alone without the component i, what will it be?
thanks!
Last edited:
quietrain
erm i sort of understand point 3 already but the rest ><
quietrain
anybody?
1)why does EijkElmn = det [ 9 terms of kronecker delta]
how do we know its those 9 terms? like dil dim din etc
Brute force, although you can get a little heuristic insight
by thinking about the symmetries of a determinant.
E.g., swapping two rows flips the sign of the det, as does
exchanging the corresponding indices in the eps's.
2)and since they use det, does it mean EijkElmn is a cross product?
No. It's a 6-th rank tensor (since there are no contracted indices).
3)[...] instead of bm ak ck, why doesn't it become bmancn - ...
Look up "summation convention", i.e., implicit summation over repeated indices.
(The latter are called dummy indices.)
Both these expressions are the same.
is it right to think that the subscript i is the x-plane in cartesian form?
Not exactly -- it's any particular component in a cartesian frame.
Staff Emeritus
Gold Member
I will only answer the questions about the very basic things here, because there's no point worrying about the difficult stuff before you know the basics. The m and the i you're asking about refers to the mth component of the vector and the ith component of the vector respectively. For example, when m=2, (5,6,7)m=6.
Summation convention: We can drop all the summation sigmas without ambiguity, because there's always a sum over indices that appear exactly twice, and there's never a sum over any of the others.
The ith component of x is of course the number $x_i$ in $x=x_i e_i$, where $e_i$ is the ith basis vector.
The dot product and cross product are defined by
$$x\cdot y=x_i y_i$$
$$(x\times y)_i=\varepsilon_{ijk}x_jy_k$$
The Levi-Civita symbol is defined by $\varepsilon_{123}=1$ and the requirement that swapping any two indices changes the sign. (For example, $\varepsilon_{ijk}=\varepsilon_{ikj}$, regardless of the values of j and k).
The Kronecker delta is defined by $\delta_{ij}=1$ when i=j, and $\delta_{ij}=0$ when i≠j.
I suggest you do the following (easy) exercises:
Exercise 1: Use these definitions to find $(x\times y)_2$.
Exercise 2: Show that if $S_{ij}=S_{ji}$ and $A_{ij}=-A_{ji}$, then $S_{ij}A_{ij}=0$. (This result is extremely useful).
Exercise 3: Show that $\nabla\cdot(\nabla\times F)=0$.
Last edited:
quietrain
Brute force, although you can get a little heuristic insight
by thinking about the symmetries of a determinant.
E.g., swapping two rows flips the sign of the det, as does
exchanging the corresponding indices in the eps's.
No. It's a 6-th rank tensor (since there are no contracted indices).
Look up "summation convention", i.e., implicit summation over repeated indices.
(The latter are called dummy indices.)
Both these expressions are the same.
Not exactly -- it's any particular component in a cartesian frame.
i see thank you!
quietrain
I will only answer the questions about the very basic things here, because there's no point worrying about the difficult stuff before you know the basics. The m and the i you're asking about refers to the mth component of the vector and the ith component of the vector respectively. For example, when m=2, (5,6,7)m=6.
Summation convention: We can drop all the summation sigmas without ambiguity, because there's always a sum over indices that appear exactly twice, and there's never a sum over any of the others.
The ith component of x is of course the number $x_i$ in $x=x_i e_i$, where $e_i$ is the ith basis vector.
The dot product and cross product are defined by
$$x\cdot y=x_i y_i$$
$$(x\times y)_i=\varepsilon_{ijk}x_jy_k$$
The Levi-Civita symbol is defined by $\varepsilon_{123}=1$ and the requirement that swapping any two indices changes the sign. (For example, $\varepsilon_{ijk}=\varepsilon_{ikj}$, regardless of the values of j and k).
The Kronecker delta is defined by $\delta_{ij}=1$ when i=j, and $\delta_{ij}=0$ when i≠j.
I suggest you do the following (easy) exercises:
Exercise 1: Use these definitions to find $(x\times y)_2$.
Exercise 2: Show that if $S_{ij}=S_{ji}$ and $A_{ij}=-A_{ji}$, then $S_{ij}A_{ij}=0$. (This result is extremely useful).
Exercise 3: Show that $\nabla\cdot(\nabla\times F)=0$.
x X y = e123x2y3
so the 2nd component means 231-213?
so its x3y1 - x1y3?
for the 2nd exercise, i am not sure? are those matrix? or are they kronecker deltas? what does it mean by Sij = Sji? meaning S12 = S21? this is not levicivita symbols right?
for exercise 3
E123 ∇1∇2F3
so its 123-132 + 231-213 + 312 - 321 ?
so its ∇∇F - ∇F∇ + ∇F∇ - ∇∇F + F∇∇ - F∇∇
so we get 0 ?
btw on the last expression, is it right to swop ∇F∇ to ∇∇F? meaning are they equal? or do i have to introduce signs when not equal?
oh and with regards to one more thing
i notice that the triple vector product is given by the b(a.c) - c(a.b) rule
but i know a.c and a.b are the dot product. but what about the b()? is just just multiplication?
but since all are vectors, then is b() a cross or dot or ?
thanks!
Staff Emeritus
Gold Member
x X y = e123x2y3
so the 2nd component means 231-213?
so its x3y1 - x1y3?
That's the correct answer, but I can't tell if you used the definition of the Levi-Civita symbol or a definition of the cross product that doesn't use the Levi-Civita symbol. You should note that the sum $$\varepsilon_{2ij}x_iy_j$$ has nine terms, but only two that are non-zero. Do you see how the fact that the other seven are zero follows from my definition of the Levi-Civita symbol?
for the 2nd exercise, i am not sure? are those matrix? or are they kronecker deltas? what does it mean by Sij = Sji? meaning S12 = S21? this is not levicivita symbols right?
For each value of i and j, Sij and Aij are real numbers. I should perhaps have stated explicitly that the equalities I included in the problem statement hold for all values of i and j. So there are nine equalities involving the Sij and nine involving the Aij.
You can think of Sij and Aij as the components of two 3×3 matrices if you want to, but you can also choose not to.
This is just an exercise in using the summation convention, so it doesn't involve the Kronecker delta or the Levi-Civita symbol. (But the result allows you to simplify many expressions that do).
for exercise 3
E123 ∇1∇2F3
It's quite hard to read this notation. You should at least use "sub" tags for the indices. An alternative is to use LaTeX. If you click the quote button, you can see how I did the math in this post. (If you decide to try LaTeX you need to be aware that there's a bug that makes old images show up in previews. The workaround is to refresh and resend after each preview).
$$\nabla\cdot(\nabla\times F)=\partial_i\varepsilon_{ijk}\partial_j F_k$$
and then use (a simple generalization of) the result of exercise 2 to immediately conclude that the right-hand side is =0.
btw on the last expression, is it right to swop ∇F∇ to ∇∇F? meaning are they equal? or do i have to introduce signs when not equal?
What does the first expression mean?
i notice that the triple vector product is given by the b(a.c) - c(a.b) rule
but i know a.c and a.b are the dot product. but what about the b()? is just just multiplication?
but since all are vectors, then is b() a cross or dot or ?
It can't be a cross product or a dot product, because the thing in parentheses is a number, not a vector.
Last edited:
quietrain
That's the correct answer, but I can't tell if you used the definition of the Levi-Civita symbol or a definition of the cross product that doesn't use the Levi-Civita symbol. You should note that the sum $$\varepsilon_{2ij}x_iy_j$$ has nine terms, but only two that are non-zero. Do you see how the fact that the other seven are zero follows from my definition of the Levi-Civita symbol?
For each value of i and j, Sij and Aij are real numbers. I should perhaps have stated explicitly that the equalities I included in the problem statement hold for all values of i and j. So there are nine equalities involving the Sij and nine involving the Aij.
You can think of Sij and Aij as the components of two 3×3 matrices if you want to, but you can also choose not to.
This is just an exercise in using the summation convention, so it doesn't involve the Kronecker delta or the Levi-Civita symbol. (But the result allows you to simplify many expressions that do).
It's quite hard to read this notation. You should at least use "sub" tags for the indices. An alternative is to use LaTeX. If you click the quote button, you can see how I did the math in this post. (If you decide to try LaTeX you need to be aware that there's a bug that makes old images show up in previews. The workaround is to refresh and resend after each preview).
$$\nabla\cdot(\nabla\times F)=\partial_i\varepsilon_{ijk}\partial_j F_k$$
and then use (a simple generalization of) the result of exercise 2 to immediately conclude that the right-hand side is =0.
What does the first expression mean?
It can't be a cross product or a dot product, because the thing in parentheses is a number, not a vector.
for part 1 ,
the 7 other terms are 0 because of the repeated indices? but i don't know why a repeated indices means 0? definition?
for part 2, i made it a 2x2 matrix, and i realise wow it all cancels out if i let say,
S=
(1,1)
(1,1)
A=
(1,1)
(-1,-1)
am i right to let them have such values?
for part 3, how do you generalise the above to proof so?
for part 4, by cross product, it says like 132-123 is not interchangeable i assume?
but if the value example, 1i+2j+3k X 2i+3j+5k, then (2)(5) - 3(3) i + ...
if i swop 2and 5 it wouldn't matter right?
also, since ∇∇F are all vectors does it mean they are cross products or dot products? or just multiplication?
btw how do i quote here and quote there? breaking up the quotes like you did?
thanks!
Staff Emeritus
Gold Member
for part 1 ,
the 7 other terms are 0 because of the repeated indices? but i don't know why a repeated indices means 0? definition?
The 27 components of the Levi-Civita symbol are defined by the following two rules:
a) The 123 component is =1.
b) If you swap two indices, this changes the sign but not the absolute value.
Note that these two rules tell you the values of all 27 components. They tell you e.g. that $\varepsilon_{312}=-\varepsilon_{132}=\varepsilon_{123}=1$. The second rule answers my question about why those seven terms are =0:
The fact that $\varepsilon_{ijk}=-\varepsilon_{ikj}$, regardless of the values of i,j,k, implies in particular that when j=k, we have $\varepsilon_{ijk}=0$. So $\varepsilon_{i11}=\varepsilon_{i22}=\varepsilon_{i33}=0$ for all i.
for part 2, i made it a 2x2 matrix,
It has to be 3×3, since the indices go from 1 to 3.
and i realise wow it all cancels out if i let say,
S=
(1,1)
(1,1)
A=
(1,1)
(-1,-1)
am i right to let them have such values?
The idea was to prove the identity $A_{ij}S_{ij}=0$ for all values of the components that are consistent with the equalities I specified, so you can't choose specific values. The values you chose for A also contradict the condition $A_{ij}=-A_{ji}$.
I'll just do this one for you:
$$A_{ij}S_{ij}=A_{ji}S_{ji}=-A_{ij}S_{ij}\quad\Rightarrow\quad A_{ij}S_{ij}=0$$
Can you at least figure out why these equalities hold?
for part 3, how do you generalise the above to proof so?
The generalization I had in mind is that my proof works even if there are additional indices on A and/or S, for example
$$A_{ijkl}S_{ijkmn}=A_{jikl}S_{jikmn}=-A_{ijkl}S_{ijkmn}$$
The conclusion is that if the entire expression is symmetric (doesn't change sign) under the exchange of two variables that are summed over, and antisymmetric (changes sign) under the exchange of the same two variables at the other place they appear, then the whole expression is zero. It doesn't even matter if the indices are distributed over one, two or more symbols.
This is an extremely useful result. A good way to remember it is "symmetric times antisymmetric equals zero". It solves exercise 3, because $\nabla\cdot(\nabla\times F)=\partial_i\varepsilon_{ijk}\partial_j F_k$ is symmetric under the exchange of i and j in the partial derivative, and antisymmetric under the exchange of i and j in the Levi-Civita symbol. So we immediately see that the whole expression is =0, without any more calculations.
for part 4, by cross product, it says like 132-123 is not interchangeable i assume?
I don't understand what you're asking here.
but if the value example, 1i+2j+3k X 2i+3j+5k, then (2)(5) - 3(3) i + ...
if i swop 2and 5 it wouldn't matter right?
You can swap them on the right, because real numbers commute. But 2i+3j+5k is obviously different from 5i+3j+2k.
also, since ∇∇F are all vectors does it mean they are cross products or dot products? or just multiplication?
I don't understand what you mean by ∇∇F. There are two different derivative operators used in exercise 3. They are often written as div and curl. The latter is defined by
$$\operatorname{curl} F=\varepsilon_{ijk}\frac{\partial F_k}{\partial x_j}$$
Here F is a function from ℝ3 into ℝ3. The partial derivative operator is often abbreviated $\partial_j$. If we define $\nabla=(\partial_1,\partial_2,\partial_3)$, we have
$$\operatorname{curl} F=\varepsilon_{ijk}\partial_j F_k=\nabla\times F$$
I'll let you figure out "div" on your own.
btw how do i quote here and quote there? breaking up the quotes like you did?
I just copy and paste the quote tags that are created automatically.
Last edited:
quietrain
I'll just do this one for you:
$$A_{ij}S_{ij}=A_{ji}S_{ji}=-A_{ij}S_{ij}\quad\Rightarrow\quad A_{ij}S_{ij}=0$$
Can you at least figure out why these equalities hold?
oh so you mean Aij and Sij are levi civita symbols? so that when i change ij to ji i introduce minus sign. but since got 2, then it cancels to become the 2nd expression? for the 3rd expression we use the ij and ji defined by you . so 2AijSij = 0, so AijSij=0?
The conclusion is that if the entire expression is symmetric (doesn't change sign) under the exchange of two variables that are summed over, and antisymmetric (changes sign) under the exchange of the same two variables at the other place they appear, then the whole expression is zero. It doesn't even matter if the indices are distributed over one, two or more symbols.
This is an extremely useful result. A good way to remember it is "symmetric times antisymmetric equals zero". It solves exercise 3, because $\nabla\cdot(\nabla\times F)=\partial_i\varepsilon_{ijk}\partial_j F_k$ is symmetric under the exchange of i and j in the partial derivative, and antisymmetric under the exchange of i and j in the Levi-Civita symbol. So we immediately see that the whole expression is =0, without any more calculations.
but i and j are indices right? or are they variables? i thought variables are del, F etc?
I don't understand what you're asking here.
You can swap them on the right, because real numbers commute. But 2i+3j+5k is obviously different from 5i+3j+2k.
I don't understand what you mean by ∇∇F. There are two different derivative operators used in exercise 3. They are often written as div and curl. The latter is defined by
$$\operatorname{curl} F=\varepsilon_{ijk}\frac{\partial F_k}{\partial x_j}$$
Here F is a function from ℝ3 into ℝ3. The partial derivative operator is often abbreviated $\partial_j$. If we define $\nabla=(\partial_1,\partial_2,\partial_3)$, we have
$$\operatorname{curl} F=\varepsilon_{ijk}\partial_j F_k=\nabla\times F$$
I'll let you figure out "div" on your own.
oh, so if del = d1+d2+d3
then del . (delXF)
= d1(d2F3-d3F2) + ... and so on? like cross product except the ijk unit vectors are replaced by the first del?
so del . (delXF) = div of the curl of F?
but is there a simple reason as to why the div of the curl is always 0? or the only way to proof is the levi civita way?
i saw an example from here
http://tutorial.math.lamar.edu/Classes/CalcIII/CurlDivergence.aspx
but it doesn't say why its 0.
i was thinking about total differtials, where if you differientiate wrt to x first, then y they will be the same for an expression if it is a total differiential. so does it mean that F is always a total differiential expression?
Last edited:
Staff Emeritus
Gold Member
oh so you mean Aij and Sij are levi civita symbols?
No, the Levi-Civita symbol is specifically $\varepsilon_{ijk}$, defined as described in my previous posts. A=$\{A_{ij}\in\mathbb R|i,j\in\{1,2,3\}\}$ is a set of 9 real numbers (not necessarily distinct). $A_{ij}$ is a member of that set, "the ij component". The nine equations $A_{ij}=-A_{ji}$ (one for each choice of i and j) describe how the nine members of the set are related to each other. You should be able to see that there are exactly three independent components of A, and exactly six independent components of S.
so that when i change ij to ji i introduce minus sign. but since got 2, then it cancels to become the 2nd expression? for the 3rd expression we use the ij and ji defined by you . so 2AijSij = 0, so AijSij=0?
I'm not sure I understand what you're saying, but it looks like you have correctly explained why the two equalities
$$A_{ij}S_{ij}=A_{ji}S_{ji}=-A_{ij}S_{ij}$$
together imply that $A_{ij}S_{ij}=0$, but failed to explain why the two equalities hold. You really shouldn't give up until you see it. I'll state the problem again:
Suppose that the equalities $A_{ij}=-A_{ji}$ and $S_{ij}=S_{ji}$ hold for all i and all j. Show that
$$A_{ij}S_{ij}=A_{ji}S_{ji}$$
and that
$$A_{ji}S_{ji}=-A_{ij}S_{ij}$$
but i and j are indices right? or are they variables? i thought variables are del, F etc?
A variable is just a symbol that represents a member of a set. A constant is a variable that always represents the same member. So F and i are both variables, and the partial derivative operators can be thought of as constants.
oh, so if del = d1+d2+d3
It's not. You need to be more careful with the notation.
If $f:\mathbb R^3\rightarrow\mathbb R$, then
$$\operatorname{grad} f=(\partial_1 f,\partial_2f,\partial_3f)$$
This is called the gradient of f. We're interested in the divergence of a vector field $E:\mathbb R^3\rightarrow\mathbb R^3$ defined by
$$\operatorname{div} E=\partial_i E_i$$.
The right-hand sides of these two definitions are often written as $\nabla f$ and $\nabla\cdot E$ respectively. To make sense of that, you need to think of $\nabla$ as a vector with 3 components: $\nabla=(\partial_1,\partial_2,\partial_3)$.
then del . (delXF)
= d1(d2F3-d3F2) + ... and so on? like cross product except the ijk unit vectors are replaced by the first del?
so del . (delXF) = div of the curl of F?
Yes.
but is there a simple reason as to why the div of the curl is always 0? or the only way to proof is the levi civita way?
The way I proved it is by far the easiest. That's one of the reasons you should make sure to understand exercise 2. But it's certainly not the only way. Just use any definitions of div and curl (that are equivalent to these), and you will see that every term cancels another term.
i was thinking about total differtials, where if you differientiate wrt to x first, then y they will be the same for an expression if it is a total differiential. so does it mean that F is always a total differiential expression?
No, F is an arbitrary vector field.
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quietrain
No, the Levi-Civita symbol is specifically $\varepsilon_{ijk}$, defined as described in my previous posts. A=$\{A_{ij}\in\mathbb R|i,j\in\{1,2,3\}\}$ is a set of 9 real numbers (not necessarily distinct). $A_{ij}$ is a member of that set, "the ij component". The nine equations $A_{ij}=-A_{ji}$ (one for each choice of i and j) describe how the nine members of the set are related to each other. You should be able to see that there are exactly three independent components of A, and exactly six independent components of S.
ok, so i get A11,A22,A33,A12,A21 and so on and get 9 terms?
but what do you mean by 3 independent components of A? are you talking about the 11 22 and 33? so that they all have to be 0 to fulfil the equation Aij=-Aji?
so for S, i assume its the 12 ,21 13,31 etc that are the 6 terms of independent? again why for S are they independent? i mean more specifically independent from what?
I'm not sure I understand what you're saying, but it looks like you have correctly explained why the two equalities
$$A_{ij}S_{ij}=A_{ji}S_{ji}=-A_{ij}S_{ij}$$
together imply that $A_{ij}S_{ij}=0$, but failed to explain why the two equalities hold. You really shouldn't give up until you see it. I'll state the problem again:
Suppose that the equalities $A_{ij}=-A_{ji}$ and $S_{ij}=S_{ji}$ hold for all i and all j. Show that
$$A_{ij}S_{ij}=A_{ji}S_{ji}$$
and that
$$A_{ji}S_{ji}=-A_{ij}S_{ij}$$
oh , so for the first equation, its just that the set of 9 numbers determined by Aij is equal to the set of 9 numbers determined by Aji just that the ordering is different only? so they are all equal
so the 2nd equation follows from the condition you stated ?
Staff Emeritus
Gold Member
ok, so i get A11,A22,A33,A12,A21 and so on and get 9 terms?
but what do you mean by 3 independent components of A? are you talking about the 11 22 and 33? so that they all have to be 0 to fulfil the equation Aij=-Aji?
so for S, i assume its the 12 ,21 13,31 etc that are the 6 terms of independent? again why for S are they independent? i mean more specifically independent from what?
To specify A is to specify the values of its nine components. To specify an A that's consistent with the condition Aij=-Aji, we have to "choose" nine real numbers, but only the first three choices can be made independently (of each other and the other six "choices"). So we're really only choosing three. The other six are fixed by the condition Aij=-Aji.
I understand that it's a bit confusing to just say that "A has three independent components", but this terminology is pretty standard, at least in physics books.
Yes, one of the things I wanted you to realize was that the "diagonal" components A11, A22, A33 are all zero.
oh , so for the first equation, its just that the set of 9 numbers determined by Aij is equal to the set of 9 numbers determined by Aji just that the ordering is different only? so they are all equal
so the 2nd equation follows from the condition you stated ?
You explained the second equality correctly (except for leaving out the "s" at the end of "conditions"...there's one for A and one for S), and you're correct that exactly the same terms appear on both sides of the first equality. But I'm not sure how you came to that conclusion, because you still don't seem to have understood what this equality is about. The terms do not appear in different order on the two sides of the = sign. Both sides have exactly the same terms, in exactly the same order.
Now do you see that the same thing works no matter how many indices there are and no matter how many symbols they're attached to? For example, if T is symmetric (unchanged) under the exchange of its 2nd and 4th index, and antisymmetric (changes sign) under the exchange of its 1st and 6th index, then we can immediately conclude that
$$T_{ijkjmiknopq}=0$$
for all values of m,n,o,p,q. When you understand this, you should have another look at the expression
$$\partial_i\varepsilon_{ijk}\partial_j F_k$$
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quietrain
To specify A is to specify the values of its nine components. To specify an A that's consistent with the condition Aij=-Aji, we have to "choose" nine real numbers, but only the first three choices can be made independently (of each other and the other six "choices"). So we're really only choosing three. The other six are fixed by the condition Aij=-Aji.
I understand that it's a bit confusing to just say that "A has three independent components", but this terminology is pretty standard, at least in physics books.
Yes, one of the things I wanted you to realize was that the "diagonal" components A11, A22, A33 are all zero.
You explained the second equality correctly (except for leaving out the "s" at the end of "conditions"...there's one for A and one for S), and you're correct that exactly the same terms appear on both sides of the first equality. But I'm not sure how you came to that conclusion, because you still don't seem to have understood what this equality is about. The terms do not appear in different order on the two sides of the = sign. Both sides have exactly the same terms, in exactly the same order.
Now do you see that the same thing works no matter how many indices there are and no matter how many symbols they're attached to? For example, if T is symmetric (unchanged) under the exchange of its 2nd and 4th index, and antisymmetric (changes sign) under the exchange of its 1st and 6th index, then we can immediately conclude that
$$T_{ijkjmiknopq}=0$$
for all values of m,n,o,p,q. When you understand this, you should have another look at the expression
$$\partial_i\varepsilon_{ijk}\partial_j F_k$$
the matrix i had in mind was
A=
(0 1 2)
(-1 0 3)
(-2 -3 0)
so the 3 independent terms are 1,2 and 3? the 0s and the minus are all dude to the condition?
for S, it can be everything except the diagonal have to be 0, so 6 terms of independent?
on the part of proving the equations
do you mean that Aij = Aji = the matrix of A above?
so Aij and Aji are not really matrix on their own , but are conditions so to speak to relate to the matrix A?
oh so for T, if i index are symmetric, and j are anti, then its like
Tij.... = -Tji....
so i assume the rest of the index can be represented by Smnop....?
so its = 0 ? by the AS example you gave?
so the last expression of del , E and F will just mean all the same likewise as above = 0 ?
Staff Emeritus
Gold Member
the matrix i had in mind was
A=
(0 1 2)
(-1 0 3)
(-2 -3 0)
so the 3 independent terms are 1,2 and 3? the 0s and the minus are all dude to the condition?
for S, it can be everything except the diagonal have to be 0, so 6 terms of independent?
If you interpret Aij as row i, column j of a matrix (which is 100% OK, but not necessary for what we've been doing), then the condition Aij=-Aji does indeed restrict it to the form
$$A=\begin{pmatrix}0 & a & b\\ -a & 0 & c\\ -b & -c & 0\end{pmatrix}$$
But you're wrong about S. It would look like this:
$$A=\begin{pmatrix}d & a & b\\ a & e & c\\ b & c & f\end{pmatrix}$$
do you mean that Aij = Aji = the matrix of A above?
This is like asking if 100=-100=(100,-100,0). You seem to be extremely careless with the notation, and to have failed to understand that the expression "Aij" always refers to a number. Please read this again:
A=$\{A_{ij}\in\mathbb R|i,j\in\{1,2,3\}\}$ is a set of 9 real numbers (not necessarily distinct). $A_{ij}$ is a member of that set, "the ij component". The nine equations $A_{ij}=-A_{ji}$ (one for each choice of i and j) describe how the nine members of the set are related to each other.
so Aij and Aji are not really matrix on their own , but are conditions so to speak to relate to the matrix A?
How could a real number be a "condition"? If we define a matrix A by saying that for each i and each j, Aij is the number on row i, column j of A, then A23 is the number on row 2, column 3, and A32 is the number on row 3 column 2. If A23=5, then A32=-5, because Aij=-Aji holds for all values of i and j, including i=2 and j=3.
I don't think I want to discuss $T_{ijkjmiknopq}$ or $\nabla\cdot(\nabla\times F)$ until I'm 100% sure that you understand why $A_{ij}S_{ij}=A_{ji}S_{ji}$.
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quietrain
If you interpret Aij as row i, column j of a matrix (which is 100% OK, but not necessary for what we've been doing), then the condition Aij=-Aji does indeed restrict it to the form
$$A=\begin{pmatrix}0 & a & b\\ -a & 0 & c\\ -b & -c & 0\end{pmatrix}$$
But you're wrong about S. It would look like this:
$$A=\begin{pmatrix}d & a & b\\ a & e & c\\ b & c & f\end{pmatrix}$$
This is like asking if 100=-100=(100,-100,0). You seem to be extremely careless with the notation, and to have failed to understand that the expression "Aij" always refers to a number. Please read this again:
How could a real number be a "condition"? If we define a matrix A by saying that for each i and each j, Aij is the number on row i, column j of A, then A23 is the number on row 2, column 3, and A32 is the number on row 3 column 2. If A23=5, then A32=-5, because Aij=-Aji holds for all values of i and j, including i=2 and j=3.
I don't think I want to discuss $T_{ijkjmiknopq}$ or $\nabla\cdot(\nabla\times F)$ until I'm 100% sure that you understand why $A_{ij}S_{ij}=A_{ji}S_{ji}$.
i don't understand,
if i were to let
A=
(0 1 2)
(-1 0 3)
(-2 -3 0)
S=
(1 4 5)
(4 2 6)
(5 6 3)
then if i do a matrix multiplication of AxS, i get 9 numbers which has opposite signs to if i were to do -AxS.
but if i do the dot product
A11B11+A12B12+...
then i get 0 for both.
so why do i get 2 type of results ? the matrix multiplication is a cross product right?
so your notation of AijSij is a dot product?
but if i can visualize them as matrix, then how come i get different results?
in any case, if Aij = 5, Aji = -5,
how can 5S = -5S? unless you sum them all up using the dot product way and get 0 ?
Staff Emeritus
Gold Member
Edit: I wrote this post before I saw your post above. I will read it now.
OK, I'll just tell you the answer. The equality $A_{ij}S_{ij}=A_{ji}S_{ji}$ holds because both expressions are by definition exactly the same thing. There are no calculations involved, no properties of A or S, no properties of real numbers. These expressions are equal for the same reason that 5+3 is equal to 5+3.
Consider e.g. the equalities
$$\sum_{i=1}^2 x_i = x_1+x_2= \sum_{j=1}^2 x_i$$
Earlier in the thread, you indicated that you understand that it never matters what symbol you're using as the summation index. But for some reason you never applied that principle to the expression $A_{ij}S_{ij}$.
$$\sum_{i=1}^3\sum_{j=1}^3A_{ij}S_{ij}=\sum_{k=1}^3\sum_{l=1}^3A_{kl}S_{kl}=\sum_{j=1}^3\sum_{i=1}^3A_{ji}S_{ji}$$
Staff Emeritus
Gold Member
i don't understand,
if i were to let
A=
(0 1 2)
(-1 0 3)
(-2 -3 0)
S=
(1 4 5)
(4 2 6)
(5 6 3)
then if i do a matrix multiplication of AxS, i get 9 numbers which has opposite signs to if i were to do -AxS.
Aren't you saying something entirely trivial here, i.e. that with these specific choices of A and S, we have -(-AS)=AS? This identity is obviously true for all matrices A and S. (By the way, you shouldn't write the product as A×S. A product of matrices is always written without a product symbol).
but if i do the dot product
A11B11+A12B12+...
then i get 0 for both.
The dot product of what? The dot product isn't defined for matrices. The sum you typed is equal to the dot product of the first row of A and the first row of S. Why are you calculating that? If you're doing it as a part of a calculation of AS, then you're doing it wrong. AS is defined by $(AS)_{ij}=A_{ik}S_{kj}$, so the entry on row i, column j of AS is the dot product of the ith row of A and the jth column of S.
the matrix multiplication is a cross product right?
No, it has nothing to do with the cross product.
so your notation of AijSij is a dot product?
No. (What two vectors would it be a dot product of?)
but if i can visualize them as matrix, then how come i get different results?
I can't answer that, because I don't understand what you're doing, or why you're doing it.
in any case, if Aij = 5, Aji = -5,
how can 5S = -5S?
