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(The following is not really an answer, or just a very partial one, but it's definitely relevant and too long for a comment.)
There is a theorem of Richard Friedberg ("The uniqueness of finite division for recursive equivalence types",
Math. Z. 75 (1961), 3–7) which goes as follows (all of this is in classical logic):
For $A$ and $B$ subsets of $\mathbb{N}$, define $A \sim B$ when there exists a partial computable function $f:\mathbb{N}\rightharpoonup\mathbb{N}$ that is one-to-one on its domain and defined at least on all of $A$ such that $f(A) = B$. (One also says that $A$ and $B$ are
computably equivalent or recursively equivalent, and it is indeed an equivalence relation, not to be confused with "computably/recursively isomorphic", see here.) Then [Friedberg's theorem states]: if $n$ is a positive integer then $(n\times A) \sim (n \times B)$ implies $A\sim B$ (here, $n\times A$ is the set of natural numbers coding pairs $(i,k)$ where $0\leq i<n$ and $k\in A$ for some standard coding of pairs of natural numbers by natural numbers).
To make this assertion closer to the question asked here, subsets of $\mathbb{N}$ can be considered as objects, indeed subobjects of $\mathcal{N}$, in the effective topos (an elementary topos with n.n.o. $\mathcal{N}$ such that all functions $\mathcal{N}\to\mathcal{N}$ are computable), in fact, these subobjects are exactly those classified by maps $\mathcal{N} \to \Omega_{\neg\neg}$ where $\Omega_{\neg\neg} = \nabla 2$ is the subobject of the truth values $p\in\Omega$ such that $\neg\neg p = p$; moreover, to say that two such objects are isomorphic, or internally isomorphic, in the effective topos, is equivalent to saying that $A$ and $B$ are computably isomorphic as above. So Friedberg's result can be reinterpreted by saying that if $A$ and $B$ are such objects of the effective topos and if $n\times A$ and $n\times B$ are isomorphic then $A$ and $B$ are.
I'm not sure how much this can be internalized (e.g., does the effective topos validate "if $A$ and $B$ are $\neg\neg$-stable sets of natural numbers and $n\times A$ is isomorphic to $n\times B$ then $A$ is isomorphic to $B$" for explicit $n$? and how about for $n$ quantified inside the topos?) or generalized (do we really need $\neg\neg$-stability?). But this may be worth looking into, and provides at least a positive kind-of-answer to the original question.
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this was given as an exercise: Prove that if $\sum_{n=1}^{\infty} |a_n|$ converges and $(b_n)^{\infty}_{n=1}$ is a bounded sequence, then $\sum_{n=1}^{\infty} |a_nb_n|$ converges
This is what i was thinking:
since $\displaystyle\sum a_n = \lim_{k\to\infty}\sum_{n<k} a_n$, and when multiplying by a constant it can jump into the limit.
then If $b_n\le c$ then $\sum a_nb_n\le \sum ca_n\le c\sum a_n$, and then $\sum_{n=1}^{\infty} |a_nb_n|$ converges.
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Let $X$ be a compact Kähler manifold of dimension $n$ with a given Kähler metric $\omega$. Let $L$ be a hermitian holomorphic line bundle on $X$ whose metric is positive. Let $x_0\in X$.
I would like to construct a section $s\in H^0(X,L)$, so that $s(x_0)=0$ and $ds(x_0)=\alpha\neq 0$ for given $\alpha$. Moreover, I want to control the $L^2$ norm of $s$ by some function of $|\alpha|$. What would be the appropriate conditions for such sections to exist?
Here $ds$ is the differential of $s$, since $x_0$ is a zero of $s$, $ds(x_0):T_{x_0}X \rightarrow L_{x_0}$ is intrinsically defined.
The question is similar to a special case of Ohsawa-Takegoshi theorem, where the prescribed information is only $s(x_0)$. If there is a good solution to my question, then I would like to know if it is possible to replace the single point $x_0$ by a more general analytic subset of $X$?
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Introduction
Chromatic aberration (CA) is one of several aberrations that degrade lens performance. (Others include coma, astigmatism, spherical aberration, and curvature of field.) It occurs because the index of refraction of glass varies with the wavelength of light, i.e., glass bends different colors by different amounts. This phenomenon is called
. It appears as color fringing, most visibly on tangential edges near the boundaries of the image. It is sometimes confused with another effect, which we call dispersion — a color channel offset that is typically uniform over the sensor and can be caused by physical misalignment of multi-chip sensors or demosaicing errors. pixel shift
Minimizing chromatic aberration is one of the traditional goals of lens design. It is accomplished by combining glass elements with different dispersion properties. But it remains a problem in several lens types, most notably ultrawide lenses, long telephoto lenses, and extreme zooms. In many designs intended for digital cameras, CA is no longer a major consideration because it can be completely corrected in the digital domain if the correction is applied
before demosaicing. Lateral and Longitudinal CA; Tangential and radial lines
Pixel shift is best measured on near-vertical and horizontal edges near the center of the image where CA is minimum. Measurement tip— Lateral chromatic aberration is best measured using (nearly) tangential edges near the sides or corners of the image, for example, edge B (above). It is not visible on radial edges such as A. Imatest applies a correction if the edge is not perfectly tangential.
CA cannot be measured reliably if the center of the region of interest (ROI) is less than 30% of the distance from the center to the corners. In this case, the chart will be displayed in pale colors and CA in % will be omitted.
ROI for CA measurement
The two types of chromatic aberration are illustrated above.
Longitudinal chromatic aberrationcauses different wavelengths to focus on different image planes. It cannot be measured from a single image by ; it causes a degradation of MTF response– by differing amounts for different colors. Imatest Lateral chromatic aberrationis the color fringing that occurs because the magnification of the image differs with wavelength. It tends to be far more visible than longitudinal CA. measures Lateral CA. Imatest It is not visible on radial edges. Lateral chromatic aberration is best measured on a tangential (or nearly tangential) edge near the sides or corners of an image. Tangential curvesand radial lines(which differ by 90 degrees) are shown in burgundyand bluein the Test chartillustration, above. Because typical Imatest SFR edges have angles in the range of 3 to 7 degrees with respect to vertical and horizontal, the best edges for measuring CA are near-vertical edges on the left and right of the image. The ISO 12233:2000 chart (shown on the right) has a very limited number of appropriate edges. One is rectangle B, above. SFRpluscharts, which are available in a wide variety of sizes and media, are much better for measuring CA.
The thumbnail on the right is from a 12 megapixel compact digital camera with fairly high chromatic aberration. The selected area is shown below. Red fringing, the result of lateral CA, is clearly visible. The black-to-white edge to the right side of this rectangle has equally vivid green fringing. Imatest analyzes the edge and produces a number that indicates the severity of the lateral chromatic aberration.
Imatest chromatic aberration measurement
The average transitions for the R, G, and B color channels, calculated by Imatest SFR for the above edge, are shown in the figure below. The edges have been normalized to have asymptotic limits of 0 and 1, i.e., they are dimensionless, approaching 0 and 1 at large distances from the transition center. Note that the three edges are not simply shifted, as you might expect if the focal lengths for the three colors were slightly different. They are distorted due to demosaicing (RAW conversion), as discussed below. When Bayer RAW images (undemosaiced) are analyzed, you do indeed see simple shifts.
The visibility of the chromatic aberration is proportional to the
between the edges with highest amplitude (in this case, area red) and the lowest (in this case, blue). This area can be expressed by the following integral.
\(\displaystyle CA (\text{area}) = \int \left[ Edge_{max}(x) – Edge_{min}(x) \right] dx\)
Since
has dimensions of distance in pixels and x is dimensionless, Edge CA (area)has units of units of distance even though it is an area. pixels— defined by this equation is called the area chromatic aberration. It is displayed in CA magentain the figure on the right. ( ). CA(area) = 1.88 pixels
The distance between the crossings (the centers of the transitions) is also shown. It is proportional to the
(the actual chromatic aberration of the lens, unless it has been corrected during raw conversion), but it is less visually significant. optical CA Chromatic Aberration figure (relatively high CA) (shown reversed: displayed levels always increasefrom left to right)
Although area in pixels is a good measure of perceptual
, it has some shortcomings. CA It penalizes cameras with high pixel counts. The result depends strongly on the measurement location. The chromatic aberration in most lenses is roughly proportional to the distance from the image center.
To deal with these issues Chromatic aberration is also expressed in percentage of the distance from the image center to the Region of Interest (ROI), corrected for the angle of the ROI with respect to the center. In the above example, CA (area) = 0.110% of the distance from the image center to the ROI. The correction is described in the
green box, below. This measurement gives the best overall results, since it’s relatively independent of the measurement location and the number of pixels. A table below presents rough guidelines for the severity of CA. Measurements displayed on the right of the figure are summarized in the following table.
10-90% rise distance (original; uncorrected) in pixels and rises per picture height (PH). CA (area): Chromatic aberration area in pixels. An indicator of the visibility of CA. The area between the channels with the highest and lowest levels. In units of pixels because the x-axis is in pixels and the y-axis is normalized to 1. Explained in the page on Chromatic aberration. Measured in along the axis indicated by on the upper-left. Profile Meaning (now obsolete): Under 0.5; insignificant. 0.5-1: minor; 1-1.5: moderate; 1.5 and over: serious. CA (area) as a percentage of the distance from the image center to the ROI. A better indicator than pixels (above) because it tends to be relatively consistent (at least in raw images).
Equal to 100% * (area between the channels with the highest and lowest levels) / (distance from center to the ROI in pixels), corrected for the angle of the ROI. This number is relatively independent of the ROI location because CA tends to be proportional to the distance from the image center. Explained in the page on
Chromatic aberration. Measured along the radial line from the image center to the edge. Meaning: Under 0.04; insignificant. 0.04-0.08: minor; 0.08-0.15: moderate; over 0.15: serious. CA (crossing). Chromatic aberration based on the most widely separated edge centers (positions where the edges cross 0.5). Tends to be less indicative of CA visibility than CA (area) but more indicative of the actual lens chromatic aberration. Measured two ways: (A) in pixels along the axis indicated by in the upper left, and (B) in percentage of the distance from the image center to center of the ROI along the line from the image center. Profile R-G, B-G Red−Green and Blue −Green crossing shift expressed as percentage of the distance from the image center to ROI, measured along the radial line from the image center. R-G is r(R cross)- r(G cross), where r is radius (from the image center). R-G, B-G Red−Green and Blue −Green crossing shift expressed in pixels, measured along the axis indicated by in the upper left. R-G is Profile x(R)- x(G) for horizontal profiles or y(R)- y(G) for vertical profiles, where x and y are distances along the horizontal and vertical axes, respectively. The sign may be different from the sign in the percentage measurement, depending on the measurement quadrant. x-pixel shift R-G, B-G Pixel shift in the x-direction (for near-vertical edges); y-pixel shift is shown for near-horizontal edges. This is most useful for measuring pixel shift from causes other than optical chromatic aberration, and is best measured near the center of the lens. Average pixel levels in the dark and light areas. Clipping can occur if they are too close to 0 or 255. Because Chromatic Aberration cannot be measured accurately near the image center, the chart is rendered in pale colors with the Region of Interest (ROI) is less than 30% of the distance from the center to the corner.
Chromatic Aberration in
percentage of distance
from the image center
Severity 0-0.04 Insignificant 0.04-0.08 Low. Hard to see unless you look for it. 0.08-0.15 Moderate. Somewhat visible at high print magnifications. over 0.15 Strong. Highly visible at high print magnifications.
Purple fringing
is
chromatic aberration, though it is often mistaken for it. It’s a saturation phenomenon in the sensor, also known as “bloom,” caused by the overflow of electrons from highly saturated pixel sites to nearby unsaturated sites. It tends to be worst in cameras with tiny pixels (e.g., 8+ megapixel compact digital cameras). It has everything to do with the sensor and nothing to do with the lens. not Imatest modules
The following Imatest modules can be used for chromatic aberration measurement. All but Dot Pattern measure Lateral Chromatic Aberration from (near) tangential slanted-edges, and also measure MTF. Some also measure optical distortion. Auto-detection slanted-edge modules are compared in
www.imatest.com/docs/#sharpness.
Module Comments SFRplus Measures a great many image quality factors. eSFR ISO Measures a great many image quality factors, including noise. Limited distortion measurements. Checkerboard Can operate over a wide range of distances and/or Fields of View. Very accurate distortion measurements. Dot Pattern Also measures Optical Distortion (but not MTF). Corresponds to several standards. SFR Any slanted-edge, selected manually. SFRreg Works with ultra-wide Fields of View. Does not measure distortion. Demosaicing
Demosaicing is the process of converting Bayer RAW images, which have one color per pixel (RGRGRG…; GBGBGB…), to standard images, which have three colors per pixel. In the process of demosaicing, missing detail for each channel is inferred from detail in other channels. This is especially significant for
Red and Blue pixels, which are half as common as Green. Demosaicing algorithms can be very mathematically sophisticated, but all of them can perform poorly in the presence of lateral CA, where detail is shifted from its expected location.
Demosaicing explains the shifted edges shown in the examples on this page. Lateral CA cannot be reliably corrected
after demosaicing, but it can be corrected to near-perfection prior to demosaicing, and this is frequently done on modern cameras and camera phones. Correction coefficients can be calculated with Imatest Master, which can analyze Bayer RAW files created by converting manufacturer’s Camera RAW files. Details are in the page on RAW files.
Here is an example illustrating the same region for a RAW and demosaiced file. Canon EOS-40D, 17-85mm IS lens, 50mm, f/8.
R-G and
B-G are the CA correction coefficients. They are the spacing between the Red and Green and Blue and Green crossings, respectively, expressed as percentage of the center-to-ROI distance. This measurement is relatively independent of the location of the measurement.
Corrected chromatic aberration measurements
In
Imatest, edge profiles are measured along horizontal or vertical lines. The blue line ( x) on the right is an example. But chromatic aberration takes place along radial lines– lines from the center of the image to the region of measurement (shown in red on the right). Unless this line is vertical or horizontal, there will be a measurement error that must be corrected. The correction is illustrated on the right for a near-vertical edge, where the profile, and hence CA, is measured horizontally. In this illustration, \(x = x_1 + x_2\) is the measured chromatic aberration, along a horizontal row of pixels. \( C =\) the true chromatic aberration, along the radial line (angle = \(\theta\)). \( \phi\) is the angle of the edge relative to vertical.
\(x_1 = C \cos \theta \: ; \quad y = C \sin \theta\)
\(x_2 = y \tan \phi \) (\(x_2\) may be negative if \(\phi\) is negative.)
\(x = x_1 + x_2 = C \cos \theta + y \tan \phi = C (\cos \theta + \sin \theta \tan \phi)\)
\(\displaystyle C = \frac{x}{\cos \theta + \sin \theta \tan \phi}\)
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This 18 page article seems pretty good as a historical account of who was responsible for what.
In general, the push for rigor is usually in response to a failure to be able to demonstrate the kinds of results one wishes to. It's usually relatively easy to demonstrate that there exist objects with certain properties, but you need precise definitions to prove that no such object exists. The classic example of this is non-computable problems and Turing Machines. Until you sit down and say "this precisely and nothing else is what it means to be solved by computation" it's impossible to prove that something isn't a computation, so when people start asking "is there an algorithm that does $\ldots$?" for questions where the answer "should be" no, you suddenly need a precise definition. Similar things happened with real analysis.
In real analysis, as mentioned in an excellent comment, there was a shift in what people's conception of the notion of a function was. This broadened conception of a function suddenly allows for a number of famous "counter example" functions to be constructed. These often that require a reasonably rigorous understanding of the topic to construct or to analyze. The most famous is the everywhere continuous nowhere differentiable Weierstrass function. If you don't have a very precise definition of continuity and differentiability, demonstrating that that function is one and not the other is extremely hard. The quest for weird functions with unexpected properties and combinations of properties was one of the driving forces in developing precise conceptions of those properties.
Another topic that people were very interested in was infinite series. There are lots of weird results that can crop up if you're not careful with infinite series, as shown by the now famously cautionary theorem:
Theorem (Summation Rearrangement Theorem): Let $a_n$ be a sequence such that $\sum a_n$ converges conditionally. Then for every $x$ there is some $b_n$ that is a reordering of $a_n$ such that $\sum b_n=x$.
This theorem means you have to be very careful dealing with infinite sums, and for a long time people weren't and so started deriving results that made no sense. Suddenly the usual free-wheeling algebraic manipulation approach to solving infinite sums was no longer okay, because sometimes doing so changed the value of the sum. Instead, a more rigorous theory of summation manipulation, as well as concepts such as uniform and absolute convergence had to be developed.
Here's an example of an problemsurrounding an infinite product created by Euler:
Consider the following formula:
$$x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right)$$
Does this expression even make sense? Assuming it does, does this equal $\sin(x)$ or $\sin(x)e^x$? How can you tell (notice that both functions have the same zeros as this sum, and the same relationship to their derivative)? If it doesn't equal $\sin(x)e^x$ (which it doesn't, it really does equal $\sin(x)$) how can we modify it so that it does?
Questions like this were very popular in the 1800s, as mathematicians were notably obsessed with infinite products and summations. However, most questions of this form require a very sophisticated understanding of analysis to handle (and weren't handled particularly well by the tools of the previous century).
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We next recall a general principle that will later be applied to distance-velocity-acceleration problems, among other things. If $F(u)$ is an anti-derivative of $f(u)$, then $\ds \int_a^bf(u)\,du=F(b)-F(a)$. Suppose that we want to let the upper limit of integration vary, i.e., we replace $b$ by some variable $x$. We think of $a$ as a fixed starting value $x_0$. In this new notation the last equation (after adding $F(a)$ to both sides) becomes: $$ F(x)=F(x_0)+\int_{x_0}^xf(u)\,du. $$ (Here $u$ is the variable of integration, called a "dummy variable,'' since it is not the variable in the function $F(x)$. In general, it is not a good idea to use the same letter as a variable of integration and as a limit of integration. That is, $\ds \int_{x_0}^xf(x)dx$ is bad notation, and can lead to errors and confusion.)
An important application of this principle occurs when we are interested in the position of an object at time $t$ (say, on the $x$-axis) and we know its position at time $\ds t_0$. Let $s(t)$ denote the position of the object at time $t$ (its distance from a reference point, such as the origin on the $x$-axis). Then the net change in position between $\ds t_0$ and $t$ is $\ds s(t)-s(t_0)$. Since $s(t)$ is an anti-derivative of the velocity function $v(t)$, we can write $$ s(t)=s(t_0)+\int_{t_0}^tv(u)du. $$ Similarly, since the velocity is an anti-derivative of the acceleration function $a(t)$, we have $$ v(t)=v(t_0)+\int_{t_0}^ta(u)du. $$
Example 8.2.1 Suppose an object is acted upon by a constant force $F$. Find $v(t)$ and $s(t)$. By Newton's law $F=ma$, so the acceleration is $F/m$, where $m$ is the mass of the object. Then we first have $$ v(t)=v(t_0)+\int_{t_0}^t{F\over m}\,du=v_0+ \left.{F\over m}u\right|_{t_0}^t=v_0+{F\over m}(t-t_0), $$ using the usual convention $\ds v_0=v(t_0)$. Then $$\eqalign{ s(t)&=s(t_0)+\int_{t_0}^t\left(v_0+{F\over m}(u-t_0)\right)du=s_0+ \left.(v_0u+{F\over2m}(u-t_0)^2)\right|_{t_0}^t\cr &=s_0+v_0(t-t_0)+{F\over2m}(t-t_0)^2.\cr }$$ For instance, when $F/m=-g$ is the constant of gravitational acceleration, then this is the falling body formula (if we neglect air resistance) familiar from elementary physics: $$s_0+v_0(t-t_0)-{g\over2}(t-t_0)^2,$$ or in the common case that $\ds t_0=0$, $$s_0+v_0t-{g\over2}t^2.$$
Recall that the integral of the velocity function gives the
net distancetraveled, that is, the displacement. If you want to know the totaldistance traveled, you must find out where the velocity functioncrosses the $t$-axis, integrate separately over the time intervals when$v(t)$ is positive and when $v(t)$ is negative, and add up the absolutevalues of the different integrals. For example, if an object is thrownstraight upward at 19.6 m/sec, its velocity function is$v(t)=-9.8t+19.6$, using $g=9.8$ m/sec$^2$ for the force of gravity.This is a straight line which is positive for $t< 2$ and negative for $t>2$.The net distance traveled in the first 4 seconds is thus$$\int_0^4(-9.8t+19.6)dt=0,$$ while the total distance traveled in the first4 seconds is$$ \int_0^2(-9.8t+19.6)dt+\left|\int_2^4(-9.8t+19.6)dt\right|=19.6+|-19.6|=39.2$$ meters, $19.6$ meters up and $19.6$ meters down.
Example 8.2.2 The acceleration of an object is given by $a(t)=\cos(\pi t)$, and its velocity at time $t=0$ is $1/(2\pi)$. Find both the net and the total distance traveled in the first 1.5 seconds.
We compute $$ v(t)=v(0)+\int_0^t\cos(\pi u)du={1\over 2\pi}+\left.{1\over\pi} \sin(\pi u)\right|_0^t={1\over\pi}\bigl({1\over2}+\sin(\pi t)\bigr). $$ The {\it net} distance traveled is then $$\eqalign{ s(3/2)-s(0)&=\int_0^{3/2}{1\over\pi}\left({1\over2}+\sin(\pi t)\right)\,dt\cr &=\left.{1\over\pi}\left({t\over2}-{1\over\pi}\cos(\pi t)\right) \right|_0^{3/2}={3\over4\pi}+{1\over\pi^2}\approx 0.340 \hbox{ meters.}\cr }$$ To find the {\it total} distance traveled, we need to know when $(0.5+\sin(\pi t))$ is positive and when it is negative. This function is 0 when $\sin(\pi t)$ is $-0.5$, i.e., when $\pi t=7\pi/6$, $11\pi/6$, etc. The value $\pi t=7\pi/6$, i.e., $t=7/6$, is the only value in the range $0\le t\le 1.5$. Since $v(t)>0$ for $t< 7/6$ and $v(t)< 0$ for $t>7/6$, the total distance traveled is $$\eqalign{ \int_0^{7/6}&{1\over \pi}\left({1\over2}+\sin(\pi t)\right)\,dt+ \Bigl|\int_{7/6}^{3/2} {1\over \pi}\left({1\over2}+\sin(\pi t)\right)\,dt\Bigr|\cr &={1\over \pi}\left( {7\over 12}+{1\over \pi}\cos(7\pi/6)+{1\over \pi}\right)+ {1\over \pi}\Bigl|{3\over 4}-{7\over 12} +{1\over \pi}\cos(7\pi/6)\Bigr|\cr &={1\over \pi}\left( {7\over 12}+{1\over \pi}{\sqrt3\over2}+{1\over \pi}\right)+ {1\over \pi}\Bigl|{3\over 4}-{7\over 12} +{1\over \pi}{\sqrt3\over2}.\Bigr| \approx 0.409 \hbox{ meters.}\cr }$$
Exercises 8.2
For each velocity function find both the net distance and the total distance traveled during the indicated time interval (graph $v(t)$ to determine when it's positive and when it's negative):
Ex 8.2.1$v=\cos(\pi t)$, $0\le t\le 2.5$(answer)
Ex 8.2.2$v=-9.8t+49$, $0\le t\le 10$(answer)
Ex 8.2.3$v=3(t-3)(t-1)$, $0\le t\le 5$(answer)
Ex 8.2.4$v=\sin(\pi t/3)-t$, $0\le t\le 1$(answer)
Ex 8.2.5An object is shot upwards from ground level with an initialvelocity of 2 meters per second; it is subject only to the force ofgravity (no air resistance). Find its maximum altitude and the time atwhich it hits the ground.(answer)
Ex 8.2.6An object is shot upwards from ground level with an initialvelocity of 3 meters per second; it is subject only to the force ofgravity (no air resistance). Find its maximum altitude and the time atwhich it hits the ground.(answer)
Ex 8.2.7An object is shot upwards from ground level with an initialvelocity of 100 meters per second; it is subject only to the force ofgravity (no air resistance). Find its maximum altitude and the time atwhich it hits the ground.(answer)
Ex 8.2.8An object moves along a straight line with acceleration given by$a(t) = -\cos(t)$, and $s(0)=1$ and$v(0)=0$. Find the maximum distance the object travels from zero, andfind its maximum speed. Describe the motion of the object.(answer)
Ex 8.2.9An object moves along a straight line with acceleration given by$a(t) = \sin(\pi t)$. Assume that when $t=0$, $s(t)=v(t)=0$. Find$s(t)$, $v(t)$, and the maximum speed of the object. Describe themotion of the object.(answer)
Ex 8.2.10An object moves along a straight line with acceleration given by$a(t) = 1+\sin(\pi t)$. Assume that when $t=0$, $s(t)=v(t)=0$. Find$s(t)$ and $v(t)$.(answer)
Ex 8.2.11An object moves along a straight line with acceleration given by$a(t) = 1-\sin(\pi t)$. Assume that when $t=0$, $s(t)=v(t)=0$. Find$s(t)$ and $v(t)$.(answer)
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For a discrete abelian cancellative semigroup$\def \xmlpi #1{}\def \mathsfbi #1{\boldsymbol {\mathsf {#1}}}\let \le =\leqslant \let \leq =\leqslant \let \ge =\geqslant \let \geq =\geqslant \def \Pr {\mathit {Pr}}\def \Fr {\mathit {Fr}}\def \Rey {\mathit {Re}}S$with a weight function$\omega $and associated multiplier semigroup$M_\omega (S)$consisting of$\omega $-bounded multipliers, the multiplier algebra of the Beurling algebra of$(S,\omega )$coincides with the Beurling algebra of$M_\omega (S)$with the induced weight.
Towards an involutive analogue of a result on the semisimplicity of${\ell }^{1} (S)$by Hewitt and Zuckerman, we show that, given an abelian$\ast $-semigroup$S$, the commutative convolution Banach$\ast $-algebra${\ell }^{1} (S)$is$\ast $-semisimple if and only if Hermitian bounded semicharacters on$S$separate the points of$S$; and we search for an intrinsic separation property on$S$equivalent to$\ast $-semisimplicity. Very many natural involutive analogues of Hewitt and Zuckerman’s separation property are shown not to work, thereby exhibiting intricacies involved in analysis on$S$.
The Beurling algebra L1(G,ω)on a locally compact Abelian group G with a measurable weight ω is shown to be semisimple. This gives an elementary proof of a result that is implicit in the work of M.C. White (1991), where the arguments are based on amenable (not necessarily Abelian) groups.
Banach and Fréchet algebras with a Laurent series generator are investigated leading, via the discrete Beurling algebras, to functional analytic characterisations of the holomorphic function algebras on the annulus as well as the C∞-algebra on the unit circle.
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Let define $f(x)=(\tan x)^{\sin 2x}$ for $x \in (0, \frac{\pi}{2})$
Please help me prove, that $f$ reaches its lower bound in only one point $x_1$ and reaches its upper bound $x_2$ also in only one point of domain.
Calculate $x_1 + x_2$.
What I have done:
I've checked the derivative, but it's looking horrible:
The key to find the extremum is to calculate $x$, that satisfy $\cos 2x \cdot \ln (\tan x) = -1$
I don't have any idea how to do this. All help and hints are appreciated. Thanks in advance!
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These exercises are not tied to a specific programming language. Example implementations are provided under the Code tab, but the Exercises can be implemented in whatever platform you wish to use (e.g., Excel, Python, MATLAB, etc.).
### Exercise 1: Error Propagation for Simple Addition/Subtraction
The template Error Propagation.ipynb is pre-set for this problem. Note that the template reports a ridiculous number of digits, so you will need to properly round off the reported uncertainties and mean values yourself.
Suppose that you have measured two distances, $x_1$ = 5.00 ± 0.20 m and $x_2$ = 6.50 ± 0.10 m.
- If you add the two distances, what is the uncertainty in that sum?
- If you find the difference $x_2 - x_1$, what is the uncertainty in that difference?
- What difference do you see between adding and subtracting?
- Explore the role of N: make it much smaller and much larger, so that you can see how the Monte Carlo results converge to the traditional results. Python is quite capable of handling a million-point array!
### Exercise 2: Error Propagation for Simple Multiplication/Division
You will need to make more modifications to the Jupyter notebook. If you want to preserve your earlier work separately, go to the $File$ menu and select $Make a copy...$ and then $Rename...$ it.
For this exercise, you could still use the same two parameters, $x_1$ = 5.00 ± 0.20 m and $x_2$ = 6.50 ± 0.10 m.
- Modify the block of code that calculates the traditional value for the function and uncertainty for the function $f(x_1,x_2) = x_2 * x_1$.
- Modify the block of code that performs the Monte Carlo method.
- Compare the average and uncertainty for the Monte Carlo method with the traditional method, they should agree very well.
- Modify for the function $f(x_1,x_2) = x_2 / x_1$ and again compare the two methods.
- What difference do you see between multiplication and division?
- Calculate the relative uncertainty for both situations, multiplication and division, and reconsider the previous question.
### Exercise 3: Error Propagation for a Simple Compound Function
A compound function involves both addition and multiplication, so that the simple quick methods for error propagation cannot be applied. In this exercise, you will predict the velocity of an object given the initial velocity, acceleration and elapsed time.
$$ v = v_0 + a * t $$
where $v_0$ = -2.40 ± 0.20 $m/s$,$a$ = 1.55 ± 0.15 $m/s^2$ and $t$ = 12.0 ± 1.0 $s$.
- Modify the block of code that calculates the traditional value for the velocity and uncertainty of the velocity.
- Modify the block of code that performs the Monte Carlo method.
- Compare the average and uncertainty for the Monte Carlo method with the traditional method.
- The uncertainty in the predicted velocity is rather large. If you could only improve the precision of $one$ of the parameters, which one would have the most impact on the final precision? Which would have the least impact? (One way to explore this to divide each uncertainty by 10, one by one.) This type of exploratory error analysis can help in deciding where to focus resources in improving the precision of an experiment.
### Exercise 4: Exploring nonlinearity
The standard approach to error propagation relies on the approximation that the function is approximately linear over the range $\overline a ± \sigma_{\overline a}$. For many situations encountered in experimental physics, this is a very good approximation. In this exercise you will push the limits.
Consider the function $f(r) = \frac{1}{r^4}$ which you might encounter in a measurement of the elastic moduli of a cylinrical rod. For the purpose of this exercise, start out with $\overline r$ = 5.00 ± 0.01 mm.
- Modify the block of code that calculates the traditional value and uncertainty of the function.
- Modify the block of code that performs the Monte Carlo method.
- Compare the average and uncertainty for the Monte Carlo method with the traditional method. Increase the standard error of $r$ to ± 0.50 mm and repeat the comparison. Also look at the shape of the histogram of the simulations, is it still symmetric?
- It is easy to calculate the derivative for this function, and evaluate it at the mean value of r and at the mean ± sigma. Use a per cent difference to describe how much the slope varies compared to the derivative at the center. Examine how the relative uncertainty in r relates to the variation in the slope, looking for where the traditional approach deviates noticeably from the Monte Carlo approach.
### Extension: Range of a Projectile
If a projectile is launched from zero altitude with initial speed $v_0$ at an angle of $\theta$ above horizontal, it will return to zero altitude at a range given by
$$R = \frac{v^2_0}{g} sin(2 \theta)$$
Do you want to input your angle in degrees? You'll need to be able to convert degrees to radians in the program!
Consider an initial speed of 25 ± 1 m/s and an angle of 35 ± 2 degrees.
- This function is a little more complicated to do the traditional error propagation, complicated enough that you might start to see some value in the Monte Carlo approach versus the traditional approach. Do the traditional error propagation and code it.
- Compare the Monte Carlo and traditional results. Is the function sufficiently linear with the given uncertainties? How can you tell?
- Which parameter is most crucial for a precise prediction of the range? Describe how you determined your answer.
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Uplifting cardinals Uplifting cardinals were introduced by Hamkins and Johnstone in [1], from which some of this text is adapted.
An inaccessible cardinal $\kappa$ is
uplifting if and only if for every ordinal $\theta$ it is $\theta$-uplifting, meaning that there is an inaccessible $\gamma>\theta$ such that $V_\kappa\prec V_\gamma$ is a proper elementary extension.
An inaccessible cardinal is
pseudo uplifting if and only if for every ordinal $\theta$ it is pseudo $\theta$-uplifting, meaning that there is a cardinal $\gamma>\theta$ such that $V_\kappa\prec V_\gamma$ is a proper elementary extension, without insisting that $\gamma$ is inaccessible.
Being
strongly uplifting (see further) is boldface variant of being uplifting.
It is an elementary exercise to see that if $V_\kappa\prec V_\gamma$ is a proper elementary extension, then $\kappa$ and hence also $\gamma$ are $\beth$-fixed points, and so $V_\kappa=H_\kappa$ and $V_\gamma=H_\gamma$. It follows that a cardinal $\kappa$ is uplifting if and only if it is regular and there are arbitrarily large regular cardinals $\gamma$ such that $H_\kappa\prec H_\gamma$. It is also easy to see that every uplifting cardinal $\kappa$ is uplifting in $L$, with the same targets. Namely, if $V_\kappa\prec V_\gamma$, then we may simply restrict to the constructible sets to obtain $V_\kappa^L=L^{V_\kappa}\prec L^{V_\gamma}=V_\gamma^L$. An analogous result holds for pseudo-uplifting cardinals.
Contents 1 Consistency strength of uplifting cardinals 2 Uplifting cardinals and $\Sigma_3$-reflection 3 Uplifting Laver functions 4 Connection with the resurrection axioms 5 Strongly Uplifting 6 Weakly superstrong cardinal 7 References Consistency strength of uplifting cardinals Theorem.
1. If $\delta$ is a Mahlo cardinal, then $V_\delta$ has a proper class of uplifting cardinals.
2. Every uplifting cardinal is pseudo uplifting and a limit of pseudo uplifting cardinals.
3. If there is a pseudo uplifting cardinal, or indeed, merely a pseudo $0$-uplifting cardinal, then there is a transitive set model of ZFC with a reflecting cardinal and consequently also a transitive model of ZFC plus Ord is Mahlo.
Proof. For (1), suppose that $\delta$ is a Mahlo cardinal. By the Lowenheim-Skolem theorem, there is a club set $C\subset\delta$ of cardinals $\beta$ with $V_\beta\prec V_\delta$. Since $\delta$ is Mahlo, the club $C$ contains unboundedly many inaccessible cardinals. If $\kappa<\gamma$ are both in $C$, then $V_\kappa\prec V_\gamma$, as desired. Similarly, for (2), if $\kappa$ is uplifting, then $\kappa$ is pseudo uplifting and if $V_\kappa\prec V_\gamma$ with $\gamma$ inaccessible, then there are unboundedly many ordinals $\beta<\gamma$ with $V_\beta\prec V_\gamma$ and hence $V_\kappa\prec V_\beta$. So $\kappa$ is pseudo uplifting in $V_\gamma$. From this, it follows that there must be unboundedly many pseudo uplifting cardinals below $\kappa$. For (3), if $\kappa$ is inaccessible and $V_\kappa\prec V_\gamma$, then $V_\gamma$ is a transitive set model of ZFC in which $\kappa$ is reflecting, and it is thus also a model of Ord is Mahlo. QED
Uplifting cardinals and $\Sigma_3$-reflection Every uplifting cardinal is a limit of $\Sigma_3$-reflecting cardinals, and is itself $\Sigma_3$-reflecting. If $\kappa$ is the least uplifting cardinal, then $\kappa$ is not $\Sigma_4$-reflecting, and there are no $\Sigma_4$-reflecting cardinals below $\kappa$.
The analogous observation for pseudo uplifting cardinals holds as well, namely, every pseudo uplifting cardinal is $\Sigma_3$-reflecting and a limit of $\Sigma_3$-reflecting cardinals; and if $\kappa$ is the least pseudo uplifting cardinal, then $\kappa$ is not $\Sigma_4$-reflecting, and there are no $\Sigma_4$-reflecting cardinals below $\kappa$.
Uplifting Laver functions
Every uplifting cardinal admits an ordinal-anticipating Laver function, and indeed, a HOD-anticipating Laver function, a function $\ell:\kappa\to V_\kappa$, definable in $V_\kappa$, such that for any set $x\in\text{HOD}$ and $\theta$, there is an inaccessible cardinal $\gamma$ above $\theta$ such that $V_\kappa\prec V_\gamma$, for which $\ell^*(\kappa)=x$, where $\ell^*$ is the corresponding function defined in $V_\gamma$.
Connection with the resurrection axioms
Many instances of the (weak) resurrection axiom imply that ${\frak c}^V$ is an uplifting cardinal in $L$:
RA(all) implies that ${\frak c}^V$ is uplifting in $L$. RA(ccc) implies that ${\frak c}^V$ is uplifting in $L$. wRA(countably closed)+$\neg$CH implies that ${\frak c}^V$ is uplifting in $L$. Under $\neg$CH, the weak resurrection axioms for the classes of axiom-A forcing, proper forcing, semi-proper forcing, and posets that preserve stationary subsets of $\omega_1$, respectively, each imply that ${\frak c}^V$ is uplifting in $L$.
Conversely, if $\kappa$ is uplifting, then various resurrection axioms hold in a corresponding lottery-iteration forcing extension.
Theorem. (Hamkins and Johnstone) The following theories are equiconsistent over ZFC: There is an uplifting cardinal. RA(all) RA(ccc) RA(semiproper)+$\neg$CH RA(proper)+$\neg$CH for some countable ordinal $\alpha$, RA($\alpha$-proper)+$\neg$CH RA(axiom-A)+$\neg$CH wRA(semiproper)+$\neg$CH wRA(proper)+$\neg$CH for some countable ordinal $\alpha$, wRA($\alpha$-proper})+$\neg$CH wRA(axiom-A)+$\neg$CH wRA(countably closed)+$\neg$CH Strongly Uplifting
(Information in this section comes from [2])
Strongly uplifting cardinals are precisely strongly pseudo uplifting ordinals, strongly uplifting cardinals with weakly compact targets, superstrongly unfoldable cardinals and almost-hugely unfoldable cardinals.
Definitions
An ordinal is
strongly pseudo uplifting iff for every ordinal $θ$ it is strongly $θ$-uplifting, meaning that for every $A⊆V_κ$, there exists some ordinal $λ>θ$ and an $A^*⊆V_λ$ such that $(V_κ;∈,A)≺(V_λ;∈,A^*)$ is a proper elementary extension.
An inaccessible cardinal is
strongly uplifting iff for every ordinal $θ$ it is strongly $θ$-uplifting, meaning that for every $A⊆V_κ$, there exists some inaccessible(*) $λ>θ$ and an $A^*⊆V_λ$ such that $(V_κ;∈,A)≺(V_λ;∈,A^*)$ is a proper elementary extension. By replacing starred "inaccessible" with "weakly compact" and other properties, we get strongly uplifting with weakly compact etc. targets.
A cardinal $\kappa$ is
$\theta$-superstrongly unfoldable iff for every $A\subseteq\kappa$, there is some transitive $M$ with $A\in M\models\text{ZFC}$ and some $j:M\rightarrow N$ an elementary embedding with critical point $\kappa$ such that $j(\kappa)\geq\theta$ and $V_{j(\kappa)}\subseteq N$.
A cardinal $\kappa$ is
$\theta$-almost-hugely unfoldable iff for every $A\subseteq\kappa$, there is some transitive $M$ with $A\in M\models\text{ZFC}$ and some $j:M\rightarrow N$ an elementary embedding with critical point $\kappa$ such that $j(\kappa)\geq\theta$ and $N^{<j(\kappa)}\subseteq N$.
$κ$ is then called
superstrongly unfoldable (resp. almost-hugely unfoldable) iff it is $θ$-strongly unfoldable (resp. $θ$-almost-hugely unfoldable) for every $θ$; i.e. the target of the embedding can be made arbitrarily large. Equivalence
For any ordinals $κ$, $θ$, the following are equivalent:
$κ$ is strongly pseudo $(θ+1)$-uplifting. $κ$ is strongly $(θ+1)$-uplifting. $κ$ is strongly $(θ+1)$-uplifting with weakly compact targets. $κ$ is strongly $(θ+1)$-uplifting with totally indescribable targets, and indeed with targets having any property of $κ$ that is absolute to all models $V_γ$ with $γ > κ, θ$.
For any cardinal $κ$ and ordinal $θ$, the following are equivalent:
$κ$ is strongly $(θ+1)$-uplifting. $κ$ is superstrongly $(θ+1)$-unfoldable. $κ$ is almost-hugely $(θ+1)$-unfoldable. For every set $A ∈ H_{κ^+}$ there is a $κ$-model $M⊨\mathrm{ZFC}$ with $A∈M$ and $V_κ≺M$ and a transitive set $N$ with an elementary embedding $j:M→N$ having critical point $κ$ with $j(κ)> θ$ and $V_{j(κ)}≺N$, such that $N^{<j(κ)}⊆N$ and $j(κ)$ is inaccessible, weakly compact and more in $V$. $κ^{<κ}=κ$ holds, and for every $κ$-model $M$ there is an elementary embedding $j:M→N$ having critical point $κ$ with $j(κ)> θ$ and $V_{j(κ)}⊆N$, such that $N^{<j(κ)}⊆N$ and $j(κ)$ is inaccessible, weakly compact and more in $V$. Relations to other cardinals If $δ$ is a subtle cardinal, then the set of cardinals $κ$ below $δ$ that are strongly uplifting in $V_δ$ is stationary. If $0^♯$ exists, then every Silver indiscernible is strongly uplifting in $L$. In $L$, $κ$ is strongly uplifting iff it is unfoldable with cardinal targets. Every strongly uplifting cardinal is strongly uplifting in $L$. Every strongly $θ$-uplifting cardinal is strongly $θ$-uplifting in $L$. Every strongly uplifting cardinal is strongly unfoldable of every ordinal degree $α$ and a stationary limit of cardinals that are strongly unfoldable of every ordinal degree and so on. Relation to boldface resurrection axiom
The following theories are equiconsistent over $\mathrm{ZFC}$:
There is a strongly uplifting cardinal. The boldface resurrection axiom for all forcing, for proper forcing, for semi-proper forcing and for c.c.c. forcing. The weak boldface resurrection axioms for countably-closed forcing, for axiom-$A$ forcing, for proper forcing and for semi-properforcing, respectively, plus $¬\mathrm{CH}$. Weakly superstrong cardinal
(Information in this section comes from [3])
Hamkins and Johnstone called an inaccessible cardinal $κ$
weakly superstrong if for every transitive set $M$ of size $κ$ with $κ∈M$ and $M^{<κ}⊆M$, a transitive set $N$ and an elementary embedding $j:M→N$ with critical point $κ$, for which $V_{j(κ)}⊆N$, exist.
It is called
weakly almost huge if for every such $M$ there is such $j:M→N$ for which $N^{<j(κ)}⊆N$.
(As usual one can call $j(κ)$ the target.)
A cardinal is superstrongly unfoldable if it is weakly superstrong with arbitrarily large targets, and it is almost hugely unfoldable if it is weakly almost huge with arbitrarily large targets.
If $κ$ is weakly superstrong, it is $0$-extendible and $\Sigma_3$-extendible. Weakly almost huge cardinals also are $\Sigma_3$-extendible. Because $\Sigma_3$-extendibility always can be destroyed, all these cardinal properties (among others) are never Lever indestructible.
References Hamkins, Joel David and Johnstone, Thomas A. Resurrection axioms and uplifting cardinals., 2014. www arχiv bibtex Hamkins, Joel David and Johnstone, Thomas A. Strongly uplifting cardinals and the boldface resurrection axioms., 2014. arχiv bibtex Bagaria, Joan and Hamkins, Joel David and Tsaprounis, Konstantinos and Usuba, Toshimichi. Superstrong and other large cardinals are never Laver indestructible.Archive for Mathematical Logic 55(1-2):19--35, 2013. www arχiv DOI bibtex
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I have a large unbalanced panel data with 460 firms and 1259 days. The model I would like to run is below
$$ Y_{it} = \beta X_{it} + \alpha Z_{t} + \epsilon_{it} $$
where $Y_{it}$ is stock return, and $Z_{t}$ are Fama French 3 factors, and $X_{it}$ are variables of interest.
I run Fama Macbeth (FM) and double clustering to correct for the standard error, but two models give inconsistent results,i.e., $\beta$ is significant in one model, and not in the other.
I understand that Fama-MacBeth techniquewas developed to account for correlation between observations on different firms in the same time point,not to account for correlation between observations on the same firm in different time points. Traditionally, it should run cross section regression at each time point, then average estimates along time $T$. But in my case, due to the inclusion of $Z_t$, I have to run time series regression first since otherwise, $Z_{t}$ are not identifiable. In this case, does FM actually correct for correlation between observations on the same firm in different time points?
More importantly, does the inconsistent result mean that my results are not robust? Under my case, can I argue one is more appropriate than the other? Can the unbalanced data structure contribute to the inconsistent results?
I'm using fm and cluster2 command on this page Stata command
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Now showing items 1-10 of 23
Production of Σ(1385)± and Ξ(1530)0 in proton–proton collisions at √s = 7 TeV
(Springer, 2015-01-10)
The production of the strange and double-strange baryon resonances ((1385)±, Ξ(1530)0) has been measured at mid-rapidity (|y|< 0.5) in proton–proton collisions at √s = 7 TeV with the ALICE detector at the LHC. Transverse ...
Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV
(Springer, 2015-05-20)
The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ...
Inclusive photon production at forward rapidities in proton-proton collisions at $\sqrt{s}$ = 0.9, 2.76 and 7 TeV
(Springer Berlin Heidelberg, 2015-04-09)
The multiplicity and pseudorapidity distributions of inclusive photons have been measured at forward rapidities ($2.3 < \eta < 3.9$) in proton-proton collisions at three center-of-mass energies, $\sqrt{s}=0.9$, 2.76 and 7 ...
Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV
(Springer, 2015-06)
We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ...
Measurement of pion, kaon and proton production in proton–proton collisions at √s = 7 TeV
(Springer, 2015-05-27)
The measurement of primary π±, K±, p and p¯¯¯ production at mid-rapidity (|y|< 0.5) in proton–proton collisions at s√ = 7 TeV performed with a large ion collider experiment at the large hadron collider (LHC) is reported. ...
Two-pion femtoscopy in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(American Physical Society, 2015-03)
We report the results of the femtoscopic analysis of pairs of identical pions measured in p-Pb collisions at $\sqrt{s_{\mathrm{NN}}}=5.02$ TeV. Femtoscopic radii are determined as a function of event multiplicity and pair ...
Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV
(Springer, 2015-09)
Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ...
Charged jet cross sections and properties in proton-proton collisions at $\sqrt{s}=7$ TeV
(American Physical Society, 2015-06)
The differential charged jet cross sections, jet fragmentation distributions, and jet shapes are measured in minimum bias proton-proton collisions at centre-of-mass energy $\sqrt{s}=7$ TeV using the ALICE detector at the ...
Centrality dependence of high-$p_{\rm T}$ D meson suppression in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Springer, 2015-11)
The nuclear modification factor, $R_{\rm AA}$, of the prompt charmed mesons ${\rm D^0}$, ${\rm D^+}$ and ${\rm D^{*+}}$, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at a centre-of-mass ...
K*(892)$^0$ and $\Phi$(1020) production in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2015-02)
The yields of the K*(892)$^0$ and $\Phi$(1020) resonances are measured in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV through their hadronic decays using the ALICE detector. The measurements are performed in multiple ...
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Hey guys! I built the voltage multiplier with alternating square wave from a 555 timer as a source (which is measured 4.5V by my multimeter) but the voltage multiplier doesn't seem to work. I tried first making a voltage doubler and it showed 9V (which is correct I suppose) but when I try a quadrupler for example and the voltage starts from like 6V and starts to go down around 0.1V per second.
Oh! I found a mistake in my wiring and fixed it. Now it seems to show 12V and instantly starts to go down by 0.1V per sec.
But you really should ask the people in Electrical Engineering. I just had a quick peek, and there was a recent conversation about voltage multipliers. I assume there are people there who've made high voltage stuff, like rail guns, which need a lot of current, so a low current circuit like yours should be simple for them.
So what did the guys in the EE chat say...
The voltage multiplier should be ok on a capacitive load. It will drop the voltage on a resistive load, as mentioned in various Electrical Engineering links on the topic. I assume you have thoroughly explored the links I have been posting for you...
A multimeter is basically an ammeter. To measure voltage, it puts a stable resistor into the circuit and measures the current running through it.
Hi all! There is theorem that links the imaginary and the real part in a time dependent analytic function. I forgot its name. Its named after some dutch(?) scientist and is used in solid state physics, who can help?
The Kramers–Kronig relations are bidirectional mathematical relations, connecting the real and imaginary parts of any complex function that is analytic in the upper half-plane. These relations are often used to calculate the real part from the imaginary part (or vice versa) of response functions in physical systems, because for stable systems, causality implies the analyticity condition, and conversely, analyticity implies causality of the corresponding stable physical system. The relation is named in honor of Ralph Kronig and Hans Kramers. In mathematics these relations are known under the names...
I have a weird question: The output on an astable multivibrator will be shown on a multimeter as half the input voltage (for example we have 9V-0V-9V-0V...and the multimeter averages it out and displays 4.5V). But then if I put that output to a voltage doubler, the voltage should be 18V, not 9V right? Since the voltage doubler will output in DC.
I've tried hooking up a transformer (9V to 230V, 0.5A) to an astable multivibrator (which operates at 671Hz) but something starts to smell burnt and the components of the astable multivibrator get hot. How do I fix this? I check it after that and the astable multivibrator works.
I searched the whole god damn internet, asked every god damn forum and I can't find a single schematic that converts 9V DC to 1500V DC without using giant transformers and power stage devices that weight 1 billion tons....
something so "simple" turns out to be hard as duck
In peskin book of QFT the sum over zero point energy modes is an infinite c-number, fortunately, it's experimental evidence doesn't appear, since experimentalists measure the difference in energy from the ground state. According to my understanding the zro pt energy is the same as the ground state, isn't it?
If so, always it is possible to substract a finite number (higher exited state for e.g.) from this zero point enrgy (which is infinte), it follows that, experimentally we always obtain infinte spectrum.
@AaronStevens Yeah, I had a good laugh to myself when he responded back with "Yeah, maybe they considered it and it was just too complicated". I can't even be mad at people like that. They are clearly fairly new to physics and don't quite grasp yet that most "novel" ideas have been thought of to death by someone; likely 100+ years ago if it's classical physics
I have recently come up with a design of a conceptual electromagntic field propulsion system which should not violate any conservation laws, particularly the Law of Conservation of Momentum and the Law of Conservation of Energy. In fact, this system should work in conjunction with these two laws ...
I rememeber that Gordon Freeman's thesis was "Observation of Einstein-Podolsky-Rosen Entanglement on Supraquantum Structures by Induction Through Nonlinear Transuranic Crystal of Extremely Long Wavelength (ELW) Pulse from Mode-Locked Source Array "
In peskin book of QFT the sum over zero point energy modes is an infinite c-number, fortunately, it's experimental evidence doesn't appear, since experimentalists measure the difference in energy from the ground state. According to my understanding the zro pt energy is the same as the ground state, isn't it? If so, always it is possible to substract a finite number (higher exited state for e.g.) from this zero point enrgy (which is infinte), it follows that, experimentally we always obtain infinte spectrum.
@ACuriousMind What confuses me is the interpretation of Peskin to this infinite c-number and the experimental fact
He said, the second term is the sum over zero point energy modes which is infnite as you mentioned. He added," fortunately, this energy cannot be detected experm., since the experiments measure only the difference between from the ground state of H".
@ACuriousMind Thank you, I understood your explanations clearly. However, regarding what Peskin mentioned in his book, there is a contradiction between what he said about the infinity of the zero point energy/ground state energy, and the fact that this energy is not detectable experimentally because the measurable quantity is the difference in energy between the ground state (which is infinite and this is the confusion) and a higher level.
It's just the first encounter with something that needs to be renormalized. Renormalizable theories are not "incomplete", even though you can take the Wilsonian standpoint that renormalized QFTs are effective theories cut off at a scale.
according to the author, the energy differenc is always infinite according to two fact. the first is, the ground state energy is infnite, secondly, the energy differenc is defined by substituting a higher level energy from the ground state one.
@enumaris That is an unfairly pithy way of putting it. There are finite, rigorous frameworks for renormalized perturbation theories following the work of Epstein and Glaser (buzzword: Causal perturbation theory). Just like in many other areas, the physicist's math sweeps a lot of subtlety under the rug, but that is far from unique to QFT or renormalization
The classical electrostatics formula $H = \int \frac{\mathbf{E}^2}{8 \pi} dV = \frac{1}{2} \sum_a e_a \phi(\mathbf{r}_a)$ with $\phi_a = \sum_b \frac{e_b}{R_{ab}}$ allows for $R_{aa} = 0$ terms i.e. dividing by zero to get infinities also, the problem stems from the fact that $R_{aa}$ can be zero due to using point particles, overall it's an infinite constant added to the particle that we throw away just as in QFT
@bolbteppa I understand the idea that we need to drop such terms to be in consistency with experiments. But i cannot understand why the experiment didn't predict such infinities that arose in the theory?
