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[1] G. Gromadzki and X. Zhao. Free degree of periodic self-homeomorphisms of compact bordered orientable surfaces. Contemporary Mathematics 629 (2014) 203-223. Book volume table of contents View Article: PDF [2] Andrew Niedermaier, Douglas Rizzolo and Francis Edward Su. A Tree Sperner Lemma. Contemporary Mathematics 625 (2014) 77-92. Book volume table of contents View Article: PDF [3] Pedro L. Q. Pergher. A coincidence theorem for commuting involutions. Proc. Amer. Math. Soc. 140 (2012) 2537-2541. Abstract, references, and article information View Article: PDF [4] Ku Yong Ha, Jong Bum Lee and Pieter Penninckx. Anosov theorem for coincidences on special solvmanifolds of type $(\mathrm{R})$. Proc. Amer. Math. Soc. 139 (2011) 2239-2248. MR 2775401. Abstract, references, and article information View Article: PDF [5] J. P. Boronski. Fixed points and periodic points of orientation-reversing planar homeomorphisms. Proc. Amer. Math. Soc. 138 (2010) 3717-3722. MR 2661570. Abstract, references, and article information View Article: PDF [6] Vesta Coufal. A parametrized fixed point theorem. Proc. Amer. Math. Soc. 137 (2009) 3939-3942. MR 2529904. Abstract, references, and article information View Article: PDF This article is available free of charge [7] Márton Elekes. On a converse to Banach's Fixed Point Theorem. Proc. Amer. Math. Soc. 137 (2009) 3139-3146. MR 2506473. Abstract, references, and article information View Article: PDF This article is available free of charge [8] Grzegorz Graff and Jerzy Jezierski. On the growth of the number of periodic points for smooth self-maps of a compact manifold. Proc. Amer. Math. Soc. 135 (2007) 3249-3254. MR 2322756. Abstract, references, and article information View Article: PDF This article is available free of charge [9] Catherine Lee. A minimum fixed point theorem for smooth fiber preserving maps. Proc. Amer. Math. Soc. 135 (2007) 1547-1549. MR 2276665. Abstract, references, and article information View Article: PDF This article is available free of charge [10] Daciberg Gonçalves and Peter Wong. Wecken property for roots. Proc. Amer. Math. Soc. 133 (2005) 2779-2782. MR 2146228. Abstract, references, and article information View Article: PDF This article is available free of charge [11] Ikumitsu Nagasaki. Isovariant Borsuk-Ulam results for pseudofree circle actions and their converse. Trans. Amer. Math. Soc. 358 (2006) 743-757. MR 2177039. Abstract, references, and article information View Article: PDF This article is available free of charge [12] Jan Jaworowski. Bundles with periodic maps and mod $p$ Chern polynomial. Proc. Amer. Math. Soc. 132 (2004) 1223-1228. MR 2045442. Abstract, references, and article information View Article: PDF This article is available free of charge [13] Jean-Paul Penot. A fixed-point theorem for asymptotically contractive mappings. Proc. Amer. Math. Soc. 131 (2003) 2371-2377. MR 1974633. Abstract, references, and article information View Article: PDF This article is available free of charge [14] Robert S. Simon, Stanislaw Spiez and Henryk Torunczyk. Equilibrium existence and topology in some repeated games with incomplete information. Trans. Amer. Math. Soc. 354 (2002) 5005-5026. MR 1926846. Abstract, references, and article information View Article: PDF This article is available free of charge [15] Daciberg L. Gonçalves, Jan Jaworowski and Pedro L. Q. Pergher. $G$-coincidences for maps of homotopy spheres into CW-complexes. Proc. Amer. Math. Soc. 130 (2002) 3111-3115. MR 1908937. Abstract, references, and article information View Article: PDF This article is available free of charge [16] Marc Bonino. Lefschetz index for orientation reversing planar homeomorphisms. Proc. Amer. Math. Soc. 130 (2002) 2173-2177. MR 1896055. Abstract, references, and article information View Article: PDF This article is available free of charge [17] Sol Schwartzman. Parallel tangent hyperplanes. Proc. Amer. Math. Soc. 130 (2002) 1457-1458. MR 1879969. Abstract, references, and article information View Article: PDF This article is available free of charge [18] Núria Fagella and Jaume Llibre. Periodic points of holomorphic maps via Lefschetz numbers. Trans. Amer. Math. Soc. 352 (2000) 4711-4730. MR 1707699. Abstract, references, and article information View Article: PDF This article is available free of charge [19] Alexander Fel′shtyn. Dynamical zeta functions, Nielsen theory and Reidemeister torsion. Memoirs of the AMS 147 (2000) MR 1697460. Book volume table of contents [20] Simeon T. Stefanov. Yang index of the deleted product. Proc. Amer. Math. Soc. 128 (2000) 885-891. MR 1707531. Abstract, references, and article information View Article: PDF This article is available free of charge [21] Roman Srzednicki. A generalization of the Lefschetz fixed point theorem and detection of chaos. Proc. Amer. Math. Soc. 128 (2000) 1231-1239. MR 1691005. Abstract, references, and article information View Article: PDF This article is available free of charge [22] Kapil D. Joshi. Mistake in Hirsch's proof of the Brouwer Fixed Point Theorem. Proc. Amer. Math. Soc. 128 (2000) 1523-1525. MR 1646193. Abstract, references, and article information View Article: PDF This article is available free of charge [23] Marek Izydorek. Bourgin-Yang type theorem and its application to $Z_2$-equivariant Hamiltonian systems. Trans. Amer. Math. Soc. 351 (1999) 2807-2831. MR 1467470. Abstract, references, and article information View Article: PDF This article is available free of charge [24] Joyce Wagner. An algorithm for calculating the Nielsen number on surfaces with boundary. Trans. Amer. Math. Soc. 351 (1999) 41-62. MR 1401531. Abstract, references, and article information View Article: PDF This article is available free of charge [25] Lucien Guillou. A simple proof of P. Carter's theorem. Proc. Amer. Math. Soc. 125 (1997) 1555-1559. MR 1372031. Abstract, references, and article information View Article: PDF This article is available free of charge [26] Charles L. Hagopian. The fixed-point property for simply connected plane continua. Trans. Amer. Math. Soc. 348 (1996) 4525-4548. MR 1344207. Abstract, references, and article information View Article: PDF This article is available free of charge [27] Gregory Lupton and John Oprea. Fixed points and powers of self-maps of $H$-spaces . Proc. Amer. Math. Soc. 124 (1996) 3235-3239. MR 1328360. Abstract, references, and article information View Article: PDF This article is available free of charge [28] Sehie Park and Kwang Sik Jeong. A general coincidence theorem on contractible spaces. Proc. Amer. Math. Soc. 124 (1996) 3203-3206. MR 1343718. Abstract, references, and article information View Article: PDF This article is available free of charge [29] Sehie Park. Fixed points of approximable maps. Proc. Amer. Math. Soc. 124 (1996) 3109-3114. MR 1343717. Abstract, references, and article information View Article: PDF This article is available free of charge [30] O. Davey, E. Hart and K. Trapp. Computation of Nielsen numbers for maps of closed surfaces. Trans. Amer. Math. Soc. 348 (1996) 3245-3266. MR 1370638. Abstract, references, and article information View Article: PDF This article is available free of charge
Results: 1 to 30 of 75 found Go to page: 1 2 3 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9153050184249878, "perplexity": 2059.5599180649247}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644064362.36/warc/CC-MAIN-20150827025424-00183-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://zbmath.org/?q=an:07010542 | ×
## Homogenization of parabolic equations with an arbitrary number of scales in both space and time.(English)Zbl 1406.35140
Summary: The main contribution of this paper is the homogenization of the linear parabolic equation $$\partial_tu^{\varepsilon}(x,t)-\nabla\cdot(a(x/\varepsilon^{q_1},\ldots,x/\varepsilon^{q_n},t/\varepsilon^{r_1},\ldots,t/\varepsilon^{r_m})\nabla u^{\varepsilon}(x,t))=f(x,t)$$ exhibiting an arbitrary finite number of both spatial and temporal scales. We briefly recall some fundamentals of multiscale convergence and provide a characterization of multiscale limits for gradients, in an evolution setting adapted to a quite general class of well-separated scales, which we name by jointly well-separated scales (see appendix for the proof). We proceed with a weaker version of this concept called very weak multiscale convergence. We prove a compactness result with respect to this latter type for jointly well-separated scales. This is a key result for performing the homogenization of parabolic problems combining rapid spatial and temporal oscillations such as the problem above. Applying this compactness result together with a characterization of multiscale limits of sequences of gradients we carry out the homogenization procedure, where we together with the homogenized problem obtain $$n$$ local problems, that is, one for each spatial microscale. To illustrate the use of the obtained result, we apply it to a case with three spatial and three temporal scales with $$q_1=1$$, $$q_2=2$$, and $$0<r_1<r_2$$.
### MSC:
35K10 Second-order parabolic equations 35B27 Homogenization in context of PDEs; PDEs in media with periodic structure
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http://en.wikipedia.org/wiki/Ternary_plot | # Ternary plot
A ternary plot, ternary graph, triangle plot, simplex plot, or de Finetti diagram is a barycentric plot on three variables which sum to a constant. It graphically depicts the ratios of the three variables as positions in an equilateral triangle. It is used in physical chemistry, petrology, mineralogy, metallurgy, and other physical sciences to show the compositions of systems composed of three species. In population genetics, it is often called a Gibbs triangle or a de Finetti diagram. In game theory, it is often called a simplex plot.[citation needed]
Approximate colours of Ag–Au–Cu alloys in jewellery making
In a ternary plot, the proportions of the three variables a, b, and c must sum to some constant, K. Usually, this constant is represented as 1.0 or 100%. Because a + b + c = K for all substances being graphed, any one variable is not independent of the others, so only two variables must be known to find a sample's point on the graph: for instance, c must be equal to K − a − b. Because the three proportions cannot vary independently - there are only two degrees of freedom - it is possible to graph the intersection of all three variables in only two dimensions.[citation needed]
## Reading values on the ternary plot
The advantage of using a ternary plot for depicting compositions is that three variables can be conveniently plotted in a two-dimensional graph. Ternary plots can also be used to create phase diagrams by outlining the composition regions on the plot where different phases exist.
Every point on a ternary plot represents a different composition of the three components. There are three common methods used to determine the ratios of the three species in the composition. The first method is an estimation based upon the phase diagram grid. The concentration of each species is 100% (pure phase) in each corner of the triangle and 0% at the line opposite it. The percentage of a specific species decreases linearly with increasing distance from this corner, as seen in figures 3–8. By drawing parallel lines at regular intervals between the zero line and the corner (as seen in the images), fine divisions can be established for easy estimation of the content of a species. For a given point, the fraction of each of the three materials in the composition can be determined by the first.
For phase diagrams that do not possess grid lines, the easiest way to determine the composition is to set the altitude of the triangle to 100% and determine the shortest distances from the point of interest to each of the three sides. The distances (the ratios of the distances to the total height of 100%) give the content of each of the species, as shown in figure 1.
The third method is based upon a larger number of measurements, but does not require the drawing of perpendicular lines. Straight lines are drawn from each corner, through the point of interest, to the opposite side of the triangle. The lengths of these lines, as well as the lengths of the segments between the point and the corresponding sides, are measured individually. Ratios can then be determined by dividing these segments by the entire corresponding line as shown in the figure 2. (The sum of the ratios should add to 1).
## Derivation from Cartesian coordinates
Derivation of a ternary plot from Cartesian coordinates
Figure (1) shows an oblique projection of point P(a,b,c) in a 3-dimensional Cartesian space with axes a, b and c, respectively.
If a + b + c = K (a positive constant), P is restricted to a plane containing A(K,0,0), B(0,K,0) and C(0,0,K). If a, b and c each cannot be negative, P is restricted to the triangle bounded by A, B and C, as in (2).
In (3), the axes are rotated to give an isometric view. The triangle, viewed face-on, appears equilateral.
In (4), the distances of P from lines BC, AC and AB are denoted by a' , b' and c' , respectively.
For any line l = s + t in vector form ( is a unit vector) and a point p, the perpendicular distance from p to l is $\| (\mathbf{s}-\mathbf{p}) - ((\mathbf{s}-\mathbf{p}) \cdot \mathbf{\hat{n}})\mathbf{\hat{n}} \|$ .
In this case, point P is at $\mathbf{p} = \begin{pmatrix}a\\b\\c\end{pmatrix}$ .
Line BC has $\mathbf{s} = \begin{pmatrix}0\\K\\0\end{pmatrix}$ and $\mathbf{\hat{n}} = \frac{\Big(\begin{smallmatrix}0\\K\\0\end{smallmatrix}\Big) - \Big(\begin{smallmatrix}0\\0\\K\end{smallmatrix}\Big)}{\Big|\Big|\Big(\begin{smallmatrix}0\\K\\0\end{smallmatrix}\Big) - \Big(\begin{smallmatrix}0\\0\\K\end{smallmatrix}\Big)\Big|\Big|} = \frac{\Big(\begin{smallmatrix}0\\K\\-K\end{smallmatrix}\Big)}{\sqrt{0^2+K^2+(-K)^2}} = \begin{pmatrix}0\\\;\;1/\sqrt{2}\\-1/\sqrt{2}\end{pmatrix}$ .
Using the perpendicular distance formula,
\begin{align} a' & = \bigg|\bigg| \Big(\begin{smallmatrix}-a\\K-b\\-c\end{smallmatrix}\Big) - \bigg( \Big(\begin{smallmatrix}-a\\K-b\\-c\end{smallmatrix}\Big) \cdot \Big(\begin{smallmatrix}0\\\;\;1/\sqrt{2}\\-1/\sqrt{2}\end{smallmatrix}\Big) \bigg) \Big(\begin{smallmatrix}0\\\;\;1/\sqrt{2}\\-1/\sqrt{2}\end{smallmatrix}\Big) \bigg|\bigg| \\ & = \bigg|\bigg| \Big(\begin{smallmatrix}-a\\K-b\\-c\end{smallmatrix}\Big) - \Big( 0 + \tfrac{K-b}{\sqrt{2}} + \tfrac{c}{\sqrt{2}} \Big) \Big(\begin{smallmatrix}0\\\;\;1/\sqrt{2}\\-1/\sqrt{2}\end{smallmatrix}\Big) \bigg|\bigg| \\ & = \bigg|\bigg| \bigg(\begin{smallmatrix}-a\\K-b-\tfrac{K-b+c}{2}\\-c+\tfrac{K-b+c}{2}\end{smallmatrix}\bigg) \bigg|\bigg| = \bigg|\bigg| \bigg(\begin{smallmatrix}-a\\\tfrac{K-b-c}{2}\\\tfrac{K-b-c}{2}\end{smallmatrix}\bigg) \bigg|\bigg| \\ & = \sqrt{(-a)^2 + \big(\tfrac{K-b-c}{2}\big)^2 + \big(\tfrac{K-b-c}{2}\big)^2} = \sqrt{a^2 + \tfrac{(K-b-c)^2}{2}} \\ \end{align}
Substituting K = a + b + c,
$a' = \sqrt{a^2 + \tfrac{(a+b+c-b-c)^2}{2}} = \sqrt{a^2 + \tfrac{a^2}{2}} = a\sqrt{\tfrac{3}{2}}$ .
Similar calculation on lines AC and AB gives
$b' = b\sqrt{\tfrac{3}{2}}$ and $c' = c\sqrt{\tfrac{3}{2}}$ .
This shows that the distance of the point from the respective lines is linearly proportional to the original values a, b and c.[1]
## Plotting a ternary plot
Cartesian coordinates are useful for plotting points in the triangle. Consider an equilateral ternary plot where $a=100\%$ is placed at $(x,y)=(0,0)$ and $b=100\%$ at $(1,0)$. Then $c=100\%$ is $\left(\tfrac{1}{2},\tfrac{\sqrt{3}}{2}\right)$, and the triple $(a,b,c)$ is $\left(\tfrac{1}{2}\tfrac{2b+c}{a+b+c},\tfrac{\sqrt{3}}{2}\tfrac{c}{a+b+c}\right).$
## Example
This example shows how this works for a hypothetical set of three soil samples:
Sample # Organic matter Clay Sand Notes
Sample 1 80% 10% 10% Because organic matter and clay make up 90% of this sample, the proportion of sand must be 10%.
Sample 2 50% 40% 10% The proportion of sand is 10% in this sample too, but the proportions of organic matter and clay are different.
Sample 3 10% 40% 50% This sample has the same proportion of clay as in Sample 2 does, but because it has a smaller proportion of organic matter, the proportion of sand must be larger, because all samples' proportions must sum to 100%.
## Software
Here is a list of software that help enable the creation of ternary plots
## References
Vaughan, Will (September 5, 2010). "Ternary plots". Retrieved September 7, 2010. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 16, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9339085221290588, "perplexity": 573.2031384606228}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131296383.42/warc/CC-MAIN-20150323172136-00242-ip-10-168-14-71.ec2.internal.warc.gz"} |
https://www.meritnation.com/cbse-class-11-science/physics/physics-part-i-ncert-solutions/motion-in-a-straight-line/ncert-solutions/41_4_1335_181_55_4343 | NCERT Solutions for Class 11 Science Physics Chapter 3 Motion In A Straight Line are provided here with simple step-by-step explanations. These solutions for Motion In A Straight Line are extremely popular among Class 11 Science students for Physics Motion In A Straight Line Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 11 Science Physics Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 11 Science Physics are prepared by experts and are 100% accurate.
#### Question 3.1:
In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
(a) The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object.
(b) The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track.
(c) The size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object.
(d) The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.
#### Question 3.2:
The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below;
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/twice).
(a) A lives closer to school than B.
(b) A starts from school earlier than B.
(c) B walks faster than A.
(d) A and B reach home at the same time.
(e) B overtakes A once on the road.
Explanation:
(a) In the given xt graph, it can be observed that distance OP < OQ. Hence, the distance of school from the A’s home is less than that from B’s home.