I have no idea what you're talking about here. I haven't said anything that suggests that 5S=-5S, and it certainly isn't implied by Aij=-Aji.
the condition Aij=-Aji does indeed restrict it to the form
$$A=\begin{pmatrix}0 & a & b\\ -a & 0 & c\\ -b & -c & 0\end{pmatrix}$$
But you're wrong about S. It would look like this:
$$A=\begin{pmatrix}d & a & b\\ a & e & c\\ b & c & f\end{pmatrix}$$
The second equality here is a typo, but you probably understood that. I meant to type "S=", not "A=". Note that Aij=-Aji implies that the diagonal entries of A are =0, but Sij=Sji says nothing at all about the diagonal entries of S.
quietrain
ok this is what i have in mind
AijSij =
(0 1 2)(1 4 5)
(-10 3)(4 2 6)
(-2-30)(5 6 3)
which gives me
(14 14 12)
(14 14 4)
(-14-14-28)
now AjiSji =
(0-1-2)(1 4 5)
(1 0 -3)(4 2 6)
(2 3 0)(5 6 3)
which gives me
(-14-14-12)
(-14-14-4)
(14 14 28)
so i see that the AijSij = - AjiSji
but is not equal to AjiSji ?
this is because of the condition Aij = -Aji ?
but you are telling me they are all dummy variables? but from the matrix it shows that AijSij is different from AjiSji
><
Staff Emeritus
Gold Member
We can choose to interpret the Aij and Sij for i,j=1,2,3 as the components of two matrices A and S, but even if we do, there's no way AijSij can be a matrix. It's clearly just a sum of real numbers.
You're still not thinking about what definitions you need to use. If you want to treat this expression as something involving a product of matrices, this is what you need:
Definition of matrix multiplication: (XY)ij=XikYkj
Definition of transpose: (XT)ij=Xji
Definition of trace: Tr X=Xii
These definitions tell us that
$$A_{ij}S_{ij}=(A^T)_{ji}S_{ij}=(A^TS)_{jj}=\operatorname{Tr}(A^TS)$$
Alternatively,
$$A_{ij}S_{ij}=A_{ij}(S^T)_{ji}=(AS^T)_{ii}=\operatorname{Tr}(AS^T)$$
You can also reverse the order of the matrices in the trace, because Tr(XY)=Tr(YX) for all matrices X and Y. (This is very easy to prove using the definitions above).
You computed AS and ATST. The conditions I put on the components of A and S translate to AT=-A and ST=S, so ATST=(-A)S=-AS.
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quietrain
so, in AiSj , theres no summation cos the indices do not repeat? so is there any meaning to this notation? or does it mean A1S1 + A2S1 +A3S1 +A1S2+ .. .
but in AiSi it is a dot product where A1S1 + A2S2 + A3S3 ?
so the way to compute AijSij = A11S11 + A11S12 + A11S13 + A21S11 +A21S12+ AA21S13 +. .. . .
so i get 81 terms? 9terms of A x 9 terms of S? or does it follow the kronecker delta where the non-identical indices = 0 and so we are left with the 11 22 and 33 ? so i assume AijSij is not equal to the matrix AS i computed since you said its just a number summed?
on 2nd thought,
if AiSi = A1S1 +A2S2 + ...
then does it mean AijSij = A11S11 +A12S12 + A13S13+A21S21+...
that means i cannot have a A11S12 where the indices of A, 11, is not equal to S, 12 ? so i get 9 terms instead of 81?
in any case, if i so if AijSij = AjiSji
(this is just dummy variables right? its not meaning that i swop the Aij=A12 to a Aji=A21 right? )
then it means AijSij = -AijSij from the condtion you set,
so 2AijSij=0
so AijSij = 0 ?
thus
[PLAIN]https://www.physicsforums.com/latex_images/31/3102234-8.png [Broken]
so, the deltas are the symmetric S and the levi civita are the antisymmetric A, and F is the other components that you generalized? am i right to say that there is summation here due to the levi civita symbol? or else like at the top of this post, AiSj would have no meaning without a summation symbol "$$\sum$$" placed to the left of the expression ?
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Staff Emeritus
Gold Member
so, in AiSj , theres no summation cos the indices do not repeat? so is there any meaning to this notation? or does it mean A1S1 + A2S1 +A3S1 +A1S2+ .. .
If i=1 and j=1, then it means A1S1.
If i=2 and j=1, then it means A2S1
etc.
I have to say, it's pretty frustrating that you are still stuck on the meaning of (Something)something this late in the thread. I have spent many hours trying to explain this to you. One last time: We're dealing with finite sets whose members are identified by the values of one or more indices that take integer values from 1 to 3. The members of the set are usually real numbers, but they can also be e.g. basis vectors or partial derivative operators.
but in AiSi it is a dot product where A1S1 + A2S2 + A3S3 ?
Yes. A person who understands that (Something)something refers to a member of an indexed set of real numbers, and who understands the summation convention, can take the following as the definition of the dot product:
$$(A_1,A_2,A_3)\cdot (S_1,S_2,S_3)=A_i S_i$$
so the way to compute AijSij = A11S11 + A11S12 + A11S13 + A21S11 +A21S12+ AA21S13 +. .. . .
so i get 81 terms? 9terms of A x 9 terms of S? or does it follow the kronecker delta where the non-identical indices = 0 and so we are left with the 11 22 and 33 ? so i assume AijSij is not equal to the matrix AS i computed since you said its just a number summed?
on 2nd thought,
if AiSi = A1S1 +A2S2 + ...
then does it mean AijSij = A11S11 +A12S12 + A13S13+A21S21+...
that means i cannot have a A11S12 where the indices of A, 11, is not equal to S, 12 ? so i get 9 terms instead of 81?
It's also frustrating to see this, because it means that you still haven't understood that the summation convention is nothing more than the practice to not bother to write the summation sigmas when an index occurs twice. Or is the problem that you don't understand the sigma notation for sums? Do you not see that you just asked me if
$$\sum_{i=1}^3\sum_{j=1}^3 X_{ij}$$
has 81 terms?
in any case, if i so if AijSij = AjiSji
(this is just dummy variables right? its not meaning that i swop the Aij=A12 to a Aji=A21 right? )
Right. Dummy variables. No swapping.
then it means AijSij = -AijSij from the condtion you set,
so 2AijSij=0
so AijSij = 0 ?
Correct.
thus
[PLAIN]https://www.physicsforums.com/latex_images/31/3102234-8.png [Broken]
so, the deltas are the symmetric S and the levi civita are the antisymmetric A, and F is the other components that you generalized?
Yes.
$$\partial_i\varepsilon_{ijk}\partial_j F_k=\partial_j\varepsilon_{jik}\partial_i F_k=-\partial_i\varepsilon_{ijk}\partial_j F_k\quad\Rightarrow\quad \partial_i\varepsilon_{ijk}\partial_j F_k=0$$
am i right to say that there is summation here due to the levi civita symbol? or else like at the top of this post, AiSj would have no meaning without a summation symbol "$$\sum$$" placed to the left of the expression ?
There's a summation over i because i occurs twice (and for no other reason).
There's a summation over j because j occurs twice (and for no other reason).
There's a summation over k because k occurs twice (and for no other reason).
Without the summation convention, the right-hand side is written as
$$\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\partial_i\varepsilon_{ijk}\partial_j F_k$$
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quietrain
ok i better start from scratch
if (must i put in a summation here?since they are not repeated indices?)$$\sum$$Aij , i,j from 1 to 3
means A11+A12+A13+21+22+23.... 9 terms
if i impose a condition of Aij=-Aji (why not just put Aij=-Aij since they are just dummy variable?)
then the value of Aij = 0 summed since they all cancel each other out right?
so now if i multiply in a Sij term which is governed by the condition = Sji
then (i drop summation symbol due to convention due to repeated indices)AijSij = A11S11+A12S12+A13S13+...
so i get 9 terms now and they all add up to a number. and so this number = 0 because of the equations AijSij=AjiSji=-AijSij=0
and the rest follows...
well i thank you for your patience
Staff Emeritus
Gold Member
if (must i put in a summation here?since they are not repeated indices?)$$\sum$$Aij , i,j from 1 to 3
means A11+A12+A13+21+22+23.... 9 terms
Correct.
if i impose a condition of Aij=-Aji (why not just put Aij=-Aij since they are just dummy variable?)
then the value of Aij = 0 summed since they all cancel each other out right?
Yes, the sum of all the components of A add up to zero. As for your other question, the alternative condition that you're suggesting clearly just says that all the components are zero. Your question shows that you have completely misunderstood the concept of dummy variables.
so now if i multiply in a Sij term which is governed by the condition = Sji
then (i drop summation symbol due to convention due to repeated indices)AijSij = A11S11+A12S12+A13S13+...
so i get 9 terms now and they all add up to a number. and so this number = 0 because of the equations AijSij=AjiSji=-AijSij=0
Correct.
quietrain
Correct.
Yes, the sum of all the components of A add up to zero. As for your other question, the alternative condition that you're suggesting clearly just says that all the components are zero. Your question shows that you have completely misunderstood the concept of dummy variables.
Correct.
oh haha... i see thank you
Staff Emeritus
Gold Member
Just to be clear, the trivial aspect of "dummy variables" that you somehow managed to overlook, is that the fact that i is a dummy variable in
$$\sum_i x_i$$
doesn't mean that it's a dummy variable in e.g. xiyj. In the former expression, we can replace i by k, precisely because we're summing over all the values. In the latter expression, we can't make that replacement, because what if e.g. i=1 and k=2? Then we have replaced x1yj with x2yj.
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quietrain
Just to be clear, the trivial aspect of "dummy variables" that you somehow managed to overlook, is that the fact that i is a dummy variable in
$$\sum_i x_i$$
doesn't mean that it's a dummy variable in e.g. xiyj. In the former expression, we can replace i by k, precisely because we're summing over all the values. In the latter expression, we're can't make that replacement, because what if e.g. i=1 and k=2? Then we have replaced x1yj with x2yj.
oh,... now it makes more sense...
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# AAPG Bulletin
Abstract
AAPG Bulletin, V. 97, No. 7 (July 2013), P. 11811207.
DOI:10.1306/11011212076
## Linking process, dimension, texture, and geochemistry in dolomite geobodies: A case study from Wadi Mistal (northern Oman)
### Veerle Vandeginste,1 Cedric M. John,2 Tina van de Flierdt,3 John W. Cosgrove4
1Department of Earth Science and Engineering, Imperial College London, SW7 2BP, London, United Kingdom; v.vandeginste@imperial.ac.uk
2Department of Earth Science and Engineering, Imperial College London, SW7 2BP, London, United Kingdom; cedric.john@imperial.ac.uk
3Department of Earth Science and Engineering, Imperial College London, SW7 2BP, London, United Kingdom; tina.vandeflierdt@imperial.ac.uk
4Department of Earth Science and Engineering, Imperial College London, SW7 2BP London, United Kingdom; j.cosgrove@imperial.ac.uk
## ABSTRACT
Understanding the distribution and geometry of reservoir geobodies is crucial for net-to-gross estimates and to model subsurface flow. This article focuses on the process of dolomitization and resulting geometry of diagenetic geobodies in an outcrop of Jurassic host rocks from northern Oman. Field and petrographic data show that a first phase of stratabound dolomite is crosscut by a second phase of fault-related dolomite. The stratabound dolomite geobodies are laterally continuous for at least several hundreds of meters (1000 ft) and probably regionally and are one-half meter (1.6 ft) thick. Based on petrography and geochemistry, a process of seepage reflux of mesosaline or hypersaline fluids during the early stages of burial diagenesis is proposed for the formation of the stratabound dolomite. In contrast, the fault-related dolomite geobodies are trending along a fault that can be followed for at least 100 m (328 ft) and vary in width from a few tens of centimeters to as much as 10 m (1–33 ft). Petrography, geochemistry, and high homogenization temperature of fluid inclusions all point to the formation of the dolomite along a normal fault under deep burial conditions during the Middle to Late Cretaceous. The high 87Sr/86Sr ratio in the dolomite and the high salinity measured in fluid inclusions indicate that the dolomitizing fluids are deep basinal brines that interacted with crystalline basement. The dolomitization styles have an impact on the dimension, texture, and geochemistry of the different dolomite geobodies, and a modified classification scheme (compared to the one from Jung and Aigner, 2012) is proposed to incorporate diagenetic geobodies in future reservoir modeling.
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2014-09-18 13:40:24
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https://math.stackexchange.com/questions/51444/find-largest-integer-x-such-that-3x-is-a-factor-of-275
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# Find largest integer $x$ such that $3^x$ is a factor of $27^5$
Is the following solvable using just arithmetic rather than a calculator, and if so, how?
Which of the following numbers is the greatest positive integer x such that $3^x$ is a factor of $27^5~$?
• a) $5$
• b) $8$
• c) $10$
• d) $15$
• e) $19$
The answer is a), but I'm not sure how to get to that conclusion. We're not supposed to use a calculator.
• a) is wrong. The answer is d). – Aryabhata Jul 14 '11 at 15:51
There is a great power at play here called Unique Factorization. Every number can be written uniquely (up to ordering) as a product of primes. $27 = 3^3$, so $27^5 = (3^3)^5 = 3^{15}$.
Now how do you go about finding out if a number divides $3^{15}$?
And why do you think a is the correct answer? Perhaps it's not.
• This has nothing to do with unique factorization but, rather, only that $\rm\:3\ne \pm1,0\:.\:$ Indeed, in any ring, for a nonunit non-zero-divisor $\rm\:c\:,\:$ and for $\rm\:n,k\in \mathbb N,\ \ c^{\:n+k}\ |\ c^n\ \iff\ k = 0\:.$ – Bill Dubuque Jul 14 '11 at 16:42
• You are quite right, the answer was in fact d). Thanks for the explanation! – Joe W Jul 14 '11 at 16:47
Big hint: $$27^5 = 3^{5 \times 3}$$
• I rewrote your tex to make it render - I hope you don't mind. – davidlowryduda Jul 14 '11 at 15:53
The fundamental fact that leads to quickly solving this without a calculator is to realize that $27 = 3^3$ and, therefore, $27^x$ = $(3^3)^x$ = $3^{(3x)}$. Therefore $27^5 = 3^{15}$.
And which of the answers given is a factor of $3^{15}$? All are factors of $3^{15}$, except answer 'e.' Which of answers 'a.' through 'd.' is the largest? Answer 'd.'
HINT $\rm\ \ 27 = 3^3\ \Rightarrow\ 27^{\:5} = 3^{15}\:,\$ so $\rm\ 3^{m}\ |\ 3^{15} \iff\ m \:\le\: \cdots$
NOTE $\$ Despite a remark elsewhere, this has little to do with unique factorization of integers. Rather,$\:$ it is a consequence of $\rm\ c\: =\: 3 \:\ne\: \pm1\:,\:0\$ implies that $\rm\ c\$ is a nonunit and $\rm\ c\$ is cancellable, i.e. $\rm\ a\:c = b\:c\ \Rightarrow\ a = b\:.\:$ These two properties easily imply $\rm\ c^m\ |\ c^{\:n}\ \iff\ m\:\le \:n\:,\:$ namely
LEMMA $\rm\ \ c^{\:m}\ |\ c^{\:n} \iff\ m\le n\:\$ for a cancellable nonunit $\rm\:c\:$ in a ring $\rm\:R\:,\ m,n\in\mathbb N$
Proof $\ (\Leftarrow)\$ Clear. $\ (\Rightarrow)\$$\rm\ \ m>n,\:\ c^{m}\ |\ c^n\:$ in $\rm\:R\ \Rightarrow\ \exists\: d\in R:\ d\ c^{m}\! =\: c^{\:n}\:.\:$ Cancelling $\rm\:n\:$ factors of $\rm\:c\:$ yields $\rm\ d\:c^{m-n} = 1\$ so $\rm\ m > n\ \Rightarrow\ c\:|\:1\:,\:$ contra $\rm\:c\:$ nonunit. Hence $\rm\ m\le n\:.$
As a corollary we infer that the above lemma is true for $\rm\:c\:$ being any nonzero nonunit element of a domain (in particular for any nonzero prime element of a UFD = unique factorization domain). But the essence of the matter has nothing at all to do with the property of an element being prime, or with the existence or uniqueness of factorizations into irreducibles.
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2019-08-25 20:21:49
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http://www.lastfm.se/user/xPunkyMandax/library/music/The+Strokes/_/The+End+Has+No+End?from=1303830161&rangetype=year&setlang=sv
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# Bibliotek
Musik » The Strokes »
## The End Has No End
24 spelade låtar | Gå till låtsida
Från tisdag 26 april 2011 till torsdag 26 april 2012 Alltid
Låtar (24)
Låt Album Längd Datum
The End Has No End 3:05 5 apr 2012, 20:52
The End Has No End 3:05 30 dec 2011, 17:57
The End Has No End 3:05 30 nov 2011, 10:48
The End Has No End 3:05 16 nov 2011, 23:35
The End Has No End 3:05 16 nov 2011, 23:35
The End Has No End 3:05 7 nov 2011, 21:57
The End Has No End 3:05 1 nov 2011, 23:33
The End Has No End 3:05 15 okt 2011, 07:00
The End Has No End 3:05 12 okt 2011, 08:13
The End Has No End 3:05 24 sep 2011, 09:23
The End Has No End 3:05 12 sep 2011, 03:15
The End Has No End 3:05 7 sep 2011, 21:43
The End Has No End 3:05 7 sep 2011, 19:04
The End Has No End 3:05 1 sep 2011, 22:57
The End Has No End 3:05 26 aug 2011, 23:45
The End Has No End 3:05 24 aug 2011, 01:10
The End Has No End 3:05 20 aug 2011, 23:56
The End Has No End 3:05 20 aug 2011, 23:53
The End Has No End 3:05 20 aug 2011, 14:01
The End Has No End 3:05 2 aug 2011, 06:02
The End Has No End 3:05 2 aug 2011, 04:30
The End Has No End 3:05 27 jul 2011, 15:55
The End Has No End 3:05 16 maj 2011, 14:35
The End Has No End 3:05 16 maj 2011, 11:56
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2013-05-24 02:05:46
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http://scicomp.stackexchange.com/questions/3199/demonstrating-that-the-time-step-size-is-small-enough-in-a-code-with-automatic-s
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# Demonstrating that the time step size is small enough in a code with automatic step size selection
I recently inherited a large body of legacy code that solves a very stiff, transient problem. I would like to demonstrate that the spatial and temporal step sizes are small enough that the qualitative nature of the computed solution would not change if they were decreased. In other words, I'd like to show that the solution is "converged" in a qualitative sense. Since I can explicitly set the spatial mesh size, that part is straightforward. However, as the code uses automatic time-step size control, I cannot set the time step size directly.
The algorithm changes the time step between two bounds based on the number of jacobian iterations needed to reach an error tolerance during the last $n$ time steps. The fact that it uses jacobian iteration makes me fairly certain that it is some sort of implicit scheme, but I cannot be absolutely certain. It does not account for the error it is seeing in the current time step, which leads to it running into the iteration limit on occasion (maybe a dozen times over the course of several thousand time steps, almost always during the most dynamic portions of the simulation). The current runs I am completing I am setting the time-step bounds two and a half orders of magnitude apart ($10^{-13}$ to $5 \cdot 10^{-11}$).
In the runs, I have control over the time-step bounds, the number of past time-steps it looks at to choose the current time step, the maximum change in the time step (ratios), the target number of jacobian iterations, the maximum number of iterations, and the error bound. I would like if someone could put me on the right path to analyzing the time-step independence, or at the very least figuring out what algorithm is used.
-
Are you saying that you think it will be easier to reverse-engineer the time stepping algorithm than to just read the code? – David Ketcheson Sep 4 '12 at 1:41 the code is approximately 50k lines of fortran written over the past 20 years, so hunting down the details of the main loop is non-trivial. I believe that it is an implicit method, which is enough information for my purposes. I am more interested in what I need to change in separate runs to establish that my time-step is sufficiently small. – Godric Seer Sep 4 '12 at 2:26 I've tried to clarify what is being asked; please correct it if I've misinterpreted. Note that the solution cannot be literally "time-step independent", since the local errors will always depend on the time step. You can only hope that the errors are small enough for your purposes. – David Ketcheson Sep 4 '12 at 5:16
The purpose of automatic error estimation and step size control is to free you from the problem of determining manually what a sufficiently small step size is. So your question is a bit like asking "somebody gave me this automatic transmission car; how can I tell what gear I'm in?" The point is that you shouldn't need to know. Of course, if the transmission is faulty, then you might need to take it apart and fix it, but that's a much bigger problem.
In your case, typically the right approach is to determine what kind of error is acceptable and impose that through the automatic step size control. It's imperfect because error control in this sense is usually only local error control, so you don't directly control the global error, which is what you probably care about.
One thing you could easily do if you're in doubt, is to run the simulation with a sequence of increasingly tight (i.e. small) error tolerances. Once the solution seems insensitive (in whatever your metric is) to decreasing the tolerance, you can stop.
Addendum: Regarding the issue of the maximum iteration limit being reached (which leads to a local error exceeding the specified tolerance), I suggest the following.
Apparently the code thinks that if it exceeds the maximum number of iterations, the right thing to do is to accept the step. I would say that is the wrong thing. A better approach is to reject the step and start that step again with a reduced step size. Of course, there is the danger of the step being reduced below the minimum step size. In that case, the right thing is to abort the simulation. But if you believe that a wrong solution is better than no solution, you could just accept the step and go on if both conditions are met: the minimum step size is reached and the maximum number of iterations is exceeded.
In a well-designed code, making these types of changes is trivial, but in an arbitrary code it may be arbitrarily difficult.
-
The last paragraph is the important one from a practical point of view. The asymptotic regime is reached once the plot of relevant solution characteristics against specified error tolerances looks regular enough. – Arnold Neumaier Sep 4 '12 at 7:17 That is good news, however I am still unsure how to handle those few occasions when it reaches the maximum iteration limit. I have seen it happily take the maximum time step, reaching $10^{-7}$ local error in half a dozen iterations, then the next time step take 100 iterations and only reach a local error of $10^2$. The time step after this it reduces the step size and reaches the error bound fine, but I still have that one large error propagating through the solution. – Godric Seer Sep 4 '12 at 11:13 I just noticed the addendum. Your description of what the code actually does agrees with what I see from logfiles. Hopefully the change you suggest is possible without too many headaches. – Godric Seer Sep 5 '12 at 16:17
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2013-05-19 23:02:02
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https://www.imj-prg.fr/gestion/evenement/affSeance/1735
|
# Séminaires : Séminaire Groupes Réductifs et Formes Automorphes
Equipe(s) : fa, tn, Responsables : Alexis Bouthier, Benoît Stroh Email des responsables : alexis.bouthier@imj-prg.fr, benoit.stroh@imj-prg.fr Salle : Adresse : Description
Orateur(s) Yannan QIU - Columbia-IHES, Titre Local integrals of matrix coefficients Date 05/12/2011 Horaire 10:30 à 11:30 Résume The Refined Global Gross-Prasad Conjecture expresses periods of tempered representations as the product of certain $L$-values and certain local integrals of matrix coefficients. When the global representations are non-tempered, these local integrals could be divergent and we suggest that they be regularized using spherical height functions. As an example, we study the $SO(4)$-periods of $SO(4) \times SO(5)$ representations when the $SO(5)$-representations are non-tempered and of Saito-Kurokawa type; using theta lifting and our regularization technique, we obtain two precise global period formulas. Salle Adresse
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2020-02-17 19:29:43
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https://byjus.com/ncert-solutions-class-6-maths/chapter-10-measurement-and-mensuration/
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# NCERT Solutions For Class 6 Maths Chapter 10
## NCERT Solutions For Class 6 Maths Chapter 10 PDF Free Download
NCERT Solutions For Class 6 Maths Chapter 10 is an important resource for students to prepare for Mensuration topic. Several times, in exams, a majority of questions are asked from this section making it one of the most crucial chapters of class 6th maths subject.
The NCERT Solutions for Class 6 Maths Mensuration chapter is prepared by our expert as per CBSE syllabus(2018-2019), to develop a strong conceptual base among students. These solved questions will help students to resolve their difficulties while solving the Mensuration problems, present in the NCERT textbook and also while solving sample papers and previous year question papers.
## Class 6 Maths NCERT Solutions for Mensuration
The students can refer the exercise-wise Solutions for class 6 maths chapter 10 and also can download the pdf file of these to study offline. The solutions provide a detailed explanation for all the concepts with the best possible methods. These works as a worksheet for students, which they can practice well and score good marks in the final exams, 2019.
Some of these major concepts in this chapter of Mensuration deals with problems related to finding the perimeter of a variety of different shapes such as;
• ‘L’ shapes
• Parallelograms
• Pentagons And
• other geometric faces
The entire chapter of measurement and mensuration deals with these concepts. Check out the chapter-wise solutions for class 6 maths chapter 10 below:
### NCERT Questions For Mensuration
#### Exercise 10.1
Q.1 Find the perimeter of each of the following figures
(a)
Ans.
Perimeter is the distance covered along the boundary of a closed figure.
Perimeter = Sum of the distances or lengths of all the sides
= PQ + QR + RS + SP.
= 3 cm + 2 cm + 1 cm + 4 cm.
= (3 + 2 + 1 + 4) cm
= 10 cm
The perimeter of the given figure is 10 cm.
(b)
Ans.
Perimeter = Sum of the distances or lengths of all the sides
= EF + FG + GH + HE.
= 20 cm + 30 cm + 20 cm + 45 cm.
= (20 + 30 + 20 + 45) cm
= 105 cm
The perimeter of the given figure is 105 cm.
(c)
Ans.
Perimeter = Sum of the distances or lengths of all the sides.
= LM + MN + NO + OL
= 20 cm + 20 cm + 20 cm + 20 cm
= (20 + 20 + 20 + 20) cm
= 80 cm
The perimeter of the given figure is 80 cm.
(d)
Ans.
Perimeter = Sum of the distances or lengths of all the sides.
= AB + BC + CD + DE + EA
= 5 cm + 5 cm + 5 cm + 5 cm + 5 cm
= (5 + 5 + 5 + 5 + 5) cm
= 25 cm
The perimeter of the given figure is 25 cm.
(e)
Perimeter = Sum of the distances or lengths of all the sides.
We are simplifying the problem by considering a single ‘L’ shaped section and finding its perimeter. There are four such ‘L’ shaped sections and the total perimeter is four times the perimeter of a single section.
Perimeter of the ‘L’ shaped section = AB + BC + CD + DE + EF
= 3 + 4 + 1 + 3 + 2
= 13 cm
The perimeter of a single ‘L’ section is 13 cm
The perimeter of the entire figure = 4 x perimeter of each ‘L’ section.
= 4 x 13 cm
= 52 cm
The perimeter of the entire figure is 52 cm.
Q2. The lid of a rectangular box of sides 50 cm by 10 cm is sealed all around with a tape. What is the length of the tape required?
Ans.
Dimensions of the given box:
Length of the rectangular box (l) = 50 cm
Breadth of the rectangular box (b) = 10 cm
Perimeter of a rectangle = 2 x (l + b)
= 2 x (50 cm + 10 cm)
= 2 x (60 cm)
= 120 cm
The perimeter of the rectangular box = Length of the tape required to measure the box = 120 cm
Q3. A table-top measures 3 m 35 cm by 1 m 40 cm. What is the perimeter of the table-top?
Ans.
The length of the rectangle should be greater than the breadth (l $>$ b)
Length of the rectangle (l) = 3 m 35 cm
= (300 + 35) cm (∵ 1 m = 100 cm)
= 335 cm
Breadth of the rectangle (b) = 1 m 40 cm
= (100 + 40) cm (∵ 1 m = 100 cm)
= 140 cm
Perimeter of the table top = Perimeter of the rectangle
= 2 x (length + breadth)
= 2 x (l + b)
= 2 x (335 cm + 140 cm)
= 2 x (475 cm)
= 950 cm
The perimeter of the table top = 950 cm = 9.5 m
Q4. A photo of sides 17.8 cm by 12.7 cm is to be framed with wood. What is the length of wooden strip required?
Ans:
Length of photograph (I) = 17.8 cm
Breadth of photograph (b) = 12.7 cm
Length of required wooden strip = Perimeter of Photograph
= 2 x (/ + b)
= 2 x (17.8 + 12.7)
= 2 x 30.5 = 61 cm
Q5. Each side of rectangular piece of land which measures 0.8km by 0.7km is to be fenced with 5 rows of wires. How much length of wire is needed?
Ans:
Length of land (I) = 0.8 km
Breadth of land (b) = 0.7 km
Perimeter = 2 x (I + b) = 2 x (0.8 + 0.7) = 2 x 1.5 = 3.0 km
Length of wire required = 5 x 3 = 15 km
Q6. Find the perimeter of the following shapes:
(a) An equilateral triangle of side 5 cm.
(b) An Isosceles triangle with equal sides 6 cm each and the third side 5 cm.
(c) A triangle of sides 6 cm, 3 cm and 2 cm
Ans:
(a) Perimeter = of an equilateral triangle = 3 x Side of triangle
= (3 x 5) cm = 15 cm
(b) Perimeter = (6 cm + 3 cm + 2 cm) = 11 cm.
(c) Perimeter = (2 x 6) + 5 = 17 cm
Q7. A triangle has sides measuring 15 cm, 12 cm and 13 cm. Find its perimeter.
Ans:
Perimeter of the triangle = Sum of the lengths of all sides of the triangle
Perimeter = 15 + 12 + 13 = 40 cm
Q8. A regular hexagon of each side measuring 6 m is given. Find its perimeter.
Ans:
Perimeter of regular hexagon = 6 x Side of regular hexagon
Perimeter of the regular hexagon = 6 x 6 = 36 m
Q9. Find the side of the square whose perimeter is 16 m.
Ans:
Perimeter of square = 4 x Side
16 = 4 x Side
$Side = \frac{16}{4} = 4\ cm$
Side = 4
Q10. Find the side of a regular pentagon if its perimeter is 150 cm.
Ans:
Perimeter of regular pentagon = 5 x Length of side
150 = S x Side
$Side = \frac{150}{5} = 30\ cm$
Therefore, Side = 30 cm
Q11. A piece of string is 120 cm long. What will be the length of each side if the string is used to form
(a) A regular hexagon?
(b) A square?
(c) An equilateral triangle?