These $e_a/R_{aa}$ terms in the big sum are called self-energy terms, and are infinite, which means a relativistic electron would also have to have infinite mass if taken seriously, and relativity forbids the notion of a rigid body so we have to model them as point particles and can't avoid these $R_{aa} = 0$ values.
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Note: In the following i'm going to use t as a time variable [s] and x as a spatial variable [m].
I'm currently working with the Discrete Fourier Transform (DFT), in order to get frequency information about my input signal. To be more precise, i am using the Fast Fourier Transform (FFT) for computational efficiency, using pythons numpy.fft package to do that.
When applying a DFT to a discrete signal of N-point, one transforms those N signal points to N transformed points. Simply speaking, each of those complex numbers we obtain from the DFT represent the a certain frequency of the signal, so one thing one has to be aware of is to properly calculate the spacing in the frequency space, in order to correctly correlate the DFT coefficients to specific frequencies.
For example: When i have a time signal f(t) at N discrete points, each with a constant spacing of dt, then i can calculate the spacing in frequency space $\Delta\omega = f_s/N$, where $f_s= 1/dt$ is the 'sampling rate' and N the number of signal points. This can be achieved easily with the fft-built in function (python) 'fftfreq': https://docs.scipy.org/doc/numpy/reference/generated/numpy.fft.fftfreq.html which outputs an array in frequency domain with the above mentioned correct frequency spacing up to the 'Nyquisit Limit', when being applied to a time domain vector.
My issue comes in, when i do not have a time signal f(t), but rather a spatial signal f(x). Now, after applying a Fourier Transform (or rather a DFT), the signal does not gets transformed to frequency space $\omega, [1/s]$, but to reciprocal space $k, [1/m]$, correct? That is, since we have a signal in space, the resulting 'frequencies' now measure the number of periods per space unit [1/m]. But how am i dealing with this fact considering my DFT calculation?
Looking at the Nyquisit Limit yields: $\omega_{Ny} = \frac{1}{2 dt}$ or in k-space $k_{Ny} = \frac{\pi}{\Delta x}$, respectively. So obviously, i have to calculate the k-domain (i.e. the spacing in k-sapce) differently when having a f(x) signal, instead of f(t). How do i calculate the k-spacing? Given the fact, that the Nyquisit-Limits only differ by a factor $2\pi$, i would naively say, that the k-spacing can by calculated with: $$ \Delta k = \Delta\omega\cdot 2\pi = \frac{2\pi}{dt\cdot N} $$
Am i missing something, or is this the correct k-spacing when transforming a spatial signal f(x)? In that case, the corresponsing python fft-method 'fftfreq' would have to be adjusted in terms of that factor, right?
Thanks for any help.
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Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
(Elsevier, 2017-12-21)
We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ...
Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV
(American Physical Society, 2017-09-08)
The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ...
Online data compression in the ALICE O$^2$ facility
(IOP, 2017)
The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ...
Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(American Physical Society, 2017-09-08)
In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ...
J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(American Physical Society, 2017-12-15)
We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($|y| < 0.9$) ...
Highlights of experimental results from ALICE
(Elsevier, 2017-11)
Highlights of recent results from the ALICE collaboration are presented. The collision systems investigated are Pb–Pb, p–Pb, and pp, and results from studies of bulk particle production, azimuthal correlations, open and ...
Event activity-dependence of jet production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV measured with semi-inclusive hadron+jet correlations by ALICE
(Elsevier, 2017-11)
We report measurement of the semi-inclusive distribution of charged-particle jets recoiling from a high transverse momentum ($p_{\rm T}$) hadron trigger, for p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in p-Pb events ...
System-size dependence of the charged-particle pseudorapidity density at $\sqrt {s_{NN}}$ = 5.02 TeV with ALICE
(Elsevier, 2017-11)
We present the charged-particle pseudorapidity density in pp, p–Pb, and Pb–Pb collisions at sNN=5.02 TeV over a broad pseudorapidity range. The distributions are determined using the same experimental apparatus and ...
Photoproduction of heavy vector mesons in ultra-peripheral Pb–Pb collisions
(Elsevier, 2017-11)
Ultra-peripheral Pb-Pb collisions, in which the two nuclei pass close to each other, but at an impact parameter greater than the sum of their radii, provide information about the initial state of nuclei. In particular, ...
Measurement of $J/\psi$ production as a function of event multiplicity in pp collisions at $\sqrt{s} = 13\,\mathrm{TeV}$ with ALICE
(Elsevier, 2017-11)
The availability at the LHC of the largest collision energy in pp collisions allows a significant advance in the measurement of $J/\psi$ production as function of event multiplicity. The interesting relative increase ...
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The functions $$f_k(x)=\frac{x+k}{e^x}$$ are given.
Let $A(u)$ be the area that is bounded by $f_1, f_3$, the $x$-axis und the line $x=u$.
I want to check the area if $u\rightarrow \infty$.
$$$$
To calculate the area $A(u)$ do we calculate the area that is bounded by $f_1$ with endpoints the intersection point of that function with the $x$-axis and $x=u$ and the the area that is bounded by $f_2$ with endpoints the intersection point of that function with the $x$-axis and $x=u$ and then we subtract these two areas?
But in that way we haven't taken into consideration that the area has to be bounded by the $x$-axis, do we?
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Given a positive integer $n$ which is not a perfect square, it is well-known that Pell's equation $a^2 - nb^2 = 1$ is always solvable in non-zero integers $a$ and $b$.
Question:Let $n$ be a positive integer which is not a perfect square. Is there always a polynomial $D \in \mathbb{Z}[x]$ of degree $2$, an integer $k$ and nonzero polynomials $P, Q \in \mathbb{Z}[x]$ such that $D(k) = n$ and $P^2 - DQ^2 = 1$, where $a = P(k)$, $b = Q(k)$ is the fundamental solution of the equation $a^2 - nb^2 = 1$?
If yes, is there an upper bound on the degree of the polynomials $P$ and $Q$ -- and if so, is it even true that the degree of $P$ is always $\leq 6$?
Example: Consider $n := 13$.Putting $D_1 := 4x^2+4x+5$ and $D_2 := 25x^2-14x+2$, we have $D_1(1) = D_2(1) = 13$.Now the fundamental solutions of the equations$P_1^2 - D_1Q_1^2 = 1$ and $P_2^2 - D_2Q_2^2 = 1$ are given by
$P_1 := 32x^6+96x^5+168x^4+176x^3+120x^2+48x+9$,
$Q_1 := 16x^5+40x^4+56x^3+44x^2+20x+4$
and
$P_2 := 1250x^2-700x+99$,
$Q_2 := 250x-70$,
respectively. Therefore $n = 13$ belongs to at least $2$ different series whose solutions have ${\rm deg}(P) = 6$ and ${\rm deg}(P) = 2$, respectively.
Examples for all non-square $n \leq 150$ can be found here.
Added on Feb 3, 2015: All what remains to be done in order to turnLeonardo's answers into a complete answer to the question is to find out which valuesthe index of the group of units of $\mathbb{Z}[\sqrt{n}]$ in the group of unitsof the ring of integers of the quadratic field $\mathbb{Q}(\sqrt{n})$can take. This part is presumably not even really MO level, but it's justnot my field -- maybe someone knows the answer? Added on Feb 14, 2015: As nobody has completed the answer so far, it seemsthis may be less easy than I thought on a first glance. Added on Feb 17, 2015: Leonardo Zapponi has given now a complete answer tothe question in this note.
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Case I: The 2 qubits are not entangled.
You can write the states of the two qubits (say $\mathrm{A}$ and $\mathrm{B}$) as $|\psi_\mathrm{A}\rangle=a|0\rangle+b|1\rangle$ and $|\psi_\mathrm{B}\rangle = c|0\rangle+d|1\rangle$ where $a,b,c,d\in\Bbb{C}$.
The individual qubits reside in two dimensional complex vector spaces $\Bbb{C}^2$ (over a $\Bbb{C}$ field). But the state of the system is a
vector (or point) residing in a four dimensional complex vector space $\Bbb{C}^4$(over a $\Bbb {C}$ field).
The state of the system can be written as a tensor product $|\psi_\mathrm{A}\rangle\otimes|\psi_\mathrm{B}\rangle$ i.e. $ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle$.
Naturally, $|ac|^2+|ad|^2+|bc|^2+|bd|^2=1$ since the state vector has to be normalized. The reason as to why the square of the amplitude of a basis state gives the probability of that basis state occurring when measured in the corresponding basis lies in the Born's rule of quantum mechanics (some physicists consider it to be a basic postulate of quantum mechanics). Now, probability of $|0\rangle$ occuring when the first qubit is measured is $|ac|^2+|ad|^2$. Similarly, probability of $|1\rangle$ occuring when the first qubit is measured is $|bc|^2+|bd|^2$.
Now, what happens if we apply a quantum gate without performing any measurement on the previous state of the system? Quantum gates are unitary gates. Their action can be written as action of an unitary operator $U$ on the initial state of the system i.e. $ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle$ to produce a new state $A|00\rangle+B|01\rangle+C|10\rangle+D|11\rangle$ (where $A,B,C,D\in\Bbb{C}$). The magnitude of this new state vector: $|A|^2+|B|^2+|C|^2+|D|^2$ again equates to $1$, since the applied gate was
unitary. When the first qubit is measured, probability of $|0\rangle$ occurring is $|A|^2+|B|^2$ and similarly you can find it for occurrence of $|1\rangle$.
But if we did perform a measurement, before the action of the unitary gate the result would be different. For example of you had measured the first qubit and it turned out to be in $|0\rangle$ state the intermediate state of the system would have
collapsed to $\frac{ac|00\rangle + ad|01\rangle}{\sqrt{(ac)^2+(ad)^2}}$ (according to the Copenhagen interpretation). So you can understand that applying the same quantum gate on this state would have given a different final result. Case II: The 2 qubits are entangled.
In case the state of the system is something like $\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle$ , you cannot represent it as a tensor product of states of two individual qubits (try!). There are plenty more such examples. The qubits are said to entangled in such a case.
Anyway, the basic logic still remains same. The probability of $|0\rangle$ occuring when the first qubit is measured is $|1/\sqrt{2}|^2=\frac{1}{2}$ and $|1\rangle$ occuring is $\frac{1}{2}$ too. Similarly you can find out the probabilities for measurement of the second qubit.
Again if you apply a unitary quantum gate on this state, you'd end up with something like $A|00\rangle+B|01\rangle+C|10\rangle+D|11\rangle$, as before. I hope you can now yourself find out the probabilities of the different possibilities when the first and second qubits are measured.
Note: Normally the basis states of the 2-qubit sytem $|00\rangle,|01\rangle,|10\rangle,|11\rangle$ are considered as the four $4\times 1$ column vectors like $\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$, etc. by mapping the four basis vectors to the standard basis of $\Bbb{R}^4$. And, the unitary transformations $U$ can be written as $4\times 4$ matrices which satisfy the property $UU^{\dagger}=U^{\dagger}U=I$.
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Following Qiaochu hint, I'll try to elaborate a bit.
Note: This is a complete rewrite of the proof to fix a flaw pointed out by Qiaochu and make it overall clearer.
First some notation. Let $S^1 = [0, 2\pi)$ with 0 and $2\pi$ identified. For two points $a, b \in S^1$ we'll denote by $a \oplus b = a + b \mod 2\pi$ (and likewise for $\ominus$) and let $A_n = \{n \mod 2\pi\}_{n=1}^{\infty}$. Also denote by $U_{\varepsilon}(p)$ a punctured $\varepsilon$-neighborhood of $p \in S^1$.
Our strategy will be to show that $\{A_n\}$ has at least one limit point in $S^1$. From this we will conclude that 0 is also a limit point and finally we will use this fact to show that every point in $S_1$ is a limit point of $A_n$ (this means that $\{A_n\}$ is dense in S^1). Having established this, we will use continuity of $\sin$ to finally resolve the problem.
Proof
As a preparation we will note the relation $A_{n+m} = A_n \oplus A_m \quad (1)$. As a corollary of this we have that the sequence $\{A_n\}$ is injective (for if not, there would exist $n, m \in \mathbb{N}, k \in \mathbb{Z} \quad n>m$ such that $A_n = A_m$ and so $n - m = 2 k \pi$, a contradiction with irrationality of $\pi$). This implies simple but crucial fact that the image of the sequence contains infinitely many points.
Now, to establish the density of ${A_n}$ in $S^1$ we will first show that there exists at least one limit point $p \in S^1$. This is established by a standard argument: consider intervals $I = [0, \pi]$ and $J = [\pi, 2\pi]$ and take the one which contains infinitely many points of $\{A_n\}$ (if both do, take the "bottom" one, i.e. $I$). Call this interval $I_0$. Now again divide it in two intervals of same length and let $I_1$ be the one with infinitely many points. Continue in this way to obtain an infinite sequance of intervals $I_0 \supset I_1 \supset \cdots$. Then the set $K = \cap_{n=0}^{\infty} I_n$ is non-empty (this should be covered in standard calculus course, I hope) and any point $p \in K$ is surely a limit point (by construction of $\{I_n\}$).
Thanks to the above we now know that for each $\varepsilon > 0$ there exist infinitely many points in $U_\varepsilon(p)$, i.e. infinitely many $n, m \quad n > m$ such that $|A_n - A_m - 2k\pi| < 2\varepsilon$ for some $k \in \mathbb{Z}$. But from (1) we get that $|A_{n-m} - 2k\pi| < 2\varepsilon$. Because we identify 0 and $2\pi$ in $S^1$ we can see that these differences are concentrated around 0. Therefore we have shown that 0 is also a limit point.
Now we will show that every $p \in S^1$ is a limit point. Suppose we are given $\varepsilon > 0$. We just take any $A_k$ from $U_{\varepsilon}(0)$ and note that for $l = \lfloor {p \over A_k} \rfloor$ we have $A_{lk} \in U_{\varepsilon}(p)$.
To conclude the answer we will show that any $r \in [-1, 1]$ is indeed a limit point of $\{\sin(n)\}$. Take $s = \arcsin (r) \mod 2\pi$. Then by the above, we know that $s$ is a limit point of $\{A_n\}$ so there is a subsequence $\{A_{n_k}\} \to s$ and observe by continuity of $\sin$ that $\{\sin(A_{n_k})\} = \{\sin(n_k)\} \to sin(s) = r$.
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Remarkable cardinal
Remarkable cardinals were introduced by Schinder in [1] to provide precise consistency strength of the statement that $L(\mathbb R)$ cannot be modified by proper forcing.
Contents Definitions
A cardinal $\kappa$ is remarkable if for each regular $\lambda>\kappa$, there exists a countable transitive $M$ and an elementary embedding $e:M\rightarrow H_\lambda$ with $\kappa\in \text{ran}(e)$ and also a countable transitive $N$ and an elementary embedding $\theta:M\to N$ such that:
the critical point of $\theta$ is $e^{-1}(\kappa)$, $\text{Ord}^M$ is a regular cardinal in $N$, $M=H^N_{\text{Ord}^M}$, $\theta(e^{-1}(\kappa))>\text{Ord}^M$.
Remarkable cardinals could be called virtually supercompact, because the following alternative definition is an exact analogue of the definition of supercompact cardinals by Magidor [Mag71]:
A cardinal $κ$ is remarkable iff for every $η > κ$, there is $α < κ$ such that in a set-forcing extension there is an elementary embedding $j : V_α → V_η$ with $j(\mathrm{crit}(j)) = κ$.[2]
Equivalently (theorem 2.4[3])
For every $η > κ$ and every $a ∈ V_η$, there is $α < κ$ such that in $V^{Coll(ω,<κ)}$ there is an elementary embedding $j : V_α → V_η$ with $j(crit(j)) = κ$ and $a ∈ range(j)$. For every $η > κ$ in $C^{(1)}$ and every $a ∈ V_η$, there is $α < κ$ also in $C^{(1)}$ such that in $V^{Coll(ω,<κ)}$ there is an elementary embedding $j : V_α → V_η$ with $j(crit(j)) = κ$ and $a ∈ range(j)$. There is a proper class of $η > κ$ such that for every $η$ in the class, there is $α < κ$ such that in $V^{Coll(ω,<κ)}$ there is an elementary embedding $j : V_α → V_η$ with $j(crit(j)) = κ$
Note: the existence of any such elementary embedding in $V^{Coll(ω,<κ)}$ is equivalent to the existence of such elementary embedding in any forcing extension (see Elementary_embedding#Absoluteness).[3].
Results
Remarkable cardinals and the constructible universe:
Remarkable cardinals are downward absolute to $L$. [1] If $0^\sharp$ exists, then every Silver indiscernible is remarkable in $L$. [1]
Relations with other large cardinals:
Strong cardinals are remarkable. [1] A $2$-iterable cardinal implies the consistency of a remarkable cardinal: Every $2$-iterable cardinal is a limit of remarkable cardinals. [4] Remarkable cardinals imply the consistency of $1$-iterable cardinals: If there is a remarkable cardinal, then there is a countable transitive model of ZFC with a proper class of $1$-iterable cardinals. [4] Remarkable cardinals are totally indescribable. [1] Remarkable cardinals are totally ineffable. [1] Virtually extendible cardinals are remarkable limits of remarkable cardinals.[2] If $κ$ is virtually measurable, then either $κ$ is remarkable in $L$ or $L_κ \models \text{“there is a proper class of virtually measurables”}$.[5] Remarkable cardinals are strategic $ω$-Ramsey limits of $ω$-Ramsey cardinals.[5] Remarkable cardinals are $Σ_2$-reflecting.[6]
Relation to various set-theoretic principles:
Equiconsistency with the weak Proper Forcing Axiom:[3] If there is a remarkable cardinal, then $\text{wPFA}$ holds in a forcing extension by a proper poset. If $\text{wPFA}$ holds, then $ω_2^V$ is remarkable in $L$. It is relatively consistent that ZFC and the generic Vopěnka scheme holds, yet $Ord$ is not definably Mahlo and not even $∆_2$-Mahlo. In such a model, there can be no $Σ_2$-reflecting cardinals and therefore also no remarkable cardinals:[7] If $0^♯$ exists, then there is a class-forcing extension $L[G]$ of the constructible universe in which the generic Vopěnka principle holds (so $gVP(κ, \mathbf{Σ_{n+1}})$ and $gVP(Π_n)$ hold for any $κ$ and $n$), but there are no $Σ_2$-reflecting cardinals and hence no remarkable cardinals (or $n$-remarkable cardinals). Weakly remarkable cardinals
(this section from [6])
A cardinal $κ$ is weakly remarkable iff for every $η > κ$, there is $α$ such that in a set-forcing extension there is an elementary embedding $j : V_α → V_η$ with $j(\mathrm{crit}(j)) = κ$. (the condition $α < κ$ is dropped)
A cardinal is remarkable iff it is weakly remarkable and $Σ_2$-reflecting.
The existence of non-remarkable weakly remarkable cardinals is equiconsistent to the existence of an $ω$-Erdős cardinal (equivalent assuming $V=L$; Baumgartner definition of $ω$-Erdős cardinals):
Every $ω$-Erdős cardinal is a limit of non-remarkable weakly remarkable cardinals. If $κ$ is a non-remarkable weakly remarkable cardinal, then some ordinal greater than $κ$ is an $ω$-Erdős cardinal in $L$. $n$-remarkable cardinals
$1$-remarkability is equivalent to remarkability. A cardinal is virtually $C^{(n)}$-extendible iff it is $n + 1$-remarkable (virtually extendible cardinals are virtually $C^{(1)}$-extendible). A cardinal is called
completely remarkable iff it is $n$-remarkable for all $n > 0$. Other definitions and properties in Extendible#Virtually extendible cardinals.[3]
References Schindler, Ralf-Dieter. Proper forcing and remarkable cardinals.Bull Symbolic Logic 6(2):176--184, 2000. www DOI MR bibtex Gitman, Victoria and Shindler, Ralf. Virtual large cardinals.www bibtex Bagaria, Joan and Gitman, Victoria and Schindler, Ralf. Generic {V}opěnka's {P}rinciple, remarkable cardinals, and the weak {P}roper {F}orcing {A}xiom.Arch Math Logic 56(1-2):1--20, 2017. www DOI MR bibtex Gitman, Victoria and Welch, Philip. Ramsey-like cardinals II.J Symbolic Logic 76(2):541--560, 2011. www arχiv MR bibtex Nielsen, Dan Saattrup and Welch, Philip. Games and Ramsey-like cardinals., 2018. arχiv bibtex Wilson, Trevor M. Weakly remarkable cardinals, Erdős cardinals, and the generic Vopěnka principle., 2018. arχiv bibtex Gitman, Victoria and Hamkins, Joel David. A model of the generic Vopěnka principle in which the ordinals are not Mahlo., 2018. arχiv bibtex
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@DavidReed the notion of a "general polynomial" is a bit strange. The general polynomial over a field always has Galois group $S_n$ even if there is not polynomial over the field with Galois group $S_n$
Hey guys. Quick question. What would you call it when the period/amplitude of a cosine/sine function is given by another function? E.g. y=x^2*sin(e^x). I refer to them as variable amplitude and period but upon google search I don't see the correct sort of equation when I enter "variable period cosine"
@LucasHenrique I hate them, i tend to find algebraic proofs are more elegant than ones from analysis. They are tedious. Analysis is the art of showing you can make things as small as you please. The last two characters of every proof are $< \epsilon$
I enjoyed developing the lebesgue integral though. I thought that was cool
But since every singleton except 0 is open, and the union of open sets is open, it follows all intervals of the form $(a,b)$, $(0,c)$, $(d,0)$ are also open. thus we can use these 3 class of intervals as a base which then intersect to give the nonzero singletons?
uh wait a sec...
... I need arbitrary intersection to produce singletons from open intervals...
hmm... 0 does not even have a nbhd, since any set containing 0 is closed
I have no idea how to deal with points having empty nbhd
o wait a sec...
the open set of any topology must contain the whole set itself
so I guess the nbhd of 0 is $\Bbb{R}$
Btw, looking at this picture, I think the alternate name for these class of topologies called British rail topology is quite fitting (with the help of this WfSE to interpret of course mathematica.stackexchange.com/questions/3410/…)
Since as Leaky have noticed, every point is closest to 0 other than itself, therefore to get from A to B, go to 0. The null line is then like a railway line which connects all the points together in the shortest time
So going from a to b directly is no more efficient than go from a to 0 and then 0 to b
hmm...
$d(A \to B \to C) = d(A,B)+d(B,C) = |a|+|b|+|b|+|c|$
$d(A \to 0 \to C) = d(A,0)+d(0,C)=|a|+|c|$
so the distance of travel depends on where the starting point is. If the starting point is 0, then distance only increases linearly for every unit increase in the value of the destination
But if the starting point is nonzero, then the distance increases quadratically
Combining with the animation in the WfSE, it means that in such a space, if one attempt to travel directly to the destination, then say the travelling speed is 3 ms-1, then for every meter forward, the actual distance covered by 3 ms-1 decreases (as illustrated by the shrinking open ball of fixed radius)
only when travelling via the origin, will such qudratic penalty in travelling distance be not apply
More interesting things can be said about slight generalisations of this metric:
Hi, looking a graph isomorphism problem from perspective of eigenspaces of adjacency matrix, it gets geometrical interpretation: question if two sets of points differ only by rotation - e.g. 16 points in 6D, forming a very regular polyhedron ...
To test if two sets of points differ by rotation, I thought to describe them as intersection of ellipsoids, e.g. {x: x^T P x = 1} for P = P_0 + a P_1 ... then generalization of characteristic polynomial would allow to test if our sets differ by rotation ...
1D interpolation: finding a polynomial satisfying $\forall_i\ p(x_i)=y_i$ can be written as a system of linear equations, having well known Vandermonde determinant: $\det=\prod_{i<j} (x_i-x_j)$. Hence, the interpolation problem is well defined as long as the system of equations is determined ($\d...
Any alg geom guys on? I know zilch about alg geom to even start analysing this question
Manwhile I am going to analyse the SR metric later using open balls after the chat proceed a bit
To add to gj255's comment: The Minkowski metric is not a metric in the sense of metric spaces but in the sense of a metric of Semi-Riemannian manifolds. In particular, it can't induce a topology. Instead, the topology on Minkowski space as a manifold must be defined before one introduces the Minkowski metric on said space. — baluApr 13 at 18:24
grr, thought I can get some more intuition in SR by using open balls
tbf there’s actually a third equivalent statement which the author does make an argument about, but they say nothing about substantive about the first two.
The first two statements go like this : Let $a,b,c\in [0,\pi].$ Then the matrix $\begin{pmatrix} 1&\cos a&\cos b \\ \cos a & 1 & \cos c \\ \cos b & \cos c & 1\end{pmatrix}$ is positive semidefinite iff there are three unit vectors with pairwise angles $a,b,c$.
And all it has in the proof is the assertion that the above is clearly true.
I've a mesh specified as an half edge data structure, more specifically I've augmented the data structure in such a way that each vertex also stores a vector tangent to the surface. Essentially this set of vectors for each vertex approximates a vector field, I was wondering if there's some well k...
Consider $a,b$ both irrational and the interval $[a,b]$
Assuming axiom of choice and CH, I can define a $\aleph_1$ enumeration of the irrationals by label them with ordinals from 0 all the way to $\omega_1$
It would seemed we could have a cover $\bigcup_{\alpha < \omega_1} (r_{\alpha},r_{\alpha+1})$. However the rationals are countable, thus we cannot have uncountably many disjoint open intervals, which means this union is not disjoint
This means, we can only have countably many disjoint open intervals such that some irrationals were not in the union, but uncountably many of them will
If I consider an open cover of the rationals in [0,1], the sum of whose length is less than $\epsilon$, and then I now consider [0,1] with every set in that cover excluded, I now have a set with no rationals, and no intervals.One way for an irrational number $\alpha$ to be in this new set is b...
Suppose you take an open interval I of length 1, divide it into countable sub-intervals (I/2, I/4, etc.), and cover each rational with one of the sub-intervals.Since all the rationals are covered, then it seems that sub-intervals (if they don't overlap) are separated by at most a single irrat...
(For ease of construction of enumerations, WLOG, the interval [-1,1] will be used in the proofs) Let $\lambda^*$ be the Lebesgue outer measure We previously proved that $\lambda^*(\{x\})=0$ where $x \in [-1,1]$ by covering it with the open cover $(-a,a)$ for some $a \in [0,1]$ and then noting there are nested open intervals with infimum tends to zero.
We also knew that by using the union $[a,b] = \{a\} \cup (a,b) \cup \{b\}$ for some $a,b \in [-1,1]$ and countable subadditivity, we can prove $\lambda^*([a,b]) = b-a$. Alternately, by using the theorem that $[a,b]$ is compact, we can construct a finite cover consists of overlapping open intervals, then subtract away the overlapping open intervals to avoid double counting, or we can take the interval $(a,b)$ where $a<-1<1<b$ as an open cover and then consider the infimum of this interval such that $[-1,1]$ is still covered. Regardless of which route you take, the result is a finite sum whi…
W also knew that one way to compute $\lambda^*(\Bbb{Q}\cap [-1,1])$ is to take the union of all singletons that are rationals. Since there are only countably many of them, by countable subadditivity this give us $\lambda^*(\Bbb{Q}\cap [-1,1]) = 0$. We also knew that one way to compute $\lambda^*(\Bbb{I}\cap [-1,1])$ is to use $\lambda^*(\Bbb{Q}\cap [-1,1])+\lambda^*(\Bbb{I}\cap [-1,1]) = \lambda^*([-1,1])$ and thus deducing $\lambda^*(\Bbb{I}\cap [-1,1]) = 2$
However, what I am interested here is to compute $\lambda^*(\Bbb{Q}\cap [-1,1])$ and $\lambda^*(\Bbb{I}\cap [-1,1])$ directly using open covers of these two sets. This then becomes the focus of the investigation to be written out below:
We first attempt to construct an open cover $C$ for $\Bbb{I}\cap [-1,1]$ in stages:
First denote an enumeration of the rationals as follows:
$\frac{1}{2},-\frac{1}{2},\frac{1}{3},-\frac{1}{3},\frac{2}{3},-\frac{2}{3}, \frac{1}{4},-\frac{1}{4},\frac{3}{4},-\frac{3}{4},\frac{1}{5},-\frac{1}{5}, \frac{2}{5},-\frac{2}{5},\frac{3}{5},-\frac{3}{5},\frac{4}{5},-\frac{4}{5},...$ or in short:
Actually wait, since as the sequence grows, any rationals of the form $\frac{p}{q}$ where $|p-q| > 1$ will be somewhere in between two consecutive terms of the sequence $\{\frac{n+1}{n+2}-\frac{n}{n+1}\}$ and the latter does tends to zero as $n \to \aleph_0$, it follows all intervals will have an infimum of zero
However, any intervals must contain uncountably many irrationals, so (somehow) the infimum of the union of them all are nonzero. Need to figure out how this works...
Let's say that for $N$ clients, Lotta will take $d_N$ days to retire.
For $N+1$ clients, clearly Lotta will have to make sure all the first $N$ clients don't feel mistreated. Therefore, she'll take the $d_N$ days to make sure they are not mistreated. Then she visits client $N+1$. Obviously the client won't feel mistreated anymore. But all the first $N$ clients are mistreated and, therefore, she'll start her algorithm once again and take (by suposition) $d_N$ days to make sure all of them are not mistreated. And therefore we have the recurence $d_{N+1} = 2d_N + 1$
Where $d_1$ = 1.
Yet we have $1 \to 2 \to 1$, that has $3 = d_2 \neq 2^2$ steps.
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Performs Infinite Impulse Response digital filtering on supplied vector. This function performs the operation:
\[ y_n = x_n c_0 + y_{n-1}c_1 + y_{n-2}c_2 \dots \]
???MATH???x???MATH??? is the input vector (argument 1) ???MATH???c???MATH??? is the coefficient vector (argument 2) ???MATH???y???MATH??? is the result (returned value)
The operation of this function (and also the function FIR ) is simple but its application can be the subject of several volumes! In principle an almost unlimited range of IIR filtering operations may be performed using this function. Any text on Digital Signal Processing will provide further details.
User's should note that using this function applied to raw transient analysis data will not produce meaningful results as the values are unevenly spaced. If you apply this function to simulation data, you must either specify that the simulator outputs at fixed intervals (select the Output at .PRINT step option in the Simulator > Choose Analysis... dialog box) or you must interpolate the results using the function Interp .
Number
Type
Compulsory
Default
Description
1 real array Yes Vector to be filtered 2 real array Yes Coefficients 3 real array No zero Initial conditions
Return type: real array
The following graph shows the result of applying a simple first order IIR filter to a step: The coefficients used give a time constant of 10 * the sample interval. In the above the sample interval was ???MATH???1\mu???MATH???Sec so giving a ???MATH???10\mu???MATH???Sec time constant. As can be seen a first order IIR filter has exactly the same response as an single pole RC network. A general first order function is: \[ y_n = x_nc_0 + y_{n-1}c_1 \] where ???MATH???c_0 = 1 - \exp\left(\frac{-T}{\tau}\right)???MATH??? and ???MATH???c_1 = \exp\left(\frac{-T}{\tau}\right)???MATH??? and ???MATH???\tau = \text{time constant}???MATH??? and ???MATH???T = \text{sample interval}???MATH??? The above example is simple but it is possible to construct much more complex filters using this function. While it is also possible to place analog representations on the circuit being simulated, use of the IIR function permits viewing of filtered waveforms after a simulation run has completed. This is especially useful if the run took a long time to complete.
▲Function Summary▲ ◄ IffV im ▶
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Let's consider a sequence of functions $f_n:[a,b] \rightarrow \mathbb{R}, f_n(x)=e^{-n|1-sin(x)|}$. Show that $f_n$ converges to $f=0$ in measure.
Attempt/Thoughts: To prove $f_n$ converges to $f=0$ in measure, I have to show that for every $\epsilon >0$, $\mu (x:|e^{-n|1-sin(x)|}| \geq \epsilon) \rightarrow 0$ as $n \rightarrow \infty$. This is clear to me pictorially as seen when we graph the sequence of functions, as n gets larger and larger, the set of $x$ for which $f_n(x)$ is positive grows smaller and smaller. The peak grows narrower and narrower. Thus for me, it is clear pictorially that $\mu (x:|f_n(x)-f(x)| \geq \epsilon) \rightarrow 0$ as $n \rightarrow \infty$. exactly sure how to prove this formally, any
However, I'm not exactly sure how to prove this formally/ where to begin.
I do know that $0 \leq |e^{-n|1-sin(x)|}|=|\frac{1}{e^{n|1-sin(x)|}}| \leq |1|$, but this doesn't show convergence in measure. Any help would be much appreciated, thanks.
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Now showing items 1-10 of 18
J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-02)
Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Centrality, rapidity and transverse momentum dependence of the J/$\psi$ suppression in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2014-06)
The inclusive J/$\psi$ nuclear modification factor ($R_{AA}$) in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76TeV has been measured by ALICE as a function of centrality in the $e^+e^-$ decay channel at mid-rapidity |y| < 0.8 ...
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2014-01)
In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured ...
Multi-strange baryon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2014-01)
The production of ${\rm\Xi}^-$ and ${\rm\Omega}^-$ baryons and their anti-particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been measured using the ALICE detector. The transverse momentum spectra at ...
Measurement of charged jet suppression in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2014-03)
A measurement of the transverse momentum spectra of jets in Pb–Pb collisions at √sNN = 2.76TeV is reported. Jets are reconstructed from charged particles using the anti-kT jet algorithm with jet resolution parameters R ...
Two- and three-pion quantum statistics correlations in Pb-Pb collisions at √sNN = 2.76 TeV at the CERN Large Hadron Collider
(American Physical Society, 2014-02-26)
Correlations induced by quantum statistics are sensitive to the spatiotemporal extent as well as dynamics of particle-emitting sources in heavy-ion collisions. In addition, such correlations can be used to search for the ...
Exclusive J /ψ photoproduction off protons in ultraperipheral p-Pb collisions at √sNN = 5.02TeV
(American Physical Society, 2014-12-05)
We present the first measurement at the LHC of exclusive J/ψ photoproduction off protons, in ultraperipheral proton-lead collisions at √sNN=5.02 TeV. Events are selected with a dimuon pair produced either in the rapidity ...
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So this is a pension framework. I am trying to code a system and I don't want to have to brute force this answer, but I can't figure out a clean solution.
$$Fund = \sum_{i=1}^t [\cfrac{I\cdot e^{\frac{\pi i}{12K}}}{12K} \cdot C \cdot e^{\frac{Ri}{12K}}]$$
$I = $, annual income, $K = $ pay periods per month, $C =$ Contribution Rate (%), $R =$ expected annualized return (continuous), $\pi =$ expected annual income growth (continuous)
Solving for the derivatives:
$$ \cfrac{dFund}{dC} = \sum_{i=1}^t [\cfrac{I\cdot e^{\frac{\pi i}{12K}}}{12K} \ \cdot e^{\frac{Ri}{12K}}]$$
$$\cfrac{dFund}{dR} = \sum_{i=1}^t [\cfrac{I\cdot e^{\frac{\pi i}{12K}}}{12K} \cdot C \cdot e^{\frac{Ri}{12K}} \cdot \frac{i}{12K}]$$
If $\Delta C = 0.01$, $\Delta Fund_{C} = \Delta C \cdot \cfrac{dFund}{dC}$
How do I solve for $\Delta R$ if I want $\Delta R \cdot \cfrac{dFund}{dR} = \Delta Fund_{C} = \Delta C \cdot \cfrac{dFund}{dC}$?
Basically, is there a way to extract the value of the $\cfrac{i}{12K}$ term within the summation so that it can be expressed outside the summation?
Edit:
The goal is that by doing so, the problem would easily simplify to $\Delta R \cdot C \cdot \Sigma \frac{i}{12K} \cdot \frac{dFund}{dC} = \Delta Fund_C$, such that I could just solve for $\Delta R = (C \cdot \Sigma \frac{i}{12K})^{-1}$. Currently I using my code to calculate $\Delta Fund_R$ for a large sequence of $\Delta R$ values and then matching the closest $\Delta Fund_R$ to $\Delta Fund_C$. Incredibly inefficient from a resource standpoint.
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Given triples of $n$ floating point values
$$(\min_1, \max_1, w_1), \dots, (\min_n, \max_n, w_n)$$
and a value $V$, what is a good algorithhm to assign values $v_i$ to each of the triples such that the following conditions hold?
$\min_i \le v_i \le \max_i$. $\displaystyle\sum_{i=1}^n v_i = V$. $\dfrac{v_i}{V}$ is as close as possible to $\dfrac{w_i}{W}$, where $W = \displaystyle\sum_{i=1}^n w_i$, i.e., minimize $\displaystyle\sum_{i=1}^n \left| \frac{v_i}{V} - \frac{w_i}{W} \right|$.
I understand the above can be solved via an LP solver but am looking for an algorithm that may not return the optimal assignment but that is deterministic, returns close to the optimum solutions, runs in $O(n)$ time, and returns solutions that are stable in the sense that if two instances of the problem definition are "close to each other" then the solutions tend to be close to each other.
It seems like there should be a greedy approach the performs sufficiently in which the first step is assigning the minima to $v_i$ and then proportioning out the "slack" but I am not sure how to deal with the maxima while doing this.
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Chinese Remainder Theorem =========================
Suppose are positive integers and coprime in pair. For any sequence of integers , there exists an integer x solving the following system of congruence equations:
There exists an unique modulo solution of the system of simultaneous congruences above:
in which:
M &= m_1 \cdots m_k \\ M_1 &= \frac{M}m_1 , \cdots, M_k = \frac{M}m_k \\ y_1 &\equiv (M_1)^{-1} \pmod{m_1}, \cdots , y_k\equiv (M_k)^{-1}\pmod{m_k} \end{aligned}" />
Continue reading Basic concepts of Chinese Remainder Theorem with respect of RSA/AES
THRESHOLD SIGNATURE SCHEME Introduction
Assuming there are 20 employees in a company, and if each employee has his or her own copy of the secret key then it is hard to assure on individuals due to compromise and machine break down. In the other hand, if there is a valid signature requires all 20 employees’ signature in the company then it will be very secure but not be easy to use. Therefore we can implement a scheme which requires only sign 5 or more out of 20 employees then it will be valid and that is exactly what a (5,20) threshold signature scheme tries to achieve. In addition, if a threat agent wants to compromise the system and obtain a message, he must compromise at least 5 people in the scheme and that is a harder thing to do compared to a traditional public scheme.
Continue reading Introduction to Threshold signature scheme
Deciphering
Ciphertext: “ VaqrprzoreoeratraWhyvhfraJnygreUbyynaqreqrgjrroebrefinaRqvguZnetbganne
NzfgreqnzNaarjvytenntzrrxbzraznnezbrgabtrrarragvwqwrovwbzn
oyvwiraBznmnyurgzbrvyvwxuroorabzNaarabtrracnnejrxraqnnegrubhqrafpuev
wsgRqvguSenaxvarraoevrsnnaTregehqAnhznaauhaiebrtrerohhezrvfwrvaSenaxs hegnzZnva”
The given ciphertext has only letters without space, punctuation or separated key, there are two classic cipher systems such as substitution cipher and transposition cipher which are known to be easy to attack by using frequency analysis or bruteforce techniques. Continue reading Deciphering Ceasar basic concept
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Let a (free) particle move in $[0,a]$ with cyclic boundary condition $\psi(0)=\psi(a)$. The solution of the Schrödinger-equation can be put in the form of a plane wave. In this state the standard deviation of momentum is $0$, but $\sigma_x$ must be finite. So we find that $\sigma_x\sigma_p=0$. Is something wrong with the uncertainty principle?
This is what happens if one cares not for the subtlety that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space.
Let's set $a=1$ for convenience. The operator $p =-\mathrm{i}\hbar\partial_x$ acting on wavefunctions with periodic boundary conditions defined on $D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land \psi'\in L^2([0,1])\}$ is self-adjoint, that is, on the domain of definition of $p$, we have $p=p^\dagger$, and $p^\dagger$ admits the same domain of definition. The self-adjointness of $p$ follows from the periodic boundary conditions killing the surface terms that appear in the $L^2$ inner product $$\langle \phi,p\psi\rangle - \langle p^\dagger \phi,\psi\rangle = \int\overline{\phi(x)}\mathrm{i}\hbar\partial_x\psi(x) - \overline{\mathrm{i}\hbar\partial_x\phi(x)}\psi(x) = 0$$ for every $\psi\in D(p)$ and every $\phi\in D(p^\dagger) = D(p)$, but not for $\phi$ with $\phi(0)\neq\phi(1)$.
Now, for the question of the commutator: the multplication operator $x$ is defined on the entire Hilbert space, since for $\psi\in L^2([0,1])$ $x\psi$ is also square-integrable. For the product of two operators $A,B$, we have the rule$$ D(AB) = \{\psi\in D(B)\mid B\psi\in D(A)\}$$and $$ D(A+B) = D(A)\cap D(B)$$so we obtain\begin{align}D(px) & = \{\psi\in L^2([0,1])\mid x\psi\in D(p)\} \\D(xp) & = D(p)\end{align}and $x\psi\in D(p)$ means $0\cdot \psi(0) = 1\cdot\psi(1)$, that is, $\psi(1) = 0$. Hence we have$$ D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1]) \land \psi(1) = 0\}$$and finally$$ D([x,p]) = D(xp)\cap D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1])\land \psi(0)=\psi(1) = 0\}$$meaning the plane waves $\psi_{p_0}$ do
not belong to the domain of definition of the commutator $[x,p]$ and you cannot apply the naive uncertainty principle to them. However, for self-adjoint operators $A,B$, you may rewrite the uncertainty principle as$$ \sigma_\psi(A)\sigma_\psi(B)\geq \frac{1}{2} \lvert \langle \psi,\mathrm{i}[A,B]\rangle\psi\rvert = \frac{1}{2}\lvert\mathrm{i}\left(\langle A\psi,B\psi\rangle - \langle B\psi,A\psi\rangle\right)\rvert$$where the r.h.s. and l.h.s. are now both defined on $D(A)\cap D(B)$. Applying this version to the plane waves yields no contradiction.
Notice that $\psi(x)$ is defined on a circle of circumference $a$. Multiplying $x$ on this circle is really multiplying a periodic extension of $x$, i.e., the sawtooth function $x - a\lfloor x/a\rfloor$, where $\lfloor y\rfloor$ means the largest integer not greater than $y$. So, the commutator of the position and momentum operators involves the derivative of not only $x$ but also the discontinuous part $-a\lfloor x/a\rfloor$. Therefore, \begin{equation} \sigma_{x} \sigma_p \geq \frac{1}{2}\Big|\langle \psi|\,[\hat{x},\hat{p}]\,|\psi\rangle\Big| = \frac{\hbar}{2}\Bigg|\Big\langle\psi\,\Big|\frac{d}{dx}\big(x - a\lfloor x/a\rfloor\big)\Big|\,\psi\Big\rangle\Bigg| = \frac{\hbar}{2}\Big|1-a|\psi(0)|^{2}\Big|. \end{equation} For a plane wave $\psi(x) = e^{ikx}/\sqrt{a}$, the above reduces to $\sigma_{x} \sigma_p\ge0$, as desired.
There are two ways to interpret the boundary conditions you are imposing.
The first case is that of a system which is infinite in extent, but has a periodic regularity. This is like an electron in an idealised 1D crystal, where the periodic boundary condition is imposed by the presence of nuclei regularly spaced. In this case, the plane wave solution has $\sigma_p$ = 0 but $\sigma_x$ is infinite.
The second case, is that of a particle in a ring. In this case, you can imagine the particle as being constrained within the ring by a infinitely deep potential well. The system is not actually 1D, it is 2D. Now you have to consider both $\sigma_x \sigma_{p_x}$ and $\sigma_y \sigma_{p_y}$, and even though $\sigma_x = \sigma_y \sim a$, the uncertainty in momentum will be imposed by the thickness of the ring. The plane wave solution will in fact represent angular momentum eigenstates.
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1,340 30
I know that ##\frac{\partial}{\partial (\partial_{\mu}\phi)} \big( \partial_{\mu} \phi\ \partial^{\mu} \phi \big) = \partial_{\mu} \phi##.
Now, I need to prove this to myself. So, here goes nothing. ##\frac{\partial}{\partial (\partial_{\mu}\phi)} \big( \partial_{\mu} \phi\ \partial^{\mu} \phi \big)## ## = \frac{\partial}{\partial (\partial_{\mu}\phi)} \big( \eta^{\mu\nu}\partial_{\mu} \phi\ \partial_{\nu} \phi \big)## ##= \eta^{\mu\nu}\ \partial_{\nu} \phi + \eta_{\mu\nu} \eta^{\mu\nu}\ \partial_{\mu} \phi##, where I first differentiated the factor ##\partial_{\mu}\phi## with respect to ##\partial_{\mu}\phi## and then I differentiated the factor ##\partial_{\nu}\phi## with respect to ##\partial_{\mu}\phi##. Am I correct so far?
Now, I need to prove this to myself.
So, here goes nothing.
##\frac{\partial}{\partial (\partial_{\mu}\phi)} \big( \partial_{\mu} \phi\ \partial^{\mu} \phi \big)##
## = \frac{\partial}{\partial (\partial_{\mu}\phi)} \big( \eta^{\mu\nu}\partial_{\mu} \phi\ \partial_{\nu} \phi \big)##
##= \eta^{\mu\nu}\ \partial_{\nu} \phi + \eta_{\mu\nu} \eta^{\mu\nu}\ \partial_{\mu} \phi##,
where I first differentiated the factor ##\partial_{\mu}\phi## with respect to ##\partial_{\mu}\phi## and then I differentiated the factor ##\partial_{\nu}\phi## with respect to ##\partial_{\mu}\phi##.
Am I correct so far?
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The impulse response and frequency response are two attributes that are useful for characterizing linear time-invariant (LTI) systems. They provide two different ways of calculating what an LTI system's output will be for a given input signal. A continuous-time LTI system is usually illustrated like this:In general, the system $H$ maps its input signal $x(...
Negative frequency doesn't make much sense for sinusoids, but the Fourier transform doesn't break up a signal into sinusoids, it breaks it up into complex exponentials (also called "complex sinusoids" or "cisoids"):$$F(\omega) = \int_{-\infty}^{\infty} f(t) \color{Red}{e^{- j\omega t}}\,dt$$These are actually spirals, spinning around in the complex plane:...
Let's say you had a spinning wheel. How would you describe how fast it is spinning? You'd probably say it's spinning at X revolutions per minute (rpm). Now how do you convey in what direction it's spinning with this number? It's the same X rpm if it's spinning clockwise or anti-clockwise. So you scratch your head and say oh well, here's a smart idea: I'll ...
FFT is actually not a great way of making a tuner. FFT has inherently a finite frequency resolution and it's not easy to detect very small frequency changes without making the time window extremely long which makes it unwieldy and sluggish.Better solutions can be based on phase-locked loops, delay-locked loops, auto correlation, zero crossing detection ...
Currently, my viewpoint (it is subject to change) is the followingFor sinusoidal repetition only positive frequencies makes sense. The physical interpretation is clear.For complex exponential repetition both positive and negative frequencies makes sense. It may be possible to attach a physical interpretation to negative frequency. That physical ...
Gabor filters are orientation-sensitive filters, used for texture analysis.The typically travel in packs, one for each direction. A gabor filter set with a given direction gives a strong response for locations of the target images that have structures in this given direction. For instance, if your target image is made of a periodic grating in a diagonal ...
The spectrum of a continuous tone is, as you said, of the form $\delta(f-f_0) + \delta(f+f_0)$: 2 impulses at frequencies $f_0$ and $-f_0$.As a lowpass signal, this is said to have bandwidth $f_0$ (the one-sided spectrum has components up to $f_0$).As a bandpass signal, it has zero bandwidth (there's nothing around the carrier frequency $f_0$).If you ...
Analog and digitalFirst off, you need to understand what an "analog" signal is, and what a "digital" signal is, how they are different, and how they are similar.The term "analog" comes from the old distinction between "analog" and "digital" computers. A "digital computer", even a very primitive one of decades ago, has always been more or less what we ...
I would do a normalized autocorrelation to determine periodicity. If it is periodic with period $P$ you should see peaks at every $P$ samples in the result. A normalized result of "1" implies perfect periodicity, "0" implies no periodicity at all at that period, and values in between imply imperfect periodicity. Subtract the data sequence's mean from the ...
One method that works if there's a relatively strong drum beat is to take the magnitude of the STFT of the waveform, and then auto-correlate it in only the time dimension. The peak of the auto-correlation function will be the beat, or a submultiple of it.This is equivalent to breaking up the signal into a lot of different frequency bands, finding the ...
Bang on something sharply once and plot how it responds in the time domain (as with an oscilloscope or pen plotter). That will be close to the impulse response.Get a tone generator and vibrate something with different frequencies. Some resonant frequencies it will amplify. Others it may not respond at all. Plot the response size and phase versus the ...
Figure 1.(c) shows the Test image reconstructed from MAGNITUDE spectrum only. We can say that the intensity values of LOW frequency pixels are comparatively more than HIGH frequency pixels.Actually, this is not correct. The phase values determine the shift in the sinusoid components of the image. With zero phase, all the sinusoids are centred at the same ...
TL;DR? Google Scholar for harmonic partial separation.A good starting point would be sinusoidal modeling techniques that separate the signal into sines+noise (deterministic and stochastic) components. The deterministic component, made up of sines, can be resynthesized convincingly:http://mtg.upf.edu/files/projectsweb/sms-piano-original.wavhttp://mtg....
The FFT can only be performed over a limited chunk of data. The basic math is based on the assumption that the time domain signal is periodic, i.e. your chunk of data is repeated in time. That typically results in a major discontinuity at the edges of the chunk. Let's look at a quick example: FFT size = 1000 points, Sample Rate = 1000 Hz, Frequency ...
An FFT reports spectrum frequency peak or peaks (quantized by FFT bin size), which is different from musical pitch. It's possible for the perceived pitch frequency to be completely missing from an FFT spectrum.Some of the simplest guitar tuners just used low-pass or band-pass filtering and measured the time between zero-crossings. The reciprocal gives a ...
suppose, Carrier signal frequency = 2800KHz message signal frequency = 3KHzThen you will get a signal that looks like this in the frequency plane.Obviously this is not to scale, but you get the idea.but what will happen if it is reversed ?i.e Message signal frequency = 2800KHz Carrie signal frequency = 3KHzplease explain would happen here ?Then ...
Posted for anyone who may find this useful...I created a picture that shows DFT frequency bin spacing for odd and even cases of N where N is the number of samples. FFTs usually operate on an even number of samples (the algorithm works by repeatedly breaking the problem into halves), so only the even case applies. The DC component (0*fs) is always part of ...
The impulse response is the response of a system to a single pulse of infinitely small duration and unit energy (a Dirac pulse). The frequency response shows how much each frequency is attenuated or amplified by the system.The frequency response of a system is the impulse response transformed to the frequency domain. If you have an impulse response, you ...
A Gabor filter is some parametrization of the idea of edges. This combines two somewhat contradictory ideas: an abrupt transition AND some fuzzy idea of where it is localized.It is mathematically a clever idea as it translates well in the Fourier domain: the Fourier transform of a Gabor is a Gaussian in Fourier space, and a Gaussian blob is the most ...
Two remarks:I am assuming you are plotting the real (or imaginary) part of the Fourier transform. It is much more common to work with the magnitude or squared magnitude (power spectrum).The peak in the spectrum is a very poor measure of fundamental frequency (pitch). Take a piano note at 440 Hz, apply a notch filter to it to remove the 440 Hz component. ...
A beamformer is basically a spatial filter. It can be passive, just like a temporal filter.Instead of samples separated by time, they are separated by space. A passive temporal filter can be a bandpass that is "aimed" or "steered" at a particular frequency. For passive spatial filters (i.e. beamformers), the filter can be steered towards a particular ...
Well, first of all the Sound Level Pressure decreases by $6 \; \mathtt{dB}$ when doubling the distance - this plays a big role. We do also have sound attenuation coming from our medium - air. Let's take a closer look onto sound absorption coefficient for different frequencies:Knowing that human speech is mostly concentrated at the range of $300\;\mathtt{Hz}...
In many common applications negative frequencies have no direct physical meaning at all. Consider a case where there is an input and an output voltage in some electrical circuit with resistors, capacitors, and inductors. There is simply a real input voltage with one frequency and there is a single output voltage with the same frequency but different ...
It is an edge detector. It just applies the Gabor Transform. The Gabor filter is basically a Gaussian (with variances sx and sy along x and y-axes respectively) modulated by a complex sinusoid (with centre frequencies U and V along x and y-axes respectively). See an example here.