(b) In the given graph, it can be observed that for x = 0, t = 0 for A, whereas for x = 0, t has some finite value for B. Thus, A starts his journey from school earlier than B.
(c) In the given xt graph, it can be observed that the slope of B is greater than that of A. Since the slope of the xt graph gives the speed, a greater slope means that the speed of B is greater than the speed A.
(d) It is clear from the given graph that both A and B reach their respective homes at the same time.
(e) B moves later than A and his/her speed is greater than that of A. From the graph, it is clear that B overtakes A only once on the road.
#### Question 3.3:
A woman starts from her home at 9.00 am, walks with a speed of 5 km h–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h–1. Choose suitable scales and plot the x-t graph of her motion.
Speed of the woman = 5 km/h
Distance between her office and home = 2.5 km
It is given that she covers the same distance in the evening by an auto.
Now, speed of the auto = 25 km/h
The suitable x-t graph of the motion of the woman is shown in the given figure.
#### Question 3.4:
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Distance covered with 1 step = 1 m
Time taken = 1 s
Time taken to move first 5 m forward = 5 s
Time taken to move 3 m backward = 3 s
Net distance covered = 5 – 3 = 2 m
Net time taken to cover 2 m = 8 s
Drunkard covers 2 m in 8 s.
Drunkard covered 4 m in 16 s.
Drunkard covered 6 m in 24 s.
Drunkard covered 8 m in 32 s.
In the next 5 s, the drunkard will cover a distance of 5 m and a total distance of 13 m and falls into the pit.
Net time taken by the drunkard to cover 13 m = 32 + 5 = 37 s
The x-t graph of the drunkard’s motion can be shown as:
#### Question 3.5:
A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of 1500 km h–1 relative to the jet plane. What is the speed of the latter with respect to an observer on ground?
Speed of the jet airplane, vjet = 500 km/h
Relative speed of its products of combustion with respect to the plane,
vsmoke = – 1500 km/h
Speed of its products of combustion with respect to the ground = vsmoke
Relative speed of its products of combustion with respect to the airplane,
vsmoke = vsmokevjet
– 1500 = vsmoke – 500
vsmoke = – 1000 km/h
The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.
#### Question 3.6:
A car moving along a straight highway with a speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Initial velocity of the car, u = 126 km/h = 35 m/s
Final velocity of the car, v = 0
Distance covered by the car before coming to rest, s = 200 m
Retardation produced in the car = a
From third equation of motion, a can be calculated as:
From first equation of motion, time (t) taken by the car to stop can be obtained as:
#### Question 3.7:
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h–1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m/s2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?
For train A:
Initial velocity, u = 72 km/h = 20 m/s
Time, t = 50 s
Acceleration, aI = 0 (Since it is moving with a uniform velocity)
From second equation of motion, distance (sI)covered by train A can be obtained as:
= 20 × 50 + 0 = 1000 m
For train B:
Initial velocity, u = 72 km/h = 20 m/s
Acceleration, a = 1 m/s2
Time, t = 50 s
From second equation of motion, distance (sII)covered by train A can be obtained as:
Hence, the original distance between the driver of train A and the guard of train B is 2250 – 1000 = 1250 m
#### Question 3.8:
On a two-lane road, car A is travelling with a speed of 36 km h–1. Two cars B and C approach car A in opposite directions with a speed of 54 km h–1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Velocity of car A, vA = 36 km/h = 10 m/s
Velocity of car B, vB = 54 km/h = 15 m/s
Velocity of car C, vC = 54 km/h = 15 m/s
Relative velocity of car B with respect to car A,
vBA = vBvA
= 15 – 10 = 5 m/s
Relative velocity of car C with respect to car A,
vCA = vC – (– vA)
= 15 + 10 = 25 m/s
At a certain instance, both cars B and C are at the same distance from car A i.e.,
s = 1 km = 1000 m
Time taken (t) by car C to cover 1000 m =
Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s.
From second equation of motion, minimum acceleration (a) produced by car B can be obtained as:
#### Question 3.9:
Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h–1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?
Let V be the speed of the bus running between towns A and B.
Speed of the cyclist, v = 20 km/h
Relative speed of the bus moving in the direction of the cyclist
= Vv = (V – 20) km/h
The bus went past the cyclist every 18 min i.e., (when he moves in the direction of the bus).
Distance covered by the bus = … (i)
Since one bus leaves after every T minutes, the distance travelled by the bus will be equal to
Both equations (i) and (ii) are equal.
Relative speed of the bus moving in the opposite direction of the cyclist
= (V + 20) km/h
Time taken by the bus to go past the cyclist
From equations (iii) and (iv), we get
Substituting the value of V in equation (iv), we get
#### Question 3.10:
A player throws a ball upwards with an initial speed of 29.4 m s–1.
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s–2 and neglect air resistance).
(a) Downward
(b) Velocity = 0, acceleration = 9.8 m/s2
(c) x > 0 for both up and down motions, v < 0 for up and v > 0 for down motion, a > 0 throughout the motion
(d) 44.1 m, 6 s
Explanation:
(a) Irrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth.
(b) At maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., 9.8 m/s2.
(c) During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive.
(d) Initial velocity of the ball, u = 29.4 m/s
Final velocity of the ball, v = 0 (At maximum height, the velocity of the ball becomes zero)
Acceleration, a = – g = – 9.8 m/s2
From third equation of motion, height (s) can be calculated as:
From first equation of motion, time of ascent (t) is given as:
Time of ascent = Time of descent
Hence, the total time taken by the ball to return to the player’s hands = 3 + 3 = 6 s.
#### Question 3.11:
Read each statement below carefully and state with reasons and examples, if it is true or false;
A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration mustbe speeding up.
(a) True
(b) False
(c) True
(d) False
Explanation:
(a) When an object is thrown vertically up in the air, its speed becomes zero at maximum height. However, it has acceleration equal to the acceleration due to gravity (g) that acts in the downward direction at that point.
(b) Speed is the magnitude of velocity. When speed is zero, the magnitude of velocity along with the velocity is zero.
(c) A car moving on a straight highway with constant speed will have constant velocity. Since acceleration is defined as the rate of change of velocity, acceleration of the car is also zero.
(d) This statement is false in the situation when acceleration is positive and velocity is negative at the instant time taken as origin. Then, for all the time before velocity becomes zero, there is slowing down of the particle. Such a case happens when a particle is projected upwards.
This statement is true when both velocity and acceleration are positive, at the instant time taken as origin. Such a case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.
#### Question 3.12:
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Ball is dropped from a height, s = 90 m
Initial velocity of the ball, u = 0
Acceleration, a = g = 9.8 m/s2
Final velocity of the ball = v
From second equation of motion, time (t) taken by the ball to hit the ground can be obtained as:
From first equation of motion, final velocity is given as:
v = u + at
= 0 + 9.8 × 4.29 = 42.04 m/s
Rebound velocity of the ball, ur =
Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
v = ur + at
0 = 37.84 + (– 9.8) t
Total time taken by the ball = t + t′ = 4.29 + 3.86 = 8.15 s
As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time.
The velocity with which the ball rebounds from the floor
Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s
The speed-time graph of the ball is represented in the given figure as:
#### Question 3.13:
Explain clearly, with examples, the distinction between:
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first.
When is the equality sign true? [For simplicity, consider one-dimensional motion only].
(a) The magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle.
The total path length of a particle is the actual path length covered by the particle in a given interval of time.
For example, suppose a particle moves from point A to point B and then, comes back to a point, C taking a total time t, as shown below. Then, the magnitude of displacement of the particle = AC.
Whereas, total path length = AB + BC
It is also important to note that the magnitude of displacement can never be greater than the total path length. However, in some cases, both quantities are equal to each other.
(b)
Since (AB + BC) > AC, average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line.
#### Question 3.14:
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h –1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]
Time taken by the man to reach the market from home,
Time taken by the man to reach home from the market,
Total time taken in the whole journey = 30 + 20 = 50 min
Time = 50 min =
Net displacement = 0
Total distance = 2.5 + 2.5 = 5 km
Speed of the man = 7.5 km
Distance travelled in first 30 min = 2.5 km
Distance travelled by the man (from market to home) in the next 10 min
=
Net displacement = 2.5 – 1.25 = 1.25 km
Total distance travelled = 2.5 + 1.25 = 3.75 km
#### Question 3.15:
In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Instantaneous velocity is given by the first derivative of distance with respect to time i.e.,
Here, the time interval dt is so small that it is assumed that the particle does not change its direction of motion. As a result, both the total path length and magnitude of displacement become equal is this interval of time.
Therefore, instantaneous speed is always equal to instantaneous velocity.
#### Question 3.16:
Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
(a)
(b)
(c)
(d)
(a) The given x-t graph, shown in (a), does not represent one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time.
(b) The given v-t graph, shown in (b), does not represent one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.
(c) The given v-t graph, shown in (c), does not represent one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative.
(d) The given v-t graph, shown in (d), does not represent one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.
#### Question 3.17:
Figure 3.21 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.
(Fig 3.21)
The x-t graph of a particle moving in a straight line for t < 0 and on a parabolic path for t > 0 cannot be shown as the given graph. This is because, the given particle does not follow the trajectory of path followed by the particle as t = 0, x = 0. A physical situation that resembles the above graph is of a freely falling body held for sometime at a height
#### Question 3.18:
A police van moving on a highway with a speed of 30 km h–1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h–1. If the muzzle speed of the bullet is 150 m s–1, with what speed does the bullet hit the thief’s car ? (Note: Obtain that speed which is relevant for damaging the thief’s car).
Speed of the police van, vp = 30 km/h = 8.33 m/s
Muzzle speed of the bullet, vb = 150 m/s
Speed of the thief’s car, vt = 192 km/h = 53.33 m/s
Since the bullet is fired from a moving van, its resultant speed can be obtained as:
= 150 + 8.33 = 158.33 m/s
Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief’s car can be obtained as:
vbt = vbvt
= 158.33 – 53.33 = 105 m/s
#### Question 3.19:
Suggest a suitable physical situation for each of the following graphs (Fig 3.22):
(a)
(b)
(c)
(Fig: 3.22)
(a)The given x-t graph shows that initially a body was at rest. Then, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Then, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately gets stopped after sometime.
(b)In the given v-tgraph, the sign of velocity changes and its magnitude decreases with a passage of time. A similar situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.
(c)The given a-t graph reveals that initially the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This indicates that the body again starts moving with the same constant velocity. A similar physical situation arises when a hammer moving with a uniform velocity strikes a nail.
#### Question 3.20:
Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.
(Fig: 3.23)
Negative, Negative, Positive (at t = 0.3 s)
Positive, Positive, Negative (at t = 1.2 s)
Negative, Positive, Positive (at t = –1.2 s)
For simple harmonic motion (SHM) of a particle, acceleration (a) is given by the relation:
a = – ω2x ω → angular frequency … (i)
t = 0.3 s
In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Therefore, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive.
t = 1.2 s
In this time interval, x is positive. Thus, the slope of the x-t plot will also be positive. Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative.
t = – 1.2 s
In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Since both x and t are negative, the velocity comes to be positive. From equation (i), it can be inferred that the acceleration of the particle will be positive.
#### Question 3.21:
Figure 3.24 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.
(Fig: 3.24)
Interval 3 (Greatest), Interval 2 (Least)
Positive (Intervals 1 & 2), Negative (Interval 3)
The average speed of a particle shown in the x-t graph is obtained from the slope of the graph in a particular interval of time.
It is clear from the graph that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Therefore, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.
#### Question 3.22:
Figure 3.25 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?
(Fig: 3.25)
Average acceleration is greatest in interval 2
Average speed is greatest in interval 3
v is positive in intervals 1, 2, and 3
a is positive in intervals 1 and 3 and negative in interval 2
a = 0 at A, B, C, D
Acceleration is given by the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time.
Since the slope of the given speed-time graph is maximum in interval 2, average acceleration will be the greatest in this interval.
Height of the curve from the time-axis gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Hence, average speed of the particle is the greatest in interval 3.
In interval 1:
The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.
In interval 2:
The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.
In interval 3:
The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval.
Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C, and D, acceleration of the particle is zero.
#### Question 3.23:
A three-wheeler starts from rest, accelerates uniformly with 1 m s–2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?
Straight line
Distance covered by a body in nth second is given by the relation
Where,
u = Initial velocity
a = Acceleration
n = Time = 1, 2, 3, ..... ,n
In the given case,
u = 0 and a = 1 m/s2
This relation shows that:
Dnn … (iii)
Now, substituting different values of n in equation (iii), we get the following table:
n 1 2 3 4 5 6 7 8 9 10 Dn 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
The plot between n and Dn will be a straight line as shown:
Since the given three-wheeler acquires uniform velocity after 10 s, the line will be parallel to the time-axis after n = 10 s.
#### Question 3.24:
A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m/s. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m/s and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
Initial velocity of the ball, u = 49 m/s
Acceleration, a = – g = – 9.8 m/s2
Case I:
When the lift was stationary, the boy throws the ball.
Taking upward motion of the ball,
Final velocity, v of the ball becomes zero at the highest point.
From first equation of motion, time of ascent (t) is given as:
But, the time of ascent is equal to the time of descent.
Hence, the total time taken by the ball to return to the boy’s hand = 5 + 5 = 10 s.
Case II:
The lift was moving up with a uniform velocity of 5 m/s. In this case, the relative velocity of the ball with respect to the boy remains the same i.e., 49 m/s. Therefore, in this case also, the ball will return back to the boy’s hand after 10 s.
#### Question 3.25:
On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h–1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h–1. For an observer on a stationary platform outside, what is the
(a) speed of the child running in the direction of motion of the belt ?.
(b) speed of the child running opposite to the direction of motion of the belt ?
(c) time taken by the child in (a) and (b) ?
Which of the answers alter if motion is viewed by one of the parents?
(Fig: 3.26)
3.25
(a) Speed of the belt, vB = 4 km/h
Speed of the boy, vb = 9 km/h
Since the boy is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:
vbB = vb + vB = 9 + 4 = 13 km/h
(b) Since the boy is running in the direction opposite to the direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:
vbB = vb + (– vB) = 9 – 4 = 5 km/h
(c) Distance between the child’s parents = 50 m
As both parents are standing on the moving belt, the speed of the child in either direction as observed by the parents will remain the same i.e., 9 km/h = 2.5 m/s.
Hence, the time taken by the child to move towards one of his parents is .
(d) If the motion is viewed by any one of the parents, answers obtained in (a) and (b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., 9 km/h.
For this reason, it can be concluded that the time taken by the child to reach any one of his parents remains unaltered.
#### Question 3.26:
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m/s2. Give the equations for the linear and curved parts of the plot.
For first stone:
Initial velocity, uI = 15 m/s
Acceleration, a = –g = – 10 m/s2
Using the relation,
When this stone hits the ground, x1 = 0
∴– 5t2 + 15t + 200 = 0
t2 – 3t – 40 = 0
t2 – 8t + 5t – 40 = 0
t (t – 8) + 5 (t – 8) = 0
t = 8 s or t = – 5 s
Since the stone was projected at time t = 0, the negative sign before time is meaningless.
t = 8 s
For second stone:
Initial velocity, uII = 30 m/s
Acceleration, a = –g = – 10 m/s2
Using the relation,
At the moment when this stone hits the ground; x2 = 0
– 5t2 + 30 t + 200 = 0
t2 – 6t – 40 = 0
t2 – 10t + 4t + 40 = 0
t (t – 10) + 4 (t – 10) = 0
t (t – 10) (t + 4) = 0
t = 10 s or t = – 4 s
Here again, the negative sign is meaningless.
t = 10 s
Subtracting equations (i) and (ii), we get
Equation (iii) represents the linear path of both stones. Due to this linear relation between (x2 x1) and t, the path remains a straight line till 8 s.
Maximum separation between the two stones is at t = 8 s.
(x2x1)max = 15× 8 = 120 m
This is in accordance with the given graph.
After 8 s, only second stone is in motion whose variation with time is given by the quadratic equation:
x2x1 = 200 + 30t – 5t2
Hence, the equation of linear and curved path is given by
x2 x1 = 15t (Linear path)
x2 – x1 = 200 + 30t – 5t2 (Curved path)
#### Question 3.27:
The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.
(Fig. 3.28)
What is the average speed of the particle over the intervals in (a) and (b)?
(a) Distance travelled by the particle = Area under the given graph
Average speed =
(b) Let s1 and s2 be the distances covered by the particle between time
t = 2 s to 5 s and t = 5 s to 6 s respectively.
Total distance (s) covered by the particle in time t = 2 s to 6 s
s = s1 + s2 … (i)
For distance s1:
Let u′ be the velocity of the particle after 2 s and a′ be the acceleration of the particle in t = 0 to t = 5 s.
Since the particle undergoes uniform acceleration in the interval t = 0 to t = 5 s, from first equation of motion, acceleration can be obtained as:
v = u + at
Where,
v = Final velocity of the particle
12 = 0 + a′ × 5
Again, from first equation of motion, we have
v = u + at
= 0 + 2.4 × 2 = 4.8 m/s
Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s
For distance s2:
Let a″ be the acceleration of the particle between time t = 5 s and t = 10 s.
From first equation of motion,
v = u + at (where v = 0 as the particle finally comes to rest)
0 = 12 + a″ × 5
Distance travelled by the particle in 1s (i.e., between t = 5 s and t = 6 s)
From equations (i), (ii), and (iii), we get
s = 25.2 + 10.8 = 36 m
#### Question 3.28:
The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:
Which of the following formulae are correct for describing the motion of the particle over the time-interval t2 to t1?
(a) x(t2) = x(t1) + v(t1)(t2–t1) + ()a(t2–t1)2
(b) v(t2)= v(t1) + a(t2–t1)
(c) vAverage = (x(t2) – x(t1)) / (t2 – t1)
(d) aAverage = (v(t2) – v(t1)) / (t2 – t1)
(e) x(t2) = x(t1) + vAverage(t2 t1) + ()aAverage(t2 t1)2
(f) x(t2) – x(t1) = area under the v–t curve bounded by the t-axis and the dotted line shown.