Ans:
(a) Perimeter = 4 x Side
120 = 4 x Side
$Side = \frac{120}{4} = 30\ cm$
(b) Perimeter = 3 x Side
120 = 3 x Side
$Side = \frac{120}{3} = 40\ cm$
(c) Perimeter = 6 x Side
120 = 6 x Side
$Side = \frac{120}{6} = 20\ cm$
Q12: What is the third side of a triangle if two of its sides are 15 cm and 12 cm, whose perimeter is 40 cm?
Ans:
Perimeter of triangle = Sum of all sides of the triangle
40 = 15 + 12 + Side
40 = 27 + Side
Side = 40 – 27 = 13 cm
Hence, the third side of the triangle is 13 cm.
Q13: A square park of side 240m is to be fenced. Find the cost of fencing if the rate of fencing per meter is Rs.10.
Ans:
Length of fence required = Perimeter of the square park
= 4 x Side = 4 x 240 = 960 m
Cost for fencing 1 m of square park = Rs 10
Cost for fencing 960 m of square park = 960 x 10 = Rs 9600
Q14: A rectangular park of length 150 and breadth 175 is to be fenced. Find the cost if the fencing rate is Rs.15 per meter.
Ans:
Length of rectangular park (I) = 150 m
Breadth of rectangular park (b) = 175 m
Length of wire required for fencing the park = Perimeter of the park
= 2 x (/ + b) = 2 x (150 + 175) = 2 x 325 = 650 m
Cost for fencing 1 m of the park = Rs 15
Cost for fencing 650 m of the square park = 650 x 15 = Rs 9750
Q15. Adam runs around a square park of side 80 m. Eve runs around a rectangular park with length 70 m and breadth 40 m. Who covers a comparatively lesser distance?
Ans:
Distance covered by Adam = 4 x Side of square park
= 4 X 80 = 240 m
Distance covered by Eve = 2 x (70 + 40)
= 2 x 110 = 220 m
Therefore, Eve covers less distance.
Q16: What is the perimeter of each of the following figures? What do you infer from the answers?
Ans:
(a) Perimeter of square = 4 x 50 = 200 cm
(b) Perimeter of rectangle = 2 x (20 + 80) = 200 cm
(c) Perimeter of rectangle = 2 x (40 + 60) = 200 cm
(d) Perimeter of triangle = 60 + 60 + 80 = 200 cm
It can be inferred that all the figures have the same perimeter.
Q17: Ram buys 9 square slabs, each having a side of 2 m. Initially, he lays them in the form of a square.
(a) Looking at the figure, find out the perimeter of his arrangement.
(b) His friend Mani looked at the figure and asked him to lay them out in the shape of a cross. Looking at the second figure, find out the perimeter of the cross.
(c) Out of the two shapes, find out which one has a larger perimeter.
(d) Mani suggests that there’s a way to get an even greater perimeter. Can you guess a way of doing it? (The paving slabs must meet along complete edges. That is, they cannot be broken.)
Ans:
(a) Side of the square = (3 x 2) m = 6 m
Perimeter of the square = (4 x 3) m = 12 m
(b) Perimeter of the cross = 2+4+4+2+4+4+2+4+4+2+4+4 = 40m
(c) The cross formation has the greatest perimeter of all arrangement.
(d) Formation with perimeter which is greater than 40 m can’t be determined.
#### Exercise-10.2
Q1. Count the squares and find out the area of the following diagrams:
Ans:
(a) The figure contains 9 fully filled squares only. Therefore, the area of this figure will be 9 square units.
(b) The figure contains 5 fully filled squares only. Therefore, the area of this figure will be S square units.
(c) The figure contains 2 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 4 square units.
(d) The figure contains 8 fully filled squares only. Therefore, the area of this figure will be 8 square units.
(e) The figure contains 10 fully filled squares only. Therefore, the area of this figure will be 10 square units.
(f) The figure contains 2 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 4 square units.
(g) The figure contains 4 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 6 square units.
(h) The figure contains 5 fully filled squares only. Therefore, the area of this figure will be S square units.
(I) The figure contains 9 fully filled squares only. Therefore, the area of this figure will be 9 square units.
(j) The figure contains 2 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 4 square units.
(k) The figure contains 4 fully filled squares and 2 half-filled squares. Therefore, the area of this figure will be 5 square units.
(l) From the given figure, it can be observed that,
Covered Area Number Area estimate ( sq units ) Fully filled squares 2 2 Half filled squares – – More than half filled squares 6 6 Less than half filled squares 6 0
Total area = 6 + 2 = 8 sq units.
(m) From the above figure, it is observed that,
Covered Area Number Area estimate ( sq units ) Fully filled squares 5 5 Half filled squares – – More than half filled squares 10 10 Less than half filled squares 9 0
Total area = 9 + 5 = 14 sq units
(n) From the above figure, it is observed that,
Covered Area Number Area estimate ( sq units ) Fully filled squares 8 8 Half filled squares – – More than half filled squares 10 10 Less than half filled squares 9 0
Total area = 10 + 8 = 18 sq units.
#### Exercise-10.3
Q1. What is the area of the following rectangles whose sides’ measure:
(a) 2 cm and 5 cm
(b) 15 m and 20 m
(c) 4 km and 5 km
(d) 3 m and 60 cm
Ans: It is known that, Area of a rectangle = Length (l) x Breadth (b)
(a) l = 2 cm
b = 5 cm
Area = l x b = 2 x 5 = 12 cm2
(b) l = 15 m
b= 20 m
Area = l x b= 20 x 15 = 300 m2
(c) l = 4 km
b= 5 km
Area = l x b = 4 x 5 = 20 km2
(d) l = 3 m
b = 60 cm = 0.60 m
Area = l x b=2 x 0.60 = 1.20 m2
Q2. Find the areas of the squares whose sides are: (a) 8 cm (b) 11 cm (c) 6 m
Ans:
It is known that, Area of a square = (Side)2
(a) Side = 8 cm Area = (8)2 =64 cm2
(b) Side = 11 cm Area = (11)2 = 121 cm2
(c) Side = 6 m Area = (6)2 = 36 m2
Q3. Which one has the largest and smallest area among the 3 rectangles whose dimensions measure:
(a) 5 m by 6 m
(b) 12m by 4 m
(c) 2 m by 10 m
Ans: We know that, Area of rectangle = Length x Breadth
(a) l = 5m
b = 6m
Area = l x b = 5 x 6 = 30 m2
(b) l = 12m
b = 4m
Area = l x b = 12 x 4 = 48 m2
(c) l = 2m
b = 10m
Area = l x b = 2 x 10 = 20 m2
From the above results, we come to know that (b) has the largest area and rectangle (c) has the smallest area.
Q5: What is the cost of tiling a rectangular plot of land 400 m long and 300 m wide at the rate of Rs 5 per hundred sq in?
Ans:
Area of rectangular plot = 400 x 300 = 120000 m2
Cost of tiling per 100 m2 = Rs 5
Cost of tiling per 120000 m2 = $\frac{5}{100}\times 120000 = Rs.6000$
Q6: What is the area of table whose sides measure 1m 20 cm in length and 2m 50 cm in breadth?
Ans:
Length (l) = 1 m 20cm = $(1+\frac{20}{100})\ m = 1.2\ m$
Breadth (b) 2 m 50 cm = $(2+\frac{50}{100})\ m = 2.5\ m$
Area = l x b = 1.2 x 2.5 = 3 m2
Q7: How many sq meters of carpet is needed to cover the floor of a room of sides measuring 3 m in length and 4 m 20cm in breadth?
Ans:
Length (l) = 3 m
Breadth (b) = 4 m 20 cm = $(4+\frac{20}{100})\ m = 4.2\ m$
Area = l x b = 3 x 4.2 = 12.6 m2
Q8: A square carpet is laid on the floor of sides measuring 4 m in a floor of sides 4 m and 5 m. Find the remaining area of the floor that is not carpeted.
Ans:
Length (I) = 4 m
Breadth (b) = 5 m
Area of floor = l x b = 4 x 5 = 20 m2
Area covered by the carpet = (Side) 2 = (4)2 = 16 m2
Area not covered by the carpet = 20 – 16 = 4 m2
Q9: From a piece of land measuring 4 m long and 6 m wide, 4 sq flower beds of side 2m each are dug out. Find out the remaining area of the land.
Ans:
Area of the land = 4 x 6 = 24 m2
Area occupied by 4 flower beds = 4 x (Side)2 = 4x (2)2 = 16m2
Area of the remaining part = 24 – 16 = 8 m2
Q10: Split the below figures into multiple rectangles and find out their area. All dimensions are given in centimeters.
Ans:
(a) The above figures can be split in to multiple rectangles as follows:
Area of 1st rectangle = 8 x 4 = 32 cm2
Area of 2nd rectangle = 12 x 2 = 24 cm2
Area of 3rd rectangle = 6 x 4 = 24 cm2
Area of 4th rectangle = 8 x 4 = 32 cm2
Total area of the whole figure = 32 + 24 + 24 + 32 = 112 cm2
(b) The above figures can be split in to multiple rectangles as follows:
Area of 1st rectangle = 6 x 2 = 12 cm2
Area of 2nd rectangle = 6 x 2 = 12 cm2
Area of 3rd rectangle = 6 x 2 = 12 cm2
Total area of the whole figure = 12 + 12 + 12 = 36 cm2
Q11. Split the below figures into multiple rectangles and find out their area. All dimensions are given in centimeters.
Ans:
(a) The above figures can be split in to multiple rectangles as follows:
Area of 1st rectangle = 24 x 4 = 96 cm2
Area of 2nd rectangle = 16 x 4 = 64 cm2
Total area of the whole figure = 96 + 64 = 160 cm2
(b) The above figures can be split in to multiple rectangles as follows:
Area of 1st rectangle = 42 x 14 = 588 cm2
Area of 2nd rectangle = 14 x 14 = 196 cm2
Area of 3rd rectangle = 14 x 14 = 196 cm2
Total area of the whole figure = 588 + 196 + 196 = 980 cm2
(c) The above figures can be split in to multiple rectangles as follows:
Area of 1st rectangle = 10 x 2 = 20 cm2
Area of 2nd rectangle = 8 x 2 = 16 cm2
Total area of the whole figure = 20 + 16 = 36 cm2
Q12. Tiles of sides measuring 10 cm by 4 cm each are to be fit in a rectangular region whose length and breadth are respectively:
(a) 144 cm and 100 cm
(b) 36 cm and 70 cm
Ans. (a) Total area of the region = 144 x 100 = 14400 cm2
Area of one tile = 10 x 4 = 40 cm2
Number of tiles required = $\frac{14400}{40}$ = 360
Therefore, 360 tiles are required.
(b) Total area of the region = 36 x 70 = 2520 cm2
Area of one tile = 40 cm2
Number of tiles required = $\frac{2520}{40}$ = 63
Therefore, 63 tiles are required.
BYJU’S provides online learning materials for all classes so that students could do the preparation for the exams. You can also check out the NCERT solutions of class 6 which provides solutions for the different types of question. Some of the major chapters are Practical Geometry, Symmetry, Ratio and Proportion, Algebra, Mensuration and Data Handling, whose solutions are provided in an explicit way. Apart from these, there are notes and preparation tips available on our website, which students can make use of it.
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2019-08-18 10:43:16
|
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|
http://mathhelpforum.com/algebra/95992-negative-rational-powers-print.html
|
# Negative and rational powers
Printable View
• July 24th 2009, 04:24 PM
Mr Rayon
Negative and rational powers
Simplify the following, expressing your answer with positive index numbers
$
\frac{(-2)^{-3}\times2^{-4}}{2^{-3}}
$
Please show complete working out. Any help will be appreciated!
• July 24th 2009, 04:30 PM
yeongil
For even exponents n, $(-a)^n = a^n$.
For odd exponents n, $(-a)^n = -a^n$.
You should also know the properties of exponents, like
$a^m \times a^n = a^{m + n}$
and
$\frac{a^m}{a^n} = a^{m - n}$.
Apply the above to the problem. (I get $-\frac{1}{2^4}$ or $-\frac{1}{16}$ as my answer.)
01
• July 24th 2009, 04:52 PM
HallsofIvy
Quote:
Originally Posted by Mr Rayon
Simplify the following, expressing your answer with positive index numbers
$
\frac{(-2)^{-3}\times2^{-4}}{2^{-3}}
$
Please show complete working out. Any help will be appreciated!
First, $(-2)^{-3}= (-1(2))^{-3}= (-1)^{-3}2^{-3}$ and $(-1)^3= -1$
$a^{-3}= 1/a^3$ so this is the same as $-(2^{-3})(2^{-4})(2^3)$. Now use the fact that $a^na^m= a^{n+m}$.
|
2015-06-30 08:56:14
|
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http://mathhelpforum.com/number-theory/24009-find-remainder.html
|
1. ## find remainder
Find r when 319^566 is divided by 37.
I can get it down to (11^2)^283*(29^2)^283.
2. Hello, frankdent1!
Are you allowed to use Modulo Arithmetic?
Find the remainder when $\displaystyle 319^{566}$ is divided by $\displaystyle 37.$
We find that: .$\displaystyle 319^6 \:\equiv \:-1 \pmod{37}$
Then: .$\displaystyle 319^{566} \:=\:319^{6\cdot94 + 2} \:=\: (319^6)^{94}\cdot319^2$
Hence: .$\displaystyle (319^6)^{94}\cdot319^2 \:\equiv \-1)^{94}\cdot319^2 \pmod{37}$
. . . . . . . . . . . . . . .$\displaystyle \equiv\:319^2 \pmod{37}$
. . . . . . . . . . . . . . .$\displaystyle \equiv\:101,761 \pmod{37}$
. . . . . . . . . . . . . . .$\displaystyle \equiv \:11 \pmod{37}$
|
2018-04-25 22:53:00
|
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https://hal.inria.fr/hal-03070702v2
|
# On a rank factorisation problem arising in gearbox vibration analysis
Abstract : Given a field $K$, $r$ matrices $D_i \in K^{n \times n}$, a matrix $M \in K^{n \times m}$ of rank at most $r$, in this paper, we study the problem of factoring $M$ as follows $M=\sum_{i=1}^r D_i \, u \, v_i$, where $u \in K^{n \times 1}$ and $v_i \in K^{1 \times m}$ for $i=1, \ldots, r$. This problem arises in modulation-based mechanical models studied in gearbox vibration analysis (e.g., amplitude and phase modulation). We show how linear algebra methods combined with linear system theory ideas can be used to characterize when this polynomial problem is solvable and if so, how to explicitly compute the solutions.
Keywords :
Document type :
Conference papers
Domain :
https://hal.inria.fr/hal-03070702
Contributor : Alban Quadrat Connect in order to contact the contributor
Submitted on : Wednesday, December 16, 2020 - 11:41:58 AM
Last modification on : Friday, January 21, 2022 - 3:17:33 AM
### File
IFAC_2020_final.pdf
Files produced by the author(s)
### Identifiers
• HAL Id : hal-03070702, version 2
### Citation
Elisa Hubert, Axel Barrau, Yacine Bouzidi, Roudy Dagher, Alban Quadrat. On a rank factorisation problem arising in gearbox vibration analysis. IFAC 2020 - 21st World Congress, Jul 2020, Berlin / Virtual, Germany. ⟨hal-03070702v2⟩
### Metrics
Les métriques sont temporairement indisponibles
|
2022-01-27 00:45:13
|
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|
https://socratic.org/questions/57fa89f311ef6b54be5697de
|
# Question 697de
Oct 10, 2016
62%
#### Explanation:
The first thing to do here is to figure out the theoretical yield of the reaction, which is essentially what you get for a reaction that has an 100% yield.
${\text{C"_ 2"H"_ (4(g)) + "H"_ 2"O"_ ((g)) -> "CH"_ 3"CH"_ 2"OH}}_{\left(g\right)}$
Notice that every mole of ethene that takes part in the reaction produces $1$ mole of ethanol. In your case, you have
4.8 color(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_4)/(28.05color(red)(cancel(color(black)("g")))) = "0.1711 moles C"_2"H"_4
The $1 : 1$ mole ratio that exists between ethene and ethanol tells you that the reaction can theoretically produce $0.1711$ moles of ethenol, the equivalent of
0.1711 color(red)(cancel(color(black)("moles CH"_3"CH"_2"OH"))) * "46.07 g"/(1color(red)(cancel(color(black)("mole CH"_3"CH"_2"OH")))) = "7.88 g"
As you can see, the actual yield of the reaction is lower than the theoretical yield. To find the percent yield simply use
color(purple)(bar(ul(|color(white)(a/a)color(black)("% yield" = "actual yield"/"theoretical yield" xx 100%)color(white)(a/a)|)))
Plug in your values to find
"% yield" = (4.9 color(red)(cancel(color(black)("g"))))/(7.88color(red)(cancel(color(black)("g")))) xx 100% = color(green)(bar(ul(|color(white)(a/a)color(black)(62%)color(white)(a/a)|)))#
The answer is rounded to two sig figs.
|
2021-06-21 22:42:38
|
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https://www.thejournal.club/c/paper/204061/
|
#### Grounding Natural Language Commands to StarCraft II Game States for Narration-Guided Reinforcement Learning
##### Nicholas Waytowich, Sean L. Barton, Vernon Lawhern, Ethan Stump, Garrett Warnell
While deep reinforcement learning techniques have led to agents that are successfully able to learn to perform a number of tasks that had been previously unlearnable, these techniques are still susceptible to the longstanding problem of {\em reward sparsity}. This is especially true for tasks such as training an agent to play StarCraft II, a real-time strategy game where reward is only given at the end of a game which is usually very long. While this problem can be addressed through reward shaping, such approaches typically require a human expert with specialized knowledge. Inspired by the vision of enabling reward shaping through the more-accessible paradigm of natural-language narration, we investigate to what extent we can contextualize these narrations by grounding them to the goal-specific states. We present a mutual-embedding model using a multi-input deep-neural network that projects a sequence of natural language commands into the same high-dimensional representation space as corresponding goal states. We show that using this model we can learn an embedding space with separable and distinct clusters that accurately maps natural-language commands to corresponding game states . We also discuss how this model can allow for the use of narrations as a robust form of reward shaping to improve RL performance and efficiency.
arrow_drop_up
|
2022-12-06 18:13:39
|
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|
http://www.ck12.org/geometry/Supplementary-and-Complementary-Angle-Pairs/lesson/Identify-Angle-Pairs/
|
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Supplementary and Complementary Angle Pairs
## Find missing angle measures for supplementary or complementary angles.
Estimated3 minsto complete
%
Progress
Practice Supplementary and Complementary Angle Pairs
Progress
Estimated3 minsto complete
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Identify Angle Pairs
Credit: DVIDSHUB
Source: https://www.flickr.com/photos/dvids/5531710013/in/photolist-9qPsEz-efonGa-dM5eSi-edUmxA-7w4yYu-6riGM8-e3mqZ4-snYgWD-9qSspN-bRTM66-9qSsqf-7vZK3r-7y5Gp9-apKWEm-skE4G7-bn795t-rqWQqQ-9iuB73-e3s7Df-9qPsFD-rr97rz-9qPsMF-s6mHrN-9iGm5C-s6nTWf-mGNh4V-pJqLBi-9sbiEf-s6uHYz-dmUnsw-snX8nr-snYgs2-snURfr-9qPsLa-9qSszG-aqVwac-9qSsxL-7ykD1J-9QWXEF-9qSsv5-9qPsHt-9iGmcy-dqY5t4-s6uEnB-9qSsoY-9qPsCk-8dgeFG-akLD5s-ioAf9b-8TGPHj
Joey is a member of the Search and Rescue team. They receive a lot of calls to find people who get lost while hiking on the mountain trails. His German shepherd dog has a keen sense of smell so Joey wants to train him to follow the scent trail left by a human being. He plans out the following tracking trail to use for training Rufus.
Credit: CK-12 Foundation
Source: CK-12 Foundation
Joey looks at the tracking trail and wonders what kind of angle pairs he has created in the trail. How can he figure out the relationship between the angle pairs indicated on the trail?
In this concept, you will learn to identify angle pairs as being complementary, supplementary or neither.
### Guidance
An angle is the intersection of two non-collinear rays at a common endpoint. The rays are called the sides or arms of the angle and the common endpoint is called the vertex. Here is an angle with its parts labelled.
A single angle that measures 180° is called a straight angle. The following diagram shows a straight angle with its vertex B.
If sides BA and BC were extended such that they intersected at B, then the diagram would look like this:
The intersecting lines have created four angles with each angle having the intersection point its vertex. The angles are numbered \begin{align*}\angle 1\end{align*}, \begin{align*}\angle 2\end{align*}, \begin{align*}\angle 3\end{align*}, \begin{align*}\angle 4\end{align*}. The intersection of the two lines has created angle pairs. Angle pairs are two angles that share a unique relationship. The angle pairs in this diagram have a measure that equals 180° which is the measure of a straight angle. Angle pairs that have a sum of 180° are called supplementary angles.
In the above diagram the supplementary angles are:
\begin{align*}\angle 1\end{align*} and \begin{align*}\angle 3\end{align*}, \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 4\end{align*}, \begin{align*}\angle 2\end{align*} and \begin{align*}\angle 3\end{align*}, \begin{align*}\angle 2\end{align*} and \begin{align*}\angle 4\end{align*}. Therefore,
\begin{align*}\angle 1 + \angle 3 = 180^\circ, \ \angle 1 + \angle 4 = 180^\circ, \ \angle 2 + \angle 3 = 180^\circ, \ \angle 2 + \angle 4 = 180^\circ\end{align*}
The following diagram shows supplementary angles:
A single angle that measures 90° is called a right angle. The following diagram shows a right angle with its vertex B.
The angle pair shown in the following diagram have a measure equal to 90° which is the measure of a right angle. Angle pairs that have a sum of 90° are called complementary angles. In the following diagram the complementary angles are \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 2\end{align*}. Therefore, \begin{align*}\angle 1 + \angle 2 = 90^\circ\end{align*}.
The following diagram shows complementary angles:
Now you can identify supplementary or complementary angles based on the sum of the angle pairs. Identify the following angle pairs as either complementary or supplementary.
In the first diagram the sum of the angle pair is \begin{align*}30^\circ + 60^\circ = 90^\circ\end{align*}. Therefore, the diagram shows complementary angles.
In the second diagram the sum of the angle pair is \begin{align*}125^\circ + 55^\circ = 180^\circ\end{align*}. Therefore, the diagram shows supplementary angles.
In the third diagram the sum of the angle pair is \begin{align*}34^\circ + 87^\circ = 121^\circ\end{align*}. Therefore, the diagram shows neither supplementary nor complementary angles.
### Guided Practice
Identify the following angles as being complementary, supplementary or neither.
#1 - The sum of the angle pair is \begin{align*}23^\circ + 67^\circ = 90^\circ\end{align*}. The angles are complementary.
#2 - The sum of the angle pair is \begin{align*}115^\circ + 112^\circ = 227^\circ\end{align*}. The sum of the angle pair \begin{align*}\ne 90^\circ\end{align*} and \begin{align*}\ne 180^\circ\end{align*}. The angle pair is neither complementary nor supplementary.
#3 - The sum of the angle pair is \begin{align*}126^\circ + 54^\circ = 180^\circ\end{align*}. The angles are supplementary.
### Examples
#### Example 1
Find the measure of two complementary angles, \begin{align*}\angle A\end{align*} and \begin{align*}\angle B\end{align*}, if the difference in the measure of the angle pair is 18°.
First, let ‘\begin{align*}x\end{align*}’ represent \begin{align*}m \angle A\end{align*}.
Next, write an equation to model the measure of the complementary angles.
Next, solve the equation for \begin{align*}m \angle B\end{align*} by subtracting ‘\begin{align*}x\end{align*}’ from both sides of the equation.
Next, write an equation to model the difference between the measures of the complementary angles.
Next, substitute in the expression for \begin{align*}m \angle B\end{align*} and the variable for \begin{align*}m \angle A\end{align*}.
Next, multiply \begin{align*}1(90-x)\end{align*} to clear the parenthesis.
Next, simplify the equation.
Next, isolate the variable by subtracting 90 from both sides of the equation. Simplify the equation.
Then, solve the equation for ‘\begin{align*}x\end{align*}’ by dividing both sides of the equation by -2.
The answer is \begin{align*}x=36\end{align*}.
The \begin{align*}m \angle A = 36^\circ\end{align*}.
Then, use the value to determine \begin{align*}m \angle B\end{align*}.
#### Example 2
Find the measure of two supplementary angles, \begin{align*}\angle A\end{align*} and \begin{align*}\angle B\end{align*}, if the difference in the measure of the angle pair is 32°.
First, let ‘\begin{align*}x\end{align*}’ represent \begin{align*}m \angle A\end{align*}.
Next, write an equation to model the measure of the complementary angles.
Next, solve the equation for \begin{align*}m \angle B\end{align*} by subtracting ‘\begin{align*}x\end{align*}’ from both sides of the equation.
Next, write an equation to model the difference between the measures of the complementary angles.
Next, substitute in the expression for \begin{align*}m \angle B\end{align*} and the variable for \begin{align*}m \angle A\end{align*}.
Next, multiply \begin{align*}1(180-x)\end{align*} to clear the parenthesis.
Next, simplify the equation.
Next, isolate the variable by subtracting 90 from both sides of the equation. Simplify the equation.
Then, solve the equation for ‘\begin{align*}x\end{align*}’ by dividing both sides of the equation by -2.
The answer is \begin{align*}x=74\end{align*}.
The \begin{align*}m \angle A = 74^\circ\end{align*}.
Then, use the value to determine \begin{align*}m \angle B\end{align*}.
Credit: O. Vincent
Source: https://www.flickr.com/photos/o_vincent/14042706040/in/photolist-noUxgo-7nQiSc-7nQiLz-7nUduq-7nQjiz-7nQj8k-7nQibi-7nUd7b-7nUcpm-7nTWd9-7nQ2uT-7nQiXx-7nQiE4-7nTVPY-7nUcWw-7nQir4-7nUdhs-7nUcbs-7nQgcn-7nU9k9-7wrGHC-7nQg7Z-7nQgGM-7nU9P1-7nQgPF-7nU9rA-7nQg1a-7nQhzn-7nQgsz-7nQhFZ-7nUaUC-7nQgkR-7nUawy-7nUbYf-7nU9FG-7nQhP8-7nUb1U-7nQhck-7nUdoN-dEMfY6-dESEc5-QqeKv-QqeLp-QqeKk-7nQjLx-9qdAkd-9LPY5o-63AjaY-7CkQiD-6QRQ1J
Remember Joey who was training his dog? He has marked three pairs of angles on his tracking trail and is wondering what the relationship between the marked angle pairs is.
First, remember if the sum of the measures of the angles is 90° then the angles are complementary. If the sum of the measures of the angles is 180° then the angles are supplementary.
Next, identify the first angle pair marked on the tracking trail to the right of the ‘Start’ point.
This angle pair has a sum of 90° since the right angle is indicated by the square. These angles are complementary.
Next, identify the second angle pair marked on the first downward trail of the training track. This angle pair has a sum of 180° since the angles are on a straight line. These angles are supplementary.
Next, identify the third angle pair marked on the third downward trail of the training track. This angle pair has a sum of 180° since the angles are on a straight line. These angles are supplementary.
### Explore More
If the following angle pairs are complementary, then what is the measure of the missing angle?
1. \begin{align*}\text{If } \angle A = 55^\circ \text{ then } \angle B = ?\end{align*}
2. \begin{align*}\text{If } \angle C = 33^\circ \text{ then } \angle D = ?\end{align*}
3. \begin{align*}\text{If } \angle E = 83^\circ \text{ then } \angle F = ?\end{align*}
4. \begin{align*}\text{If } \angle G = 73^\circ \text{ then } \angle H = ?\end{align*}
If the following angle pairs are supplementary, then what is the measure of the missing angle?
5. \begin{align*}\text{If } \angle A = 10^\circ \text{ then } \angle B = ?\end{align*}
6. \begin{align*}\text{If } \angle A = 80^\circ \text{ then } \angle B = ?\end{align*}
7. \begin{align*}\text{If } \angle C = 30^\circ \text{ then } \angle F = ?\end{align*}
8. \begin{align*}\text{If } \angle D = 15^\circ \text{ then } \angle E = ?\end{align*}
9. \begin{align*}\text{If } \angle M = 112^\circ \text{ then } \angle N = ?\end{align*}
10. \begin{align*}\text{If } \angle O = 2^\circ \text{ then } \angle P = ?\end{align*}
State whether the following statements are True or False.
11. Complementary angles are equal to 180°.
12. Complementary angles are equal to 90°.
13. Supplementary angles are equal to 90°.
14. Supplementary angles are equal to 180°.
15. Angle pairs less than 90 degrees are neither supplementary nor complementary.
### Explore More
Sign in to explore more, including practice questions and solutions for Supplementary and Complementary Angle Pairs.
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2016-02-06 16:47:23
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https://math.stackexchange.com/questions/2530888/the-quotient-of-a-finitely-generated-group-is-a-finite-group
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# The quotient of a finitely generated group is a finite group.
Let $K$ be a number field , $\mathcal{O}_K$ the ring of integers , and $\mathcal{O}_K ^\times$ the unit group of $\mathcal{O}_K$.
Then using the Dirichlet's theorem , we know that $\mathcal{O}_K ^\times$ is finitely generated group , and so $(\mathcal{O}_K ^\times )^m$ is also finitely generated.
In this conditions , is $\mathcal{O}_K ^\times / (\mathcal{O}_K ^\times )^m$ finite group?
I'm sorry that my question is very elementary.
I think the claim holds more generally , but I don't know the proof. Thank you.
• Do you know the structure theorem for finitely generated abelian groups? [That's more than needed, it suffices to know that a finitely generated abelian torsion group is finite.] – Daniel Fischer Nov 21 '17 at 15:47
• Yes , but I don't know how to use. – GreenTea Nov 22 '17 at 0:48
Yes, under these conditions the quotient is finite. By Dirichlet's unit theorem we have $\mathcal{O}_K^\times \cong \mathbb{Z}^{r+s-1} \times F$ where $r$ is the number of embeddings $K \hookrightarrow \mathbb{R}$, $s$ is the number of conjugate pairs of embeddings $K \hookrightarrow \mathbb{C}$, and $F$ is the finite group of roots of unity in $K$.