Make sure that your frequency doesn't reach values below 0 or above the half of your sample rate.Please post more information/code about how you generate your waveform! Chances are you are not doing it correctly.For example, if you want to generate a sine wave with a time-varying frequency $f(t)$ (for example to implement frequency modulation), ...
You are right that the repetition is around 650 by how exactly do I compute that automatically? Seems like a peak-picking problem to me? Or is there some other methods that can be used?Yes, it's just peak-picking. Your period is the x value of the first strong peak:Your peaks are all similar in height, probably because you're doing the autocorrelation ...
You can make a positive frequency spectrum quite simply (where fs is the sampling rate and NFFT is the number of fft bins). In the Matlab implementation of the FFT algorithm, the first element is always the DC component, hence why the array starts from zero. THis is true for odd and even values of NFFT.%//Calculate frequency axisdf = fs/NFFT;fAxis = 0:df:...
This comes from music terminology. The name "octave" comes from the fact that in the heptatonic musical scales (which are the prevalent scales in western music), the note with a 2:1 frequency ratio is the eighth note in the scale.For example, in the C major scale (C D E F G A B C) the eighth note is one octave above / has a 2:1 frequency ratio with the ...
One trick, for even-length signals, is what to do with the "middle" sample. Here, I've split it half and half between each side of the FFT.The other trick is to ensure that you have the right amplitudes in the resampled signal. Here's it's a factor of 2.Try this in scilab:x = rand(1,100,'normal');X = fft(x);XX = 2*[X(1:50) X(51)/2 zeros(1,99) X(51)/...
Phase Noise and Frequency Noise are not two different noise sources, they are artifacts of the same noise, it is just a matter of what units you want to use. Frequency and Phase are directly related as frequency is phase changing with time, so if you have one you will always have the other; frequency and phase are related by derivatives and integrals: the ...
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Basic reproduction number
... the
basic reproduction numberis R0 = β/γ In cases of diseases with varying latent periods, the basic reproduction numbercan ... In epidemiology, the basic reproduction number(sometimes called basicreproductive ratio, or incorrectly basicreproductive ... The basic reproduction numberis affected by several factors including the duration of infectivity of affected patients, the ... Free text O Diekmann; J.A.P. Heesterbeek; J.A.J. Metz (1990). "On the definition and the computation of the basic reproduction... Force of infection Basic reproduction numberEpidemic Mathematical modelling of infectious disease. ... λ = numberof new infections numberof susceptible persons exposed × average duration of exposure {\displaystyle \lambda ={\ ... frac {\mbox{ numberof new infections}}{{\mbox{ numberof susceptible persons exposed}}\times {\mbox{average duration of exposure ... Compartmental models in epidemiology
... the so-called
basic reproduction number(also called basic reproductionratio). This ratio is derived as the expected numberof ... For this model, the basic reproduction numberis: R 0 = a μ + a β μ + γ . {\displaystyle R_{0}={\frac {a}{\mu +a}}{\frac {\beta ... The role of the basic reproduction numberis extremely important. In fact, upon rewriting the equation for infectious ... The model consists of three compartments- S for the numbersusceptible, I for the numberof infectious, and R for the number... Infectivity Basic reproduction number( basicreproductive rate, basicreproductive ratio, R0, or r nought) Stewart, AD; Logsdon, JM; Kelley ... This means that as a pathogen's ability to infect a greater numberof hosts increases, so does the level of harm it brings to ... Aedes aegypti
New research has attempted to estimate the
basic reproduction number, or R0 value, for Zika virus in several locations. ... "Preliminary estimation of the basic reproduction numberof Zika virus infection during Colombia epidemic, 2015-2016". Travel ... Since the latter is an uncritical reproductionof the former, they are both considered to antedate the starting point for ... In the environment, the antidote is unavailable to rescue mosquito reproduction, so the pest population is suppressed. The ... Mathematical modelling of infectious disease
... the
basic reproduction numbercan be calculated as the sum of the reproduction numberfor each transition time into the disease ... This quantity is called the basic reproduction number, denoted by R0, which can be defined as the numberof secondary ... the basic reproduction numberis R0 = β/γ This value quantifies the transmission potential of a disease. If the basic... As a consequence of this lower basic reproduction number, the average age of infection A will also change to some new value Aq ... Endemic (epidemiology)
Assuming a completely susceptible population, that means that the
basic reproduction number(R0) of the infection must equal 1 ... the basic reproduction numbermultiplied by the proportion of susceptible individuals in the population (S) must be 1. This ... depending on a numberof factors, including the virulence of the disease and its mode of transmission. If a disease is in ... the infection neither dies out nor does the numberof infected people increase exponentially but the infection is said to be in ... Enterovirus 71
Ma E, Fung C, Yip SH, Wong C, Chuang SK, Tsang T (Aug 2011). "Estimation of the
basic reproduction numberof enterovirus 71 and ... The basicreproductive number(R0) for enterovirus 71 (EV71) was estimated to a median of 5.48 with an interquartile range of ... Total numberof cases of Hand Foot and Mouth Disease (HFMD) and herpangina in Taiwan was 129,106. The outbreaks of the ... Xinhua reported the numberof people infected also rose from 4,000 to 15,799. Enteroviruses were isolated from a total of 1,892 ... Viral phylodynamics
... is given by the equilibrium
numberof infected individuals, I {\displaystyle I} . The mean basic reproduction number, averaged ... The mean and variance of these individual basic reproduction numbers, E [ ν ] {\displaystyle \mathrm {E} [\nu ]} and V a r [ ν ... This finding aided in making an early estimate of the basic reproduction numberR 0 {\displaystyle R_{0}} of the pandemic. ... These differences can be acknowledged by assuming that the basic reproduction numberis a random variable ν {\displaystyle \nu ... Transmission risks and rates
The
basic reproduction numberincludes all secondary cases infected by a primary case, while x is only the numberof secondary ... x is generally smaller than the basic reproduction numberfor the disease. That is defined as the numberof individuals each ... If the numberof susceptibles in the group is n and the numberof secondary cases is x, then an estimation of the transmission ... There are a numberof difficulties in using this relation. The first is that it is very difficult to measure contact rates ... Super-spreader
The
basic reproduction numberR0 is the average numberof secondary infections caused by a typical infective person in a ... The basicreproductive numberis found by multiplying the average numberof contacts by the average probability that a ... R0 = Numberof contacts X Shedding potential The individual reproductive numberrepresents the numberof secondary infections ... The greater the numberof immunized individuals, the less likely an outbreak can occur because there are fewer susceptible ... Robert Killick-Kendrick
... incidence and
basic reproduction numbercalculated from a cross-sectional serological survey on the island of Gozo, Malta. ... Next-generation matrix
In epidemiology, the next-generation matrix is a method used to derive the
basic reproduction number, for a compartmental model ... 1990) and van den Driessche and Watmough (2002). To calculate the basic reproduction numberby using a next-generation matrix, ... West African Ebola virus epidemic
The
basic reproduction number, R0, is a statistical measure of the average numberof people expected to be infected by one ... Althaus, Christian L. (2014). "Estimating the Reproduction Numberof Ebola Virus (EBOV) During the 2014 Outbreak in West Africa ... for this particular epidemic differs from the actual ratio between the numberof deaths and the numberof cases. Current ... The numberof cases eventually plateaued at 1 or 2 cases per week after the beginning of August, and no new cases occurred from ... Reed-Frost model
... the
basic reproduction number-in a large population when the initial numberof infecteds is small, an infected individual is ... The following parameters are set initially: - Size of the population - Numberof individuals already immune - Numberof cases ( ... Let I t {\displaystyle I_{t}} represent the numberof cases of infection at time t {\displaystyle t} . Assume all cases recover ... Let S t {\displaystyle S_{t}} represent the numberof susceptible individuals at time t {\displaystyle t} . Let B ( x ) {\ ... BRN
... an insurgent movement in Southern Thailand
Basic reproduction number, a measure of the speed in which an epidemic spreads Berne ... ISO 3-letter country code Bulk Richardson number, meteorological numberrelating vertical stability and vertical shear Bunte ... Contagion (film)
Other scientists determine the virus is spread by fomites, with a
basic reproduction numberof four when the virus mutates, ... To move the equipment for the casino scenes to the on-the-water location, producers hired a numberof locals to carry out the ... "DVD Sales: Moneyball Doesn't Hit a Home Run, but gets a Solid Double". The Numbers. Nash Information Services. January 24, 2012 ... "Blu-ray Sales: Overall Sales are in Remission". The Numbers. Nash Information Services. January 17, 2012. Retrieved June 29, ... Transmissibility
In medicine, transmissibility is a synonym for
basic reproduction numberTransmitter, a device for propagating electronic ... R0
... or R00 may refer to: Haplogroup R0, a gene R0, the Brussels Ring-road R0,
basic reproduction number, an important concept in ... List of MeSH codes (G03)
...
basic reproduction numberMeSH G03.850.505.400.975.525.375 --- incidence MeSH G03.850.505.400.975.525.750 --- prevalence MeSH ... basic reproduction numberMeSH G03.850.520.308.985.525.375 --- incidence MeSH G03.850.520.308.985.525.750 --- prevalence MeSH ... basic reproduction numberMeSH G03.850.310.300.320 --- disease transmission, patient-to-professional MeSH G03.850.310.300.330 ... Herd immunity
This threshold can be calculated by taking R0, the
basic reproduction number, or the average numberof new infections caused by ... Althaus, C. L. (2014). "Estimating the Reproduction Numberof Ebola Virus (EBOV) During the 2014 Outbreak in West Africa". PLoS ... Biggerstaff, M; Cauchemez, S; Reed, C; Gambhir, M; Finelli, L (2014). "Estimates of the reproduction numberfor seasonal, ... To account for this, the effective reproductive numberRe, also written as Rt, or the average numberof infections caused at ... List of MeSH codes (L01)
...
basic reproduction numberMeSH L01.280.975.525.375 - incidence MeSH L01.280.975.525.750 - prevalence MeSH L01.280.975.550 - ... List of MeSH codes (N01)
...
basic reproduction numberMeSH N01.224.935.597.500 --- incidence MeSH N01.224.935.597.750 --- prevalence MeSH N01.224.935.698 ... List of MeSH codes (E05)
...
basic reproduction numberMeSH E05.318.308.985.525.375 --- incidence MeSH E05.318.308.985.525.750 --- prevalence MeSH E05.318. ... Luna Sea (Luna Sea album)
... and a
reproductionshirt from their 1991 Under the New Moon Episode I-III tours. The regular edition reached numbersix on the ... After ten years, we have to recreate ourselves, and going back to something as original and basicas our first album is part of ... and a premium which includes a reproductioncassette of their 1989 demo "Shade" ...
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So in order to understand why this must be true, we have to consider what "equals" means in this context.
Assume you were talking to someone who had never heard of division before in his life. How would you explain why you can write $$\frac{1}{5}=\frac{5}{25}?$$ These fractions are equal precisely because $$25\cdot 1 = 5 \cdot 5.$$ In abstract algebra, this is how we
define fractions: the equivalency classes of pairs of numbers $(a,b)$ with $b\not= 0$ under the relation $(a,b)\sim(c,d)\Leftrightarrow ad=cb$. (Note: I use the word "numbers" here loosely; see the last paragraph for more info.) We write these pairs $(a,b)$ as $a/b$ or $\frac{a}{b}$ simply because it's convenient.
So, you want to know whether the division property is a special case of the multiplicative property. Let's see if it is.
Assume for the remainder of the answer that $x\not= 0 $. The multiplicative property is that $$\frac{a}{b}=\frac{c}{d} \hspace{10pt}\Leftrightarrow \hspace{10pt}\frac{xa}{b}=\frac{xc}{d}.$$First, let's check if that's true, based on the definition of fractions above. $$\frac{a}{b}=\frac{c}{d}\hspace{10pt} \Leftrightarrow \hspace{10pt}ad=cb \hspace{10pt}\Leftrightarrow\hspace{10pt} xad=xcb\hspace{10pt} \Leftrightarrow \hspace{10pt}\frac{xa}{b}=\frac{xc}{d}.$$Alright, I buy that. Now what about the division property?$$\frac{a}{b}=\frac{c}{d} \hspace{10pt}\Leftrightarrow \hspace{10pt}\frac{a}{xb}=\frac{c}{xd}.$$How do we check if this is true?$$\frac{a}{b}=\frac{c}{d}\hspace{10pt} \Leftrightarrow \hspace{10pt}ad=cb \hspace{10pt}\Leftrightarrow\hspace{10pt} axd=cxb\hspace{10pt} \Leftrightarrow \hspace{10pt}\frac{a}{xb}=\frac{c}{xd}.$$You can see that this is
exactly the same thing. The only difference in the proofs are that $xad=xcb$ and $axd=cxb$, so the two properties are equivalent precisely because multiplication is a commutative operation (that is, $ab=ba$ always). In fact, the division property implies the multiplication property too:$$\frac{xa}{b}=\frac{xc}{d}\hspace{10pt}\Leftrightarrow \hspace{10pt}xad=xcb\hspace{10pt} \Leftrightarrow \hspace{10pt}axd=cxb\hspace{10pt} \Leftrightarrow \hspace{10pt}\frac{a}{xb}=\frac{c}{xd}.$$
Optional Note: All this comes from the concept of a field of fractions, which you can read about on Wikipedia if you're feeling ambitious. In short, any time you have a set of elements that can be added and multiplied, where every $a,b$ satisfy $ab=ba$ and such that no nonzero $a,b$ satisfy $ab=0$, you can define fractions of that set which work exactly as they do with regular numbers.The word for making fractions like this is "localization," and you can do it with all kinds of stuff other than numbers, like functions, for example.
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Motivation:
Many interesting
irrational numbers (or numbers believed to be irrational) appear as answers to natural questions in mathematics. Famous examples are $e$, $\pi$, $\log 2$, $\zeta(3)$ etc. Many more such numbers are described for example in the wonderful book "Mathematical Constants" by Steven R. Finch. The question:
I am interested in theorems where a "special"
rational number makes a surprising appearance as an answer to a natural question. By a special rational number I mean one with a large denominator (and preferably also a large numerator, to rule out numbers which are simply the reciprocals of large integers, but I'll consider exceptions to this rule). Please provide examples.
For illustration, here are a couple of nice examples I'm aware of:
The average geodesic distance between two random points of the Sierpinski gasket of unit side lengths is equal to $\frac{466}{885}$. This is also equivalent to a natural discrete math fact about the analysis of algorithms, namely that the average number of moves in the Tower of Hanoi game with $n$ disks connecting a randomly chosen initial state to a randomly chosen terminal state with a shortest number of moves, is asymptotically equal to $\frac{466}{885}\times 2^n$. See here and here for more information.
The answer to the title question of the recent paper ""The density of primes dividing a term in the Somos-5 sequence" by Davis, Kotsonis and Rouse is $\frac{5087}{10752}$.
Rules:
1) I won't try to define how large the denominator and numerator need to be to for the rational number to qualify as "special". A good answer will maximize the ratio of the number's information theoretic content to the information theoretic content of the statement of the question it answers. (E.g., a number like 34/57 may qualify if the question it answers is simple enough.) Really simple fractions like $3/4$, $22/7$ obviously do not qualify.
2) The question the number answers needs to be natural. Again, it's impossible to define what this means, but avoid answers in the style of "what is the rational number with smallest denominator solving the Diophantine equation [some arbitrary-sounding, unmotivated equation]".
Edit: a lot of great answers so far, thanks everyone. To clarify my question a bit, while all the answers posted so far represent very beautiful mathematics and some (like Richard Stanley's and Max Alekseyev's answers) are truly astonishing, my favorite type of answers involve questions that are conceptual in nature (e.g., longest increasing subsequences, tower of Hanoi, Markov spectrum, critical exponents in percolation) rather than purely computational (e.g., compute some integral or infinite series) and to which the answer is an exotic rational number. (Note that someone edited my original question changing "exotic" to "special"; that is fine, but "exotic" does a better job of signaling that numbers like 1/4 and 2 are not really what I had in mind. That is, 2 is indeed quite a special number, but I doubt anyone would consider it exotic.)
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Let $\Omega \subset \mathbb{R}^n$ be a $C^{1}$ bounded domain.
It is possible to define the space $L^1(\partial\Omega)$ as the set of functions $u\colon \partial\Omega \to \mathbb{R}$ with the finite norm $$\lVert u \rVert_{L^1(\partial\Omega)} := \sum_i\lVert {\phi_i(u\circ g_i)}\rVert_{L^1(B_i)}$$ where $\phi_i$ partition of unity, $g_i$ is a $C^1$ diffeomorphism and $B_i$ are subsets of $\mathbb{R}^{n}$.
There are lots of definitions like this (eg. see Renardy and Rogers, Krylov, James Robinson's Infinite Dimensional Dynamical System) or with small variations.
There is also the surface integral of a function $v\colon \partial\Omega \to \mathbb{R}$: $$|u| := \int_{\partial \Omega}fdS$$ where $dS$ is the surface density. To compute this quantity, we need to use a parametrisation.
Now my question, is it always the case that $$\lVert u \rVert_{L^1(\partial\Omega)} \qquad\text{and}\qquad |u|$$ are equivalent? Does this hold for all the little variations of the $L^1$ norm in the first equation??
Now when we have a Lipschitz surface, these norms are equivalent (see Necas). BUT their definition of the $L^1$ norm is different so let us not use that result here.
Edit: hmm, it seems $C^1$ domains are not a subset of Lipschitz domains.
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It is well know that if $X_1,X_2,.., X_n$ are all independent and sampled from $Poisson(\lambda)$, then $\sum_{i=1}^n{X_i}\sim Poisson(n\lambda)$.
I have a following situation which is similar to the one hererelated discussion, however, I can not comment so I can not ask for a clarification on a specific term used in the discussion. The term is
exposure. The situation is the following: I have an ab-test setup with two groups of users. Each about 3000 users. Then, in each group 45 and 53 cases are converted (the 0/1 event happened). I am treating these as rare events. Then we have poisson distribution for each group with $\hat{\lambda}=\frac{events}{n}$, where $n$ is number of , say, users per fixed time unit. Obviously, both lambdas will be small: (45/3000) and (53/3000), respectively. In the related discussion as well as in rate ratio test poisson this $n$ is called exposure. And now the question, if I follow the property of sum of poissons, I will get $\hat{\lambda n}=\frac{events}{n}\times n=43$ for group 1 and similarly for the other group. In this case a poisson with 43 events on average per unit of time is not rare anymore, but approximately normal. But in reality I know that 43 out of waiting 3000 "user-times" is rare. Isn't it a contradiction , is it? This exposure makes my individual rare count are now summed to a distribution with large lambda, not suitable for rare counts data.
Sorry for a bit of a messy question. May be I am missing some lambda scaling or other things.
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You and a friend lost each other on the line to a concert, and neither is sure which of you is further ahead. Formally, each is at some integer coordinate and may only walk towards a higher coordinate or stay in place.
Assuming you and your friend are following the exact same algorithm (and no, you may not say "if (name=="R B") do something :) ), and the initial distance between the two of you was $x$ (which is not known to you).
What is the algorithm which minimizes the expected walking distance till you and your friend meet?
You may assume both your friend and yourself are moving in the same constant speed.
An example simple algorithm would be something like:
At stage $n$ (starting from $0$):
Walk $3^n$ steps to the right w.p. $\frac{1}{2}$ or wait $3^n$ time units otherwise.
To see this algorithm makes the friends meet with probability 1 consider what happens at stage $(\log_3 x+1)$. Even if the friend that was $x$ step ahead always walked and the other always stayed in place, the distance between the two would be: $$x+1+3+9+\ldots+3^{\log_3 x}=2x+\frac{x-1}{2}\leq 3x$$
Therefore, at the $\log_3 x+1$ iteration, the friend which chooses to walk will cover distance of $3^{\log_3 x+1}=3x$, hence with probability $\frac{1}{4}$, the friend which is behind will catch up and they will meet.
A simple optimization (to reduce walking distance) would be, instead of walking $3^x$ steps, walking $c^x$ steps, where $c$ is given by: $$2+\frac{1}{c-1}=c$$
Hence the optimal $c$
for this algorithm is $c=\frac{3+\sqrt 5}{2}\approx 2.618$
Unfortunately, while this algorithm guarantees that the friends will meet with probability 1, the expected walking distance is infinite, which is something I'd like to avoid if possible.
Is there a more efficient algorithm?
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Some notes, Paul:
The LT1006 is weak. (Output impedance seems about \$300\:\Omega\$.) But that's not the cause of the clipping you see when supplying \$V_{PEAK}=1.5\:\textrm{V}\$. You are driving your circuit directly with an uncommon input signal that has a nice DC bias exactly half-way between your rails. But that's obviously just a convenience.
So let's cut to the chase. The proximate cause for the clipping when supplying even so little as \$V_{PEAK}=1.5\:\textrm{V}\$ is \$R_2\$ and \$R_3\$. Those are the devices that are supplying base currents to your two output transistors, \$Q_1\$ and \$Q_3\$. I'll analyze this situation below, too.
Hopefully you can see that the LT1006 isn't driving your output BJTs. It can't forward bias either BJT, because in both cases it's output is blocked from reaching and forward biasing either BJT by one or the other of those two diodes in the circuit. All the LT1006 can do is to sink added current from \$R_2\$ or source added current into \$R_3\$. That's how it acts to move the BJT bases around.
But the base current supply isn't coming from the opamp. It's coming from the resistors.
Let's look at the situation with your worst case \$V_{PEAK}=4\:\textrm{V}\$. This will help caricature the situation more bleakly and make the obvious stand out even more. Then you can return to the question of \$V_{PEAK}=1.5\:\textrm{V}\$, which will then be clear to see.
(Keep in mind that we are talking about AC frequencies fast enough that \$C_2\$ can be treated as nearly a dead short compared to your load resistance, \$R_1\$. If you were to operate this at very low frequency -- say \$1\:\textrm{Hz}\$ -- then some of this problem would disappear because \$C_2\$ would present a high impedance and reduce the loading which will resolve some of the problem ahead.)
With \$V_{PEAK}=4\:\textrm{V}\$, your input range to the opamp is from \$V_+=2\:\textrm{V}\$ to \$V_+=10\:\textrm{V}\$ (centered, of course, on \$6\:\textrm{V}\$.) With \$V_+=2\:\textrm{V}\$, the output also needs to match this so that the \$V_-\$ terminal can have the same value. But this means that the base of \$Q_3\$ must be perhaps \$850\:\textrm{mV}\$ lower, or about \$1.15\:\textrm{V}\$. But that only allows \$R_3\$ to sink \$1.15\:\textrm{mA}\$ of base current. (The LT1006 will be itself sinking \$\frac{12\:\textrm{V}-2.5\:\textrm{V}}{1\:\textrm{k}\Omega}-1.15\:\textrm{mA}\approx 8.35\:\textrm{mA}\$ in order to force sufficent drop across \$R_2\$.) With that low of a base current via \$R_3\$, \$Q_3\$ can't handle much collector current.
As you can see, as the base voltage on \$Q_3\$ drives too close to ground, \$R_3\$ can sink less and less base current for \$Q_3\$ and this means lower current compliance out of \$Q_3\$. The same happens on the other side, where \$Q_2\$'s base is driven up towards the upper voltage rail and there is less and less base current available for \$Q_2\$ via \$R_2\$.
Before we look at the situation with \$V_{PEAK}=1.5\:\textrm{V}\$, where you still saw clipping, let's just try and figure out where that clipping should occur. We'll do this by seeing just how low the output voltage can be, while at the same time providing adequate current for the load.
The output voltage at the node with the shared emitters, \$V_{OUT}\$, less a forward biased BE junction \$V_{BE}=850\:\textrm{mV}\$, is the base voltage for \$Q_3\$. This base voltage, divided by \$R_3\$ is the total base current available. This base current, times an assumed \$\beta=100\$, provides the collector current. So the following equation applies:
$$6\:\textrm{V}-R_3\cdot\frac{V_{OUT}-V_{BE}}{R_2}\cdot\beta=V_{OUT}$$
(The \$6\:\textrm{V}\$ present above is the quiescent output voltage.) That solves out as:
$$V_{OUT}=\frac{6\:\textrm{V}\cdot R_2+V_{BE}\cdot \beta\cdot R_3}{R_2+\beta\cdot R_3}$$
And if you plug in your values, you should get the result of \$V_{OUT}\approx 4.53\:\textrm{V}\$. Which, after accounting for the quiescent output voltage, translates to \$V_{PEAK}\approx 1.47\:\textrm{V}\$. And that is about what you found, too, for the point at which clipping started occurring.
So it is predictable!! Good thing. (I guess there's no need now to go look at the situation where \$V_{PEAK}=1.5\:\textrm{V}\$ since we just predicted that this is where the problems will start.)
A solution to this problem is not necessarily a beefier opamp (well, the LT1006 is weak, so I'd pick something stronger regardless.) What you need is an adequate constant current supply. You can do this with an actual constant current supply built out of parts or else you can use bootstrapping (cheaper, easier.) The simplest fix is bootstrapping. It doesn't create a good current source but it does create a
good-enough one to get by.
simulate this circuit – Schematic created using CircuitLab
The idea here is that \$C_2\$ will charge up to a DC bias that is about half your rail voltage of \$12\:\textrm{V}\$. So about \$6\:\textrm{V}\$ across \$C_2\$. In theory, discounting some minor details, this means that \$R_2\$ should have a constant \$5.2\:\textrm{V}\$ across it. (The emitters of the BJTs and the output of the opamp should have about the same voltage; and there is about \$800\:\textrm{mV}\$ across \$D_2\$ and, as already mentioned, about \$6\:\textrm{V}\$ across \$C_2\$.) This helps to create that constant current source I mentioned you needing there. (It's not perfect, but it is very much better than you had before despite the fact that it's mostly just a wiring change to get there.) In this case, the constant current is about \$\frac{5.2\:\textrm{V}}{180\:\Omega}\approx 29\:\textrm{mA}\$. Of course, this now means your opamp needs to be able to divert a fair portion of that current. So you need an opamp with output current compliance better than that. I've used such in the above circuit.
Let's assume, for a moment, that you can only expect (at most) a \$\beta=100\$ from your BJTs at these currents. If you wanted to drive the above circuit with \$V_{PEAK}=4\:\textrm{V}\$, you'd need \$I_{PEAK}=1\:\textrm{A}\$ of collector current. This would mean at least \$10\:\textrm{mA}\$ of base current, which means at least \$1.8\:\textrm{V}\$ across those \$180\:\Omega\$ resistors (\$R_2\$ and \$R_3\$.)
At those currents, of course, your small signal BJTs would be well beyond their specification range. But let's assume they are fine. We'd still have to expect more than \$1\:\textrm{V}\$ across their BE junction. So the output cannot swing lower than about \$2.8\:\textrm{V}\$. Which basically means you won't be able to drive this circuit using \$V_{PEAK}\gt 3.2\:\textrm{V}\$. Probably less, actually. And that's assuming these BJTs could operate with those collector currents. And we know they can't. But that gives you an idea of how to estimate the maximum swing to expect, even with bootstrapping added.
Try the above circuit out in your simulator and use \$V_{PEAK}\le 3\:\textrm{V}\$. (It can't do better, so why try?)
You'll probably also need to replace those BJTs, when you get around to thinking more practically. And there are other circuit changes to make, anyway. But that's for another day.
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Usually in texts about Physics that uses tensors defines them as multilinear maps. So if $V$ is a vector space over the field $F$, a tensor is a multilinear mapping:
$$T:V\times\cdots\times V\times V^\ast\times\cdots\times V^\ast\to F.$$
In texts about multilinear algebra, however, a tensor is defined differently. They consider a collection $V_1,\dots,V_k$ of vector spaces over the same field, consider the free vector space $\mathcal{M}=F(V_1\times\cdots\times V_k)$, consider the subspace $\mathcal{M}_0$ genereated by vectors of the form
$$(v_1,\dots,v_i+v_i',\dots,v_k)-(v_1,\dots,v_i,\dots,v_k)-(v_1,\dots,v_i',\dots,v_k)$$
$$(v_1,\dots,kv_i,\dots,v_k)-k(v_1,\dots,v_i,\dots,v_k)$$
And then define the tensor product $V_1\otimes\cdots\otimes V_k = \mathcal{M}/\mathcal{M}_0$ and define tensors as elements of such space, which are equivalence classes of functions with finite support in $V_1\times\cdots\times V_k$.
Now, is there some cases in Physics where it's better to think as tensors as such equivalence classes rather than multilinear mappings? If so, how then we get some physical intuition behind those objects?
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Definition
An equivalent way of writing it, is $[A]_{\mathcal{E}}$ is symmetric iff there exists an orthogonal basis for $\mathbb{R}^n$ consisting of eigenvectors of A.
The reason they are equivalent can be recalled from the change-of-basis formula. $$[A]_\mathcal{E} = \underset{E \leftarrow B}{P}[A]_\mathcal{B}\underset{E \leftarrow B}{P}^{-1}$$ If we rewrite the first definition in a slightly different way, we get $$A = QDQ^{-1} = QDQ^T$$ In order for a change-of-basis transformation to be orthonormal, there must be an orthonormal basis! Thus, they are equivalent.
The Following Are Equivalent (TFAE)
$$\text{1. T is symmetric}$$ $$\text{2. There exists an orthonormal basis such that } [T]_\mathcal{B} \text{ is symmetric}$$ $$\text{3. For every orthonormal basis C in } \mathbb{R}^n, [T]_\mathcal{C} \text{ is symmetric}$$ We can change the order of our orthonormal basis, it doesn't matter. It will still be symmetric. $$\text{4. } T(x) \cdot y = x \cdot T(y)$$ This one is easy to prove. $Ax \cdot y = (xA)^Ty = x^T A^Ty = x \cdot A^Ty = x \cdot Ay$. This is only true if $A^T = A$
Lagrange’s Revenge
If A is symmetric, then A has a real eigenvalue.
In case it wasn't clear, not all matrices have real eigenvalues. $$\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix} \text{ has eigenvalues of } \pm i $$ This is mainly important in the proof for the Spectral Theorem, because we prove the Spectral Theorem with induction, and we use Lagrange's Revenge to choose an arbitrary eigenvector, which we wouldn't be able to do without justification.
In case it wasn't clear, not all matrices have real eigenvalues. $$\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix} \text{ has eigenvalues of } \pm i $$ This is mainly important in the proof for the Spectral Theorem, because we prove the Spectral Theorem with induction, and we use Lagrange's Revenge to choose an arbitrary eigenvector, which we wouldn't be able to do without justification.
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in my math class we were given a list of indefinite integrals, and one of them was:
$$\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}$$
My working:
$$\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}=\int \frac{dx}{(x+2)\sqrt{(x+2)^2-1}}$$
Then I used the substitution $x+2=\sec t$ to get:
$$\int \frac{\tan t}{\sqrt{\sec^2 t-1}}dt=\int \frac{\tan t}{|\tan t|}dt= t\,\text{sgn}\, (\tan t)+C...$$
Then I checked the answer sheet, and this is what they did:
$$\int \frac{\tan t}{\sqrt{\sec^2 t-1}}dt=\int dt=t+C=\text{arcsec}(x+2)+C$$
What I don't understand is, why are they allowed to say $\sqrt{\sec^2 t-1}=\tan t?$ I tried to put some values in and I have found that:
$$\int_{\sec \left(\frac{8}{5}\right)-2}^{\sec \left(\frac{9}{5}\right)-2} \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}<0$$ but according to the answer sheet I would get $\dfrac{1}{5}$
My answer looks wrong, I would be happy if someone could explain what the problem is, and also why we are allowed to simplify like they did.
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Let's recast the first example from the previous section. Suppose thatthe speed of the object is $3t$ at time $t$. How far does the objecttravel between time $t=a$ and time $t=b$? We are no longer assumingthat we know where the object is at time $t=0$ or at any othertime. It is certainly true that it is
somewhere, so let'ssuppose that at $t=0$ the position is $k$. Then just as in theexample, we know that the position of the object at any time is $\ds 3t^2/2+k$. This means that at time $t=a$ the position is $\ds 3a^2/2+k$ and at time $t=b$ the position is $\ds 3b^2/2+k$. Therefore thechange in position is $\ds 3b^2/2+k-(3a^2/2+k)=3b^2/2-3a^2/2$. Notice thatthe $k$ drops out; this means that it doesn't matter that we don'tknow $k$, it doesn't even matter if we use the wrong $k$, we get thecorrect answer. In other words, to find the change in position betweentime $a$ and time $b$ we can use any antiderivative of thespeed function $3t$—it need not be the one antiderivative thatactually gives the location of the object.
What about the second approach to this problem, in the new form? Wenow want to approximate the change in position between time $a$ andtime $b$. We take the interval of time between $a$ and $b$, divide itinto $n$ subintervals, and approximate the distance traveled duringeach. The starting time of subinterval number $i$ is now $a+(i-1)(b-a)/n$, which we abbreviate as $\ds t_{i-1}$, so that $\ds t_0=a$, $\ds t_1=a+(b-a)/n$, and so on. The speed of the object is$f(t)=3t$, and each subinterval is $(b-a)/n=\Delta t$ seconds long.The distance traveled during subinterval number$i$ is approximately $\ds f(t_{i-1})\Delta t$, and the total change indistance is approximately$$ f(t_0)\Delta t+f(t_1)\Delta t+\cdots+f(t_{n-1})\Delta t.$$The exact change in position is the limit of this sum as $n$ goes toinfinity. We abbreviate this sum using
sigma notation:$$ \sum_{i=0}^{n-1} f(t_i)\Delta t =f(t_0)\Delta t+f(t_1)\Delta t+\cdots+f(t_{n-1})\Delta t.$$The notation on the left side of the equal sign uses a large capitalsigma, a Greek letter, and the left side is an abbreviation for theright side. The answer we seek is$$ \lim_{n\to\infty}\sum_{i=0}^{n-1} f(t_i)\Delta t.$$Since this must be the same as the answer we have already obtained, weknow that $$ \lim_{n\to\infty}\sum_{i=0}^{n-1} f(t_i)\Delta t={3b^2\over 2}-{3a^2\over 2}.$$The significance of $\ds 3t^2/2$, into which we substitute $t=b$ and$t=a$, is of course that it is a function whose derivative is $f(t)$.As we have discussed, by the time we know that we want to compute$$ \lim_{n\to\infty}\sum_{i=0}^{n-1} f(t_i)\Delta t,$$it no longer matters what $f(t)$ stands for—it could be a speed, orthe height of a curve, or something else entirely. We know thatthe limit can be computed by finding any function with derivative$f(t)$, substituting $a$ and $b$, and subtracting. We summarize thisin a theorem. First, we introduce some new notation and terms.
We write$$ \int_a^b f(t)\,dt = \lim_{n\to\infty}\sum_{i=0}^{n-1} f(t_i)\Delta t$$ if the limit exists. That is, the left hand side means, or is anabbreviation for, the right hand side. The symbol $\int$ is called an
integral sign, and the wholeexpression is read as "the integral of $f(t)$ from $a$ to $b$.'' Whatwe have learned is that this integral can be computed by finding afunction, say $F(t)$, with the property that $F'(t)=f(t)$, and thencomputing $F(b)-F(a)$. The function $F(t)$ is called an antiderivative of $f(t)$. Now the theorem:
Let's rewrite this slightly: $$ \int_a^x f(t)\,dt = F(x)-F(a). $$ We've replaced the variable $x$ by $t$ and $b$ by $x$. These are just different names for quantities, so the substitution doesn't change the meaning. It does make it easier to think of the two sides of the equation as functions. The expression $$ \int_a^x f(t)\,dt $$ is a function: plug in a value for $x$, get out some other value. The expression $F(x)-F(a)$ is of course also a function, and it has a nice property: $$ {d\over dx} (F(x)-F(a)) = F'(x) = f(x), $$ since $F(a)$ is a constant and has derivative zero. In other words, by shifting our point of view slightly, we see that the odd looking function $$ G(x)=\int_a^x f(t)\,dt $$ has a derivative, and that in fact $G'(x)=f(x)$. This is really just a restatement of the Fundamental Theorem of Calculus, and indeed is often called the Fundamental Theorem of Calculus. To avoid confusion, some people call the two versions of the theorem "The Fundamental Theorem of Calculus, part I'' and "The Fundamental Theorem of Calculus, part II'', although unfortunately there is no universal agreement as to which is part I and which part II. Since it really is the same theorem, differently stated, some people simply call them both "The Fundamental Theorem of Calculus.''
We have not really proved the Fundamental Theorem. In a nutshell, we gave the following argument to justify it: Suppose we want to know the value of $$ \int_a^b f(t)\,dt = \lim_{n\to\infty}\sum_{i=0}^{n-1} f(t_i)\Delta t. $$ We can interpret the right hand side as the distance traveled by an object whose speed is given by $f(t)$. We know another way to compute the answer to such a problem: find the position of the object by finding an antiderivative of $f(t)$, then substitute $t=a$ and $t=b$ and subtract to find the distance traveled. This must be the answer to the original problem as well, even if $f(t)$ does not represent a speed.
What's wrong with this? In some sense, nothing. As a practical matter it is a very convincing argument, because our understanding of the relationship between speed and distance seems to be quite solid. From the point of view of mathematics, however, it is unsatisfactory to justify a purely mathematical relationship by appealing to our understanding of the physical universe, which could, however unlikely it is in this case, be wrong.
A complete proof is a bit too involved to include here, but we will indicate how it goes. First, if we can prove the second version of the Fundamental Theorem, theorem 7.2.2, then we can prove the first version from that:
Proof of Theorem 7.2.1. We know from theorem 7.2.2 that $$ G(x)=\int_a^x f(t)\,dt $$ is an antiderivative of $f(x)$, and therefore any antiderivative $F(x)$ of $f(x)$ is of the form $F(x)=G(x)+k$. Then $$\eqalign{ F(b)-F(a)=G(b)+k-(G(a)+k) &= G(b)-G(a)\cr &=\int_a^b f(t)\,dt-\int_a^a f(t)\,dt.\cr }$$ It is not hard to see that $\ds \int_a^a f(t)\,dt=0$, so this means that $$ F(b)-F(a)=\int_a^b f(t)\,dt, $$ which is exactly what theorem 7.2.1 says.
So the real job is to prove theorem 7.2.2. We will sketch the proof, using some facts that we do not prove. First, the following identity is true of integrals: $$ \int_a^b f(t)\,dt = \int_a^c f(t)\,dt + \int_c^b f(t)\,dt. $$ This can be proved directly from the definition of the integral, that is, using the limits of sums. It is quite easy to see that it must be true by thinking of either of the two applications of integrals that we have seen. It turns out that the identity is true no matter what $c$ is, but it is easiest to think about the meaning when $a\le c\le b$.
First, if $f(t)$ represents a speed, then we know that the three integrals represent the distance traveled between time $a$ and time $b$; the distance traveled between time $a$ and time $c$; and the distance traveled between time $c$ and time $b$. Clearly the sum of the latter two is equal to the first of these.
Second, if $f(t)$ represents the height of a curve, the three integrals represent the area under the curve between $a$ and $b$; the area under the curve between $a$ and $c$; and the area under the curve between $c$ and $b$. Again it is clear from the geometry that the first is equal to the sum of the second and third.
Proof sketch forTheorem 7.2.2. We want to compute $G'(x)$, so we start with the definition of the derivative in terms of a limit: $$\eqalign{ G'(x)&=\lim_{\Delta x\to0}{G(x+\Delta x)-G(x)\over\Delta x}\cr &=\lim_{\Delta x\to0}{1\over \Delta x}\left( \int_a^{x+\Delta x} f(t)\,dt - \int_a^x f(t)\,dt\right)\cr &=\lim_{\Delta x\to0}{1\over \Delta x}\left( \int_a^{x} f(t)\,dt + \int_x^{x+\Delta x} f(t)\,dt - \int_a^x f(t)\,dt\right)\cr &=\lim_{\Delta x\to0}{1\over \Delta x}\int_x^{x+\Delta x} f(t)\,dt.\cr }$$ Now we need to know something about $$ \int_x^{x+\Delta x} f(t)\,dt $$ when $\Delta x$ is small; in fact, it is very close to $\Delta x f(x)$, but we will not prove this. Once again, it is easy to believe this is true by thinking of our two applications: The integral $$ \int_x^{x+\Delta x} f(t)\,dt $$ can be interpreted as the distance traveled by an object over a very short interval of time. Over a sufficiently short period of time, the speed of the object will not change very much, so the distance traveled will be approximately the length of time multiplied by the speed at the beginning of the interval, namely, $\Delta x f(x)$. Alternately, the integral may be interpreted as the area under the curve between $x$ and $x+\Delta x$. When $\Delta x$ is very small, this will be very close to the area of the rectangle with base $\Delta x$ and height $f(x)$; again this is $\Delta x f(x)$. If we accept this, we may proceed: $$ \lim_{\Delta x\to0}{1\over \Delta x}\int_x^{x+\Delta x} f(t)\,dt =\lim_{\Delta x\to0}{\Delta x f(x)\over \Delta x}=f(x), $$ which is what we wanted to show.
It is still true that we are depending on an interpretation of the integral to justify the argument, but we have isolated this part of the argument into two facts that are not too hard to prove. Once the last reference to interpretation has been removed from the proofs of these facts, we will have a real proof of the Fundamental Theorem.
Now we know that to solve certain kinds of problems, those that lead to a sum of a certain form, we "merely'' find an antiderivative and substitute two values and subtract. Unfortunately, finding antiderivatives can be quite difficult. While there are a small number of rules that allow us to compute the derivative of any common function, there are no such rules for antiderivatives. There are some techniques that frequently prove useful, but we will never be able to reduce the problem to a completely mechanical process.
Because of the close relationship between an integral and anantiderivative, the integral sign is also used to mean"antiderivative''. You can tell which is intended by whether thelimits of integration are included:$$ \int_1^2 x^2\,dx$$is an ordinary integral, also called a
definite integral,because it has a definite value, namely$$ \int_1^2 x^2\,dx={2^3\over3}-{1^3\over3}={7\over3}.$$We use$$ \int x^2\,dx$$to denote the antiderivative of $\ds x^2$, also called an indefinite integral.So this is evaluated as$$ \int x^2\,dx = {x^3\over 3}+C.$$It is customary to include the constant $C$ to indicate that there arereally an infinite number of antiderivatives. We do not need this $C$to compute definite integrals, but in other circumstances we will needto remember that the $C$ is there, so it is best to get into the habitof writing the $C$.When we compute a definite integral, we first find an antiderivativeand then substitute. It is convenient to first display theantiderivative and then do the substitution; we need a notationindicating that the substitution is yet to be done. A typical solutionwould look like this:$$ \int_1^2 x^2\,dx=\left.{x^3\over 3}\right|_1^2 = {2^3\over3}-{1^3\over3}={7\over3}.$$The vertical line with subscript and superscript is used to indicatethe operation "substitute and subtract'' that is needed to finish theevaluation.
Exercises 7.2
Find the antiderivatives of the functions:
Ex 7.2.1$\ds 8\sqrt{x}$(answer)
Ex 7.2.2$\ds 3t^2+1$(answer)
Ex 7.2.3$\ds 4/\sqrt{x}$(answer)
Ex 7.2.4$\ds 2/z^2$(answer)
Ex 7.2.5$\ds (5x+1)^2$(answer)
Ex 7.2.6$\ds (x-6)^2$(answer)
Ex 7.2.7$\ds x^{3/2}$(answer)
Ex 7.2.8$\ds {2\over x\sqrt x}$(answer)
Ex 7.2.9$\ds |2t-4|$(answer)
Compute the values of the integrals:
Ex 7.2.10$\ds \int_1^4 t^2+3t\,dt$(answer)
Ex 7.2.11$\ds \int_0^\pi \sin t\,dt$(answer)
Ex 7.2.12$\ds \int_0^3 x^3\,dx$(answer)
Ex 7.2.13$\ds \int_1^2 x^5\,dx$(answer)
Ex 7.2.14Find the derivative of $\ds G(x)=\int_1^x t^2-3t\,dt$(answer)
Ex 7.2.15Find the derivative of $\ds G(x)=\int_1^{x^2} t^2-3t\,dt$(answer)
Ex 7.2.16Find the derivative of $\ds G(x)=\int_1^x \tan(t^2)\,dt$(answer)
Ex 7.2.17Find the derivative of $\ds G(x)=\int_1^{x^2} \tan(t^2)\,dt$(answer)
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We have so far integrated "over'' intervals, areas, and volumes with single, double, and triple integrals. We now investigate integration over or "along'' a curve—"line integrals'' are really "curve integrals''.
As with other integrals, a geometric example may be easiest to understand. Consider the function $f=x+y$ and the parabola $y=x^2$ in the $x$-$y$ plane, for $0\le x\le 2$. Imagine that we extend the parabola up to the surface $f$, to form a curved wall or curtain, as in figure 18.2.1. What is the area of the surface thus formed? We already know one way to compute surface area, but here we take a different approach that is more useful for the problems to come.
As usual, we start by thinking about how to approximate the area. We pick some points along the part of the parabola we're interested in, and connect adjacent points by straight lines; when the points are close together, the length of each line segment will be close to the length along the parabola. Using each line segment as the base of a rectangle, we choose the height to be the height of the surface $f$ above the line segment. If we add up the areas of these rectangles, we get an approximation to the desired area, and in the limit this sum turns into an integral.
Typically the curve is in vector form, or can easily be put in vector form; in this example we have ${\bf v}(t)=\langle t,t^2\rangle$. Then as we have seen in section 15.3 on arc length, the length of one of the straight line segments in the approximation is approximately $ds=|{\bf v}'|\,dt=\sqrt{1+4t^2}\,dt$, so the integral is $$\int_0^2 f(t,t^2)\sqrt{1+4t^2}\,dt =\int_0^2 (t+t^2)\sqrt{1+4t^2}\,dt= {167\over48}\sqrt{17}-{1\over12}-{1\over64}\ln(4+\sqrt{17}). $$ This integral of a function along a curve $C$ is often written in abbreviated form as $$\int_C f(x,y)\,ds.$$
Example 18.2.1 Compute $\ds\int_C ye^x\,ds$ where $C$ is the line segment from $(1,2)$ to $(4,7)$.
We write the line segment as a vector function: ${\bf v}=\langle 1,2\rangle + t\langle 3,5\rangle$, $0\le t\le 1$, or in parametric form $x=1+3t$, $y=2+5t$. Then $$\int_C ye^x\,ds = \int_0^1 (2+5t)e^{1+3t}\sqrt{3^2+5^2}\,dt ={16\over 9}\sqrt{34}e^4-{1\over9}\sqrt{34}\,e. $$
All of these ideas extend to three dimensions in the obvious way.
Example 18.2.2 Compute $\ds\int_C x^2 z \,ds$ where $C$ is the line segment from $(0,6,-1)$ to $(4,1,5)$.
We write the line segment as a vector function: ${\bf v}=\langle 0,6,-1\rangle + t\langle 4,-5,6\rangle$, $0\le t\le 1$, or in parametric form $x=4t$, $y=6-5t$, $z=-1+6t$. Then $$\int_C x^2 z\,ds = \int_0^1 (4t)^2 (-1+6t)\sqrt{16+25+36}\,dt =16\sqrt{77}\int_0^1 -t^2+6t^3\,dt={56\over3}\sqrt{77}. $$
Now we turn to a perhaps more interesting example. Recall that in the simplest case, the work done by a force on an object is equal to the magnitude of the force times the distance the object moves; this assumes that the force is constant and in the direction of motion. We have already dealt with examples in which the force is not constant; now we are prepared to examine what happens when the force is not parallel to the direction of motion.
We have already examined the idea of components of force, in example 14.3.4: the component of a force $\bf F$ in the direction of a vector $\bf v$ is $${{\bf F}\cdot {\bf v}\over|{\bf v}|^2}{\bf v},$$ the projection of $\bf F$ onto $\bf v$. The length of this vector, that is, the magnitude of the force in the direction of $\bf v$, is $${{\bf F}\cdot {\bf v}\over|{\bf v}|},$$ the scalar projection of $\bf F$ onto $\bf v$. If an object moves subject to this (constant) force, in the direction of $\bf v$, over a distance equal to the length of $\bf v$, the work done is $${{\bf F}\cdot {\bf v}\over|{\bf v}|}|{\bf v}|={\bf F}\cdot {\bf v}.$$ Thus, work in the vector setting is still "force times distance'', except that "times'' means "dot product''.
If the force varies from point to point, it is represented by a vector field $\bf F$; the displacement vector $\bf v$ may also change, as an object may follow a curving path in two or three dimensions. Suppose that the path of an object is given by a vector function ${\bf r}(t)$; at any point along the path, the (small) tangent vector ${\bf r}'\,\Delta t$ gives an approximation to its motion over a short time $\Delta t$, so the work done during that time is approximately ${\bf F}\cdot{\bf r}'\,\Delta t$; the total work over some time period is then $$\int_{t_0}^{t_1} {\bf F}\cdot{\bf r}'\,dt.$$ It is useful to rewrite this in various ways at different times. We start with $$\int_{t_0}^{t_1} {\bf F}\cdot{\bf r}'\,dt=\int_C {\bf F}\cdot\,d{\bf r},$$ abbreviating ${\bf r}'\,dt$ by $d{\bf r}$. Or we can write $$\int_{t_0}^{t_1} {\bf F}\cdot{\bf r}'\,dt= \int_{t_0}^{t_1} {\bf F}\cdot{{\bf r}'\over|{\bf r}'|}|{\bf r}'|\,dt= \int_{t_0}^{t_1} {\bf F}\cdot{\bf T}\,|{\bf r}'|\,dt= \int_{C} {\bf F}\cdot{\bf T}\,ds,$$ using the unit tangent vector $\bf T$, abbreviating $|{\bf r}'|\,dt$ as $ds$, and indicating the path of the object by $C$. In other words, work is computed using a particular line integral of the form we have considered. Alternately, we sometimes write $$\eqalign{ \int_C {\bf F}\cdot{\bf r}'\,dt&= \int_C \langle f,g,h\rangle\cdot\langle x',y',z'\rangle\,dt= \int_C \left(f{dx\over dt}+g{dy\over dt}+h{dz\over dt}\right)\,dt\cr &= \int_C f\,dx+g\,dy+h\,dz= \int_C f\,dx+\int_C g\,dy+\int_C h\,dz,\cr}$$ and similarly for two dimensions, leaving out references to $z$.
Example 18.2.3 Suppose an object moves from $(-1,1)$ to $(2,4)$ along the path ${\bf r}(t)=\langle t,t^2\rangle$, subject to the force ${\bf F}=\langle x\sin y,y\rangle$. Find the work done.
We can write the force in terms of $t$ as $\langle t\sin(t^2),t^2\rangle$, and compute ${\bf r}'(t)=\langle 1,2t\rangle$, and then the work is $$\int_{-1}^2 \langle t\sin(t^2),t^2\rangle\cdot\langle 1,2t\rangle\,dt = \int_{-1}^2 t\sin(t^2) + 2t^3\,dt={15\over2}+{\cos(1)-\cos(4)\over2}. $$
Alternately, we might write $$\int_C x\sin y\,dx+\int_C y\,dy= \int_{-1}^2 x\sin(x^2)\,dx + \int_1^4 y\,dy= -{\cos(4)\over 2}+{\cos(1)\over 2}+{16\over2}-{1\over2}$$ getting the same answer.