The correct formulae describing the motion of the particle are (c), (d) and, (f)
The given graph has a non-uniform slope. Hence, the formulae given in (a), (b), and (e) cannot describe the motion of the particle. Only relations given in (c), (d), and (f) are correct equations of motion.
View NCERT Solutions for all chapters of Class 11 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8090594410896301, "perplexity": 632.6206375173484}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400213454.52/warc/CC-MAIN-20200924034208-20200924064208-00579.warc.gz"} |
http://mathhelpforum.com/calculus/172761-integration.html | Math Help - Integration
1. Integration
$\int^{4}_{0} \log(1+\tan x)dx$
2. Originally Posted by kjchauhan
$\int^{4}_{0} \log(1+\tan x)dx$
note that $y = \log(1+\tan{x})$ has an infinite discontinuity between $x = 0$ and $x = 4$, specifically, at $x = \dfrac{\pi}{2}$, making it improper. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9922023415565491, "perplexity": 1336.4228996851864}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375095775.68/warc/CC-MAIN-20150627031815-00236-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/123448/what-is-the-standard-basis-for-fields-of-complex-numbers | # What is the “standard basis” for fields of complex numbers?
What is the "standard basis" for fields of complex numbers?
For example, what is the standard basis for $\Bbb C^2$ (two-tuples of the form: $(a + bi, c + di)$)? I know the standard for $\Bbb R^2$ is $((1, 0), (0, 1))$. Is the standard basis exactly the same for complex numbers?
P.S. - I realize this question is very simplistic, but I couldn't find an authoritative answer online.
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@Sid: I don't see what that has to do with anything. I assume $\mathbb{C}^2$ is to be understood as a complex vector space. – Qiaochu Yuan Mar 22 '12 at 22:05
@QiaochuYuan, yes, sorry, that wasn't a particularly relevant response! – Sid Raval Mar 22 '12 at 22:11
The title still sounds vague. Will someone please edit it? – user21436 Mar 23 '12 at 5:09
The "most standard" basis is also $\left\lbrace(1,0),\, (0,1)\right\rbrace$. You just take complex combinations of these vectors.
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Makes good sense, just didn't realize if that was considered the "standard". – Casey Patton Mar 22 '12 at 22:07
Yes. In fact, that is "the standard basis" for $\mathbb{F}^2$ where $\mathbb{F}$ is any field: $\mathbb{F}=\mathbb{R},\mathbb{C},\mathbb{Q},\mathbb{Z}_p,$ etc. – Bill Cook Mar 22 '12 at 22:11
@JuanBermejoVega How is the set ${(1,0), (0,1)}$ a basis for $\mathbb{C}$? It only spans the real part of each of the complex plane. – krismath Oct 9 '14 at 16:30
@krismath Are you only taking real combinations of those vectors? In a complex vector space you should take complex combinations. Does this answer your question? – Juan Bermejo Vega Oct 10 '14 at 20:18
@JuanBermejoVega Oh I see. Thanks. – krismath Oct 11 '14 at 0:23
Just to be clear, by definition, a vector space always comes along with a field of scalars $F$. It's common just to talk about a "vector space" and a "basis"; but if there is possible doubt about the field of scalars, it's better to talk about a "vector space over $F$" and a "basis over $F$" (or an "$F$-vector space" and an "$F$-basis").
Your example, $\mathbb{C}^2$, is a 2-dimensional vector space over $\mathbb{C}$, and the simplest choice of a $\mathbb{C}$-basis is $\{ (1,0), (0,1) \}$.
However, $\mathbb{C}^2$ is also a vector space over $\mathbb{R}$. When we view $\mathbb{C}^2$ as an $\mathbb{R}$-vector space, it has dimension 4, and the simplest choice of an $\mathbb{R}$-basis is $\{(1,0), (i,0), (0,1), (0,i)\}$.
Here's another intersting example, though I'm pretty sure it's not what you were asking about:
We can view $\mathbb{C}^2$ as a vector space over $\mathbb{Q}$. (You can work through the definition of a vector space to prove this is true.) As a $\mathbb{Q}$-vector space, $\mathbb{C}^2$ is infinite-dimensional, and you can't write down any nice basis. (The existence of the $\mathbb{Q}$-basis depends on the axiom of choice.)
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I'm not sure that this is what you want, but under the usual Argand-Gauss identification $\Bbb C=\Bbb R^2$ the standard basis of $\Bbb C$ would be $\{1,i\}$, the standard basis of $\Bbb C^2$ would be $\{(1,0),(i,0),(0,1),(0,i)\}$ and so on.
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This is a little confusing, because the previous answer gave me a basis of dimension 2 and this answer gives me a basis of dimension 4. How can this be possible? – Casey Patton Mar 22 '12 at 22:28
What is not clear (to me, at least) from your question is that you consider $\Bbb C^2$ as a real or complex vector space. As a complex vector space it has dimension $2$, as a real vector space it has dimension $4$. – Andrea Mori Mar 22 '12 at 22:37
Ah gotcha. Well....being a student in an introductory Linear Algebra class, I haven't actually learned what those terms mean yet! Hence the confusing question. – Casey Patton Mar 23 '12 at 23:04
## protected by Community♦May 14 at 17:58
Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9468768835067749, "perplexity": 327.62502221902037}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988718.8/warc/CC-MAIN-20150728002308-00220-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://lambda-the-ultimate.org/node/811 | ## Lambda-mu
Either I cannot search, or the term has web-unfriendly name, but it's pretty tough to look for lambda-mu calculus, even in scope of LtU only.
For example, did we discuss this paper or related?
Control Categories and Duality: on the Categorical Semantics of the Lambda-Mu Calculus Just one of the results:
As a corollary, we obtain a syntactic duality result: there exist syntactic translations between call-by-name and call-by-value which are mutually inverse and which preserve the operational semantics.
...and...
It is interesting to compare this with Filinski’s work, in which he obtains a duality result by working with a larger and more symmetric syntax, in which the dual of a term is essentially its mirror image.
Also, is Parigot's Lambda-mu-calculus: an algorithmic interpretation of classical natural deduction available online anywhere (except ACM)?
[on edit: aha, found one reference (actually a pair forming one reference): Call-by-Value is Dual to Call-by-Name and Call-by-value is Dual to Call-by-name, Reloaded ("We consider the relation of the dual calculus of Wadler to the lambda-mu-calculus of Parigot")] | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8558422923088074, "perplexity": 2119.7917885489364}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398468233.50/warc/CC-MAIN-20151124205428-00003-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/phase-change-problem.131641/ | # Phase Change Problem!
1. Sep 11, 2006
### DanielT
I am working on a proble with this basic set up:
500g of some material starts at -22'C in a 400g container. 90g of steam (water) starts at 120'C and is mixed with the unknown material. Find the final temp. and amount of material and water in each state.
Unkown
c(solid) = 0.4 cal/g'c
Melting = 20'c
c(liquid) = 0.9 cal/g'c
Boiling = 80'c
c(vapor) = 0.6 cal/g'c
Hf = 100 cal/g
Hv = 600 cal/g
***Specific Heat of container is 0.2 cal/g'c
Can someone point me in the right direction, OR work it out if you are so inclined!
2. Sep 11, 2006
### Staff: Mentor
Well its a matter of starting with the sample at 500 g and the container of 400 g, at -22°C and bringing both to 20°C, so delta-T=2°C. Calculate that energy by mass * specific heat of each. That heat comes from the steam, which cools from 120° to some temperature. If above 100°C then it is still superheated steam. If the T is less than 100°C (assuming 1 atm of pressure), then set T=100°C, so delta-T = 20°C, and the remainder of the heat would come from condensation from vapor to liquid.
As vapor cools to liquid, it does so at constant temperature, in the process of condensation. The opposite, vaporization, also occurs at constant temperature. One has to determine the amount of steam (vapor) and liquid. The fraction of steam is called the quality.
As a substance melts from solid to liquid, it does so at constant temperature. Melting is the opposite of freezing or solidification, and heat of fusion is the thermal property.
Where the sample melts at 20°C, assume all heat is absorbed by the unknown, while the container stays at 20°C. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.907504141330719, "perplexity": 1860.8961920570555}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806509.31/warc/CC-MAIN-20171122065449-20171122085449-00225.warc.gz"} |
http://tex.stackexchange.com/questions/104211/contour-around-multi-line-tikz-text | # Contour around multi-line tikz text
I am trying to produce a text inside a node which has a contour. Problem is: when using contour, text width seems to be ignored, so I can not use auto multi-line text.
I discovered contours in tikz halo around text, and I can use it correctly as in:
\draw (0, 0) node {\contour{white}{A very long title that we mean to split on multiple lines}};
but when adding text width it is not multi-lined:
\draw (0, 0) node[text width=5cm] {\contour{white}{A very long title that we mean to split on multiple lines}};
And without contour it works correctly:
\draw (0, 0) node[text width=5cm] {A very long title that we mean to split on multiple lines};
How can I fix this?
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Apparently contour puts everything in a box. Just try this, out of TikZ picture. \contour{red}{A very long text that hopefully breaks at the end of the this line otherwise I need to write more stuff}. So I think, the pragmatic approach is to break manually. – percusse Mar 25 '13 at 13:59
Same problem with shadowtext package. Very annoying. It seems there is no way of getting wrapping contour or shaded multiline text inside a TikZ node. – mmj Apr 15 '13 at 14:08
Judging from the output of the pdfrender package, it seems that it does the same as the following TikZ code (which is coded as the opposite of optimal) and as the contour package.
## Code
\documentclass[tikz]{standalone}
\tikzset{
\pgfkeysalso{/tikz/.cd,#1}%
\foreach \angle in {0,5,...,359}{
\node[#1,text=white] at ([shift={(\angle:.5pt)}] #4){#5};
}
}
}
\begin{document}
\begin{tikzpicture}
\fill[gray, step=.5mm] (-1.6,-1) rectangle (1.6,1);
\node[
text shadow={[align=center,text width=3cm] at (0,0) {A very long title that we mean to split on multiple lines \ldots}}]
at (0,0) {A very long title that we mean to split on multiple lines \ldots};
\end{tikzpicture}
\end{document}
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.982823371887207, "perplexity": 3029.0268337491093}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398460519.28/warc/CC-MAIN-20151124205420-00339-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/110983/why-was-quantum-mechanics-regarded-as-a-non-deterministic-theory | # Why was quantum mechanics regarded as a non-deterministic theory?
It seems to be a wide impression that quantum mechanics is not deterministic, e.g. the world is quantum-mechanical and not deterministic.
I have a basic question about quantum mechanics itself. A quantum-mechanical object is completely characterized by the state vector. The time-evolution of state vector is perfectly deterministic. The system, equipment, environment, and observer are part of the state vector of universe. The measurements with different results are part of state vector at different spacetime. The measurement is a complicated process between system and equipment. The equipment has $10^{23}$ degrees of freedom, the states of equipment we neither know nor able to compute. In this sense, the situation of QM is quite similar with statistical physics. Why can't the situation just like statistical physics, we introduce an assumption to simply calculation, that every accessible microscopic state has equal probability? In QM, we also introduce an assumption about the probabilistic measurement to produce the measurement outcome.
PS1: If we regarded non-deterministic is intrinsic feature of quantum mechanics, then the measurement has to disobey the Schrödinger picture.
PS2: The bold phase argument above does not obey the Bell's inequality. In the local hidden variable theory from Sakurai's modern quantum mechanics, a particle with $z+$, $x-$ spin measurement result corresponds to $(\hat{z}+,\hat{x}-)$ "state". If I just say the time-evolution of universe is $$\hat{U}(t,t_0) \lvert \mathrm{universe} (t_0) \rangle = \lvert \mathrm{universe} (t) \rangle.$$ When the $z+$ was obtained, the state of universe is $\lvert\mathrm{rest} \rangle \lvert z+ \rangle$. Later the $x-$ was obtained, the state of universe is $\lvert\mathrm{rest}' \rangle \lvert x- \rangle$. It is deterministic, and does not require hidden-variable setup as in Sakurai's book.
PS3: My question is just about quantum mechanics itself. It is entirely possible that the final theory of nature will require drastic modification of QM. Nevertheless it is outside the current question.
PS4: One might say the state vector is probabilistic. However, the result of measurement happens in equipment, which is a part of total state vector. Given a probabilistic interpretation in a deterministic theory is logical inconsistent.
• Quantum mechanics is deterministic, but it is also probabilistic -- i.e. you can deterministically calculate the probability of a random event happening. This is to distinguish it from non-deterministic (i.e. stochastic) systems where you do not generally have "one" solution but an entire family of solutions depending on random variables. – webb May 2 '14 at 22:33
• If I know the wavefunction, or state vector, more generally, of the universe, then I don't need the probability anymore – user26143 May 3 '14 at 7:34
• If you know the state vector of the universe, then this still doesn't give you information about exact outcome of any quantum experiment — only probabilities. – Ruslan May 3 '14 at 7:56
• If the equipment and system are governed by the Schrodinger picture, there is no (strict, by means of not in the sense happened in statistical mechanics) probability. If there is (strict) probability, then the Schrodinger picture is incomplete. – user26143 May 3 '14 at 8:22
• It is not clear what you are asking. Quantum theory is non-deterministic in the sense that it works with objects ($\psi$ functions, kets) that can be used to calculate probabilities, not the actual results. It is the same as in statistical physics, only probabilistic statements can be derived. – Ján Lalinský May 3 '14 at 9:53
I agree with much of what you write in your question. Whether quantum mechanics is considered to be deterministic is a matter of interpretation, summarised in this wiki comparison of interpretations. The wiki definition of determinism is this context, which I think is entirely satisfactory, is
Determinism is a property characterizing state changes due to the passage of time, namely that the state at a future instant is a function of the state in the present (see time evolution). It may not always be clear whether a particular interpretation is deterministic or not, as there may not be a clear choice of a time parameter. Moreover, a given theory may have two interpretations, one of which is deterministic and the other not.
In, for example, many-worlds interpretation, time evolution is unitary and is governed entirely by Schrödinger’s equation. There is nothing like the "collapse of the wave-function" or a Born rule for probabilities.
In other interpretations, for example, Copenhagen, there is a Born rule, which introduces a non-deterministic collapse along with the deterministic evolution of the wave-function by Schrödinger’s equation.
In your linked text, the author writes that quantum mechanics is non-deterministic. I assume the author rejects the many-worlds and other deterministic interpretations of quantum mechanics. Aspects of such interpretations remain somewhat unsatisfactory; for example, it is difficult to calculate probabilities correctly without the Born rule.
• The problem is that in many-worlds interpretation there is no deterministic connection between the state and the observed behavior since theoretically all branches co-exist, but in practice only one is observed. "Determinism" is a linguistic sleight of hand. Bohmian mechanics is indeed deterministic, but it involves faster than light signals and ephemeral entities (Bohmian particles) unobservable in principle, like ether. For that matter Everett's branches are much like ether as well, and play the same role as Bohmian particles. – Conifold Apr 28 '16 at 1:46
Quantum mechanics is non deterministic of actual measurements even in a gedanken experiment because of the Heisenberg Uncertainty Principle, which in the operator representation appears as non commuting operators. It is a fundamental relation of quantum mechanics:
If you measure the position accurately, the momentum is completely undefined.
The interpretation of the solutions of Schrodinger's equation as predicting the behavior of matter depends on the postulates: the state function determined by the equation is a probability distribution for finding the system under observation with given energy and coordinates. This does not change if large ensembles are considered except computationally. The probabilistic nature will always be there as long as the theory is the same.
• You are wrong. The HUP is not optional. The total universe obeys the HUP postulate so as far as the theory of quantum mechanics goes, which is what you are asking, it will always be indeterminate by construction of the theory. It was constructed to fit observations and if you extrapolate to the total universe it makes no difference. (You said you are not considering other theories ) – anna v May 3 '14 at 12:02
• When measuring one particle's x then going to the next, their momentum will be indeterminate and "next" will have a whole phase space to be chosen from because momentum determines the next probability of x, not a point but a probability of being found at that point, whether 1 2 3 or infinite number of particles. – anna v May 3 '14 at 12:48
• The HUP is a postulate incorporated into the mathematics of commutators. – anna v May 3 '14 at 12:50
• No, the schrodinger picture gives a probability of finding any measurement value, not a fixed value of the momentum. One has to operate on the schrodinger state function, with the momentum operator to get the momentum, and the operation.measurement will give a value within the probability envelope. – anna v May 3 '14 at 13:05
• HUP isn't critical to determinism, the key point is born rule/wave function collapse. i think this answer is off target. – innisfree May 3 '14 at 19:11
The difference between statistical physics and quantum mechanics is that, in statistical physics, it is always reasonable to either measure a quantity, or demonstrate that the effect of that quantity can be bundled into an easy to work with random variable, often through the use of the Central Limit Theorem. In such situations, it can be shown that the answer will be a deterministic answer plus a small perturbation from the random variables with a 0 expectation and a very small variance.
In quantum mechanics, the interesting properties show up in situations where its not possible to measure a quantity and not plausible to bundle it up into a random variable using the central limit theorem. Sometimes you can, of course: in particular, this approach works well in modeling an quantum mechanic system which is already well modeled in classical physics. For the most part, we don't observe many quantum effects in day to day life! However, quantum mechanics is focused on the more interesting regions where those unmeasurable quantities have an important impact on the outcome of the system.
As an example, in many entanglement scenarios, you can get away with ignoring the correlation between the states of the particles. This is good, because in theory, there's some small level of entanglement between all particles that have interacted, and its good to know that we can often get away with ignoring this, and treating the values as simple independent and identically distributed variables. However, in the entanglement cases quantum mechanics are interested in, we intentionally explore situations where the entanglement is strong enough that that correlation can't just be handwaved away and still yield experimentally validated results. We are obliged to carry it through our equations if we want to provide a good model of reality.