Note, that the group law on the left-hand side is multiplication while the group law in $\mathbb{Z}^{r+s-1}$ is addition. (I am keeping it this way since it is usually easiest to think of a free abelian group as a power of $\mathbb{Z}$.)
Thus $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^m \cong \mathbb{Z}^{r+s-1} / m \mathbb{Z}^{r+s-1} \times F/F^m$. Any quotient of $F$ is finite and $\mathbb{Z}^{r+s-1} / m \mathbb{Z}^{r+s-1} \cong (\mathbb{Z}/m\mathbb{Z})^{r+s-1}$ is of order $m^{r+s-1}$ whence also finite.
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2021-04-23 07:25:14
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http://hmemcpy.com/
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# Debug symbols, dotPeek and long paths in Visual Studio
In my previous post, I explained how to use the symbol server in dotPeek 1.2 to debug any assembly in Visual Studio, allowing you to set breakpoints and step into any method (provided it was decompiled by dotPeek).
While this is great, I noticed that there was one particular method I couldn’t step into – the moment I tried I got the sadly familiar Source Not Found page:
Clicking the Browse and find… link did nothing, and the Source search information dropdown appeared below. Expanding it, I could see where Visual Studio attempted to load the source file from:
Locating source for 'C:\Users\Igal\AppData\Local\JetBrains\dotPeek\v1.2\SymbolCache\CSharp\Microsoft.VisualStudio.ProjectSystem.VS.Implementation.pdb\538887009A094E419882756878C69B2A1\Microsoft.VisualStudio.ProjectSystem.VS.Implementation\VisualStudio\ProjectSystem\VS\Implementation\Package\Automation\OAProjectItems.cs'. (No checksum.)
The file 'C:\Users\Igal\AppData\Local\JetBrains\dotPeek\v1.2\SymbolCache\CSharp\Microsoft.VisualStudio.ProjectSystem.VS.Implementation.pdb\538887009A094E419882756878C69B2A1\Microsoft.VisualStudio.ProjectSystem.VS.Implementation\VisualStudio\ProjectSystem\VS\Implementation\Package\Automation\OAProjectItems.cs' does not exist.
Looking in script documents for 'C:\Users\Igal\AppData\Local\JetBrains\dotPeek\v1.2\SymbolCache\CSharp\Microsoft.VisualStudio.ProjectSystem.VS.Implementation.pdb\538887009A094E419882756878C69B2A1\Microsoft.VisualStudio.ProjectSystem.VS.Implementation\VisualStudio\ProjectSystem\VS\Implementation\Package\Automation\OAProjectItems.cs'...
Looking in the projects for 'C:\Users\Igal\AppData\Local\JetBrains\dotPeek\v1.2\SymbolCache\CSharp\Microsoft.VisualStudio.ProjectSystem.VS.Implementation.pdb\538887009A094E419882756878C69B2A1\Microsoft.VisualStudio.ProjectSystem.VS.Implementation\VisualStudio\ProjectSystem\VS\Implementation\Package\Automation\OAProjectItems.cs'.
...
And so on. Quick search for OAProjectItems.cs using my most favorite tool, Everything, revealed that it was indeed present in that location, so why couldn’t Visual Studio open it? I decided to open the file manually by pasting its full path into the Start – Run dialog (Win-R), but then I got the following error:
Finally, I tried to go to the file location using the cmd, and I got my answer – the path was simply too long for Windows (and therefore, Visual Studio) to handle!
Windows has an unfortunate MAX_PATH limitation at 260 characters is the source of great pain, and I hope Microsoft fixes it one day. Meanwhile, here’s how to “work around” this particular issue: dotPeek generates debug symbols under %LOCALAPPDATA%\JetBrains\dotPeek\v1.2\SymbolCache\CSharp for C# code, and unfortunately, it isn’t possible to configure. Luckily, it is possible to configure where Visual Studio looks for debug symbols. This is a little-known page in the Solution properties called Debug Source Files:
What I did was copy the entire contents of dotPeek’s SymbolCache\CSharp directory into a local directory d:\sym, and added it to the search list (pictured above). Also, I made sure to delete everything from the bottom list (Do not look for these source files) – if Visual Studio is unable, for any reason, to open a source file, it will add it to this blacklist. It was filled with the files I needed, so I removed them from the list.
After doing this, I could go back to debugging, was was able to step into methods that were previously unavailable!
As the comments below mention, this is also possible to do without copying, by creating a (symbolic) link between the directories! From an elevator command shell, run:
mklink /D d:\sym %localappdata%\JetBrains\dotPeek\v1.2\SymbolCache\CSharp
Thanks guys!
Happy debugging!
# How to debug anything with Visual Studio and JetBrains dotPeek v1.2!
Sometimes, we wish we could just step into some 3rd party library, to figure out how it works, but we either don’t have the source code, or otherwise just can’t. Fortunately, this is made possible by dotPeek v1.2 that was just released, which can act as a symbol server for decompiled assemblies!
So let’s suppose we want to put a breakpoint inside Console.WriteLine (or any other method in any other assembly). Here’s what we need to do:
• Open dotPeek, add the required assemblies to the Assembly Explorer, and press the Start Symbol Server button.
You can configure the port and the symbol generation settings in Tools – Options. The default address is http://localhost:33417/
• In Visual Studio, go to Tools – Options, then navigate to Debugging – Symbols. Add the location of the dotPeek symbol server.
In addition, make sure that Just My Code (in General) is unchecked, and press OK. Some symbols will be loaded, this might take a few moments.
• Next, we need to set a breakpoint inside the method which we’re interested in. This can be done with a little-known Visual Studio trick, allowing you to create a breakpoint at any function. Go to Debug – New Breakpoint – Break at Function, and in the dialog enter the fully qualified method name, e.g. System.Console.WriteLine.
After pressing OK, you’ll get a message saying “IntelliSense could not find the specified location. Do you still want to set the breakpoint?”. It’s fine – press Yes.
• Finally, start your application with the debugger (F5) and you will stop at the breakpoint! You can use all familiar debugging options, such as stepping over/into, watch, autos and the datatip.
Happy Debugging!
# How I Improved Sharing My Blog Posts On Twitter With This One Simple Trick!
(I’m so sorry for writing this title! They made me do it, I swear! I won’t ever do it again!)
I came across a wonderful blog post the other day, titled Cassandra vs MongoDB vs CouchDB vs Redis vs Riak vs HBase vs Couchbase vs Neo4j vs Hypertable vs ElasticSearch vs Accumulo vs VoltDB vs Scalaris comparison, written by Kristof Kovacs (@kkovacs). Being the twitter addict I am, I immediately felt the need to share this post, and I was met with the eternal twitter dilemma – which part of the title/credit I omit, due to the 140 character limit. I ended up doing this:
Which is what most people would do (chop enough of the title to fit, use a URL shortener for the link, the twitter way). Which didn’t leave room to give credit to the original author of the post (a fact pointed to me by @omervk, thank you very much!)
I was wondering if there was a way to encode this information (author’s twitter handle) within the post itself, using html meta tags. Then I realized that most online tech blogs have this feature available to them – any time you paste a link from, say, Engadget or The Verge on twitter, a preview of the article is visible in the twitter stream.
Turns out, this is available not just for tech blogs, but to everyone! Enter Twitter Cards.
Twitter Cards is exactly what it sounds like – you can embed media and other information to your own tweets! There are many types of cards to choose from, but for my purposes I wanted the Summary card – a simple card including a title, description, thumbnail and the twitter account information. This requires adding various html meta tags to your page, all detailed in the URL above.
To enable Twitter Cards, you will need to Apply for the service, after adding the meta tags on your blog/site. The application process supposedly takes up to 72 hours, but in my case I was approved within minutes! Your mileage may vary.
###### Here is what I did to enable this for my WordPress blog:
1. Install Twitter Cards Meta plugin
2. After installing and activating, go to Settings – Twitter Cards Meta
5. Select desired card type (I selected Summary)
6. Click the Validate & Apply tab, and paste a link to any one of your blog posts
7. If you’re satisfied with the results, press Go and after seeing the preview, apply for approval
It might take up to 72 hours to get approved, but usually this happens within minutes (as it was in my case). You will get an email from twitter when you’re approved:
And voilà! Now, every time someone shares a link to a blog post I wrote, it will contain a preview, and my twitter handle!
Turning on Twitter Cards will also give you much better Twitter Analytics – with much more detailed information about clicks and retweets of your cards!
# Migrating from TFS to git to Visual Studio Online–the survival guide
This is a step-by-step recount of my attempt to migrate an existing 3-year old TFS (TFVC) repository to git, while keeping all the history, and then moving it to Visual Studio Online, a TFS in the cloud. This wasn’t an easy task, as there is no direct export-import built into either tools. I also ran into some problems during this lengthy process, and I describe the steps to solve them. Set aside a few hours of your time, brew some coffee (or tea), and let’s get started!
# Step 1: TFS to git
We first need to export the entire TFS repository to git. This is achieved by “cloning” the entire TFS repository with git-tfs or git-TF, both open source tools. While designed to do the same, mainly, providing a bridge between git and TFS, the former is an older, more mature project, and the latter is a tool created by Microsoft for the same purpose. I initially tried to use Microsoft’s git-TF, but after more than 24 hours of waiting for the clone to end, it died with a Java exception (the tool by Microsoft is written in Java).
First, install git-tfs. It’s best installed with Chocolatey by using cinst gittfs. Next, we need to get the exact name of the project we want to clone. Assuming your TFS server is https://tfs.contoso.com, first, run the following command to list all the branches:
git tfs list-remote-branches https://tfs.contoso.com/tfs/DefaultCollection
This will output all the branches that exist in the DefaultCollection. Branches marked with [*] are the root branches, and these are what we want to clone. Run the following command to clone the required branch, e.g.:
git tfs clone https://tfs.contoso.com/tfs/DefaultCollection \$/Project/MyApp/Dev [<directory>]
This will clone the Dev branch (in this case) to a new directory specified in [<directory>]. If [<directory>] is not specified, a new directory called Dev will be created.
The clone operation might take up to several hours, depending on the size of your TFS repository. The cloning operation with git-tfs will pull each individual check-in, and will apply it as a series of git operations, recreating the history exactly as it happened! Git-tfs will also run a git-gc operation every 100 commits, so the new repository size will be kept in check, and obsolete files will be removed.
After the clone operation is complete, you will have a git representation of the TFS repository on your local disk.
# Step 2: git to Visual Studio Online
After setting up a free account in Visual Studio Online (VSO), we first need to create a new project, and make sure Git is selected as the source control. After the project had been created, we can go to the CODE tab, and be presented with instructions to either clone an empty repository, or push an existing one. We want to use the second option, so first, set the remote URL in our cloned git repository to point to VSO.
In a command prompt (or use any one of the 3rd-party visual Git clients), go to the cloned git directory and paste the first command, e.g.:
git remote add origin https://contoso.visualstudio.com/DefaultCollection/_git/MyProj
This will set the origin (the remote repository address) to point to our git repository in VSO.
The most important step is next: unfortunately VSO has a timeout limit of 1 hour for any connection, meaning that an attempt to push the entire repository might fail, if the repository is too big! What’s worse, pushing is an atomic operation, it cannot be resumed in case of a timeout, you will have to start from the beginning.
Luckily, there’s a workaround I was managed to find at the bottom of a similar issue in MSDN forums. Basically, we can push our repository in smaller chunks, thus minimizing the chance of a timeout. This is not the ultimate solution, but it worked for me. Our problem is currently being able to push, subsequent cloning should be much faster, since download bandwidth is almost always faster than the upload.
Disclaimer: This is a somewhat advanced git usage, and I must admit, I don’t quite understand it myself fully. Below is a series of commands I ran, based on the mentioned issue, and it worked in my case. YMMV. If you can explain this, please leave a comment below!
First, the issue tells us, we need to get an idea of the number of commits we have, by running the following:
git rev-list --all --count
In my case, the number was 4012, which I assume is the total number of commits.
Next, however, the issue says to run this:
git rev-list master --first-parent --count
Which is the “depth of the ‘first-parent’ lineage of master”, according to the post author. Not quite sure what this meant (the documentation wasn’t very helpful, either), running this command produced a number which was about half the previous one: 2512.
Not knowing exactly what the numbers meant, I decided to try and follow the post’s advice, and split that number into 5 sections, of 500 commits each, and I pushed them in the following order:
git push origin master~2500:refs/heads/master
And finally:
git push -u origin master
To push the remaining commits, and create a remote tracking branch for master. To make extra sure, I ran this last command again, and got “Everything up-to date” message.
Refreshing the CODE tab in VSO revealed that all my source code was uploaded successfully! I cloned the repository locally, and ran a diff between the two directories in Beyond Compare, just to make absolutely certain that everything was copied properly. It was!
# Step 3: VSO to FogBugz
Few more things left to migrate, and those are all the issues on FogBugz to VSO. In the meantime, though, we need to keep our source control integration with FogBugz. Turns out this is also not a simple task. But, this is a blog post for another day. Stay tuned!
# When good permissions gone bad–a case of a failed build
I was called over to see if I could help solve a strange issue – every time the build script (Ant) for the client’s Android app ran – certain files that were modified by the build script (a .properties file, few others), were suddenly inaccessible to other people logging to the machine – only the user who initiated the build could still write to the files. Looking at the file permissions tab proved as much: only the current user and the Administrators group could access the file!
My initial investigation into Ant’s build.xml led me to an interesting discovery – all the files that lost their permissions were modified using the ReplaceRegExp task for Ant – a task that could replace text in a file using regular expressions. Quick Google search for “replaceregexp ant file permissions” led me to this similar issue, which was, unfortunately, closed as wontfix.
I decided to investigate for myself. Using my go-to tool for this task, Process Monitor, I decided to trace all activity of java.exe (which runs the Ant tasks), looking for anything that has to do with setting permissions or writing the file, filtering just the file I was interested in (project.properties). After running the specific Ant task that did a simple regex replace in the file, it was indeed changed, and its permissions were changed as well.
However, looking at the File Summary window of Process Monitor (Tools – File Summary), I saw that not only there were no new ACL permissions set (Set ACL was called 0 times), no actual bytes were written!
So how was it possible that the file was modified, but nothing was written? Looking back at the events in Process Monitor I also could not see any calls to WriteFile.
Feeling confused, I then started looking for anything that could appear relevant, until one entry in particular caught my eye: a call to SetDispositionInformationFile with a flag Delete: True.
I checked who was calling this API from the Stack tab of this event’s properties, and saw that it was a call to deleteFile, originating in java.dll. This confirmed my suspicion that the file was not written directly at all – but replaced with another file, possibly from a temp directory.
My suspicions proved to be correct – looking at the source code for ReplaceRegExp task I saw that it was exactly how it did it: by using
FILE_UTILS.createTempFile("replace", ".txt", null, true, true)
to create a temporary file, and then later renaming that temporary file with the original file name!
FILE_UTILS turned out to be just a wrapper calling Java’s File.createTempFile. The 3rd argument, which was null in this case, tells createTempFile where to create the file. If null is passed, it will use the default %TEMP% environment variable.
Which ended up explaining the problem exactly – by default, it uses the local user’s %TEMP% directory to create the temp file that replaces the original one. The default per-user %TEMP% directory located in the %LOCALAPPDATA% directory of the user’s profile – meaning only the current user (and local Administrators) can access it! This means, that any file created in this directory will inherit the same permissions! In our case, ReplaceRegExp’s implementation caused the per-user temp file to overwrite the file that was in a public folder, causing it to lose all permissions!
A quick workaround for the problem was to set the TEMP directory to c:\temp, temporarily, during the build.
Next time your Ant script causes issues with file permissions – make sure your files are not replaced under your nose.
It’s funny how a random twitter rant can yield valid solutions. One such rant between @omervk (who co-runs plaintextoffenders.com) and myself regarding enabling Two Factor Authentication (2FA, or “login verification”) on twitter being unavailable for people in Israel, caught the attention of Per Thorsheim, an independent security consultant and founder of the Passwords conference. Per, turns out, was interested in this problem because of another matter, that Twitter seemingly turned off login verification for people who do not have their phone numbers associated with Twitter! Per wrote about his own experiences trying to enable 2FA on twitter here.
So what is the problem exactly? In order to enable login verifications on Twitter, turns out, you need to either associate your mobile phone number with twitter, or enable sending login verifications to your iOS or Android twitter app. The latter, turns out also requires your phone number!
Here’s how it supposed to work: you go to the Security and privacy settings in your twitter account. If no phone number is associated, you will see these options:
Notice that both “Send…” options are disabled. Pressing the add a phone link takes you to a page where you can enter your phone number (after selecting your country and carrier), however in my case, I got this message after submitting:
This is the part where I initially gave up, but it turned out that clicking Add phone takes you to the mobile twitter page (in your phone’s browser), where you can enter your phone number, and in my case twitter accepted it, and sent me a verification SMS!
The UX is confusing at this point, after adding my number, I got an SMS with a 6 digit verification code (and short t.co URL to continue verification), but no place in the page to enter it! In fact, I was asked to remove the phone number at this point:
Very odd. However, pressing the manage link above (in the yellow bar) took me to the actual verification page where I was asked to enter the code from the SMS message:
And after pressing Verify, I got a message confirming that my phone number was successfully added!
At this point, after going back to the Security and privacy settings on twitter, I saw that my phone number was now added, and could turn on login verifications!
Also, going to the Security settings on my Android twitter app were now also working!
Good luck, stay secure!
# How to remap calc.exe to scriptcs.exe
Lately I found myself launching ScriptCS more and more to do simple calculations. I half-jokingly said on twitter that I’d better remap calc.exe to scriptcs.exe on my machine. However it seems that my joke tweet was taken seriously by some people, and I was asked how this was done. So here goes!
For this next trick I will use my most favorite Windows trick – the Image File Execution Options (IFEO). I’ve blogged about IFEO in the past, it’s generally used to allow attaching a debugger to a process before it starts, but can also do other useful things, such as replacing the Windows Task Manager with Process Explorer, or even disabling some processes launching, which is useful to prevent Narrator in Windows 8 from launching via the Win-Enter key.
So here is how to remap calc.exe to launch ScriptCS instead. First, locate on your machine where scriptcs.exe is installed, as we need the full path. You can use the command where scriptcs.exe in CMD to find it. If you don’t have scriptcs.exe in path, best install it via Chocolatey (you’re welcome!)
1. Open the registry editor (regedit.exe), and navigate to:
HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows NT\CurrentVersion\Image File Execution Options
2. Create a new key called calc.exe
3. Inside the newly created key, create a new String Value called Debugger
4. Double-click Debugger, and set c:\path\to\scriptcs.exe -repl as the value
That’s it, from now on when you launch calc.exe, scriptcs.exe will open instead! To undo this, simply delete the key calc.exe from the registry path above.
P.S. if you have Windows SDK installed, you can use the utility gflags.exe to do this instead:
1. Launch gflags.exe (need to be launched elevated)
2. Go to the Image File tab
3. In the Image text box, write calc.exe and press TAB (I know)
4. Down at the bottom, under Debugger, write c:\path\to\scriptcs.exe -repl and press OK
Bonus: now replace devenv.exe with scriptcs.exe!
Happy hacking!
|
2014-08-30 06:28:08
|
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https://psspy.org/psse-help-forum/answers/5253/revisions/
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# Revision history [back]
Some ideas:
Do not mix running simulation with plotting. Use a second script (or loop) to perform the plotting. Plotting within a loop, if not coded properly, is known to eat memory.
Run the code outside PSSe GUI. Increase bus allocation to its max. -> psspy.psseinit(buses=200000).
Minimize commands within the main loop. save the case after:
ierr = psspy.tysl(0) #solving the converted case
ierr = psspy.sav('case_cnv.sav') #as example
save a snap file after defining the channels:
ierr = psspy.bus_frequency_channel([-1, bus])
ierr = psspy.snap([-1,-1,-1,-1,-1],'case.snp') #as example
and move all that code before the main loop, since it does not change per iteration. Within the loop, load the converted case and load the snp file. This way previous memory allocation and temporary files in the background may be deleted.
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2019-08-18 10:59:28
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{"extraction_info": {"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, "math_score": 0.3272051513195038, "perplexity": 10566.375284299524}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027313803.9/warc/CC-MAIN-20190818104019-20190818130019-00415.warc.gz"}
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https://mi2datalab.github.io/auditor/reference/plotModelRanking.html
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Radar plot with model scores. Scores are scaled to [0,1], each score is inversed and divided by maximum score value.
plotModelRanking(object, ..., scores = c("MAE", "MSE", "REC", "RROC"),
new.score = NULL, table = TRUE)
## Arguments
object An object of class ModelAudit. Other modelAudit objects to be plotted together. Vector of score names to be plotted. A named list of functions that take one argument: object of class ModelAudit and return a numeric value. The measure calculated by the function should have the property that lower score value indicates better model. Logical. Specifies if rable with score values should be plotted
## Value
ggplot object
plot.modelAudit
## Examples
library(car)
lm_model <- lm(prestige~education + women + income, data = Prestige)
lm_au <- audit(lm_model, data = Prestige, y = Prestige$prestige) library(randomForest) rf_model <- randomForest(prestige~education + women + income, data = Prestige) rf_au <- audit(rf_model, data = Prestige, y = Prestige$prestige)
plotModelRanking(lm_au, rf_au)#> Coordinate system already present. Adding new coordinate system, which will replace the existing one.
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2019-03-25 16:35:22
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https://physics.stackexchange.com/questions/521447/a-doubt-in-trigonometric-approximation-used-in-the-derivation-of-mirror-formula
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# A doubt in trigonometric approximation used in the derivation of mirror formula
The following text is from Concepts of Physics by Dr. H.C.Verma, from chapter "Geometrical Optics", page 387, topic "Relation between $$u$$,$$v$$ and $$R$$ for Spherical Mirrors":
If the point $$A$$ is close to $$P$$, the angles $$\alpha$$,$$\beta$$ and $$\gamma$$ are small and we can write
$$\alpha\approx\frac{AP}{PO},\ \beta=\frac{AP}{PC}\ \ \ \text{and} \ \ \gamma \approx\frac{AP}{PI}$$
As $$C$$ is the centre of curvature, the equation for $$\beta$$ is exact whereas the remaining two are approximate.
The terms on the R.H.S. of the equations for the angles $$\alpha$$,$$\beta$$ and $$\gamma$$, are the tangents of the respective angles. We know that, when the angle $$\theta$$ is small, then $$\tan\theta\approx\theta$$. In the above case, this can be imagined as, when the angle becomes smaller, $$AP$$ becomes more and more perpendicular to the principal axis. And thus the formula for the tangent could be used.
But, how can this approximation result in a better accuracy for $$\beta$$ when compared to $$\alpha$$ and $$\gamma$$? I don't understand the reasoning behind the statement: "As $$C$$ is the centre of curvature, the equation for $$\beta$$ is exact whereas the remaining two are approximate." I can see the author has used "$$=$$" instead of "$$\approx$$" for $$\beta$$ and he supports this with that statement. But why is this so? Shouldn't the expression for $$\beta$$ be also an approximation over equality? Is the equation and the following statement really correct?
I believe the author is assuming that you recognize that AP is an arc by the context of the image. The main point here then is that C is the center of curvature of the mirror while O and I are not. As C is the center, AC = PC, and the usual arc length formula can be applied exactly with angle $$\beta$$ even if $$\beta$$ isn't assumed small.
On the other hand, O and I are not centers of curvatures. Thus, OA $$\neq$$ PO and AI $$\neq$$ PI, so the arc length formula does not apply exactly. However, as you note, AP becomes more tangent to the horizontal axis as the angles become smaller. For small angles, arc AP is approximately a straight line perpendicular to the horizontal axis and the angles are approximately the tangents of the angles, so you can apply the tangent definition to relate AP to PO and AP to PI by the angles in this case.
For the angle $$\beta$$ the author is apparantly thinking of AP as being the arc length along the circle rather than the length of the straight line joining A to P. Then $${\rm arclength}({\rm AP}) = \beta R$$ exactly.
• But then that is also the case with $\alpha$ and $\gamma$.
– user243267
Dec 26, 2019 at 14:51
• @FakeMod Yes. I think so. The book is not being very precise, and I agree with M. Guru Vishnu's complaint. Dec 26, 2019 at 14:56
• @FakeMod Not so. As C is the center of curvature, AC = PC, so the arclength relation holds exactly as with a circle. For the other cases, OA is not equal to OP and AI is not equal to IP except approximately when the angles are small. Dec 26, 2019 at 14:56
• @JeqHar I definitely missed that point! Thanks for clarifying.
– user243267
Dec 26, 2019 at 14:58
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2022-12-07 03:02:33
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https://doc.cgal.org/4.7/AABB_tree/classAABBTraits.html
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CGAL 4.7 - 3D Fast Intersection and Distance Computation (AABB Tree)
AABBTraits Concept Reference
## Definition
The concept AABBTraits provides the geometric primitive types and methods for the class CGAL::AABB_tree<AABBTraits>.
Has Models:
CGAL::AABB_traits<AABBGeomTraits,AABBPrimitive>
CGAL::AABB_traits<AABBGeomTraits,AABBPrimitive>
CGAL::AABB_tree<AABBTraits>
AABBPrimitive
## Types
enum Axis
typedef unspecified_type FT
Value type of the Squared_distance functor.
typedef unspecified_type Point_3
Type of a 3D point.
typedef unspecified_type Primitive
Type of primitive. More...
typedef unspecified_type Bounding_box
Bounding box type.
typedef std::pair< Point_3,
Primitive::Id >
Point_and_primitive_id
typedef std::pair< Object,
Primitive::Id >
Object_and_primitive_id
template<typename Query >
using Intersection_and_primitive_id = unspecified_type
A nested class template providing as a pair the intersection result of a Query object and a Primitive::Datum, together with the Primitive::Id of the primitive intersected. More...
## Splitting
During the construction of the AABB tree, the primitives are sorted according to some comparison functions related to the $$x$$, $$y$$ or $$z$$ coordinate axis:
typedef unspecified_type Split_primitives_along_x_axis
A functor object to split a range of primitives into two sub-ranges along the X-axis. More...
typedef unspecified_type Split_primitives_along_y_axis
A functor object to split a range of primitives into two sub-ranges along the Y-axis. More...
typedef unspecified_type Split_primitives_along_z_axis
A functor object to split a range of primitives into two sub-ranges along the Z-axis. More...
typedef unspecified_type Compute_bbox
A functor object to compute the bounding box of a set of primitives. More...
## Intersections
The following predicates are required for each type Query for which the class CGAL::AABB_tree<AABBTraits> may receive an intersection detection or computation query.
typedef unspecified_type Do_intersect
A functor object to compute intersection predicates between the query and the nodes of the tree. More...
typedef unspecified_type Intersect
A functor object to compute the intersection of a query and a primitive. More...
## Distance Queries
The following predicates are required for each type Query for which the class CGAL::AABB_tree<AABBTraits> may receive a distance query.
typedef unspecified_type Compare_distance
A functor object to compute distance comparisons between the query and the nodes of the tree. More...
typedef unspecified_type Closest_point
A functor object to compute closest point from the query on a primitive. More...
typedef unspecified_type Squared_distance
A functor object to compute the squared distance between two points. More...
## Operations
Split_primitives_along_x_axis split_primitives_along_x_axis_object ()
Returns the primitive splitting functor for the X axis.
Split_primitives_along_y_axis split_primitives_along_y_axis_object ()
Returns the primitive splitting functor for the Y axis.
Split_primitives_along_z_axis split_primitives_along_z_axis_object ()
Returns the primitive splitting functor for the Z axis.
Compute_bbox compute_bbox_object ()
Returns the bounding box constructor.
Do_intersect do_intersect_object ()
Returns the intersection detection functor.
Intersect intersect_object ()
Returns the intersection constructor.
Compare_distance compare_distance_object ()
Returns the distance comparison functor.
Closest_point closest_point_object ()
Returns the closest point constructor.
Squared_distance squared_distance_object ()
Returns the squared distance functor.
## Primitive with Shared Data
In addition, if Primitive is a model of the concept AABBPrimitiveWithSharedData, the following functions are part of the concept:
template<class... T>
void set_shared_data (T...t)
the signature of that function must be the same as the static function Primitive::construct_shared_data. More...
## Member Typedef Documentation
A functor object to compute closest point from the query on a primitive.
Provides the operator: Point_3 operator()(const Query& query, const Primitive& primitive, const Point_3 & closest); which returns the closest point to query, among closest and all points of the primitive.
A functor object to compute distance comparisons between the query and the nodes of the tree.
Provides the operators:
• bool operator()(const Query & query, const Bounding_box& box, const Point & closest); which returns true iff the bounding box is closer to query than closest is
• bool operator()(const Query & query, const Primitive & primitive, const Point & closest); which returns true iff primitive is closer to the query than closest is
A functor object to compute the bounding box of a set of primitives.
Provides the operator: Bounding_box operator()(Input_iterator begin, Input_iterator beyond); Iterator type InputIterator must have Primitive as value type.
A functor object to compute intersection predicates between the query and the nodes of the tree.
Provides the operators:
• bool operator()(const Query & q, const Bounding_box & box); which returns true iff the query intersects the bounding box
• bool operator()(const Query & q, const Primitive & primitive); which returns true iff the query intersects the primitive
A functor object to compute the intersection of a query and a primitive.
Provides the operator: boost::optional<Intersection_and_primitive_id<Query>::Type > operator()(const Query & q, const Primitive& primitive); which returns the intersection as a pair composed of an object and a primitive id, iff the query intersects the primitive.
Note on Backward Compatibility
Before the release 4.3 of CGAL, the return type of this function used to be boost::optional<Object_and_primitive_id>.
template<typename Query >
A nested class template providing as a pair the intersection result of a Query object and a Primitive::Datum, together with the Primitive::Id of the primitive intersected.
The type of the pair is Intersection_and_primitive_id<Query>::Type.
typedef std::pair AABBTraits::Object_and_primitive_id
Deprecated:
This requirement is deprecated and is no longer needed.
Type of primitive.
Must be a model of the concepts AABBPrimitive or AABBPrimitiveWithSharedData.