Exercises 18.2
Ex 18.2.1Compute $\ds\int_C xy^2\,ds$ along the line segment from$(1,2,0)$ to $(2,1,3)$.(answer)
Ex 18.2.2Compute $\ds\int_C \sin x\,ds$ along the line segment from$(-1,2,1)$ to $(1,2,5)$.(answer)
Ex 18.2.3Compute $\ds\int_C z\cos(xy)\,ds$ along the line segment from$(1,0,1)$ to $(2,2,3)$.(answer)
Ex 18.2.4Compute $\ds\int_C \sin x\,dx+\cos y\,dy$ along the top halfof the unit circle, from $(1,0)$ to $(-1,0)$.(answer)
Ex 18.2.5Compute $\ds\int_C xe^y\,dx+x^2y\,dy$ along the line segment$y=3$, $0\le x\le 2$.(answer)
Ex 18.2.6Compute $\ds\int_C xe^y\,dx+x^2y\,dy$ along the line segment$x=4$, $0\le y\le 4$.(answer)
Ex 18.2.7Compute $\ds\int_C xe^y\,dx+x^2y\,dy$ along the curve$x=3t$, $y=t^2$, $0\le t\le1$.(answer)
Ex 18.2.8Compute $\ds\int_C xe^y\,dx+x^2y\,dy$ along the curve $\langle e^t,e^t\rangle$, $-1\le t\le1$.(answer)
Ex 18.2.9Compute $\ds\int_C \langle \cos x,\sin y\rangle\cdot d{\bf r}$ along the curve $\langle t,t\rangle$, $0\le t\le1$.(answer)
Ex 18.2.10Compute $\ds\int_C \langle 1/xy,1/(x+y)\rangle\cdotd{\bf r}$ along the path from $(1,1)$ to $(3,1)$ to $(3,6)$ using straight line segments.(answer)
Ex 18.2.11Compute $\ds\int_C \langle 1/xy,1/(x+y)\rangle\cdotd{\bf r}$ along the curve $\langle 2t,5t\rangle$, $1\le t\le 4$.(answer)
Ex 18.2.12Compute $\ds\int_C \langle 1/xy,1/(x+y)\rangle\cdot d{\bf r}$ along the curve $\langle t,t^2\rangle$, $1\le t\le 4$.(answer)
Ex 18.2.13Compute $\ds\int_C yz\,dx+xz\,dy+xy\,dz$ along the curve$\langle t,t^2,t^3\rangle$, $0\le t\le1$.(answer)
Ex 18.2.14Compute $\ds\int_C yz\,dx+xz\,dy+xy\,dz$ along the curve$\langle \cos t,\sin t,\tan t\rangle$, $0\le t\le\pi$.(answer)
Ex 18.2.15An object moves from $(1,1)$ to$(4,8)$ along the path ${\bf r}(t)=\langle t^2,t^3\rangle$,subject to the force ${\bf F}=\langle x^2,\sin y\rangle$. Find the workdone. (answer)
Ex 18.2.16An object moves along the line segment from $(1,1)$ to $(2,5)$,subject to the force ${\bf F}=\langlex/(x^2+y^2),y/(x^2+y^2)\rangle$. Find the work done.(answer)
Ex 18.2.17An object moves along the parabola ${\bf r}(t)=\langlet,t^2\rangle$, $0\le t\le1$, subject to the force ${\bf F}=\langle1/(y+1),-1/(x+1)\rangle$. Find the work done.(answer)
Ex 18.2.18An object moves along the line segment from $(0,0,0)$ to$(3,6,10)$,subject to the force ${\bf F}=\langle x^2,y^2,z^2\rangle$. Find the workdone. (answer)
Ex 18.2.19An object moves along the curve ${\bf r}(t)=\langle\sqrt{t},1/\sqrt{t},t\rangle$ $1\le t\le4$, subject to the force ${\bf F}=\langle y,z,x\rangle$. Find the work done.(answer)
Ex 18.2.20An object moves from $(1,1,1)$ to$(2,4,8)$ along the path ${\bf r}(t)=\langle t,t^2,t^3\rangle$,subject to the force ${\bf F}=\langle \sin x,\sin y,\sin z\rangle$. Find the workdone. (answer)
Ex 18.2.21An object moves from $(1,0,0)$ to$(-1,0,\pi)$ along the path ${\bf r}(t)=\langle \cos t,\sin t, t\rangle$,subject to the force ${\bf F}=\langle y^2,y^2,xz\rangle$. Find the workdone. (answer)
Ex 18.2.22Give an example of a non-trivial force field ${\bf F}$ andnon-trivial path ${\bf r}(t)$ for which the total work done moving alongthe path is zero.
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As xnor points out in his answer, this question is basically asking for the way to most evenly distribute $6^n$ results among $100$ bins, and gives a very brief description of the solution. I'll go into a
bit more detail here. If you're not interested in the proofs, skip to section 2.3 Summary
In order to construct a "best" algorithm for converting rolls of one die into another, there are two types of optimality to consider:
Minimum "score." BmyGuest's metric:$$\sum_i ( p_i - 1/100 )^2$$is basically the chi-squared statistic for goodness-of-fit, between the distribution resulting from the algorithm and a uniform distribution. I'll call the metric applied to a particular algorithm that algorithm's score. Minimum number of rolls. I'll consider only the expected number of rolls for an algorithm.
Note I'll be diverging a from typical RPG terminology: when I refer to $\mathsf{3d6}$, for example, I mean that you should roll three six-sided dice
without adding them, keeping track of the order the dice were rolled in (usually the dice are treated as indistinguishable from each other, i.e. rolled all at once). This is important because, as you noted, the results in a traditional roll are not equiprobable.
I also consider a $\mathsf{d}n$ to be numbered from $0$ to $n-1$, to make some of the math a little easier. If you're following my algorithms with real dice, just map $n\to 0$ for the input and vice-versa for the output.
1 Optimal Scoring Algorithms
In the general case, we want to map $m$ "results" into $n$ "bins." (In the original case, we have $m=6^k$, $n=100$.) Consider an algorithm that maps $r_i$ results to bin $i$.
The probability of landing in bin $i$ is $r_i/m$, so the score is:
$$\sum_{i=0}^{n-1}\left(\frac{r_i}{m} - \frac{1}{n}\right)^2$$
1.1 Distribution of an Optimal Algorithm
Since the score is a sum of independent terms we can easily isolate the effect of changing a pair of elements of the solution, $(r_i,r_j)$. Let's decrease $r_i$ by $1$ and increase $r_j$ by $1$, and see under what conditions the score decreases:
$$\begin{align}\left(\frac{r_i}{m} - \frac{1}{n}\right)^2 + \left(\frac{r_j}{m} - \frac{1}{n}\right)^2 &> \left(\frac{r_i-1}{m} - \frac{1}{n}\right)^2 + \left(\frac{r_j+1}{m} - \frac{1}{n}\right)^2 \\\frac{r_i^2 + r_j^2}{m^2} - 2\frac{r_i + r_j}{m n} + \frac{2}{n^2} &> \frac{r_i^2 + r_j^2 - 2(r_i - r_j) + 2}{m^2} - 2\frac{r_i + r_j}{m n} + \frac{2}{n^2} \\0 &> \frac{2}{m^2}(1 - r_i + r_j) \\r_i &> 1 + r_j\end{align}$$
This means that if the solution contains two values which differ from each other by more than one, it cannot be optimal (since moving the two values closer together will decrease the score). Therefore, in an optimal solution $r_i$ will take on at most two values, separated from each other by $1$. Let's call them $r$ and $r+1$. Calling the number of bins with $r$ results $x$, and the number of bins with $r+1$ results $x'$, we can write a system of equations:
$$n = x + x' \\m = rx + (r+1)x'$$
We can solve for $x$ and $x'$ in terms of $r$:
$$x' = n - x \\m = rx + (r+1)(n-x) \\m = rx + rn - rx + n - x \\m = (r+1)n - x \\x = (r+1)n-m$$
Given that $0\leq x \leq n$:
$$0 \leq (r+1)n-m \leq n \\0 \leq (r+1) - \frac{m}{n} \leq 1 \\\frac{m}{n} \leq r+1 \leq \frac{m}{n} + 1 \\\frac{m}{n} - 1 \leq r \leq \frac{m}{n}$$
Since $r$ must be an integer, we have:
$$r = \left\lfloor \frac{m}{n} \right\rfloor$$
Substituting back in, we see that:
$$\begin{align}r + 1 &= \left\lfloor \frac{m}{n} \right\rfloor + 1 \\x &= \left(\left\lfloor \frac{m}{n} \right\rfloor +1 \right)n - m \\x' &= m - \left\lfloor \frac{m}{n} \right\rfloor n\end{align}$$
1.2 Score of an Optimal Algorithm
The score of the optimal algorithm described above gives us a way to compute the minimum possible score for a given $m$ and $n$:
$$\sum_{i=0}^{n-1}\left(\frac{r_i}{m} - \frac{1}{n}\right)^2 = x\left(\frac{r}{m}-\frac{1}{n}\right)^2+x'\left(\frac{r+1}{m}-\frac{1}{n}\right)^2 \\= \left[\left(\left\lfloor \frac{m}{n} \right\rfloor +1 \right)n - m\right]\left(\frac{\lfloor m/n\rfloor}{m}-\frac{1}{n}\right)^2+\left[m - \left\lfloor \frac{m}{n} \right\rfloor n\right]\left(\frac{\lfloor m/n\rfloor + 1}{m}-\frac{1}{n}\right)^2$$
To simplify this expression, we can make use of the identity:
$$\left\lfloor \frac{m}{n} \right\rfloor = \frac{m}{n} - \frac{m\ \mathrm{mod}\ n}{n}$$
To prevent the equations from getting too long, I'll use $m' = m\ \mathrm{mod}\ n$:
$$\left[\left(\frac{m-m'}{n} +1 \right)n - m\right]\left(\frac{m-m'}{mn}-\frac{1}{n}\right)^2+\left[m - \frac{m-m'}{n} n\right]\left(\frac{m-m'+n}{mn}-\frac{1}{n}\right)^2 \\= (m-m' + n - m)\left(\frac{m-m' - m}{mn}\right)^2+(m - m+m')\left(\frac{m-m'+n-m}{mn}\right)^2 \\= \frac{(n-m')(-m')^2+(m')(n-m')^2}{(mn)^2} \\= \frac{(n-m')(m')}{(mn)^2}(m' + n-m') \\= \frac{1}{n}\left(\frac{m'}{m}\right)\left(\frac{n-m'}{m}\right)$$
Thus we can see that the score decreases as $m$ increases, and has local minima when $m'$ or $n-m'$ are close to zero (i.e. $m$ close to divisible by $n$). This agrees with our intuition.
1.3 Optimal Scores for Mapping a $k\mathsf{d6}$ to a $\mathsf{d100}$
Using the expression above, Here's a table of the best possible scores for each $k \le 10$:
k m (m mod n) score
---------------------------
0 1 1 0.990
1 6 6 0.157
2 36 36 0.0178
3 216 16 2.88e-4
4 1,296 96 2.29e-6
5 7,776 76 3.02e-7
6 46,656 56 1.13e-8
7 279,936 36 2.94e-10
8 1,679,616 16 4.76e-12
9 10,077,696 96 3.78e-14
10 60,466,176 76 4.99e-15
An unstated assumption was that the assignment of results to bins was the same on each roll. Some algorithms achieve a better average score by allowing the results of one roll to influence the next. However, the score of each individual roll is limited by the same maximum.
2 Optimal Expected Number of Rolls
An algorithm can have an expected number of rolls less than the maximum number of rolls, while maintaining the same score. In fact, an algorithm can have a finite expected number of rolls even with an infinite maximum number. This is possible when the algorithm allows "early termination."
An algorithm maps each input roll (previously "result") to a corresponding output (previously "bin"). If an algorithm maps all rolls beginning with a certain prefix sequence to the same output, once that prefix has been rolled we know the output will be the same regardless of the subsequent rolls and we can stop rolling.
Pay careful attention to the terminology in the following sections: a roll is the roll of a single die, and a sequence or input is a list of rolls input to an algorithm
2.1 Bounding the Expected Number of Rolls
In order to compute the minimum expected number of rolls, I make the assumption that each roll is made with the same type of die. That is, if the first roll is a $\mathsf{d6}$, all the subsequent rolls will also be $\mathsf{d6}$s. Essentially, instead of rolling a large number of different dice, we roll one die over and over. (Note this restriction means that we cannot analyze algorithms that use another source of randomness such as the orientation of the die.) Assume we are using a $\mathsf{d}b$.
The expected number of rolls is the sum, over all inputs, of the probability of each input times its length. If the length of sequence $i$ is $k_i$, then the expected number of rolls $E$ is:
$$E = \sum_i k_i b^{-k_i}$$
Consider replacing a roll $\{x_1, x_2, x_3, \ldots, x_k\}$ with $b$ rolls all mapping to the same output as the original roll:
$$\{x_1, x_2, x_3, \ldots, x_k, 0\} \\\{x_1, x_2, x_3, \ldots, x_k, 1\} \\\vdots \\\{x_1, x_2, x_3, \ldots, x_k, b-1 \}$$
The algorithm is still valid, since all possible rolls are mapped to outputs, and the probability of each output is the same. The change in $E$ is:
$$b\left[(k+1)b^{-(k+1)}\right]- k b^{-k} \\= (k+1) b^{-k} - k b^{-k} \\= b^{-k}$$
Thus, splitting an input increases $E$. Conversely, merging the set of inputs with a given prefix will decrease $E$. Therefore, choosing how many inputs of each length to assign to a given output can be solved by a greedy algorithm.
Imagine that the output space of the algorithm is represented by an interval of length $1$, and that each output is a sub-interval of length $p_i$ (the probability of that output). Each input is an interval of length $b^{-k}$, and the problem is to cover each output interval with a number of input intervals, with no overlap. The greedy algorithm does so optimally by repeatedly inserting the largest input (smallest $k$) that fits in the remaining space in the output.
Example: find the optimal distribution of inputs for $b=6$, $k=4$, $n=100$. We have:
$$ m = 1296 \\ \left\lfloor \frac{m}{n} \right\rfloor = 12 \\ m' = 96 $$
Remember that $p_i=r_i/m$, and $r_i=\lfloor m/n \rfloor + 0\text~{or}~1$, so:
$$ p = \frac{12}{1296}~\text{or}~\frac{13}{1296} $$
$6^{-1}=1/6$ and $6^{-2}=1/36$ are both too big, but $6^{-3}=1/216$ is small enough to fit in both $p_i$:
$$ \begin{align} p &= 2\times 6^{-3} + \frac{0}{1296} \\
&~\text{or}~2\times 6^{-3} + \frac{1}{1296} \end{align} $$
Finally, we fill up the remainder with multiples of $6^{-4}$:
$$ \begin{align} p &= 2\times 6^{-3} \\ &~\text{or}~2\times 6^{-3} +
1\times 6^{-4} \end{align} $$
We can see that this process is just the expression of $p$ as a decimal in base $b$! With this in mind, I'll skip an analysis of the minimum $E$ for optimally-scoring algorithms of finite $k$, and jump straight to algorithms where $k$ is infinite.
2.2 Optimal Algorithm with Exact Distribution
With an infinite maximum $k$, the value of $m$ is also infinite, and the score of the algorithm drops to zero. This means that we can achieve an exactly uniform output distribution. This allows us to forget about $m$, $m'$, $x$, and other values that complicated the analysis. $p=1/n$ for all the outputs. We can find the decimal expansion of $p$ by using long division to divide $n$ into $1$.
Example: $b=6$, $n=100=244_6$. The long division is done in base-6.
.0020543...
__________
244 ) 1
.532
----
.024
.02152
------
.00204
.001504
-------
.000132
.0001220
--------
.0000100
The final row is a power of the base, signifying that the decimal expansion repeats. Note that if we round the expansion to four digits, we get the same result as in the example for $k=4$.
With the expansion in hand, we can write an expression for $E$:
$$
E = n\left[\sum_k d_k \times k \times b^{-k} \right] + (e+ k_r)\times b^{-k_r} \\
E = 100\left[2\times 3\times 6^{-3} + 5\times 5\times 6^{-5} + 4\times 6\times 6^{-6} + 3\times 7\times 6^{-7} \right] + (E+5)\times 6^{-5} \\
E = 14738/4665 \approx 3.159\ldots
$$
The factor of $n$ appears because there are $n$ identical outputs contributing to $E$. $d_k$ is the $k$-th digit in the base-$b$ expansion of $1/n$, and $k_r$ is the number of repeating digits.
I don't know of a way to simplify the formula for $E$, since the decimal expansion can vary wildly with $b$ and $n$. For example, although $1/100=0.00\overline{20543}_6$ is a fairly short expansion, changing $n$ by $1$ results in $1/101=0.\overline{0020455351}_6$.
2.3 A Implementation of an Optimal Algorithm
Here I will describe one class of optimal algorithm, based on the interpretation of the input as a base-$b$ decimal such that the $k$-th roll is the $k$-th digit after the decimal point. For example, the input:
$$\{0, 4, 2, 2, 3\}$$
(with $b=6$) is equivalent to the decimal:
$$.04223_6$$
Note that inputs are uniformly distributed over the interval $[0,1)$.
Unlike the previous section where we grouped the inputs together based on their output, here I'll group the inputs based on size. The total amount of the interval occupied by inputs of length $k$ is $n d_k b^{-k}$. Note that this quantity is equivalent to the amount that we subtracted from the remainder at step $k$ of the long division, for example:
.0020543...
__________
244 ) 1
.532 <= 100 * 2 * 6^-3
----
.024
.02152 <= 100 * 5 * 6^-5
------
.00204
.001504 <= 100 * 4 * 6^-6
-------
.000132
.0001220 <= 100 * 3 * 6^-7
--------
.0000100
Thus, the remainder at each stage is simply the amount of the interval remaining after all the inputs of length $k$ or less have been removed. This leads us to the algorithm, which I'll demonstrate for $b=6$, $n=100$.
Setup: put the remainders from dividing $n$ into $1$ in base $b$ into
a table, along with the number of new digits for each step.
roll 3
.024 => 24 <--+
roll 2 |
.00204 => 204 |
roll 1 |
.000132 => 132 |
roll 1 |
.0000100 => 100 |
roll 1 --+
For every output you need, start at the top of the table. Alternate
between two steps:
Roll the number of dice listed, and append their digits to the current value (which starts at zero).
If the current value is greater than the number on the current row, return the difference of the two $\mathrm{mod}\ n$; otherwise continue
to the next row.
If you reach the end of the table, go back to the beginning, skipping the first row. (Note that if the decimal expansion terminates, the last remainder will be $0$ and you will never reach the end of the table.)
An example:
Start with a current value of $0$.
The first row is
roll 3, so roll $\mathsf{3d6}$ and append their digits: $$ \{6, 1, 5\} \to 015_6 $$
The next row is
24. Since $15_6 < 24_6$, continue to the next row.
The next row is
roll 2, so roll $\mathsf{2d6}$: $$ \{4, 6\} \to 1540_6 $$
The next row is
204. Since $1540_6 \geq 204_6$, return the difference between the two, $\text{mod}\ 100$: $$ 1540_6 - 204_6 = 1332_6 = 344 \to \boxed{44} $$
Side note: The probability that you have to repeat the table at least once is equal to the last remainder: $.00001_6=1/7776\approx 0.013\%$. Assume that the DM and your entire party of five (six people total) make one roll per minute, every minute, for the next 10 years. That's a total of around $3\cdot 10^{7}$ rolls. Therefore the number of times you'll loop through the table will be around $\log_{7776}3\cdot 10^{7} \approx 1.9$!
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In the CAPM theory Beta of asset $i$ are estimated in this way:
$ \beta_i = \frac{\sigma_{im}}{\sigma^2_m} $ where $\sigma_{im} = \rho_{im} \sigma_i \sigma_m$
But all these data are historical data. So, I'm wondering what if I use
$\sigma^2_m$ <-
Implied volatilityof SP500 (VIX)
$\sigma_{im}$ <-
implied volatilityfor the asset $i$ using the at-the-money call option with a 1-month maturity. $\rho_{im}$ will be statistically estimated.
This way is a better estimation of the $\beta_{i}$ for the next month?
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I'm using Area Under Curve ROC as a performance measure of my classification algorithms (logistic regressions). Since I'm going to choose the model that maximize the Area Under Curve ROC, I would like to know if AUC penalizes somehow models with too many regressors (for example, like BIC information criterion).
You mention in the comments that you are computing the AUC using a 75-25 train-test split, and you are puzzled why AUC is maximized when training your model on only 8 of your 30 regressors. From this you have gotten the impression that AUC is somehow penalizing complexity in your model.
In reality there is something penalizing complexity in your model, but it is not the AUC metric.
Train-test splitting is what makes it possible to use pretty much any metric, even AUC, for model selection, even if they have no inherent penalty on model complexity. It is the train-test split.
As you probably know, we do not measure performance on the same data that we train our models on, because the training data error rate is generally an overly optimistic measure of performance in practice (see Section 7.4 of the ESL book). But this is not the most important reason to use train-test splits. The most important reason is to avoid overfitting with excessively complex models.
Given two models A and B such that B "contains A" (the parameter set of B contains that of A) the training error is mathematically
guaranteed to favor model B, if you are fitting by optimizing some fit criterion and measuring error by that same criterion. That's because B can fit the data in all the ways that A can, plus additional ways that may produce lower error than A's best fit. This is why you were expecting to see lower error as you added more predictors to your model.
However, by splitting your data into two reasonably independent sets for training and testing, you guard yourself against this pitfall. When you fit the training data aggressively, with many predictors and parameters, it doesn't necessarily improve the test data fit. In fact, no matter what the model or fit criterion, we can generally expect that a model which has overfit the training data will not do well on an independent set of test data which it has never seen. As model complexity increases into overfitting territory, test set performance will generally worsen as the model picks up on increasingly spurious training data patterns, taking its predictions farther and farther away from the actual trends in the system it is trying to predict. See for example slide 4 of this presentation, and sections 7.10 and 7.12 of ESL.
If you still need convincing, a simple thought experiment may help. Imagine you have a dataset of 100 points with a simple linear trend plus gaussian noise, and you want to fit a polynomial model to this data. Now let's say you split the data into training and test sets of size 50 each and you fit a polynomial of degree 50 to the training data. This polynomial will interpolate the data and give zero training set error, but it will exhibit wild oscillatory behavior carrying it far, far away from the simple linear trendline. This will cause extremely large errors on the test set, much larger than you would get using a simple linear model. So the linear model will be favored by CV error. This will also happen if you compare the linear model against a more stable model like smoothing splines, although the effect will be less dramatic.
In conclusion, by using train-test splitting techniques such as CV, and measuring performance on the test data, we get an implicit penalization of model complexity, no matter what metric we use, just because the model has to predict on data it hasn't seen. This is why train-test splitting is universally used in the modern approach to evaluating performance in regression and classification.
There is a good reason why the regression coefficients in logistic regression are estimated by maximizing the likelihood or penalized likelihood. This leads to certain optimality properties. The concordance probability ($c$-index; AUROC) is a useful supplemental measure for describing the final model's predictive discrimination, but it is not sensitive enough for the use you envisioned nor would it lead to an optimal model. This is quite aside from the overfitting issue, which affects both the $c$-index and the (unpenalized) likelihood.
The $c$-index can reach its maximum with a misleadingly small number of predictors, even though it does not penalize for model complexity, because the concordance probability does not reward extreme predictions that are "correct". $c$ uses only the rank order of predictions and not the absolute predicted values. $c$ is not sensitive enough to be used to compare two models.
Seeking a model that does not use the entire list of predictors is often not well motivated. Model selection brings instability and extreme difficulty with co-linearities. If you want optimum prediction, using all candidate features and incorporating penalization will work best in most situations you are likely to encounter. The data seldom have sufficient information to allow one to make correct choices about which variables are "important" and which are worthless.
This should help clarify a few things, in as few words as possible:
AUC= measure of model's actualpredictive performance BIC= estimateof model's predictive performance Performance Measures, like AUC, are something you would use to evaluate a model's predictions on data it has never seen before. Information Criteria, like BIC, on the other hand, attempt to guess at how well a model would make predictions by using how well the model fit the training data AND the number of parameters used to make that fit as a penalty (using the number of parameters makes for better guesses).
Simply put, BIC (and other information criteria), approximate what performance measures, like AUC, give you directly. To be more precise:
Information criteria attempt to approximate out-of-sample deviance using only training data, and make better approximations when accounting for the number of parameters used. Direct performance measures, like deviance or AUC, are used to asses how well a model makes predictions on validation/test data. The number of parameters is irrelevant to them because they're illustrating performance in the most straightforward way possible.
I thought the link between information criteria and performance measures was hard to understand at first, but it's actually quite simple. If you were to use deviance instead of AUC as a performance measure then BIC would basically tell you what deviance you could expect if you actually made predictions with your model, and then measured their deviance.
This begs the question, why use information criteria at all? Well you shouldn't if you're just trying to build the most accurate model possible.
Stick to AUC because models that have unnecessary predictors are likely to make worse predictions (so AUC doesn't penalize them per se, they just happen to have less predictive power).
In logistic regression (I do it univariate for easier typing) you try to explain a binary outome $y_i \in \{0,1\}$ by assuming that it is the outcome of a Bernouilli random variable with a success probability $p_i$ that depends on your explanatory variable $x_i$, i.e. $p_i=P(y_i=1|_{x_i})=f(x_i)$, where $f$ is the logistic function: $f(x)=\frac{1}{1+e^{-(\beta_0+\beta_1 x)}}$. The parameters $\beta_i$ are estimated by maximum likelihood. This works as follows: for the $i$-th observation you observe the outcome $y_i$ and the success probability is $p_i=f(x_i)$, the probability to observe $y_i$ for a Bernouilli with success probability $p_i$ is $p_i^{y_i}(1-p_i)^{(1-y_i)}$. So, for all the observations in the sample, assuming independence between observations, the probability of observing $y_i, i=1,2, \dots n$ is $\prod_{i=1}^np_i^{y_i}(1-p_i)^{(1-y_i)}$. Using the above definition of $p_i=f(x_i)$ this becomes $\prod_{i=1}^nf(x_i)^{y_i}(1-f(x_i))^{(1-y_i)}=$. As the $y_i$ and $x_i$ are observed values, we can see this as a function of the unknown parameters $\beta_i$, i.e. $\mathcal{L}(\beta_0, \beta_1)=\prod_{i=1}^n\left(\frac{1}{1+e^{-(\beta_0+\beta_1 x_i)}}\right)^{y_i}\left(1-\frac{1}{1+e^{-(\beta_0+\beta_1 x_i)}}\right)^{(1-y_i)}$. Maximimum likelihood finds the values for $\beta_i$ that maximise $\mathcal{L}(\beta_0, \beta_1)$. Let us denote this maximum $(\hat{\beta}_0, \hat{\beta}_1)$, then the value of the likelihood in this maximum is $\mathcal{L}(\hat{\beta}_0, \hat{\beta}_1)$.
In a similar way, if you would have used two explanatory variables $x_1$ and $x_2$, then the likelihood function would have had three parameters $\mathcal{L}'(\beta_0, \beta_1, \beta_2)$ and the maximum would be $(\hat{\beta}'_0, \hat{\beta}'_1, \hat{\beta}'_2)$ and the value of the likelihood would be $\mathcal{L}'(\hat{\beta}'_0, \hat{\beta}'_1, \hat{\beta}'_2)$. Obviously it would hold that $\mathcal{L}'(\hat{\beta}'_0, \hat{\beta}'_1, \hat{\beta}'_2) > \mathcal{L}(\hat{\beta}_0, \hat{\beta}_1)$, whether the incerase in likelihood is significant has to be 'tested' with e.g. a likelihood ratio test.
So likelihood ratio tests allow you te 'penalize' models with too many regressors.
This is not so for AUC ! In fact AUC does not even tell you whether your 'success probabilities' are well predicted ! If you take all possible couples $(i,j)$ where $y_i=1$ and $y_j=0$ then AUC will be equal to the fraction of all these couples that have $p_i < p_j$. So AUC has to do with (1) how good your model is in distinguishing between '0' and '1' (it tells you about couples with one 'zero' and one 'one'), it does not say anything about how good your model is in predicting the probabilities ! and (2) it is only based on the 'ranking' ($p_i < p_j$) of the probabilities.
If adding 1 explanatory variable does not change anything to the ranking of the probabilities of the subjects, then AUC will not change by adding an explanatory variable. So the first question you have to ask is what you want to predict: do you want to distinguish between zeroes and ones or do you want to have 'well predicted probabilities' ? Only after you have answered this question you can look for the most parsimonious technique.
If you want to distinguish between zeroes and ones then ROC/AUC may be an option, if you want well predicted probabilities you should take a look at Goodness-of-fit test in Logistic regression; which 'fit' do we want to test?.
As Marc said, AUC is only a measure of performance, just like missclassification rate. It does not require any information about the model.
Conversely, BIC, AIC, need to know the number of parameters of your model to be evaluated.
There is no good reason, if all of your predictors are relevant, that the missclassification rate or the AUC decreases when removing variables.
However, it is quite common that combining a learning algorithm, an importance measure of the variables and variable selection (based on the importance the algorithm grants them) will perform better than fitting the model on the whole data set.
You have an implementation of this method for Random Forests in the R RFauc package.
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Answer
The solution set is $$\{0.8751+2\pi n, 2.2665+2\pi n,3.5908+2\pi n, 5.834+2\pi n, n\in Z\}$$
Work Step by Step
$$\sin x(3\sin x-1)=1$$ 1) Solve the equation over the interval $[0,2\pi)$ $$\sin x(3\sin x-1)=1$$ $$3\sin^2x-\sin x-1=0$$ Consider the equation as a quadratic formula, with $a=3, b=-1, c=-1$ - Calculate $\Delta$: $\Delta=b^2-4ac=(-1)^2-4\times3\times(-1)=1+12=13$ - Find out $\sin x$: $$\sin x=\frac{-b\pm\sqrt\Delta}{2a}=\frac{1\pm\sqrt{13}}{6}$$ - For $\sin x=\frac{1+\sqrt{13}}{6}$. $x$ would be $$x=\sin^{-1}\frac{1+\sqrt{13}}{6}$$ $$x\approx0.8751$$ (Be careful with degrees and radians. Here we need to use radians.) However, over the interval $[0,2\pi)$, there is one more value of $x$ where $\sin x=\frac{1+\sqrt{13}}{6}$, which is $x=\pi-0.8751\approx2.2665$ In total, $x\in\{0.8751, 2.2665\}$ - For $\sin x=\frac{1-\sqrt{13}}{6}$. $x$ would be $$x=\sin^{-1}\frac{1-\sqrt{13}}{6}$$ $$x\approx-0.4492$$ In the interval $[0,2\pi)$, that equals to $x\approx-0.4492+2\pi\approx5.834$ However, over the interval $[0,2\pi)$, there is one more value of $x$ where $\sin x=\frac{1-\sqrt{13}}{6}$, which is $x=\pi-(-0.4492)\approx3.5908$ In total, $x\in\{3.5908, 5.834\}$ Therefore, overall, $$x=\{0.8751, 2.2665,3.5908, 5.834\}$$ 2) Solve the equation for all solutions - The integer multiples of the period of the sine function is $2\pi$. - We apply it to each solution found in step 1. The results are the solution set. So the solution set is $$\{0.8751+2\pi n, 2.2665+2\pi n,3.5908+2\pi n, 5.834+2\pi n, n\in Z\}$$
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9.1. Attention Mechanism¶
In Section 8.14, we encode the source sequence input information in the recurrent unit state and then pass it to the decoder to generate the target sequence. A token in the target sequence may closely relate to some tokens in the source sequence instead of the whole source sequence. For example, when translating “Hello world.” to “Bonjour le monde.”, “Bonjour” maps to “Hello” and “monde” maps to “world”. In the seq2seq model, the decoder may implicitly select the corresponding information from the state passed by the decoder. The attention mechanism, however, makes this selection explicit.
Attention is a generalized pooling method with bias alignment over inputs. The core component in the attention mechanism is the attention layer, or called attention for simplicity. An input of the attention layer is called a query. For a query, the attention layer returns the output based on its memory, which is a set of key-value pairs. To be more specific, assume a query \(\mathbf{q}\in\mathbb R^{d_q}\), and the memory contains \(n\) key-value pairs, \((\mathbf{k}_1, \mathbf{v}_1), \ldots, (\mathbf{k}_n, \mathbf{v}_n)\), with \(\mathbf{k}_i\in\mathbb R^{d_k}\), \(\mathbf{v}_i\in\mathbb R^{d_v}\). The attention layer then returns an output \(\mathbf o\in\mathbb R^{d_v}\) with the same shape as a value.
To compute the output, we first assume there is a score function \(\alpha\) which measures the similarity between the query and a key. Then we compute all \(n\) scores \(a_1, \ldots, a_n\) by
Next we use softmax to obtain the attention weights
The output is then a weighted sum of the values
Different choices of the score function lead to different attentionlayers. We will discuss two commonly used attention layers in the restof this section. Before diving into the implementation, we firstintroduce a masked version of the softmax operator and explain aspecialized dot operator
nd.batched_dot.
import mathfrom mxnet import ndfrom mxnet.gluon import nn
The masked softmax takes a 3-dim input and allows us to filter out some elements by specifying valid lengths for the last dimension. (Refer to Section 8.12 for the definition of a valid length.)
# Save to the d2l package.def masked_softmax(X, valid_length): # X: 3-D tensor, valid_length: 1-D or 2-D tensor if valid_length is None: return X.softmax() else: shape = X.shape if valid_length.ndim == 1: valid_length = valid_length.repeat(shape[1], axis=0) else: valid_length = valid_length.reshape((-1,)) # fill masked elements with a large negative, whose exp is 0 X = nd.SequenceMask(X.reshape((-1, shape[-1])), valid_length, True, axis=1, value=-1e6) return X.softmax().reshape(shape)
Construct two examples, where each example is a 2-by-4 matrix, as the input. If we specify the valid length for the first example to be 2, then only the first two columns of this example are used to compute softmax.
masked_softmax(nd.random.uniform(shape=(2,2,4)), nd.array([2,3]))
[[[0.488994 0.511006 0. 0. ] [0.43654838 0.56345165 0. 0. ]] [[0.28817102 0.3519408 0.3598882 0. ] [0.29034293 0.25239873 0.45725834 0. ]]]<NDArray 2x2x4 @cpu(0)>
The operator
nd.batched_dot takes two inputs \(X\) and \(Y\)with shapes \((b, n, m)\) and \((b, m, k)\), respectively. Itcomputes \(b\) dot products, with
Z[i,:,:]=dot(X[i,:,:], Y[i,:,:] for \(i=1,\ldots,n\).
nd.batch_dot(nd.ones((2,1,3)), nd.ones((2,3,2)))
[[[3. 3.]] [[3. 3.]]]<NDArray 2x1x2 @cpu(0)>
9.1.1. Dot Product Attention¶
The dot product assumes the query has the same dimension as the keys, namely \(\mathbf q, \mathbf k_i \in\mathbb R^d\) for all \(i\). It computes the score by an inner product between the query and a key, often then divided by \(\sqrt{d}\) to make the scores less sensitive to the dimension \(d\). In other words,
Assume \(\mathbf Q\in\mathbb R^{m\times d}\) contains \(m\) queries and \(\mathbf K\in\mathbb R^{n\times d}\) has all \(n\) keys. We can compute all \(mn\) scores by
Now let’s implement this layer that supports a batch of queries and key-value pairs. In addition, it supports randomly dropping some attention weights as a regularization.
# Save to the d2l package.class DotProductAttention(nn.Block): def __init__(self, dropout, **kwargs): super(DotProductAttention, self).__init__(**kwargs) self.dropout = nn.Dropout(dropout) # query: (batch_size, #queries, d) # key: (batch_size, #kv_pairs, d) # value: (batch_size, #kv_pairs, dim_v) # valid_length: either (batch_size, ) or (batch_size, xx) def forward(self, query, key, value, valid_length=None): d = query.shape[-1] # set transpose_b=True to swap the last two dimensions of key scores = nd.batch_dot(query, key, transpose_b=True) / math.sqrt(d) attention_weights = self.dropout(masked_softmax(scores, valid_length)) return nd.batch_dot(attention_weights, value)
Now we create two batches, and each batch has one query and 10 key-valuepairs. We specify through
valid_length that for the first batch, wewill only pay attention to the first 2 key-value pairs, while for thesecond batch, we will check the first 6 key-value pairs. Therefore,though both batches have the same query and key-value pairs, we obtaindifferent outputs.
atten = DotProductAttention(dropout=0.5)atten.initialize()keys = nd.ones((2,10,2))values = nd.arange(40).reshape((1,10,4)).repeat(2,axis=0)atten(nd.ones((2,1,2)), keys, values, nd.array([2, 6]))
[[[ 2. 3. 4. 5. ]] [[10. 11. 12.000001 13. ]]]<NDArray 2x1x4 @cpu(0)>
9.1.2. Multilayer Perceptron Attention¶
In multilayer perceptron attention, we first project both query and keys into \(\mathbb R^{h}\).
To be more specific, assume learnable parameters \(\mathbf W_k\in\mathbb R^{h\times d_k}\), \(\mathbf W_q\in\mathbb R^{h\times d_q}\), and \(\mathbf v\in\mathbb R^{p}\). Then the score function is defined by
This concatenates the key and value in the feature dimension and feeds them into a single hidden layer perceptron with hidden layer size \(h\) and output layer size \(1\). The hidden layer activation function is tanh and no bias is applied.
# Save to the d2l package.class MLPAttention(nn.Block): def __init__(self, units, dropout, **kwargs): super(MLPAttention, self).__init__(**kwargs) # Use flatten=True to keep query's and key's 3-D shapes. self.W_k = nn.Dense(units, activation='tanh', use_bias=False, flatten=False) self.W_q = nn.Dense(units, activation='tanh', use_bias=False, flatten=False) self.v = nn.Dense(1, use_bias=False, flatten=False) self.dropout = nn.Dropout(dropout) def forward(self, query, key, value, valid_length): query, key = self.W_k(query), self.W_q(key) # expand query to (batch_size, #querys, 1, units), and key to # (batch_size, 1, #kv_pairs, units). Then plus them with broadcast. features = query.expand_dims(axis=2) + key.expand_dims(axis=1) scores = self.v(features).squeeze(axis=-1) attention_weights = self.dropout(masked_softmax(scores, valid_length)) return nd.batch_dot(attention_weights, value)
Despite
MLPAttention containing an additional MLP model, given thesame inputs with identical keys, we obtain the same output as for
DotProductAttention.
atten = MLPAttention(units=8, dropout=0.1)atten.initialize()atten(nd.ones((2,1,2)), keys, values, nd.array([2, 6]))
[[[ 2. 3. 4. 5. ]] [[10. 11. 12.000001 13. ]]]<NDArray 2x1x4 @cpu(0)>
9.1.3. Summary¶ An attention layer explicitly selects related information. An attention layer’s memory consists of key-value pairs, so its output is close to the values whose keys are similar to the query.
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From Daron Acemoglu's Introduction to Modern Economic Growth, proposition 9.4 is that:
In the overlapping-generations model with two-period lived households, Cobb-Douglas technology and CRRA preferences, there exists a unique steady-state equilibrium with the capital-labor ratio k* given by (9.15) and as long as $\theta \geq 1$, this steady-state equilibrium is globally stable for all k (0) > 0.
where (9.15) is: $$(1+n)[1+\beta^{-\frac{1}{\theta}}(\alpha(k^*)^{\alpha-1})^{\frac{\theta-1}{\theta}}] = (1-\alpha)(k^*)^{\alpha - 1}$$
My question is why $\theta$ has to be greater than or equal to 1 for the steady-state equilibrium to be globally stable?
As the textbook derives in (9.17): $$k(t+1)=\frac{(1-\alpha)k(t)^\alpha}{(1+n)[1+\beta^{-\frac{1}{\theta}}(\alpha k(t+1)^{\alpha-1})^{\frac{\theta-1}{\theta}}]}$$
We can rearrange to get: $$\begin{align} k(t)&=\big[ \frac{1+n}{1-\alpha} [k(t+1)+\beta^{-\frac{1}{\theta}}\alpha^{\frac{\theta-1}{\theta}}k(t+1)^{(\alpha-1)(1-\frac{1}{\theta})+1}] \big]^\frac{1}{\alpha} \text{ .....(1)} \end{align} $$
Let $n = 0.01$, $\alpha = 0.25$, $\beta = 0.75$.
The blue line is equation (1) where $\theta = 1$ and the red line is 45-degrees line. It can be seen that for all k > 0, k will converge to steady-state k*. The steady-state equilibrium is globally stable.
The case is similar for $\theta > 1$, in which the steady-state equilibrium is globally stable.
The graph is similar to graphs for the case that $\theta \geq 1$. The steady-state equilibrium is still globally stable.
I can't find a case where $\theta < 1$, but the steady-state equilibrium is not globally stable. It seems that $\frac{1}{\alpha} > 1$ for $\alpha \in (0,1)$ determines the shape of equation (1), making the steady-state equilibrium globally stable. It would be good if someone could show me a counter-example where $\theta < 1$, but the steady-state equilibrium is not globally stable. It would be better if someone could show me how to prove proposition 9.4 formally.
Acknowledgement: The graphs are modified from those generated by Wolframalpha.
Edit (Apr 19, 2017):Case $\theta = 0$:Note that when the textbook derives (9.17), it implicitly assumes that $\theta \neq 0$ (for derivation of Euler equation for consumption in P.333 of 2009 edition of the textbook). When $\theta = 0$, equation (1) no longer applies. Returning to the utility maximization problem with $\theta = 0$:
$$ \text{max } U(t) = c_1(t) + \beta(c_2(t+1)) \text{ such that } c_1(t) + \frac{c_2(t+1)}{R(t+1)} = w(t) \\ \Leftrightarrow \text{max } U(t) = c_1(t) + \beta(w(t) - c_1(t))R(t+1) = c_1(t) (1 - \beta R(t+1)) + \beta R(t+1)w(t) \\ \text{ ...Should treat R(t+1) as given as consumer's own optimization problem} $$
s(t) has to be non-negative for $k(t+1) = \frac{s(t)}{1+n}$ and k(t+1) is non-negative. $$ c_1(t)^* = \begin{cases} w(t)\text{, for }\beta R(t+1) <1 \\ [0, w(t)]\text{, for }\beta R(t+1) = 1 \\ 0\text{, for }\beta R(t+1) > 1 \\ \end{cases} $$ $$ s(t)^* = \begin{cases} 0\text{, for }\beta R(t+1) <1 \\ w(t) - c_1(t)^* \in [0, w(t)]\text{, for }\beta R(t+1) = 1 \\ w(t)\text{, for }\beta R(t+1) > 1 \\ \end{cases} $$ For $R(t+1) = f'(k(t+1)) = \alpha k(t+1)^{\alpha - 1}$, $$ k(t+1) = \frac{s(t)}{1+n} = \begin{cases} 0\text{, for }\beta R(t+1) <1 \Leftrightarrow k(t+1) < (\alpha \beta)^{\frac{1}{1 - \alpha}} \\ \frac{w(t) - c_1(t)}{1+n} \in [0, \frac{w(t)}{1+n}]\text{, for }\beta R(t+1) = 1 \Leftrightarrow k(t+1) = (\alpha \beta)^{\frac{1}{1 - \alpha}}\\ \frac{w(t)}{1+n} = \frac{k(t)^\alpha - k(t)\alpha k(t)^{\alpha -1}}{1+n} = \frac{1-\alpha}{1+n}k(t)^\alpha\text{, otherwise} \Leftrightarrow k(t) > [\frac{1+n}{1 - \alpha} (\alpha \beta)^{\frac{1}{1 - \alpha}}]^{\frac{1}{\alpha}} \\ \end{cases} $$
Cases:
Case 1: $\beta R(t+1) < 1 \Leftrightarrow R(t+1) < \frac{1}{\beta}$: As the production function $f(k)$ is Cobb-Douglas, it satisfies the Inada condition: $lim_{k(t) \to 0} f'(k(t)) = \infty$. But as $f'(k(t)) = R(t)$, $lim_{k(t) \to 0} R(t) < \infty$ for $R(t) < \frac{1}{\beta} < \infty$ as $\beta \in (0,1)$, violating the Inada condition. This contradiction means this case is impossible.
Case 2: $\beta R(t+1) = 1 \Leftrightarrow \beta \alpha k(t+1)^{\alpha-1} = 1 \Leftrightarrow k(t+1)^{\alpha-1} = \frac{1}{\alpha \beta} \Leftrightarrow k(t+1)^{1-\alpha} = \alpha \beta$: Denote saving rate at t as $\mathcal{S}(t) = \frac{s(t)}{w(t)}$. $k(t+1) = \frac{s(t)}{1+n} = \frac{\mathcal{S}(t)w(t)}{1+n} = \frac{\mathcal{S}(t)(1-\alpha)k(t)^\alpha}{1+n}$. At steady state, $k^* = \frac{\mathcal{S}^*(1-\alpha){k^*}^\alpha}{1+n}$, meaning $\mathcal{S}^* = \frac{1+n}{1-\alpha}{k^*}^{1-\alpha} = \frac{1+n}{1-\alpha} \alpha \beta$. For $\mathcal{S}^* > 1 \Leftrightarrow (1+n)\alpha\beta > 1 - \alpha \Leftrightarrow \beta > \frac{1 - \alpha}{\alpha (1+n)}$, which is possible. As saving rate cannot be larger than 1, this contradiction means this case is impossible.
Case 3: $\beta R(t+1) > 1$:This case is possible.$$k(t+1) = \frac{1-\alpha}{1+n} k(t)^\alpha$$We can draw the graph:
The red line is 45-degrees line. The blue line is $k(t+1) = \frac{1-\alpha}{1+n} k(t)^\alpha$ where $0 < \alpha < 1$. For all k(0) > 0, k will converge to steady state $k^* = \frac{1-\alpha}{1+n} {k^*}^\alpha \Leftrightarrow k^* = [\frac{1-\alpha}{1+n}]^{\frac{1}{1-\alpha}}$. The steady-state equilibrium is globally stable.
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This is really just a comment, but is way too long. Since in my opinion it's in fact extremely far from an actual answer, I've made it community wiki - I don't think a reputation bonus for it would be appropriate. EDIT: apparently I don't know how to do that (my recollection is that there used to be a cw checkbox, but I don't see it now) - if someone does know how to make this cw, please do.
Here's one way to assign a sheaf to a name which captures the entire structure of the name.
First, the intuitive version. When $\nu$ is a name and $p$ is a condition, we get a class $Elt^*_\nu(p)$ of names given by $$Elt^*_\nu(p)=\{\mu: p\Vdash \mu\in\nu\}$$ (where the set-brackets are used a tad abusively).
This assignment is "sheafy" in the following sense. Put the order topology on $\mathbb{P}$, so that basic open sets are those of the form $$\mathbb{P}_p:=\{q\in\mathbb{P}: q\le p\}$$ for $p\in\mathbb{P}$. Then $p\mapsto Elt^*_\nu(p)$ literally is a sheaf valued in, well, okay fine it's not a sheaf since it's valued in
classes, but meh. In particular, restriction = inclusion.
A couple minor remarks:
We can make things even nicer by replacing $\mathbb{P}$ formally with its
finite completion: $\mathbb{P}^+_{pre}$ is the preorder with underlying set consisting of all finite $F\subseteq\mathbb{P}$ admitting a common extension, ordered by $F_0\le F_1$ iff every condition extending every condition in $F_0$ also extends every condition in $F_1$. We then let $\mathbb{P}^+$ be the induced partial order (although forcing with preorders is also perfectly fine). This makes no substantive difference, but does make things a little easier to write on occasion.
We could also use instead the standard topology on the set of maximal $\mathbb{P}$-filters, but I think that's a bit further away from the Grothendieck topology idea, since the Grothendieck topology lives on $\mathbb{P}$ itself.
So that's the idea. Now we need to fix it. Luckily, this isn't hard (hence the "meh"):
For names $\nu,\mu$ there is a name $\hat{\mu}$ such that $(i)$ the rank of $\hat{\mu}$ (thought of as a normal set) is less than that of $\nu$ and $(ii)$ for every condition $q$ forcing $\mu\in\nu$ we also have $q\Vdash \mu=\hat{\mu}$.
This means we can safely restrict attention to a small class - in particular, a
set - of names and preserve the idea above. Specifically we define the right map $Elt_\nu$ as $$Elt_\nu(p)=\{\mu\in V_{rk(\nu)}: p\Vdash\mu\in\nu\}.$$ This is now genuinely a sheaf, valued in names. Moreover, we can recover $\nu$ up to $\mathbb{1}$-forced equivalence from $Elt_\nu(p)$, so in a precise sense names are sheaves on $\mathbb{P}$ valued in names - an unsurprisingly recursive notion.
It's now worth noting that the construction above satisfies a certain universality property. Namely, fixing a forcing notion $\mathbb{P}$, call a class $\mathcal{C}$ of set-valued sheaves on $\mathbb{P}$ a
nameslike class if (playing a bit fast and loose with proper classes - if you want, we're working in NBG or MK instead of vanilla ZFC):
Every element of the image of a sheaf $S\in\mathcal{C}$ is a subclass (and necessarily a
set) of $\mathcal{C}$.
There is a class function $\rho:\mathcal{C}\rightarrow Ord$ or $S\in\mathcal{C}$ and $R\in im(S)$ we have $\rho(R)<\rho(S)$
(this essentially ensures both well-foundedness and set-likeness). OK fine, this is materially redundant (no function can have an element in its image of higher rank than itself), but it's spiritualy non-redundant: the more general picture is to consider an arbitrary class relation $R\subseteq \mathbb{P}\times V^2$ satisfying certain properties, with $R(p,a,b)$ being interpreted as $p\Vdash a\in b$. In principle, the appropriate structures $V[G]_R$ (see below) could be ill-founded, or well-founded yet "taller" than $V$ itself. This seems potentially interesting to me, but unnecessarily general right now, so I want to make it clear that we're ruling it out.
Given any nameslike class $\mathcal{C}$ and any $G$ which is $\mathbb{P}$-generic over $V$, we get a class-sized $\in$-structure $V[G]_\mathcal{C}$ given by essentially taking the extensional collapse of the structure generated by setting $S_0\in S_1$ whenever $S_0\in S_1(p)$ for some $p\in G$. The point now is:
In a precise and strong sense, $V[G]_\mathcal{C}$ always is a
sub-thing of $V[G]_{V^\mathbb{P}}$ (= $V[G]$ in the usual sense).
Formalizing and proving this is not hard (although it is a bit tedious).
That's all very classical, and while I've never seen it explicitly I'm sure it's folklore; the next step would be to lift it to the categorial context. The key points should be:
Every Grothendieck topology $\tau$ on $\mathbb{P}$ gives rise naturally to an association to each name $\nu$ of a sheaf $Elt_\nu^\tau$ on $(\mathbb{P},\tau)$ valued in (lower-rank) names.
Taking $\tau$ to be the double negation topology, we get the construction above (or something equivalent to it).
An analogue of the universality property above should hold.
At a glance, it seems this idea will still work, but I'm not familiar enough with the topic to say more about it.
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Let $t \in (a, \infty )$
if $f(t) \leq g(t)$ for all $t$, then $\sup f(t) \leq \sup g(t)$
Is there a simple proof for this?
Thank you!!
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$\sup g$ is an upper bound for $g$. Since $g(t) \ge f(t)$ for all $t$, $\sup g$ is also an upper bound for $f$. Therefore $\sup g\ge \sup f$ since $\sup f$ is the smallest upper bound of $f$.
$$f(t)\leq g(t)\leq sup\ g,\ \forall t\in(a,\infty) \Rightarrow f(t) \leq\ sup\ g$$ meaning $sup\ g$ is an upper quota for $f(t)$. Therefore, by definition of $sup$, $$sup\ f \leq sup\ g$$
Let $A=(a,\infty)$, and let $f(t)\leq g(t)$. With the definition of supremum we have:$$f(t) \leq g(t) \leq \sup\{g(t): t\in A\}$$So we have $f(t) \leq \sup\{g(t): t\in A\}$. Because $\sup\{g(t): t\in A\}$ is "an" upper bound of $f(t)$ and $\sup\{f(t): t\in A\}$ is the
smallest upper bound, and remembering that the supremum is a number. It follows that:
$$ f(t) \leq \sup\{f(t): t\in A\} \leq \sup\{g(t): t\in A\}. $$ From which it follows that: $\sup\{f(t): t\in A\} \leq \sup\{g(t): t\in A\}$.
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Sure. It's a little bit lengthy though, so it might take some work to read it:P(m)=\frac{\sqrt{2 \pi n}n^ne^{-n}}{\sqrt{2 \pi m}m^me^{-m}*\sqrt{2 \pi (n-m)}(n-m)^{(n-m)}e^{-(n-m)}}p^m(1-p)^{n-m}=\frac{n^{n+1}}{\sqrt{2 \pi n}*m^{m+\frac{1}{2}}*(n-m)^{(n-m)+\frac{1}{2}}}p^m(1-p)^{n-m}...
I want to show that the binomial distribution:P(m)=\frac{n!}{(n-m)!m!}p^m(1-p)^{n-m}using Stirling's formula:n!=n^n e^{-n} \sqrt{2\pi n}reduces to the normal distribution:P(m)=\frac{1}{\sqrt{2 \pi n}} \frac{1}{\sqrt{p(1-p)}}exp[-\frac{1}{2}\frac{(m-np)^2}{np(1-p)}]...
The link I'm watching from is:http://www.uccs.edu/~math/vidarchive.htmlIt requires free registration, but is available to anyone. The course for some reason is called "real analysis", but it's entirely a course on measure theory.It's kind of boring: basically the professor just writes...
Wait a second. How does that work? Some subsets of the Cantor set are not measurable. But the Cantor set is itself measurable (with zero measure). The problem is the fact that the Borel sigma algebra is not complete. So if you go to a complete one, like the Lebesgue sigma algebra, you introduce...
That's intuitive enough. The intersection of B with A-compliment, which is in the sigma algebra, must have zero measure if A and B have the same measure. So if you take all subsets of regions of zero measure, then you can build any subset inside B (but containing all of A).I'm not exactly...
I just saw a demonstration of a subset on the real line that is non-measurable.The proof relied on the fact that the measure of the union of a countable number of disjoint sets is additive, and also that the measure is unchanged by translation.But I thought the Banach-Tarski paradox said...