There are many ways to do this, and one of the dividing lines regarding the topic is the line drawn between the different interpretations of QM. Some of them hold to a deterministic model, others hold to non-deterministic arguments (the Copenhagen interpretation being an example). In general, the models which are deterministic have to give up something else which is valued by physicists. The many-worlds theory gets away with being deterministic by arguing that every possible outcome of every classical observation occurs, in its own universe. This is consistent with the equations that we believe are a good model of quantum mechanics, but comes with strange side effects when applied to the larger world (quantum suicide, for instance). The Copenhagen interpretation is, in my opinion, the most natural interpretation in that it dovetails with the way we do classical physics smoothly, without any pesky alternate realities. I have found that mere mortals are most comfortable with the intuitive leaps of the Copenhagen interpretation, as compared to the intuitive leaps of other interpretations. However, the Copenhagen interpretation is decidedly non-deterministic. Because this one seems easier to explain to many people, it has achieved a great deal of notoriety, so its non-determinism gets applied to all of quantum mechanics via social mechanisms (which are far more complicated than any quantum mechanisms!)
So you can pick any interpretation you please. If you like determinism, there are plenty of options. However, one cannot use many of the basic tools of statistical mechanics to handle quantum scenarios because the basic physics of quantum mechanics leads to situations where the basic assumptions of statistical mechanics become untenable. Your example of the result of the measurement happening in the equipment is an excellent example. Like in statistical physics, the state of the measurement equipment can be modeled as a state vector, and it turns out that it's a very reasonable assumption to assume that it is randomly distributed. However, equipment designed to measure quantum effects is expressly designed to strongly correlate with the state of the particle under observation before measurement began. When the measurement is complete, the distribution of the state of the measurement equipment is decidedly poorly modeled as a state plus a perturbation with a small variance. The distribution is, instead, a very multimodal distribution, because it was correlated to the state of the particle, and most of the interesting measurements we want to take are those of a particle whose [unmeasured] state is well described by a multimodal distribution.
If you learn Quantum Mechanics you will see that the observables of any quantum system depend on the state of the system(final, initial, ground state or excited state). In theory, there are a number of interpretations of Quantum Mechanics wiki, link.
The mathematical formulation of quantum mechanics is built onto the notions of an operators. When you do a measurement you perturb the system state by applying an operator on it. The eigenvalue of the operator corresponds to the measured value of the system observable. However, each eigenvalue have a certain probability, and therefore by measuring(applying) an operator on the state system there will be a finite(or infinite) number of final states, each of them with a given probability. This is the essence of non-deterministic in quantum mechanics.
The next question arises:how the non-deterministic applies on large scale universe and the "length" of the not-deterministic" phenomena in the universe?
Because in classical theory(like general relativity, electromagnetism), you have for example the Einstein equations which govern the dynamics and they are full deterministic.
Forget interpretations. The predictions of quantum mechanics - which agree with all interpretations (by definition of 'interpretation')- does not allow prediction of experimental/observational outcomes no matter how much information is gathered about initial conditions. (You can't even get the classical information needed in classical physics because of the uncertainty principle). None of the interpretations challenge this, not even in principle. According to the math, which is wildly successful in it's predictions, a given present does not determine the future. That's why quantum mechanics is said to be indeterministic, not because of any interpretation. It doesn't matter if you believe in wave function collapse or not or other worlds or not or whatever. Saying the theory is deterministic because of some math involved in the calculation isn't related to the fact that experimental outcomes cannot be predicted, The present does not determine the future.
The quantum state of a system is completely characterized by a state vector only when the system is a pure state. The state vector evolves in two different ways described by two postulates: the Schrödinger postulate (valid when there is no measurements) and the measurement postulate. The Schrödinger postulate describes a deterministic and reversible evolution $U$. The measurement postulate describes a non-deterministic and irreversible evolution $R$.
$R$ is not derivable from $U$. In fact $R$ is incompatible with $U$, and that is the reason why the founder fathers introduced two evolution postulates in QM. Indeed, assuming an initial superposition of two states for the composite supersystem (system + apparatus + environment)
$$|\Psi\rangle = a |A\rangle + b |B\rangle$$
the result of a measurement is either $|A\rangle$ or $|B\rangle$, but because these states are orthogonal, they cannot both have evolved from a single initial state by a deterministic, unitary evolution, since that $|A\rangle = U |\Psi\rangle$ and $|B\rangle = U |\Psi\rangle$ implies $\langle A|B\rangle = \langle\Psi |U^{*} U | \Psi\rangle = 1$, which is incompatible with the requirement of ortohogonality.
So, if the result of the measurement was $|B\rangle$, the evolution was $|B\rangle = R |\Psi\rangle$.
The fact that QM is probabilistic and not deterministic is forced by the 4 rules stated below. This rules can not coexist logically to provide determinism. They lead without effort to the probablistic interpretation.
Yes, unfortunately (for me) I am not a physicist. So take this with a grain of salt.
@Quantum world:
1) Entities have a 'spread' existence. (A kind of 'field of energy' which tries to 'fill' all space).
2) Entities have some 'oscilatory' existence. (Which gives rise to 'interference' phenomena).
3) Interactions between entities are 'discrete'. (They exchange 'quanta' of somestuff).
4) Interactions use the 'minimum amount' of some 'energy stuff'.
The interplaying of these facts is what gives rise to the non-determinism (probability) in QM.
Let's think of a simple example:
Suppose you have 3 entities A, B and C (a 1 sender & 2 receivers scenario), where A is the source of some perturbation to be sent to B and C at the 'same time'. Let's think of the perturbation in practical terms (i.e.: money) and assign it a unit of measure (dollars).
Now how would A send 2 dollars total to both of them (B & C)?
Well, A should give them 1 dollar each and problem solved!!!. However, there is a constraint here (remember #4) and that is: Interactions are only done with minimun currency!!!'.
With that in mind, how can A give B and C one cent (minimun currency) at the same time? Well, .. It can't!!!
At each time (interaction) A must choose between B or C to give away every cent until completes the 2 dollars to both of them. And if you think a little bit about it, you realize that the only objective solution for A must be to throw an imaginary coin each time to decide whom will receive the 1 cent!. [Of course, for this 1 sender & 2 receivers situation, a coin with 2 faces fits rigth!. But for others scenarios, the coin or dice will have to change.]
In the analog world of classical mechanics, A would send an infinite small amount of money to both of them (no minimum currency constraint and at the same time!) and what we will see is a beautiful continuous growing of B and C money pockets. No need to deal with probabilities!!!!.
If you think carefully, in plain simple terms, probability arise from the discrete nature of interactions between entities. This is the real deal which turns everything so strange and interesting.
[Hope this general and somewhat vague answer gives you a clue about why probability arise in the description offered by QM]
The question now is: Why it has to be like that? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.888778567314148, "perplexity": 390.75294686429095}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986726836.64/warc/CC-MAIN-20191020210506-20191020234006-00530.warc.gz"} |
http://aimsciences.org/search/author?author=Jiahui%20%20Zhu | # American Institue of Mathematical Sciences
## Journals
DCDS-B
We study a class of abstract nonlinear stochastic equations of hyperbolic type driven by jump noises, which covers both beam equations with nonlocal, nonlinear terms and nonlinear wave equations. We derive an Itô formula for the local mild solution which plays an important role in the proof of our main results. Under appropriate conditions, we prove the non-explosion and the asymptotic stability of the mild solution.
keywords: Itô formula. local and global mild solution Stochastic nonlinear beam equation Lyapunov function Poisson random measure
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http://www.isc.senshu-u.ac.jp/~the0774/nyumon.htm | 2016Nxi28NxjCw[~i[̗Cɂ 2016N621XV yPz@Ȗڂ̓BڕW uCẘwƂĂ̎oACwŊwԂƂ̈ӖjwԂƂƂɁAwŊwԂ߂̊{IȋZ@AȂ킿A̎W@Em[ĝƂEW̍쐬@E_|[g̍쐬@KBviwu`vxjB yQz@ƂŎgޓe iPjw_ \ wŊwԂƂ̈Ӌ`w_Ȃǂ̍u`B iQjCw̗j \ DVDƐCw̗jizzj𗘗pāACw̗jwԁB iRj }كcA[F}ٗp̌ʂAvAЂ̌A}ّSʂ̊p@ @@}و̈ēɂ莑̔zˏꏊ̎{݂wAe{݂̗p@A[ɂ鏑ЁEGEv̌Eo@wK܂B @ iSj[KFȊwZ^[[̌oσf[^փANZXÅp@ @@ȓvǁEvZ^[ (http://www.stat.go.jp/)փANZXāAoσf[^̉{A_E[hs܂BɃGNZgāAt@C̕ҏWAOt̍쐬Aă[pẴt@C̑t@wK܂B iӁj@[ḰAQNXŎ{܂BPEQJuNX͂XUQ[i9قUKjARJuNX͒[DiXقSKjōs܂ [K̎͂ł yRz@75̃R[X̋@ NEW! NXɂĈقȂ܂̂ŒӂĂB 11g`14giPjFSQUiSقQKj 15g`18giPjFSQXiSقQKj 19g`22giQjFPQPiPقQKj 23g`26giQjFPQQiPقQKj 27g`30giRjFPQPiPقQKj 31g`34giRjFPQQiPقQKj ySz@Sҕʂ̐}ٌȂǂ̓ }كcA[ A[K̓́AS҂ɂĈقȂ܂B̗\\ŊmFĉB 28NxCw[~i[\\ioϊwoϊwȁj 1 Q 3 4 5 6 7 8 9 10 11 12 13 14 15 S NX j 4/12 4/19 4/26 5/3 5/10 5/17 5/24 5/31 6/7 6/14 6/21 6/28 7/5 7/12 7/19 Vc@ 11g 1 [1] [2] [4] @ @ @ [3] @ @ @ @ @ [5] @ [6] zc@ 12g [1] [2] [4] @ @ @ [3] @ @ @ @ @ [5] @ [6] P@Y 13g [1] [2] @ @ [4] [3] @ @ @ @ @ @ [5] @ [6] R@_ 14g [1] [2] @ @ [4] [3] @ @ @ @ @ @ [5] @ [6] @Ĝ 15g [1] [3] [2] @ @ [4] @ @ @ @ @ @ [5] @ [6] c@O 16g [1] [2] @ @ @ [4] [3] @ @ @ @ @ [5] @ [6] ec@mF 17g [1] [2] @ @ @ [3] [4] @ @ @ @ @ [5] @ [6] @͏r 18g [1] [2] @ @ @ [3] [4] @ @ @ @ @ [5] @ [6] r@ 19g 2 [1] [2] [4] @ @ [3] @ @ @ @ @ @ [5] @ [6] @` 20g [1] [2] [4] @ @ [3] @ @ @ @ @ @ [5] @ [6] N@GY 21g [1] [2] @ @ [4] @ [3] @ @ @ @ @ [5] @ [6] @ 22g [1] [2] @ @ [4] @ [3] @ @ @ @ @ [5] @ [6] @וv 23g [1] [2] @ @ @ [4] [3] @ @ @ @ @ [5] @ [6] i]@a 24g [1] [2] @ @ @ [4] [3] @ @ @ @ @ [5] @ [6] ꎓ@j 25g [1] [2] @ @ [3] @ [4] @ @ @ @ @ [5] @ [6] {{@ 26g [1] [2] @ @ [3] @ [4] @ @ @ @ @ [5] @ [6] c@Ĉ 27g 3 [1] [2] [3] @ [4] @ @ @ @ @ @ @ [5] @ [6] @ޕ 28g [1] [2] [3] @ [4] @ @ @ @ @ @ @ [5] @ [6] c@V 29g [1] [2] @ @ [3] @ [4] @ @ @ @ @ [5] @ [6] c@v 30g [1] [2] @ @ [3] @ [4] @ @ @ @ @ [5] @ [6] R@ 31g [1] [2] [4] @ @ [3] @ @ @ @ @ @ [5] @ [6] @Sg 32g [1] [2] [4] @ @ [3] @ @ @ @ @ @ [5] @ [6] @ 33g [1] [2] @ @ @ [4] [3] @ @ @ @ @ [5] @ [6] @K 34g [1] [2] @ @ @ [4] [3] @ @ @ @ @ [5] @ [6] [1] Ƃ̐iߕEwɊւKC_XiNXƁj [2] Cw̗jieNXƁj [3] }كcA[ [4] [KiQNXŎ{j [5] R[XIɊւKC_XiNXŎ{j [6] ꂩ̑w̉߂R[XIɊւeNXł̃KC_X | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9602989554405212, "perplexity": 3469.28863936907}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171053.19/warc/CC-MAIN-20170219104611-00287-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://aimsciences.org/search/author?author=Fritz%20Gesztesy | # American Institute of Mathematical Sciences
## Journals
DCDS
We discuss the algebro-geometric initial value problem for the Ablowitz-Ladik hierarchy with complex-valued initial data and prove unique solvability globally in time for a set of initial (Dirichlet divisor) data of full measure. To this effect we develop a new algorithm for constructing stationary complex-valued algebro-geometric solutions of the Ablowitz-Ladik hierarchy, which is of independent interest as it solves the inverse algebro-geometric spectral problem for general (non-unitary) Ablowitz-Ladik Lax operators, starting from a suitably chosen set of initial divisors of full measure. Combined with an appropriate first-order system of differential equations with respect to time (a substitute for the well-known Dubrovin-type equations), this yields the construction of global algebro-geometric solutions of the time-dependent Ablowitz-Ladik hierarchy.
The treatment of general (non-unitary) Lax operators associated with general coefficients for the Ablowitz-Ladik hierarchy poses a variety of difficulties that, to the best of our knowledge, are successfully overcome here for the first time. Our approach is not confined to the Ablowitz-Ladik hierarchy but applies generally to $(1+1)$-dimensional completely integrable soliton equations of differential-difference type.
keywords: initial value problem. complex-valued solutions Ablowitz-Ladik hierarchy
IPI
We extend the classical spectral estimation problem to the infinite-dimensional case and propose a new approach to this problem using the Boundary Control (BC) method. Several applications to inverse problems for partial differential equations are provided.
keywords: Spectral estimation inverse problems boundary control method.
DCDS
A detailed description of the model Hilbert space $L^2(\mathbb{R}; d\Sigma; K)$, where $K$ represents a complex, separable Hilbert space, and $\Sigma$ denotes a bounded operator-valued measure, is provided. In particular, we show that several alternative approaches to such a construction in the literature are equivalent.
These spaces are of fundamental importance in the context of perturbation theory of self-adjoint extensions of symmetric operators, and the spectral theory of ordinary differential operators with operator-valued coefficients.
keywords: Direct integrals of Hilbert spaces model Hilbert spaces.
[Back to Top] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8935586810112, "perplexity": 447.7105975492223}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891813883.34/warc/CC-MAIN-20180222022059-20180222042059-00107.warc.gz"} |
http://zbmath.org/?q=an:0762.15016 | # zbMATH — the first resource for mathematics
##### Examples
Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.
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Linear preserver problems: A brief introduction and some special techniques. (English) Zbl 0762.15016
Let $M$ be any one of the following matrix spaces: the set of all $m×n$ matrices over the field $𝔽$, where usually $𝔽$ is $ℝ$ or $ℂ$; the set of all $n×n$ symmetric matrices over $𝔽$; the set of all $n×n$ skew-symmetric matrices over $𝔽$; the set of all Hermitian matrices. The typical linear preserving problems are:
1. let $F$ be a (scalar-valued, vector-valued, or set-valued) function on $M$. Characterize those linear operators $\varphi$ on $M$ that satisfy $F\left(\varphi \left(A\right)\right)=F\left(A\right)$ for all $A\in M$;
2. let $S\subset M$. Characterize those linear operators $\varphi$ on $M$ that satisfy $\varphi \left(S\right)=S$ or $\subset S$;
3. let $\sim$ be a relation or an equivalence relation on $M$. Characterize those linear operators $\varphi$ on $M$ that satisfy $\varphi \left(A\right)\sim \varphi \left(B\right)$ whenever $A\sim B$ (or iff $A\sim B\right)$;
4. given a transform $F:M\to M$, characterize those linear operators $\varphi$ on $M$ that satisfy $F\left(\varphi \left(A\right)\right)=\varphi \left(F\left(A\right)\right)$ for all $A\in M$.
This paper is a survey which gives a gentle introduction to these problems.
##### MSC:
15A72 Vector and tensor algebra, theory of invariants 15-02 Research monographs (linear algebra) 15A57 Other types of matrices (MSC2000) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 33, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9090225100517273, "perplexity": 3518.7339846089167}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999676283/warc/CC-MAIN-20140305060756-00068-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://math-mprf.org/journal/articles/id1152/ | Determinant Solution for the TASEP with Particle-dependent Hopping Probabilities on a Ring
#### S. Poghosyan, V.B. Priezzhev
2008, v.14, Issue 2, 233-254
ABSTRACT
We consider the totally asymmetric exclusion process on a ring in discrete time with the backward-ordered sequential update and particle-dependent hopping probabilities. Using a combinatorial treatment of the Bethe ansatz, we derive the determinant expression for the non-stationary probability of transitions between particle configurations. In the continuous-time limit, we find a generalization of the recent result, obtained by A. Rakos and G.M. Schuetz for infinite lattice, to the case of ring geometry.
Keywords: totally asymmetric exclusion process,periodic boundaryconditions,backward sequential update,Bethe ansatz | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9428178071975708, "perplexity": 2598.5428297841636}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570767.11/warc/CC-MAIN-20220808061828-20220808091828-00235.warc.gz"} |
http://informationtransfereconomics.blogspot.com/2016/01/eff-035.html | ## Tuesday, January 12, 2016
### EFF ≈ 0.35%
OK, this isn't specific to the information equilibrium model, but I think I did a pretty awesome job of estimating what the effective Fed funds rate would be after the Fed's rate rise. I predicted a value of 0.35% (scenario C, a 0.25% rise in both the ceiling and floor rate) and we get ...