A functor object to split a range of primitives into two sub-ranges along the X-axis.
Provides the operator: void operator()(InputIterator first, InputIterator beyond); Iterator type InputIterator must be a model of RandomAccessIterator and have Primitive as value type. The operator is used for determining the primitives assigned to the two children nodes of a given node, assuming that the goal is to split the X-dimension of the bounding box of the node. The primitives assigned to this node are passed as argument to the operator. It should modify the iterator range in such a way that its first half and its second half correspond to the two children nodes.
A functor object to split a range of primitives into two sub-ranges along the Y-axis.
See Split_primitives_along_x_axis for the detailed description.
A functor object to split a range of primitives into two sub-ranges along the Z-axis.
See Split_primitives_along_x_axis for the detailed description.
A functor object to compute the squared distance between two points.
Provides the operator: FT operator()(const Point& query, const Point_3 & p); which returns the squared distance between p and q.
## Member Function Documentation
template<class... T>
void AABBTraits::set_shared_data ( T... t)
the signature of that function must be the same as the static function Primitive::construct_shared_data.
The type Primitive expects that the data constructed by a call to Primitive::construct_shared_data(t...) is the one given back when accessing the reference point and the datum of a primitive.
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2019-07-22 10:52:07
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http://haskell.1045720.n5.nabble.com/Preventing-sharing-td5825124.html
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# Preventing sharing
17 messages
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## Preventing sharing
The old article Preventing memoization in (AI) search problems http://okmij.org/ftp/Haskell/index.html#memo-offdeals with the problem, explaining the trick to deliberately confuse GHC so that it won't perform memoization (sharing). Yes, I know how bad this confusing of GHC sounds: which is part of my argument that lazy evaluation by default was a mistake. _______________________________________________ Haskell-Cafe mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
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## Re: Preventing sharing
On Mon, Dec 21, 2015 at 08:17:10PM +0900, Oleg wrote: > The old article > Preventing memoization in (AI) search problems > http://okmij.org/ftp/Haskell/index.html#memo-off> > deals with the problem, explaining the trick to deliberately confuse > GHC so that it won't perform memoization (sharing). Yes, I know how > bad this confusing of GHC sounds: which is part of my argument that > lazy evaluation by default was a mistake. Hi Oleg, As I explained here https://mail.haskell.org/pipermail/haskell-cafe/2013-February/106673.html-fno-full-laziness fixes the space leak issue in your iterative deepening example. This isn't a problem with laziness. It's a problem with performing a time optimization which is a space pessimization. In the absence of the "optimization" there is no problem. Tom _______________________________________________ Haskell-Cafe mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
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## Re: Preventing sharing
In reply to this post by oleg-30 > -fno-full-laziness fixes the space leak issue in your iterative deepening > example. Yes, and I think it has been mentioned that the flag is a blunt weapon as it affects the whole module... > This isn't a problem with laziness. It's a problem with performing a time > optimization which is a space pessimization. In the absence of the > "optimization" there is no problem. How come it isn't the problem with laziness?! Recall, that pure call-by-name calculus is observationally undistinguishable from the call-by-need (i.e., lazy) calculus. The only reason to have laziness is to avoid recomputations of argument computations should an argument be used more than once -- at the cost of taking memory to store the result of the first evaluation. Thus "performing a time optimization which is a space pessimization" is exactly what laziness is all about -- as the article mentioned earlier argued. Laziness isn't an absolute good -- it is a time-space trade-off, which is not always beneficial. _______________________________________________ Haskell-Cafe mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
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## Re: Preventing sharing
On Mon, Dec 21, 2015 at 08:55:05PM +0900, Oleg wrote: > > -fno-full-laziness fixes the space leak issue in your iterative deepening > > example. > Yes, and I think it has been mentioned that the flag is a blunt weapon > as it affects the whole module... Agreed. > > This isn't a problem with laziness. It's a problem with performing a time > > optimization which is a space pessimization. In the absence of the > > "optimization" there is no problem. > > How come it isn't the problem with laziness?! Recall, that pure > call-by-name calculus is observationally undistinguishable from the > call-by-need (i.e., lazy) calculus. The only reason to have laziness > is to avoid recomputations of argument computations should an argument > be used more than once -- at the cost of taking memory to store the > result of the first evaluation. Thus "performing a time optimization > which is a space pessimization" is exactly what laziness is all about > -- as the article mentioned earlier argued. Laziness isn't an absolute > good -- it is a time-space trade-off, which is not always beneficial. I don't agree at all. To my mind you are assigning blame to the wrong thing. The operational semantics of f () = let x = in ... are perfectly clear. x is reallocated each time, and is free to be released between calls to f. It's only when an "optimization" rewrites this to x = f () = ... that there is a space leak. Exactly the same applies if the language is strict. Tom _______________________________________________ Haskell-Cafe mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
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## Re: Preventing sharing
In reply to this post by oleg-30 On Mon, 21 Dec 2015 12:55:05 +0100, Oleg <[hidden email]> wrote: : > The only reason to have laziness > is to avoid recomputations of argument computations should an argument > be used more than once -- at the cost of taking memory to store the > result of the first evaluation. Thus "performing a time optimization > which is a space pessimization" is exactly what laziness is all about > -- as the article mentioned earlier argued. Laziness isn't an absolute > good -- it is a time-space trade-off, which is not always beneficial. In paper "Why Functional Programming Matters"[0], John Hughes shows how lazy functional programming can be used for better modularity. A more precise title for the paper would be "Why Lazy Functional Programming Matters". Regards, Henk-Jan van Tuyl [0] http://www.cse.chalmers.se/~rjmh/Papers/whyfp.pdf-- Folding@home What if you could share your unused computer power to help find a cure? In just 5 minutes you can join the world's biggest networked computer and get us closer sooner. Watch the video. http://folding.stanford.edu/http://Van.Tuyl.eu/http://members.chello.nl/hjgtuyl/tourdemonad.htmlHaskell programming -- _______________________________________________ Haskell-Cafe mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
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## Re: Preventing sharing
Administrator On Mon, Dec 21, 2015 at 10:50 PM, Henk-Jan van Tuyl wrote:In paper "Why Functional Programming Matters"[0], John Hughes shows how lazy functional programming can be used for better modularity. A more precise title for the paper would be "Why Lazy Functional Programming Matters".This is Oleg. He's perfectly aware of the paper.The point he's making is not that laziness is bad, but that it shouldn't be the default.And if you note the recent work on -XStrict, there are good arguments about bolting laziness on top of strictness and not doing a nasty -- and thus necessarily heroic -- shoehorn in the reverse direction. See:https://www.reddit.com/r/programming/comments/3sux1d/strict_haskell_xstrict_has_landed/However, the record remains that Oleg has offered little by way of elegant bolting. His lazy programs based on a strict language tend to be cluttered with lazy and force functions that uglify previously elegant code.His arguments would persuade many more folks if, for instance, he could offer lazy-over-strict translations of Doug McIlroy's power serious one-liners with no loss in elegance:http://www.cs.dartmouth.edu/~doug/powser.html-- Kim-Ee _______________________________________________ Haskell-Cafe mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
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## Re: Preventing sharing
Administrator His lazy programs based on a strict language tend to be cluttered with lazy and force functions that uglify previously elegant code.* I think the pair of functions are called "delay" and "force", I forget the precise names.-- Kim-Ee _______________________________________________ Haskell-Cafe mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
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## Re: Preventing sharing
In reply to this post by Kim-Ee Yeoh I have a cunning plan. Default to call-by-name. _______________________________________________ Haskell-Cafe mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
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## Re: Preventing sharing
In reply to this post by Tom Ellis > > Thus "performing a time optimization > > which is a space pessimization" is exactly what laziness is all about > > -- as the article mentioned earlier argued. Laziness isn't an absolute > > good -- it is a time-space trade-off, which is not always beneficial. > > I don't agree at all. To my mind you are assigning blame to the wrong > thing. The operational semantics of > > f () = let x = in ... > > are perfectly clear. x is reallocated each time, and is free to be released > between calls to f. It's only when an "optimization" rewrites this to > > x = > > f () = ... > > that there is a space leak. Exactly the same applies if the language is > strict. Let us take a step back. The article on my web page noted the great difficulty of writing AI search program in Haskell because the search tree is evaluated lazily: whenever a node is evaluated, its result is memoized for further use. That is precisely the wrong thing to do for such problems. Again, this problem is precisely of lazy evaluation (as defined in the book below). The obvious way to avoid memoization was to introduce thunks -- which didn't work. The article then developed a more involved solution. Yes, -no-full-laziness would have worked just as well. However, the solution in the article works on the case-by-case basis whereas -no-full-laziness affects the whole compilation unit. It is for that reason that I pointed out the article in this discussion. Let us turn to the problem of the "optimization" that we are discussing. Does it have anything to do with laziness? Yes, it does. Please see Chap 15 of the excellent book http://research.microsoft.com/en-us/um/people/simonpj/papers/slpj-book-1987/which explains the full laziness. Once I mention the book I must point out Chap 23 of the book (near the end). It should be the required read. The introduction to the section contains the following emphasized statement (emphasized by the author, Simon Peyton Jones): A major weakness of functional languages is the difficulty of reasoning about their space and time behavior. The last paragraph of the introduction says No good solutions are yet known to most of these problems.'' This is true even today. Sec 23.3 Space behavior'' is the excellent collection of the problems with lazy evaluation and full laziness. The book was published in 1987. The problems are still with us. _______________________________________________ Haskell-Cafe mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
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## Re: Preventing sharing
In reply to this post by Henk-Jan van Tuyl Henk-Jan van Tuyl wrote: > In paper "Why Functional Programming Matters"[0], John Hughes shows how > lazy functional programming can be used for better modularity. A more > precise title for the paper would be "Why Lazy Functional Programming > Matters". The first paragraph of our paper (published 3 years ago) http://okmij.org/ftp/ftp/continuations/PPYield/index.html#introductionis as follows Lazy evaluation is regarded as one of the main reasons why functional programming matters \cite{hughes:matters-cj}. Lazy evaluation lets us write \emph{producers} and \emph{consumers} separately, whereas the two are inextricably intertwined in a call-by-value language. This separation allows a modular style of programming, in which a variety of producers, consumers, and transformers can readily be plugged together.'' Lazy evaluation is also an elegant implementation of a form of coroutines, suspending and resuming computations based on the demand for values, giving us memory-efficient, incremental computation `for free' \cite{McIlroy:1999:PSP:968592.968597,Bird:1984:UCP,AG-embed}. But do read the next paragraph and the rest of the paper, and other articles on the web site http://okmij.org/ftp/continuations/PPYield/index.htmlOur conclusion is that the modularity benefit of lazy evaluation can be held without lazy evaluation, gaining the predictability of the space and time behavior. _______________________________________________ Haskell-Cafe mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
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## Re: Preventing sharing
Am 22.12.2015 um 13:31 schrieb Oleg: > > But do read the next paragraph and the rest of the paper, and other > articles on the web site > http://okmij.org/ftp/continuations/PPYield/index.htmlI have to say I almost stopped reading at "worst in lazy evaluation: its incompatibility with effects". Incompatibility with side effects is a good thing, so that's not "worst", and there are frameworks for doing effects (most notably IO), so there goes "incompatibility". > Our conclusion is that the modularity benefit of lazy evaluation can > be held without lazy evaluation, gaining the predictability of the > space and time behavior. I just skimmed the paper, so correct me if I'm wrong. It seems to show how one can transform a specific class of lazy functions into generators. This seems to miss the point of laziness. Having lazy evaluation means that when writing a function, you don't know (and don't care) how much of the returned data structure is ever explored, that's the caller's decision. This means you do not ever transform your code as a generator, because you don't need to. As I said, I didn't do more than a quick scan, and I might be in error thinking it's a transformation just for a specific class of lazy functions. More importantly, I might have missed some essential point about how this isn't really a transformation, or that one does not need to transform the code to get a transformer out. So... please correct the omissions :-) Regards, Jo _______________________________________________ Haskell-Cafe mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
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## Re: Preventing sharing
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## Re: Preventing sharing
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## Re: Preventing sharing
Am 23.12.2015 um 04:00 schrieb wren romano: > To play the celestial advocate: the ability to not care about > strictness/laziness when writing a function is precisely what causes > it to be hard to reason about the space/time costs of that function. Sure. I'm not disputing well-known facts, I'm just wondering that the paper highlights just the problems and does not put them into proportion to the advantages. _______________________________________________ Haskell-Cafe mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
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2019-12-15 21:20:08
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http://acm.hznu.edu.cn/OJ/problem.php?id=1220
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# Stars
Tags:
Time Limit: 1 s Memory Limit: 128 MB
Submission：3 AC：0 Score：99.96
## Description
On a clear moon-less night, you can see millions of stars glimmering in the sky. Faced with this overwhelming number, the Greeks started nearly 2,000 years ago to bring some order to the chaos. They identified groups of stars, called constellations, and gave them names, mostly from the Greek mythology, that are still in use today. Examples are Ursa Minor', Pisces', Cancer', and many others.
Given a sketch of the constellation, it is not easy for the amateur to actually find the constellation in the sky. Moreover, simple constellations, such as Triangulum' (triangle,) which consists of only three stars, may appear several times in the sky. Again, singling out the correct' occurrence is not easy.
Traditionally, maps were printed for just this purpose. But in this problem, we will see how the computer can help us find constellations in the sky.
You will be given a star map; for simplicity this will be a collection of points in the plane, each having a certain brightness associated with it. Then, given a constellation, also as a set of points in the plane, you are to determine:
• the number of occurrences of the constellation in the star map, and
• the position of the brightest occurrence, if one exists. (The rationale behind this is as follows: if a constellation seems to appear several times in the sky, the brightest one is most likely to be the real one, since it is the most eye-catching one.)
An occurrence is a subset of stars from the map that forms a (possibly) arbitrarily rotated and/or scaled copy of the stars in the constellation.
The brightness of an occurrence is the average brightness of the stars it consists of, i.e. the sum of individual brightnesses divided by the number of stars in the constellation.
## Input
On a clear moon-less night, you can see millions of stars glimmering in the sky. Faced with this overwhelming number, the Greeks started nearly 2,000 years ago to bring some order to the chaos. They identified groups of stars, called constellations, and gave them names, mostly from the Greek mythology, that are still in use today. Examples are Ursa Minor'', Pisces'', Cancer'', and many others.
Given a sketch of the constellation, it is not easy for the amateur to actually find the constellation in the sky. Moreover, simple constellations, such as Triangulum'' (triangle,) which consists of only three stars, may appear several times in the sky. Again, singling out the correct'' occurrence is not easy.
Traditionally, maps were printed for just this purpose. But in this problem, we will see how the computer can help us find constellations in the sky.
You will be given a star map; for simplicity this will be a collection of points in the plane, each having a certain brightness associated with it. Then, given a constellation, also as a set of points in the plane, you are to determine:
• the number of occurrences of the constellation in the star map, and
• the position of the brightest occurrence, if one exists. (The rationale behind this is as follows: if a constellation seems to appear several times in the sky, the brightest one is most likely to be the real one, since it is the most eye-catching one.)
An occurrence is a subset of stars from the map that forms a (possibly) arbitrarily rotated and/or scaled copy of the stars in the constellation.
The brightness of an occurrence is the average brightness of the stars it consists of, i.e. the sum of individual brightnesses divided by the number of stars in the constellation.
## Output
For each star map first output the number of the map (Map #1', Map #2', etc.) on a line of its own.
For each constellation, in the same order as in the input, output first its name and how many times it occurs in the map on one line, as shown in the output sample.
If there is at least one occurrence, output the position of the brightest occurrence by listing the positions of the stars that form the brightest occurrence. The star positions have to be printed in ascending x-order. Positions having the same x-coordinates must be sorted in ascending y-order. If there are several equally bright solutions, output only one of them. Adhere to the format shown in the sample output.
Output a blank line before each constellation and a line of 5 dashes (`-----') after every star map.
## Samples
input
6 1 2 1 2 1 4 2 4 3 3 2 1 4 1 5 4 3 2 2 3 Triangulum 1 1 3 1 2 4 4 Cancer 1 3 4 3 6 1 7 5 0
output
Map #1 Triangulum occurs 2 time(s) in the map. Brightest occurrence: (1,2) (4,1) (4,3) Cancer occurs 0 time(s) in the map. -----
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2021-04-15 08:49:20
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http://www.emis.de/classics/Erdos/cit/59605027.htm
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## Zentralblatt MATH
Publications of (and about) Paul Erdös
Zbl.No: 596.05027
Autor: El-Zahar, M.; Erdös, Paul
Title: On the existence of two non-neighboring subgraphs in a graph. (In English)
Source: Combinatorica 5, 295-300 (1985).
Review: There is raised the following question: Is there a minimal integer f(r,n) such that each graph G with \chi(G) \geq f(r,n) and which does not contain a complete subgraph of order r must contain two non-neighboring n-chromatic subgraphs? It is known that f(r,2) exists. There is shown that for a fixed n, an upper bound for f(r,n), r > n is given in terms of f(r,n), r \leq n. From f(3,3) \leq 8 is deduced an upper bound for f(r,3) and proved that a vertex critical 4-chromatic graph which does not contain two independent edges has order \leq 13.
Reviewer: J.Fiamcik
Classif.: * 05C15 Chromatic theory of graphs and maps
05C35 Extremal problems (graph theory)
Keywords: non-neighboring n-chromatic subgraphs
© European Mathematical Society & FIZ Karlsruhe & Springer-Verlag
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2015-11-27 19:04:51
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https://www.esaral.com/q/if-the-lines-x-a-y-b-z-c-y-d-21869
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# If the lines x=a y+b, z=c y+d
Question:
If the lines $x=a y+b, z=c y+d$ and $x=a^{\prime} z+b^{\prime}$, $\mathrm{y}=\mathrm{c}^{\prime} \mathrm{z}+\mathrm{d}^{\prime}$ are perpendicular, then:
1. $c^{\prime}+a+a^{\prime}=0$
2. $\mathrm{aa}^{\prime}+\mathrm{c}+\mathrm{c}^{\prime}=0$
3. $a b^{\prime}+b c^{\prime}+1=0$
4. $\mathrm{bb}^{\prime}+\mathrm{cc}^{\prime}+1=0$
Correct Option: , 2
Solution:
Line $x=a y+b, z=c y+d \Rightarrow \frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}$
Line $x=a^{\prime} z+b^{\prime}, y=c^{\prime} z+d^{\prime}$
$\Rightarrow \frac{\mathrm{x}-\mathrm{b}^{\prime}}{\mathrm{a}^{\prime}}=\frac{\mathrm{y}-\mathrm{d}^{\prime}}{\mathrm{c}^{\prime}}=\frac{\mathrm{z}}{1}$
Given both the lines are perpendicular $\Rightarrow \mathrm{aa}^{\prime}+\mathrm{c}^{\prime}+\mathrm{c}=0$
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2023-02-05 21:03:40
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https://wiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_20&oldid=864
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# 005 Sample Final A, Question 20
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Question Consider the following rational function,
${\displaystyle f(x)={\frac {x^{2}+x-2}{x^{2}-1}}}$
a. What is the domain of f?
b. What are the x and y-intercepts of f?
c. What are the vertical and horizontal asymptotes of f, if any? Does f have any holes?
d. Graph f(x). Make sure to include the information you found above.
Foundations:
1) What points are not in the domain of f(x)?
2) How do you find the intercepts?
3) How do you find the asymptotes and zeros?
4) How do you determine if f has any holes?
1) The point that are not in the domain of f(x) are zeros of the denominator.
2) To find the x-intercepts set y = 0 and solve for x. For y-intercepts set x = 0 and simplify.
3) For zeros, find the zeros of the numerator. Vertical asymptotes correspond to zeros of the denominator. Horizontal asymptotes correspond to taking the limit as x goes to ${\displaystyle \infty }$
4) Holes occur when a single value of x is a zero of both the numerator and denominator.
Solution:
Step 1:
We start by finding the zeros of the denominator since this will give us information about vertical asymptotes and the domain. The zeros of the denominator are x = -1, 1. This tells us the domain is ${\displaystyle (-\infty ,-1)\cup (-1,1)\cup (1,\infty )}$ and the potential vertical asymptotes are x = -1 and x = 1.
Step 2:
Now we should find the y-intercepts(zeros) to determine if f has any holes, which are zeros of numerator and denominator. Thus, y-intercepts correspond to zeros of the numerator x = -2, 1. Now we know we have a hole at x = 1, and a y-intercept at ${\displaystyle (-2,0).}$
Step 3:
For the horizontal asymptote take the limit as x goes to ${\displaystyle \infty .}$ This tells us that the horizontal asymptote is y = 1.
Domain: ${\displaystyle (-\infty ,-1)\cup (-1,1)\cup (1,\infty )}$ The x-intercept is ${\displaystyle (-2,0)}$ and y-intercept is ${\displaystyle (0,2)}$
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2022-01-23 08:48:58
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http://qmviews.blogspot.com/2011/06/jmol-002-control-over-two-structures.html
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## Mittwoch, 22. Juni 2011
### Jmol 002: Control over Two Structures
I loaded two structures into a jmolApplet by the following code:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN""http://www.w3.org/TR/html4/loose.dtd"><html><head><title>UI Controls example</title><script src="./jmolDir/Jmol.js" type="text/javascript"></script></head><body> <script type="text/javascript"> jmolInitialize("./jmolDir"); jmolApplet(400, "load 1AVD.pdb;\ load APPEND nwSurface.xyz;\ axes on;\ boundbox on;"); </script><br /></body></html>
This results in the figure below.
A protein structure is placed above an array of carbon atoms which are supposed to illustrate the surface of a nanowire. What I would like to do now is to be able to rotate the protein with respect to the surface (i.e. keeping the surface in the same orientation regarding the indicated coordinate system).
Unfortunately, I dont know how to include Jmol on Blogspot, otherwise I naturally would have included it directly here as well.
The input from Bob Hanson at the Jmol Mailing list is actually pointing me into exactly the correct direction. This script was tested in a local version of Jmol but it should just as well be working in a browser.
load 1PDB.pdbload APPEND nwSurface.xyzselect proteinset allowRotateSelected Trueset dragSelected True
Now using <ALT>+<Left_Mouse> the user can rotate the protein structure without changing the surface and <ALT>+<SHIFT>+<Left_Mouse> to drag the structure over the surface.
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2019-02-17 15:33:32
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https://en.zdam.xyz/problem/12732/
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#### Problem 96E
96. The biomass $B(t)$ of a fish population is the total mass of the members of the population at time $t$. It is the product of the number of individuals $N(t)$ in the population and the average mass $M(t)$ of a fish at time $t$. In the case of guppies, breeding occurs continually. Suppose that at time $t=4$ weeks the population is 820 guppies and is growing at a rate of 50 guppies per week, while the average mass is $1.2 \mathrm{~g}$ and is increasing at a rate of $0.14 \mathrm{~g} /$ week. At what rate is the biomass increasing when $t=4 ?$
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2022-10-07 12:29:49
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https://www.gradesaver.com/textbooks/math/calculus/calculus-with-applications-10th-edition/chapter-r-algebra-reference-r-6-exponents-r-6-exercises-page-r-25/40
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## Calculus with Applications (10th Edition)
Published by Pearson
# Chapter R - Algebra Reference - R.6 Exponents - R.6 Exercises - Page R-25: 40
1
#### Work Step by Step
$\displaystyle \frac{3^{-5/2}\cdot 3^{3/2}}{3^{7/2}\cdot 3^{-9/2}}$=...use: $a^{m}\cdot a^{n}=a^{m+n}$ = $\displaystyle \frac{3^{-5/2+3/2}}{3^{7/2-9/2}}$ =$\displaystyle \frac{3^{-2/2}}{3^{-2/2}}$ =$\displaystyle \frac{3^{-1}}{3^{-1}}$ = 1
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.
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2019-05-26 06:18:21
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https://search.r-project.org/CRAN/refmans/energy/html/edist.html
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edist {energy} R Documentation
## E-distance
### Description
Returns the E-distances (energy statistics) between clusters.
### Usage
edist(x, sizes, distance = FALSE, ix = 1:sum(sizes), alpha = 1,
method = c("cluster","discoB"))
### Arguments
x data matrix of pooled sample or Euclidean distances sizes vector of sample sizes distance logical: if TRUE, x is a distance matrix ix a permutation of the row indices of x alpha distance exponent in (0,2] method how to weight the statistics
### Details
A vector containing the pairwise two-sample multivariate \mathcal{E}-statistics for comparing clusters or samples is returned. The e-distance between clusters is computed from the original pooled data, stacked in matrix x where each row is a multivariate observation, or from the distance matrix x of the original data, or distance object returned by dist. The first sizes[1] rows of the original data matrix are the first sample, the next sizes[2] rows are the second sample, etc. The permutation vector ix may be used to obtain e-distances corresponding to a clustering solution at a given level in the hierarchy.
The default method cluster summarizes the e-distances between clusters in a table. The e-distance between two clusters C_i, C_j of size n_i, n_j proposed by Szekely and Rizzo (2005) is the e-distance e(C_i,C_j), defined by
e(C_i,C_j)=\frac{n_i n_j}{n_i+n_j}[2M_{ij}-M_{ii}-M_{jj}],
where
M_{ij}=\frac{1}{n_i n_j}\sum_{p=1}^{n_i} \sum_{q=1}^{n_j} \|X_{ip}-X_{jq}\|^\alpha,
\|\cdot\| denotes Euclidean norm, \alpha= alpha, and X_{ip} denotes the p-th observation in the i-th cluster. The exponent alpha should be in the interval (0,2].
The coefficient \frac{n_i n_j}{n_i+n_j} is one-half of the harmonic mean of the sample sizes. The discoB method is related but with different ways of summarizing the pairwise differences between samples. The disco methods apply the coefficient \frac{n_i n_j}{2N} where N is the total number of observations. This weights each (i,j) statistic by sample size relative to N. See the disco topic for more details.
### Value
A object of class dist containing the lower triangle of the e-distance matrix of cluster distances corresponding to the permutation of indices ix is returned. The method attribute of the distance object is assigned a value of type, index.
### Author(s)
Maria L. Rizzo mrizzo@bgsu.edu and Gabor J. Szekely
### References
Szekely, G. J. and Rizzo, M. L. (2005) Hierarchical Clustering via Joint Between-Within Distances: Extending Ward's Minimum Variance Method, Journal of Classification 22(2) 151-183.
doi: 10.1007/s00357-005-0012-9
M. L. Rizzo and G. J. Szekely (2010). DISCO Analysis: A Nonparametric Extension of Analysis of Variance, Annals of Applied Statistics, Vol. 4, No. 2, 1034-1055.
doi: 10.1214/09-AOAS245
Szekely, G. J. and Rizzo, M. L. (2004) Testing for Equal Distributions in High Dimension, InterStat, November (5).
Szekely, G. J. (2000) Technical Report 03-05, \mathcal{E}-statistics: Energy of Statistical Samples, Department of Mathematics and Statistics, Bowling Green State University.
energy.hclust eqdist.etest ksample.e disco
### Examples
## compute cluster e-distances for 3 samples of iris data
data(iris)
edist(iris[,1:4], c(50,50,50))
## pairwise disco statistics
edist(iris[,1:4], c(50,50,50), method="discoB")
## compute e-distances from a distance object
data(iris)
edist(dist(iris[,1:4]), c(50, 50, 50), distance=TRUE, alpha = 1)
## compute e-distances from a distance matrix
data(iris)
d <- as.matrix(dist(iris[,1:4]))
edist(d, c(50, 50, 50), distance=TRUE, alpha = 1)
[Package energy version 1.7-10 Index]
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2022-05-25 19:35:23
|
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https://statkat.com/stattest.php?t=5&t2=44&t3=8&t4=1
|
# One sample z test for the mean - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
One sample $z$ test for the mean
Binomial test for a single proportion
Two sample $z$ test
$z$ test for a single proportion
Independent variableIndependent variableIndependent/grouping variableIndependent variable
NoneNoneOne categorical with 2 independent groupsNone
Dependent variableDependent variableDependent variableDependent variable
One quantitative of interval or ratio levelOne categorical with 2 independent groupsOne quantitative of interval or ratio levelOne categorical with 2 independent groups
Null hypothesisNull hypothesisNull hypothesisNull hypothesis
H0: $\mu = \mu_0$
Here $\mu$ is the population mean, and $\mu_0$ is the population mean according to the null hypothesis.
H0: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis.
H0: $\mu_1 = \mu_2$
Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2.
H0: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
H1 two sided: $\mu_1 \neq \mu_2$
H1 right sided: $\mu_1 > \mu_2$
H1 left sided: $\mu_1 < \mu_2$
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
AssumptionsAssumptionsAssumptionsAssumptions
• Scores are normally distributed in the population
• Population standard deviation $\sigma$ is known
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Within each population, the scores on the dependent variable are normally distributed
• Population standard deviations $\sigma_1$ and $\sigma_2$ are known
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
• Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb:
• Significance test: $N \times \pi_0$ and $N \times (1 - \pi_0)$ are each larger than 10
• Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures in sample are each 15 or more
• Plus four 90%, 95%, or 99% confidence interval: total sample size is 10 or more
• Sample is a simple random sample from the population. That is, observations are independent of one another
If the sample size is too small for $z$ to be approximately normally distributed, the binomial test for a single proportion should be used.
Test statisticTest statisticTest statisticTest statistic
$z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
Here $\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to the null hypothesis, $\sigma$ is the population standard deviation, and $N$ is the sample size.
The denominator $\sigma / \sqrt{N}$ is the standard deviation of the sampling distribution of $\bar{y}$. The $z$ value indicates how many of these standard deviations $\bar{y}$ is removed from $\mu_0$.
$X$ = number of successes in the sample$z = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $\sigma^2_1$ is the population variance in population 1, $\sigma^2_2$ is the population variance in population 2, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis.
The denominator $\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}}$ is the standard deviation of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $z$ value indicates how many of these standard deviations $\bar{y}_1 - \bar{y}_2$ is removed from 0.
Note: we could just as well compute $\bar{y}_2 - \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$.