What's the point of defining \sigma algebras, i.e., why must we assign measures only to elements of a \sigma algebra?Second, can you give an example of a set contained in the Lebesgue \sigma algebra, but not the Borel \sigma algebra? Also, is there a set not contained even in the...
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Introduction
I am trying to obtain the eigenvectors of a unitary matrix $M(k)$ which depends on a parameter k.
This matrix $M(k)$ has dimension 6, and while for general matrices of dimension 6 it's not possible to write down algebraic expressions for their eigenvalues, because the characteristic polynomial will also be of order 6, for the case of $M(k)$ it's characteristic polynomial is such that it is possible to write its eigenvalues in algebraic form.
When I use the Eigenvectors[] function in Mathematica, it gives me eigenvectors which
Cannot be written algebraically, i.e. it involves expressions with roots and # when it should be possible to avoid this since the eigenvalues are algebraic. Discontinuous when I don't think they should be for this matrix.
So basically I don't trust how Eigenvectors[] is working for this matrix. I would like to use another way of calculating the eigenvectors.
Question
I have tried computing the eigenvectors $v$ of a matrix $M(k)$ of dimension 6 in Mathematica by using Solve[] on this equation
$(M(k) - aI)v = 0$
where a is an eigenvalue of $M(k)$.
Why does Mathematica only give me the trivial solution $v=0$?
I checked that the determinant of $(M(k) - aI)$ is zero, and so if I denote an eigenvector $v$ as $v = (v_1,v_2,v_3,v_4,v_5,v_6)$ then there should be a solution where $v_2, v_3, v_4, v_5,$ and $v_6$ are written solely in terms of $v_1$, but this solution doesn't appear.
I tried doing the exact same procedure as above for a different matrix of dimension 2 and
IT DID GIVE ME the non-trivial solutions, so I'm not sure why it can't do it for $M(k)$.
This is the matrix
$ \frac{1}{\sqrt{2}}\begin{bmatrix} 0 & 0 & 1 & i & 0 & 0\\ 0 & 0 & 0 & 0 & ie^{-ik} & e^{-ik} \\ 0 & 0 & 0 & 0 & e^{i\frac{2\pi}{3}} & ie^{i\frac{2\pi}{3}} \\ ie^{i\frac{2\pi}{3}} & e^{i\frac{2\pi}{3}} & 0 & 0 & 0 & 0\\ e^{ik}e^{-i\frac{2\pi}{3}}& ie^{ik}e^{-i\frac{2\pi}{3}} & 0 & 0 & 0 & 0 \\ 0 & 0 & ie^{-i\frac{2\pi}{3}} & e^{-i\frac{2\pi}{3}} & 0 & 0 \\ \end{bmatrix}$
{{0,0,1/Sqrt[2],I/Sqrt[2],0,0}, {0,0,0,0,(I E^(-I k))/Sqrt[2],E^(-I k)/Sqrt[2]}, {0,0,0,0,E^((2 I π)/3)/Sqrt[2],(I E^((2 I π)/3))/Sqrt[2]}, {(I E^((2 I π)/3))/Sqrt[2],E^((2 I π)/3)/Sqrt[2],0,0,0,0}, {E^(I k-(2 I π)/3)/Sqrt[2],(I E^(I k-(2 I π)/3))/Sqrt[2],0,0,0,0}, {0,0,(I E^(-((2 I π)/3)))/Sqrt[2],E^(-((2 I π)/3))/Sqrt[2],0,0}}
I need to find the eigenvectors of this matrix
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Inspired by Digitangular numbers.
The triangle of a natural number is given by
$$\triangle(n)=\frac{n(n+1)}2$$
The digitangular counterpart of a natural number is given by the sum of the triangles of the digits of the number, e.g. $\newcommand{rdig}{{\rm digi}\triangle~}$
$$\rdig(613)=\sum_{k=6,1,3}\triangle(k)=21+1+6=28$$
A digitangular sequence of a natural number is given by repeated applications of the digitangular counterpart of that number (and for the sake of consistency, we'll include the original number too) until we reach $1$. Beginning with $613$, the digitangular sequence is given by
$$613,\rdig(613),\rdig(\rdig(613)),\dots\\613,28,39,51,16,22,6,21,4,10,1$$
I then proved that this sequence always terminates i.e. we'll always reach $1$ eventually with computer assistance. First show that the sequence always reaches $1$ for any starting natural number in $[1,135]$. For $x>135$, let $n=1+\lfloor\log_{10}(x)\rfloor$ be the number of digits in $x$. The largest value $\rdig(x)$ could be is if all of $x$'s digits were $9$'s, providing us with the bound
$$\rdig(x)\le45n=45(1+\lfloor\log_{10}(x)\rfloor)$$
And for any $x>135$, we have
$$45(1+\lfloor\log_{10}(x)\rfloor)<x$$
Hence,
$$x>135\implies\rdig(x)<x$$
So the sequence eventually goes below $135$, and that's where we brute force the proof.
I'm searching for (possibly more elegant) alternative proofs, that preferably do not require computer assistance, that this sequence, for every natural $n$, always goes down to $1$.
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Answer
$ v = \frac{\omega A}{\sqrt{2}}$
Work Step by Step
We find: $ \frac{1}{2}mv^2 = \frac{1}{2}(\frac{1}{2}m\omega^2A^2)$ $ \frac{1}{2}mv^2 =(\frac{1}{4}m\omega^2A^2)$ $ v = \frac{\omega A}{\sqrt{2}}$
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Learning Objectives
Make sure you thoroughly understand the following essential ideas:
Describe the major reasons that solutions are so important in the practicalaspects of chemistry. Explain why expressing a concentration as " x-percent" can be ambiguous. Explain why the molarity of a solution will vary with its temperature, whereas molality and mole fraction do not. Given the necessary data, convert (in either direction) between any two concentration units, e.g. molarity - mole fraction. Show how one can prepare a given volume of a solution of a certain molarity, molality, or percent concentration from a solution that is more concentrated (expressed in the same units.) Calculate the concentration of a solution prepared by mixing given volumes to two solutions whose concentrations are expressed in the same units.
Solutions are homogeneous (single-phase) mixtures of two or more components. For convenience, we often refer to the majority component as the solvent; minority components are solutes; there is really no fundamental distinction between them. Solutions play a very important role in Chemistry because they allow intimate and varied encounters between molecules of different kinds, a condition that is essential for rapid chemical reactions to occur. Several more explicit reasons can be cited for devoting a significant amount of effort to the subject of solutions:
For the reason stated above, most chemical reactions that are carried out in the laboratory and in industry, and that occur in living organisms, take place in solution. Solutions are so common; very few pure substances are found in nature. Solutions provide a convenient and accurate means of introducing known small amounts of a substance to a reaction system. Advantage is taken of this in the process of titration, for example. The physical properties of solutions are sensitively influenced by the balance between the intermolecular forces of like and unlike (solvent and solute) molecules. The physical properties of solutions thus serve as useful experimental probes of these intermolecular forces.
We usually think of a solution as a liquid made by adding a gas, a solid or another liquid
solute in a liquid solvent. Actually, solutions can exist as gases and solids as well.
Solid solutions are very common; most natural minerals and many metallic alloys are solid solutions.
Still, it is liquid solutions that we most frequently encounter and must deal with. Experience has taught us that sugar and salt dissolve readily in water, but that “oil and water don’t mix”. Actually, this is not strictly correct, since all substances have at least a slight tendency to dissolve in each other. This raises two important and related questions: why do solutions tend to form in the first place, and what factors limit their mutual solubilities?
Understanding Concentrations
Concentration is a general term that expresses the quantity of solute contained in a given amount of solution. Various ways of expressing concentration are in use; the choice is usually a matter of convenience in a particular application. You should become familiar with all of them.
Parts-per concentration
In the consumer and industrial world, the most common method of expressing the concentration is based on the quantity of solute in a fixed quantity of solution. The “quantities” referred to here can be expressed in weight, in volume, or both (i.e., the
weight of solute in a given volume of solution.) In order to distinguish among these possibilities, the abbreviations (w/w), (v/v) and (w/v) are used.
In most applied fields of Chemistry, (w/w) measure is often used, and is commonly expressed as weight-percent concentration, or simply "percent concentration". For example, a solution made by dissolving 10 g of salt with 200 g of water contains "1 part of salt per 20 g of water".
Cent" is the Latin-derived prefix relating to the number 100 (L. centum), as in centuryor centennial. It also denotes 1/100th (from L. centesimus) as in centimeterand the monetary unit cent. It is usually more convenient to express such concentrations as "parts per 100", which we all know as "percent". So the solution described above is a "5% (w/w) solution" of NaCl in water. In clinical chemistry, (w/v) is commonly used, with weight expressed in grams and volume in mL (Example 8.1.1).
Example 8.1.1
The normal saline solution used in medicine for nasal irrigation, wound cleaning and intravenous drips is a 0.91% (w/v) solution of sodium chloride in water. How would you prepare 1.5 L of this solution?
SOLUTION
The solution will contain 0.91 g of NaCl in 100 mL of water, or 9.1 g in 1 L. Thus you will add (1.5 × 9.1g) = 13.6 g of NaCl to 1.5 L of water.
Percent means parts per 100; we can also use parts per thousand (ppt) for expressing concentrations in grams of solute per kilogram of solution. For more dilute solutions, parts per million (ppm) and parts per billion (10
9; ppb) are used. These terms are widely employed to express the amounts of trace pollutants in the environment.
Example 8.1.2
Describe how you would prepare 30 g of a 20 percent (w/w) solution of KCl in water.
SOLUTION
The weight of potassium chloride required is 20% of the total weight of the solution, or 0.2 × (3 0 g) = 6.0 g of KCl. The remainder of the solution (30 – 6 = 24) g consists of water. Thus you would dissolve 6.0 g of KCl in 24 g of water.
Weight/volume and volume/volume basis
It is sometimes convenient to base concentration on a fixed volume, either of the solution itself, or of the solvent alone. In most instances, a 5% by volume solution of a solid will mean 5 g of the solute dissolved in 100 ml of the solvent.
Example 8.1.3
Fish, like all animals, need a supply of oxygen, which they obtain from oxygen dissolved in the water. The minimum oxygen concentration needed to support most fish is around 5 ppm (w/v). How many moles of O
2 per liter of water does this correspond to? SOLUTION
5 ppm (w/v) means 5 grams of oxygen in one million mL (1000 L) of water, or 5 mg per liter. This is equivalent to (0.005 g) / (32.0 g mol
–1) = 1.6 × 10 –4 mol.
If the solute is itself a liquid, volume/volume measure usually refers to the volume of solute contained in a fixed volume of
solution (not solvent). The latter distinction is important because volumes of mixed substances are not strictly additive. These kinds of concentration measure are mostly used in commercial and industrial applications. The "proof" of an alcoholic beverage is the (v/v)-percent, multiplied by two; thus a 100-proof vodka has the same alcohol concentration as a solution made by adding sufficient water to 50 ml of alcohol to give 100 ml of solution. Molarity: mole/volume basis
This is the method most used by chemists to express concentration, and it is the one most important for you to master. Molar concentration (molarity) is the number of moles of solute per liter of solution.
The important point to remember is that the volume of the
solution is different from the volume of the solvent; the latter quantity can be found from the molarity only if the densities of both the solution and of the pure solvent are known. Similarly, calculation of the weight-percentage concentration from the molarity requires density information; you are expected to be able to carry out these kinds of calculations, which are covered in most texts.
Example 8.1.4
How would you make 120 mL of a 0.10 M solution of potassium hydroxide in water?
SOLUTION
The amount of KOH required is
(0.120 L) × (0.10 mol L
–1) = 0.012 mol.
The molar mass of KOH is 56.1 g, so the weight of KOH required is
\[(0.012\; mol) \times (56.1\; g \;mol^{-1}) = 0.67\; g\]
We would dissolve this weight of KOH in a volume of water that is less than 120 mL, and then add sufficient water to bring the volume of the solution up to 120 mL.
Note
: if we had simply added the KOH to 120 mL of water, the molarity of the resulting solution would not be the same. This is because volumes of different substances are not strictly additive when they are mixed. Without actually measuring the volume of the resulting solution, its molarity would not be known.
Example 8.1.5
Calculate the molarity of a 60-% (w/w) solution of ethanol (C
2H 5OH) in water whose density is 0.8937 g mL –1. SOLUTION
One liter of this solution has a mass of 893.7 g, of which
\[0.60 \times (893.7\; g) = 536.2\; g\]
consists of ethanol. The molecular weight of C
2H 5OH is 46.0, so the number of moles of ethanol present in one liter (that is, the molarity) will be
\[ \dfrac{\dfrac{536.2\;g}{46.0\;g\;mol^{-1}}}{1 L} =11.6\; mol\,L^{-1}\]
Normality and Equivalents Normality is a now-obsolete concentration measure based on the number of equivalents per liter of solution. Although the latter term is now also officially obsolete, it still finds some use in clinical- and environmental chemistry and in electrochemistry. Both terms are widely encountered in pre-1970 textbooks and articles.
The
equivalent weight of an acid is its molecular weight divided by the number of titratable hydrogens it carries. Thus for sulfuric acid H 2SO 4, one mole has a mass of 98 g, but because both hydrogens can be neutralized by strong base, its equivalent weight is 98/2 = 49 g. A solution of 49 g of H 2SO 4per liter of water is 0.5 molar, but also "1 normal" (1 N = 1 eq/L). Such a solution is "equivalent" to a 1 M solution of HCl in the sense that each can be neutralized by 1 mol of strong base. 3is said to be "3 normal" (3 N) because it dissociates into three moles/L of chloride ions.
Although molar concentration is widely employed, it suffers from one serious defect: since volumes are temperature-dependent (substances expand on heating), so are molarities; a 0.100 M solution at 0° C will have a smaller concentration at 50° C. For this reason, molarity is not the preferred concentration measure in applications where physical properties of solutions and the effect of temperature on these properties is of importance.
Mole fraction: mole/mole basis
This is the most fundamental of all methods of concentration measure, since it makes no assumptions at all about volumes. The mole fraction of substance
i in a mixture is defined as
\[ X_i= \dfrac{n_i}{\sum_j n_j}\]
in which
n j is the number of moles of substance j, and the summation is over all substances in the solution. Mole fractions run from zero (substance not present) to unity (the pure substance). The sum of all mole fractions in a solution is, by definition, unity:
\[\sum_i X_i=1\]
Example 8.1.6
What fraction of the molecules in a 60-% (w/w) solution of ethanol in water consist of H
2O? SOLUTION
From the previous problem, we know that one liter of this solution contains 536.2 g (11.6 mol) of C
2H 5OH. The number of moles of H 2O is
( (893.7 – 536.2) g) / (18.0 g mol
–1) = 19.9 mol.
The mole fraction of water is thus
\[\dfrac{19.9}{19.9+11.6} = 0.63\]
Thus 63% of the molecules in this solution consist of water, and 37% are ethanol.
In the case of ionic solutions, each kind of ion acts as a separate component.
Example 8.1.7
Find the mole fraction of water in a solution prepared by dissolving 4.5 g of CaBr
2 in 84.0 mL of water. SOLUTION
The molar mass of CaBr
2 is 200 g, and 84.0 mL of H 2O has a mass of very close to 84.0 g at its assumed density of 1.00 g mL –1. Thus the number of moles of CaBr 2 in the solution is
\[\dfrac{4.50\; g}{200\; g/mol} = 0.0225 \;mol\]
Because this salt is completely dissociated in solution, the solution will contain 0.268 mol of Ca
2 + and (2 × .268) = 0.536 of Br –. The number of moles of water is
(84 g) / (18 g mol
–1) = 4.67 mol.
The mole fraction of water is then
\[\dfrac{0.467\; \cancel{mol}}{0.268 + 0.536 + 4.67\; \cancel{mol}} = \dfrac{0.467}{5.47} = 0.854\]
Thus H
2O constitutes 85 out of every 100 molecules in the solution. Molality: mole/weight basis
A 1-molal solution contains one mole of solute per 1 kg of solvent. Molality is a hybrid concentration unit, retaining the convenience of mole measure for the solute, but expressing it in relation to a temperature-independent mass rather than a volume. Molality, like mole fraction, is used in applications dealing with certain physical properties of solutions; we will see some of these in the next lesson.
Example 8.1.8
Calculate the molality of a 60-% (w/w) solution of ethanol in water.
SOLUTION
From the above problems, we know that one liter of this solution contains 11.6 mol of ethanol in
(893.7 – 536.2) = 357.5 g
of water. The molarity of ethanol in the solution is therefore
(11.6 mol) / (0.3575 kg) = 32.4 mol kg
–1. Conversion between Concentration Measures
Anyone doing practical chemistry must be able to convert one kind of concentration measure into another. The important point to remember is that any conversion involving molarity requires a knowledge of the
density of the solution.
Example 8.1.9
A solution prepared by dissolving 66.0 g of urea (NH
2) 2CO in 950 g of water had a density of 1.018 g mL –1. Express the concentration of urea in weight-percent mole fraction molarity molality SOLUTION a) The weight-percent of solute is (100%) –1 (66.0 g) / (950 g) = 6.9%
The molar mass of urea is 60, so the number of moles is
(66 g) /(60 g mol
–1) = 1.1 mol.
The number of moles of H
2O is
(950 g) / (18 g mol
–1) = 52.8 mol.
b) Mole fraction of urea:
(1.1 mol) / (1.1 + 52.8 mol) = 0.020
c) molarity of urea: the volume of 1 L of solution is
(66 + 950)g / (1018 g L
–1)= 998 mL.
The number of moles of urea (from a) is 1.1 mol.
Its molarity is then
(1.1 mol) / (0.998 L) = 1.1 mol L
–1.
d) The molality of urea is (1.1 mol) / (.066 + .950) kg = 1.08 mol kg
–1.
Example 8.1.10
Ordinary dry air contains 21% (v/v) oxygen. About many moles of O
2 can be inhaled into the lungs of a typical adult woman with a lung capacity of 4.0 L? SOLUTION
The number of molecules (and thus the number of moles) in a gas is directly proportional to its volume (Avogadro's law), so the mole fraction of O
2 is 0.21. The molar volume of a gas at 25° C is
(298/271) × 22.4 L mol
–1 = 24.4 L mol –1
so the moles of O
2 in 4 L of air will be
(4 / 24.4) × (0.21 mol) × (24.4 L mol
–1) = 0.84 mol O 2. Dilution calculations
These kinds of calculations arise frequently in both laboratory and practical applications. If you have a thorough understanding of concentration definitions, they are easily tackled. The most important things to bear in mind are
Concentration is inversely proportional to volume; Molarity is expressed in mol L –1, so it is usually more convenient to express volumes in liters rather than in mL; Use the principles of unit cancelations to determine what to divide by what.
Example 8.1.11
Commercial hydrochloric acid is available as a 10.17 molar solution. How would you use this to prepare 500 mL of a 4.00 molar solution?
SOLUTION
The desired solution requires (0.50 L) × (4.00 M L
–1) = 2.0 mol of HCl. This quantity of HCl is contained in (2.0 mol) / (10.17 M L –1) = 0.197 L of the concentrated acid. So one would measure out 197 mL of the concentrated acid, and then add water to make the total volume of 500 mL.
Example 8.1.12
Calculate the molarity of the solution produced by adding 120 mL of 6.0 M HCl to 150 mL of 0.15 M HCl. What important assumption must be made here?
SOLUTION
The assumption, of course, is that the density of HCl within this concentration range is constant, meaning that their volumes will be additive.
Moles of HCl in first solution:
(0.120 L) × (6.0 mol L
–1) = 0.72 mol HCl
Moles of HCl in second solution:
(0.150 L) × (0.15 mol L
–1) = 0.02 mol HCl
Molarity of mixture:
(0.72 + 0.02) mol / (.120 + .150) L = 4.3 mol L
–1.
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The following is essentially a contribution to point 2. Non-commutative (which can be specialized to matrices or scalars) continued fraction are used in enumeration and language theories. Two examples : Dyck and Motzkin paths.
These are two lattice paths drawn on $\mathbb{N}^2$ (the first quarter-plane).
Dyck Paths Steps are $a=(1,1)$ (north-east step) and $b=(1,-1)$ (south-east step) Dyck paths are defined as paths that start at $(0,0)$ end at $(0,2n)$ (for some $n$) always stand above the $x$-axis
They can be coded by words $w$ (Dyck words) in the alphabet $\{a,b\}$ such that, if $w=uv$ one has $|u|_a\geq |u|_b$ (the path is always in $\mathbb{N}^2$) and, at the end $|w|_a=|w|_b$ (it returns to the $x$-axis). In what preceeds $|u|_a$ (resp. $|u|_b$) stand for the number of occurences of $a$ (resp. $b$) in the word $u$.
Let $D$ be the set of Dyck words. It can be shown that it is a free monoid with alphabet (irreducible Dyck) $A$ (the paths which return to the $x$-axis only at the end). Indentifying them with their characteristic series $$D=\sum_{w\atop\small{Dyck\ word}}\,w\ ; A=\sum_{w\atop\small{Irreducible\ Dyck\ word}}\,w $$(computed in $\mathbb{Z}\langle\langle a,b\rangle\rangle$) one has $A=aDb$ and $D=(1-A)^{-1}$, then
$$A=a(1-A)^{-1}b=a\frac{1}{1-A}b=a\frac{1}{1-a\frac{1}{1-A}b}b=a\frac{1}{1-a\frac{1}{1-a\frac{1}{1-A}b}b}b=\ldots $$with suitable structures, one can show that this continued fraction converges.
Motzkin Paths Similarly, but with steps are $a=(1,1)$ (north-east step), $c=(1,0)$ (east step, i.e. constant) $b=(1,-1)$ (south-east step) Motzkin paths are defined as paths that start at $(0,0)$ end at $(0,n)$ (for some $n$) always stand above the $x$-axis
They are coded by Motzkin words which form again a free monoid $M$ with alphabet, say $B$ of (irreducible Motzkin words). One has $B=(c+aMb)$ and $M=(1-B)^{-1}$ thus $$B=c+aMb=c+a\frac{1}{(1-B)}b=\ldots$$Hope this helps. Do not hesitate to interact.
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Contents Ray-Triangle Intersection: Geometric Solution
In the previous paragraphs we learned how to compute the plane's normal (which is the same as the triangle's normal). Next what we need to find out is the position of point P (for some illustrations we also used Phit), the point where the ray intersects the plane.
Step 1: Finding P
We know that P is somewhere on the ray defined by its origin \(O\) and its direction \(R\) (we used \(D\) in the previous lesson but we wont used it in the lesson to avoid confusion with the term \(D\) from the plane equation). The ray parametric equation is (equation 1):$$P=O + tR.$$
Where \(t\) is distance from the ray origin \(O\) to P. To find P we need to find \(t\) (figure 1). What else do we know? We know the plane's normal which we have already computed and the plane equation (2) which is (check the chapter on the ray-plane intersection in previous lesson for more information on this topic):$$Ax + By + Cz + D = 0.$$
Where A, B, C can be seen as the components (or coordinates) of the normal to the plane (\(N_{plane}=(A, B, C)\)) and \(D\) is the distance from the origin (0, 0, 0) to the plane (if we trace a line from the origin to the plane, parallel to the plane's normal). The variables x, y and z are the coordinates of any point that lies on this plane.
We know the plane normal and we know than any of the three triangle's vertices (V0, V1, V2) lie in the plane. It is therefore possible to compute \(D\). Any of the three vertices can be chosen but we use the first one V0 by convention:
We also know that the point P which is the intersection point of the ray and the plane lies in the plane. Consequently we can substitute P (from equation 1) to (x, y, z) in equation 2 and solve for t (equation 3):$$ \begin{array}{l} P = O + tR\\ Ax + By + Cz + D = 0\\ A * P_x + B * P_y + C * P_z + D = 0\\ A * (O_x + tR_x) + B * (O_y + tR_y) + C * (O_z + tR_z) + D = 0\\ A * O_x + B * O_y + C * O_z + A * tR_x + B * tR_y + C * tR_z + D = 0\\ t * (A * R_x + B * R_y + C * R_z) + A * O_x + B * O_y + C * O_z + D = 0\\ t = -{\dfrac{A * O_x + B * O_y + C * O_z + D}{A * R_x + B * R_y + C * R_z}}\\ t = -{\dfrac{ N(A,B,C) \cdot O + D}{N(A,B,C) \cdot R}} \end{array} $$
Be careful though when it comes to implementing this technique. By default, our camera is oriented along the negative z-axis, thus all primary rays (assuming the camera is in its default position) have negative z-coordinates. This is also mean that when the normal of a triangle faces the camera, the dot product of the normal with primary rays will be negative. Because of that, the negative sign in front of the equation will be cancelled out by the negative sign of the denominator. Practically this mean that you need to negate the denominator and therefore we end up with:$$ \begin{array}{l} t &=& -{\dfrac{ N(A,B,C) \cdot O + D}{-N(A,B,C) \cdot R}}\\ &=&{\dfrac{ N(A,B,C) \cdot O + D}{N(A,B,C) \cdot R}} \end{array} $$
Check the actual code at the end of this chapter for a practical example.
We now have computed \(t\) which we can use to compute the position of P:
There are two very important cases though we need to look at before we check if the point is inside the triangle.
The Ray And The Triangle Are Parallel
If the ray and the plane are parallel they wont intersect (figure 2). For robustness we need to handle that case if it happens. This is very simple. if the triangle is parallel to the ray direction, it means that the triangle's normal and the ray's direction should be perpendicular.
We have learned that the dot product of two perpendicular vectors is 0. If you look at the denominator of equation 3 (the term below the line) we do already compute the dot product between the triangle's normal N and the ray direction \(D\). In fact our code is not very robust because this term can potentially be 0 and we should always catch a possible division by 0. In turns out that when this term is equal to 0 the ray is parallel to the triangle. Since they don't intersect in that case there's therefore no need to compute \(t\) anyway. Conclusion: before we compute \(t\) in our code, we will compute the result of the term \(N \cdot R\) first and if the result is 0, the function will return the value false (no intersection).
The Triangle Is "Behind" The Ray
So far we have assumed that the triangle was always in front of the ray. But what happens if the triangle is behind the ray with the ray still looking in the same direction? Normally the triangle shouldn't be visible. Even when the triangle is "behind" the ray, we do get a valid value with equation 3. In that case, \(t\) is negative which causes the intersection point to be in the opposition direction than the ray's direction. If we don't catch this "error" though, the triangle will show up in the final image which we don't want. Therefore we need to check for the sign of \(t\) before we can decide if the intersection is valid or not. If \(t\) is lower than 0, the triangle is behind the ray's origin (with regards to the ray's direction) and is not visible. There is no intersection and we can return from the function. If \(t\) is greater than 0, the triangle is visible for that ray and we can proceed to the next step.
Step 2: Is P Inside Or Outside The Triangle?
Now that we have found the point P which is the point where the ray and the plane intersection we still have to find out if P is inside the triangle (in which case the ray intersects the triangle) or if P is outside (in which case the rays misses the triangle). Figure 2 illustrates these two cases.
The solution to this problem is simple and is also called the "inside-outside" test (we have already used this term in the lesson on rasterization, though while the two tests have for function to find if a point lies in a triangle, in rasterization the test works for 2D triangles while the method we will explain now works in 3D). Imagine that you have a vector A which is aligned with the x-axis (figure 4). Let's imagine that this vector is actually aligned with one edge of our triangle (the edge defined by the two vertices V0V1). Now, the second edge B, is defined by the vertices V0 and V2 of the triangles as shown in figure 4. Let's compute the cross product of these two vectors. As expected the result is a vector which is pointing in the same direction as the z-axis and the normal of the triangle.$$ \begin{array}{l} A=(1, 0, 0)\\ B=(1, 1, 0)\\ C_x = A_y B_z - A_z B_y = 0\\ C_y = A_z B_x - A_x B_z = 0\\ C_z = A_x B_y - A_y B_x = 1 * 1 - 0 * 1 = 1\\ C = (0, 0, 1) \end{array} $$
Now let's imagine that rather than having the coordinates (1, 1, 0) the vertex V2 has the coordinates (1, -1, 0). In other words we have mirrored its position about the x-axis. If we now compute the cross product AxB' we get the result C'=(0, 0, -1).
$$ \begin{array}{l} A=(1, 0, 0)\\ B=(1, -1, 0)\\ C_x = A_y B_z - A_z B_y = 0\\ C_y = A_z B_x - A_x B_z = 0\\ C_z = A_x B_y - A_y B_x = 1 * -1 - 0 * 1 = -1\\ C = (0, 0, -1) \end{array} $$
Because C and N are pointing in the same direction, their dot product returns a value greater than 0 (positive). However because C' and N are pointing in opposite directions, their dot product returns a value lower than 0 (negative). What does that test tell us? We know that the point where the ray intersects the triangle and the triangle are in the same plane. We also know from the test we have just made that if a point P which is in the triangle's plane (such as the vertex V2 or the intersection point) is on the left side of vector A, then the dot product between the triangle's normal and vector C is positive (C is the result of the cross product between A and B. In this scenario, A = (V1 - V0) and B = (P - V0)). However if P is on the right side of A (as with V2') then this dot product is negative. You can see in figure 5, that point P is inside the triangle when it is located on the left side of A. To apply the technique we have just described to the ray-triangle intersection problem, we need to repeat the left/right test for each edge of the triangle. If for each one of triangle's edges we find that point P is on the left side of vector C (where C is defined as V1-V0, V2-V1 and V0-V2 respectively for each edge of the triangle), then we know for sure that P is inside the triangle. If the test fails for any of the triangle edges, then P lies outside the triangle's boundaries. This process is illustrated in figure 6.
Here is an example of the inside-outside test in pseudocode:
Let's write the code of complete ray-triangle intersection test routine. First we will compute the triangle's normal, then test if the ray and the triangle are parallel. If they are, the intersection test fails. If they are not parallel, we compute \(t\) from which we can compute the intersection point P. If the inside-out test succeeds (we test if P is on the left side of each one of the triangle's edges) then the ray intersects the triangle and P is inside the triangle's boundaries. The test is successful.
The "inside-outside" technique we have just described works for any convex polygon. Repeat the technique we have used for triangles for each edge of the polygon. Compute the cross product of the vector defined by the two edges' vertices and the vector defined by the first edge's vertex and the point. Compute the dot product of the resulting vector and the polygon's normal. The sign of the resulting dot product determines if the point is on the right or left side of that edge. Iterate through each edge of the polygon. There's no need to test the other edges if one fails passing the test.
Note that this technique can be made faster if the triangle's normal as well as the value \(D\) from the plane equation are precomputed and stored in memory for each triangle of the scene.
What's Next?
In this chapter we have presented a technique to compute the ray-triangle intersection test, using simple geometry. However there is much more to the ray-triangle intersection test which we haven't considered yet such as whether the ray hits the triangle from the front or from the back. We can also compute what we call the barycentric coordinates. These coordinates are necessary for doing things such as applying a texture to the triangle.
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In recent times a great amount of progress has been achieved in symplectic and contact geometry, leading to the development of powerful invariants of 3-manifolds such as Heegaard Floer homology and embedded contact homology. These invariants are based on holomorphic curves and moduli spaces, but in the simplest cases, some of their structure reduces to some elementary combinatorics and algebra which may be of interest in its own right. In this note, which is essentially a light-hearted exposition of some previous work of the author, we give a brief introduction to some of the ideas of contact topology and holomorphic curves, discuss some of these elementary results, and indicate how they arise from holomorphic invariants.
Joint with Eric Schoenfeld.
We show that the sutured Floer homology of a sutured 3-manifold of the form [latex](D^2 \times S^1, F \times S^1)[/latex] can be expressed as the homology of a string-type complex, generated by certain sets of curves on [latex](D^2, F)[/latex] and with a differential given by resolving crossings. We also give some generalisations of this isomorphism, computing “hat” and “infinity” versions of this string homology. In addition to giving interesting elementary facts about the algebra of curves on surfaces, these isomorphisms are inspired by, and establish further, connections between invariants from Floer homology and string topology.
We construct an elementary, combinatorial kind of topological quantum field theory, based on curves, surfaces, and orientations. The construction derives from contact invariants in sutured Floer homology and is essentially an elaboration of a TQFT defined by Honda–Kazez–Matic. This topological field theory stores information in binary format on a surface and has “digital” creation and annihilation operators, giving a toy-model embodiment of “it from bit”.
We use the theory of sutured TQFT to classify contact elements in the sutured Floer homology, with Z coefficients, of certain sutured manifolds of the form $(\Sigma \times S^1, F \times S^1)$ where $\Sigma$ is an annulus or punctured torus. Using this classification, we give a new proof that the contact invariant in sutured Floer homology with Z coefficients of a contact structure with Giroux torsion vanishes. We also give a new proof of Massot’s theorem that the contact invariant vanishes for a contact structure on $(\Sigma \times S^1, F \times S^1)$ described by an isolating dividing set.
We introduce a notion of the twist of an isometry of the hyperbolic plane. This twist function is defined on the universal covering group of orientation-preserving isometries of the hyperbolic plane, at each point in the plane. We relate this function to a function defined by Milnor and generalised by Wood. We deduce various properties of the twist function, and use it to give new proofs of several well-known results, including the Milnor–Wood inequality, using purely hyperbolic-geometric methods. Our methods express inequalities in Milnor’s function as equalities, with the deficiency from equality given by an area in the hyperbolic plane. We find that the twist of certain products found in surface group presentations is equal to the area of certain hyperbolic polygons arising as their fundamental domains.
We consider the relationship between hyperbolic cone-manifold structures on surfaces, and algebraic representations of the fundamental group into a group of isometries. A hyperbolic cone-manifold structure on a surface, with all interior cone angles being integer multiples of [latex]2\pi[/latex], determines a holonomy representation of the fundamental group. We ask, conversely, when a representation of the fundamental group is the holonomy of a hyperbolic cone-manifold structure. In this paper we build upon previous work with punctured tori to prove results for higher genus surfaces. Our techniques construct fundamental domains for hyperbolic cone-manifold structures, from the geometry of a representation. Central to these techniques are the Euler class of a representation, the group [latex]\widetilde{PSL_2\mathbb{R}}[/latex], the twist of hyperbolic isometries, and character varieties. We consider the action of the outer automorphism and related groups on the character variety, which is measure-preserving with respect to a natural measure derived from its symplectic structure, and ergodic in certain regions. Under various hypotheses, we almost surely or surely obtain a hyperbolic cone-manifold structure with prescribed holonomy.
We define a “sutured topological quantum field theory”, motivated by the study of sutured Floer homology of product 3-manifolds, and contact elements. We study a rich algebraic structure of suture elements in sutured TQFT, showing that it corresponds to contact elements in sutured Floer homology. We use this approach to make computations of contact elements in sutured Floer homology over Z of sutured manifolds [latex](D^2 \times S^1, F \times S^1)[/latex] where F is finite. This generalises previous results of the author over Z_2 coefficients. Our approach elaborates upon the quantum field theoretic aspects of sutured Floer homology, building a non-commutative Fock space, together with a bilinear form deriving from a certain combinatorial partial order; we show that the sutured TQFT of discs is isomorphic to this Fock space.
We consider the relationship between hyperbolic cone-manifold structures on surfaces, and algebraic representations of the fundamental group into a group of isometries. A hyperbolic cone-manifold structure on a surface, with all interior cone angles being integer multiples of [latex]2\pi[/latex], determines a holonomy representation of the fundamental group. We ask, conversely, when a representation of the fundamental group is the holonomy of a hyperbolic cone-manifold structure. In this paper we prove results for the punctured torus; in the sequel, for higher genus surfaces. We show that a representation of the fundamental group of a punctured torus is a holonomy representation of a hyperbolic cone-manifold structure with no interior cone points and a single corner point if and only if it is not virtually abelian. We construct a pentagonal fundamental domain for hyperbolic structures, from the geometry of a representation. Our techniques involve the universal covering group [latex]\widetilde{PSL_2\mathbb{R}}[/latex] of the group of orientation-preserving isometries of [latex]\mathbb{H}^2[/latex] and Markoff moves arising from the action of the mapping class group on the character variety.
We consider contact elements in the sutured Floer homology of solid tori with longitudinal sutures, as part of the (1+1)-dimensional topological quantum field theory defined by Honda–Kazez–Mati\'{c}. The Z_2 SFH of these solid tori forms a “categorification of Pascal’s triangle”, and contact structures correspond bijectively to chord diagrams, or sets of disjoint properly embedded arcs in the disc. Their contact elements are distinct and form distinguished subsets of SFH of order given by the Narayana numbers. We find natural “creation and annihilation operators” which allow us to define a QFT-type basis of each SFH vector space, consisting of contact elements. Sutured Floer homology in this case reduces to the combinatorics of chord diagrams. We prove that contact elements are in bijective correspondence with comparable pairs of basis elements with respect to a certain partial order, and in a natural and explicit way. The algebraic and combinatorial structures in this description have intrinsic contact-topological meaning. In particular, the QFT-basis of SFH and its partial order have a natural interpretation in pure contact topology, related to the contact category of a disc: the partial order enables us to tell when the sutured solid cylinder obtained by “stacking” two chord diagrams has a tight contact structure. This leads us to extend Honda’s notion of contact category to a “bounded” contact category, containing chord diagrams and contact structures which occur within a given contact solid cylinder. We compute this bounded contact category in certain cases. Moreover, the decomposition of a contact element into basis elements naturally gives a triple of contact structures on solid cylinders which we regard as a type of “distinguished triangle” in the contact category. We also use the algebraic structures arising among contact elements to extend the notion of contact category to a 2-category.
This is a previous version of the article
Chord diagrams, contact-topological quantum field theory, and contact categories. It contains less content, in particular about contact categories, but is less terse (or more prolix!) and contains more background. It might be useful for some readers, and so I retain it here, even though it has been superseded by that article.
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Set-theoretic arguments often make use of the fact that a particular property $\varphi$ is
local, in the sense that instances of the property can be verified by checking certain facts in only a bounded part of the set-theoretic universe, such as inside some rank-initial segment $V_\theta$ or inside the collection $H_\kappa$ of all sets of hereditary size less than $\kappa$. It turns out that this concept is exactly equivalent to the property being $\Sigma_2$ expressible in the language of set theory.
Theorem. For any assertion $\varphi$ in the language of set theory, the following are equivalent: $\varphi$ is ZFC-provably equivalent to a $\Sigma_2$ assertion. $\varphi$ is ZFC-provably equivalent to an assertion of the form “$\exists \theta\, V_\theta\models\psi$,” where $\psi$ is a statement of any complexity. $\varphi$ is ZFC-provably equivalent to an assertion of the form “$\exists \kappa\, H_\kappa\models\psi$,” where $\psi$ is a statement of any complexity.
Just to clarify, the $\Sigma_2$ assertions in set theory are those of the form $\exists x\,\forall y\,\varphi_0(x,y)$, where $\varphi_0$ has only bounded quantifiers. The set $V_\theta$ refers to the rank-initial segment of the set-theoretic universe, consisting of all sets of von Neumann rank less than $\theta$. The set $H_\kappa$ consists of all sets of hereditary size less than $\kappa$, that is, whose transitive closure has size less than $\kappa$.
Proof. ($3\to 2$) Since $H_\kappa$ is correctly computed inside $V_\theta$ for any $\theta>\kappa$, it follows that to assert that some $H_\kappa$ satisfies $\psi$ is the same as to assert that some $V_\theta$ thinks that there is some cardinal $\kappa$ such that $H_\kappa$ satisfies $\psi$.
($2\to 1$) The statement $\exists \theta\, V_\theta\models\psi$ is equivalent to the assertion $\exists\theta\,\exists x\,(x=V_\theta\wedge x\models\psi)$. The claim that $x\models\psi$ involves only bounded quantifiers, since the quantifiers of $\psi$ become bounded by $x$. The claim that $x=V_\theta$ is $\Pi_1$ in $x$ and $\theta$, since it is equivalent to saying that $x$ is transitive and the ordinals of $x$ are precisely $\theta$ and $x$ thinks every $V_\alpha$ exists, plus a certain minimal set theory (so far this is just $\Delta_0$, since all quantifiers are bounded), plus, finally, the assertion that $x$ contains every subset of each of its elements. So altogether, the assertion that some $V_\theta$ satisfies $\psi$ has complexity $\Sigma_2$ in the language of set theory.
($1\to 3$) This implication is a consequence of the following absoluteness lemma.
Lemma. (Levy) If $\kappa$ is any uncountable cardinal, then $H_\kappa\prec_{\Sigma_1} V$.
Proof. Suppose that $x\in H_\kappa$ and $V\models\exists y\,\psi(x,y)$, where $\psi$ has only bounded quantifiers. Fix some such witness $y$, which exists inside some $H_\gamma$ for perhaps much larger $\gamma$. By the Löwenheim-Skolem theorem, there is $X\prec H_\gamma$ with $\text{TC}(\{x\})\subset X$, $y\in X$ and $X$ of size less than $\kappa$. Let $\pi:X\cong M$ be the Mostowski collapse of $X$, so that $M$ is transitive, and since it has size less than $\kappa$, it follows that $M\subset H_\kappa$. Since the transitive closure of $\{x\}$ was contained in $X$, it follows that $\pi(x)=x$. Thus, since $X\models\psi(x,y)$ we conclude that $M\models \psi(x,\pi(y))$ and so hence $\pi(y)$ is a witness to $\psi(x,\cdot)$ inside $H_\kappa$, as desired. QED
Using the lemma, we now prove the remaining part of the theorem. Consider any $\Sigma_2$ assertion $\exists x\,\forall y\, \varphi_0(x,y)$, where $\varphi_0$ has only bounded quantifiers. This assertion is equivalent to $\exists\kappa\, H_\kappa\models\exists x\,\forall y\,\varphi_0(x,y)$, simply because if there is such a $\kappa$ with $H_\kappa$ having such an $x$, then by the lemma this $x$ works for all $y\in V$ since $H_\kappa\prec_{\Sigma_1}V$; and conversely, if there is an $x$ such that $\forall y\, \varphi_0(x,y)$, then this will remain true inside any $H_\kappa$ with $x\in H_\kappa$.
QED
In light of the theorem, it makes sense to refer to the $\Sigma_2$ properties as the
locally verifiable properties, or perhaps as semi-local properties, since positive instances of $\Sigma_2$ assertions can be verified in some sufficiently large $V_\theta$, without need for unbounded search. A truly local property, therefore, would be one such that positive and negative instances can be verified this way, and these would be precisely the $\Delta_2$ properties, whose positive and negative instances are locally verifiable.
Tighter concepts of locality are obtained by insisting that the property is not merely verified in
some $V_\theta$, perhaps very large, but rather is verified in a $V_\theta$ where $\theta$ has a certain closeness to the parameters or instance of the property. For example, a cardinal $\kappa$ is measurable just in case there is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$, and this is verified inside $V_{\kappa+2}$. Thus, the assertion “$\kappa$ is measurable,” has complexity $\Sigma^2_1$ over $V_\kappa$. One may similarly speak of $\Sigma^n_m$ or $\Sigma^\alpha_m$ properties, to refer to properties that can be verified with $\Sigma_m$ assertions in $V_{\kappa+\alpha}$. Alternatively, for any class function $f$ on the ordinals, one may speak of $f$-local properties, meaning a property that can be checked of $x\in V_\theta$ by checking a property inside $V_{f(\theta)}$.
This post was made in response to a question on MathOverflow.
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Euclidean Geometry • Equitable Distributing • Linear Programming • Set Theory • Nonstandard Analysis • Advice • Topology • Number Theory • Computation of Time (Previous | Next)
In the following section, we presuppose Set Theory and Nonstandard Analysis. The exponential simplex and the polynomial intex method (
inter-/ extrapolation) solve linear programmes (LPs).
Diameter theorem for polytopes: The diameter of an \(n\)-dimensional polytope defined by \(m\) constraints with \(m, n \in {}^{\omega}\mathbb{N}_{\ge2}\) is at most \(2(m + n - 3)\).
Proof: We can assemble at most \(\acute{m}\) hyperplanes into an incomplete cycle (of dimension 2) and have to consider \(n - 2\) alternatives sidewards (in the remaining dimensions). Since we can pass each section with a maximum of two edges, the factor is 2. This theorem can be extended to polyhedra analogously by dropping the requirement of finiteness.\(\square\)
Definition: Let \(\omega\) be the limit, to which all variables in Landau notation tend, and \(\vartheta := \ell \omega\). A method is
polynomial ( exponential) if its computation time in seconds and (or) its memory consumption in bits is \(\mathcal{O}({\omega}^{\mathcal{O}(1)}) (\mathcal{O}({e}^{|\mathcal{O}(\omega)|}))\).
Theorem: The simplex method is exponential.
Proof and algorithm: Let \(P := \{x \in {}^{\omega}\mathbb{R}^{n} : Ax \le b, b \in {}^{\omega}\mathbb{R}^{m}, A \in {}^{\omega}\mathbb{R}^{m \times n}, m, n \in {}^{\omega}\mathbb{N}^{*}\}\) be the feasible domain of the LP max \(\{{d}^{T}x : d \in {}^{\omega}\mathbb{R}^{n}, x \in P\}\). By taking the dual or setting \(x := {x}^{+} - {x}^{-}\) with \({x}^{+}, {x}^{-} \ge 0\), we obtain \(x \ge 0\). We first solve max \(\{-z : Ax - b \le {(z, ..., z)}^{T} \in {}^{\omega}\mathbb{R}^{m}, z \ge 0\}\) to obtain a feasible \(x\) when \(b \ge 0\) does not hold. Initial and target value are \(z := |\text{min } \{{b}_{1}, ..., {b}_{m}\}|\) and \(z = 0\). We begin with \(x := 0\) as in the first case. Pivoting if necessary, we may assume that \(b \ge 0\).
Let \(i, j, k \in {}^{\omega}\mathbb{N}^{*}\) and let \({a}_{i}^{T}\) the \(i\)-th row vector of \(A\). If \({d}_{j} \le 0\) for all \(j\), the LP is solved. If for some \({d}_{j} > 0 \; ({d}_{j} = 0)\), \({a}_{ij} \le 0\) for all \(i\), the LP is positively unbounded (for now, we may drop \({d}_{j}\) and \({A}_{.j}\) as well as \({b}_{i}\) and \({a}_{i}\), but only when \({a}_{ij} < 0\) holds). The inequality \({a}_{ij}{x}_{j} \ge 0 > {b}_{i}\) for all \(j\) has no solution, too. If necessary, divide all \({a}_{i}^{T}x \le {b}_{i}\) by \(||{a}_{i}||\) and all \({d}_{j}\) and \({a}_{ij}\) by the minimum of \(|{a}_{ij}|\) such that \({a}_{ij} \ne 0\) for each \(j\). This will be reversed later. If necessary, renormalise by \(||{a}_{i}||\).
We may always remove redundant constraints (with \({a}_{i} \le 0\)). Select for each \({d}_{j} > 0\) and non-base variable \({x}_{j}\) the minimum ratio \({b}_{k}/{a}_{kj}\) for \({a}_{ij} > 0\). The variables with \({}^{*}\) are considered in the next step. The next potential vertex is given by \({x}_{j}^{*} = {x}_{j} + {b}_{k}/{a}_{kj}\) for feasible \({x}^{*}\). To select the steepest edge, select the pivot \({a}_{kj}\) corresponding to \({x}_{j}\) that maximises \({d}^{T}\Delta x/||\Delta x||\) or \({d}_{j}^{2}/(1 + ||{A}_{.j}{||}^{2})\) for \(\Delta x := {x}^{*} - x\) in the \(k\)-th constraint.
If there are multiple maxima, select max\({}_{k,j} {d}_{j}{b}_{k}/{a}_{kj}\) according to the rule of best pivot value or alternatively maybe less well the smallest angle min \({(1, ..., 1)}^{T}d^{*}/||(\sqrt{n}) d^{*}||\). If we cannot directly maximise the objective function, we perturb, which means that we relax, the constraints with \({b}_{i} = 0\) by the same, minimal modulus. These do not need to be written into the tableau: We simply set \({b}_{i} = ||{a}_{i}||\).
If another multiple vertex is encountered, despite this being unlikely, simply increase the earlier \({b}_{i}\) by \(||{a}_{i}||\). The cost of eliminating a multiple vertex, after which we revert the relaxation, corresponds to an LP with \(d > 0\) and \(b = 0\). Along the chosen path, the objective function increases otherwise strictly monotonically. We can then simply calculate \({d}_{j}^{*}, {a}_{ij}^{*}\) and \({b}_{i}^{*}\) using the rectangle rule (cf. [775], p. 63).
In the worst-case scenario, the simplex method is not polynomial despite the diameter theorem for polytopes under any given set of pivoting rules, since an exponential "drift" can be constructed with Klee-Minty or Jeroslow polytopes, or others, creating a large deviation from the shortest path by forcing the selection of the least favourable edge. This is consistent with existing proofs. The result follows.\(\square\)
Theorem: The intex method solves every solvable LP in \(\mathcal{O}({\vartheta}^{3})\).
Proof and algorithm: First, we normalise and scale \({b}^{T}y - {d}^{T}x \le 0, Ax \le b\) and \({A}^{T}y \ge d\). Let the
height \(h\) and \(v := ({{x}^{T}, {y}^{T})}^{T} \in {}^{\omega}\mathbb{R}^{m+n}\) have the initial values \({h}_{0} := |\text{min } \{{b}_{1}, ..., {b}_{m}, {-d}_{1}, ..., {-d}_{n}\}| + s\) for sufficient clearance \(s \in {}^{\omega}\mathbb{R}_{>0}\) and 0. We compute the LP min \(\{h \in [0, {h}_{0}] : 0 \le x \in {}^{\omega}\mathbb{R}^{n}, 0 \le y \in {}^{\omega}\mathbb{R}^{m}, {b}^{T}y - {d}^{T}x \le h, Ax - b \le (h, ..., h)^{T} \in {}^{\omega}\mathbb{R}^{m}, d - {A}^{T}y \le (h, ..., h)^{T} \in {}^{\omega}\mathbb{R}^{n}\}\) via the (dual) programme min \(\{{b}^{T}y : 0 \le y \in {}^{\omega}\mathbb{R}^{m}, {A}^{T}y \ge d\}\) for the (primal) programme max \(\{{d}^{T}x : d \in {}^{\omega}\mathbb{R}^{n}, x \in {P}_{\ge 0}\}\).
We successively interpolate all \({v}_{k}^{*} := (\text{max } {v}_{k} + \text{min } {v}_{k})/2\) until all \(\Delta{v}_{k}\) are sufficiently small in the point \(({v}^{T}, h)^{T}\), and repeat this in the height \({h}^{*} := (h_0 + \text{min } h)/2\) for the point \(({v}^{*T}, {h}^{*})^{T}\). Then, we extrapolate \(({v}^{T}, h)^{T}\) via \(({v}^{*T}, {h}^{(*)})^{T}\) stopping just before the boundary of the polytope. There, we start over until \(x\) and \(y\) are optimal or \(h\) cannot be minimised anymore. Since \(h\) at least roughly halves itself for each iteration step in \(\mathcal{O}({\omega\vartheta}^{2})\), the claim follows by the strong duality theorem ([775], p. 60 - 65).\(\square\)
Corollary: If neither a primally feasible \(x\) nor a dual solution \(y\) needs to be computed, the runtime of the LP can roughly be halved by setting \(h := {d}^{T}x.\square\)
Remarks: Simplex method and face algorithm ([916], p. 580 f.) may solve the LP faster for small \(m\) and \(n\). We can easily change the current stock of constraints or variables, because the intex method is a non-transforming method and faster than all known (worst-case) LP-solving algorithms in \(\mathcal{O}({\ell\omega}^{4.5})\). We simply have to adjust \(h\) if necessary, because we can initialise additional variables with 0. Increasing the precision can make sense.
Corollary: Every solvable linear system (LS) \(Ax = b\) for \(x \in {}^{\omega}\mathbb{R}^{n}\) can be solved as LP min \(\{h \in [0, \text{max } \{|{b}_{1}|, ..., |{b}_{m}|\}] : \pm(Ax - b) \le (h, ..., h)^{T} \in {}^{\omega}\mathbb{R}^{m}\}\) in \(\mathcal{O}({\vartheta}^{3}).\square\)
Corollary: Every solvable LP min \(\{h \in [0, 1] : \pm(Ax - \lambda x) \le (h, ..., h)^{T} \in {}^{\omega}\mathbb{R}^{n}\}\) can determine an eigenvector \(x \in {}^{\omega}\mathbb{R}^{n} \setminus \{0\}\) of the matrix \(A \in {}^{\omega}\mathbb{R}^{n \times n}\) for the eigenvalue \(\lambda \in {}^{\omega}\mathbb{R}\) in \(\mathcal{O}({\vartheta}^{3}).\square\)
Corollary: Let \({\alpha }_{j}\) the \(j\)-th column vector of the matrix \({A}^{-1} \in {}^{\omega}\mathbb{R}^{n \times n}\) and let \({\delta}_{ij}\) the Kronecker delta. Every LS \({A \alpha }_{j} = {({\delta}_{1j}, ..., {\delta}_{nj})}^{T}\) to determine the inverse \({A}^{-1}\) of the regular matrix \(A\) can be solved for \(j = 1, ..., n\) in \(\mathcal{O}({\vartheta}^{3})\). Whether \(A\) is regular can be also determined in \(\mathcal{O}({\vartheta}^{3}).\square\)
Corollary: If only finite floating-point numbers are used and if \(q \in [0, 1]\) is the density of \(A, \mathcal{O}({\vartheta}^{3})\) can above be everywhere replaced by max \(\{\mathcal{O}(qmn), \mathcal{O}(m + n)\}\) or max \(\{\mathcal{O}(q{n}^{2}), \mathcal{O}(n)\}.\square\)
Remarks: The four corollaries can be easily transferred to complex ones. The intex method is numerically very stable, since we can keep rounding errors small also by resorting to the initial data and a Kahan-Babuška-Neumaier summation modified according to Klein, especially near the end. It can be solved in \({}^{c}\mathbb{R}^{c}\) if modified by distributed computing in \(\mathcal{O}(1)\). It is also well-suited for (mixed) integer problems and (non-)convex (Pareto) optimisation.