Note that halfway in between 0.25 and 0.5 would be a bit higher at 0.38%.
...
Update 13 January 2016
Here's an update with a bit more analysis of the log midpoint vs midpoint versions of the estimated EFF. Here are both estimates with data from 2011-present along with error and the error distribution:
I don't think this is necessarily the proper comparison, however. There are two periods where the EFF data is significantly above the blue line. Most of 2012 and most of 2015. I think these periods might represent expectations of a rate hike -- and therefore aren't random errors, but systematic. Most of 2015 was spent asking when the Fed would finally "lift-off" the zero bound. And 2012 was the last time core PCE inflation had any quarterly data points above 2%. That optimism seems to have been terminated by the disastrous RGDP growth numbers from 2012 Q3 and Q4 (0.5% and 0.1%, respectively). In that case, the deviation above the estimate would be systematic, un-modeled effects rather than measurement error.
1. Great! How did your competitors do? There must be somebody out there who gave it a go. The SPF guys maybe?
1. No idea. I saw no predictions for the EFF.
2. I asked both Allan Gregory and Dave Giles if they are aware of any such forecasts. I'll let you know if I hear back.
3. It's not really a forecast ... it's a prior (estimate of an input parameter). The monetary base forecast is based on the model and an estimate of what the EFF would be after the rate rise.
In a sense, it was an application of Benford's law (uniform distribution over a logarithmic scale).
Jim Hamilton seems to think the jump of that size was expected/intended by the Fed:
http://econbrowser.com/archives/2015/12/managing-the-feds-balance-sheet | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.865254819393158, "perplexity": 1630.6123431224814}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676596336.96/warc/CC-MAIN-20180723110342-20180723130342-00083.warc.gz"} |
http://calculator.mathcaptain.com/simplifying-expressions-calculator.html | Algebraic Expression is an expression consists of numbers, constants, variables and algebraic operations like addition, subtraction, multiplication etc.
We can simplify the algebraic expression:
• By combining the like terms and grouping them.
• By expanding the multiplication.
• Reduced the expression into simplest form.
Before solving Algebraic Expression, we need to know about variables, parentheses, brackets, order of operations, exponents, division, fractions, square root, radicals.
## How to Simplify Expressions
Steps for simplifying algebraic expressions:
Step 1: Categorize the given algebraic expression.
Step 2: Simplify the equation and reduce the solution into simplest form.
### Equation Calculator
How to Simplify an Expression How to Simplify Expression Algebra Function Algebra Graph Algebra Quadrants Algebra Quotient
How to Simplify Rational Expressions Calculator Simplify Boolean Expression Calculator Simplify Variable Expressions Calculator Use the Product Rule to Simplify the Expression Calculator Simplifying Algebraic Expression Algebra Factor Calculator | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9415702819824219, "perplexity": 1828.3504951143907}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936461332.16/warc/CC-MAIN-20150226074101-00228-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://socratic.org/questions/what-are-the-horizontal-and-vertical-asymptotic-if-any-of-the-curve-f-x-12x-52-3#566189 | Precalculus
Topics
# What are the horizontal and vertical asymptotic (if any) of the curve f(x)=(12x+52)/(3x^2+2x-1)?
## Find the horizontal and vertical asymptotic if any of the curve $f \left(x\right) = \frac{12 x + 52}{3 {x}^{2} + 2 x - 1}$
Mar 6, 2018
Verticsl asymptote at $x = \frac{1}{3} \mathmr{and} x = - 1$
Horizontal asymptote $y = 0$
#### Explanation:
You can find the vertical asymptote if you equate the denominator to 0
$3 {x}^{2} + 2 x - 1 = 0$
$\left(3 x - 1\right) \left(x + 1\right) = 0$
$x = \frac{1}{3} , x = - 1$
Horizontal asymptote is at y=0 as the highest degree ( power on the x) on the numerator is less than the highest degree on the denominator
Mar 6, 2018
$\text{vertical asymptotes at "x=-1" and } x = \frac{1}{3}$
$\text{horizontal asymptote at } y = 0$
#### Explanation:
$\text{the denominator of f(x) cannot be zero as this would}$
$\text{make f(x) undefined. Equating the denominator to }$
$\text{zero and solving gives the values that x cannot be and}$
$\text{if the numerator is non-zero for these values then they}$
$\text{are vertical asymptotes}$
$\text{solve } 3 {x}^{2} + 2 x - 1 = 0 \Rightarrow \left(3 x - 1\right) \left(x + 1\right) = 0$
$\Rightarrow x = - 1 \text{ and "x=1/3" are the asymptotes}$
$\text{horizontal asymptotes occur as}$
${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{( a constant)}$
$\text{divide terms on numerator/denominator by the }$
$\text{highest power of x that is } {x}^{2}$
$f \left(x\right) = \frac{\frac{12 x}{x} ^ 2 + \frac{52}{x} ^ 2}{\frac{3 {x}^{2}}{x} ^ 2 + \frac{2 x}{x} ^ 2 - \frac{1}{x} ^ 2} = \frac{\frac{12}{x} + \frac{52}{x} ^ 2}{3 + \frac{2}{x} - \frac{1}{x} ^ 2}$
$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{0 + 0}{3 + 0 - 0}$
$\Rightarrow y = 0 \text{ is the asymptote}$
graph{(12x+52)/(3x^2+2x-1) [-20, 20, -10, 10]}
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https://www.hpmuseum.org/forum/showthread.php?mode=threaded&tid=1844&pid=16373 | Modified Newton method for small slopes near the roots
07-17-2014, 03:01 PM (This post was last modified: 07-17-2014 04:04 PM by Namir.)
Post: #1
Namir Senior Member Posts: 688 Joined: Dec 2013
Modified Newton method for small slopes near the roots
I am reading the new book Practical Numerical Methods For Chemical Engineers by Richard A. Davis. This book uses a lot of Excel VBA code and has fantastic library of numerical routines. In the chapter about nonlinear equations, Davis lists the following formula for modifying Newtons method when f(x) has a very small slope near the root. This new formula can also help solve equations like f(x)=x^2 (where f(x) does not cross the x-axis) and f(x)=x^3:
x = x - f(x) f'(x)/([f'(x)]^2 + delta)
Where f'(x) is the first derivative of f(x) and delta is a small value (like 1e-7). I find the above equation to be a clever variant since the denominator calculates the square of the slope (making sure the result is always zero or greater) and adds a positive number. The result is a denominator that always has positive value. Placing the derivative in the numerator assures that the sign of that derivative is not lost and the value of the derivative is not mutilated by squaring the slope in the denominator. The above equation is a clever variant of the following simpler and very vulnerable version:
x = x - f(x) / (f'(x) + delta)
The first equation works BUT IS SLOW. The slow convergence sure beats NO SOLUTION AT ALL.
MAKE SURE THAT YOU CALCULATE THE DERIVATIVE USING A CENTRAL DIFFERENCE FORMULA LIKE:
f'(x) = (f(x+h) - f(x-h)) / 2h
Where h is something like h = 0.01*(1+|x|). Using the following forward difference approximation SLOWS CONVERGENCE EVEN MORE:
f'(x) = (f(x+h) - f(x)) / h
Have fun!
Namir
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Messages In This Thread Modified Newton method for small slopes near the roots - Namir - 07-17-2014 03:01 PM
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https://physics.stackexchange.com/tags/causality/hot | # Tag Info
5
You seem to be conflating "causally linked" with "in the same reference frame". This is not correct. Two points in the spacetime are "causally linked" if there is a causal (timelike or null) curve that connects them (meaning you can reach one point from the other without moving faster than $c$), so two observers can be ...
2
The choice of Riemannian metric $h_{\mu \nu}$ is itself arbitrary, since there are multiple inequivalent rank-2 non-degenerate tensors on an arbitrary space; and different choices of $h_{\mu \nu}$ will lead to different "preferred time directions." For example, consider the following two rank-2 tensors on Minkowski space with coordinates $t$, $x$, ...
2
There is a fundamental problem with this scheme: constructing a warp drive spacetime. The papers exploring such spacetimes all work by (1) assuming some negative energy density exotic matter to hold together the bubble, and (2) that the bubble already exists. (1) is problematic since we have never seen such exotic matter, but (2) is more fundamental: there ...
2
The principle is that information cannot move from Point 1 to Point 2 faster than the speed of light. In the example you give, information moves from the initial positions of the two participants to Planet B at less than the speed of light, so the principle is not violated.
1
Some work along these lines has been done by Bob Wald and his colleagues over the years: Wald, R. Dynamics in nonglobally hyperbolic, static space‐times. J. Math. Phys. 21, 2802 (1980). Ishibashi, A. & R. Wald. Dynamics in non-globally-hyperbolic static spacetimes: II. General analysis of prescriptions for dynamics. Class. Quantum Grav. 20, 3815 (...
1
The limit here is about how far the information can travel between two points. If people who originate at those points move while the message is propagating, that can reduce the time it takes for the people to get the message, of course, because it cuts the distance that the message has to travel to reach them. (But it does not reduce the minimum time ...
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https://homework.cpm.org/category/CCI_CT/textbook/int3/chapter/9/lesson/9.1.7/problem/9-94 | Home > INT3 > Chapter 9 > Lesson 9.1.7 > Problem9-94
9-94.
What is the equation of a third degree polynomial that has roots $x = 3, 2$, and $–1$ and passes through the point $\left(1, 1\right)$?
Third-degree means that the highest power of $x$ will be $3$.
The roots are $3, 2$, and $−1$. These three values make the polynomial equal to $0$.
The corresponding polynomial is $y = a\left(x − 3\right)\left(x − 2\right)\left(x + 1\right)$.
Substitute the point $\left(1, 1\right)$ into the polynomial equation to find the value of $a$.
$y=\frac{1}{4}(x-3)(x-2)(x+1)$ | {"extraction_info": {"found_math": true, "script_math_tex": 12, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8651924133300781, "perplexity": 206.60873157607455}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710808.72/warc/CC-MAIN-20221201085558-20221201115558-00520.warc.gz"} |
https://infoproc.blogspot.com/2007/06/curved-space-and-monsters.html | ## Monday, June 25, 2007
### Curved space and monsters
New paper!
http://arxiv.org/abs/0706.3239
A simple question: how many different black holes can there be with mass M? Conventional wisdom: of order exp(A), where A is the surface area of the hole and scales as M^2.
Using curved space, we construct objects of ADM mass M with far more than exp(A) microstates. These objects have pathological properties, but, as far as we can tell, can be produced via quantum tunneling from ordinary (non-pathological) initial data. Our results suggest that the relation between black hole entropy and the number of microstates of the hole is more subtle than perhaps previously appreciated.
Update! Rafael Sorkin was kind enough to inform us of his earlier related work with Wald and Zhang. We've added the following end-note to the paper.
Note added: After this work was completed we were informed of related results obtained by Sorkin, Wald and Zhang [25]. Those authors investigated monster-like objects as well as local extrema of the entropy S subject to an energy constraint, which correspond to static configurations and obey $A^{3/4}$ scaling. For example, in the case of a photon gas the local extrema satisfy the Tolman--Oppenheimer--Volkoff equation of hydrostatic equilibrium. In considering monster configurations, Sorkin et al. show that requiring a configuration to be no closer than a thermal wavelength $\lambda \sim \rho^{-1/4}$ from its Schwarzschild radius imposes the bound $S < A$. While this may be a reasonable criterion that must be satisfied for the assembly of an initial configuration, it does not seem to apply to states reached by quantum tunneling. From a global perspective configurations with $S > A^{3/4}$ are already black holes in the sense that the future of parts of the object does not include future null infinity.
Black hole entropy, curved space and monsters
Stephen D.H. Hsu, David Reeb
(Submitted on 21 Jun 2007)
We investigate the microscopic origin of black hole entropy, in particular the gap between the maximum entropy of ordinary matter and that of black holes. Using curved space, we construct configurations with entropy greater than their area in Planck units. These configurations have pathological properties and we refer to them as monsters. When monsters are excluded we recover the entropy bound on ordinary matter $S < A^{3/4}$. This bound implies that essentially all of the microstates of a semiclassical black hole are associated with the growth of a slightly smaller black hole which absorbs some additional energy. Our results suggest that the area entropy of black holes is the logarithm of the number of distinct ways in which one can form the black hole from ordinary matter and smaller black holes, but only after the exclusion of monster states.
#### 3 comments:
Anonymous said...
Freaky, I read this post just as my playlist switched to the song "Monsters" by Band of Horses. Now there's an idea for an interesting MP3 player: one which observes what your surfing and adjusts your playlist.
Kea said...
Was your choice of the term monster a deliberate reference to moonshine, or not?
Steve Hsu said...
No, that is an unfortunate coincidence. I say unfortunate because it's misleading and (as far as I know) there is no connection between our monsters and the monster group.
We just called them monsters because they have pathological properties. Note these configurations have so much entropy (number of possible states) that they can't be accommodated in any kind of holographic dual description. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8208266496658325, "perplexity": 626.4287082595228}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178358976.37/warc/CC-MAIN-20210227144626-20210227174626-00328.warc.gz"} |
http://mathoverflow.net/questions/54799/on-a-proof-of-the-existence-of-tubular-neighborhoods | # On a proof of the existence of tubular neighborhoods.
Studying analysis on manifolds, I have found, in the proof of the existence of tubular neighborhoods, a reference to theorem 3.1.2 in "Topologie algebrique et theorie des faisceaux" of Godement. Without going through the machinery of the sheaves, at least now, is it possible to bypass the Godement's result? And, if yes, what is an accessible (not sheaf-theoretic) route?
This is the initial setting: $J:N\rightarrow M$ is a smooth embedding. $\pi:E\rightarrow N$ is a vector bundle, and $s_0:N\rightarrow E$ is the zero section of $\pi$. $\psi:U\rightarrow M$ is a smooth map from a neighborhood $U$ of $s_0(N)$ in $E.$ $\psi$ is a local diffeomorphism in each point of $s_0(N),$ and $\psi\circ s_0=j$.
At this point there is a reference to the argument of Godement in order to prove that: (*)There exists a neighborhood $V$ of $s_0(N)$ in $U$ such that $\psi|_V$ is a diffeomorphism.
What is the argument (differential geometric, not sheaf-theoretic) in order to conclude (*)?
Added by Mariano: I now have a copy in my hands. Theorem 3.1.2 reads (my translation):
Let $$0\to\mathscr L'\to \mathscr L\to\mathscr L''\to0$$ be a short exact sequence of sheaves of abelian groups. If $\mathscr L'$ is flasque, then for all open sets $U$ there is a short exact sequence $$0\to\mathscr L'(U)\to \mathscr L(U)\to\mathscr L''(U)\to0$$
He remarks that we therefore have a short exact sequence of pre-sheaves.
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If you told us what the result says... – Mariano Suárez-Alvarez Feb 8 '11 at 18:22
You use lots of \mapstos in weird places! :) – Mariano Suárez-Alvarez Feb 8 '11 at 20:05
A differential-geometric proof is available in Guillemin and Pollack's "Differential Topology". It's an exercise, but they set you up so that it's relatively straight-forward. The exercise is broken into two parts, 1st for manifolds in Euclidean space, then for submanifolds of manifolds. – Ryan Budney Feb 9 '11 at 19:35
It's also written out nicely in Milnor-Stasheff's Characteristic Classes (and probably most other texts on differential geometry topology). – Dave Anderson Feb 9 '11 at 19:38
Right, Hirsch's "Differential Topology" would be another source. Kosinski's "Differential Manifolds", Bredon's "Geometry and Topology", etc. – Ryan Budney Feb 9 '11 at 19:59
In the finite-dimensional setting, it's possible to construct tubular neighborhoods without anything like Godement's lemma. Many sources simply rely on a point-set topology argument that's based on the same idea as Godement's lemma (to be precise, I'm talking about the argument on p. 109 of Lang's book Differential and Riemannian Manifolds, which he says follows Godement). I'll explain another approach.
The idea is to use a Riemannian metric on the manifold $M$, which also induces a Riemannian metric on $TM$ (viewed as a manifold in its own right). The geodesic distance then gives a (topological) metric on $TM$. If $Y\subset M$ is a (not necessarily closed) submanifold, then a simple metric geometry argument can then be used to find a neighborhood of the zero section of $N(Y)$ (thought of as the perpendicular complement of $TY$ inside $TM$) on which the exponential map is injective. The key fact about the exponential map $f$ is that every point in the zero section of $N(Y)$ has a neighborhood on which $f$ is a diffeomorphism onto an open subset of $M$. (Edit: Note that in the finite dimensional setting, $N(Y)$ is automatically a locally trivial vector bundle. This does not seem to be the case for arbitrary infinite dimensional Riemannian manifolds, as discussed here: Orthogonal complements in Hilbert bundles. Hence the discussion that follows does not work in as great generality as arguments based on Godement's lemma.)
The general metric geometry fact is this:
Consider a metric space $T$ and subspaces $X, Y$, and $D$ such that $Y \subset X$ and $Y\subset D$. (Think: $T = TM$, $X$ is the zero section of $TM$, $Y$ is a submanifold of $M$, and $D$ is the domain of the exponential map, lying inside $NY$.) Let $f: D\to X$ be a continuous map that restricts to the identity on $Y$ (think: $f$ is the exponential map). Assume further that for each $y\in Y$ there exists $\epsilon_y >0$ such that $f$ restricted to $B_{\epsilon (y)} (y, D) = \{z\in D \,:\, d(z,y) < \epsilon(y)\}$ is a homeomorphism onto an open subset of $X$. Then there exists a subspace $D'$, open in $D$, on which $f$ is injective.