$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
Here $p$ is the sample proportion of successes: $\dfrac{X}{N}$, $N$ is the sample size, and $\pi_0$ is the population proportion of successes according to the null hypothesis.
Sampling distribution of $z$ if H0 were trueSampling distribution of $X$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $z$ if H0 were true
Standard normal distributionBinomial($n$, $P$) distribution.
Here $n = N$ (total sample size), and $P = \pi_0$ (population proportion according to the null hypothesis).
Standard normal distributionApproximately the standard normal distribution
Significant?Significant?Significant?Significant?
Two sided:
Right sided:
Left sided:
Two sided:
• Check if $X$ observed in sample is in the rejection region or
• Find two sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
Right sided:
• Check if $X$ observed in sample is in the rejection region or
• Find right sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
Left sided:
• Check if $X$ observed in sample is in the rejection region or
• Find left sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
Two sided:
Right sided:
Left sided:
Two sided:
Right sided:
Left sided:
$C\%$ confidence interval for $\mu$n.a.$C\%$ confidence interval for $\mu_1 - \mu_2$Approximate $C\%$ confidence interval for $\pi$
$\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).
The confidence interval for $\mu$ can also be used as significance test.
-$(\bar{y}_1 - \bar{y}_2) \pm z^* \times \sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).
The confidence interval for $\mu_1 - \mu_2$ can also be used as significance test.
Regular (large sample):
• $p \pm z^* \times \sqrt{\dfrac{p(1 - p)}{N}}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
With plus four method:
• $p_{plus} \pm z^* \times \sqrt{\dfrac{p_{plus}(1 - p_{plus})}{N + 4}}$
where $p_{plus} = \dfrac{X + 2}{N + 4}$ and the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
Effect sizen.a.n.a.n.a.
Cohen's $d$:
Standardized difference between the sample mean and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{\sigma}$$ Cohen's $d$ indicates how many standard deviations $\sigma$ the sample mean $\bar{y}$ is removed from $\mu_0.$
---
Visual representationn.a.Visual representationn.a.
--
n.a.n.a.n.a.Equivalent to
---
• When testing two sided: goodness of fit test, with a categorical variable with 2 levels.
• When $N$ is large, the $p$ value from the $z$ test for a single proportion approaches the $p$ value from the binomial test for a single proportion. The $z$ test for a single proportion is just a large sample approximation of the binomial test for a single proportion.
Example contextExample contextExample contextExample context
Is the average mental health score of office workers different from $\mu_0 = 50$? Assume that the standard deviation of the mental health scores in the population is $\sigma = 3.$Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$?Is the average mental health score different between men and women? Assume that in the population, the standard devation of the mental health scores is $\sigma_1 = 2$ amongst men and $\sigma_2 = 2.5$ amongst women.Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$? Use the normal approximation for the sampling distribution of the test statistic.
n.a.SPSSn.a.SPSS
-Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
• Put your dichotomous variable in the box below Test Variable List
• Fill in the value for $\pi_0$ in the box next to Test Proportion
-Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
• Put your dichotomous variable in the box below Test Variable List
• Fill in the value for $\pi_0$ in the box next to Test Proportion
If computation time allows, SPSS will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution
n.a.Jamovin.a.Jamovi
-Frequencies > 2 Outcomes - Binomial test
• Put your dichotomous variable in the white box at the right
• Fill in the value for $\pi_0$ in the box next to Test value
• Under Hypothesis, select your alternative hypothesis
-Frequencies > 2 Outcomes - Binomial test
• Put your dichotomous variable in the white box at the right
• Fill in the value for $\pi_0$ in the box next to Test value
• Under Hypothesis, select your alternative hypothesis
Jamovi will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution
Practice questionsPractice questionsPractice questionsPractice questions
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2022-12-09 15:58:13
|
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https://socratic.org/questions/an-object-with-a-mass-of-15-kg-is-revolving-around-a-point-at-a-distance-of-16-m#370645
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# An object with a mass of 15 kg is revolving around a point at a distance of 16 m. If the object is making revolutions at a frequency of 15 Hz, what is the centripetal force acting on the object?
Jan 29, 2017
I got:
${F}_{c} = 1 , 932 , 407 N$
#### Explanation:
We start remembering that centripetal force is:
${F}_{c} = m {v}^{2} / r$
To evaluate the velocity we use frequency and we consider that our object is making 15 revolutions in 1 second so we get that it takes:
$\frac{1}{15} = 0.07 s$ to complete one revolution that represents a distance along a circle of $2 \pi r = 2 \cdot 3.14 \cdot 16 = 100.53 \approx 100.5 m$.
Velocity will then be: $v = \frac{100.5}{0.07} = 1435.7 \frac{m}{s}$
So we get:
${F}_{c} = 15 \cdot {\left(1435.7\right)}^{2} / 16 = 1 , 932 , 407 N$
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2022-08-18 10:14:28
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https://math.stackexchange.com/questions/2226993/find-length-of-triangle-as-well-as-radius-of-inscribed-circle
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# Find length of triangle as well as radius of inscribed circle
Question: In triangle ABC, angle A = 60°, AC = 21, BC = 19. The circle inscribed in triangle ABC touches the sides AB, BC, and CA at points P, Q, R respectively. Find all possible lengths of all AB, BQ and the radius of the inscribed circle.
Using the sine law, I have found that angle $B = 73.17355^\circ$ and angle $C = 46.82645^\circ$. I also found that the missing side length $C$ is $16$.
However, I don't understand how I can somehow use this knowledge to find the radius of the circle inscribed within the triangle. In addition, how can there be more than one length for $AB$ and $BQ$?
$$AB^2+AC^2-BC^2=2\cdot AB\cdot AC\cdot \cos A \implies AB = 5 \text{ or } 16$$ The area of $\triangle ABC$ can be written as the following two ways: $$S_{ABC}=\frac{1}{2}\cdot AB\cdot AC\cdot \sin A = \frac{1}{2}\cdot(AB+AC+BC)\cdot r$$ where $r$ is the radius of the incircle of $\triangle ABC$. This will give us $$r=\frac{AB\cdot AC\cdot \sin60^\circ}{AB+AC+BC}$$ By plugging in the values of $AB, AC, BC$ and $\sin A$, we can get $r$. Further, because $BP=BQ,\ CQ=CR,\ AR=AP$, and $$BP+AP=AB\\BQ+CQ=BC\\AR+RC=AC$$ we know that $$BQ = \frac{BA+BC-AC}{2}$$Therefore, the following two cases are:
Case 1: $AB=5$, then $r=\frac{7\sqrt{3}}{6}$ and $BQ=\frac{3}{2}$.
Case 2: $AB = 16$, then $r = 3\sqrt{3}$ and $BQ = 7$.
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2022-01-17 02:57:46
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https://www.ideals.illinois.edu/handle/2142/100827
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Files in this item
FilesDescriptionFormat
application/pdf
1100633.pdf (7MB)
PresentationPDF
application/pdf
2999.pdf (17kB)
AbstractPDF
Description
Title: COSMIC RAY-DRIVEN RADIATION CHEMISTRY IN COLD INTERSTELLAR ENVIRONMENTS Author(s): Shingledecker, Christopher N. Contributor(s): Herbst, Eric; Le Gal, Romane; Tennis, Jessica D. Subject(s): Astronomy Abstract: The physiochemical impact of cosmic rays on interstellar regions is widely known to be significant \footnote{Indriolo, N. \& McCall, B. J.,\textit{Chem. Soc. Rev.}, 42, 7763-7773, 2013}. Indeed, the cosmic ray-driven formation of H$_3^+$ via the ionization of H$_2$ was shown to be of key importance in even the first astrochemical models \footnote{Herbst, E. \& Klemperer, W., \textit{Ap.J.}, 185, 505-534, 1973}. Later, cosmic rays were implicated in the collisional excitation of H$_2$, which leads to the production of internally produced UV photons that also have profound effects on the chemistry of molecular clouds \footnote{Prasad, S. S. \& Tarafdar, S. P.,\textit{Ap.J.}, 267, 603-609, 1983}. Despite these key findings, though, attempts at a more complete consideration of interstellar radiation chemistry have been stymied by the lack of a general method suitable for use in astrochemical models and capable of preserving the salient macroscopic phenomena that emerge from a large number of discrete microscopic events. Recently, we have developed a theoretical framework which meets these criteria and allows for the estimation of the decomposition pathways, yields, and rate coefficients of radiation-chemical reactions \footnote{Shingledecker, C. N. \& Herbst, E., \textit{Phys. Chem. Chem. Phys.}, 20, 5359-5367, 2018}. In this talk, we present preliminary results illustrating the effect of solid-phase radiation chemistry on models of TMC-1 in which we consider the radiolysis of the primary ice-mantle constituents of dust grains. We further discuss how the inclusion of this non-thermal chemistry can lead to the formation of complex organic molecules from simpler ice-mantle constituents, even under cold core conditions. Issue Date: 06/20/18 Publisher: International Symposium on Molecular Spectroscopy Citation Info: APS Genre: Conference Paper / Presentation Type: Text Language: English URI: http://hdl.handle.net/2142/100827 DOI: 10.15278/isms.2018.WL04 Other Identifier(s): WL04 Date Available in IDEALS: 2018-08-172018-12-12
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2020-10-27 10:31:45
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http://www.ams.org/mathscinet-getitem?mr=316377
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MathSciNet bibliographic data MR316377 10B10 (14G25 14H25) Mohanty, S. P. A note on Mordell's equation \$y\sp{2}=x\sp{3}+k\$$y\sp{2}=x\sp{3}+k$. Proc. Amer. Math. Soc. 39 (1973), 645–646. Article
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.
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2016-06-01 07:29:17
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https://stats.stackexchange.com/questions/102759/differences-between-logistic-regression-and-perceptrons/284013
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# Differences between logistic regression and perceptrons
As I understand, a perceptron/single-layer artificial neural network with a logistic sigmoid activation function is the same model as logistic regression. Both models are given by the equation:
$F(x) = \frac{1}{1-e^{-\beta X}}$
The perceptron learning algorithm is online and error-driven, whereas the parameters for logistic regression could be learned using a variety of batch algorithms, including gradient descent and Limited-memory BFGS, or an online algorithm, like stochastic gradient descent. Are there any other differences between logistic regression and a sigmoid perceptron? Should the results of a logistic regressor trained with stochastic gradient descent be expected to be similar to the perceptron?
• Looks like this question is similar, and it seems to contain better responses :) – Ralph Tigoumo May 7 '16 at 9:34
You mentioned already the important differences. So the results should not differ that much.
• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – Xi'an Mar 1 '15 at 17:14
• Actually I tried to answer both questions: 1) "Are there any other differences between logistic regression and a sigmoid perceptron?" and 2) "Should the results of a logistic regressor trained with stochastic gradient descent be expected to be similar to the perceptron?" – Michael Dorner Mar 1 '15 at 17:27
• That's a reasonable position, @MichaelDorner. Would you mind expanding your answer a little to make that clearer? – gung Mar 1 '15 at 17:44
I believe one difference you're missing is the fact that logistic regression returns a principled classification probability whereas perceptrons classify with a hard boundary.
This is mentioned in the Wiki article on Multinomial logistic regression.
There is actually a big substantial difference, which is related to the technical differences that you mentioned. Logistic regression models a function of the mean of a Bernoulli distribution as a linear equation (the mean being equal to the probability p of a Bernoulli event). By using the logit link as a function of the mean (p), the logarithm of the odds (log-odds) can be derived analytically and used as the response of a so-called generalised linear model. Parameter estimation on this GLM is then a statistical process which yields p-values and confidence intervals for model parameters. On top of prediction, this allows you to interpret the model in causal inference. This is something that you cannot achieve with a linear Perceptron.
The Perceptron is a reverse engineering process of logistic regression: Instead of taking the logit of y, it takes the inverse logit (logistic) function of wx, and doesn't use probabilistic assumptions for neither the model nor its parameter estimation. Online training will give you exactly the same estimates for the model weights/parameters, but you won't be able to interpret them in causal inference due to the lack of p-values, confidence intervals, and well, an underlying probability model.
Long story short, logistic regression is a GLM which can perform prediction and inference, whereas the linear Perceptron can only achieve prediction (in which case it will perform the same as logistic regression). The difference between the two is also the fundamental difference between statistical modelling and machine learning.
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2019-08-24 20:18:54
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https://eccc.weizmann.ac.il/keyword/19338/
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Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > KEYWORD > MATRIX SPACES:
Reports tagged with Matrix Spaces:
TR16-145 | 16th September 2016
Markus Bläser, Gorav Jindal, Anurag Pandey
#### Greedy Strikes Again: A Deterministic PTAS for Commutative Rank of Matrix Spaces
Revisions: 2
We consider the problem of commutative rank computation of a given matrix space, $\mathcal{B}\subseteq\mathbb{F}^{n\times n}$. The problem is fundamental, as it generalizes several computational problems from algebra and combinatorics. For instance, checking if the commutative rank of the space is $n$, subsumes problems such as testing perfect matching in graphs ... more >>>
ISSN 1433-8092 | Imprint
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2022-10-07 09:02:32
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https://www.intel.com/content/www/us/en/developer/articles/technical/how-to-install-the-python-version-of-intel-daal-in-linux.html
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# How to Install the Python* Version of Intel® Data Analytics Acceleration Library (Intel® DAAL) in Linux*
Published: 05/11/2016
Last Updated: 05/11/2016
## Introduction
The Intel® Data Analytics Acceleration Library (Intel® DAAL) 1, 2 is a software solution for data analytics. It provides building blocks for data preprocessing, transformation, modeling, predicting, and so on.
The beta version of Intel DAAL 2017 provides support for the Python* scripting language.
This article shows you a simple way to install Intel Distribution for Python and Intel DAAL in Linux.
## Installation
There is more than one way to install the Python version of Intel DAAL in Linux*. Intel DAAL is included with the Intel® Distribution for Python* and in the Intel® Parallel Studio XE 2017 beta. However, for the Intel Parallel Studio 2017 beta 3, users will need to compile the Python source before it can be used.
Thanks to Zhang for showing a simple way to install Intel DAAL through the Anaconda* 4 distribution.
## Step-by-Step Instructions
The following steps show you how to install Intel Distribution for Python and Intel DAAL:
1. Install the Anaconda distribution.
• Go to the Anaconda download website 4, and then download either the 2.7 or 3.5 Python version of Anaconda for 64-bit Linux.
• Execute the following command to install Anaconda:
bash Anaconda2-4.0.0-Linux-x86_64.sh (for Python 2.7)
or
bash Anaconda3-4.0.0-Linux-x86_64.sh (for Python 3.5)
2. Add the Intel channel to Anaconda.
1. Before adding the Intel channel into Anaconda, make sure that Anaconda has been correctly installed. One way to do that is to run Python at the command prompt to see if this Python version is from Anaconda:
2. If step 2.1 is correct, use the following command to add the Intel channel to Anaconda using the conda utility included in Anaconda:
conda config --add channels intel
3. Check to ensure that the Intel channel has been added using the following command:
conda config --get channels
If the Intel channel has been added, the result will look something like this:
3. Install Python version of Intel® DAAL under Anaconda.
Using the following command to create a new environment for the Intel version Python installation and to install the Intel Distribution for Python and Intel DAAL:
conda create -n intelpython --override-channels --channel intel python=3.5 intelpython scipy pydaal
where:
intelpython (after the switch '-n'): The name of the Intel version Python environment. This can be any name. This name will be used to switch to the Intel version Python environment later on.
Channel intel: the Intel channel.
python=3.5: the Intel version Python installation uses Python version 3.5, not 2.7.
intelpython: the Intel version Python installation name. This cannot be changed.
pydaal: Python version of Intel DAAL.
scipy: pydaal needs this package.
4. Check to make sure the Python's environment exists.
Use the following command to ensure the Intel version Python environment “intelpython” has been created:
conda env list
The screen should look like this:
5. Switch to the Intel Python environment.
To use the Python version of Intel DAAL, you need to be in the Intel version Python environment. Use the following command to switch to the Intel version Python environment “intelpython” created above:
Source activate intelpython
If the environment is switched successfully, you will see something like this:
To switch back to the default environment, issue the following command:
Source deactivate
6. Check to make sure the Intel version Python installation is selected.
Before using Intel Distribution for Python, we need to switch to the Intel version Python environment as shown above. Run the following command to ensure the Intel version Python installation is selected, not the version of Python that comes with Anaconda:
Python3.5
The screen should look like this:
Note: The Python version of Intel DAAL only works with Python 3.5 and above. It does not work with Python 2.7.
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2022-09-30 20:53:11
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http://sacha.work/thoughts/
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# A way to think about thought
### Preface
This is not a mathematical piece, this is just a piece that uses math to help us grok the concept of thought. It is pseudomathematical in both terminology and implementation—sometimes by design, other times by my ignorance. In either event, this is just the way I thought through everything and this piece is my attempt to bring you into that process.
Also, this essay ended up being more of a mind dump than a careful and calculated exposition about my framework. Please keep that in mind. Feel free to fork this and make whatever changes you think make sense—and, of course, to let me know when you do so. I hope you enjoy :)
### Overview/Terminology
In trying to articulate something for another thought piece I was writing, I inadvertently devised a new framework I found even more exciting…
Consider two ideologies, A and B, projected into some n-dimensional idea or thought space. Each dot is an individual in that thought space, each with their own ideological composition that places them in a certain position in this thought space.
That weird shape “A” in my figure is an ideology, which I’ll call a “subspace”—that is, the minimum bounded thought space that encapsulates all individuals belonging to that ideology. People (again, those dots) with minor deviations from the accepted school of thought push the bounds of that ideology outward into new pockets of thought space. The morphology of the subspace is therefore determined by the ideologies of its constituent individuals.
“B” is an ideological antagonist to A. That is, there is no set of shared coordinates in thought space between A and B (i.e. A ⋂ B = Ø.) “x” is an individual whose current ideology is not readily defined (i.e. there is no publicly accepted label* for that individual’s ideas and therefore their position in thought space.)
*This is slightly misleading and so the nuance is captured below.
### Expanded model
The proximity of x in thought space to a given subspace is an important component in determinining whether x will be subsumed by that subspace. One can think of it as the extent of ideological alignment—someone whose opinions may be aligned with Socialism but who has yet to classify themselves as such.
However, an important distinction is that it is not actual proximity in thought space to a given subspace but is instead the proximity to the nearest known subspace that is the relevant factor. If one is unaware of the existence of Buddhism but arrives at conclusions similar to Buddhist tenets, one will not become a Buddhist until it appears on their map of known thought space. They are only then even remotely likely to classify themselves as a Buddhist.
Now it only naturally follows that the likelihood of someone being sucked into that ideology, or subspace, is directly proportional to their proximity in thought space to that subspace. That is, the more aligned with a given ideology a person is, the more likely they are to convert. A corollary to this is that the likelihood of an individual not joining any defined ideology is directly proportional to the distance to their nearest ideology (again, not globally, but in known thought space.)
It might be conceptually convenient to think of each subspace in known thought space as having a certain graviational pull on all individuals. That gravitational pull seems most likely to be a function of its proximity to a given individual in thought space as well as its size (i.e. how many people are contained within the subspace, or ideology.)
Where the numerator $m_{A}$ is the number of known individuals in subspace A and the denominator $d_{xA}$ is the Euclidean distance in thought space between person x and the centroid of subspace A.
It is important to underscore the fact that both variables are determined by the known thought space—that is, the ideologies, the number of individuals in each ideology, and the nuances of each ideology known to the individual. In instances where people’s average awareness of other ideologies is limited, a certain ideology can very readily claim dominance.
Another factor that feels important to capture in this model is the innate malleability of any individual, given their age. We can likely agree (I hope definitively) that one’s adolescent years are more formative than one’s late 60’s. This is not to say that ideological conversion is impossible at a later stage, it is just to acknowledge that there is an innate bias that predisposes people to being more receptive to ideologies at certain ages than at others.
You might argue with the shape of this curve (and we’d definitely both disagree about its relative proportions) but the important thing is that this malleability curve, let’s say M(t), is a nonlinear function normalized to the peak malleability of an individual throughout their lifetime. This serves as a weighting function for our “gravitational pull” equation. People are more malleable, or receptive, to the pull of ideologies at certain points in their life. We’re just looking for a way to express this “mathematically”.
So now to update the original equation…
With the only difference being that we are now also weighting the influence of an ideology by the malleability/receptiveness of that individual based on their age at the time of exposure.
Peak likelihood of indoctrination is likely during adolescence when the known thought space is minimal and the malleability is high. This basically affords an ideology unfettered influence over that individual, especially if the exposure to other ideologies is negligible.
Peak likelihood of conversion might come during college, where exposure to more ideologies is extreme (i.e. the known thought space broadens) and malleability is high.
### Outliers
Given the information above, it follows that there must be some threshold of dissonance between the individual’s beliefs and the “mean ideology” (i.e. the centroid of the subspace to which that individual belongs) that causes a fracturing between the two (i.e. x no longer is part of A.) I see this as the point at which the individual would openly dissociate from the label of that subspace.
As nodes (individuals) start to cluster around a particular region, especially near the boundaries of a subspace, it makes sense that they might collectively split from the subspace with enough momentum (ideological meiosis?.)
Their likelihood of splitting is similar to the equation outlined above. The further on the periphery of an ideology they are and the more frequent and raw the exposure to each other’s deviations from that mean ideology, the more likely they are to split and form their own subspace. Think: denominations of Christianity that formed and continue to form.
Requisite for these types of subspaces to form is some medium through which these individuals/nodes can be made aware of one another. Prior to Gutenberg’s invention of the movable-type printing press, new subspaces were not able to form as easily given that the set of nodes whose position was known to other nodes was limited. That is, the known thought space for the average node was significantly more limited than it is today (duh.)
As communication technologies emerged they made more people more aware of other people and their respective ideologies. The average known thought space was broadened.
History has provided us with many such communication technologies (of all scales.) From books/literacy to cities to IRC to 4chan to Pioneer and more, substrates have and continue to emerge to facilitate quicker development of known thought space. As more nodes become more aware of global thought space more subspaces and conversions emerge.
As more nodes join each subspace, they strengthen its existence by broadening its boundaries (in both physical and thought spaces!.) This reflexively increases the likelihood of subsuming new nodes.
With increasing means of exposing individuals to global thought space, we can expect to see more ideologies/communities continue to emerge. This is already expected, of course, it is just pleasing to arrive to the same conclusion using this framework.
### Nascent subspaces and their formation
Martin Luther would not have been able to spark the Reformation without making people aware that other people were aware of his nascent ideology (early use of network effects.) Communication tools serve as a substrate through which the average known thought space of its participants broadens.
### Real-world implications
I believe there are some real-world implications for this framework. Consider for instance the targeting of youth for ideologies of various kinds (e.g. Christian Camp, Hitler Youth, Boy Scouts, etc.) Indoctrination 101… This framework potentially offers a crude way of modeling the various factors that drive the adoption of those ideologies—that is:
• the innate malleability or plasticity of people of a certain age
• the number of people they are exposed to who believe that ideology
• the extent of their awareness of other ideologies
• the extent of ideological alignment with other ideologies of which the individual is aware
This also provides a convenient way to think about how limited exposure to other ideologies breeds blind dogma—think: the feedback loop of social media echo chambers, people in remote regions who have limited exposure to people of the “opposite” political persuasion, etc. Conversely, we can also see how exposure breeds intellectual diversity—in this framework, new subspaces (or ideologies) invariably emerge from exposure (i.e. expanding known thought space.) People who live in big cities likely have broader known thought spaces than those who live in rural regions. For politically inclined adults who move to big cities but still are blinded by dogma, this is also explained: their M(t) (malleability) is so low that the gravitational pull of another ideology that would otherwise exist is significantly reduced. This is not at all to say that it is impossible to have conviction in a big city—it is just to say that the constant exposure will on average erode that hypothetical max degree of conviction.
### Final thoughts
This framework is incomplete (and possibly incoherent..?) I will plan to expand on it as both time and my headspace allows. For now, I’m posting prematurely to feel the pressure of random eyeballs as well as to get feedback in the process. Thanks for reading!
# My secret side project
On a gorgeous Hawaiian day in late December I whipped together a little script. For various reasons, I am keeping the purposes of this script secret. What you need to know is that this little script was birthed from the confluence of my boredom and my randomly reading API docs of a popular mobile app. I thought of a cool feature one could build using the API and built out a pretty basic service in a few hours. I sent my MVP to a close friend in consumer social. 10sec later he texts back…
He gives me the idea for an improved derivative of this product and suggests we productize and ship that new product in the next 2 weeks. We partner up with one of his colleagues to build out a website and productionize my little MVP.
I whip together a sufficiently scalable backend and clean up my script. My friend designs a simple website which his colleague builds with a basic API layer to interface with our db. We reach out to some TikTokers to post and within a day we’ve locked in the campaign.
After some basic stress tests, we feel we are ready and the first TikTok post goes out on Jan. 16th. Within the first 22 days we get the following data:
TL;DR this project is still in its infancy and we initially got obliterated by the demand (crashing servers, blocking/delaying services, etc..) We have a lot of conviction in the project and are going to start properly ramping up the growth. This was all just a very prelimenary test-run.
I’ve realized in pursuing this side project the power of rapid exploration. This flies in the face of what most founders are taught to believe. People in tech overindex on the merits of shipping products the conventional way (i.e. come up with an idea, make a pitch deck, reach out to investors, build out an MVP after hiring the 10x engineers you convinced to join, tell everyone on LinkedIn you’re the CEO of a future unicorn despite the fact that all you’ve done thus far is incorporate with Stripe Atlas, etc..) Even as kids we are trained to think so definitively about things—”I will be an X one day”. Why not just give it a shot and reassess in a couple of weeks?
This approach is not per se a prescription for anyone considering pursuing a startup but moreso a recommendation to those who are capable enough to build products and curious enough to indulge that capability. Just build something casually for 2 weeks and make sure to get it to market fast enough to determine whether it’s worth pursuing further.
If you are toying with an idea, I suggest you:
1. Quickly (1 week MAX) build an MVP (if you are technical—if not, then make a Figma protoype)
2. Text the absolute minimum number of people you need to see this project through to completion to collaborate with you (2 other people MAX, in my opinion)
3. Make a compelling landing page with at least a sign up field
4. Launch using TikTok and make an assessment in 2 weeks as to whether the product is worth pursuing further (spend no more than 2,000) The opportunity cost is completely overshadowed by the potential upside for these side projects. I want to advocate heavily for lowering the mental friction of starting a project. Start putting together something cool tomorrow and let me know how it goes (DM’s open on Twitter!) ### March 23rd, 2021 # Change vs. Progress It is incredibly important not to conflate change with progress. Progress exists at the intersection between change (state B ≠ state A) and growth (some optimization towards the objective). Change can exist without growth (e.g. a new presidency, binary fission of a bacterium, etc.), but growth cannot exist without change (some sense of modification is requisite for optimization). Without a driving force towards growth, the action or change will be what we can call “yieldless”. For example, if I choose to go running, the act of running itself may in fact be the objective, and that will hold under any context in which I am running. However, if my goal has shifted to using running as a means of achieving something else (like moving from point A to point B), then the actions that progress me towards that goal become contextual—running on a treadmill achieves the same action without any progress (as defined by the context). By the same token, science has the bad habit of feigning progress where, in most instances, we are simply just running in place.1 It is very easy to see changes within science and be excited by the carrot of progress, but how have those changes progressed us towards the initial objective? In medicine, as an admitted generalization, the objective function is easily defined: are we closer to curing the disease in question? Without proper heuristics for determining this, yiedless changes can safely hide under the veil of progress.2 I suspect that we would benefit greatly from inoculating ourselves to the allure of change and no longer misconstruing it as progress. Who would have thought that using the ultimate goal to guide the action could be so hard?? ##### 1The iPhone, as another example, has arguably undergone many yieldless changes over the years (compare the growth rate of the product quality in the early years of the iPhone to now), yet the illusion of continued progress has been upheld by the consistency of these iterations. ##### 2Collison and Cowen’s 2019 piece on this (here) outlines the urgency of establishing these heurestics across industries. It’s a great read. ### Novemeber 3rd, 2020 # Reflexivity in politics Distributing poll results to a population before an election degrades the predictiveness of that population by a significant margin. Is there a psephological law for this? I’m not sure, but there should be… In 2016, I warned my family and friends that distributing polls that predict a landslide in Clinton’s favor would result in mass complacency, and that this complacency would manifest as poor Democractic turnout. If my community of friends is at all a microcosm to the younger population in the United States, in 2016, ~10 of them (likely more—I only asked a small handful) did not vote because “Hilary [was] going to win in a landslide… No need to go through the pain of voting”. 3 of those 10 friends were from Florida, a crucial swing state in that election. I fear that the same thing may happen with this election today. While I am not confident in the final outcome of this election, I do feel comfortable positing that the final result will be much closer than predicted by any leading pollsters. The point here is that I think that there is a lot about reflexivity that remains unexplored and believe that finding elegant ways to mathematically model reflexivity in different systems would be of immeasurable benefit to statistics, economics, neuroscience, and beyond. ### June 1st, 2020 # Some questions… ##### A few random things that I’ve been thinking about quite a bit. Some date back years, others in the past few hours #### Why does it feel like Hungary disproportionately produced some of the brightest and most prolific minds of the 20th century? #### How can we quantify the changes to a predicted outcome after that prediction is disseminated to the population whose behavior is being predicted? • Think: election polls, distributions of wait times, etc. Is this sufficient to describe the glaring errors of all the 2016 US Presidential polls? #### What city in a Third World country will bear the greatest intellectual fruits in the next 25 years? • Lagos? CDMX? #### Why has the wash time of restaurant-grade dishwashers (~1min) been so disparate from its domestic counterpart for so long? • I’m not convinced that this “industry-domestic discrepancy” exists to the same extent for other applicable technologies, especially not for so long (the processor in your iPhone is 5 orders of magnitude more powerful than that of Apollo 11) #### Is globalization always a good thing? And to what extent does the globalization of technology align with (and also engender) cultural globalization? • China has reaped the rewards of Western innovation without the “burden” of democracy (great convo on this here), yet it does not feel that this adoption has been at the cost of any cultural identity. On the other hand, the physical presence of Turks in Germany has had profound influence on the culture (and cuisine) of the latter, though their relationship dates back to at least WWI (before US-China relations) so maybe the comparison isn’t fair. ### May 9th, 2020 # My take on the COVID-19 market ##### Disclaimer: I am not recommending any specific stock purchases here, nor am I licensed to do so. I merely want to share the thought processes that helped me profit during the volatility effectuated by COVID-19 People have been scared. It makes sense, how could you not be when there exists the incessant echo chamber that is contemporary media? Call it ignorance, but I usually never listen to stock recommendations or follow markets through any particular media outlet. My investment approach could not be simpler: I think about the market, I substantiate or uproot my hypothesis with market data, and update my internal model of that market using these data. A reasonable response would be to laugh. What the hell do you know, punk? Well, not much, but enough to consistently beat out some of the weirdest markets in the past 3 years and hold a constant >30% returns for that time. I have helped my family, other families, and my friends invest in the market, because at its best, it’s a lot of fun. I’m no guru, I just love contrarian thinking. Thankfully, this translates well in investing: flying in the face of conventional wisdom, in late 2018, I shorted FB and AAPL based on a theory I had about their self-imposed pigeon holing. If this sound stupid, I encourage you to check out their stock prices for that year (P.S. I made >350% returns from those puts). Big money is made when you are right and the majority is wrong (see Burry, Dalio, or Soros). If everyone says I am wrong about a prediction, it’s often an incentive to chase the idea further. Back to 2020… COVID-19 has taken a huge chunk out of the economy, a chunk which is almost perfectly distributed across each industry. On March 10th, I was debating with my mom about this investment climate, as she was consumed by anxiety about the stock market and how the experts were saying it would be a recession (I remain pretty convinced that an imminent threat of such magnitude to the economy that’s universally accepted as likely will in fact beget solutions that prevent it from actually materializing. These things really are reflexive in that way and every major economic recession has blindsided the majority, that’s why they happen! That’s probably for another blog post, though). My mom, in line with the rest of America, was convinced that everything would plummet and that it would be a year for any of these stocks to rebound. I thought this all over… Just looking at the tech industry, it would make sense that hardware companies would be severely impacted—a lot of these tech behemoths got a cold splash of water exposing their overdependence on China (as if the ongoing trade wars weren’t enough). These hardware companies (e.g. AAPL, NVDA, AMD) now have to rethink their entire supply chain. The software companies (e.g. TWLO, OKTA, SHOP), however, should be much more nimble and unaffected by the downturn, right? Teams of programmers should still be able to deploy code from their bedrooms and people should still consume their products remotely. As it turned out, the aforementioned industries (and individual stocks) were affected equally, and I just knew that the software stocks would rebound earlier. I’ll let the data speak for themselves: Stock returns from 3/10/20 - 5/9/20 (green = software, red = hardware) Now before you think that I retroactively changed my selection of stocks to sound smart in a blog post, these were the actual 3 that I recommended. I should probably say that I’m still invested in TWLO, and have been for years (first bought at76/share), but I sold off all my other stocks about 5 months ago when it became too time consuming.