Corollary: Every solvable convex programme min \(\{{f}_{1}(x) : x \in {}^{\omega}\mathbb{R}^{n}, {({f}_{2}(x), ..., {f}_{m}(x))}^{T} \le 0\}\) where the \({f}_{i} \in {}^{\omega}\mathbb{R}\) are convex functions for \(i = 1, ..., m\) may be solved by the intex method and two-dimensional bisection or Newton's methods in polynomial runtime, if the number of operands \({x}_{j}\) of the \({f}_{i}\) is \(\le {\omega}^{c-3}\) and if an \(x\) exists so that \({f}_{i}(x) < 0\) for all \(i > 1\) (see [939], p. 589 ff.).\(\square\)
code of simplex method
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Learning Objectives
A brief look at series and parallel circuits. Also defines voltage and current divider.
Fig. 3.6.1 The circuit shown is perhaps the simplest circuit that performs a signal processing function. The input is provided by the voltage source vin and the output is the voltage across the resistor labeled
The results shown in other modules (circuit elements, KVL and KCL, interconnection laws) with regard to the above circuit and the values of other currents and voltages in this circuit as well, have profound implications.
Resistors connected in such a way that current from one must flow
only into another—currents in all resistors connected this way have the same magnitude—are said to be connected in series. For the two series-connected resistors in the example, the voltage across one resistor equals the ratio of that resistor's value and the sum of resistances times the voltage across the series combination. This concept is so pervasive it has a name: voltage divider.
The
input-output relationship for this system, found in this particular case by voltage divider, takes the form of a ratio of the output voltage to the input voltage.
\[\frac{v_{out}}{v_{in}} = \frac{R_{2}}{R_{1}+R_{2}}\]
In this way, we express how the components used to build the system affect the input-output relationship. Because this analysis was made with ideal circuit elements, we might expect this relation to break down if the input amplitude is too high (Will the circuit survive if the input changes from 1 volt to one million volts?) or if the source's frequency becomes too high. In any case, this important way of expressing input-output relationships—as a ratio of output to input—pervades circuit and system theory.
The current is the current flowing out of the voltage source. Because it equals
we have that i2,
\[\frac{v_{in}}{i_{1}} = R_{1}+R_{2}\]
Resistors in series
The series combination of two resistors acts, as far as the voltage source is concerned, as a single resistor having a value equal to the sum of the two resistances.
This result is the first of several equivalent service ideas: In many cases, a complicated circuit when viewed from its terminals (the two places to which you might attach a source) appears to be a single circuit element (at best) or a simple combination of elements at worst. Thus, the equivalent circuit for a series combination of resistors is a single resistor having a resistance equal to the sum of its component resistances.
Fig. 3.6.2 The resistor (on the right) is equivalent to the two resistors (on the left) and has a resistance equal to the sum of the resistances of the other two resistors.
Thus, the circuit the voltage source "feels" (through the current drawn from it) is a single resistor having resistance Note that in making this equivalent circuit, the output voltage can no longer be defined: The output resistor labeled
R 2 no longer appears. Thus, this equivalence is made strictly from the voltage source's viewpoint.
Fig. 3.6.3 A simple parallel circuit
One interesting simple circuit in Fig. 3.6.3 has two resistors connected side-by-side, what we will term a
parallel connection, rather than in series. Here, applying KVL reveals that all the voltages are identical: v 1 =v and v 2= v. This result typifies parallel connections. To write the KCL equation, note that the top node consists of the entire upper interconnection section. The KCL equation is:
\[i_{in}-i_{1}-i_{2}=0\]
Using the
v-i relations, we find that
\[i_{out} = \frac{R_{1}}{R_{1}+R_{2}}i_{in}\]
Exercise \(\PageIndex{1}\)
Suppose that you replaced the current source in Fig. 3.6.3 by a voltage source. How would
be related to the source voltage? Based on this result, what purpose does this revised circuit have? iout Solution
Replacing the current source by a voltage source does not change the fact that the voltages are identical. Consequently,
\[v_{in}=R_{2}i_{out}\; \; or\; \; i_{out}=\frac{v_{in}}{R_{2}}\]
This result does not depend on the resistor which means that we simply have a resistor (
R 2) across a voltage source. The two-resistor circuit has no apparent use.
This circuit highlights some important properties of parallel circuits. You can easily show that the parallel combination of
R has the 2 v-irelation of a resistor having resistance
\[\left ( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right )^{-1}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}\]
A shorthand notation for this quantity is
\[R_{1}\parallel R_{2}\]
As the reciprocal of resistance is conductance, we can say that
for a parallel combination of resistors, the equivalent conductance is the sum of the conductances.
Fig. 3.6.4 Parallel Resistors
Similar to voltage divider in fig. 3.6.1 for series resistances, we have
current divider for parallel resistances. The current through a resistor in parallel with another is the ratio of the conductance of the first to the sum of the conductances. Thus, for the depicted circuit,
\[i_{2}=\frac{G_{2}}{G_{1}+G_{2}}i\]
Expressed in terms of resistances, current divider takes the form of the resistance of the
other resistor divided by the sum of resistances:
\[i_{2}=\frac{R_{1}}{R_{1}+R_{2}}i\]
Fig. 3.6.5 The Simple Attenuator Circuit is attached to an oscilloscope's input. The input-output relation for the above circuit without a load is:
\[v_{out}=\frac{R_{2}}{R_{1}+R_{2}}v_{in}\]
Suppose we want to pass the output signal into a voltage measurement device, such as an oscilloscope or a voltmeter. In system-theory terms, we want to pass our circuit's output to a sink. For most applications, we can represent these measurement devices as a resistor, with the current passing through it driving the measurement device through some type of display. In circuits, a sink is called a
load; thus, we describe a system-theoretic sink as a load resistance Thus, we have a complete system built from a cascade of three systems: a source, a signal processing system (simple as it is), and a sink.
We must analyze afresh how this revised circuit, shown in Fig. 3.6.5 works. Rather than defining eight variables and solving for the current in the load resistor, let's take a hint from other analysis (series rules, parallel rules). Resistors
R 2 and Rare in a L parallelconfiguration: The voltages across each resistor are the same while the currents are not. Because the voltages are the same, we can find the current through each from their v-irelations:
\[i_{2}=\frac{v_{out}}{R_{2}}\; \: and\; \; i_{L}=\frac{v_{out}}{R_{L}}\]
Considering the node where all three resistors join, KCL says that the sum of the three currents must equal zero. Said another way, the current entering the node through
R 1 must equal the sum of the other two currents leaving the node. Therefore,
\[i_{1}=i_{2}+i_{L}\\ i_{1}=v_{out}\left ( \frac{1}{R_{2}}+\frac{1}{R_{L}} \right )\]
Let
R eq denote the equivalent resistance of the parallel combination of Using 's v-irelation, the voltage across it is
\[v_{1}=\frac{R_{1}v_{out}}{R_{eq}}\]
The KVL equation written around the leftmost loop has
\[v_{in}=v_{1}+v_{out}\]
Substituting for
v 1, we find
\[v_{in}=v_{out}\left ( \frac{R_{1}}{R_{eq}} +1\right )\]
\[\frac{v_{out}}{v_{in}}=\frac{R_{1}}{R_{eq}} +1\]
Thus, we have the input-output relationship for our entire system having the form of voltage divider, but it does
not equal the input-output relation of the circuit without the voltage measurement device. We can not measure voltages reliably unless the measurement device has little effect on what we are trying to measure. We should look more carefully to determine if any values for the load resistance would lessen its impact on the circuit. Comparing the input-output relations before and after, what we need is
\[R_{eq}\approx R_{2}\]
As
\[R_{eq}=\left ( \frac{1}{R_{2}}+\frac{1}{R_{L}} \right )^{-1}\]
The approximation would apply if
\[\frac{1}{R_{2}}\gg \frac{1}{R_{L}}\; \; or\\ R_{2}\ll R_{L}\]
This is the condition we seek:
Voltage Measurement
Voltage measurement devices must have large resistances compared with that of the resistor across which the voltage is to be measured.
Exercise \(\PageIndex{1}\)
Let's be more precise: How much larger would a load resistance need to be to affect the input-output relation by less than 10%? by less than 1%?
Solution
\[R_{eq}=\frac{R_{2}}{1+\frac{R_{2}}{R_{L}}}\]
Thus, a 10% change means that the ratio
\[\frac{R_{2}}{R_{L}}< 0.1\]
A 1% change means that
\[\frac{R_{2}}{R_{L}}< 0.01\]
Example \(\PageIndex{1}\):
We want to find the total resistance of the example circuit. To apply the series and parallel combination rules, it is best to first determine the circuit's structure: What is in series with what and what is in parallel with what at both small- and large-scale views. We have
R 2 in parallel with this combination is in series with This series combination is in parallel with Note that in determining this structure, we started awayfrom the terminals, and worked toward them. In most cases, this approach works well; try it first. The total resistance expression mimics the structure:
\[R_{T}=R_{1}\parallel (R_{2}\parallel R_{3}+R_{4})\\ R_{T}=\frac{R_{1}R_{2}R_{3}+R_{1}R_{2}R_{4}+R_{1}R_{3}R_{4}}{R_{1}R_{2}+R_{2}R_{3}+R_{2}R_{4}+R_{3}R_{4}}\]
Such complicated expressions typify circuit "simplifications." A simple check for accuracy is the units: Each component of the numerator should have the same units (here
Ω 3) as well as in the denominator( Ω). The entire expression is to have units of resistance; thus, the ratio of the numerator's and denominator's units should be ohms. Checking units does not guarantee accuracy, but can catch many errors. 2
Another valuable lesson emerges from this example concerning the difference between cascading systems and cascading circuits. In system theory, systems can be cascaded without changing the input-output relation of intermediate systems. In cascading circuits, this ideal is rarely true
unless the circuits are so designed. Design is in the hands of the engineer; he or she must recognize what have come to be known as loading effects. In our simple circuit, you might think that making the resistance R L large enough would do the trick. Because the resistors Rand 1 Rcan have virtually any value, you can never make the resistance of your voltage measurement device big enough. Said another way, 2 a circuit cannot be designed in isolation that will work in cascade with all other circuits. Electrical engineers deal with this situation through the notion of specifications: Under what conditions will the circuit perform as designed? Thus, you will find that oscilloscopes and voltmeters have their internal resistances clearly stated, enabling you to determine whether the voltage you measure closely equals what was present before they were attached to your circuit. Furthermore, since our resistor circuit functions as an attenuator, with the attenuation (a fancy word for gains less than one) depending only on the ratio of the two resistor values:
\[\frac{R_{2}}{R_{1}+R_{2}}=\left ( 1+\frac{R_{1}}{R_{2}} \right )^{-1}\]
We can select
any values for the two resistances we want to achieve the desired attenuation. The designer of this circuit must thus specify not only what the attenuation is, but also the resistance values employed so that integrators—people who put systems together from component systems—can combine systems together and have a chance of the combination working.
Fig. 3.6.6 below, summarizes the series and parallel combination results. These results are easy to remember and very useful. Keep in mind that for series combinations, voltage and resistance are the key quantities, while for parallel combinations current and conductance are more important. In series combinations, the currents through each element are the same; in parallel ones, the voltages are the same.
Fig. 3.6.6 Series and parallel combination rules
Exercise \(\PageIndex{1}\)
Contrast a series combination of resistors with a parallel one. Which variable (voltage or current) is the same for each and which differs? What are the equivalent resistances? When resistors are placed in series, is the equivalent resistance bigger, in between, or smaller than the component resistances? What is this relationship for a parallel combination?
Solution
In a series combination of resistors, the current is the same in each; in a parallel combination, the voltage is the same. For a series combination, the equivalent resistance is the sum of the resistances, which will be larger than any component resistor's value; for a parallel combination, the equivalent conductance is the sum of the component conductances, which is larger than any component conductance. The equivalent resistance is therefore smaller than any component resistance.
Contributor ContribEEOpenStax
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The field and its conjugate momentum
are operators. They act on the Hilbert space of the theory, which is not the space of square-integrable functions of position, as it is in single particle quantum mechanics. Rather, the configuration space of the theory is the set of all field configurations and the wavefunction is actually a functional, which ascribes an amplitude $\Psi[\phi(x^\mu)]$ for the field to be in a configuration $\phi(x^\mu)$.
QFT is just quantum mechanics applied to an infinite collection of degrees of freedom labeled by positions $x$. Once you get used to the switch it comes to seem pretty simple, but the set of all field configurations makes a huge space so naturally it is very difficult to make things mathematically rigorous. For the most part physicists can live without rigourously defining the configuration space.
The dictionary from ordinary quantum mechanics to quantum field theory is
$$\begin{array}{lcl}\mathrm{QM} && \mathrm{QFT}\\t,i &\to& \left(t,x,y,z\right)\equiv\left(t,{\bf x}\right)\equiv x^{\mu}\\q_{i}\left(t\right) &\to& \phi\left(x^{\mu}\right)\\p_{i}\left(t\right) &\to& \pi\left(x^{\mu}\right)\\\psi\left(t,q_{1},q_{2},\cdots,q_{N}\right) &\to& \Psi\left[\phi\left(x^{\mu}\right)\right]\\\left[q_{i}\left(t\right),\ p_{j}\left(t\right)\right]~=~i\hbar\delta_{ij} &\to& \left[\phi\left(t, {\bf x}\right),\ \pi\left(t,{\bf y} \right)\right]=i\hbar\delta^3\left({\bf x}-{\bf y}\right)\\\hat{H}\left(t\right) &\to& \hat{H}\left(t\right)=\int\mathrm{d}^{3}x\ \hat{\mathcal{H}}\left(x^{\mu}\right)\end{array}$$
You can represent the field momenta explicitly by functional derivatives:
$$ \pi(x^\mu) ~=~ - i\hbar \frac{\delta}{\delta \phi(x^\mu)}, $$
and check that this satisfies the commutation relationship.
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We often face the problem of having to sell an asset within a specified period of time. This problem worsens if we also lack information about our asset’s historical prices. Under this circumstance,
when should we sell our asset?
In this post I’ll resolve this dilemma based on the classic
Secretary Problem – also known as The Sultan’s Dowry Problem. This problem owes its name to the situation that a manager encounters when he wants to hire the best applicant for the vacant secretary position in his firm. He sequentially interviews each applicant for the job. Immediately after each interview, the manager must decide whether to reject the applicant or hire them. Once rejected, an applicant cannot be recalled.
Our problem is identical. Let’s imagine I’m in possession of a share whose prices have gotten so unstable that
its historical price information is no longer useful. My financial advisors have warned me against trading it, at least until the dust has finally settled. However, it seems that this situation will linger much longer than I’m willing to be involved in it for. So I’ve come to a decision: I’m selling this share within the next 100 days.
Each day, my share takes a different price and I can decide to sell it, or hold in hopes of finding a better price. I don’t want to settle for less, so I’m aiming to sell it for
the best price of all among the next 100 prices. Thus, just as the manager from the Secretary Problem, I have to decide when to stop looking through new applicant prices and sell my asset. Let’s start recruiting!
On
, I check my share price \(p_{1}\). I feel that if I sell it right now – and taking into account our hypotheses – I’m betting that none of the next 99 prices will beat the current one. I’d better wait until tomorrow. day 1
On
, I gladly find that today’s price \(p_{2}\) beats \(p_{1}\). “ day 2 Good news I didn’t sell yesterday!” I say to myself. Should I sell now? Should I hold? I feel I’m on a roll so I hold, at least, until tomorrow.
On
, I get a dose of reality when I find that my share’s price has dropped even lower than day 3 day 1price. I can’t go on with this game. It’s only day 3and I might have lost the chance of selling the share for the best price. Time to put our math to work
At this point, I notice several aspects of my problem:
Selling too soonreduces the chance of finding the best price. Waiting too longincreases the chance of looking past the best price and rejecting it.
The solution to this problem consists of checking the first \(k\) prices, \(p_{1}\), \(p_{2}\),…, \(p_{k}\) and rejecting them, regardless. After that, we should sell our share for the first price that beats those. In case none of the latter beats them – this happens when the best price is among the first \(k\) prices – we’ll have to sell it for the last price \(p_{N}\). The aim of our problem is to choose the optimal number of prices, \(k\), that we have to reject no matter what, in order to maximise the odds of choosing the best price of all.
In short, the steps we’ll follow are:
Begin checking and rejecting the first \(k\) prices. After the \(k\)-th price is known: If the current price beats all the previous ones: Sell the share. Else: Reject current price and keep checking. If we reach the last price, we’ll have to sell our share for that price.
In order to compute the optimal \(k\), we’ll denote the probability of choosing the best price among the \(N\) available, rejecting prices up until \(k\) just as mentioned before by \(P_{N}(S_{k})\). From the formula of total probability:
$$P_{N}(S_{k}) = \sum_{i=1}^{N} P(A_{i})\cdot P_{N}(S_{k}|A_{i})$$
Where \(A_{i}\) is the event of the best price being in the \(i\)-th position, and \(N\) is the maximum number of prices we can check. In our case \(N\)=100.
We’ll assume that the events \(A_{i}\) are equally possible for every \(i\). Thus,\(P(A_{i})={\frac{1}{N}} \forall i\). We also know that we’ll only choose the best price if it’s posterior to the \(k\)-th position, so \(P_{N}(S_{k}|A_{i}) = 0\) for \(i=1,…,k\). This means that:
$$P_{N}(S_{k}) = {\frac{1}{N}} \sum_{i=k+1}^{N} P_{N}(S_{k}|A_{i})$$
The best price located at the \(i\)-th position (\(i>k\)) will be chosen if, and only if, the highest price prior to it has been rejected, i.e. if the highest of the \(i-1\) already seen prices is among the \(k\) first prices. The odds of this happening are \({\frac{k}{i-1}}\). Substituting in the previous formula:
$$P_{N}(S_{k}) = {\frac{1}{N}} \sum_{i=k+1}^{N} {\frac{k}{i-1}} = {\frac{k}{N}}\sum_{i=k+1}^{N} {\frac{1}{i-1}} = {\frac{k}{N}} [ \, {\frac{1}{k}}+{\frac{1}{k+1}}+…+{\frac{1}{N-1}}] \,$$
Back to our problem
If we plot this probability with \(N=100\), for every \(k=1,…,99\) we get:
We can see that \(k=37\) maximizes the chances of success. According to my calculations,
I’ll choose the best price with a probability of 37.10% if I follow the strategy of rejecting the first 37 prices. Not bad!
Nevertheless, I can’t help noticing that my hypotheses imply that prices follow no pattern and this wild behaviour will remain for at least 100 days.
In which other situations could I benefit from this strategy?
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The Stochastic portfolio theory (SPT), a relatively new portfolio management theory, was first introduced in 1999 by Robert Fernholz. It can be combined with Machine Learning and Bayesian statistics. This allows the investor to generate trading strategies. It’s a very attractive theory for several reasons: it’s theoretical, it’s not very well known and, most importantly, it’s cool.
Let’s begin this post with a gentle introduction to the topic.
This theory is all about defining and studying investment strategies from the perspective of stochastic calculus. This branch of mathematics studies the characteristics of stochastic processes. So a good starting point is:
What’s a stochastic process?
A stochastic process, as fancy as it sounds, is just a collection of random variables, usually indexed in a certain way, (frequently, by time). So at each point of time we have a realisation of a random variable. As an example, a geometric Brownian motion is a stochastic process widely used in finance. Under this model, asset returns are a normal random walk. As such, the dynamics of the process can by specified by the following stochastic differential equation:
\( dS = \mu S dt + \sigma S dX \)
Where \(S\) is the stock price process with mean return \(\mu\) and volatility \(\sigma\).
If we simulate several realisations of this process, we get:
As you can see, the results actually look much like what you’d expect in a financial asset in terms of feel-and-look, but of course, these simulations have a lot of differences and limitations in terms of statistical properties (check out this post on Levy flights).
Diversity-weighted portfolios
The SPT theory assumes that stock market capitalisations (or prices, if we assume that the number of outstanding shares is one), follow a general stochastic process that fulfils a set of mathematical conditions.
The theory proposes ways of generating portfolios of these assets (modelled as stochastic processes) and then analyses the characteristics of these theoretical portfolios.
The best example of these portfolios are diversity-weighted portfolios. These portfolios are built by assigning weights according to the following function:
\( \pi_i = \frac{\mu_i^p}{\sum \mu_i^p}\)
Where \(\mu_i(t) := \frac{X_i(t)}{X_1(t) + … + X_n(t)}\) is the capitalisation share of asset i, \(\pi_i\) is the new weight assigned to the asset and \(p\) is a hyper-parameter of the investment function that is crucial for the performance of the strategy.
In layman terms, the diversity-weighted approach (for the case of negative p), assigns weights inversely proportional to market capitalization, overweighting small companies over big ones.
Surprisingly enough, under certain conditions, Fernholz proves that a portfolio built in such a way outperforms the capitalisation-weighted index –wait for it… – with probability one!
As appealing as it might sound, this portfolio just exploits a mix of a mean-reversion operative at the market level, and the effect of the small minus big factor.
All in all, this result is extremely interesting for an investment strategist, and also poses the question of how to solve the inverse problem. That is, of finding general investment functions that match our investment objectives.
The inverse problem
Fortunately enough, Kom Samo and Vervuurt introduce, in a recent paper, a Bayesian approach to learning this kind of functions for stock portfolios using random gaussian processes to model the objective investment strategy.
The idea behind their approach is that, by assigning a specific probability distribution to the results we want to obtain, we can force the distributions of the learnt parameters to meet our investment objective.
This is a very interesting idea that can be generalised without limits! To carry out a small test let’s imagine building a trading strategy over only one series (EURCHF currency cross for this example).
We want to find a function that maps the value of some features (macroeconomic, financial and technical indicators) at a given point in time to the position that should be kept for the next day.
\(X^d \rightarrow f\) with \( f\in [-1, 1] \)
where \(f\) is the daily position of the strategy.
To train that function, a Bayesian neuron with
tanh activation function is perfect, because it’s naturally bounded between -1 and 1:
\(f = tanh(\beta ^T \cdot X)\)
We can now set normal priors on the betas and uninformative priors on the means of the betas. The last step to train our investment strategy would be computing our objective investment function from the positions and placing a prior on it.
A good investment objective is the well-known Sharpe Ratio, that with a small modification can be used to try to match some objective performance:
\( SR = \frac{mean – obj}{std} \cdot \sqrt{261} \)
Finally, we place a (very greedy) normal prior distribution over the Sharpe ratio with mean=3, standard deviation=1, using a 30% annualised return as an objective.
Running 2500 simulations with the pyMC3 package for python, we obtain the following posterior distributions over the variables of our model:
As we can see, the results aren’t very encouraging at first glance. The uncertainty of the estimations on the \(\beta\) is huge, and the posterior of the objective function (SR) takes negative values!
However, it’s worth noting that our objective was extremely greedy, so there’s still room for some hope.
The results of the learnt strategy
By taking the average of the means of our coefficients (the betas), we can get the cumulative performace of the strategy:
It looks great, but playing in-sample is cheating, so let’s check the out-sample performance:
As usual, out-sample performance of the learnt strategy is worse.
Finally, we can check the robustness of our results. The nice thing about having many sample strategies of our posterior distribution is that we can take a look at the different realisations of the trading function we have learnt:
The realisations of our strategy are quite random, but not very volatile. They have just enough positive bias to light the candle of hope in our little profit-seeking hearts…
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Answer
The testimony is going to hurt their case.
Work Step by Step
We multiply the density of the beer (which is equal to that of water) by the increase in height and by the cross-sectional area of the keg to obtain: $ \Delta m = 1000 \times .012 \times \pi \times (.2^2)=1.51kg\approx 53 \ oz$ This is more than the amount that the person would need to be above the legal limit, so this is going to hurt their case.
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I am trying to compute the dBFS value of a group of samples (stereo wave file), according to this formula:
$$p_{RMS} = \sqrt{\frac{ x_1^2 + x_2^2 + \ldots}n } $$ $$dbFS = 20\log_{10}\frac{ p_{RMS}}{p_{max}}$$
The value I get is wrong, and not stable (I am using a stereo constant -18dbFS 1000H sinewave audio file). I know the input/output of the program is working, because I did an fft on a music file and the output is fine.
float get_db(Header *header, unsigned int *buffer_size, int32_t left[], int32_t right[]) { unsigned int i = 0; float dbfs = 0; //From -144 to 0dbfs double pmoy = 0; double pmax = 0; pmax = maxint(header->bits_per_sample); //Max positive range of sample, here it is 8388608 for (i=0; i < *buffer_size; i++){ //Buffer size of 4096 here pmoy =+ pow(left[i], 2); //Just the left channel for now } printf("sum= %f\n", pmoy); //temporary sum pmoy = sqrt( pmoy / *buffer_size); printf("pmoy= %f\n", pmoy); //pmoy rms of 4096 samples dbfs = 20 * log10( pmoy / pmax); return dbfs; }
I really don't understand why I don't get a constant value of -18dBFS. I am still learning C, there should be some obvious reasons and I hope you could help me. Is there a normalized definition of a dBFS scale for discrete values, and a particular integration time that should be used ?
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I have a question about calculating the maximum allowable noise for an ADC signal conditioning stage.
I have an existing signal conditioning stage as well as an ADC on a microcontroller.
I would like to use the existing stage and calculate the maximum allowable noise on the input signal in order to maintain an error of less than 1 LSB.
However, I am having a bit of trouble with the math.
The ADC has the following specs:
ENOB: 10 bits
SNR: 60 dB
Input Range: 0 - 5V
Max Sampling Rate: 1 Msps
The input signal has the following characteristics but the noise
vary depending on the sensor used: will
BW: 150 Hz
FS Amplitude: 9V pk-pk
The existing signal conditioning stage has an attenuation of .5.
So far I have done the following math using this Analog Devices technical article:
\begin{equation} \text{ADC Input}_{RMS} = \frac{4.5}{2\sqrt{2}} = 883mV \end{equation} \begin{equation} \text{ADC Noise}_{RMS} = \frac{883mv}{10^\frac{60}{10}} = 883 \mu V \end{equation}
\begin{equation} \text{ADC Noise Density} = \frac{883 \mu V}{\sqrt{\frac{1}{2}}f_{sample}} = \frac{883 \mu V}{\sqrt{500KHz}} = 1.24 \frac{\mu V}{\sqrt{Hz}} \end{equation}
Now the signal conditioning stage has the following non-linear spectral noise density found using an LTSpice noise simulation:
Integrating this over the 150 Hz BW gives 926.67nV.
My question from here is how can I relate these two values and calculate the maximum allowable input noise?
Can I simply assume the input noise must be an order of magnitude higher than the root sum square (RSS) of these two noise densities? Am I missing something critical?
If there is any other information I can provide to help please let me know.
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Euclidean Geometry • Equitable Distributing • Linear Programming • Set Theory • Nonstandard Analysis • Advice • Topology • Number Theory • Computation of Time (Previous | Next)
The following section presupposes the chapters on Set Theory and Nonstandard Analysis. Let \(m, n \in {}^{\omega}\mathbb{N}\) and \(k \in \mathbb{N}\).
Bounding theorem for \(\omega\)-transcendental numbers: Every non-zero complex number whose imaginary or real part has absolute value is \(\le \hat{\omega}\) or \(\ge \omega\) is automatically \(\omega\)-transcendental.
Proof: In a polynomial or series equation, set \({a}_{m} = 1\) and \({a}_{k} = -{\acute{\omega}}\) for \(k < m\), then the claim in the real case follows from the geometric series formula after taking the reciprocal. We can find the exact limit value by replacing \(\omega\) by \({\omega}(m) = \omega - \acute{\omega}/{\omega(m)}^{m}\). The complex cases are solved by setting \(x = (1 + ib)\omega\) for \(b \in {}^{\omega }{\mathbb{R}}\).\(\square\)
Coefficient theorem for \(\omega\)-transcendental numbers: Every normalised irreducible polynomial and series such that that \(|{a}_{k}| \ge \omega\) for at least one \({a}_{k}\) only has \(\omega\)-transcendental zeros.
Proof: The zeros of normalised irreducible polynomials and series are pairwise distinct and uniquely determined. Since they are not \(\omega\)-algebraic, they must be \(\omega\)-transcendental.\(\square\)
Approximation theorem for \(\omega\)-algebraic numbers: Every real \(\omega\)-algebraic number of degree \(n > 1\) may be approximated by a real \(\omega\)-algebraic number of degree \(e^\nu := m < n\) with an average asymptotic error of \(\hat{\nu}\iota \zeta(\grave{m}){|{}^{\omega }\mathbb{Z}|}^{-m}\).
Proof: On the conventionally real axis, the number of \(\omega\)-algebraic numbers approximately evenly distributed between fixed limits increases by a factor of approximately \(|{}^{\omega }\mathbb{Z}|\) per degree. The error corresponds to the distance between \(\omega\)-algebraic numbers. The non-real \(\omega\)-algebraic numbers are less dense.\(\square\)
Conclusion: Two distinct real \(\omega\)-algebraic numbers have an average distance of at least \(\hat{\varphi}\pi{|{}^{\omega }\mathbb{Z}|}^{-\acute{\omega}}\). Determining this minimum distance exactly requires an infinite non-linear non-convex optimisation problem to be solved. Therefore, the \(c\)-algebraic numbers have an approximate order of \(\mathcal{O}(c)\). This disproves Roth's theorem, which does not prove more than the (trivial) minimum distance between two rational numbers. Thus, the abc conjecture is wrong, but not Liouville's result.
Theorem: The maximum distance between two neighbouring real \(\omega\)-algebraic numbers is \(\Omega/\acute{\omega}\) for the \(\omega\)-transcendental omega constant \(\Omega = e^{-\Omega} = W(1)\) (see below Lambert-W function).
Proof: The distance between two real \(\omega\)-algebraic numbers is largest around the points \(\pm 1\). The number 1 may be approximated by an real \(\omega\)-algebraic x that satisfies the polynomial or series equation \(\acute{x}x^{\acute{m}}\acute{\omega} = 1\) for \(x > 1\) or \(x^m = -\acute{x}\acute{\omega}\) for \(x < 1.\square\)
Theorem: For every number \(z \in \mathbb{Q}+i\mathbb{Q}\) that is neither 0 nor root of unity, the geometric series \(\sum\limits_{n=0}^{\omega}{{{z}^{n}}}=\widehat{1-z}(1-{{z}^{\grave{\omega}}})\) is already \(\omega\)-transcendental.
Proof: The modulus of either the numerator or denominator \({z}^{\grave{\omega}}\) is \(>{2}^{\omega/2}.\square \)
Theorem: Euler's number \(e\) is \(\omega\)-transcendental.
Proof: If we accept the exponential series as a representation of \(e\), it follows that \(e = (k{\omega} + 1)/\omega!\) for \(k > \omega\). Therefore, the numerator and the denominator of this fraction must be \(> \omega\), since neither \(\omega\) nor a prime divisor of \(k\) in the numerator simplifies with \(\omega!\). However, if we accept the representation \({(1 + \hat{\omega})}^{\omega}\) for \(e\), the claim is trivial. Note that these two representations give different numbers.\(\square\)
The greatest-prime criterion for \(\omega\)-transcendental numbers: If a real number may be represented as an irreducible fraction \(\widehat{ap}b \pm \hat{s}t\), where \(a, b, s\), and \(t\) are natural numbers, \(abst \ne 0\), \(a + s > 2\), and the (second-)greatest prime number \(p \in {}^{\omega }\mathbb{P}, p \nmid b\) and \(p \nmid s\), then \(r\) is \(\omega\)-transcendental.
Proof: The denominator \(\widehat{ap s} (bs \pm apt)\) is \(\ge 2p \ge 2\omega - \mathcal{O}(\ell) > \omega\) by the prime number theorem.\(\square \)
Theorem: Pi \(\pi\) is \(\omega\)-transcendental.
Proof: This follows from its Wallis product representation, or its product representation using the gamma function with value \(-\hat{2}\), provided that we accept these representations. It should be noted that these two representations yield distinct numbers. Alternatively, we can apply the greatest-prime criterion to the Leibniz series, or the Taylor series of arcsin\((x)\) at \(x = 1\).\(\square\)
Theorem: The constants of \(({C}_{Artin})\), Baxter \(({C}_{2})\), Chaitin \(({\Omega}_{F})\), Champernowne \(({C}_{10})\), Copeland-Erdős \(({C}_{CE})\), Erd\H{o}s-Borwein \((E)\), Feller-Tornier \(({C}_{FT})\), Flajolet and Richmond \((Q)\), Glaisher-Kinkelin \((A)\), Heath-Brown-Moroz \(({C}_{HBM})\), Landau-Ramanujan \((K)\), Liouville \(({£}_{Li})\), Murata \(({C}_{M})\), Pell \(({P}_{Pell})\), Prouhet-Thue-Morse \((\tau)\), Sarnak \(({C}_{sa})\) and Stephen \(({C}_{S})\) as well as the Euler resp. Landau totient constant \((ET\) resp. \(LT)\), the twin prime constant \(({C}_{2})\) and the carefree constants \(({K}_{1}, {K}_{2}\) and \({K}_{3})\) are \(\omega\)-transcendental, since an existing (large) power of a small or very large prime cannot be removed from numerator or denominator by simplifying.\(\square\)
Remark: The claim for \({C}_{CE}\) clearly also holds for every base from \({}^{c}\mathbb{N}^{*}\).
Theorem: The constants of Catalan \((G)\), Gieseking \((\pi \, \ln \, \beta)\), Smarandache \(({S}_{1})\) and Taniguchi \(({C}_{T})\) are \(\omega\)-transcendental because of the greatest-prime criterion.\(\square\)
Theorem: The trigonometric and hyperbolic functions and their inverse functions, the digamma function \(\psi\), the Lambert-\(W\)-function, the \(Ein\) function, the (hyperbolic) sine integral \(S(h)i\), the Euler's Beta function \(B\), and, for positive natural numbers \(s\) and \(u\) and natural numbers \(t\), the generalised error function \({E}_{t}\), the hypergeometric function \({}_{0}{F}_{t}\), the Fresnel integrals \(C\) and \(S\) and the Bessel function \({I}_{t}\) and the Bessel function of the first kind \({J}_{t}\), the Legendre function \({\chi}_{t}\), the polygamma function \({\psi}_{t}\), the generalised Mittag-Leffler function \({E}_{s,t}\), the Dirichlet series \(\sum\limits_{n=1}^{\omega}{{\hat{n}^{s}f(n)}\;}\) with maximally finite rational \(|f(n)|\), the prime zeta function \(P(s)\), the polylogarithm \({Li}_{s}\) and the Lerch zeta-function \(\Phi(q, s, r)\) always yield \(\omega\)-transcendental values for rational arguments and maximal finite rational \(|q|\) and \(|r|\) at points where their Taylor series converge.
Proof: The claim follows from the greatest-prime criterion, the Dirichlet prime number theorem, and the Wallis product. For the digamma function, the claim follows from the proof of \(\omega\)-transcendence of Euler's constant below.\(\square\)
Theorem: The gamma function \(\Gamma(z) := k! \, {k}^{z}/(z(z + 1) ... (z + k))\), where \(k = {\omega}^{{\omega}^{2}}\) and \(z \in {}^{\omega }\mathbb{C} \setminus -{}^{\omega }\mathbb{N}\), is \(\omega\)-transcendental for \(z \in {}^{\omega }\mathbb{Q}\) and for suitable supersets of \({}^{\omega }\mathbb{N}\) resp. \({}^{\omega }\mathbb{Q}\).
Proof: The values of \(\Gamma(z)\) are the zeros of minimal polynomials or series with infinite integer coefficients.\(\square\)
Theorem: For \(x \in {}^{\omega }{\mathbb{R}}\), let be \(s(x) := \sum\limits_{n=1}^{\omega}{\hat{n}{{x}^{n}}}\) and \(\gamma := s(1) - \ln \, \omega = \int\limits_{1}^{\omega}{\left( \widehat{\left\lfloor x \right\rfloor} - \hat{x} \right)dx} \in \; ]0, 1[\) (can be seen by rearranging) Euler's constant. If we accept \(s(\hat{2})\ell\) as a representation of ln \(\omega, \gamma\) is therefore with a precision \(\mathcal{O}({2}^{-\omega}\hat{\omega}\ell)\) \(\omega\)-transcendental.
Proof: We obtain \(-\ln(-\acute{x}) = s(x) + \mathcal{O}(\hat{\omega}{x}^{\grave{\omega}}/\acute{x}) + t(x)dx\) for \(x \in [-1, 1 - \hat{c}]\) and a real function \(t(x)\) such that \(|t(x)| < {\omega}\) by (exact) integration (see Nonstandard Analysis) of the geometric series. After applying Fermat's little theorem to the numerator of \(\hat{p}(1 - 2^{-p}\ell)\) for \(p = \max\, {}^{\omega}\mathbb{P}\), the greatest-prime criterion yields the claim.\(\square\)
Remark: If \(\omega\) is replaced by an arbitrary \(k \in {}^{\omega }\mathbb{N}_{\ge\omega/2}\), the preceding proof is barely more difficult.
Definition: When two numbers \(x, y \in {}^{\omega }\mathbb{C}^{*}\) or their reciprocals do not satisfy any polynomial or series equation \(p(x, y) = 0\), so they are called \(\omega\)-
algebraically independent.
Theorem: The greatest-prime criterion, with \(e = {(1 + \hat{p})}^{p}\) for maximal \(p \in {}^{\omega }\mathbb{P}\) and \(\pi\) as Wallis product, yields pairwise \(\omega\)-algebraically independent representations of \(A, {C}_{2}, \gamma, e, K\) and \(\pi.\square\)
Theorem: The BBP series \(\sum\limits_{n=1}^{\omega}{p(n)\widehat{q(n){{b}^{n}}}}\) for \(b \in {}^{\omega }\mathbb{N}_{\ge 2}\) and integer polynomials resp. series \(p\) and \(q \in {}^{\omega }\mathbb{Z}\) with \(q(n) \ne 0\) and \(\deg(p) < \deg(q)\) only yield \(\omega\)-transcendental values.
Proof: We can reduce the sum to a smallest common denominator \(d \ge {b}^{k} > \omega\) with \(d, k \in \mathbb{N}^{*}.\square\)
Definition: A rational number \(\ne 0\) is said to be
power-free if it cannot be represented as the power of a rational number with integer exponent \(\ne \pm 1\). Let \(||\cdot|{{|}_{d}}\) be the distance to the nearest integer.
Theorem: For any power-free \(q \in Q := {\mathbb{Q}}_{>0}\), we have that \({q}^{x} \in Q\) for real \(x\) if and only if \(x \in {}^{\omega }\mathbb{Z}\) and \(|x|\) is not excessively large.
Proof: Let wlog \(x > 0\). Since there is no contradiction for \(x \in {}^{\omega }\mathbb{N}\), assume \(x \in Q \setminus {}^{\omega }{\mathbb{N}}^{*}\). Since this implies \({q}^{x} \in {}^{\omega }{\mathbb{A}}_{R} \setminus Q\), assume \(x := k/d \in {}^{\omega }\mathbb{R}_{>0} \setminus Q\) for \(d, k \in {\mathbb{N}}^{*}\) and gcd\((d, k) = 1\). This implies \({q}^{k} = {r}^{d}\) for an \(r \in Q\). The fundamental theorem of arithmetic yields a numerator or denominator of \(q\) or \(r\) greater than \(2^{\omega}\). This contradiction results in the claim.\(\square\)
Remark: This theorem proves the Alaoglu and Erdős conjecture, which states that \({p}^{x}\) and \({q}^{x}\) are \(c\)-rational for distinct \(p, q \in {}^{c}\mathbb{P}\) if and only if \(x \in {}^{c}\mathbb{Z}\) and \(|x|\) is not excessively large. By replacing \({}^{c}\mathbb{N}\) with \({}^{\omega }\mathbb{N}\) and making the required adjustments these arguments can be extended to finite transcendental numbers. Inconcrete transcendence implies finite transcendence.
Littlewood theorem in conventional mathematics: We have for all \(a,b\in {}^{c}\mathbb{R}\) and \(n\in {}^{c}\mathbb{N}^{*}\):\[\underset{n\to \infty }{\mathop{\lim \inf }}\,n\;||na|{{|}_{d}}\;||nb|{{|}_{d}}=0.\]Proof: Let be \(k, m \in {}^{c}\mathbb{N}^{*}\) the denominators of the continued fraction of \(a\) resp. \(b\) with precision \(g \in {}^{c}\mathbb{R}_{> 0}\) and \(n\) again and again a natural multiple of \(km\). Then we have according to Dirichlet's approximation theorem (see \cite{455}, p. 63): \[\underset{n\to \infty }{\mathop{\lim \inf }}\,n||na|{{|}_{d}}||nb|{{|}_{d}}=\underset{n\to \infty }{\mathop{\lim \inf }}\,n\mathcal{O}{{(\hat{n})}^{2}}=\underset{n\to \infty }{\mathop{\lim \inf }}\,\mathcal{O}(\hat{n})=0.\square\]Refutation of the Littlewood conjecture in nonstandard mathematics: Let \(a = b := {{\omega}^{-{3}/{2}}}\). Then we have: \[\omega \;||\omega a|{{|}_{d}}\;||\omega b|{{|}_{d}}= 1 \ne 0.\square\]Theorem: The generalised Riemann hypothesis is disproved by the Dirichlet \(L\)-function \(L\left(s,\chi\right)=\sum\limits_{n=1}^{\omega}{\chi\left(n\right)n^{-s}}\), which clearly has because of the geometric series (cf. Set Theory) only zeros for \(s = 0\) and nontrivial Dirichlet characters \(\chi(n)\).\(\square\)
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How to Implement the Fourier Transformation from Computed Solutions
We previously learned how to calculate the Fourier transform of a rectangular aperture in a Fraunhofer diffraction model in the COMSOL Multiphysics® software. In that example, the aperture was given as an analytical function. The procedure is a bit different if the source data for the Fourier transformation is a computed solution. In this blog post, we will learn how to implement the Fourier transformation for computed solutions with an electromagnetic simulation of a Fresnel lens.
Fourier Transformation with Fourier Optics
Implementing the Fourier transformation in a simulation can be useful in Fourier optics, signal processing (for use in frequency pattern extraction), and noise reduction and filtering via image processing. In Fourier optics, the Fresnel approximation is one of the approximation methods used for calculating the field near the diffracting aperture. Suppose a diffracting aperture is located in the (x,y) plane at z=0. The diffracted electric field in the (u,v) plane at the distance z=f from the diffracting aperture is calculated as
where, \lambda is the wavelength and E(x,y,0), \ E(u,v,f) account for the electric field at the (x,y) plane and the (u,v) plane, respectively. (See Ref. 1 for more details.)
In this approximation formula, the diffracted field is calculated by Fourier transforming the incident field multiplied by the quadratic phase function {\rm exp}\{-i\pi (x^2+y^2)/(\lambda f)\}.
The sign convention of the phase function must follow the sign convention of the time dependence of the fields. In COMSOL Multiphysics, the time dependence of the electromagnetic fields is of the form {\rm exp}(+i\omega t). So, the sign of the quadratic phase function is negative.
Fresnel Lenses
Now, let’s take a look at an example of a Fresnel lens. A Fresnel lens is a regular plano-convex lens except for its curved surface, which is folded toward the flat side at every multiple of m \lambda/(n-1) along the lens height, where
m is an integer and n is the refractive index of the lens material. This is called an m th-order Fresnel lens.
The shift of the surface by this particular height along the light propagation direction only changes the phase of the light by 2m \pi (roughly speaking and under the paraxial approximation). Because of this, the folded lens fundamentally reproduces the same wavefront in the far field and behaves like the original unfolded lens. The main difference is the diffraction effect. Regular lenses basically don’t show any diffraction (if there is no vignetting by a hard aperture), while Fresnel lenses always show small diffraction patterns around the main spot due to the surface discontinuities and internal reflections.
When a Fresnel lens is designed digitally, the lens surface is made up of discrete layers, giving it a staircase-like appearance. This is called a multilevel Fresnel lens. Due to the flat part of the steps, the diffraction pattern of a multilevel Fresnel lens typically includes a zeroth-order background in addition to the higher-order diffraction.
Why are we using a Fresnel lens as our example? The reason is similar to why lighthouses use Fresnel lenses in their operations. A Fresnel lens is folded into m \lambda/(n-1) in height. It can be extremely thin and therefore of less weight and volume, which is beneficial for the optics of lighthouses compared to a large, heavy, and thick lens of the conventional refractive type. Likewise, for our purposes, Fresnel lenses can be easier to simulate in COMSOL Multiphysics and the add-on Wave Optics Module because the number of elements are manageable.
Modeling a Focusing Fresnel Lens in COMSOL Multiphysics®
The figure below depicts the optics layout that we are trying to simulate to demonstrate how we can implement the Fourier transformation, applied to a computed solution solved for by the
Wave Optics, Frequency Domain interface. Focusing 16-level Fresnel lens model.
This is a first-order Fresnel lens with surfaces that are digitized in 16 levels. A plane wave E_{\rm inc} is incident on the incidence plane. At the exit plane at z=0, the field is diffracted by the Fresnel lens to be E(x,y,0). This process can be easily modeled and simulated by the
Wave Optics, Frequency Domain interface. Then, we calculate the field E(u,v,f) at the focal plane at z=f by applying the Fourier transformation in the Fresnel approximation, as described above.
The figures below are the result of our computation, with the electric field component in the domains (top) and on the boundary corresponding to the exit plane (bottom). Note that the geometry is not drawn to scale in the vertical axis. We can clearly see the positively curved wavefront from the center and from every air gap between the saw teeth. Note that the reflection from the lens surfaces leads to some small interferences in the domain field result and ripples in the boundary field result. This is because there is no antireflective coating modeled here.
The computed electric field component in the Fresnel lens and surrounding air domains (vertical axis is not to scale). The computed electric field component at the exit plane. Implementing the Fourier Transformation from a Computed Solution
Let’s move on to the Fourier transformation. In the previous example of an analytical function, we prepared two data sets: one for the source space and one for the Fourier space. The parameter names that were defined in the Settings window of the data set were the spatial coordinates (x,y) in the source plane and the spatial coordinates (u,v) in the image plane.
In today’s example, the source space is already created in the computed data set, Study 1/Solution 1
(sol1){dset1}, with the computed solutions. All we need to do is create a one-dimensional data set, Grid1D {grid1}, with parameters for the Fourier space; i.e., the spatial coordinate u in the focal plane. We then relate it to the source data set, as seen in the figure below. Then, we define an integration operator
intop1 on the exit plane.
Settings for the data set for the transformation.
The intop1 operator defined on the exit plane (vertical axis is not to scale).
Finally, we define the Fourier transformation in a 1D plot, shown below. It’s important to specify the data set we previously created for the transformation and to let COMSOL Multiphysics know that u is the destination independent variable by using the
dest operator.
Settings for the Fourier transformation in a 1D plot.
The end result is shown in the following plot. This is a typical image of the focused beam through a multilevel Fresnel lens in the focal plane (see Ref. 2). There is the main spot by the first-order diffraction in the center and a weaker background caused by the zeroth-order (nondiffracted) and higher-order diffractions.
Electric field norm plot of the focused beam through a 16-level Fresnel lens. Concluding Remarks
In this blog post, we learned how to implement the Fourier transformation for computed solutions. This functionality is useful for long-distance propagation calculation in COMSOL Multiphysics and extends electromagnetic simulation to Fourier optics.
Next Steps
Download the model files for the Fresnel lens example by clicking the button below.
Read More About Simulating Wave Optics Simulating Holographic Data Storage in COMSOL Multiphysics How to Simulate a Holographic Page Data Storage System How to Implement the Fourier Transformation in COMSOL Multiphysics References J.W. Goodman, Introduction to Fourier Optics, The McGraw-Hill Company, Inc. D. C. O’Shea, Diffractive Optics, SPIE Press.
Comments (10) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science
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11 0
I have two questions I'm a bit confused on...
1st:
Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was [tex]2.87 \times 10^{6} km[/tex] from the earth and traveling at [tex]1.20 \times 10^{4} km/h[/tex] relative to the earth. a.) At this time, what was the spacecraft's kinetic energy relative to the earth? b.) What was the potential energy of the earth-spacecraft system?
Common sense will tell me that I can find one, once I figure out the other... Now, the first time I approached this, I used the law of conservation of energy... But don't think I'm going the right way in finding the correct initial kinetic and potential energy. For this type of problem, although the terms are different for finding potential energy, is the method to find kinetic energy still the same:
[tex]K = \displaystyle{\frac{1}{2}}mv^2[/tex]? If this is true, does this mean that the initial velocity is [tex]1.20 \times 10^{4} km/h[/tex]?
I understand at 10 days, the orbiter is travelling that fast... But that speed can't be constant, can it? Wouldn't it be decreasing as it escapes earth's gravitational pull?
How would I use the velocity at 10 days to find the kinetic energy? If I simply plug in the given velocity into the kinetic energy formula, I get some insane number that can't be correct...
Any pointers??
2nd:
For a problem, I found the speed of a spacecraft at which it would crash into earth, assuming it were a distance of [tex]\infty[/tex] away, and no other astronomical object's gravitational pull affected it.
[tex]s_{e}=\sqrt{\displaystyle{\frac{2M_{e}G}{R_{e}}}}[/tex]
And this is what I was then asked:
Now find the spacecraft's speed when its distance from the center of the earth is [tex]R=\alpha R_{\rm e}[/tex], where [tex]\alpha \ge 1[/tex]. Express the speed in terms of [tex]s_{e}[/tex] and [tex]\alpha[/tex].
So, this appears to be a simple algebra problem. And after a bout of apparently incorrect algebra, I got
[tex]s_{\alpha}=\displaystyle{\frac{s_{e}}{\alpha}}[/tex]
Eh... What did I do wrong here??
1st:
Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was [tex]2.87 \times 10^{6} km[/tex] from the earth and traveling at [tex]1.20 \times 10^{4} km/h[/tex] relative to the earth.
a.) At this time, what was the spacecraft's kinetic energy relative to the earth?
b.) What was the potential energy of the earth-spacecraft system?
Common sense will tell me that I can find one, once I figure out the other... Now, the first time I approached this, I used the law of conservation of energy... But don't think I'm going the right way in finding the correct initial kinetic and potential energy. For this type of problem, although the terms are different for finding potential energy, is the method to find kinetic energy still the same:
[tex]K = \displaystyle{\frac{1}{2}}mv^2[/tex]?
If this is true, does this mean that the initial velocity is [tex]1.20 \times 10^{4} km/h[/tex]?
I understand at 10 days, the orbiter is travelling that fast... But that speed can't be constant, can it? Wouldn't it be decreasing as it escapes earth's gravitational pull?
How would I use the velocity at 10 days to find the kinetic energy? If I simply plug in the given velocity into the kinetic energy formula, I get some insane number that can't be correct...
Any pointers??
2nd:
For a problem,
I found the speed of a spacecraft at which it would crash into earth, assuming it were a distance of [tex]\infty[/tex] away, and no other astronomical object's gravitational pull affected it.
[tex]s_{e}=\sqrt{\displaystyle{\frac{2M_{e}G}{R_{e}}}}[/tex]
And this is what I was then asked:
Now find the spacecraft's speed when its distance from the center of the earth is [tex]R=\alpha R_{\rm e}[/tex], where [tex]\alpha \ge 1[/tex].
Express the speed in terms of [tex]s_{e}[/tex] and [tex]\alpha[/tex].
So, this appears to be a simple algebra problem. And after a bout of apparently incorrect algebra, I got
[tex]s_{\alpha}=\displaystyle{\frac{s_{e}}{\alpha}}[/tex]
Eh... What did I do wrong here??