Proof. For each $y\in Y$, $f(B_{\epsilon (y)/2} (y, D))$ is open in $X$, hence contains $B_{\epsilon'(y)} (y, X)$, for some $\epsilon'_y < \epsilon_y/4$ (remember that $f(y) = y$). Now consider the inverse image $Z_y$ of $B_{\epsilon'_y} (y, X)$ under the restriction of $f$ to $B_{\epsilon (y)/2} (y, D)$. Since $f$ is a homeomorphism when restricted to this ball, $Z_y$ is open as a subset of $D$. Now I claim that $f$ is injective on $D' = \bigcup_{y\in Y} Z_y$. Say $f(z_1) = f(z_2) = y_0$ with $z_1 \in Z_{y_1}$ and $z_2\in Z_{y_2}$, and assume $\epsilon_{y_1} \geq \epsilon_{y_2}$. Then we have $$d(z_2, y_1) \leq d(z_2, y_2) + d(y_2, y_0) + d(y_0, y_1) < \epsilon_{y_2}/2 + \epsilon'_{y_2} + \epsilon'_{y_1}$$ $$< \epsilon_{y_1}/2 + \epsilon_{y_2}/4 + \epsilon_{y_1}/4 \leq \epsilon_{y_1}$$ (for the second inequality, note that by definition, $y_0 = f(z_i) \in f(Z_{y_i}) \subset B_{\epsilon'_{y_i}} (y_i, X)$ for $i=1, 2$). So $z_2$ and $z_1$ both lie in $B_{\epsilon_{y_1}} (y_1, D)$, and since $f$ is injective on this ball we have $z_1 = z_2$.
This argument is useful for the construction of equivariant tubular neighborhoods in certain infinite-dimensional settings. See http://arxiv.org/abs/1006.0063; I just updated it so I guess the new version will show up tomorrow. The equivariant version of the above argument is in Proposition 2.3 or 2.4, depending on the version.
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Dear Dan Ramras This argument was just what I was searching for! Thank you. – Giuseppe Mar 11 '11 at 6:45
Nice argument. It is interesting that it is valid for any embedding submanifold not necessarily closed. As far as I know in most textbooks in Riemannian geometry when presenting the tubular neighborhood theorem they assume that the submanifold is closed or compact. One exception is the book of John M. Lee but he only present it for embedding submanifolds in $\mathbb{R}^{n}$. – Coffee May 23 at 1:38
Yes, it's possible. And this answer is serious!
If the result you want to apply is (say) about "flasque sheaves", but you're working about "indefinitely differentiable functions", then it will be easy to explain the proof in your specific setting without diving in the whole theory of sheaves.
That being said, I guess it will be possible to actually explain the proof you want... when you'll have explained exactly which result you want!
EDIT: The result you ask about is just some kind of implicit function application ; in fact the french wikipedia even has a full page on this... with no english equivalent apparently.
EDIT 2: Ok, third try at answering the question ; this result on flasque sheaves is exercise 1.16(b), p67 of "Algebraic Geometry" by R.Hartshorne. You'll find the basic theory of sheaves in H.Cartan's lecture (page 6 has half the result -- the part which doesn't need the flasque hypothesis) ; the second part of the lecture has the other half, the theorem page 5 -- and you'll notice the way he defines fine sheaves (page 1) is asking for partitions of unity -- which are available in your particular setting!
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I have inserted the initial setting, and the conclusion borrowed appealing to Godement. In the book I am reading there is not the statement of Theorem 3.1.2 from Godement. – Giuseppe Feb 8 '11 at 19:39
I have recovered a restatement of the result of Godement. – Giuseppe Feb 8 '11 at 19:52
The straightening out theorem is only local. My problem is about the existence of an inverse of $\psi$ on a whole neighborhood of the zero section $s_0(N)$, not only around any one of its points. – Giuseppe Feb 8 '11 at 21:00
If you can do it around any point in the zero section, it implies you can do it around the zero section. – Deane Yang Feb 9 '11 at 19:43
Dear Deane Yang. I have to understand better what are the difficulties I have met, even reading differential topology texts, and eventually rewrite a question. Excuse me. – Giuseppe Feb 9 '11 at 21:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9471626877784729, "perplexity": 174.10499858605894}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049276964.14/warc/CC-MAIN-20160524002116-00111-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://socratic.org/questions/a-box-with-an-initial-speed-of-5-m-s-is-moving-up-a-ramp-the-ramp-has-a-kinetic--8 | Physics
Topics
# A box with an initial speed of 5 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 3/7 and an incline of (5 pi )/8 . How far along the ramp will the box go?
Jul 30, 2017
$\text{distance} = 1.17$ $\text{m}$
#### Explanation:
I'd like to point out that there can't realistically be an inclined ramp with angle of inclination $\frac{5 \pi}{8}$...it must be between $0$ and $\frac{\pi}{2}$, so I'll choose the closest angle to this, $\frac{3 \pi}{8}$...
We're asked to find the distance traveled by the box given its initial speed, coefficient of kinetic friction, and angle of inclination.
NOTE: the friction force actually acts down the incline as opposed to the image (because the box is traveling up the ramp).
The magnitude of the kinetic friction force ${f}_{k}$ is given by
${f}_{k} = {\mu}_{k} n$
where
• ${\mu}_{k}$ is the coefficient of kinetic friction ($\frac{3}{7}$)
• $n$ is the magnitude of the normal force exerted by the incline plane, equal to $m g \cos \theta$
We must first find the acceleration of the box, using Newton's second law:
$\sum F = m a$
$a = \frac{\sum F}{m}$
The net horizontal force $\sum F$ is
$\sum F = m g \sin \theta + {f}_{k} = m g \sin \theta + {\mu}_{k} m g \cos \theta$
Therefore, we have
$a = \frac{\cancel{m} g \sin \theta + \cancel{m} {u}_{k} m g \cos \theta}{\cancel{m}} = g \sin \theta + {\mu}_{k} g \cos \theta$
Plugging in known values, we have
$a = \left(9.81 \textcolor{w h i t e}{l} {\text{m/s"^2)sin((3pi)/8) + 3/7(9.81color(white)(l)"m/s}}^{2}\right) \cos \left(\frac{3 \pi}{8}\right)$
$= 10.7$ ${\text{m/s}}^{2}$
directed down the incline, so this can also be written as
a = ul(-10.7color(white)(l)"m/s"^2
Now, we can use the equation
${\left({v}_{x}\right)}^{2} = {\left({v}_{0 x}\right)}^{2} + 2 {a}_{x} \left(\Delta x\right)$
to find the distance it travels up the ramp before it comes to a stop.
Here,
${v}_{x} = 0$ (instantaneously at rest at maximum height)
• ${v}_{0 x} = 5$ $\text{m/s}$
• ${a}_{x} = - 10.7$ ${\text{m/s}}^{2}$
• $\Delta x =$ trying to find
Plugging in known values, we have
$0 = \left(5 \textcolor{w h i t e}{l} {\text{m/s")^2 + 2(-10.7color(white)(l)"m/s}}^{2}\right) \left(\Delta x\right)$
Deltax = color(red)(ul(1.17color(white)(l)"m"
##### Impact of this question
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https://docs.scipy.org/doc/scipy-0.16.1/reference/generated/scipy.optimize.fmin_tnc.html | scipy.optimize.fmin_tnc¶
scipy.optimize.fmin_tnc(func, x0, fprime=None, args=(), approx_grad=0, bounds=None, epsilon=1e-08, scale=None, offset=None, messages=15, maxCGit=-1, maxfun=None, eta=-1, stepmx=0, accuracy=0, fmin=0, ftol=-1, xtol=-1, pgtol=-1, rescale=-1, disp=None, callback=None)[source]
Minimize a function with variables subject to bounds, using gradient information in a truncated Newton algorithm. This method wraps a C implementation of the algorithm.
minimize
Interface to minimization algorithms for multivariate functions. See the ‘TNC’ method in particular.
Notes
The underlying algorithm is truncated Newton, also called Newton Conjugate-Gradient. This method differs from scipy.optimize.fmin_ncg in that
1. It wraps a C implementation of the algorithm
2. It allows each variable to be given an upper and lower bound.
The algorithm incoporates the bound constraints by determining the descent direction as in an unconstrained truncated Newton, but never taking a step-size large enough to leave the space of feasible x’s. The algorithm keeps track of a set of currently active constraints, and ignores them when computing the minimum allowable step size. (The x’s associated with the active constraint are kept fixed.) If the maximum allowable step size is zero then a new constraint is added. At the end of each iteration one of the constraints may be deemed no longer active and removed. A constraint is considered no longer active is if it is currently active but the gradient for that variable points inward from the constraint. The specific constraint removed is the one associated with the variable of largest index whose constraint is no longer active.
Return codes are defined as follows:
-1 : Infeasible (lower bound > upper bound)
0 : Local minimum reached (|pg| ~= 0)
1 : Converged (|f_n-f_(n-1)| ~= 0)
2 : Converged (|x_n-x_(n-1)| ~= 0)
3 : Max. number of function evaluations reached
4 : Linear search failed
5 : All lower bounds are equal to the upper bounds
6 : Unable to progress
7 : User requested end of minimization
References
Wright S., Nocedal J. (2006), ‘Numerical Optimization’
Nash S.G. (1984), “Newton-Type Minimization Via the Lanczos Method”, SIAM Journal of Numerical Analysis 21, pp. 770-778
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https://en.wikipedia.org/wiki/Projective_polyhedron | # Projective polyhedron
In geometry, a (globally) projective polyhedron is a tessellation of the real projective plane.[1] These are projective analogs of spherical polyhedra – tessellations of the sphere – and toroidal polyhedra – tessellations of the toroids.
Projective polyhedra are also referred to as elliptic tessellations[2] or elliptic tilings, referring to the projective plane as (projective) elliptic geometry, by analogy with spherical tiling,[3] a synonym for "spherical polyhedron". However, the term elliptic geometry applies to both spherical and projective geometries, so the term carries some ambiguity for polyhedra.
As cellular decompositions of the projective plane, they have Euler characteristic 1, while spherical polyhedra have Euler characteristic 2. The qualifier "globally" is to contrast with locally projective polyhedra, which are defined in the theory of abstract polyhedra.
Non-overlapping projective polyhedra (density 1) correspond to spherical polyhedra (equivalently, convex polyhedra) with central symmetry. This is elaborated and extended below in relation with spherical polyhedra and relation with traditional polyhedra.
## Examples
The hemi-cube is a regular projective polyhedron with 3 square faces, 6 edges, and 4 vertices.
The best-known examples of projective polyhedra are the regular projective polyhedra, the quotients of the centrally symmetric Platonic solids, as well as two infinite classes of even dihedra and hosohedra:[4]
These can be obtained by taking the quotient of the associated spherical polyhedron by the antipodal map (identifying opposite points on the sphere).
On the other hand, the tetrahedron does not have central symmetry, so there is no "hemi-tetrahedron". See relation with spherical polyhedra below on how the tetrahedron is treated.
### Hemipolyhedra
The tetrahemihexahedron is a projective polyhedron, and the only uniform projective polyhedron that immerses in Euclidean 3-space.
Further information: Hemipolyhedron
Note that the prefix "hemi-" is also used to refer to hemipolyhedra, which are uniform polyhedra having some faces that pass through the center of symmetry. As these do not define spherical polyhedra (because they pass through the center, which does not map to a defined point on the sphere), they do not define projective polyhedra by the quotient map from 3-space (minus the origin) to the projective plane.
Of these uniform hemipolyhedra, only the tetrahemihexahedron is topologically a projective polyhedron, as can be verified by its Euler characteristic and visually obvious connection to the Roman surface. It is 2-covered by the cuboctahedron, and can be realized as the quotient of the spherical cuboctahedron by the antipodal map. It is the only uniform (traditional) polyhedron that is projective – that is, the only uniform projective polyhedron that immerses in Euclidean three-space as a uniform traditional polyhedron.
## Relation with spherical polyhedra
There is a 2-to-1 covering map ${\displaystyle S^{2}\to \mathbf {RP} ^{2}}$ of the sphere to the projective plane, and under this map, projective polyhedra correspond to spherical polyhedra with central symmetry – the 2-fold cover of a projective polyhedron is a centrally symmetric spherical polyhedron. Further, because a covering map is a local homeomorphism (in this case a local isometry), both the spherical and the corresponding projective polyhedra have the same abstract vertex figure.
For example, the 2-fold cover of the (projective) hemi-cube is the (spherical) cube. The hemi-cube has 4 vertices, 3 faces, and 6 edges, each of which is covered by 2 copies in the sphere, and accordingly the cube has 8 vertices, 6 faces, and 12 edges, while both these polyhedra have a 4.4.4 vertex figure (3 squares meeting at a vertex).
Further, the symmetry group (of isometries) of a projective polyhedron and covering spherical polyhedron are related: the symmetries of the projective polyhedron are naturally identified with the rotation symmetries of the spherical polyhedron, while the full symmetry group of the spherical polyhedron is the product of its rotation group (the symmetry group of the projective polyhedron) and the cyclic group of order 2, {±I}. See symmetry group below for elaboration and other dimensions.
Spherical polyhedra without central symmetry do not define a projective polyhedron, as the images of vertices, edges, and faces will overlap. In the language of tilings, the image in the projective plane is a degree 2 tiling, meaning that it covers the projective plane twice – rather than 2 faces in the sphere corresponding to 1 face in the projective plane, covering it twice, each face in the sphere corresponds to a single face in the projective plane, accordingly covering it twice.
The correspondence between projective polyhedra and centrally symmetric spherical polyhedra can be extended to a Galois connection including all spherical polyhedra (not necessarily centrally symmetric) if the classes are extended to include degree 2 tilings of the projective plane, whose covers are not polyhedra but rather the polyhedral compound of a non-centrally symmetric polyhedron, together with its central inverse (a compound of 2 polyhedra). This geometrizes the Galois connection at the level of finite subgroups of O(3) and PO(3), under which the adjunction is "union with central inverse". For example, the tetrahedron is not centrally symmetric, and has 4 vertices, 6 edges, and 4 faces, and vertex figure 3.3.3 (3 triangles meeting at each vertex). Its image in the projective plane has 4 vertices, 6 edges (which intersect), and 4 faces (which overlap), covering the projective plane twice. The cover of this is the stellated octahedron – equivalently, the compound of two tetrahedra – which has 8 vertices, 12 edges, and 8 faces, and vertex figure 3.3.3.
## Generalizations
In the context of abstract polytopes, one instead refers to "locally projective polytopes" – see Abstract polytope: Local topology. For example, the 11-cell is a "locally projective polytope", but is not a globally projective polyhedron, nor indeed tessellates any manifold, as it not locally Euclidean, but rather locally projective, as the name indicates.
Projective polytopes can be defined in higher dimension as tessellations of projective space in one less dimension. Defining k-dimensional projective polytopes in n-dimensional projective space is somewhat trickier, because the usual definition of polytopes in Euclidean space requires taking convex combinations of points, which is not a projective concept, and is infrequently addressed in the literature, but has been defined, such as in (Vives & Mayo 1991).
## Symmetry group
The symmetry group of a projective polytope is a finite (hence discrete)[note 1] subgroup of the projective orthogonal group, PO, and conversely every finite subgroup of PO is the symmetry group of a projective polytope by taking the polytope given by images of a fundamental domain for the group.
The relevant dimensions are as follows: n-dimensional real projective space is the projectivization of (n+1)-dimensional Euclidean space, ${\displaystyle \mathbf {RP} ^{n}=\mathbf {P} (\mathbf {R} ^{n+1}),}$ so the projective orthogonal group of an n-dimensional projective space is denoted
PO(n+1) = P(O(n+1)) = O(n+1)/{±I}.
If n=2k is even (so n+1 = 2k+1 is odd), then O(2k+1) = SO(2k+1)×{±I} decomposes as a product, and thus ${\displaystyle PO(2k+1)=PSO(2k+1)\cong SO(2k+1)}$[note 2] so the group of projective isometries can be identified with the group of rotational isometries.
Thus in particular the symmetry group of a projective polyhedron is the rotational symmetry group of the covering spherical polyhedron; the full symmetry group of the spherical polyhedron is then just the direct product with reflection through the origin, which is the kernel on passage to projective space. The projective plane is non-orientable, and thus there is no distinct notion of "orientation-preserving isometries of a projective polyhedron", which is reflected in the equality PSO(3) = PO(3).
If n=2k + 1 is odd, then O(n+1) = O(2k+2) does not decompose as a product, and thus the symmetry group of the projective polytope is not simply the rotational symmetries of the spherical polytope, but rather a 2-to-1 quotient of the full symmetry group of the corresponding spherical polytope (the spherical group is a central extension of the projective group). Further, in odd projective dimension (even vector dimension) ${\displaystyle PSO(2k)\neq PO(2k)}$ and is instead a proper (index 2) subgroup, so there is a distinct notion of orientation-preserving isometries.
For example, in n = 1 (polygons), the symmetries of a 2r-gon is the dihedral group Dih2r (of order 4r), with rotational group the cyclic group C2r, these being subgroups of O(2) and SO(2), respectively. The projectivization of a 2r-gon (in the circle) is an r-gon (in the projective line), and accordingly the quotient groups, subgroups of PO(2) and PSO(2) are Dihr and Cr. Note that the same commutative square of subgroups occurs for the square of Spin group and Pin group – Spin(2), Pin+(2), SO(2), O(2) – here going up to a 2-fold cover, rather than down to a 2-fold quotient.
Lastly, by the lattice theorem there is a Galois connection between subgroups of O(n) and subgroups of PO(n), in particular of finite subgroups. Under this connection, symmetry groups of centrally symmetric polytopes correspond to symmetry groups of the corresponding projective polytope, while symmetry groups of spherical polytopes without central symmetry correspond to symmetry groups of degree 2 projective polytopes (tilings that cover projective space twice), whose cover (corresponding to the adjunction of the connection) is a compound of two polytopes – the original polytope and its central inverse.
These symmetry groups should be compared and contrasted with binary polyhedral groups – just as Pin±(n) → O(n) is a 2-to-1 cover, and hence there is a Galois connection between binary polyhedral groups and polyhedral groups, O(n) → PO(n) is a 2-to-1-cover, and hence has an analogous Galois connection between subgroups. However, while discrete subgroups of O(n) and PO(n) correspond to symmetry groups of spherical and projective polytopes, corresponding geometrically to the covering map ${\displaystyle S^{n}\to \mathbf {RP} ^{n},}$ there is no covering space of ${\displaystyle S^{n}}$ (for ${\displaystyle n\geq 2}$) as the sphere is simply connected, and thus there is no corresponding "binary polytope" for which subgroups of Pin are symmetry groups.