These differences may seem slight, but the worst performing of my recommendations is 134% of the biggest return for the aforementioned hardware companies (and TWLO returns are 519% of AAPL returns for this period!).
The point here is that intuition and calculated thought can be an important factor in distinguishing you in investment. Don’t believe me? Billionaire George Soros literally moves millions of dollars based on pain in different parts of his body. Everyone else uses financials and the same old indicators—be different.
# The crucial missing link for Artificial Neural Networks
##### I’m going to jump right into this one, so apologies if it lacks the technical introduction such a post might demand
Artificial Neural Networks (ANNs), as the name might suggest, were built with the brain as a primary influence—artificial neurons connect to each other, each conferring signals to others and each with connections that vary mostly in accordance with the alignment between the neuron’s activity and the ultimate achievement of the system’s goal (see objective functions). This was a breakthrough in the field of computer science as it completely transformed the capabilities of computers to achieve complex tasks. Computers were no longer just efficient computing devices that did simple tasks quickly and consistently—they now could improve iteratively (see Perceptron and the lesser appreciated MENACE).
Things have improved a lot with regards to computing power since those early days, but the reality is that the fundamentals of ANNs simply haven’t witnessed the same growth.
What we see is that, despite the fact that we can reach unprecedented network depth, performance just doesn’t scale. Part of this is due to what is dubbed the vanishing gradient problem, in which the changes made to the connections of neurons deep into the network are negligible and therefore less likely to contribute to the final network output. This problem was significantly alleviated by the development of ResNets, the original version of which broke numerous world records for image classification and other tasks. However, increasing the depth of this network still results in a performance plateau at a certain depth (the error difference between 50 and 100 layer ResNets, for the tasks I’ve used them for, is negligible). How has increasing the number of neurons, connections, and layers to unprecedented heights in our ANNs not brought us within reasonable proximity to the cognitive powers of the brain?
Take a look at how performance of a basic network can actually get worse by adding layers (image classification network on CIFAR-10 dataset):
Deeper version of network does worse at image classification
It’s clear that a network with connectivity as its only parameter is insufficient to generate the flexibility of the brain.
I believe that the most significant factor missing from the way we build ANNs that is present in the brain is electric fields. While artifical neurons in ANNs are mathematical representations, biological neurons (just “neurons”, from now on) occupy physical space, and thus their interactions are influenced along a spatial dimension (Note: I definitely do not consider layers of an ANNs as spatial). I also believe that the importance of the electric fields emitted by neurons is severely underappreciated in the field of neuroscience, despite the vast evidence for their significance in the biophysics of the brain as well as cognition (here and here, and I’m comfortable arguing that the breakthrough of LFPs is because they register both action potentials and the resulting electric fields).
The electric fields emitted by active neurons contribute to the complexity of the entire system by being reflexive—an active neuron emits a field which in turn influences the firing dynamics of that neuron and the surrounding neurons, which influence downstream neurons which in turn emit electric fields which in turn… you see where this going. The result is that groups of neurons in the same vicinity will collectively generate large electric fields that will modify their biophysical signature and possibly enable computations that are otherwise unachievable in a state of individual isolation. The fact that neurons are exist in the world of biology has been seen as a hindrance to their computational abilities (slow time constants, high noise, energetically expensive, etc.), but the reality is that such complexity simply could not exist on a processor and instead only in a biological system, or one that merges the two.
Anyway, the point here is that for ANNs to approximate or surpass the complexity and cognitive flexibility of the brain, they must first address this issue. I think embedding actual neural tissue in circuitry (“wetware”) is a really smart way of tackling this, and I’ll be keeping my eye on Cortical Labs, Koniku, and maybe even this nutjob (who did the inverse).
# Humanity has already found the neural seat of consciousness, science has just overlooked it
If someone put a gun to my head and made me blurt out the part of the brain that houses consciousness, I’d say “thalamus” without breaking a sweat.
The thalamus is a chunk of gray matter found in the forebrain. This is the Grand Central of the brain—it receives input from a vast number of areas and its projections innervate almost all sections of the cerebral cortex. The result: the thalamus both integrates and distributes information from countless areas. For instance, if sensory input comes in, the thalamus works its magic (transforms that sensory information along some dimension), and then spits it back into the sensory area from whence it came.
I believe quite strongly that anticipation is an evolutionary prerequisite for consciousness and that it demands the coordinated integration of the senses. Why does anticipation require the integration of the senses? Just think about it: let’s say a person is swinging a punch at you. Well, first of all, you would need to calculate the trajectory of their swing (motor cortex) which would require that you visualize their fist reaching your face (visual cortex). Significance is assigned to avoiding their fist reaching your face because you know that it will hurt (nociceptors). All of this activity coming from disparate regions of your brain before the fist has even come near your face. Without the tight and timely integration of all the information, you can bet your ass that tomorrow you’d wake up with a black eye…
Why is anticipation a requisite for consciousness? Because anticipation itself requires:
• An internal, updatable model of the world (think: Bellman Equation)
• A sense of futurity (not just living in the now, but in the next)
• Pattern identification (I would feel pretty comfortable positing that this is the product of the other two)
I am putting forward that these 3 pillars are the building blocks on which consciousness is built. If we could find regions in the brain that encode or enable these, then I’d say we are well on our way to uncovering the secrets of consciousness.
Back to the thalamus… Just based on sheer connectivity, it seems like the thalamus has always been a prime candidate for understanding neural processes on a fundamental level. Why has it taken so long for focus to shift to the thalamus? The thalamus has historically been viewed as a “low-level” (yuck!) center (yuck yuck!!), as part of a contrived framework of anatomical hierarchy in the brain. Everyone learns in their 101 courses that “the midbrain is the lower brain… and the thalamus and associated areas are part of the limbic system” (here, here, here, and here). The fact that they peddle this crap to unwitting freshmen is infuriating! “Experts” loooove to talk about the limbic system, also called your “Lizard Brain” (yes, experts literally call it that). These experts have grouped the thalamus in this category based on the antiquated idea that neural anatomy recapitulates the evolutionary development of the brain. In plain English: the deepest parts of the brain are the ones most like our evolutionary ancestors. As an unfortunate consequence of this antiquated school of thought, the thalamus has unfairly been identified as a primitive area, with little historical investigation into its role in cognition! Individuals and scientific communities are influenced deeply by egos and the perception of fundamental truths—scientists are unwilling to accept novel discoveries if they fly in the face of what the scientists fundamentally believe to be true (or if they spent most of their life committed to that which is being overturned). This has largely stifled investigation into the thalamus and acceptance of the promising results that come out of it.
Mike Halassa at MIT is one of the few scientists who gets it. He’s still relatively early in his career but I am willing to predict that he will be one of the pioneers of his generation that will unhinge our current understanding of the brain (see also Doris Tsao and more recently, Steve Ramirez). If you’re looking for a good read on the thalamus and cognition, Halassa’s paper Schmidtt et al. 2017 is seminal. In brief, the paper demonstrates that thalamic neurons engage in task-relevant computations that freely transition from object-based to task-based. This means that the thalamic neurons can encode the identity of specific stimuli as well as their significance. Importantly, these thalamic neurons were found to regulate activity in the pre-frontal cortex (PFC), which is largerly seen as the most significant center in cognition. This means that the “low-level” area was not merely relaying basic sensory information from the PFC, but in fact was responsible for conferring its cognitive flexibility!
Mike and I had a great discussion last November about this fundamental oversight in the field and how overwhelming the evidence is for the thalamus’ role in cognition. The ingredients for the thalamus coming out as the neural seat of consciousness are all there—it integrates and distributes information from the broadest reaches of the brain, it confers the “high-level brain” the ability to transition between definitions of identity (object –> significance), and has been overlooked in the field for reasons unrelated to its significance in the brain (no one has gotten far trying to find consciousness in the brain, so maybe they should start looking under their noses!). I am super excited to know that this fundamental transition in the field of neuroscience is imminent and will be keeping a close eye on it all from afar.
Defining consciousness is not something I will dive into here, despite some implicit definitions. That’s an unproductive rabbit hole to be avoided when trying to identify candidate neural substrates for consciousness
Bonus: Google Trends for “thalamus” (I promise this won’t be my only data source) shows an incredible periodicity with annual maxima every October going back to 2007. Can you guess why? :)
# Brown Gold: How our poop is the most untapped gold mine of the 21st Century
Poop, and by extension, wastewater, is a data source that will be replenished each minute of every hour of every day until the end of civilization. The microbiome is severely underappreciated in medicine and health. I’ve always been puzzled by this… The evidence supporting the significance of the microbiome is almost insurmountable (Alzheimer’s, longevity, obesity, autism. The list goes on…), so how the hell could this be almost universally overlooked?
Most recently, researchers discovered that sewage water serves as a useful proxy for the extent of COVID-19 infections in the community. This made me wonder: why haven’t we done this before?
What we have is a fundamental misalignment in the current and actual values of this asset (poop). Call it arbitrage or just plain old opportunity—the reality is clear. Wastewater is (literally) the shit that no one wants (well, a few people), and so starting to collect mass samples of wastewater wouldn’t really perturb the existing system too much.
Despite all this, public interest in the microbiome seems to be stagnant:
###### Okay, these might not be the most convincing data, but it’s a start… Clearly the public appreciates the importance of health and bacterial diversity, but why don’t people attend to the microbiome as an extension of that?
This space is going to make a select few a lot of money. The factors for success are all there: a paucity of competition, an almost infinite abundance of data, simple means of data collection, and a demonstrated role in domains relevant to health and disease… These are fertile grounds for discovery and the intrepid pioneers that get in early will no doubt be rewarded.
Let not thy wastewater be wasted… You heard it here first.
I can name a couple good companies that recognize this systemic short-sightedness: Pendulum Therapeutics and Gingko Bioworks. For the interested reader, I also recommend looking into Richard Sprague, who has a pretty extensive passion project on his own microbiome
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2022-12-01 14:40:59
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https://physics.tutorvista.com/fluid-dynamics/absorptivity.html
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Top
# Absorptivity
If the absorptivity varies over the frequency spread of the beam, the more strongly absorbed frequencies will be depleted, and the absorptivity will therefore decrease with sample path length. Chemical and physical interactions in the sample may modify the nature of the absorbing species, causing the absorptivity to vary with concentration. Thus it is necessary to check each sample to determine whether Beer's law is obeyed before attempting a quantitative analysis.
## Definition
The absorptivity is defined as the fraction of the energy incident on a body that is absorbed by the body. The incident radiation depends on the radiative conditions at the source of the incident energy. The spectral distribution of the incident radiation is independent of the temperature or physical nature of the absorbing surface unless radiation emitted from the surface is partially reflected back to the surface. Compared with emissivity, the absorptivity has additional complexities because directional and spectral characteristics of the incident radiation must be included. It is desirable to have relations between emissivity and absorptivity so that measured values of one will allow the other to be calculated.
## Formula
Absorbance A is directly proportional to the path length l and the concentration of the absorbing species c. That is
A ∝ cl
A = acl
where a is a proportionality constant called the absorptivity. When the concentration is expressed in moles/litre and the path length in centimeters, the absorptivity is called the molar absorptivity or molar extinction coefficient ε.
Thus, A = εcl
or
$\log$ $\frac{I_{0}}{I}$ = $\varepsilon\ cl$
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2019-05-21 13:15:58
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https://www.transtutors.com/questions/jade-larson-antiques-owes-20-000-on-a-truck-purchased-for-use-in-the-business-the-co-1259992.htm
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# Jade Larson Antiques owes $20,000 on a truck purchased for use in the business. The company makes Jade Larson Antiques owes$20,000 on a truck purchased for use in the business. The company makes principal payments of $5,000 each year plus interest at 8%. Which of the following is true? 1. After the first payment is made, the company owes$15,000 plus three year’s interest.
2. After the first payment, $15,000 would be shown as a long-term liability. 3. After the first payment is made,$5,000 would be shown as the current portion due on the long-term note.
4. Just before the last payment is made, \$5,000 will appear as a long-term liability on the balance sheet.
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2018-11-19 22:02:57
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https://meanderful.blogspot.com/2018/06/
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Friday, 29 June 2018
CME patent identifies speculative systems used by speculators
Paul Westmont tweeted a reference to a new CME patent, "Message processing protocol which mitigates optimistic messaging behavior" published on 28th June 2018.
Hot off the press:
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A clever person in Amsterdam examining a gem(source)
This patent identifies a few potential techniques, aka tricks, for optimistically processing on a network to improve your effective latency.
I referred to some of these in the recent "Negative Latency" piece. Some such speculative skills were also discussed a few years ago in my piece originally published by Automated Trader, and subsequently published here once the rights reverted, as, "The accidental HFT firm."
I've pulled out the fun list of the techniques CME explicitly identifies below.
Speculating on your speculation is stepping up a level.
--Matt.
____________________
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Wednesday, 27 June 2018
iex_update.md IEX update
It’s been at least fifteen minutes since I complained about IEX. It must be time for another meander.
IEX starts pricing off the SIP
Mr Brad Katsuyama’s misleading of investors would be amusing if it wasn’t so dangerous. His false and misleading statements have undermined confidence in the US National Market System over the last few years.
Do you remember the classic Lewis, Katsuyama, O’Brien CNBC debate where Mr Katsuyama accused BATS Mr William O’Brien of pricing off the SIP and disadvantaging his customers? Mr O’Brien denied it as it slipped his mind in the heat of the moment the acquired Direct Edge did that. O’Brien was no longer at BATS four months later. Despite Mr Katsuyama’s continued misrepresentations, he remains employed.
Well, IEX has decided to start using the SIP to contribute to its pricing.
File No
SR-IEX-2018-10
Date
2018-05-09
Description
Proposed rule change to amend rule 11.410(a) to update the market data source that the Exchange will use to determine the Top of Book quotation for NYSE National in anticipation of its planned re-launch.
… the Exchange proposes to amend and update the table specifying the primary and secondary sources for NYSE National (“XCIS”) in anticipation of the planned re-launch of XCIS on May 21, 2018.9 As proposed, the Exchange will use securities information processor (“SIP”) data, i.e., CQS SIP data for securities reported under the Consolidated Quotation Services and Consolidated Tape Association plans and UTDF SIP data for securities reported under the Nasdaq Unlisted Trading Privileges national market system plan, to determine XCIS Top of Book quotes. No secondary source is proposed to be specified as SIP data will be used exclusively.
Perhaps it’s not that important. In the context of Mr Katsuyama’s misdemeanours, perhaps it is.
I wish IEX would apply their own Rule 3.14.0 to staff, including Mr Katsuyama:
Rule 3.140. False Statements
No Member or applicant for membership, or person associated with a Member or applicant for membership, shall make any false statements or misrepresentations in any application, report or other communication to the Exchange. No Member or person associated with a Member shall make any false statement or misrepresentation to any Exchange committee, officer, the Board or any designated self-regulatory organization in connection with any matter within the jurisdiction of the Exchange.
IEX keeps markets simple by making its pricing more complex
IEX long ago promised to keep order types and pricing simple. This was to ensure your average punter would not be disadvantaged from not understanding the complexity. An admirable goal IEX has failed at. IEX continue to increase complexity making life worse for your average investor.
File No
SR-IEX-2018-09
Date
2018-04-20
Description
Proposed rule change to charge a more deterministic fee of $0.0003 per share for executions at or above$1.00 that result from removing liquidity with an order that is executable at the far side of the NBBO.
IEX added a Spread-Crossing Remove Fee at 3 mills, $0.0003 per share, to ameliorate part of the rape and pillage they introduced with the Crumbling Quote Remove Fee Indication of 30 mills. The Crumbling Quote Remove Fee Indicator adds great uncertainty to trading at IEX. You don’t know if you’re going to be hit with a high fee or not. IEX’s crummy crumbling quote voodoo is often wrong causing massive fee charges to appear on your statement for no real reason. It’s quite terrible you can’t determine the crummy crumbling fee a priori. You can’t even calculate with rigour ex ante. You get hit with a high fee for a trade. Is it correct? You have no way of knowing as the crummy crumbling quote is dependent on the feeds IEX gets internally within their own parameters of delay, jitter, and determinism, or lack thereof. You can’t deterministically parse chaotic information you don’t have. You can’t check your own fees. Does the SEC check the fees every month? I don’t think so. IEX had to refund customers for incorrectly charged fees earlier this year. No one knew. At least this new fee introduced is lower than the expensive dark fee you get pinged with if you stumble over a midpoint in the dark. I did like this line from the fee reasoning on page 14, Furthermore, the proposed Spread-Crossing Remove Fee is substantially lower than the fee for removing liquidity on competing exchanges with a “maker-taker” fee structure (i.e., that provide a rebate to liquidity adders and charge liquidity removers) One sided misrepresentation from IEX again. The nett fee IEX takes is much higher than other exchanges. IEX is the expensive dark exchange. Clarity in pricing The rules were so clear, IEX had to issue on the 20th June a further update to add clarity to the simple rules they have, effective 1st July. Yes, Minister. File No SR-IEX-2018-11 Date 2018-06-20 Description Proposed Rule Change to Modify the Structure of its Fee Schedule and Make Several Conforming and Clarifying Changes, pursuant to IEX Rule 15.110(A) and (C). Let’s check out the new simple and clear fee structure: The fee schedule is quite misleading as for some of your stocks the quote instability charge that you pay may be more likely when the quote is stable rather than when it is unstable, according to their own formulae. If you put it in capitals and pervert the meaning of Quote Instability by subverting its normal definition, thereby turning it into your own proper noun, then you have a a nice marketing message. Science be damned. IEX aren’t the only offender in the complex pricing stakes, but their impossible to know a priori, or audit ex ante, CQRFI fee takes it to a new level in the market of the absurd. IEX crumbling quote indicator changes IEX changed the crumbling quote formulae a while ago as they do from time to time. It is still a bad idea and a bad formula. I think many undergraduates could come up with a better toxin in a few days if they were appropriately directed with the same data. It is what it is: File No SR-IEX-2018-07 Date 2018-04-03 Description Proposed rule change to amend Rule 11.190(g) to incrementally optimize and enhance the effectiveness of the quote instability calculation in determining whether a crumbling quote exists. One funny thing about this update is the sheepish admission by IEX they have been indicating quote instability on one side it is infact the other side of the NBBO that is crumbling. And yes, you will have paid an excessive 30 mill fee for trading on the stable side of the market. You couldn’t make this stuff up if it wasn’t true. Digest this bit first from page 8, When the System determines that a quote, either the Protected NBB or the Protected NBO, is unstable, the determination remains in effect at that price level for two (2) milliseconds. The System will only treat one side of the Protected NBBO as unstable in a particular security at any given time Do you see the problem? The market could step up and then switch back around pretty easily, even in just 100 microseconds. You could end up with a variety of stocks locked into Quote Instability mode on opposite sides of the market for two milliseconds. Crazy stuff, right? IEX was woken a little from their slumber and decided that if their often false crumbling quote switches around then they will too rather than keeping it locked at the wrong side. The indicator is still often false, but at least that is something. IEX describes the changes starting on page eleven, Rule 11.190(g)(1) provides in part that when the System determines that a quote, either the Protected NBB or the Protected NBO is unstable, the determination remains in effect at that price level for two (2) milliseconds. The Exchange proposes to revise the time limitation on how long each determination remains in effect, and reorganize certain existing rule text for clarity. As proposed, when the System determines that either the Protected NBB or the Protected NBO in a particular security is unstable, the determination remains in effect at that price level for two (2) milliseconds, unless a new determination is made before the end of the two (2) millisecond period. Only one determination may be in effect at any given time for a particular security. A new determination may be made after at least 200 microseconds has elapsed since a preceding determination, or a price change on either side of the Protected NBBO occurs, whichever is first. If a new determination is made, the original determination is no longer in effect. A new determination can be at either the Protected NBB or the Protected NBO and at the same or different price level as the original determination. Based upon our analysis of market data, as described above, the Exchange believes that changes to the time limitation would provide for a more dynamic methodology for quote instability determinations thereby incrementally increasing the accuracy of the formula in predicting a crumbling quote by expanding the scope of the model to additional situations where a crumbling quote exists at a different price point, or again at the same price point within two (2) milliseconds. For example, suppose that the NBBO is currently$10.03 by $10.04 in a particular security, and the System determines that the NBB is unstable. This determination goes into effect, with an expiration time set two (2) milliseconds in the future. Now suppose that one (1) millisecond later, the NBB falls to$10.02 and the System determines that this new NBB is unstable. As proposed once the System makes a new determination that the NBB of $10.02 is unstable, even though the prior determination at$10.03 has not expired, the new determination will overwrite the old determination, and its expiration time will be set to two (2) milliseconds in the future from the time of this determination.
The dark and expensive exchange that should be an ATS and not an exchange, is still deploying some wriggle room chicanery with their 200-microsecond from the proceeding determination thing, which they really shouldn’t need given they have a 350 microsecond delay in their stupid system.
And they have to change some of the variable definitions in their dumbly inaccurate, often false, crumbling quote indicator,
The Exchange proposes to revise five of the quote stability variables currently specified in subparagraph (1)(A)(i)(b) of Rule 11.190(g). Specifically, the Exchange proposes to revise variables NC, EPosPrev, ENegPrev and Delta to be calculated over a time window looking back from the time of calculation to one (1) millisecond ago or the most recent PBBO change on the near side (rather than on either side), whichever happened more recently. Based on our analysis of market data, as described above, the Exchange identified that for each variable, considering the maximum change over the time window defined in this manner is a more accurate indicator of a crumbling quote than the current approach. Similarly, the Exchange proposes to revise variable FC to be calculated over a time window looking back from the time of calculation to one (1) millisecond ago or the most recent PBBO change on the far side (rather than on either side), whichever happened more recently. Based on our analysis of market data, as described above, the Exchange identified that for this variable, considering the maximum change over the time window described in this manner is a more accurate indicator of a crumbling quote than the current approach.
You also have to keep in mind IEX is acting in the future due to their internal lack of 350 microsecond delay implying that one millisecond look-back is really 650 microseconds of look-back from their lack of perspective.
IEX should be forced to change the term from “quote instability” or “quote stability” as that is a clear misrepresentation of the facts and misleading, regardless of their intention.
Thanks again for signing up for our “Stability Operation.” Just because you’ll be losing two perfectly good legs and be confined to a wheelchair doesn’t matter. Many people may need a wheelchair in the future, so we’re enhancing the stability of everyone. No, we don’t think it is at all misleading. Being more stable on a flat surface that takes a wheelchair will be a boon to everyone. Yes, even gymnasts and rock climbers. It was right there in the EULA you clicked the OK button on. Stop complaining. The regulator approved it.
Double squeak from double speak indeed.
Anyhow, just to sound Presidential, here is the new failing, often fake, Quote Instability Factor and friends from the failing IEX’s new logistic regression formula:
Definitions
Protected Best Offer minus Protected Best Bid
Protected Quotations, Protected NBB, Protected NBO, Protected NBBO
includes quotations from not all of the exchanges, just these ones: XNYS, ARCX, XNGS, XBOS, BATS, BATY, EDGX, EDGA.
1. N = the number of Protected Quotations on the near side of the market, i.e. Protected NBB for buy orders and Protected NBO for sell orders.
2. F = the number of Protected Quotations on the far side of the market, i.e. Protected NBO for buy orders and Protected NBB for sell orders.
1. NC = the number of Protected Quotations on the near side of the market minus the maximum number of Protected Quotations on the near side at any point since one (1) millisecond ago or the most recent PBBO change on the near side, whichever happened more recently.
2. FC = the number of Protected Quotations on the far side of the market minus the minimum number of Protected Quotations on the far side at any point since one (1) millisecond ago or the most recent PBBO change on the far side, whichever happened more recently.
3. EPos = a Boolean indicator that equals 1 if the most recent quotation update was a quotation of a protected market joining the near side of the market at the same price.
4. ENeg = a Boolean indicator that equals 1 if the most recent quotation update was a quotation of a protected market moving away from the near side of market that was previously at the same price.
5. EPosPrev = a Boolean indicator that equals 1 if the second most recent quotation update was a quotation of a protected market joining the near side of the market at the same price AND the second most recent quotation update occurred since one (1) millisecond ago or the most recent PBBO change on the near side, whichever happened more recently.
6. ENegPrev = a Boolean indicator that equals 1 if the second most recent quotation update was a quotation of a protected market moving away from the near side of market that was previously at the same price AND the second most recent quotation update occurred since one (1) millisecond ago or the most recent PBBO change on the near side, whichever happened more recently.
7. Delta = the number of these three (3) venues that moved away from the near side of the market on the same side of the market and were at the same price at any point since one (1) millisecond ago or the most recent PBBO change on the near side, whichever happened more recently: XNGS, EDGX, BATS.
Now, it’s relatively easy to get a feel for what this is doing without thinking too hard. If a DPEG is a buy order and the market’s protected bid quote volume is large, then N is big. The QIF formula will use $QIF = \frac {1}{1+e^{big}}$ resulting in one over a big number, which will result in a near zero, or low, number. It will be likely be less than 0.19, IEX’s fake Quote Instability Factor threshold. That is, lots of volume on your side of the market means the market is stable and not in a position that indicates a crumbling quote. Quite natural.
Though please remember when referencing the formulae, now is 350 microseconds into the future, as per the use of non-speed bumped data in a speed bumped exchange.
From the IEX Signal 2.0 whitepaper, IEX told us,
“On our example day of December 15, 2016, our current formula resulted in about 1 million true positives and 975,000 false positives. This new candidate formula would have produced about 2 million true positives and 2.1 million false positives.”
That’s a stupid number of false positives. It is good if you can fleece charge your customers big fat fees when you’re wrong 😇 👀.
How false is this new calibration of the fake indicator? For which stocks? IEX does not tell us. We can be sure that such a synthetic silly sausage shop standard doesn’t fit all stocks the same. QIF will be more fake for some than others. One size does not fill all. The universe of stocks IEX hinders your trading with is broad. For example, liquid and illiquid stocks behave quite differently. Somehow I don’t think you’re suprised.
Conclusion
You may be surprised to hear I’m not a fan of a dark and expensive exchange. Especially one where you can’t work out your fees in advance or even after you’re charged.
IEX impedes price discovery and harms public markets.
Fortunately, despite the harm IEX has done to the NMS, it’s marginal impact is economically small due to its small market share. Unfortunately, the qualitative burden of IEX is large due to the noise and misleading statements from IEX the industry has had to try to counter. The additional execution infrastructure costs the industry has had to bear is wasteful at best. IEX is a frustrating little baby who thinks its soiled nappy is innovation.
All of this for an exchange that leaks information to the well connected trader via both the SIP and significant clients’ holdings. No other exchange comes close to the level of bad latency arbitrage exposed by IEX’s displayed market and dark executions.
IEX also gives a greater advantage to faster traders than other exchanges as it does not have a fair co-location facility. Shorter cables and better positions help.
The SEC should send back IEX to the land of the ATS, where it belongs. There’s no shame in that. Many a good ATS serves a useful purpose. Then, perhaps, we could all get along.
–Matt.
Monday, 25 June 2018
US Finra ATS tier 1 stats update: June 25 release for June 4
UBS and Credit Suisse remained ensconced at #1 and #2 respectively. Barclays dropped a couple of slots to 6th.