Last edited:
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They're believed to be symmetric because quite often a normal approximation is used. This one works well enough in case p lies around 0.5.
binom.test on the other hand reports "exact" Clopper-Pearson intervals, which are based on the F distribution (see here for the exact formulas of both approaches). If we would implement the Clopper-Pearson interval in R it would be something like (see
): note
Clopper.Pearson <- function(x, n, conf.level){
alpha <- (1 - conf.level) / 2
QF.l <- qf(1 - alpha, 2*n - 2*x + 2, 2*x)
QF.u <- qf(1 - alpha, 2*x + 2, 2*n - 2*x)
ll <- if (x == 0){
0
} else { x / ( x + (n-x+1)*QF.l ) }
uu <- if (x == 0){
0
} else { (x+1)*QF.u / ( n - x + (x+1)*QF.u ) }
return(c(ll, uu))
}
You see both in the link and in the implementation that the formula for the upper and the lower limit are completely different. The only case of a symmetric confidence interval is when p=0.5. Using the formulas from the link and taking into account that in this case $n = 2\times x$ it's easy to derive yourself how it comes.
I personally understood it better looking at the confidence intervals based on a logistic approach. Binomial data is generally modeled using a logit link function, defined as:
$${\rm logit}(x) = \log\! \bigg( \frac{x}{1-x} \bigg)$$
This link function "maps" the error term in a logistic regression to a normal distribution. As a consequence, confidence intervals in the logistic framework are symmetric around the logit values, much like in the classic linear regression framework. The logit transformation is used exactly to allow for using the whole normality-based theory around the linear regression.
After doing the inverse transformation:
$${\rm logit}^{-1}(x) = \frac{e^x}{1+e^{x}}$$
You get an asymmetric interval again. Now these confidence intervals are actually biased. Their coverage is not what you would expect, especially at the boundaries of the binomial distribution. Yet, as an illustration they show you why it is logic that a binomial distribution has asymmetric confidence intervals.
An example in R:
logit <- function(x){ log(x/(1-x)) }
inv.logit <- function(x){ exp(x)/(1+exp(x)) }
x <- c(0.2, 0.5, 0.8)
lx <- logit(x)
upper <- lx + 2
lower <- lx - 2
logxtab <- cbind(lx, upper, lower)
logxtab # the confidence intervals are symmetric by construction
xtab <- inv.logit(logxtab)
xtab # back transformation gives asymmetric confidence intervals
note : In fact, R uses the beta distribution, but this is completely equivalent and computationally a bit more efficient. The implementation in R is thus different from what I show here, but it gives exactly the same result.
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I tried to find an even simpler product. Here's my solution:
$$ \zeta(2) =\prod _{n=1}^{\infty } \frac{1}{\left(1-\frac{1}{4 n^2}\right) \left(1-\frac{1}{36 n^2}\right)}$$
In Mathematica
Product[ 1/(1 - 1/(4 n^2)) 1/(1 - 1/(36 n^2)), {n, 1, \[Infinity]}]
(* Out[76]= \[Pi]^2/6 *)
We can derive this from the well-known product formula of the sine
Product[1 - x^2/n^2, {n, 1, \[Infinity]}]
(*
Out[85]= Sin[\[Pi] x]/(\[Pi] x)
*)
considering that
\[Pi] x/Sin[\[Pi] x] /. x -> 1/2
(*
Out[79]= \[Pi]/2
*)
and
\[Pi] x/Sin[\[Pi] x] /. x -> 1/6
(*
Out[80]= \[Pi]/3
*)
The product of these two expressions give [Pi]^2/6. Finally, it is easy to transform the corresponding infinite products into the form provided above.
Remark: Although Euler used the product formula for the sine in his famous proof that the infinite sum of the inverse squares is equal to [Pi]^2/6, normally an Euler product is a product over primes, such as the one defining the zeta function:
$$\zeta(s) =\prod _{n=1}^{\infty } \frac{1}{1-p_n^{-s}}$$
EDIT #1
It is not difficult to prove that for any positive integer m we can write$$\zeta (2 m)=\prod _{n=1}^{\infty } a (n)$$
Where the $a(n)$ are rational functions of n.
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Polynomial is Continuous Theorem Proof
From Linear Function is Continuous, setting $\alpha = 1$ and $\beta = 0$, we have that:
$\displaystyle \lim_{x \to c} \ x = c$
Repeated application of the Product Rule for Limits of Functions shows us that:
$\displaystyle \forall k \in \N: \lim_{x \to c} \ x^k = c^k$ Now let $P \left({x}\right) = a_n x^N + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$.
Now by repeated application of the Combined Sum Rule for Limits of Functions, we find that:
$\displaystyle \lim_{x \to c} \ P \left({x}\right) = P \left({c}\right)$ So whatever value we choose for $c$, we have that $P \left({x}\right)$ is continuous at $c$. From the definition of continuity on an interval, the second assertion follows.
$\blacksquare$
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Fitting a logistic regression (
LR) model (with Age, Sex and Pclass as predictors) to the survival outcome in the Titanic data yields a summary such as this one:
##
## Call:
## glm(formula = Survived ~ Age + Sex + Pclass, family = binomial(link = logit),
## data = NoMissingAge)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.7303 -0.6780 -0.3953 0.6485 2.4657
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 3.777013 0.401123 9.416 < 2e-16 ***
## Age -0.036985 0.007656 -4.831 1.36e-06 ***
## Sexmale -2.522781 0.207391 -12.164 < 2e-16 ***
## Pclass2 -1.309799 0.278066 -4.710 2.47e-06 ***
## Pclass3 -2.580625 0.281442 -9.169 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 964.52 on 713 degrees of freedom
## Residual deviance: 647.28 on 709 degrees of freedom
## AIC: 657.28
##
## Number of Fisher Scoring iterations: 5
## 1 2 3 4 5 7
## -0.4716467 0.4224490 1.0794775 0.4005642 -0.3747404 -0.8821915
## 1 2 3 4 5 7
## -0.4716467 0.4224490 1.0794775 0.4005642 -0.3747404 -0.8821915
For “normal regression” we know that the value of \(\beta_j\) simply gives us \(\Delta y\) if \(x_j\) is increased by one unit.
In order to fully understand the exact meaning of the coefficients for a LR model we need to first warm up to the definition of a
link function and the concept of probability odds.
Using linear regression as a starting point \[
y_i = \beta_0 + \beta_1 x_{1,i} + \beta_2 x_{2,i} + \ldots +\beta_k x_{k,i} + \epsilon_i \] we modify the right hand side such that (i) the model is still basically a linear combination of the \(x_j\)s but (ii) the output is -like a probability- bounded between 0 and 1. This is achieved by “wrapping” a sigmoid function \(s(z) = 1/(1+exp(z))\) around the weighted sum of the \(x_j\)s: \[ y_i = s(\beta_0 + \beta_1 x_{1,i} + \beta_2 x_{2,i} + \ldots +\beta_k x_{k,i} + \epsilon_i) \] The sigmoid function, depicted below to the left, transforms the real axis to the interval \((0;1)\) and can be interpreted as a probability.
The inverse of the sigmoid is the
logit (depicted above to the right), which is defined as \(log(p/(1-p))\). For the case where p is a probability we call the ratio \(p/(1-p)\) the probability odds. Thus, the logit is the log of the odds and logistic regression models these log-odds as a linear combination of the values of x.
Finally, we can interpret the coefficients directly: the odds of a positive outcome are multiplied by a factor of \(exp(\beta_j)\) for every unit change in \(x_j\). (In that light, logistic regression is reminiscient of linear regression with logarithmically transformed dependent variable which also leads to multiplicative rather than additive effects.)
As an example, the coefficient for Pclass 3 is -2.5806, which means that the odds of survival compared to the reference level Pclass 1 are reduced by a factor of \(exp(-2.5806) = 0.0757\);with all other input variables unchanged. How does the relative change in odds translate into probabilities? It is relatively easy to memorize the to and forth relationship between odds and probability \(p\): \[
odds = \frac{p}{1-p} \Leftrightarrow p = \frac{odds}{1 + odds} \] So the intercept 3.777 (= log-odds!) translates into odds of \(exp(3.777) = 43.685\) which yields a base probability of survival (for female Pclass1 of age 0) of \(43.685/(1 + 43.685) = 0.978\) Why make life so complicated?
I am often asked by students why we have to go through the constant trouble of (i) exponentiating the coefficients, (ii) multiplying the odds and finally (iii) compute the resulting probabilities? Can we not simply transform the coefficients from the summary table such that we can read off their effects directly – just like in linear regression?
The trouble is that there is no simple way to translate the coefficients into an additive or even a multiplicative adjustment of the probabilities. (One simple way to see this is the impossibility of keeping the output bounded between 0 and 1) The following graph shows the effect of various coefficient values on a base/reference probability.
We immediately see that there is no straightforward additive or multiplicative effect that could be quantified.
// add bootstrap table styles to pandoc tables $(document).ready(function () { $('tr.header').parent('thead').parent('table').addClass('table table-condensed'); });
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Hilbert 16th problem asks for a uniform upper bound $H(n)$ for the number of limit cycles of a polynomial vector field of degree $n$ on the plane. Here is an updated proof of the finitness part of the Hilbert 16th problem:
This problem is open even for $n=2$.
In this question, by a quadratic vector field we mean a polynomial vector field $$P(x,y)\partial_x+Q(x,y) \partial_y\;\;\;(V)$$ where $P,Q\in \mathbb{R}[x,y]$ are polynomials of degree 2 with $P(0,0)=Q(0,0)=0$.
So the question is that
(Hilbert 16th problem for n=2) Is $H(2)$ a finite number?
As a possible approach to this question we would like to look at limit cycles of a vector field as closed geodesics. Namely we would like to put a Riemannian metric on the (regular part of the ) phase space such that the trajectories of the vector field would be unparametrized geodesics. By regular part we mean the complement of singularities of the vector field. Thus each limit cycle would be a closed geodesic. Then using Gauss Bonnete theorem we try to count the number of closed geodesics. So the sign of the Gaussian curvature play a very important role. But the first problem is that, regardless of the curvature sign, is there a Riemannian metric on the regular part of the phase space such that the trajectories of the vector field would be geodesic.
This example shows that, for a polynomial vector field of arbitrary degree such metric does not necessarily existed. But for quadratic vector field the situation is different as follows. In the following $C$ is the algebraic curve $$yP(x,y)-xQ(x,y)=0\;\;\;\;\;(C)$$
Observation:There is a Riemannian metric on $\mathbb{R}^2 \setminus C$ such that all solutions of Quadratic vector field$(V)$ are geodesics. Furthermore every limit cycle of $(V) $ which surround the origin can not intersect the curve $C$
The proof of this observation is based on Proposition 6.7 and 6.8 as follows:
So we are sure that, on the complement of the above algebraic curve $C$, we have a Riemanian metric such that all solutions of $(V)$ are geodesics.
The methods of the proof of the Propositions 6.7 and 6.8 in the book Geometry of foliation suggest that we choose a Riemannian metric whose orthonormal base is the following:
$$\left\{\frac{x^2+y^2}{yP(x,y)-xQ(x,y)}V,\ \frac{1}{x^2+y^2}W\right\}.$$
where $V=P\partial_x +Q\partial_y$ is the quadratic vector field as in $(V)$ and $W$ is the radial vector field $W=x\partial_x+y\partial_y$. In fact $W$, as it is required in the proof of the above two propositions, lies in the kernel of $1\_$ form $\psi=\frac{1}{x^2+y^2}(ydx-xdy)$. This $1\_$ form is very essential to apply propositions 6.7 and 6.8 in the above reference. In fact this is a closed form which is identically equal to $1$ on the first vector of the above orthonormal frame.That is $\psi(\frac{x^2+y^2}{yP-xQ}(V))=1$.
I am really indebted to Ben McKay who suggested the $1\_$form $d\theta$ as a required $1\_$ form for possible satisfactions of proposition 6.8.
On the other hand the method of the proof of the Proposition 6.7 shows that, on the complement of $C$, all trajectories of $V$ are geodesics for the metric arising from the above orthonormal frame.
Moreover we are free to rescale the second vector of the frame, arbitrarily.
Now the main problem is that we control the sign of the curvature of such metric to make facilities to count the number of closed geodesics. In fact we use the Gauss Bonnete theorem.
Call the curvature of this metric $\kappa$.
Question: When a quadratic vector field $V$ does not have a center on the plane, is the curve $$\{(x,y)\mid \kappa(x,y)=0\}$$ transverse to $V$?
If not, what appropriate rescaling of the the second vector $ \frac{1}{x^2+y^2} W$ of the orthonormal frame would give a positive answer?
Notes:
1)A center is a singularity which is surrounded by a band of closed orbits. For quadratic vector fields they are classified at this paper.
2)
Of course existence of a positive answer to this question implies that $H(2)$ is finite. Because the number of limit cycles of $V$ which suround the origin can be uniformly bounded since $\kappa=0$ is an algebraic curve so in every connected component of this algebraic curve, we have at most on limit cycle surrounding origin.On the other hand there are at most 2 nest of limit cycles. So this would imply that $H(2)<\infty$
However the initial motivation is mentioned in page 3, item 5 of this arxiv note.
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Here's a neat one from optimization: the Alternating Direction Method of Multipliers (ADMM) algorithm.
Given an uncoupled and convex objective function of two variables (the variables themselves could be vectors) and a linear constraint coupling the two variables:
$$\min f_1(x_1) + f_2(x_2) $$$$ s.t. \; A_1 x_1 + A_2 x_2 = b $$
The Augmented Lagrangian function for this optimization problem would then be $$ L_{\rho}(x_1, x_2, \lambda) = f_1(x_1) + f_2(x_2) + \lambda^T (A_1 x_1 + A_2 x_2 - b) + \frac{\rho}{2}|| A_1 x_1 + A_2 x_2 - b ||_2^2 $$
The ADMM algorithm roughly works by performing a "Gauss-Seidel" splitting on the augmented Lagrangian function for this optimization problem by minimizing $L_{\rho}(x_1, x_2, \lambda)$ first with respect to $x_1$ (while $x_2, \lambda$ remain fixed), then by minimizing $L_{\rho}(x_1, x_2, \lambda)$ with respect to $x_2$ (while $x_1, \lambda$ remain fixed), then by updating $\lambda$. This cycle goes on until a stopping criterion is reached.
(Note: some researchers such as Eckstein discard the Gauss-Siedel splitting view in favor of proximal operators, for example see http://rutcor.rutgers.edu/pub/rrr/reports2012/32_2012.pdf )
For convex problems, this algorithm has been proven to converge - for two sets of variables. This is not the case for three variables. For example, the optimization problem
$$\min f_1(x_1) + f_2(x_2) + f_3(x_3)$$$$ s.t. \; A_1 x_1 + A_2 x_2 + A_3x_3 = b $$
Even if all the $f$ are convex, the ADMM-like approach (minimizing the Augmented Lagrangian with respect to each variable $x_i$, then updating the dual variable $\lambda$) is NOT guaranteed to converge, as was shown in this paper.
https://web.stanford.edu/~yyye/ADMM-final.pdf
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Description
We present an improved algorithm for solving symmetrically diagonally dominant linear systems. On input of an $n\times n$ symmetric diagonally dominant matrix $A$ with $m$ non-zero entries and a vector $b$ such that $A\bar{x} = b$ for some (unknown) vector $\bar{x}$, our algorithm computers a vector $x$ such that $||{x}-\bar{x}||_A < \epsilon ||\bar{x}||_A $ \footnote{$||\cdot||_A$ denotes the A-norm} in time $${\tilde O}(m\log n \log (1/\epsilon)). The solver utilizes in a standard way a `preconditioning' chain of progressively sparser graphs. To claim the faster running time we make a two-fold improvement in the algorithm for constructing the chain. The new chain exploits previously unknown properties of the graph sparsification algorithm given in [Koutis,Miller,Peng, FOCS 2010], allowing for stronger preconditioning properties. We also present an algorithm of independent interest that constructs nearly-tight low-stretch spanning trees in time $\tilde{O}(m\log{n})$, a factor of $O(\log{n})$ faster than the algorithm in [Abraham,Bartal,Neiman, FOCS 2008]. This speedup directly reflects on the construction time of the preconditioning chain.
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ADS
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Commercial flights are statistically quite safe (in terms of number of deaths per passenger-kilometer, only going to the moon is safer). But there are still reasons for precautions and safety regulations. An early such rule was the so-called “60-minute rule,” which required that a two-engine plane must always be within 60 minutes of the nearest adequate airport along its entire flight path. A variety of similar rules have existed, but at their core, they remain the same: the flight path can not take the airplane more than a certain maximum allowed distance from the nearest airport. With these restrictions, planes cannot always use a direct route for flying from one airport to another.
In this problem we will compute the shortest flight path between two airports while adhering to a maximum allowed distance rule. In the figure below, which illustrates the first sample test case, any flight route has to stay within the three circles. Thus a plane going from airport 2 to airport 3 has to detour from the direct route via the region around airport 1. Note that the plane would not necessarily have to go to airport 1 itself.
Things are further complicated by the fact that planes have limited fuel supply, and to go longer distances they may need to make a stopover at intermediate airports. Thus, depending on the fuel capacity, a plane going from airport 2 to airport 3 in the figure might have to stop over at airport 1 (or the fuel capacity might be too low even to go to airport 1, in which case the trip would be impossible to make).
We make the following simplifying assumptions:
The surface of the earth is a sphere of radius 6370 km.
Both time and fuel consumption are directly proportional to distance traveled. In other words we are interested only in total distance traveled.
The difference in distance caused by planes flying at different altitudes is negligible. Thus, effectively, we assume them to be flying along the earth’s surface.
A plane may stop for refueling at as many intermediate airports as needed, each time getting a full tank.
The first line of each test case contains two integers $N$ and $R$, where $2 \le N \le 25$ is the number of airports and $1 \le R \le 10\, 000$ is the maximum allowed flight distance (in km) from the nearest airport. Each of the next $N$ lines contains two integers $\phi $, $\theta $ satisfying $0 \le \phi < 360$ and $-90 \le \theta \le 90$, the longitude and latitude (respectively) of an airport, in degrees. The airports are numbered according to their order in the input starting from one. No two airports are at the same position.
Following this is a line containing an integer $Q$, satisfying $1 \le Q \le 100$. Each of the next $Q$ lines contains three integers $s$, $t$, $c$ satisfying $1 \le s, t \le N$, $s \ne t$, and $1 \le c \le 50\, 000$, indicating a plane going from airport $s$ to airport $t$ with a fuel capacity yielding a range of $c$ km.
For each test case, display the case number followed by one line for each query containing the length in km of the shortest flight path between airport $s$ and $t$, subject to the fuel constraint $c$. Display the length with exactly three decimal places. If there is no permissible path between the two airports, then display the word
impossible instead.
You may assume the answer is numerically stable for perturbations of up to $0.1$ km of $R$ or $c$.
Sample Input 1 Sample Output 1 3 2000 0 0 0 30 30 0 3 2 3 5000 2 3 4000 2 3 3000 2 10000 45 45 225 -45 2 1 2 50000 2 1 50000 Case 1: 4724.686 6670.648 impossible Case 2: impossible impossible
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High-jump High-jump cardinals are a certain kind of large cardinals. A cardinal $\kappa$ is high-jump if it is the critical point of an elementary embedding $j:V\to M$ such that $M$ is closed under sequences of length $\text{sup}\{j(f)(\kappa)|f:\kappa\to\kappa\}$. This closure condition is a weakening of the definition of a huge cardinal. Definition
Let $j:M\to N$ be a (nontrivial) elementary embedding. The
clearance of $j$ is the ordinal $\text{sup}\{j(f)(\kappa)$ $|$ $f:\kappa\to\kappa,f\in M\}$ where $\kappa$ is the critical point of $j$.
A cardinal $\kappa$ is
high-jump if there exists $j:V\to M$ with critical point $\kappa$ and clearance $\theta$ such that $M^\theta\subseteq M$, i.e. $M$ contains all sequences of elements of $M$ of length $\theta$. $j$ is called a high-jump embedding, and a normal fine ultrafilter on some $\mathcal{P}_\kappa(\lambda)$ generating an ultrapower embedding that is high-jump is a high-jump ultrafilter (or high-jump measure).
$\kappa$ is called
almost high-jump if $M$ is closed under sequences of length $<\theta$ instead, i.e. $M^\lambda\subseteq M$ for all $\lambda<\theta$. $j$ is then an almost high-jump embedding. This means that for all $f:\kappa\to\kappa$, $M^{j(f)(\kappa)}\subseteq M$. Shelah for supercompactness cardinals are a natural weakening of almost high-jump cardinals which allows to have one embedding per $f:\kappa\to\kappa$ rather than a single embedding for all such $f$s.
$\kappa$ is high-jump
order $\eta$ (resp. almost high-jump order $\eta$) if there exists a strictly increasing sequence of ordinals $\{\theta_\alpha:\alpha<\eta\}$ such that for all $\alpha<\eta$, there exists a high-jump embedding (resp. almost high-jump embedding) with critical point $\kappa$ and clearance $\theta_\alpha$. $\kappa$ is super high-jump (resp. super almost high-jump) if there are high-jump embeddings (resp. almost high-jump embeddings) with arbtirarily large clearance (i.e. it is "(almost) high-jump order Ord").
A high-jump cardinal $\kappa$ has
unbounded excess closure if for some clearance $\theta$, for all cardinals $\lambda\geq\theta$, there is a high-jump measure on $\mathcal{P}_\kappa(\lambda)$ generating an ultrapower embedding with clearance $\theta$.
The dual notion
high-jump-for-strongness, where the closure condition $M^\theta\subseteq M$ is weakened to $V_\theta\subseteq M$, turns out to be equivalent to superstrongness. Properties
Let $j:V\to M$ a nontrivial elementary embedding with critical point $\kappa$ and clearance $\theta$. Then there is no $f:\kappa\to\kappa$ such that $j(f)(\kappa)=\theta$. Also, $\kappa^{+}\leq cf(\theta)\leq 2^\kappa$ (see cofinality) and $\beth^M_\theta=\theta$. Moreover, $M_\theta\prec M_{j(\kappa)}$ and $M_\theta$ satisfies ZFC where $M_\theta=M\cap V_\theta$.
When $\kappa$ is almost high-jump, in both $V$ and $M$, $\theta^\kappa=\theta$, also $\theta$ is singular. Moreover, $V_\theta\prec M_{j(\kappa)}$ and $V_\theta$ satisfies ZFC.
The following statements also holds:
Suppose there is a almost high-jump cardinal. Then there are many cardinals below it that are Shelah for supercompactness. Also, in the model $V_\kappa$ there are many supercompact cardinals. Every high-jump cardinal is almost high-jump, and has order $\theta$; in fact, in the models $V_\theta$, $V_\kappa$ and $M_{j(\kappa)}$ there are many super almost high-jump cardinals. The existence of a high-jump cardinal with order $\eta$ implies that for every $\gamma<\eta$, there exists a model in which that cardinal is high-jump with order $\gamma$. The same statement holds for almost high-jump cardinals. The existence of a high-jump cardinal with unbounded excess closure is equiconsistent with the existence of a cardinal $\kappa$ such that for all sufficiently large $\lambda$, there exists a high-jump measure on $\mathcal{P}_\kappa(\lambda$). Suppose $\kappa$ is almost huge; then in the model $V_\kappa$ there are many cardinals that are high-jump with unbounded excess closure. Suppose that there exists a pair of cardinals ($\kappa$, $\theta$) such that there is a high-jump embedding $j:V\to M$ with critical point $\kappa$ and clearance $\theta$ and such that $M^{2^\theta}\subseteq M$. Then the cardinal $\kappa$ is super high-jump in the model $V_\theta$, and the cardinal $\kappa$ has high-jump order $\theta$ in $V$. Furthermore, there are many super high-jump cardinals in the models $V_\kappa$, $V_\theta$, and $M_{j(\kappa)}$. The least high-jump cardinal is not $\Sigma_2$-reflecting. In particular, it is not supercompact and not even strong. The same is true for the least almost huge cardinal, the least superstrong cardinal, the least almost-high-jump cardinal, and the least Shelah-for-supercompactness cardinal. Name origin Read More Norman Lewis Perlmutter, The large cardinals between supercompact and almost-huge[1] This article is a stub. Please help us to improve Cantor's Attic by adding information.
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I think the best way to think about this is in terms of operads.
Braided monoidal categories are representations of an operad $\Pi$ in the category of (small) categories.
The category $\Pi(n)$ has objects parenthesised permutations of $\{1,\ldots,n\}$ like $(4(23))1$. The morphisms $(\sigma) \to (\tau)$ are braids $\beta \in B_n$ so that $\beta$ maps to $\tau \sigma^{-1}$ under $B_n\to S_n$.
The operad structure $$ \Pi(n) \times \Pi(k_1) \times \ldots \times \Pi(k_n) \to \Pi(k_1+\ldots+k_n)$$is given by replacing the $i$-th sting by the braid on $k_i$-strings.
Now any representations $\Pi(n) \to \underline{Hom}(\mathcal{C}^n,\mathcal{C}) = End(\mathcal{C})(n)$ induces a braided monoïdal structure with
$\otimes$ corresponding to the object $(12) \in \Pi(2)$ the associativity morphism corresponding to the trivial braid $(12)3 \to 1(23)$ in $\Pi(3)$ the braiding corresponding to the morphism $(12) \to (21)$ in $\Pi(2)$ induced by the generator of $B_2 = \mathbb{Z}$.
MacLane's coherence theorem tells you that this is an equivalence.
All the operations you're looking at come from this operad structure.
The $\Pi(n)$ have a nice geometric interpretation. They are the fundamental groupoids of the operad of little discs $C_2(n)$ (or equivalently of $F(\mathbb{C},n)$ the spaces of configurations of $n$ points in the plane) restricted to a suitable collection of basepoints. This generalizes the classical definition $P_n = \pi_1(F(\mathbb{C},n),p)$, $B_n = \pi_1(F(\mathbb{C},n)/S_n,\overline{p})$.
I think this makes the whole picture a lot clearer because we get all of these operations as part of the same structure and we get a universal characterisation ofthat structure and a geometrical interpretation for it.
This leads to other nice considerations. For example the Grothendieck-Teichmuller group $GT$ defined by Drinfeld is the automorphism group of the (prounipotent completion) operad $\Pi$. This explains why $GT$ is universal for quasi triangular quasi Hopf algebras as their representations form braided monoïdal categories and related to the Galois group of $\mathbb{Q}$ as $GT$ appears as an automorphism group of fundamental groupoïds of the algebraic varieties $F(\mathbb{A}^1_{\mathbb{Q}},n)$.
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Reflecting cardinals
Reflection is a fundamental motivating concern in set theory. The theory of ZFC can be equivalently axiomatized over the very weak Kripke-Platek set theory by the addition of the reflection theorem scheme, below, since instances of the replacement axiom will follow from an instance of $\Delta_0$-separation after reflection down to a $V_\alpha$ containing the range of the defined function. Several philosophers have advanced philosophical justifications of large cardinals based on ideas arising from reflection.
Contents Reflection theorem
The Reflection theorem is one of the most important theorems in Set Theory, being the basis for several large cardinals. The Reflection theorem is in fact a "meta-theorem," a theorem about proving theorems. The Reflection theorem intuitively encapsulates the idea that we can find sets resembling the class $V$ of all sets.
Theorem (Reflection): For every set $M$ and formula $\phi(x_0...x_n,p)$ ($p$ is a parameter) there exists some limit ordinal $\alpha$ such that $V_\alpha\supseteq M$ such that $\phi^{V_\alpha}(x_0...x_n,p)\leftrightarrow \phi(x_0...x_n,p)$ (We say $V_\alpha$ reflects $\phi$). Assuming the Axiom of Choice, we can find some countable $M_0\supseteq M$ that reflects $\phi(x_0...x_n,p)$.
Note that by conjunction, for any finite family of formulas $\phi_0...\phi_n$, as $V_\alpha$ reflects $\phi_0...\phi_n$ if and only if $V_\alpha$ reflects $\phi_0\land...\land\phi_n$. Another important fact is that the truth predicate for $\Sigma_n$ formulas is $\Sigma_{n+1}$, and so we can find a (Club class of) ordinals $\alpha$ such that $(V_\alpha,\in)\prec_{{T_{\Sigma_n}}\restriction{V_\alpha}} (V,\in)$, where $T_{\Sigma_n}$ is the truth predicate for $\Sigma_n$ and so $ZFC\vdash Con(ZFC(\Sigma_n))$ for every $n$, where $ZFC(\Sigma_n)$ is $ZFC$ with Replacement and Separation restricted to $\Sigma_n$.
Lemma: If $W_\alpha$ is a cumulative hierarchy, there are arbitrarily large limit ordinals $\alpha$ such that $\phi^{W_\alpha}(x_0...x_n,p)\leftrightarrow \phi^W(x_0...x_n,p)$. Reflection and correctness
For any class $\Gamma$ of formulas, an inaccessible cardinal $\kappa$ is
$\Gamma$-reflecting if and only if $H_\kappa\prec_\Gamma V$, meaning that for any $\varphi\in\Gamma$ and $a\in H_\kappa$ we have $V\models\varphi[a]\iff H_\kappa\models\varphi[a]$. For example, an inaccessible cardinal is $\Sigma_n$-reflecting if and only if $H_\kappa\prec_{\Sigma_n} V$. In the case that $\kappa$ is not necessarily inaccessible, we say that $\kappa$ is $\Gamma$-correct if and only if $H_\kappa\prec_\Gamma V$ . A simple Löwenheim-Skolem argument shows that every uncountable cardinal $\kappa$ is $\Sigma_1$-correct. For each natural number $n$, the $\Sigma_n$-correct cardinals form a closed unbounded proper class of cardinals, as a consequence of the reflection theorem. This class is sometimes denoted by $C^{(n)}$ and the $\Sigma_n$-correct cardinals are also sometimes referred to as the $C^{(n)}$-cardinals. Every $\Sigma_2$-correct cardinal is a $\beth$-fixed point and a limit of such $\beth$-fixed points, as well as an $\aleph$-fixed point and a limit of such. Consequently, we may equivalently define for $n\geq 2$ that $\kappa$ is $\Sigma_n$-correct if and only if $V_\kappa\prec_{\Sigma_n} V$.
A cardinal $\kappa$ is
correct, written $V_\kappa\prec V$, if it is $\Sigma_n$-correct for each $n$. This is not expressible by a single assertion in the language of set theory (since if it were, the least such $\kappa$ would have to have a smaller one inside $V_\kappa$ by elementarity). Nevertheless, $V_\kappa\prec V$ is expressible as a scheme in the language of set theory with a parameter (or constant symbol) for $\kappa$.
Although it may be surprising, the existence of a correct cardinal is equiconsistent with ZFC. This can be seen by a simple compactness argument, using the fact that the theory ZFC+"$\kappa$ is correct" is finitely consistent, if ZFC is consistent, precisely by the observation about $\Sigma_n$-correct cardinals above.
$C^{(n)}$ are the classes of $\Sigma_n$-correct ordinals. These classes are clubs (closed unbounded). $C^{(0)}$ is the class of all ordinals. $C^{(1)}$ is precisely the class of all uncountable cardinals $α$ such that $V_\alpha=H(\alpha)$; i.e. precisely the Beth fixed points. References to the $C^{(n)}$ classes (different from just the requirement that the cardinal belongs to $C^{(n)}$) can sometimes make large cardinal properties stronger (for example $C^{(n)}$-superstrong, $C^{(n)}$-extendible, $C^{(n)}$-huge and $C^{(n)}$-I3 and $C^{(n)}$-I1 cardinals). On the other hand, every measurable cardinal is $C^{(n)}$-measurable for all $n$ and every ($λ$-)strong cardinal is ($λ$-)$C^{(n)}$-strong for all $n$.[1]
A cardinal $\kappa$ is
reflecting if it is inaccessible and correct. Just as with the notion of correctness, this is not first-order expressible as a single assertion in the language of set theory, but it is expressible as a scheme ( Lévy scheme). The existence of such a cardinal is equiconsistent to the assertion ORD is Mahlo.
If there is a pseudo uplifting cardinal, or indeed, merely a pseudo $0$-uplifting cardinal $\kappa$, then there is a transitive set model of ZFC with a reflecting cardinal and consequently also a transitive model of ZFC plus Ord is Mahlo. You can get this by taking some $\lambda\gt\kappa$ such that $V_\kappa\prec V_\lambda$.
$\Sigma_2$-correct cardinals
The $\Sigma_2$-correct cardinals are a particularly useful and robust class of cardinals, because of the following characterization: $\kappa$ is $\Sigma_2$-correct if and only if for any $x\in V_\kappa$ and any formula $\varphi$ of any complexity, whenever there is an ordinal $\alpha$ such that $V_\alpha\models\varphi[x]$, then there is $\alpha\lt\kappa$ with $V_\alpha\models\varphi[x]$. The reason this is equivalent to $\Sigma_2$-correctness is that assertions of the form $\exists \alpha\ V_\alpha\models\varphi(x)$ have complexity $\Sigma_2(x)$, and conversely all $\Sigma_2(x)$ assertions can be made in that form.
It follows, for example, that if $\kappa$ is $\Sigma_2$-correct, then any feature of $\kappa$ or any larger cardinal than $\kappa$ that can be verified in a large $V_\alpha$ will reflect below $\kappa$. So if $\kappa$ is $\Sigma_2$-reflecting, for example, then there must be unboundedly many inaccessible cardinals below $\kappa$. Similarly, if $\kappa$ is $\Sigma_2$-reflecting and measurable, then there must be unboundedly many measurable cardinals below $\kappa$.
Other facts:
Remarkable cardinals are $Σ_2$-reflecting.[2] It is relatively consistent that ZFC and the generic Vopěnka scheme holds, yet $Ord$ is not definably Mahlo and not even $∆_2$-Mahlo. In such a model, there can be no $Σ_2$-reflecting cardinals.[3] The Feferman theory
This is the theory, expressed in the language of set theory augmented with a new unary class predicate symbol $C$, asserting that $C$ is a closed unbounded class of cardinals, and every $\gamma\in C$ has $V_\gamma\prec V$. In other words, the theory consists of the following scheme of assertions: $$\forall\gamma\in C\ \forall x\in V_\gamma\ \bigl[\varphi(x)\iff\varphi^{V_\gamma}(x)\bigr]$$as $\varphi$ ranges over all formulas. Thus, the Feferman theory asserts that the universe $V$ is the union of a chain of elementary substructures $$V_{\gamma_0}\prec V_{\gamma_1}\prec\cdots\prec V_{\gamma_\alpha}\prec\cdots \prec V$$Although this may appear at first to be a rather strong theory, since it seems to imply at the very least that each $V_\gamma$ for $\gamma\in C$ is a model of ZFC, this conclusion would be incorrect. In fact, the theory does
not imply that any $V_\gamma$ is a model of ZFC, and does not prove $\text{Con}(\text{ZFC})$; rather, the theory implies for each axiom of ZFC separately that each $V_\gamma$ for $\gamma\in C$ satisfies it. Since the theory is a scheme, there is no way to prove from that theory that any particular $\gamma\in C$ has $V_\gamma$ satisfying more than finitely many axioms of ZFC. In particular, a simple compactness argument shows that the Feferman theory is consistent provided only that ZFC itself is consistent, since any finite subtheory of the Feferman theory is true by the reflection theorem in any model of ZFC. It follows that the Feferman theory is actually conservative over ZFC, and proves with ZFC no new facts about sets that is not already provable in ZFC alone.
The Feferman theory was proposed as a natural theory in which to undertake the category-theoretic uses of Grothendieck universes, but without the large cardinal penalty of a proper class of inaccessible cardinals. Indeed, the Feferman theory offers the advantage that the universes are each elementary substructures of one another, which is a feature not generally true under the universe axiom.
Maximality Principle
The existence of an inaccessible reflecting cardinal is equiconsistent with the boldface maximality principle $\text{MP}(\mathbb{R})$, which asserts of any statement $\varphi(r)$ with parameter $r\in\mathbb{R}$ that if $\varphi(r)$ is forceable in such a way that it remains true in all subsequent forcing extensions, then it is already true; in short, $\text{MP}(\mathbb{R})$ asserts that every possibly necessary statement with real parameters is already true. Hamkins showed that if $\kappa$ is an inaccessible reflecting cardinal, then there is a forcing extension with $\text{MP}(\mathbb{R})$, and conversely, whenever $\text{MP}(\mathbb{R})$ holds, then there is an inner model with an inaccessible reflecting cardinal.
$Σ_n(A)$-correct
(this section from [5])
Definitions:
An ordinal $γ$ is $Σ_n(A)$-correct, if $⟨V_γ, ∈, A ∩ V_γ⟩ ≺_{Σ_n} ⟨V, ∈, A⟩$. A cardinal $κ$ is $Σ_n(A)$-reflecting, if it is inaccessible and $Σ_n(A)$-correct.
Results:
If $κ$ is $A$-extendible for a class $A$, then $κ$ is $Σ_2(A)$-reflecting. If $κ$ is $A ⊕ C$-extendible, where $C$ is the class of all $Σ_1(A)$-correct ordinals, then $κ$ is $Σ_3(A)$-reflecting.
References Bagaria, Joan. $C^{(n)}$-cardinals.Archive for Mathematical Logic 51(3--4):213--240, 2012. www DOI bibtex Wilson, Trevor M. Weakly remarkable cardinals, Erdős cardinals, and the generic Vopěnka principle., 2018. arχiv bibtex Gitman, Victoria and Hamkins, Joel David. A model of the generic Vopěnka principle in which the ordinals are not Mahlo., 2018. arχiv bibtex Bagaria, Joan and Hamkins, Joel David and Tsaprounis, Konstantinos and Usuba, Toshimichi. Superstrong and other large cardinals are never Laver indestructible.Archive for Mathematical Logic 55(1-2):19--35, 2013. www arχiv DOI bibtex Hamkins, Joel David. The Vopěnka principle is inequivalent to but conservative over the Vopěnka scheme., 2016. www arχiv bibtex
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A
reaction engine is an engine or motor which provides propulsion (thrust) by expelling reaction mass, in accordance with Newton's third law of motion. This law of motion is most commonly paraphrased as: "For every action force there is an equal, but opposite, reaction force".
Examples include both jet engines and rocket engines, and more uncommon variations such as Hall effect thrusters, ion drives, mass drivers and nuclear pulse propulsion.
Contents Energy use 1 Propulsive efficiency 1.1 Cycle efficiency 1.2 Oberth effect 1.3 Types of reaction engines 2 See also 3 Notes and references 4 External links 5 Energy use Propulsive efficiency
For all reaction engines which carry their propellant onboard prior to use (such as rocket engines and electric propulsion drives) some energy must go into accelerating the reaction mass. Every engine will waste some energy, but even assuming 100% efficiency, the engine will need energy amounting to
\begin{matrix} \frac{1}{2} \end{matrix} MV_e^2
(where M is the mass of propellent expended and V_e is the exhaust velocity), which is simply the energy to accelerate the exhaust.
Due to energy carried away in the exhaust the energy efficiency of a reaction engine varies with the speed of the exhaust relative to the speed of the vehicle, this is called propulsive efficiency
, blue is the curve for rocket-like reaction engines, red is for air-breathing (duct) reaction engines
Comparing the rocket equation (which shows how much energy ends up in the final vehicle) and the above equation (which shows the total energy required) shows that even with 100% engine efficiency, certainly not all energy supplied ends up in the vehicle - some of it, indeed usually most of it, ends up as kinetic energy of the exhaust.
Interestingly, if the specific impulse (I_{sp}) is fixed, for a mission delta-v, there is a particular I_{sp} that minimises the overall energy used by the rocket. This comes to an exhaust velocity of about ⅔ of the mission delta-v (see the energy computed from the rocket equation). Drives with a specific impulse that is both high and fixed such as Ion thrusters have exhaust velocities that can be enormously higher than this ideal, and thus end up powersource limited and give very low thrust. Where the vehicle performance is power limited, e.g. if solar power or nuclear power is used, then in the case of a large v_{e} the maximum acceleration is inversely proportional to it. Hence the time to reach a required delta-v is proportional to v_{e}. Thus the latter should not be too large.
On the other hand, if the exhaust velocity can be made to vary so that at each instant it is equal and opposite to the vehicle velocity then the absolute minimum energy usage is achieved. When this is achieved, the exhaust stops in space
^ and has no kinetic energy; and the propulsive efficiency is 100% all the energy ends up in the vehicle (in principle such a drive would be 100% efficient, in practice there would be thermal losses from within the drive system and residual heat in the exhaust). However, in most cases this uses an impractical quantity of propellant, but is a useful theoretical consideration.
Some drives (such as VASIMR or electrodeless plasma thruster) actually can significantly vary their exhaust velocity. This can help reduce propellant usage and improve acceleration at different stages of the flight. However the best energetic performance and acceleration is still obtained when the exhaust velocity is close to the vehicle speed. Proposed ion and plasma drives usually have exhaust velocities enormously higher than that ideal (in the case of VASIMR the lowest quoted speed is around 15 km/s compared to a mission delta-v from high Earth orbit to Mars of about 4 km/s).
For a mission, for example, when launching from or landing on a planet, the effects of gravitational attraction and any atmospheric drag must be overcome by using fuel. It is typical to combine the effects of these and other effects into an effective mission delta-v. For example, a launch mission to low Earth orbit requires about 9.3–10 km/s delta-v. These mission delta-vs are typically numerically integrated on a computer.
Cycle efficiency
All reaction engines lose some energy, mostly as heat.
Different reaction engines have different efficiencies and losses. For example, rocket engines can be up to 60-70% energy efficient in terms of accelerating the propellant. The rest is lost as heat and thermal radiation, primarily in the exhaust.
Oberth effect
Reaction engines are more energy efficient when they emit their reaction mass when the vehicle is travelling at high speed.
This is because the useful mechanical energy generated is simply force times distance, and when a thrust force is generated while the vehicle moves, then:
E = F \times d \;
where F is the force and d is the distance moved.
Dividing by length of time of motion we get:
\frac E t = P = \frac {F \times d} t = F \times v
Hence:
P = F \times v \;
where P is the useful power and v is the speed.
Hence you want v to be as high as possible; and a stationary engine does no useful work.
[1] Types of reaction engines See also Notes and references ^ Note, that might seem to suggest that a stationary engine would not start to move. However, at low speeds the amount of energy needed to start to move tends to zero faster than the power does. So in practice it does move, as you would expect. External links
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The problem is to determine whether the given differential operator $L[y]$, whose domain consists of all functions that have continuous second derivatives on the interval $[0,\pi]$ and satisfy the given boundary conditions, is selfadjoint.
$$L[y]=y''+\lambda y;\;\;\;\;y(0)+y'(\pi)=0,\;\;\;\;y'(0)+y(\pi)=0$$
For an operator to be selfadjoint over an interval, it must satisfy the equation $(u,L[v])=(L[u],v)$ for all $u,v$ in the domain, where the inner product over the interval $[a,b]$ is defined as:
$$(f,g)=\int_a^b f(x)g(x)dx$$
Applying integration-by-parts twice to the integral for $(u,L[v])$ goes as follows:
$$ (u,L[v])= \int_0^\pi u(v''+\lambda v)dx = \int_0^\pi uv''\;dx + \int_0^\pi \lambda v\;dx = uv'|_0^\pi - \int_0^\pi u'v'\;dx + \int_0^\pi \lambda v\;dx $$
$$ = uv'|_0^\pi - u'v|_0^\pi + \int_0^\pi u''v\;dx + \int_0^\pi \lambda v\;dx = (uv'-u'v)|_0^\pi + (L[u],v) $$
Thus, the operator is selfadjoint iff $(uv'-u'v)|_0^\pi=0$. The boundary conditions can be used to show that $uv'|_0^\pi=-u'v|_0^\pi$, leading to the following condition for selfadjointness:
$$uv'|_0^\pi=0$$
This seems promising, but I'm not sure where to go from here. I can't even think of functions that satisfy the boundary conditions, which makes finding a counter-example difficult.
Edit:
It looks like the following is a counter-example:
$$u(x)=2\cos{2x}-\sin{2x},\;\;\;\;v(x)=\cos{x}+\sin{x}$$ $$ uv'|_0^\pi = (2\cos{2x}-\sin{2x})(\cos{x}-\sin{x})|_0^\pi = -4$$
Unfortunately, I don't find this particularly enlightening. Is there any way to show that the operator is not selfadjoint without a counter-example?
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I've just begun learning capacitance, and my lecture notes have a section on calculating capacitance for capacitors in vacuum of various shapes, e.g. two parallel plates and concentric spherical shells.
For a cylindrical capacitor, comprising of a
long cylindrical conductor with radius $r_a$ and linear charge density $+\lambda$, and a coaxial cylindrical conducting shell with radius $r_b$ and linear charge density $-\lambda$, to calculate its capacitance per unit length, the notes lay out the steps as follows: A point outside a long line of charge a distance r from axis has potential $V = \frac{\lambda}{2\pi\epsilon_0}ln\frac{r_0}{r}(1)$. Holds here also because charge on outer cylinder doesn't contribute to field between cylinders. Then $V_{ab} = V_a - V_b = \frac{\lambda}{2\pi\epsilon_0}ln\frac{r_b}{r_a}$ Total charge $Q = \lambda L$ in a length $L$, so $C = \frac{Q}{V_{ab}} = \frac{2\pi\epsilon_0L}{ln(r_b/r_a)}$ Therefore $\frac{C}{L} = \frac{2\pi\epsilon_0}{ln(r_b/r_a)}$
There are a couple of things in here which I don't understand:
The first bullet mentions "charge on outer cylinder doesn't contribute to field between cylinders". Is this true? Shouldn't the electric field strength double since there is both a positive charge $+\lambda$ and a negative charge $-\lambda$? The second bullet assigns $r_b$ as $r_0$ and $r_a$ as $r$ in equation $(1)$. Why is this so? As far as I understand, $r_0$ denotes the arbitrary distance where $V_b = 0$, which suggests that the potential on the coaxial shell is 0. Why is this not, say, the other way around? Why don't we take the potential at the edge of the inner cylinder as 0? Would that cause $r_b$ and $r_a$ to switch places in the fraction? Why is it possible to assign the potential on the coaxial shell as 0 in the first place?
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Need help! I was working on a project when I required to use a projection operator. For an example case, I have the Bell state, $$|\psi\rangle = \frac1{\sqrt2}\left(\color{blue}{|0}0\rangle+|11\rangle\right)$$ which now I want to take to the state, $$|\psi'\rangle = |\color{blue}{0}0\rangle$$ by weeding out the states with the leftmost qubit as $1$.
Another example would be, $$\frac1{2}\left(|\color{blue}{0}0\rangle+|\color{blue}{0}1\rangle+|10\rangle+|11\rangle\right)\rightarrow\frac1{\sqrt2}\left(|00\rangle+|01\rangle\right).$$
Edit: Just putting another example to make my question clear. I want to weed out states when the third (leftmost) qubit is $1$ in the following example.
Suppose we have a three qubit state, $$|\psi\rangle=\displaystyle\frac1{N}\left(|\color{blue}{0}\rangle\otimes\left[|00\rangle+|01\rangle+|10\rangle\right] + |1\rangle\otimes|11\rangle\right)$$ it should then get transformed to $$|\psi'\rangle = \frac1{N'}|\color{blue}{0}\rangle\otimes\left[|00\rangle+|01\rangle+|10\rangle\right].$$
Is this possible? If yes, how can I implement it in the IBM Quantum Experience?
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@DavidReed the notion of a "general polynomial" is a bit strange. The general polynomial over a field always has Galois group $S_n$ even if there is not polynomial over the field with Galois group $S_n$
Hey guys. Quick question. What would you call it when the period/amplitude of a cosine/sine function is given by another function? E.g. y=x^2*sin(e^x). I refer to them as variable amplitude and period but upon google search I don't see the correct sort of equation when I enter "variable period cosine"
@LucasHenrique I hate them, i tend to find algebraic proofs are more elegant than ones from analysis. They are tedious. Analysis is the art of showing you can make things as small as you please. The last two characters of every proof are $< \epsilon$
I enjoyed developing the lebesgue integral though. I thought that was cool
But since every singleton except 0 is open, and the union of open sets is open, it follows all intervals of the form $(a,b)$, $(0,c)$, $(d,0)$ are also open. thus we can use these 3 class of intervals as a base which then intersect to give the nonzero singletons?
uh wait a sec...
... I need arbitrary intersection to produce singletons from open intervals...
hmm... 0 does not even have a nbhd, since any set containing 0 is closed
I have no idea how to deal with points having empty nbhd
o wait a sec...
the open set of any topology must contain the whole set itself
so I guess the nbhd of 0 is $\Bbb{R}$
Btw, looking at this picture, I think the alternate name for these class of topologies called British rail topology is quite fitting (with the help of this WfSE to interpret of course mathematica.stackexchange.com/questions/3410/…)
Since as Leaky have noticed, every point is closest to 0 other than itself, therefore to get from A to B, go to 0. The null line is then like a railway line which connects all the points together in the shortest time
So going from a to b directly is no more efficient than go from a to 0 and then 0 to b
hmm...
$d(A \to B \to C) = d(A,B)+d(B,C) = |a|+|b|+|b|+|c|$
$d(A \to 0 \to C) = d(A,0)+d(0,C)=|a|+|c|$
so the distance of travel depends on where the starting point is. If the starting point is 0, then distance only increases linearly for every unit increase in the value of the destination
But if the starting point is nonzero, then the distance increases quadratically
Combining with the animation in the WfSE, it means that in such a space, if one attempt to travel directly to the destination, then say the travelling speed is 3 ms-1, then for every meter forward, the actual distance covered by 3 ms-1 decreases (as illustrated by the shrinking open ball of fixed radius)
only when travelling via the origin, will such qudratic penalty in travelling distance be not apply
More interesting things can be said about slight generalisations of this metric:
Hi, looking a graph isomorphism problem from perspective of eigenspaces of adjacency matrix, it gets geometrical interpretation: question if two sets of points differ only by rotation - e.g. 16 points in 6D, forming a very regular polyhedron ...
To test if two sets of points differ by rotation, I thought to describe them as intersection of ellipsoids, e.g. {x: x^T P x = 1} for P = P_0 + a P_1 ... then generalization of characteristic polynomial would allow to test if our sets differ by rotation ...
1D interpolation: finding a polynomial satisfying $\forall_i\ p(x_i)=y_i$ can be written as a system of linear equations, having well known Vandermonde determinant: $\det=\prod_{i<j} (x_i-x_j)$. Hence, the interpolation problem is well defined as long as the system of equations is determined ($\d...
Any alg geom guys on? I know zilch about alg geom to even start analysing this question
Manwhile I am going to analyse the SR metric later using open balls after the chat proceed a bit
To add to gj255's comment: The Minkowski metric is not a metric in the sense of metric spaces but in the sense of a metric of Semi-Riemannian manifolds. In particular, it can't induce a topology. Instead, the topology on Minkowski space as a manifold must be defined before one introduces the Minkowski metric on said space. — baluApr 13 at 18:24
grr, thought I can get some more intuition in SR by using open balls
tbf there’s actually a third equivalent statement which the author does make an argument about, but they say nothing about substantive about the first two.
The first two statements go like this : Let $a,b,c\in [0,\pi].$ Then the matrix $\begin{pmatrix} 1&\cos a&\cos b \\ \cos a & 1 & \cos c \\ \cos b & \cos c & 1\end{pmatrix}$ is positive semidefinite iff there are three unit vectors with pairwise angles $a,b,c$.
And all it has in the proof is the assertion that the above is clearly true.
I've a mesh specified as an half edge data structure, more specifically I've augmented the data structure in such a way that each vertex also stores a vector tangent to the surface. Essentially this set of vectors for each vertex approximates a vector field, I was wondering if there's some well k...
Consider $a,b$ both irrational and the interval $[a,b]$
Assuming axiom of choice and CH, I can define a $\aleph_1$ enumeration of the irrationals by label them with ordinals from 0 all the way to $\omega_1$
It would seemed we could have a cover $\bigcup_{\alpha < \omega_1} (r_{\alpha},r_{\alpha+1})$. However the rationals are countable, thus we cannot have uncountably many disjoint open intervals, which means this union is not disjoint
This means, we can only have countably many disjoint open intervals such that some irrationals were not in the union, but uncountably many of them will
If I consider an open cover of the rationals in [0,1], the sum of whose length is less than $\epsilon$, and then I now consider [0,1] with every set in that cover excluded, I now have a set with no rationals, and no intervals.One way for an irrational number $\alpha$ to be in this new set is b...
Suppose you take an open interval I of length 1, divide it into countable sub-intervals (I/2, I/4, etc.), and cover each rational with one of the sub-intervals.Since all the rationals are covered, then it seems that sub-intervals (if they don't overlap) are separated by at most a single irrat...
(For ease of construction of enumerations, WLOG, the interval [-1,1] will be used in the proofs) Let $\lambda^*$ be the Lebesgue outer measure We previously proved that $\lambda^*(\{x\})=0$ where $x \in [-1,1]$ by covering it with the open cover $(-a,a)$ for some $a \in [0,1]$ and then noting there are nested open intervals with infimum tends to zero.