## Notes
1. ^ Since PO is compact, finite and discrete sets are identical – infinite sets have an accumulation point.
2. ^ The isomorphism/equality distinction in this equation is because the context is the 2-to-1 quotient map ${\displaystyle O\to PO}$ – PSO(2k+1) and PO(2k+1) are equal subsets of the target (namely, the whole space), hence the equality, while the induced map ${\displaystyle SO\to PSO}$ is an isomorphism but the two groups are subsets of different spaces, hence the isomorphism rather than an equality. See (Conway & Smith 2003, p. 34) for an example of this distinction being made.
## References
### Footnotes
1. ^ Schulte, Egon; Weiss, Asia Ivic (2006), "5 Topological classification", Problems on Polytopes, Their Groups, and Realizations, pp. 9–13, arXiv:
2. ^ Coxeter, Harold Scott Macdonald (1970). Twisted honeycombs. CBMS regional conference series in mathematics (4). AMS Bookstore. p. 11. ISBN 978-0-8218-1653-0.
3. ^
4. ^ Coxeter, Introduction to geometry, 1969, Second edition, sec 21.3 Regular maps, p. 386-388 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 9, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9067932367324829, "perplexity": 773.6258539245298}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607998.27/warc/CC-MAIN-20170525044605-20170525064605-00507.warc.gz"} |
https://www.physicsforums.com/threads/calculating-the-proper-mean-lifetime-of-pions.343605/ | # Calculating the proper mean lifetime of pions
1. Oct 7, 2009
### matt_crouch
Calculating the "proper mean lifetime" of pions
1. The problem statement, all variables and given/known data
The mean lifetime of pions at rest is to=2.6x10^-8. if a beam of pions has a speed v=0.85c
a)what would their lifetime be in the laboratory?
b)how far would they travel before they decay?
d)what would the "proper mean lifetime" be?
2. Relevant equations
3. The attempt at a solution
i think done a-c right but dont really know how to do d)
a)4.94x10^-8
b)12.5 m
c)6.63m
im not sure how to calculate the proper mean lifetime can anyone help?
Can you offer guidance or do you also need help?
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# The Entropy Production Fluctuation Theorem and the Nonequilibrium Work Relation for Free Energy Differences
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### Abstract
There are only a very few known relations in statistical dynamics that are valid for systems driven arbitrarily far-from-equilibrium. One of these is the fluctuation theorem, which places conditions on the entropy production probability distribution of nonequilibrium systems. Another recently discovered far-from-equilibrium expression relates nonequilibrium measurements of the work done on a system to equilibrium free energy differences. In this paper, we derive a generalized version of the fluctuation theorem for stochastic, microscopically reversible dynamics. Invoking this generalized theorem provides a succinct proof of the nonequilibrium work relation.
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### A Mathematical Theory of Communication
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### Equation of State Calculations by Fast Computing Machines
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An expression is derived for the classical free energy difference between two configurations of a system, in terms of an ensemble of finite-time measurements of the work performed in parametrically switching from one configuration to the other. Two well-known equilibrium identities emerge as limiting cases of this result.
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### Author and article information
###### Journal
cond-mat/9901352
10.1103/PhysRevE.60.2721
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http://blog.stata.com/page/3/ | ## Flexible discrete choice modeling using a multinomial probit model, part 1
$$\newcommand{\xb}{{\bf x}} \newcommand{\betab}{\boldsymbol{\beta}} \newcommand{\zb}{{\bf z}} \newcommand{\gammab}{\boldsymbol{\gamma}}$$We have no choice but to choose
We make choices every day, and often these choices are made among a finite number of potential alternatives. For example, do we take the car or ride a bike to get to work? Will we have dinner at home or eat out, and if we eat out, where do we go? Scientists, marketing analysts, or political consultants, to name a few, wish to find out why people choose what they choose.
## Unit-root tests in Stata
$$\newcommand{\mub}{{\boldsymbol{\mu}}} \newcommand{\eb}{{\boldsymbol{e}}} \newcommand{\betab}{\boldsymbol{\beta}}$$Determining the stationarity of a time series is a key step before embarking on any analysis. The statistical properties of most estimators in time series rely on the data being (weakly) stationary. Loosely speaking, a weakly stationary process is characterized by a time-invariant mean, variance, and autocovariance.
In most observed series, however, the presence of a trend component results in the series being nonstationary. Furthermore, the trend can be either deterministic or stochastic, depending on which appropriate transformations must be applied to obtain a stationary series. For example, a stochastic trend, or commonly known as a unit root, is eliminated by differencing the series. However, differencing a series that in fact contains a deterministic trend results in a unit root in the moving-average process. Similarly, subtracting a deterministic trend from a series that in fact contains a stochastic trend does not render a stationary series. Hence, it is important to identify whether nonstationarity is due to a deterministic or a stochastic trend before applying the proper transformations.
## Multiple equation models: Estimation and marginal effects using mlexp
We continue with the series of posts where we illustrate how to obtain correct standard errors and marginal effects for models with multiple steps. In this post, we estimate the marginal effects and standard errors for a hurdle model with two hurdles and a lognormal outcome using mlexp. mlexp allows us to estimate parameters for multiequation models using maximum likelihood. In the last post (Multiple equation models: Estimation and marginal effects using gsem), we used gsem to estimate marginal effects and standard errors for a hurdle model with two hurdles and an exponential mean outcome.
We exploit the fact that the hurdle-model likelihood is separable and the joint log likelihood is the sum of the individual hurdle and outcome log likelihoods. We estimate the parameters of each hurdle and the outcome separately to get initial values. Then, we use mlexp to estimate the parameters of the model and margins to obtain marginal effects. Read more…
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## Multiple equation models: Estimation and marginal effects using gsem
Starting point: A hurdle model with multiple hurdles
In a sequence of posts, we are going to illustrate how to obtain correct standard errors and marginal effects for models with multiple steps.
Our inspiration for this post is an old Statalist inquiry about how to obtain marginal effects for a hurdle model with more than one hurdle (http://www.statalist.org/forums/forum/general-stata-discussion/general/1337504-estimating-marginal-effect-for-triple-hurdle-model). Hurdle models have the appealing property that their likelihood is separable. Each hurdle has its own likelihood and regressors. You can estimate each one of these hurdles separately to obtain point estimates. However, you cannot get standard errors or marginal effects this way.
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## Tests of forecast accuracy and forecast encompassing
$$\newcommand{\mub}{{\boldsymbol{\mu}}} \newcommand{\eb}{{\boldsymbol{e}}} \newcommand{\betab}{\boldsymbol{\beta}}$$Applied time-series researchers often want to compare the accuracy of a pair of competing forecasts. A popular statistic for forecast comparison is the mean squared forecast error (MSFE), a smaller value of which implies a better forecast. However, a formal test, such as Diebold and Mariano (1995), distinguishes whether the superiority of one forecast is statistically significant or is simply due to sampling variability.
A related test is the forecast encompassing test. This test is used to determine whether one of the forecasts encompasses all the relevant information from the other. The resulting test statistic may lead a researcher to either combine the two forecasts or drop the forecast that contains no additional information.
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## Gelman–Rubin convergence diagnostic using multiple chains
Overview
MCMC algorithms used for simulating posterior distributions are indispensable tools in Bayesian analysis. A major consideration in MCMC simulations is that of convergence. Has the simulated Markov chain fully explored the target posterior distribution so far, or do we need longer simulations? A common approach in assessing MCMC convergence is based on running and analyzing the difference between multiple chains.
For a given Bayesian model, bayesmh is capable of producing multiple Markov chains with randomly dispersed initial values by using the initrandom option, available as of the update on 19 May 2016. In this post, I demonstrate the Gelman–Rubin diagnostic as a more formal test for convergence using multiple chains. For graphical diagnostics, see Graphical diagnostics using multiple chains in [BAYES] bayesmh for more details. To compute the Gelman–Rubin diagnostic, I use an unofficial command, grubin, which can be installed by typing the following in Stata: Read more…
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## Understanding omitted confounders, endogeneity, omitted variable bias, and related concepts
Initial thoughts
Estimating causal relationships from data is one of the fundamental endeavors of researchers. Ideally, we could conduct a controlled experiment to estimate causal relations. However, conducting a controlled experiment may be infeasible. For example, education researchers cannot randomize education attainment and they must learn from observational data.
In the absence of experimental data, we construct models to capture the relevant features of the causal relationship we have an interest in, using observational data. Models are successful if the features we did not include can be ignored without affecting our ability to ascertain the causal relationship we are interested in. Sometimes, however, ignoring some features of reality results in models that yield relationships that cannot be interpreted causally. In a regression framework, depending on our discipline or our research question, we give a different name to this phenomenon: endogeneity, omitted confounders, omitted variable bias, simultaneity bias, selection bias, etc.
Below I show how we can understand many of these problems in a unified regression framework and use simulated data to illustrate how they affect estimation and inference. Read more…
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## Programming an estimation command in Stata: Consolidating your code
$$\newcommand{\xb}{{\bf x}} \newcommand{\gb}{{\bf g}} \newcommand{\Hb}{{\bf H}} \newcommand{\Gb}{{\bf G}} \newcommand{\Eb}{{\bf E}} \newcommand{\betab}{\boldsymbol{\beta}}$$I write ado-commands that estimate the parameters of an exponential conditional mean (ECM) model and a probit conditional mean (PCM) model by nonlinear least squares, using the methods that I discussed in the post Programming an estimation command in Stata: Nonlinear least-squares estimators. These commands will either share lots of code or repeat lots of code, because they are so similar. It is almost always better to share code than to repeat code. Shared code only needs to be changed in one place to add a feature or to fix a problem; repeated code must be changed everywhere. I introduce Mata libraries to share Mata functions across ado-commands, and I introduce wrapper commands to share ado-code.
This is the 27th post in the series Programming an estimation command in Stata. I recommend that you start at the beginning. See Programming an estimation command in Stata: A map to posted entries for a map to all the posts in this series.
Ado-commands for ECM and PCM models
I now convert the examples of Read more…
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## Programming an estimation command in Stata: Nonlinear least-squares estimators
$$\newcommand{\xb}{{\bf x}} \newcommand{\gb}{{\bf g}} \newcommand{\Hb}{{\bf H}} \newcommand{\Gb}{{\bf G}} \newcommand{\Eb}{{\bf E}} \newcommand{\betab}{\boldsymbol{\beta}}$$I want to write ado-commands to estimate the parameters of an exponential conditional mean (ECM) model and probit conditional mean (PCM) model by nonlinear least squares (NLS). Before I can write these commands, I need to show how to trick optimize() into performing the Gauss–Newton algorithm and apply this trick to these two problems.
This is the 26th post in the series Programming an estimation command in Stata. I recommend that you start at the beginning. See Programming an estimation command in Stata: A map to posted entries for a map to all the posts in this series.
Gauss–Newton algorithm
Gauss–Newton algorithms frequently perform better than Read more…
Categories: Programming Tags:
## ARMA processes with nonnormal disturbances
Autoregressive (AR) and moving-average (MA) models are combined to obtain ARMA models. The parameters of an ARMA model are typically estimated by maximizing a likelihood function assuming independently and identically distributed Gaussian errors. This is a rather strict assumption. If the underlying distribution of the error is nonnormal, does maximum likelihood estimation still work? The short answer is yes under certain regularity conditions and the estimator is known as the quasi-maximum likelihood estimator (QMLE) (White 1982).
In this post, I use Monte Carlo Simulations (MCS) to verify that the QMLE of a stationary and invertible ARMA model is consistent and asymptotically normal. See Yao and Brockwell (2006) for a formal proof. For an overview of performing MCS in Stata, refer to Monte Carlo simulations using Stata. Also see A simulation-based explanation of consistency and asymptotic normality for a discussion of performing such an exercise in Stata.
Simulation | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8206307888031006, "perplexity": 1259.3827999739597}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542060.60/warc/CC-MAIN-20161202170902-00082-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://www.khanacademy.org/math/differential-calculus/limit-basics-dc/limits-introduction-dc/e/two-sided-limits-from-graphs | Limits from graphs
Practice finding two sided limits by looking at graphs.
Problem
What appears to be the value of $\displaystyle\lim_{x\to 0}h(x)$?
Please choose from one of the following options. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.879206657409668, "perplexity": 771.8159209402771}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719542.42/warc/CC-MAIN-20161020183839-00218-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://getrevising.co.uk/revision-tests/chemistry-500 | # Chemistry
HideShow resource information
• Created by: emmapxc
• Created on: 26-03-16 12:21
What is the Collision Theory?
For a chemical reaction to occur the reacting particles must collide into each other with sufficient kinetic energy and with the correct alignment.
1 of 10
What occurs when you increase the concentration?
Increasing concentration, increases the rate of reaction. There are more reacting particles in a given volume so more collisions can take place.
2 of 10
What occurs when pressure is increased?
Increasing pressure, increases rate of reaction. There are more reacting particles in a given volume.
3 of 10
What occurs when surface area is increased?
Increasing surface area, increases rate of reaction. There is a larger number of particles on the surface of the solid that are able to take part in collisions.
4 of 10
What occurs when temperature is increased?
Increasing temperature, increases rate of reaction. Particles will have a greater kinetic energy so collision between reacting particles are more likely to be successful.
5 of 10
How do you measure the relative rate of reaction?
1 / time taken for reaction
6 of 10
What does a potential energy diagram show?
It shows the energy pathway for a chemical reaction, how the potential energy changes as the reactants change into products.
7 of 10
What is enthalpy?
Enthalpy (H) is a measure of the chemical potential energy obtained in a substance.
8 of 10
How do you calculate enthalpy change?
ΔH = HP - HR
9 of 10
What is an exothermic reaction?
Chemical potential energy is changed into heat. Energy is released to the surroundings.
10 of 10
## Other cards in this set
### Card 2
#### Front
What occurs when you increase the concentration?
#### Back
Increasing concentration, increases the rate of reaction. There are more reacting particles in a given volume so more collisions can take place.
### Card 3
#### Front
What occurs when pressure is increased?
### Card 4
#### Front
What occurs when surface area is increased?
### Card 5
#### Front
What occurs when temperature is increased? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8780702948570251, "perplexity": 2406.337391117503}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280410.21/warc/CC-MAIN-20170116095120-00064-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://en.wikipedia.org/wiki/Pascal's_pyramid | # Pascal's pyramid
In mathematics, Pascal's pyramid is a three-dimensional arrangement of the trinomial numbers, which are the coefficients of the trinomial expansion and the trinomial distribution. Pascal's Pyramid is the three-dimensional analog of the two-dimensional Pascal's triangle, which contains the binomial numbers and relates to the binomial expansion and the binomial distribution. The binomial and trinomial numbers, coefficients, expansions, and distributions are subsets of the multinomial constructs with the same names. Pascal's Pyramid is more precisely called "Pascal's tetrahedron", since it has four triangular surfaces. (The pyramids of ancient Egypt had five surfaces: a square base and four triangular sides.)
## Structure of the tetrahedron
Because the tetrahedron is a three-dimensional object it is difficult to display it on a piece of paper or a computer screen. Assume the tetrahedron is divided into a number of levels, or floors, or slices, or layers. The top layer (the apex) is labelled "Layer 0". Other layers can be thought of as overhead views of the Tetrahedron with the previous layers removed. The first six layers are as follows:
Layer 0 1
Layer 1 1 1 1
Layer 2 1 2 1 2 2 1
Layer 3 1 3 3 1 3 6 3 3 3 1
Layer 4 1 4 6 4 1 4 12 12 4 6 12 6 4 4 1
Layer 5 1 5 10 10 5 1 5 20 30 20 5 10 30 30 10 10 20 10 5 5 1
The layers of the Tetrahedron have been deliberately displayed with the point down so that the Tetrahedron is not confused with Pascal's triangle.
## Overview of the tetrahedron
• There is three-way symmetry of the numbers in each layer.
• The number of terms in the nth Layer is the nth triangular number: (n + 1) × (n + 2) / 2.
• The sum of the values of the numbers in the nth Layer is 3n.
• Each number in any layer is the sum of the three adjacent numbers in the layer above.
• Each number in any layer is a simple whole number ratio of the adjacent numbers in the same layer.
• Each number in any layer is a coefficient of the Trinomial Distribution and the trinomial expansion. This non-linear arrangement makes it easier to:
• display the trinomial expansion in a coherent way;
• compute the coefficients of the Trinomial Distribution;
• calculate the numbers of any Tetrahedron layer.
• The numbers along the three edges of the nth Layer are the numbers of the nth Line of Pascal's triangle. And almost all the properties listed above have parallels with Pascal's triangle and Multinomial Coefficients.