TWEB dropped 8 places to 21st with only 114 trades for the week. The 211,017 shares per trade kept it at around 1% of ATS tier 1. Normal volatility in the rankings for DealerWeb thanks to the small number of trades.
Instinet's second venue, Block-Cross was the biggest mover up in the rankings, popping up 3 slots to 18th. This gives Instinet a solid aggregate top ten viewpoint from their three venues. The menu of Instinet venues has an interesting variety of average trade sizes with 209 shares, 12,049 shares, and 3,925 shares.
Luminex jumped up a slot to 25th after being a topic of conversation last week in "Liquidity Dancing.
--Matt.
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Saturday, 23 June 2018
Liquidity dancing
Liquidity misconceptions in finance are some of the annoying itches I have wanted to scratch for a while. This is a meander that scratches that itch.
I won’t meander anything newsworthy for most market professionals here. If you’re a professional, I’d save your time. Nevertheless, this may be a reasonable reference to point to if you’re asked about liquidity.
The question of liquidity is very much a question of dance. The point of trading is to engage with another party so that two transactional needs are met. Utility is the goal. Utility is also the gaol5 constraining a liquidity hunter's options. Just as it takes one to provide a cliche for another one to read, it takes two to tango. If your trading friends, or marketplace, do not supply you with liquidity, you have to leave them behind, “Cause your friends don’t dance and if they don’t dance, Well they’re no friends of mine.
Safety Dance1
Marketplaces allow a more efficient risk transfer and price discovery process to take place. Without a marketplace, you are in danger of trading at an unfair price or being not able to trade at all. Despite all the warts of capitalism, markets provide the best way for parties to trade, despite the motivations of the participants.
Generally, none of the participants is trying to help any other participant. They trade for their own interests. The magic of the market is they work despite those motivations. The misunderstanding of this simple fact is the cause of much strangeness and inefficient enterprise in modern markets.
Asset manager versus HFT
Asset managers are wary of HFT firms, particularly private trading firms (PTFs). An asset manager views with suspicion the utility an HFT gains from trading with them. Very few losing days in the HFT bank account seems unnatural. An asset manager feels they are being ripped and feeding this continuous profit the HFT enjoys.
One way for an asset manager to avoid an HFT is to find a marketplace without an HFT. One minor trouble is such venues are difficult to find. Even at IEX the most prolific traders are HFT. If you’re an asset manager trading at IEX it is more likely than not that your counterparty is an HFT. HFTs are prepared to risk more and earn less at the best price which is why they win such trades. Buying low and selling high is still the only way to make money. An HFT is more prepared, in many ways, to roll the trading die. The law of large numbers gives them a good chance of making positive cash-flow every day even if their success rate on your individual trade with them is only 51-55%. The asset manager doesn’t realise there is a fairly high chance the HFT may lose on their individual trade.
But, but, I want to earn that spread…
OK, go build your own marketplace then and see how well that works for you.
Luminex
A bunch of asset managers get together and decide that they’d be better off without proprietary trading, especially those “nasty” HFT firms.
From Reuters 5th October 2015 story, “Luminex ‘dark pool’ enlists 73 members ahead of trading launch,”
Luminex plans to only allow institutional investors, such as mutual fund companies, to trade on its platform, snubbing the broker dealers and proprietary high-frequency trading firms that own and generate much of the trading in other private trading venues.
Luminex’s 73 subscribers, which are listed on the company’s website, include Vanguard, TIAA-CREF Investment Management, AllianceBernstein, Eaton Vance, Goldman Sachs Asset Management, Greenlight Capital Inc, and J.P. Morgan Investment Management. (luminextrading.com/)
The consortium that owns Luminex and collectively manages 40 percent of U.S. fund assets said it started the trading platform with the aim of lowering transaction costs and eliminating the types of profit driven conflicts of interest that have been seen in some existing venues. Any profits made by Luminex will be invested back into the company to further reduce trading costs.*
The Reuters article implies some motivation may have stemmed from illegal activity such as that from ITG,
Recent enforcement actions involving the trading platforms include a $20.3 million SEC settlement with brokerage firm Investment Technology Group on charges it ran a secret trading desk that profited off of confidential customer information within its dark pool. Not an unreasonable motivation. Luminex has recently touted their May 2018 results: “Luminex May 2018 Recap” Wow, that sounds impressive. The$10 Billion traded in a month sounds a big number, doesn’t it? An efficient exchange might make $0.0001 per trade. That is$20,000 per month at an average stock price of $50 for such a Luminex notional size. Not bad but it is not easy to run the technology of an ATS, cope with the regulatory requirements, and pay your staff on such a budget. The National Market System (NMS) exchanges traded 6,875 billion dollars of notional value for the same period. Luminex traded the equivalent of just 0.15% of the exchanges which is only a partial picture of the liquidity landscape. Let’s have a look at their tier one stock performance compared to the other thirty odd somewhat odd ATS peers: That doesn’t look so great. The percentage market share of the ATS space is just a paltry 0.45% with a ranking, by volume, of 26th. It’s hard to tell if that record-breaking memo is something to shout about or a job preservation exercise. Here is a chart of the changing market share landscape in ATS land: That’s right. Luminex is somewhere down there in the noise floor in the bottom of that chart. What has gone wrong at Luminex? Nothing really. This is my point. They wanted to create a mono-culture client trading pool and succeeded. It is an understandable reaction to the fear that mongers peoples’ minds against HFT and proprietary trading. However, such an enterprise is going to find success a difficulty. Success often requires heterogeneity. Dare I say it? It takes two to tango: Pearl Baily - Takes two to tango Luminex looks to have done a good job with admirable enterprise. However, it represents a fundamental misunderstanding of marketplaces and where liquidity comes from. Asset manager fears There are several notions of wrong. Firstly, anti-HFT is a misunderstanding of who your competitors are. Your competitors are other asset managers. An asset manager should be worrying more about selling a stock to another asset manager. They both think their opposing actions are a sensible medium or long term solution. Both can’t be right. If you sell to an HFT, both can be right and make money as there is a time arbitrage there. Not only does such a time difference support an asset manager’s and HFT’s difference in utility, there is a more fundamental aspect to the utility of every trade. Let’s meander through this thought. Take your passive buy at best that gets traded through. If you’re an asset manager you’ll grumble about being traded through. You’ll imagine it was one of those nasty HFT firms and be pissed off that you could have bought the stock a tick cheaper. However, you’ll be happy as you wanted the trade and shrug your shoulders as, at an average price of$50 today, a tick of one cent is just 0.02%. If you’re in the asset manager game to only make 0.02% on your medium or long term trades please send the investors their money back; you’re in the wrong game. Annoying but not crippling.
If you’re an HFT and you get traded through, that is an existential crisis. Your margins as so thin that if such adverse selection happens too often you’re simply out of business. You have to take the risk and take such trades on the chin to play at all, but this is a careful calibration. The law of large numbers that guarantees your daily profit also guarantees the immediacy of your death if you get such calibrations slightly wrong. It’s a tough game for an HFT.
Such a “traded through” case is quite different for both types of market participants. It is an existential threat to an HFT, and an annoying mosquito bite to an asset manager.
The other case of the trade not happening at all is also quite different for both. For an HFT if their bid does not get lifted, it’s simply a shrug and they move on. For an asset manager it is a huge pain as they want the stock and now they are chasing it higher and not getting their portfolio at the right price. An HFT can feel a bit of that pain if they are long inventory, but it is quite different to the pain the asset manager endures.
Again, we see different utility for the outcomes. Your trade not filled is a shrug for an HFT and quite the pain for the asset manager.
This is what makes a market. Diverse participants with differing views of utility, risk, and trading. More sophisticated traders, such as an HFT, may also offset in different markets or products bringing further diversity to the utility profile beyond the time horizon difference normally considered. More speculation in a market hurts speculators and helps investors.
A clever asset manager would not create a marketplace for just the their own competitors. They would encourage a marketplace with much diversity and competition. This would provide the most efficient price whilst those nasty speculators crush each other with their games.
Imagine you’re an HFT sitting in the market passively. What do you fear most? A big institutional order that trades through your price. That big information stick an institutional investor carries is something to fear.
What do asset managers fear? Many often fear that those pesky HFT firms with their mosquito bites and going to destroy their long term alpha with the spread friction they impede their trading with.
The real truth of the matter is that they shouldn’t fear each other they should fear their own kind. The kind of firm that will put an HFT out of business is another HFT. The kind of asset manager that will destroy another asset manager’s business is an asset manager with better returns, not an HFT firm.
In this context, you may see the fundamental philosophical tension of the Luminex model. I’m only going to allow trading with folks I’m directly competing against, with the same painful utility points on trades, and without a diverse culture of transference between other risk dimensions such as time, product, correlation, covariance, informational understanding, etcetera.
In game theoretic and evolutionary terms, a homogeneous gene pool rarely outperforms the heterogeneous. Perhaps the Luminex experiment shows this. Then again, perhaps having a worse trading pool is OK if you’re psychologically safer. I’m not sure your shareholders would have the same feeling when considering your psychological well-being versus their returns.
Smarter asset managers
If the size of assets under management counts as success, perhaps Vanguard is smarter than your average bear. Here are some historical comments from Vanguard on the issue of competing with HFT reported by Stephen Foley in the FT, “Vanguard chief defends high-frequency trading firms”
This wasn’t just a reaction for the Lewis fiction in Flash Boys as you see in this similar comment from September 2010 in Forbes, “Vanguard’s Gus Sauter Thanks High-Frequency Traders”
Sauter said that there might be a small percentage of high-frequency trading that is manipulative, but that overall this is a type of trading that benefits long-term investors like Vanguard’s customers. “Most high-frequency traders in our view are essentially electronic market makers or they’re statistical arbitrage players. In both cases, they’re working to tighten spreads and enhance liquidity. If that’s the case, then we as investors in the marketplace benefit by getting lower transaction costs when we go into the marketplace to invest for the long-term.”
How much of a benefit does Sauter attribute to market developments in the past 15 years, which include decimalization, high-frequency trading, increased fragmentation, Reg. NMS and order handling rules? “We’ve measured our transaction costs and over the past 15 years, they’ve been cut by about 60 percent,” Sauter said. “That results in hundreds of millions of dollars a year in savings to investors in our funds.”
The somewhat biased Modern Market Initiative2, a group that exists to support their private HFT trading members, makes the same points here with commentators not having such bias, “HFT = Cheaper Trading = More Money in Retirement Accounts”:
“Main Street is the great beneficiary … We are better off with high-frequency trading than we are without it.”
– Jack Bogle, Vanguard founder, pioneer of low-cost investing
“Trading has never been easier and costs never lower thanks to human intermediaries being rendered obsolete.”
– Robin Wigglesworth, Financial Times
“In the decade of migration to electronic trading and HFT arrival, transaction cost decreased by over 50% for both retail and institutional investors.”
– Albert J. Menkveld, Professor of Finance, VU University Amsterdam
“By providing so many bids and offers, high frequency trading firms have narrowed pricing spreads. Spreads for almost every financial instrument are substantially less than they were a decade ago. For example, spreads in most US equities are half of what they were 10 years ago.”
– John Servidio and Bo Harvey, McGuireWoods LLP
“From an institutional or buy-side perspective, today’s markets are more efficient than anytime in the past. Today’s markets are faster, they’re cheaper to trade, the typical buy-side desk has more choice due to competition in terms of execution venues, whether they’re block trading systems or dark pools, registered ATSs, and algorithms that we can customize that allow us to navigate electronic markets and obtain the best possible price for our fund shareholders.”
– Bill Baxter, Fidelity’s Head of Global Program Trading and Market Structure
“The world should be measured the following ways: The spread between bid and ask, and the commissions charged to do a transaction, are so dramatically smaller today — they’re measured in thousandths of a penny sometimes compared to when I was in the securities business.”
– Michael Bloomberg, founder and CEO, Bloomberg, LP
“High-frequency trading in general has been good for the retail investor.”
– Fred Tomczyk, CEO, TD Ameritrade
“Overall, HFT enhances market liquidity, reduces trading costs, and makes stock prices more efficient.”
– Charles Jones, Columbia University professor, study analyzing 30 papers on HFT
“[Electronic market making] brings tangible benefits to our clients through tighter spreads”
– BlackRock Viewpoint
“How do we feel about high-frequency trading? We think it helps us.”
– Cliff Asness, Founding Principal, AQR Capital Management
“… the increased use of electronic trading has brought many benefits, such as more efficient execution and lower spreads.”
“Numerous studies – including the recently released UK Foresight HFT project – have shown that transaction costs for both retail and institutional traders decreased substantially with the growth of high-frequency trading.”
– Larry Harris, USC Marshall School of Business, former chief economist at SEC 2002-04
Nevertheless, the fear of HFT persists.
My thesis is that such fears are primarily a misunderstanding of marketplaces and liquidity dancing.
Some liquidity dance misconceptions
Before I meander about what I feel liquidity is, let’s look at some of the itches I wish to scratch.
Providing liquidity versus taking liquidity
This drives me nuts. A trade takes place if there is a willing buyer and seller. Both wish to trade at the price implying an agreement, of sorts, in utility. Both sides provide the liquidity the other side needs. There is no taking of liquidity here. Both sides are providing the other with a service. That service is liquidity at a particular price. The buyer is providing the seller liquidity. The seller is providing the buyer liquidity.
Aggressive and passive are fair enough adjectives but the all too common vernacular of providing or taking liquidity paints an improper picture. Taking liquidity has come to have an unwarranted negative connotation. The aggressor is providing the liquidity sought at the price requested. This is a good thing. Service is provided to both parties. Taking liquidity implies removing the ability to trade at all. This is quite the opposite of enabling a trade. Next time you see or hear the all too frequent term “taking liquidity,” rebel.
Taking liquidity is perhaps better described by someone cancelling an order. The immediate ability to trade with that stock has been lost.
However, it is not really lost. Perhaps just delayed. It still exists which highlights my next point.
Most liquidity lives in peoples’ heads
Exchanges may transact much of activity, but the displayed liquidity is not where most of the liquidity lives. The prior example shows this somewhat as the aggressor is not showing liquidity before they provide it. It may have been either a hidden intention in an algorithm or in a trader’s head.
Liquidity lives in exchanges, visible as lit prices as well as being hidden in the dark folds of the exchange. Liquidity lives in ATSs and other pools. Liquidity lives in brokers internalisation pools and procedures. Liquidity lives in the intention of programmed algorithms’ and models’ automated intentions. Liquidity lives in the people who control stock. It lives in their heads. Most liquidity is in this form.
Earlier this year Barron’s reported, “The U.S. Stock Market Is Now Worth $30 Trillion” with a reference to the Russell 3000 Index, which represents approximately 98.5% of the market’s capitalisation. As noted above, the NMS exchanges traded around$6.875 trillion in May. Even over a month, most stock doesn’t trade. Someone controls whether a parcel of publicly listed stock trades. Again, the liquidity is effectively in that someone’s head. A simple thing we should remind ourselves of from time to time.
HFT pushes people out of the market risking the market itself
If that is what you think, then you are not playing the right game. If you want to trade to earn the bid-ask spread then you need to be an HFT market maker, not a trader in this technological age. The real risks to liquidity in the market are different to what you may think.
A primary liquidity risk is the growth of alternates to publicly listed stocks. To my simple soul, the massive growth in private equity is the number one existential threat to the viability of US public stock markets. Much of the returns accreted by new ventures used to come from a time after they listed. Private equity giants, such as Uber et al, have turned that thinking on its head. Much of the venture returns now exist outside public markets. How significant the disappearance of such a source of alpha from the public markets really is, remains to be seen. I suspect this may get worse before it gets better. It may be one of those one-way doors. Now that the genie is out of the bottle, it may be hard to plug the damned dam wall. Unless you’re one of those qualified investors with enough qualification to be interesting, you may be missing out on funding your retirement. Economic democratisation in the US is in reverse. It’s not just politics looking increasingly totalitarian.
If you haven’t seen the 2015 presentation from A16Z, “U.S. Technology Funding – What’s Going On?”,
then I highly recommend it. It has dated a little as some of the giants have returned decent public returns since and there has been an uptick in IPOs, but the basic case still stands. Large private equity is eating returns that used to be available to main street.
Private company growth needs government regulation to prevent government regulation from contributing to the death of investment returns and the happy retirement of perhaps ninety-nine per cent of the population. You really do want to occupy Wall Street.
Passive equity management also largely impedes market liquidity. There is simply less desire to actively trade. That is not only due to index funds but also, perhaps to a lesser extent, due to the ETF industry. Hugging a market index involves less industry from the industry. There has been somewhat of a stall in market volumes over the last few years that perhaps this mix of passive and ETF explains. Some of this is due to a lack of trading intention even though such passivity represents, somewhat paradoxically, an increasing share of actual trading. See the FT article by Robin Wigglesworth from January 2017, “ETFs are eating the US stock market”:
Similarly, dark liquidity impedes price efficiency and the risk transfer process by disguising the liquidity picture. This is not so bad if you wish not to impact the price, but not impacting price is also a form of manipulation. You’re avoiding the impact your trade may otherwise should have. Price discovery is impeded by the horror of the growing darkness. Even the public exchanges are twisting into the downward spiralling vortex of darkness with IEX being the chief horror story abusing efficiency and price discovery.
The irony of an iceberg, or a lie about the true size of your order, should be reconsidered, just as pretending your order is big when small is considered an illegal spoof3.
If everyone bought and sold the index, no prices would change. If all liquidity was dark there would be nothing to set the prices for the parasitic darkness. Efficiency be damned. There is a certain amount of good from index funds, ETFs, and darkness. Yet, too much of a good thing may kill you.
Financial transaction tax will help normal investors
Be careful what you wish for. Liquidity is not magic but sometimes a hard-won feature. It took CME nearly thirty years to build their FX future liquidity. Market liquidity and efficiency are not always easy to come by. An FTT harms market efficiency, but it is a matter of size.
People often say there needs to be a financial transaction tax (FTT) to slow the market down and make it more purposeful. Adding friction obviously harms efficiency. It should be obvious an FTT doesn’t help as damaging efficiency is probably not a good idea. Size matters. If it is large enough it will destroy a market. Sweden is the poster child for such damage. It is also little understood there is already an FTT in the US. This is the small tax in place that essentially funds the SEC from public company transactions. It is wrong that private companies evade this tax. The government should level this playing field.
That said, FTTs with too much heat may evaporate much liquidity. Pause and consider carefully any FTT proposal. The consequences are probably not what you intend.
Measuring Efficiency
Measuring efficiency is tough. Spread is often a good proxy but it also falls down for specific aspects. Efficiency is also about the trades that are wanted that do not trade. Spread doesn’t capture that. Efficiency is about the reliability of the liquidity at best over time. Efficiency is about both absorbing large trades with little impact and recognising the impact of changing information that may cause the desire to trade.
A smarter fund manager will likely trade passively when they can and just cross the spread when they feel they have too. Some efficiency measures will see such trading away from the midpoint as bad. Under such a regime, this is reported as a bad outcome for the fund manager who has just benefited. IEX is particularly guilty of such misdemeanours, as is the tardy Dr Hu’s4 SEC report. Measuring correctly is sometimes hard. Look out for measurements that may deceive. Looking at you, Dr Hu.
The liquidity dance
To me, liquidity represents the ability to execute a trade. More liquidity means a bigger trade can be executed with less price impact. It is not easy to draw comparisons between markets by simply looking at the displayed liquidity though. Often much of the liquidity exists off-market, is hidden, or its reliability exists over time.
For example, there are some interesting commodity markets where there is always a few contracts on the bid and offer with a reasonably tight spread. That is, regardless of activity, there always is a small displayed market over time. Integrated over time, the market is quite liquid even though it is instantaneously quite illiquid. All is not what it seems.
The message this kind of market delivers is important. It is not just about what is displayed but it is also the behaviour of the market.
This is one reason why faster exchanges typically have better liquidity and serve the market good with greater efficiency.
Take for example IEX, with a 350-microsecond delay, or a 700-microsecond round trip added to the processing time. Contrast this with an exchange that may serve the public in 100 microseconds. Now, this fast speed doesn’t seem to help the average punter, but it does. A fast trader can accept your order, hedge, and be ready to do the same again several times before an IEX customer even would know about the trade. This doesn’t help the individual trades directly but it helps all individual trades collectively. The liquidity over time can be much greater with less risk at faster exchanges. IEX interferes with this process and harms the market. At IEX the lit liquidity at best is usually non-existent or small but this understates the poor quality of the market as the slow speed of IEX also means that the liquidity integrated over time is even worse than its peers due to its slow speed. IEX sux.
A faster speed for arbitrage, hedging, and information transmission greatly improves a market by enabling more liquidity. The market is better. It is not about the individual trade but it affects all individual trades as a collective.
Think of it as a dance. An investor takes the price on offer. The dance begins. A market maker (MM) is likely on the other end. They may be at their inventory bounds and may pull orders from other locales. They may instantly hedge via an arb at the exchange or at another locale. This arb may be in another product. The MM may even immediately repost anticipating a likely hedge. The dance between the MM and the investors is quite something if you think about it.
Both the MM and investor can win. Even more importantly, two medium or longer term speculators may also win. It is the nature of markets that they are not zero-sum games as many people may think. I’m yet to see a market completely disconnected from the real world, or at least, disconnected from other markets.
As part of this liquidity dance, the lead may be taken by an asset manager wishing to hedge part of her $100 billion portfolio. Say she hedges 10 per cent in the market by selling$10 billion in futures. Now other traders are holding $10 billion of delta risk. If the market goes up 20 per cent the asset manager is happy as she has gained$20 billion on her portfolio and lost only $2 billion on their hedge.$18 billion in gains would make me happy. The mass of speculators, in whatever markets the delta has diffused to, are also happy as they’ve gained that \$2 billion. Markets have a purpose.
I refer to this aspect as happy losing. This is an underrated market aspect I wrote about in 2013, “Hedging – losses as profits that feed traders”.
You dance with the MM, they shuffle, twist, and continue to provide a suitable dance partner with the magic of connected financial services. It is quite a delightful and intricate dance if you think about the connectedness and impact on all of us over time.
Regulators have a responsibility to make it safe to dance. The government has a responsibility to enable all of us to dance if we want to.
There are some uncomfortable truths in the current market structure. Index funds and ETFs erode price efficiency as they grow. Likewise parasitic dark liquidity, especially at public exchanges like IEX, damage the utility of the public markets. More speed is good even if you hate it and over anthropomorphise faster than eye-blink speeds. There is no known answer to sudden bouts of market volatility. The growth of private markets is harming public markets, the public, and the SEC’s funding. Markets can still be irrational and subject to animal spirits.
Dance safely.
–Matt.
And one more time, it is catchy 😃.
1. The Safety Dance
Men Without Hats
safety!
Safety-dance!
Ah we can dance if we want to, we can leave your friends behind
Cause your friends don’t dance and if they don’t dance
Well they’re no friends of mine
I say, we can go where we want to, a place where they will never find
And we can act like we come from out of this world
Leave the real one far behind,
And we can dance
We can dance if we want to, we can leave your friends behind
Cause your friends don’t dance and if they don’t dance
Well they’re no friends of mine
I say, we can go where we want to a place where they will never find
And we can act like we come from out of this world
Leave the real one far behind
And we can dance.
Dances!
Ah we can go when we want to the night is young and so am i
And we can dress real neat from our hats to our feet
And surprise ‘em with the victory cry
I say we can act if want to if we don’t nobody will
And you can act real rude and totally removed
And i can act like an imbecile
I say we can dance, we can dance everything out control
We can dance, we can dance we’re doing it wall to wall
We can dance, we can dance everybody look at your hands
We can dance, we can dance everybody takin’ the chance
Safety dance
Oh well the safety dance
Ah yes the safety dance
Safety
Safety-dance
We can dance if we want to, we’ve got all your life and mine
As long as we abuse it, never gonna lose it
Everything’ll work out right
I say, we can dance if we want to we can leave your friends behind
Cause your friends don’t dance and if they don’t dance
Well they’re no friends of mine
I say we can dance, we can dance everything out of control
We can dance, we can dance we’re doing it wall to wall
We can dance, we can dance everybody look at your hands
We can dance, we can dance everybody’s takin’ the chance
Oh well the safety dance
Ah yes the safety dance
Oh well the safety dance
Oh well the safety dance
Oh yes the safety dance
Oh the safety dance yeah
Oh it’s the safety dance
It’s the safety dance
Well it’s the safety dance
Oh it’s the safety dance
Oh it’s the safety dance
Oh it’s the safety dance
Oh it’s the safety dance
Songwriters: Ivan Doroschuk
The Safety Dance lyrics © Universal Music Publishing Group
2. A prejudice that this author also shares.
3. For further information, or prejudice, on spoofing here is a link to my previous article “To spoof, or not to spoof, that is the question”
4. Dr Hu get’s it wrong as discussed in this meander IPCC wrong on climate change & SEC wrong on IEX
5. This is the the Australian/British "gaol" which is jail in Amercian and often now jail in Australian too though the original "gaol" is still used. I guess I could have used prison but I liked the literary twist on the preceding use of "goal". Sorry for any confusion ;-)
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2019-09-21 04:44:13
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https://bitbucket.org/pypy/extradoc/src/2c2a6d3a67ad/blog/draft/pypy-alpha-arm.rst
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# PyPy 2.0 alpha for ARM
Hello.
We're pleased to announce an alpha release of PyPy 2.0 for ARM. This is mostly a technology preview, as we know the JIT is not yet stable enough for the full release. However please try your stuff on ARM and report back.
This is the first release that supports a range of ARM devices - anything with ARMv6 (like the Raspberry Pi) or ARMv7 (like Beagleboard, Chromebook, Cubieboard, etc.) that supports VFPv3 should work. We provide builds with support for both ARM EABI variants, hard-float and for some older operating systems soft-float.
This release comes with a list of limitations, consider it alpha quality, not suitable for production:
• stackless support is missing.
• assembler produced is not always correct, but we successfully managed to run large parts of our extensive benchmark suite, so most stuff should work.
Part of the work was sponsored by the Raspberry Pi foundation.
## What is PyPy?
PyPy is a very compliant Python interpreter, almost a drop-in replacement for CPython 2.7.3. It's fast due to its integrated tracing JIT compiler.
This release supports ARM machines running Linux 32bit. Both hard-float armhf and soft-float armel builds are provided. armhf builds are created using the Raspberry Pi custom cross-compilation toolchain_ based on gcc-arm-linux-gnueabihf and should work on ARMv6 and ARMv7 devices running at least debian or ubuntu. armel builds are built using gcc-arm-linux-gnuebi toolchain provided by ubuntu and currently target ARMv7. If there is interest in other builds, such as gnueabi for ARMv6 or without requiring a VFP let us know in the comments or in IRC.
## Benchmarks
Everybody loves benchmarks. Here is a table of our benchmark suite (for ARM we don't provide it yet on http://speed.pypy.org, unfortunately).
This is a comparison of Cortex A9 processor with 4M cache and Xeon W3580 with 8M of L3 cache. The set of benchmarks is a subset of what we run for http://speed.pypy.org that finishes in reasonable time. The ARM machine was provided by Calxeda. Columns are respectively:
• benchmark name
• PyPy speedup over CPython on ARM
• PyPy speedup over CPython on x86
• speedup on Xeon vs Cortex A9, as measured on CPython
• speedup on Xeon vs Cortex A9, as measured on PyPy
• relative speedup (how much bigger the x86 speedup is over ARM speedup)
Benchmark PyPy vs CPython (arm) PyPy vs CPython (x86) x86 vs arm (pypy) x86 vs arm (cpython) relative speedup ai 3.61 3.16 7.70 8.82 0.87 bm_mako 3.41 2.11 8.56 13.82 0.62 chaos 21.82 17.80 6.93 8.50 0.82 crypto_pyaes 22.53 19.48 6.53 7.56 0.86 django 13.43 11.16 7.90 9.51 0.83 eparse 1.43 1.17 6.61 8.12 0.81 fannkuch 6.22 5.36 6.18 7.16 0.86 float 5.22 6.00 9.68 8.43 1.15 go 4.72 3.34 5.91 8.37 0.71 hexiom2 8.70 7.00 7.69 9.56 0.80 html5lib 2.35 2.13 6.59 7.26 0.91 json_bench 1.12 0.93 7.19 8.68 0.83 meteor-contest 2.13 1.68 5.95 7.54 0.79 nbody_modified 8.19 7.78 6.08 6.40 0.95 pidigits 1.27 0.95 14.67 19.66 0.75 pyflate-fast 3.30 3.57 10.64 9.84 1.08 raytrace-simple 46.41 29.00 5.14 8.23 0.62 richards 31.48 28.51 6.95 7.68 0.91 slowspitfire 1.28 1.14 5.91 6.61 0.89 spambayes 1.93 1.27 4.15 6.30 0.66 sphinx 1.01 1.05 7.76 7.45 1.04 spitfire 1.55 1.58 5.62 5.49 1.02 spitfire_cstringio 9.61 5.74 5.43 9.09 0.60 sympy_expand 1.42 0.97 3.86 5.66 0.68 sympy_integrate 1.60 0.95 4.24 7.12 0.60 sympy_str 0.72 0.48 3.68 5.56 0.66 sympy_sum 1.99 1.19 3.83 6.38 0.60 telco 14.28 9.36 3.94 6.02 0.66 twisted_iteration 11.60 7.33 6.04 9.55 0.63 twisted_names 3.68 2.83 5.01 6.50 0.77 twisted_pb 4.94 3.02 5.10 8.34 0.61
It seems that Cortex A9, while significantly slower than Xeon, has higher slowdowns with a large interpreter (CPython) than a JIT compiler (PyPy). This comes as a surprise to me, especially that our ARM assembler is not nearly as polished as our x86 assembler. As for the causes, various people mentioned branch predictor, but I would not like to speculate without actually knowing.
## How to use PyPy?
We suggest using PyPy from a virtualenv. Once you have a virtualenv installed, you can follow instructions from pypy documentation on how to proceed. This document also covers other installation schemes.
We would not recommend using PyPy on ARM just quite yet, however the day of a stable PyPy ARM release is not far off.
Cheers, fijal, bivab, arigo and the whole PyPy team
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2015-05-05 06:03:16
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