We also knew that by using the union $[a,b] = \{a\} \cup (a,b) \cup \{b\}$ for some $a,b \in [-1,1]$ and countable subadditivity, we can prove $\lambda^*([a,b]) = b-a$. Alternately, by using the theorem that $[a,b]$ is compact, we can construct a finite cover consists of overlapping open intervals, then subtract away the overlapping open intervals to avoid double counting, or we can take the interval $(a,b)$ where $a<-1<1<b$ as an open cover and then consider the infimum of this interval such that $[-1,1]$ is still covered. Regardless of which route you take, the result is a finite sum whi…
W also knew that one way to compute $\lambda^*(\Bbb{Q}\cap [-1,1])$ is to take the union of all singletons that are rationals. Since there are only countably many of them, by countable subadditivity this give us $\lambda^*(\Bbb{Q}\cap [-1,1]) = 0$. We also knew that one way to compute $\lambda^*(\Bbb{I}\cap [-1,1])$ is to use $\lambda^*(\Bbb{Q}\cap [-1,1])+\lambda^*(\Bbb{I}\cap [-1,1]) = \lambda^*([-1,1])$ and thus deducing $\lambda^*(\Bbb{I}\cap [-1,1]) = 2$
However, what I am interested here is to compute $\lambda^*(\Bbb{Q}\cap [-1,1])$ and $\lambda^*(\Bbb{I}\cap [-1,1])$ directly using open covers of these two sets. This then becomes the focus of the investigation to be written out below:
We first attempt to construct an open cover $C$ for $\Bbb{I}\cap [-1,1]$ in stages:
First denote an enumeration of the rationals as follows:
$\frac{1}{2},-\frac{1}{2},\frac{1}{3},-\frac{1}{3},\frac{2}{3},-\frac{2}{3}, \frac{1}{4},-\frac{1}{4},\frac{3}{4},-\frac{3}{4},\frac{1}{5},-\frac{1}{5}, \frac{2}{5},-\frac{2}{5},\frac{3}{5},-\frac{3}{5},\frac{4}{5},-\frac{4}{5},...$ or in short:
Actually wait, since as the sequence grows, any rationals of the form $\frac{p}{q}$ where $|p-q| > 1$ will be somewhere in between two consecutive terms of the sequence $\{\frac{n+1}{n+2}-\frac{n}{n+1}\}$ and the latter does tends to zero as $n \to \aleph_0$, it follows all intervals will have an infimum of zero
However, any intervals must contain uncountably many irrationals, so (somehow) the infimum of the union of them all are nonzero. Need to figure out how this works...
Let's say that for $N$ clients, Lotta will take $d_N$ days to retire.
For $N+1$ clients, clearly Lotta will have to make sure all the first $N$ clients don't feel mistreated. Therefore, she'll take the $d_N$ days to make sure they are not mistreated. Then she visits client $N+1$. Obviously the client won't feel mistreated anymore. But all the first $N$ clients are mistreated and, therefore, she'll start her algorithm once again and take (by suposition) $d_N$ days to make sure all of them are not mistreated. And therefore we have the recurence $d_{N+1} = 2d_N + 1$
Where $d_1$ = 1.
Yet we have $1 \to 2 \to 1$, that has $3 = d_2 \neq 2^2$ steps.
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A brief description of the 18 electron rule
A valence shell of a transition metal contains the following: 1 $s$ orbital, 3 $p$ orbitals and 5 $d$ orbitals; 9 orbitals that can collectively accommodate 18 electrons (as either bonding or nonbonding electron pairs). This means that, the combination of these nine atomic orbitals with ligand orbitals creates nine molecular orbitals that are either metal-ligand bonding or non-bonding, and when metal complex has 18 valence electrons, it is said to have achieved the same electron configuration as the noble gas in the period.
In some respect, it is similar to the octet rule for main group elements, something you might be more familiar with, and thus it may be useful to bear that in mind. So in a sense, there's not much more to it than "electron bookkeeping"
As already mentioned in the comments, 18 electron rule is more useful in the context of organometallics.
Two methods are commonly employed for electron counting:
Neutral atom method: Metal is taken as in zero oxidation state for counting purpose
Oxidation state method: We first arrive at the oxidation state of the metal by considering the number of anionic ligands present and overall charge of the complex
I think this website does a good job of explaining this: http://www.ilpi.com/organomet/electroncount.html (plus, they have some practice exercises towards the end)
Let's just focus on Neutral Atom Method (quote from the link above)
The major premise of this method is that we remove all of the ligands from the metal, but rather than take them to a closed shell state, we do whatever is necessary to make them neutral. Let's consider ammonia once again. When we remove it from the metal, it is a neutral molecule with one lone pair of electrons. Therefore, as with the ionic model, ammonia is a neutral two electron donor.
But we diverge from the ionic model when we consider a ligand such as methyl. When we remove it from the metal and make the methyl fragment neutral, we have a neutral methyl radical. Both the metal and the methyl radical must donate one electron each to form our metal-ligand bond. Therefore, the methyl group is a one electron donor, not a two electron donor as it is under the ionic formalism. Where did the other electron "go"? It remains on the metal and is counted there. In the covalent method, metals retain their full complement of d electrons because we never change the oxidation state from zero; i.e. Fe will always count for 8 electrons regardless of the oxidation state and Ti will always count for four.
Ligand Electron Contribution (for neutral atom method)
a. Neutral Terminal (eg. $\ce{CO}, \ce{PR_3}, \ce{NR_3}$) : 2 electrons
b. Anionic Terminal (eg. $\ce{X^-}, \ce{R_2P^-}, \ce{Ro^-}$) : 1 electron
c. Hapto Ligands (eg. $ \eta^2-\ce{C_2R_4}, \eta^1-\text{allyl}$): Same as hapticity
d. Bridging neutral (eg. $\mu_2-\ce{CO}$) : 2 electrons
e. Bridging anionic (eg. $\mu_2-\ce{CH_3}$) ( no lone pairs): 1 electron
f.Bridging anionic (eg. $\mu_2-\ce{Cl}, \mu_2-\ce{OR}$) (with 1 lone pair)): 3 electrons
(with 1 lone pair)
or, $\mu_2-\ce{Cl}$(with 2 lone pairs): 5 electrons
g. Bridging alkyne 4 electrons
h. NO linear 3 electrons
i. NO bent ( lone pair on nitrogen): 1 electron
j. Carbene (M=C): 2 electron
k.Carbyne (M≡C): 3 electron
Determining # Metal-Metal bonds
Step 1: Determine the total valence electrons (TVE) in the entire molecule (that is, the number of valence electrons of the metal plus the number of electrons from each ligand and the charge)-- I'll call this T (T for total, I'm making this up )
Subtract this number from $n × 18$ where $n$ is the number of metals in the complex, i.e $(n × 18) – T$ -- call this R (R for result, nothing fancy)
(a) R divided by 2 gives the total number of M–M bonds in the complex.(b) T divided by n gives the number of electrons per metal.
If the number of electrons is 18, it indicates that there is no M–M bond; if it is 17 electrons, it indicates that there is 1 M–M bond; if it is 16electrons, it indicates that there are 2 M–M bonds and so on.
At this point, let's apply this method to a few examples
a) Tungsten Hexacarbonyl (picture)
Let's use the neutral atom method, W has 6 electrons, the carbonyls donate 12 electrons and we get a total of 18. Of course there can be no metal metal bonds here.
(b) Tetracobalt dodecacarbonyl (picture), here let's figure out the no. of metal-metal bonds. T is 16, R is 12, Total # M-M bonds is 6, # electrons per metal is 15, so 3 M-M bonds.
A few examples where the "18 electron Rule" works
I.
Octahedral Complexes with strong $\pi$ - acceptor ligands
eg. $\ce{[Cr(CO)_6]}$
Here, $t_{2g}$ is strongly bonding and is filled and, $e_g$ is strongly antibonding, and empty. Complexes of this kind tend to obey the 18-electron rule irrespective of their coordination number. Exceptions exist for $d^8$, $d^{10}$ systems (see below)
II. Tetrahedral Complexes
e.g. $\ce{[Ni(PPh_3)_4]}$ ($\ce{Ni^0}$, $d^{10}$ 18-electron complex
Tetrahedral complexes cannot exceed 18 electrons because there are no lowlying MOs that can be filled to obtain tetrahedral complexes with more than 18 electrons. In addition, a transition metal complex with the maximum of 10 d-electrons, will receive 8 electrons from the ligands and end up with a total of 18 electrons.
Violations of 18 electron rule
I. Bulky ligands : (eg. $\ce{Ti(\text{neopentyl})_4} $ has 8 electrons) Bulky ligands prevent a full complement of ligands to assemble around the metal to satisfy the 18 electron rule.
Additionally, for early transition metals, (e.g in $d^0$ systems), it is often not possible to fit the number of ligands necessary to reach 18 electrons around the metal. (eg. tungsten hexamethyl, see below)
II. Square Planar $d^8$ complexes (16 electrons) and Linear $d^{10}$ complexes (14 electrons)
For square planar complexes, $d^8$ metals with 4 ligands, gives 16-electron complexes. This is commonly seen with metals and ligands high in the spectrochemical series
For instance, $\ce{Rh^+}, \ce{Ir^+}, \ce{Pd^2+}, \ce{Pt^2+}$) are square planar. Similarly, $\ce{Ni^2+}$ can be square planar, with strong $\pi$-acceptor ligands.
Similarly, $d^{10}$ metals with 2 ligands, give 14-electron complexes. Commonly seen for for $\ce{Ag^+}, \ce{Au^+}, \ce{Hg^{2+}}$
III. Octahedral Complexes which disobey the 18 electron rule, but still have fewer than 18 electrons (12 to 18)
This is seen with second and third row transition metal complexes, high in the spectrochemical series of metal ions with $\sigma$-donor or $\pi$-donor ligands (low to medium in the spectrochemical series).$t_{2g}$ is non-bonding or weakly anti-bonding (because the ligands are either $\sigma$-donor or $\pi$-donor), and $t_{2g}$ usually contains 0 to 6 electrons. On the other hand, $e_g$ are strongly antibonding, and thus are empty.
IV. Octahedral Complexes which exceed 18 electrons (12 to 22)
This is observed in first row transition metal complexes that are low in the spectrochemical series of metal ions, with $\sigma$-donor or $\pi$-donor ligands Here, the $t_{2g}$ is non-bonding or weakly anti-bonding, but the $e_g$ are only weakly antibonding, and thus can contain electrons. Thus, 18 electrons maybe exceeded.
References:
The following weblinks proved useful to me while I was writing this post, (especially handy for things like MO diagrams)
http://www.chem.tamu.edu/rgroup/marcetta/chem462/lectures/Lecture%203%20%20excerpts%20from%20Coord.%20Chem.%20lecture.pdf
http://classes.uleth.ca/201103/chem4000b/18%20electron%20rule%20overheads.pdf
http://web.iitd.ac.in/~sdeep/Elias_Inorg_lec_5.pdf
http://www.ilpi.com/organomet/electroncount.html
http://www.yorku.ca/stynes/Tolman.pdf
and obviously, the wikipedia page is a helpful guide https://en.wikipedia.org/wiki/18-electron_rule
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This is not true.
For example, consider $T\mathbb{CP}^2$ which has total Chern class $c(T\mathbb{CP}^2) = 1 + 3x + 3x^2$ where $x = c_1(\mathcal{O}(1)) \in H^2(\mathbb{CP}^2; \mathbb{Z})$ is a generator. Suppose $c(T\mathbb{CP}^2) = (1 + x_1)(1 + x_2)$ for some $x_1, x_2 \in H^2(\mathbb{CP}^2; \mathbb{Z})$, then $x_1 = kx$ and $x_2 = lx$ for some integers $k$ and $l$. Then $$1 + 3x + 3x^2 = (1 + kx)(1 + lx) = 1 + (k + l)x + klx^2.$$ This is impossible: if $k + l = 3$, one of $k$ and $l$ is even, but then $kl$ would be even, so it can't be equal to $3$.
This shows that $T\mathbb{CP}^2$ is not isomorphic to the direct sum of line bundles. Note however that if the total Chern class did factor, it does not necessarily mean the bundle is isomorphic to a direct sum; see this MathOverflow question.
What is true however is the splitting principle:
Let $\xi$ be a rank $n$ vector bundle over $X$. There exists a space $Y$ (the total space of the flag bundle of $\xi$) and a map $p : Y \to X$ such that:
the graded ring homomorphism $p^* : H^*(X, \mathbb{Z}) \to H^*(Y, \mathbb{Z})$ is injective, and
$p^*\xi = L_1\oplus\dots\oplus L_n$ where $L_1, \dots, L_n$ are complex line bundles on $Y$.
Note that
$$p^*c(\xi) = c(p^*\xi) = c(L_1\oplus\dots\oplus L_n) = c(L_1)\dots c(L_n) = (1 + x_1)\dots(1+x_n)$$
where $x_i = c_1(L_i)$.
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Suppose you have the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Suppose somewhere on the interior you have point $P(m,n)$. What is the closest point on the ellipse to the $P$? Can this be found without having to solve a 4th degree polynomial?
Because of symmetry, we can assume that $P$ is in the First Quadrant or on its boundary without loss of generality.
Lagrange multipliers come to mind. Perpendicularity. Figure most efficient approach is parameterization.
Let $x=a\cos t$, $y=b\sin t$. Let $d^2=(x-m)^2+(y-n)^2$.
Take the derivative of $d^2$ and set it to zero.
$$0=2(a\cos t-m)(-a\sin t)+2(b\sin t-n)(b\cos t)$$ $$0=(x-m)(\frac{-a}{b}y)+(y-n)(\frac{b}{a}x)$$
This represents a hyperbola passing through the origin and containing $P(m,n)$.
Expanding, then rearranging:
$$0=-2a^2\sin t\cos t+2ma\sin t+2b^2\sin t\cos t-2nb\cos t$$ $$nb\cos t=(b^2-a^2)\sin t\cos t+ma\sin t$$
Squaring both sides: $$n^2b^2\cos^2t=(b^2-a^2)^2\sin^2 t\cos^2 t+m^2a^2\sin^2 t+2ma(b^2-a^2)\sin^2 t\cos t$$
That $\cos t$ appears to the first power and $\sin t$ only to the second, this suggests swapping out $\sin^2 t$ with $(1-\cos^2 t):
$$n^2b^2\cos^2t$$ $$=(b^2-a^2)^2\cos^2t(1-\cos^2 t)+m^2a^2(1-\cos^2t)-2ma(b^2-a^2)(1-\cos^2t)\cos t$$ or,
$$n^2b^2\cos^2t=$$ $$(b^2-a^2)^2\cos^2t -(b^2-a^2)^2\cos^4t+m^2a^2-m^2a^2\cos^2t +2ma(b^2-a^2)\cos t -2ma(b^2-a^2)\cos^3t$$
Isolating $\cos t$ to one side: $$(b^2-a^2)^2\cos^4t+2ma(b^2-a^2)\cos^3t+[n^2b^2-(b^2-a^2)^2+m^2a^2]\cos^2 t-2ma(b^2-a^2)\cos t-m^2a^2=0$$
We now have a 4th order polynomial in $\cos t$. Numerical methods are available, but can it be reduced further?
I think I've made some progress. The x-y form of the equation to be solved represents a hyperbola that passes through the origin as P(m,n). So there must be at least 2 points of intersection of the hyperbola with the ellipse.
Swapping in $(z+1/z)/2=\cos t$ and $(z-1/z)/(2i)=\sin t$ and introducing $z_0$ to simply the coefficients a complex equation comes up: $$z^4+2z_0z^3-2\bar{z_0}z-1=0$$
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Say we have some equations $f_1(x)=0, \ldots f_k(x)=0$ defining a variety $X$ in ${\mathbb C}^n$ (not necessarily a minimal number of generators, and not necessarily of minimal degree), and suppose we want to know the ideal $\bar{I}$ of its closure $\overline{X}$ in ${\mathbb P}^n$. A naive question:
If all the original generators $f_i$ were of degree at most $d$, can we generate $\bar{I}$ using polynomials of degree at most $d$? If not, what degree bound $d'$ can we give for a minimal generating set of $\bar{I}$? Motivation, in short: Ideal saturation is an extremely slow operation, but solving a linear system to identify vanishing forms up to a certain degree is polynomial time, which would give a more practical way to compute $\bar{I}$ in high dimensions.
(Sorry, I'm new to computational algebra, and maybe people already know a fast way to compute projective closure...)
Motivation, in long: To compute $\bar{I}$, it's not enough to simply homogenize the generators with respect to a new variable $z$. Denote such homogenizations by $\hat{f}$, their ideal by $\hat{I}$, and their variety by $\hat{X}$. The problem is that $\hat{X}$ might contain irreducible components on the hyperplane at infinity $(z=0)$, so we need to saturate it, $\bar{I}=(\hat{I}:z^\infty)$, to eliminate these components and get the ideal of $\bar{X}$. But saturation is extremely slow! (I think it's at least doubly exponential time in the number of variables.) So I'd like to find $\bar{I}$ instead by solving a linear system on the vector space of homogeneous polynomials of degree at most $d$ to see which ones vanish, and declare that they cut out the variety $\bar{X}$, yay! But do they? If searching up to degree $d$ is not enough, how far do I have to go? A simple example: Say $f_1 = x$ and $f_2 = y-x^2$ in ${\mathbb C}^2$, so $X$ is just the origin $(0,0)$. Suppose we first homogenize the generators $f_i$. The parabola $\hat{f}_2=yz-x^2=0$ is tangent to the line at infinity at $(x:y:z)=(0:1:0)$, which also lies on $\hat{f}_1=0$. So together they cut out $\hat{X} = \{ (0:0:1) \cup (0:1:0)\}$, and we don't get $\bar{X}=(0:0:1)$ until we saturate $\hat{I}=\langle x,yz-x^2\rangle$ by $z$ to eliminate the component at infinity.
In this example, we already know the answer is the origin, and saturation is very fast for such a small system anyway. But for many variables, we don't already know what the variety looks like, and the saturation might not finish in a human lifetime, so it would be nice to solve a linear system instead.
Thanks for any help on this question!
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First, to clarify, the first Lagrangian you gave is the Lagrangian describing a non-relativistic point particle moving in a
fixed background potential $(\phi,\textbf{A})$. The second Lagrangian describes the dynamics of the electromagnetic field in the presence of a fixed background source $(\rho,\textbf{j})$.
Now, the short answer to why the minus sign is there is that, without it, the equations obtained aren't Maxwell's equations.
A slightly longer answer is that the minus sign exists to ensure relativistic invariance. Under a Lorentz transformation, the electromagnetic fields transform as
$$\textbf{E}_{\perp}\to\gamma\left(\textbf{E}_{\perp}+\textbf{v}\times\textbf{B}\right),\hspace{0.5cm}\textbf{B}_{\perp}\to\gamma\left(\textbf{B}_{\perp}-\frac{1}{c^2}\textbf{v}\times\textbf{E}\right),$$
where $\perp$ denotes the component of the field perpendicular to boost velocity $\textbf{v}$. Under this transformation, we have
$$\frac{\epsilon_0}{2}\textbf{E}^2-\frac{1}{2\mu_0}\textbf{B}^2=\left(\frac{\epsilon_0}{2}E_{\parallel}^2-\frac{1}{2\mu_0}B_{\parallel}^2\right)+\frac{\gamma^2\epsilon_0}{2}\left(\textbf{E}_{\perp}+\textbf{v}\times\textbf{B}\right)^2-\frac{\gamma}{2\mu_0}\left(\textbf{B}_{\perp}-\frac{1}{c^2}\textbf{v}\times\textbf{E}\right)^2.$$
Being careful with our cross product identities, we can show
$$(\textbf{E}_{\perp}+\textbf{v}\times\textbf{B})^2=\textbf{E}_{\perp}^2+\textbf{v}^2\textbf{B}^2-\left(\textbf{v}\cdot\textbf{B}\right)^2+2\textbf{E}\cdot\left(\textbf{v}\times\textbf{B}\right)$$$$\left(\textbf{B}_{\perp}-\frac{1}{c^2}\textbf{v}\times\textbf{E}\right)=\textbf{B}_{\perp}^2+\frac{1}{c^4}\textbf{v}^2\textbf{E}^2-\frac{1}{c^4}\left(\textbf{v}\cdot\textbf{E}\right)^2-\frac{2}{c^2}\textbf{B}\cdot\left(\textbf{v}\times\textbf{E}\right).$$
Plugging this directly into the transformation of the Lagrangian shows that the $\gamma$ factors cancel, and the Lagrangian is relativistically invariant.
The longest answer requires the language of differential forms to truly appreciate. If $A$ is the electromagnetic 1-form, and $F=\mathrm{d}A$ is its corresponding field strength, the canonical action for this $U(1)$ gauge theory is given by
$$S=-\frac{1}{2}\int_{\mathcal{M}} F\wedge\star F.$$
Now, to express this in terms of the electric and magnetic field, we note that if we have an orthonormal basis $\{\mathrm{d}x^\mu\}$ of $\Omega^{1}(\mathcal{M})$ (the space of one-forms on $\mathcal{M}$), we can define an electric field one form
$$E=E_{1}\mathrm{d}x^1+E_{2}\mathrm{d}x^2+E_{3}\mathrm{d}x^3.$$
Next, we can define a magnetic field 2-form
$$B=B_1\mathrm{d}x^2\wedge\mathrm{d}x^{3}+B_2\mathrm{d}x^3\wedge\mathrm{d}x^1+B_3\mathrm{d}x^1\wedge\mathrm{d}x^2.$$
In terms of these variables, the field strength is decomposed as
$$F=B+E\wedge\mathrm{d}x^0.$$
Now, when evaluating $F\wedge\star F$, the hodge star of $E\wedge\mathrm{d}x^0$ will have an extra minus sign relative to the $B$ term, coming from the minus sign in the metric tensor $g=\text{diag}(-1,1,1,1)$. This is the mathematical origin of the minus sign.
In summary, there are three answers to this question, with three different levels of intuitive satisfaction:
The minus sign is there to ensure that the equations of motion obtained are Maxwell's equations (not that satisfying). The minus sign is there to ensure relativistic invariance of the action (somewhat satisfying). The minus sign comes from the minus sign in the Minkowski space metric tensor (very satisfying).
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I have come across a problem which is a homework indeed, but i tried to pack this question up so that it is more theoretical.
What I want to know is: If I am allowed to write energy conservation for an atom which emits a photon (when its electron changes energy for a value $\Delta E$) like this (the atom is kicked back when it emits a photon):
\begin{align} E_1 &= E_2\\ E_{ \text{H atom 1}} &= E_{ \text{H atom 2} } + E_\gamma\\ \sqrt{ \!\!\!\!\!\!\!\!\!\!\smash{\underbrace{(E_0 + \Delta E)^2}_{\substack{\text{I am not sure about}\\\text{this part where normally}\\\text{we write only ${E_0}^2$. Should I}\\\text{put $\Delta E$ somewhere else?}}}}\!\!\!\!\!\!\!\!\!\!\!\! + {p_1}^2c^2} &= \sqrt{ {E_0}^2 + {p_2}^2c^2 } + E_\gamma \longleftarrow \substack{\text{momentum $p_1=0$ and because of}\\\text{the momentum conservation}\\\text{$p_2 = p_\gamma = E_\gamma/c$}}\\ \phantom{1}\\ \phantom{1}\\ \phantom{1}\\ \sqrt{{(E_0 + \Delta E)}^2} &= \sqrt{{E_0}^2 + {E_\gamma}^2} + E_\gamma\\\ E_0 + \Delta E &= \sqrt{{E_0}^2 + {E_\gamma}^2} + E_\gamma\\\ \end{align}
EDIT: Is this a better way? I know we get the same result this time, but what if momentum $p_1$ wasn't $0$? Then it would come out differently right?
\begin{align} E_1 &= E_2\\ E_{ \text{H atom 1}} + \Delta E &= E_{ \text{H atom 2} } + E_\gamma\\ \sqrt{{E_0}^2 + {p_1}^2c^2} + \Delta E &= \sqrt{ {E_0}^2 + {p_2}^2c^2 } + E_\gamma \longleftarrow \substack{\text{momentum $p_1=0$ and because of}\\\text{the momentum conservation}\\\text{$p_2 = p_\gamma = E_\gamma/c$}}\\ E_0 + \Delta E &= \sqrt{{E_0}^2 + {E_\gamma}^2} + E_\gamma \end{align}
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Set is Closed iff Equals Topological Closure Theorem
Let $T$ be a topological space.
Let $H \subseteq T$.
$H = \operatorname{cl} \left({H}\right)$
Let $H'$ denote the derived set of $H$.
The result follows from the definition of closure.
$\blacksquare$
Let $H^{\complement}$ denote the relative complement of $H$ in $T$.
$\forall x \in H^{\complement}: \exists U \in \tau: x \in U \subseteq H^{\complement}$
By Empty Intersection iff Subset of Complement, we have that:
$U \subseteq H^{\complement} \iff U \cap H = \varnothing$ $\forall x \in H^{\complement}: x \notin \operatorname{cl} \left({H}\right)$
From Set is Subset of its Topological Closure, we have that $H \subseteq \operatorname{cl} \left({H}\right)$.
$H = \operatorname{cl} \left({H}\right)$
$\blacksquare$
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The well-ordered replacement axiom is the scheme asserting that if $I$ is well-ordered and every $i\in I$ has unique $y_i$ satisfying a property $\phi(i,y_i)$, then $\{y_i\mid i\in I\}$ is a set. In other words, the image of a well-ordered set under a first-order definable class function is a set.
Alfredo had introduced the theory Zermelo + foundation + well-ordered replacement, because he had noticed that it was this fragment of ZF that sufficed for an argument we were mounting in a joint project on bi-interpretation. At first, I had found the well-ordered replacement theory a bit awkward, because one can only apply the replacement axiom with well-orderable sets, and without the axiom of choice, it seemed that there were not enough of these to make ordinary set-theoretic arguments possible.
But now we know that in fact, the theory is equivalent to ZF.
Theorem. The axiom of well-ordered replacement is equivalent to full replacement over Zermelo set theory with foundation.
$$\text{ZF}\qquad = \qquad\text{Z} + \text{foundation} + \text{well-ordered replacement}$$
Proof. Assume Zermelo set theory with foundation and well-ordered replacement.
Well-ordered replacement is sufficient to prove that transfinite recursion along any well-order works as expected. One proves that every initial segment of the order admits a unique partial solution of the recursion up to that length, using well-ordered replacement to put them together at limits and overall.
Applying this, it follows that every set has a transitive closure, by iteratively defining $\cup^n x$ and taking the union. And once one has transitive closures, it follows that the foundation axiom can be taken either as the axiom of regularity or as the $\in$-induction scheme, since for any property $\phi$, if there is a set $x$ with $\neg\phi(x)$, then let $A$ be the set of elements $a$ in the transitive closure of $\{x\}$ with $\neg\phi(a)$; an $\in$-minimal element of $A$ is a set $a$ with $\neg\phi(a)$, but $\phi(b)$ for all $b\in a$.
Another application of transfinite recursion shows that the $V_\alpha$ hierarchy exists. Further, we claim that every set $x$ appears in the $V_\alpha$ hierarchy. This is not immediate and requires careful proof. We shall argue by $\in$-induction using foundation. Assume that every element $y\in x$ appears in some $V_\alpha$. Let $\alpha_y$ be least with $y\in V_{\alpha_y}$. The problem is that if $x$ is not well-orderable, we cannot seem to collect these various $\alpha_y$ into a set. Perhaps they are unbounded in the ordinals? No, they are not, by the following argument. Define an equivalence relation $y\sim y’$ iff $\alpha_y=\alpha_{y’}$. It follows that the quotient $x/\sim$ is well-orderable, and thus we can apply well-ordered replacement in order to know that $\{\alpha_y\mid y\in x\}$ exists as a set. The union of this set is an ordinal $\alpha$ with $x\subseteq V_\alpha$ and so $x\in V_{\alpha+1}$. So by $\in$-induction, every set appears in some $V_\alpha$.
The argument establishes the principle: for any set $x$ and any definable class function $F:x\to\text{Ord}$, the image $F\mathrel{\text{”}}x$ is a set. One proves this by defining an equivalence relation $y\sim y’\leftrightarrow F(y)=F(y’)$ and observing that $x/\sim$ is well-orderable.
We can now establish the collection axiom, using a similar idea. Suppose that $x$ is a set and every $y\in x$ has a witness $z$ with $\phi(y,z)$. Every such $z$ appears in some $V_\alpha$, and so we can map each $y\in x$ to the smallest $\alpha_y$ such that there is some $z\in V_{\alpha_y}$ with $\phi(y,z)$. By the observation of the previous paragraph, the set of $\alpha_y$ exists and so there is an ordinal $\alpha$ larger than all of them, and thus $V_\alpha$ serves as a collecting set for $x$ and $\phi$, verifying this instance of collection.
From collection and separation, we can deduce the replacement axiom $\Box$
I’ve realized that this allows me to improve an argument I had made some time ago, concerning Transfinite recursion as a fundamental principle. In that argument, I had proved that ZC + foundation + transfinite recursion is equivalent to ZFC, essentially by showing that the principle of transfinite recursion implies replacement for well-ordered sets. The new realization here is that we do not need the axiom of choice in that argument, since transfinite recursion implies well-ordered replacement, which gives us full replacement by the argument above.
Corollary. The principle of transfinite recursion is equivalent to the replacement axiom over Zermelo set theory with foundation.
$$\text{ZF}\qquad = \qquad\text{Z} + \text{foundation} + \text{transfinite recursion}$$
There is no need for the axiom of choice.
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What is the difference between a randomly built binary search tree (using n keys )and choosing a binary search tree (of n key) from a random distribution
You want to distinguish
randomly built binary search tree
from
choosing a binary search tree from a random distribution.
Let me first tell you why this is not a well-posed query. Neither version iscompletely specified; the first misses the process by which the tree is builtas well as the used random source, the second lacks a specification of thedistribution to be used. In particular, the "randomly built" trees
are"chosen from a random distribution"!
Now I'll assume that you (or your teacher) made the "usual" assumptions and reformulate the models.
Starting with the empty tree, we insert keys $s_1, \dots, s_n \in [1..n]$ successively (with BST insertion). The sequence is chosen uniformly at random from all permutations of $[1..n]$.
This is the famous
random permutation model.
Pick a tree uniformly at random from all BSTs with keys $[1..n]$.
There are several ways to see that the two models define different random distributions.
There are $n!$ different permutations but only $C_n = \frac{1}{n+1} \cdot \binom{2n}{n}$ BSTs. Show that it is not possible to distribute the permutations evenly, i.e. $C_n$ does not (always) divide $n!$.
In particular, for $n=3$ we have six permutations but five BSTs.
For $n=3$, enumerate all trees and notice that four trees result from one permutation each while the fifth results from
two. Ergo, the random permutation model is not uniform.
Know (from earlier studies) that BSTs have average height $\Theta(\log n)$ in the random permutation model but $\Theta(\sqrt{n})$ in the uniform model.
The first is a standard result found in most textbooks on the matter. The latter, admittedly, a less known result by Flajolet and Odlyzko [1]; the method of analysis has broader relevance and the result generalises to many families of trees.
The average height of binary trees and other simple trees by P. Flajolet and A. Odlyzko (1982) Randomly built binary search tree : When binary tree is built by picking n keys one by one randomly from a set of keys.
While
binary search tree from a random distribution mean there are all possible ways of binary tree with n keys from a set of keys . and you are picking directly one of the tree.
Only way of making a random binary tree is different.
Randomly built binary search tree will be more efficient to make than the later one.
but both will result in estimated search time of O(log n)
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Pure three dimensional stream functions exist physically but at present there is no known way to represent then mathematically. One of the ways that was suggested by Yih in 1957 suggested using two stream functions to represent the three dimensional flow. The only exception is a stream function for three dimensional flow exists but only for axisymmetric flow i.e the flow properties remains constant in one of the direction (say z axis).
Advance Material
The three dimensional representation is based on the fact the continuity equation must be satisfied. In this case it will be discussed only for incompressible flow. The \(\nabla \pmb{U} = 0\) and vector identity of \(\nabla \cdot \nabla \pmb{U} = 0 \) where in this case \(\pmb{U}\) is any vector. As opposed to two dimensional case, the stream function is defined as a vector function as
\[ \label{if:eq:3DstreamFun} \pmb{B} = \psi \,\nabla \xi \tag{65} \] The idea behind this definition is to build stream function based on two scalar functions one provide the "direction'' and one provides the the magnitude. In that case, the velocity (to satisfy the continuity equation) \[ \label{if:eq:velocityVector} \pmb{U} = \boldsymbol{\nabla} \boldsymbol{\times} \left( \psi \,\boldsymbol{\nabla} \chi \right) \tag{66} \] where \(\psi\) and \(\chi\) are scalar functions. Note while \(\psi\) is used here is not the same stream functions that were used in previous cases. The velocity can be obtained by expanding equation (66) to obtained \[ \label{if:eq:UstreamFun1} \pmb{U} = \boldsymbol{\nabla}\psi \boldsymbol{\times} \boldsymbol{\nabla}\chi + \psi \,\overbrace{\boldsymbol{\nabla} \boldsymbol{\times}\left( \boldsymbol{\nabla}\chi\right)}^{=0} \tag{67} \] The second term is zero for any operation of scalar function and hence equation (67) becomes \[ \label{if:eq:UstreamFun} \pmb{U} = \boldsymbol{\nabla}\psi \boldsymbol{\times} \boldsymbol{\nabla}\chi \tag{68} \] These derivations demonstrates that the velocity is orthogonal to two gradient vectors. In another words, the velocity is tangent to the surfaces defined by \(\psi = constant\) and \(\chi = constant\). Hence, these functions, \(\psi\) and \(\chi\) are possible stream functions in three dimensions fields. It can be shown that the flow rate is \[ \label{if:eq:stream3DFlowRate} \dot{Q} = \left(\psi_2 - \psi_1\right) \left( \chi - \chi_1 \right) \tag{69} \] The answer to the question whether this method is useful and effective is that in some limited situations it could help. In fact, very few research papers deals this method and currently there is not analytical alternative. Hence, this method will not be expanded here.
End Advance Material
Contributors
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.
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The lecturer taught this method in my Optimization and Control Theory Class and I wasn't quite there when he named it. Could you help me out?
He gave the following example of the method in class:
Example: Solve $$ \text{max} [ f(x) = x_1 (30 - x_1) + x_2 (50 -2x_2) - 3x_1 - 5x_2 - 10x_3]$$Subject to the constraints: $$ x_1 + x_2 \le x_3 \text{and } x_3 \le 17.25 $$ Solution: Begin by converting constraints to form:
$$g_1(\bar x) = x_1 + x_2 - x_3 \le 0$$ $$g_2(\bar x) = x_3 - 17.25 \le 0$$ Then: $$ L(\bar x, \bar \lambda) = f(\bar x) \pm \left( \lambda_1 g_1(\bar x) + ... + \lambda_m g_m(\bar x) \right) $$
And then he proceeded as follows:
$$\begin{align} D_{\bar x} L: & \frac{\partial L}{\partial x_1} = 30 - 2x_1 - 3 - \lambda_1 = 0 \\ & \frac{\partial L}{\partial x_2} = 50 - 4x_2 - 5 - \lambda_1 = 0\\ & \frac{\partial L}{\partial x_3} = -10 + \lambda_1 - \lambda_2 = 0 \\ \end{align}$$
A system of equations is found and solved with further constraints: $\lambda_1 \ge 0$ and $\lambda_2 \ge 0$:
$$\lambda_1(x_1 + x_2 + x_3) = 0$$ $$\lambda_2 (x_3 - 17.25) = 0$$
(Through trial and error it is found that $\lambda_1 > 0 $ and $ \lambda_2 = 0$ is the best condition to solve this system)
Ultimately we get the solution: $$ x_1 = 8.5$$ $$ x_2 = 8.75$$ $$ x_3 = 17.25$$
And the equation is solved.
-- So I want to read more background on this method. I'd like to know what its called. I notice the 'L' as the name of the function. Could it be Lagrange? Or something of the sort?
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Note: The original answer has a flaw in "without loss of generality". The following one is based on Section 8.2.3 of the book "Computer Algorithms" (3rd edition) by Sara Baase and Allen Van Gelder.
The right part of the theorem is called the MST property.
MST Property: Let $T$ be any spanning tree. For any edge $e \notin T$, $e$ has the maximal weight on the cycle created by adding it to $T$.
In terms of the MST property, the theorem to prove can be restated as:
A spanning tree $T$ is an MST $\iff$ $T$ has the MST property.
We first prove the following lemma:
Lemma: If $T$ and $T'$ are spanning trees that have the MST property, then $w(T) = w(T')$.
Proof: Let $\Delta E = (E(T) \setminus E(T')) \cup (E(T') \setminus E(T))$. That is, $\Delta E$ is the set of edges that are in one of $T$ or $T'$ but not in both. $\Delta E \neq \emptyset$. Let $|\Delta E| = k$.
If $k = 0$, we are done.
Let $k > 0$ and $e$ be a minimum edge in $\Delta E$. Assume that $e \in T \setminus T'$ (the case of $e \in T' \setminus T$ is symmetrical).
$T' + e$ create a cycle $C_{T'}$. There must be an edge $e'$ on the cycle $C_{T}$ that $e' \neq e$ and $e' \in T' \setminus T$ (otherwise $T$ contains the cycle $C_{T'}$).
Since $T'$ has the MST property, $e$ has the maximum weight on the cycle $C_{T'}$. Thus, $w(e') \le w(e)$. On the other hand, $e$ by definition is a minimum edge in $\Delta E$. Thus, $w(e) \le w(e')$.
Therefore, $w(e) = w(e')$.
Add $e$ to $T'$, creating a cycle, then remove $e'$, leaving a spanning tree $T''$. Since $w(T'') = w(T')$, $T''$ is an MST. However, $T$ and $T''$ differ by $k-1$ edges.
Repeating the procedure above, we will end with an MST, say $T^{k+1}$, such that $T$ and $T^{k+1}$ differ by $0$ edges.
Therefore, we have $w(T) = w(T^{k+1}) = \cdots = w(T'') = w(T')$.
Given the lemma above, we proceed to prove the "$\Leftarrow$" direction as follows:
Assume $T$ has the MST property. Let $T_m$ be any MST. By the "$\Rightarrow$" direction, $T_m$ has the MST property. By the lemma above, $w(T) = w(T_m)$. Thus, $T$ is an MST.
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I wish to explain my younger brother: he is interested and curious, but he cannot grasp the concepts of limits and integration just yet. What is the best mathematical way to justify not allowing division by zero?
“One of the ways to look at division is as how many of the smaller number you need to make up the bigger number, right? So 20/4 means: how many groups of 4 do you need to make 20? If you want 20 apples, how many bags of 4 apples do you need to buy?
So for dividing by 0, how many bags of 0 apples would make up 20 apples in total? It’s impossible — however many bags of 0 apples you buy, you’ll never get any apples — you’ll certainly never get to 20 apples! So there’s no possible answer, when you try to divide 20 by 0.”
When we first start teaching multiplication, we use successive additions. So,
3 x 4 = 3 | 3 + 3 | 6 + 3 | 9 + 3 | 12=12
Division can be taught as successive subtractions. So 12 / 3 becomes,
12 - 3 -> 9 (1)9 - 3 -> 6 (2)6 - 3 -> 3 (3)3 - 3 -> 0 (4)
Now apply the second algorithm with zero as a divisor. Tell your brother to get back to you when he's done.
While this algorithmic approach is not rigorous, I think it is probably a good way of developing an intuitive understanding of the concept.
New story
Suppose that we can divide numbers with $0$. So if I would divide $1$ with zero i would get some new number name it $a$. Now what can we say about this number $a$?
Remember:
If I divide say $21$ with $3$ we get $7$. Why? Because $3\cdot 7 = 21$.
And similiary if I divide $36$ with $9$ we get $4$. Why? Because $9\cdot 4 = 36$.
So if I divide $1$ with $0$ and we get $a$ then we have $a\cdot 0 =1$ which is clearly nonsense since $a\cdot 0 =0$.
Old explanation:
Suppose that ${1\over 0}$ is some number $a$. So $${1\over 0} =a.$$ Remember that $$\boxed{{b\over c} = d\iff b = c\cdot d}$$ So we get $$1= a\cdot 0=0$$ a contradiction. So ${1\over 0}$ doesn't exist.
An explanation that might make sense to a fifth grader is one that gets to the heart of why we have invented these operations in the first place.
Multiplication is a trick we use to add similar things to form a sum. When we say 5 x 3, what we really mean is take five things of size three each and add them all together. We invented this trick because we are frequently in the situation where we have many of a similar thing, and we wish to know their sum.
Division is the same trick but the other way. When we say 15 / 3, we are asking the question "how many times would we have to add a thing of size three starting from nothing to make a thing of size fifteen?" We'd have to add five things of size three together to make a thing of size fifteen. Again, division is just a trick we use to answer questions about sums.
Now it becomes clear why division by zero is not defined.
There is no number of times you can add zero to itself to get a non-zero sum.
A sophisticated fifth grader would then note that 0 / 0 is by this definition defined as zero. Going into why 0 / 0 is not defined would require more work!
For non-zero divided by zero, there is no number
at all of times that you can add zero to itself to get non-zero. For zero divided by zero, every number of times you add zero to itself, you get zero, so the solution is not unique. We like our mathematical questions to have unique answers where possible and so we by convention say that 0 / 0 is also not defined.
I prefer the algebraic argument, that there is no multiplicative inverse of $0$, this would need you to explain a bit about algebra.
The argument from calculus, looking at limits of $1/x$, I find also useful, but perphaps harder to explain.
How many nothings do you need to add together to get 12?
Ask Siri.
Imagine that you have zero cookies and you split them evenly among zero friends. How many cookies does each person get? See? It doesn't make sense. And Cookie Monster is sad that there are no cookies, and you are sad that you have no friends.
You shouldn't try to do that. Instead make counter question.
"What should it be, then?" and let them think about it.
(Lengthy) justification: There are many important concepts in math you can come up with if you start experimenting with multiplication. Take for example area of a rectangle. You multiply the sides. Area of a curve? You take the integral. What is an integral? Well Riemann imagined thin thin slices, almost infinitely thin, actually. The idea that we can calculate area of these slices where one side is so tiny it almost is 0. If we disqualify limits, or the idea of multiplying something "almost 0" to be 0 then we would have a tougher time coming up with an excuse to investigate integrals, which have been veeery important to the development of modern technology.
Any kid who could come up with some new interpretation of this could be very valuable.
Division by zero is meaningless because that's what we decided division means. All you can do is explain why such a convention is a useful one for ordinary arithmetic.
It might even help to demonstrate some
other context (e.g. arithmetic in the projectively extended number line) where it can be useful to define division by zero, so that the student is able to compare and contrast the reasons why we might or might not like to define something.
Your question might be better placed on https://matheducators.stackexchange.com/
@Jack M and @greedoid probably highlight a good point: division does not exist. It's only the inverse operation of multiplication.
You could explain your brother the complete truth: dividing 20 by 5 is about finding the only answer (if it exists) to this question: what number can be multiplied by 5 to give 20?. The unique answer is easy: 4 times 5 is 20. And the division is only another phrasing to say the exact same thing: 20 divided by 5 is 4. Can you always find one and only one answer? Yup, almost always... There's only one exception... What number, multiplied by 0, gives 20? There's none. So "division" by 0 has no meaning, since we cannot find any number that satisfies our definition.
You could even draw his attention by mentioning that most grown-ups don't know there's no such thing as "division", and that's the first step to learn about "E-vector spaces", "rings" and other funny-named artefacts when he's in college... or before that!
Note: what if he raises a question about "0/0"?
OK, let's try: "what number, multiplied by 0, gives 0?" All of them! We cannot find one and only one answer, so, it's still impossible to divide 0 by 0!
I don't have kids (my wife says one 3-year-old in the house is enough for her) and it's been a while since I was in the 5th grade (although at work sometimes...), but I'll give it a go.
I know you're too old to play with blocks, but lets start with 12 blocks.
Let's start with $12/6$ - that's $2$, right? Take $6$ at a time and there are two "sets". There are $2$ sets of $6$ in $12$.
Then $12/4$ is $3$ - $3$ sets of $4$ in $12$.
Then $12/3$ is $4$ - $4$ sets of $3$ in $12$ (commutation of the last case).
Then $12/2$ is $6$ - $2$ sets of $6$ in $12$ (commutation of first case).
Then $12/1$ is $12$ - $1$ set of $12$ in $12$ (degenerate case).
Notice the size of the result set is getting bigger as the denominator (the number on the bottom) gets smaller.
Before we go to $0$ let's try something between $1$ and $0$ - $1/2$ or $0.5$. Think of just splitting each block into two (take a hatchet to the wooden blocks blocks, or just imagine it if mom doesn't want you handling a hatchet).
$12/0.5$ is $24$ - $24$ sets of $0.5$ (half-pieces) in $12$
$12/0.25$ is $48 - 48$ sets of $0.25$ (quarter-pieces) in $12$
$12/0.125$ is $96 - 96$ sets of $0.125$ (pieces of eight**) in $12$
$12/0.0625$ is $192 - 192$ sets of $0.0625$ (pieces of 16) in $12$
The close you get to zero, the larger the set you get gets.
$12/0.000000001$ (a billionth) is $12$ billion sets of a billionth of a block (aka, sawdust)
The as you approach zero, the resulting set size is too large to represent (not enough paper in this room, not enough memory on this computer) and the size of the pieces approach zero.
A cheat for "Too large to represent" is "infinity".
** pirate reference - do 5th graders still like pirates these days?
One would need to first explain what we mean by division. That is, what does $/$ mean in the expression $a/b,$ where $a$ and $b$ are integers?
Well, whatever it is, it is a way of combining two numbers. Now recall that every time we defined an operation (say addition), we always had a unique result as the product of the combination, so that we would like this to continue to hold. What else? We define $/$ indirectly, by looking at what we want $a/b$ to mean. Well, we want it to stand for the number $c$ which when multiplied together with $b$ recovers $a.$ (Recall how we similarly defined subtraction as the inverse operation of $+.$)
Therefore, in summary, if we let $a/b=c,$ then by definition this equality is equivalent to $c×b=a.$ Also, we want $c$ to be unique for all possible integers $a$ and $b.$
Now consider the expression $a/0.$ First let us take $a\ne0.$ Then if we let $a/0=c,$ it follows by definition that $c×0=a.$ But with the way we defined multiplication (remind him of this), we required that $0$ must make any number vanish, so that there simply is no such $c$ as we seek. If now we let $a=0,$ then we want a unique $c$ such that $c×0=0.$ But again, by the property $r×0=0\,\,\,\forall r$ which we've previously allowed in defining $×,$ we have infinitely many candidates for $c$ and there is no other condition we can impose to select one uniquely. We therefore do not allow ourselves to divide by $0$ in any case, in order to avoid all that mess.
Division is
: sharing
1 / 10:
10 boys in a class grab at a toy -- they rip the toy to tiny bits!
1 / 2:
2 boys fight for a toy -- they rip the toy in half!
1 / 0:
A different toy is alone -- he is a special boy!
The way I taught it, even to junior college students who were taking elementary mathematics courses, was with a calculator.
I would show them that 1/1 = 1, 1/0.1 = 10, 1/0.01 = 100, and so on. I would ask them if they saw how the numbers kept getting bigger as we divided by smaller and smaller numbers. Then I would ask them what they thought would happen when we hit zero. "We would get the biggest possible number that exists, right? But there is no biggest number. So dividing by zero gives you a number that doesn't exist. Does that make any sense? No. So we say that dividing by zero is undefined."
Number of marbles : Number of boxes = Number of marbles in each box.
20 marbles : 4 boxes = 5 marbles per box
0 marbles : 4 boxes = 0 marbles per box
20 marbles : 0 boxes = "how many marbles in each box while no box?" ---> undefined!
Because before you think about dividing something, it is more important to consider if you have someone to divide it for (he/she/it must be present, exist, etc). If you do not have anyone who can 'benefit' from the division, no point in dividing. Non rigorous, pragmatic, heuristic approach. It might pave the way for more reasoned proofs and demonstrations.
To divide means to subtract many times. So, how many times can we subtract $0$ from a given number?
It might be a duplicated answer and I apologize, in case. But, according to my experience as a teacher, this worked well.
The point, as others had observed, is what does "to divide" mean. This sometimes looked obscure to the students, whereas the concept of subtraction was more clear.
Thus, once you convey the message that "to divide" means "to subtract many times", everything becomes more clear.
How many times can we subtract $3$ from $10$? Well, usually my students got this. How many times can we subtract $0$ from $10$? Well, how many times we want!
So there is not a precise answer, because any answer is good. This made more clear the sense of "not defined", at least to my students.
Hope it helps!
Explain him the problems, don't enforce him as an "official view".
Explain him, what are the problems of the division by zero.
Let him to think about a possible solution.
You might also explain, that also the negative numbers don't have a suqare root, but this problem had a solution, the imaginary numbers. Let him try to think about a similar solution for the division by zero.
The following explanation in terms of division as the inverse of multiplication may help, as modern fifth graders should have been introduced to the idea of division as something that undoes multiplication.
6/2 = 3. Why? Because 3 * 2 = 6 and division means find the number (3) that multiplies the dividing number (3) to give the number being divided (6). To divide 6 by 2, we ask what number, when multiplied by 2, gives 6.
Ask your brother to do this exercise for 6 and 0. What number, when multiplied by 0, will give 6? He should see the problem here, because, no matter what number we try, when we multiply it by 0, we get the same answer 0.
A diagram might help to bring the problem into sharper sight. What you're doing in the following is conveying the lack of bijectivity of $x\mapsto 0 \times x$, in age appropriate words, of course ...
The left hand diagram shows the mapping $x\mapsto 2 \times x$; encourage your brother to think of multiplication as a stretch or shrink induced on the number line. The crucial property to note here is that every arrow on the diagram is reversible, meaning that you can find one and only one number that 2 multiplies to get the answer. Every answer is 2 times a unique something.
Multiplication by 2 is reversible - use this word - in the sense that we do not lose the knowledge of what has been multiplied by 2 to get the answer.
The same kind of situation holds for every nonzero multiplier - the real line is stretched or shrunken, and sometimes flipped in orientation as well, but we can always work out what was multiplied originally to arrive at the end of any given arrow.
Now have your brother look at the diagram for $x\mapsto 0 \times x$. Everything goes awry because
all the arrows wind up at the image 0. Given only our answer (0), we have no idea what we multiplied by 0 to get the answer, because it could have been any real number. Multiplication by 0 destroys the knowledge of what was multipled.
Later on, your brother might like to come back to this idea to understand the pole of $z\mapsto 1/z$ at 0 in a bit more detail: multiplication by a very small number $\epsilon$ corresponds to a very severe shrink, but, as long as the number is not nought, the arrows do not quite merge and the shrink can be undone.
0 as a multiplier is a destroyer of information: no other real number is like this and this property is why we can't invert the multiplication. One boy in my daughter's class whom I explained this to (I help out with numeracy at my daughter's school) has a particular love and encyclopoedic knowledge of Greek, Hindu and other gods (I think he may know every pantheon conceived!). He was most chuffed to learn that $0$ was the "Shiva" number.
Try to make him realize himself that there's no solution.
Take a (imaginary) pizza.
Ask him to cut the pizza into one piece.
Ask him to cut the pizza into two pieces. Ask him to cut the pizza into three pieces. Ask him to cut the pizza into zero pieces.
Just give him some questions e.g 2/0 ,5/0 ,6/0 and tell him to divide just using simple division tell him to keep on dividing till he reaches a satisfactory.Let him try for some time.And that satisfactory won't come how much me try.
Now you tell him that you will never come to a satisfactory result.Hence it's answer will be
meaningless!!!
protected by Daniel Fischer♦ Aug 15 '18 at 18:03
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From Jacod and Shiryaev's Limit Theorems for Stochastic Processes, we get the following definitions.
Definitions: A process with independent increments (abbreviated PII) $X = (X_t)_{t \geq 0}$ on a stochastic basis $(\Omega, \mathcal{F}, \mathbb{F} = (\mathcal{F}_t)_{t \geq 0}, \mathbb{P})$ is a càdlàg adapted real-valued process with $X_0 = 0$ and for all $0 \leq s \leq t < +\infty$, $X_t - X_s$ is independent of $\mathcal{F}_s$. A Lévy process (also called process with independent and stationary increments) on a stochastic basis $(\Omega, \mathcal{F}, \mathbb{F}, \mathbb{P})$ is a PII $X$ such that the distribution of the increment $X_t - X_s$ depends only on $t-s$, for all $0 \leq s \leq t$.
Lévy processes are ubiquitous in mathematical finance. For example, most models for the return of financial assets (Brownian motion, Kou, Merton, CGMY, etc...) are Lévy processes.
I would like to know if processes with independent increments which are not Lévy (i.e. not stationary) are used in finance. One possible application could be models with seasonal changes in parameters of the distribution. Thanks a lot !
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