## Trinomial expansion connection
The numbers of the Tetrahedron are derived from trinomial expansion. The nth Layer is the detached coefficient matrix (no variables or exponents) of a trinomial expression (e.g.: A + B + C) raised to the nth power. The trinomial is expanded by repeatedly multiplying the trinomial by itself:
(A + B + C)1 × (A + B + C)n = (A + B + C)n+1
Each term in the first expression is multiplied by each term in the second expression; and then the coefficients of like terms (same variables and exponents) are added together. Here is the expansion of (A + B + C)4:
1A4B0C0 + 4A3B0C1 + 6A2B0C2 + 4A1B0C3 + 1A0B0C4 +
4A3B1C0 + 12A2B1C1 + 12A1B1C2 + 4A0B1C3 +
6A2B2C0 + 12A1B2C1 + 6A0B2C2 +
4A1B3C0 + 4A0B3C1 +
1A0B4C0
Writing the expansion in this non-linear way shows the expansion in a more understandable way. It also makes the connection with the Tetrahedron obvious−the coefficients here match those of Layer 4. All the implicit coefficients, variables, and exponents, which are normally not written, are also shown to illustrate another relationship with the Tetrahedron. (Usually, "1A" is "A"; "B1" is "B"; and "C0" is "1"; etc.) The exponents of each term sum to the Layer number (n), or 4, in this case. More significantly, the value of the coefficients of each term can be computed directly from the exponents. The formula is: (x + y + z)! / (x! × y! × z!), where x, y, z are the exponents of A, B, C, respectively, and "!" means factorial (e.g.: n! = 1 × 2 ×...× n). The exponent formulas for the 4th Layer are:
$\textstyle {(4+0+0)!\over 4!\times 0!\times 0!} \ {(3+0+1)!\over 3!\times 0!\times 1!} \ {(2+0+2)!\over 2!\times 0!\times 2!} \ {(1+0+3)!\over 1!\times 0!\times 3!} \ {(0+0+4)!\over 0!\times 0!\times 4!}$
$\textstyle {(3+1+0)!\over 3!\times 1!\times 0!} \ {(2+1+1)!\over 2!\times 1!\times 1!} \ {(1+1+2)!\over 1!\times 1!\times 2!} \ {(0+1+3)!\over 0!\times 1!\times 3!}$
$\textstyle {(2+2+0)!\over 2!\times 2!\times 0!} \ {(1+2+1)!\over 1!\times 2!\times 1!} \ {(0+2+2)!\over 0!\times 2!\times 2!}$
$\textstyle {(1+3+0)!\over 1!\times 3!\times 0!} \ {(0+3+1)!\over 0!\times 3!\times 1!}$
$\textstyle {(0+4+0)!\over 0!\times 4!\times 0!}$
The exponents of each expansion term can be clearly seen and these formulae simplify to the expansion coefficients and the Tetrahedron coefficients of Layer 4.
## Trinomial distribution connection
The numbers of the Tetrahedron can also be found in the Trinomial Distribution. This is a discrete probability distribution used to determine the chance some combination of events occurs given three possible outcomes−the number of ways the events could occur is multiplied by the probabilities that they would occur. The formula for the Trinomial Distribution is:
[ n! / ( x! × y! × z!) ] × [ (PA)x × (PB)y × (PC)z]
where x, y, z are the number of times each of the three outcomes does occur; n is the number of trials and equals the sum of x+y+z; and PA, PB, PC are the probabilities that each of the three events could occur.
For example, in a three-way election, the candidates got these votes: A, 16%; B, 30%; C, 54%. What is the chance that a randomly selected four-person focus group would contain the following voters: 1 for A, 1 for B, 2 for C? The answer is:
[ 4! / ( 1! × 1! × 2!) ] × [ (16%)1 × (30%)1 × (54%)2] = 12 × 0.0140 = 17%
The number 12 is the coefficient of this probability and it is number of combinations that can fill this "112" focus group. There are 15 different arrangements of four-person focus groups that can be selected. Expressions for all 15 of these coefficients are:
$\textstyle {4!\over 4!\times 0!\times 0!} \ {4!\over 3!\times 0!\times 1!} \ {4!\over 2!\times 0!\times 2!} \ {4!\over 1!\times 0!\times 3!} \ {4!\over 0!\times 0!\times 4!}$
$\textstyle {4!\over 3!\times 1!\times 0!} \ {4!\over 2!\times 1!\times 1!} \ {4!\over 1!\times 1!\times 2!} \ {4!\over 0!\times 1!\times 3!}$
$\textstyle {4!\over 2!\times 2!\times 0!} \ {4!\over 1!\times 2!\times 1!} \ {4!\over 0!\times 2!\times 2!}$
$\textstyle {4!\over 1!\times 3!\times 0!} \ {4!\over 0!\times 3!\times 1!}$
$\textstyle {4!\over 0!\times 4!\times 0!}$
The numerator of these fractions (above the line) is the same for all expressions. It is the sample size−a four-person group−and indicates that the coefficients of these arrangements can be found on Layer 4 of the Tetrahedron. The three numbers of the denominator (below the line) are the number of the focus group members that voted for A, B, C, respectively.
Shorthand is normally used to express combinatorial functions in the following "choose" format (which is read as "4 choose 4, 0, 0", etc.).
$\textstyle {4\choose 4,0,0} \ {4\choose 3,0,1} \ {4\choose 2,0,2} \ {4\choose 1,0,3} \ {4\choose 0,0,4}$
$\textstyle {4\choose 3,1,0} \ {4\choose 2,1,1} \ {4\choose 1,1,2} \ {4\choose 0,1,3}$
$\textstyle {4\choose 2,2,0} \ {4\choose 1,2,1} \ {4\choose 0,2,2}$
$\textstyle {4\choose 1,3,0} \ {4\choose 0,3,1}$
$\textstyle {4\choose 0,4,0}$
But the value of these expression is still equal to the coefficients of the 4th Layer of the Tetrahedron. And they can be generalized to any Layer by changing the sample size (n).
This notation makes an easy way to express the sum of all the coefficients of Layer n:
$\textstyle \sum_{x,y,z} {n \choose x,y,z}$ = 3n.
## Addition of coefficients between layers
The numbers on every layer (n) of the Tetrahedron are the sum of the three adjacent numbers in the layer (n−1) "above" it. This relationship is rather difficult to see without intermingling the layers. Below are italic Layer 3 numbers interleaved among bold Layer 4 numbers:
1 4 6 4 1 1 3 3 1 4 12 12 4 3 6 3 6 12 6 3 3 4 4 1 1
The relationship is illustrated by the lower, central number 12 of the 4th Layer. It is "surrounded" by three numbers of the 3rd Layer: 6 to the "north", 3 to the "southwest", 3 to the "southeast". (The numbers along the edge have only two adjacent numbers in the layer "above" and the three corner numbers have only one adjacent number in the layer above, which is why they are always "1". The missing numbers can be assumed as "0", so there is no loss of generality.) This relationship between adjacent layers is not a magical coincidence. Rather, it comes about through the two-step trinomial expansion process.
Continuing with this example, in Step 1, each term of (A + B + C)3 is multiplied by each term of (A + B + C)1. Only three of these multiplications are of interest in this example:
Layer 3 term Multiply by Product term 6A1B1C1 1B1 6A1B2C1 3A1B2C0 1C1 3A1B2C1 3A0B2C1 1A1 3A1B2C1
(The multiplication of like variables causes the addition of exponents; e.g.: D1 × D2 = D3.)
Then, in Step 2, the summation of like terms (same variables and exponents) results in: 12A1B2C1, which is the term of (A + B + C)4; while 12 is the coefficient of the 4th Layer of the Tetrahedron.
Symbolically, the additive relation can be expressed as:
C(x,y,z) = C(x−1,y,z) + C(x,y−1,z) + C(x,y,z−1)
where C(x,y,z) is the coefficient of the term with exponents x, y, z and x+y+z = n is the layer of the Tetrahedron.
This relationship will work only if the trinomial expansion is laid out in the non-linear fashion as it is portrayed in the section on the "trinomial expansion connection".
## Ratio between coefficients of same layer
On each layer of the Tetrahedron, the numbers are simple whole number ratios of the adjacent numbers. This relationship is illustrated for horizontally adjacent pairs on the 4th Layer by the following:
1 <1:4> 4 <2:3> 6 <3:2> 4 <4:1> 1
4 <1:3> 12 <2:2> 12 <3:1> 4
6 <1:2> 12 <2:1> 6
4 <1:1> 4
1
Because the tetrahedron has three-way symmetry, the ratio relation also holds for diagonal pairs (in both directions), as well as for the horizontal pairs shown.
The ratios are controlled by the exponents of the corresponding adjacent terms of the trinomial expansion. For example, one ratio in the illustration above is:
4 <1:3> 12
The corresponding terms of the trinomial expansion are:
4A3B1C0 and 12A2B1C1
The following rules apply to the coefficients of all adjacent pairs of terms of the trinomial expansion:
• The exponent of one of the variables remains unchanged (B in this case) and can be ignored.
• For the other two variables, one exponent increases by 1 and one exponent decreases by 1.
• The exponents of A are 3 and 2 (the larger being in the left term).
• The exponents of C are 0 and 1 (the larger being in the right term).
• The coefficients and larger exponents are related:
• 4 × 3 = 12 × 1
• 4 / 12 = 1 / 3
• These equations yield the ratio: "1:3".
The rules are the same for all horizontal and diagonal pairs. The variables A, B, C will change.
This ratio relationship provides another (somewhat cumbersome) way to calculate tetrahedron coefficients:
The coefficient of the adjacent term equals the coefficient of the current term multiplied by the current-term exponent of the decreasing variable divided by the adjacent-term exponent of the increasing variable.
The ratio of the adjacent coefficients may be a little clearer when expressed symbolically. Each term can have up to six adjacent terms:
For x = 0: C(x,y,z−1) = C(x,y−1,z) × z / y C(x,y−1,z) = C(x,y,z−1) × y / z
For y = 0: C(x−1,y,z) = C(x,y,z−1) × x / z C(x,y,z−1) = C(x−1,y,z) × z / x
For z = 0: C(x,y−1,z) = C(x−1,y,z) × y / x C(x−1,y,z) = C(x,y−1,z) × x / y
where C(x,y,z) is the coefficient and x, y, z are the exponents. In the days before pocket calculators and personal computers, this approach was used as a school-boy short-cut to write out Binomial Expansions without tedious algebraic expansions or clumsy factorial computations.
This relationship will work only if the trinomial expansion is laid out in the non-linear fashion as it is portrayed in the section on the "trinomial expansion connection".
## Relationship with Pascal's triangle
It is well known that the numbers along the three outside edges of the nth Layer of the tetrahedron are the same numbers as the nth Line of Pascal's triangle. However, the connection is actually much more extensive than just one row of numbers. This relationship is best illustrated by comparing Pascal's triangle down to Line 4 with Layer 4 of the tetrahedron.
Pascal's triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
Tetrahedron Layer 4
1 4 6 4 1
4 12 12 4
6 12 6
4 4
1
Multiplying the numbers of each line of Pascal's triangle down to the nth Line by the numbers of the nth Line generates the nth Layer of the Tetrahedron. In the following example, the lines of Pascal's triangle are in italic font and the rows of the tetrahedron are in bold font.
1
× 1 =
1
1 1
× 4 =
4 4
1 2 1
× 6 =
6 12 6
1 3 3 1
× 4 =
4 12 12 4
1 4 6 4 1
× 1 =
1 4 6 4 1
The multipliers (1 4 6 4 1) compose Line 4 of Pascal's triangle.
This relationship demonstrates the fastest and easiest way to compute the numbers for any layer of the Tetrahedron without computing factorials, which quickly become huge numbers. (Extended precision calculators become very slow beyond Tetrahedron Layer 200.)
If the coefficients of Pascal's triangle are labeled C(i,j) and the coefficients of the Tetrahedron are labeled C(n,i,j), where n is the layer of the Tetrahedron, i is the row, and j is the column, then the relation can be expressed symbolically as:
C(i,j) × C(n,i) = C(n,i,j) i = 0 to n, j = 0 to i
[It is important to understand that i, j, n are not exponents here, just sequential labeling indexes.]
## Parallels to Pascal's triangle and Multinomial Coefficients
This table summarizes the properties of the trinomial expansion and the trinomial distribution, and it compares them to the binomial and multinomial expansions and distributions:
Type of polynomial bi-nomial tri-nomial multi-nomial
Order of polynomial 2 3 m
Example of polynomial A+B A+B+C A+B+C+...+M
Geometric structure[1] triangle tetrahedron m-simplex
Element structure line layer group
Symmetry of element 2-way 3-way m-way
Number of terms per element n+1 (n+1) × (n+2) / 2 (n+1) × (n+2) ×...× (n+m−1) / (m−1)
Sum of values per element 2n 3n mn
Example of term AxBy AxByCz AxByCz...Mm
Sum of exponents, all terms n n n
Coefficient equation[2] n! / (x! × y!) n! / (x! × y! × z!) n! / (x1! × x2! × x3! ×...× xm!)
Sum of coefficients "above" 2 3 m
Ratio of adjacent coefficients 2 6 m × (m−1)
^1 A simplex is the simplest linear geometric form that exists in any dimension. Tetrahedrons and triangles are examples in 3 and 2 dimensions, respectively.
^2 The formula for the binomial coefficient is usually expressed as: n! / (x! × (n−x)!); where n−x = y.
## Other properties
### Exponentional construction
Arbitrary layer n can be obtained in a single step using the following formula:
$\left(b^{d\left(n+1\right)}+b^d+1\right)^n,$
where b is the radix and d is the number of digits of any of the central multinomial coefficients, that is
$\textstyle d=1+\left\lfloor\log_b{n\choose k_1,k_2,k_3}\right\rfloor,\ \sum_{i=1}^3{k_i} = n,\ \left\lfloor\frac{n}{3}\right\rfloor \le k_i \le \left\lceil\frac{n}{3}\right\rceil,$
then wrapping the digits of its result by d(n+1), spacing by d and removing leading zeros.
This method generalised to arbitrary dimension can be used to obtain slices of any Pascal's simplex.
#### Examples
For radix b = 10, n = 5, d = 2:
$\textstyle\left(10^{12} + 10^2 + 1\right)^5$
= 10000000001015
= 1000000000505000000102010000010303010000520302005010510100501
1 1 1
000000000505 00 00 00 00 05 05 .. .. .. .. .5 .5
000000102010 00 00 00 10 20 10 .. .. .. 10 20 10
~ 000010303010 ~ 00 00 10 30 30 10 ~ .. .. 10 30 30 10
000520302005 00 05 20 30 20 05 .. .5 20 30 20 .5
010510100501 01 05 10 10 05 01 .1 .5 10 10 .5 .1
wrapped by d(n+1) spaced by d leading zeros removed
For radix b = 10, n = 20, d = 9:
$\textstyle\left(10^{189} + 10^9 + 1\right)^{20}$
Pascal's pyramid layer #20.
### Sum of coefficients of a layer by rows
Summing the numbers in each row of a layer n of Pascal's pyramid gives
$\left(b^d + 2\right)^n,$
where b is the radix and d is the number of digits of the sum of the 'central' row (the one with the greatest sum).
1 ~ 1 \ 1 ~ 1 \ 1 ~ 1 \ 1 ~ 1 \ 1 ~ 1
--- 1 \ 1 ~ 2 \ 2 \ 2 ~ 4 \ 3 \ 3 ~ 06 \ 4 \ 4 ~ 08
1 ----- 1 \ 2 \ 1 ~ 4 \ 3 \ 6 \ 3 ~ 12 \ 6 \12 \ 6 ~ 24
1 2 --------- 1 \ 3 \ 3 \ 1 ~ 08 \ 4 \12 \12 \ 4 ~ 32
1 4 4 ------------- 1 \ 4 \ 6 \ 4 \ 1 ~ 16
1 06 12 08 ------------------
1 08 24 32 16
120 121 122 1023 1024
### Sum of coefficients of a layer by columns
Summing the numbers in each column of a layer n of Pascal's pyramid gives
$\left(b^{2d} + b^d + 1\right)^n,$
where b is the radix and d is the number of digits of the sum of the 'central' column (the one with the greatest sum).
1 |1| |1| |1| | 1| | 1|
--- 1| |1 |2| |2| |3| |3| | 4| | 4| | 5| | 5|
1 ----- 1| |2| |1 |3| |6| |3| | 6| |12| | 6| |10| |20| |10|
1 1 1 --------- 1| |3| |3| |1 | 4| |12| |12| | 4| |10| |30| |30| |10|
1 2 3 2 1 ------------- 1| | 4| | 6| | 4| | 1 | 5| |20| |30| |20| | 5|
1 3 6 7 6 3 1 -------------------------- 1| | 5| |10| |10| | 5| | 1
1 04 10 16 19 16 10 04 01 --------------------------------
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1110 1111 1112 1113 101014 101015
## Usage
In genetics, it is common to use Pascal's pyramid to find out the proportion between different genotypes on the same crossing. This is done by checking the line that is equivalent to the number of phenotypes (genotypes + 1). That line will be the proportion.[more detail needed] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 22, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8635914325714111, "perplexity": 1211.4489474494305}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375095557.73/warc/CC-MAIN-20150627031815-00213-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/i-must-use-the-wave-equation-to-to-find-the-speed-of-a-wave.296983/ | # I must use the wave equation to to find the speed of a wave.
#### afcwestwarrior
457
0
1. The problem statement, all variables and given/known data
I must use the wave equation to to find the speed of a wave.
y(x,t) = (3.0mm) sin [(4.00mm^-1)x - (7.00 s^-1)t]
2. Relevant equations
Here's the wave equation. It has strange symbols.
(∂^2 y) / (∂ x^2) = (1 / v^2) ((∂^2 y)/ (∂^2 t)
3. The attempt at a solution
There's no way that I can attempt this solution if I don't know how to use this formula.
Do I just take the derivative.
Last edited:
Related Introductory Physics Homework News on Phys.org
#### afcwestwarrior
457
0
Nevermind. I figured it out. V = (w/k)
k= 4.00 mm^-1
w = 7.00 s^-1
V= 7/4 = 1.750 m/s
#### BishopUser
161
0
I don't think your relevant equation is relevant for this problem. You have the equation of the wave for y-position as a function of x-position and time. What do you know about taking derivatives of position functions with respect to time?
#### afcwestwarrior
457
0
I know how to find Velocity and Acceleration, but when I derive these equations, I have a bit of trouble.
"I must use the wave equation to to find the speed of a wave."
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• Solo and co-op problem solving | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8402893543243408, "perplexity": 2339.4597157218336}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578532050.7/warc/CC-MAIN-20190421180010-20190421202010-00208.warc.gz"} |
https://www.clutchprep.com/physics/practice-problems/40981/consider-the-circuit-shown-in-the-sketch-the-current-in-the-8-0-937-160-resistor-1 | Kirchhoff's Loop Rule Video Lessons
Concept
# Problem: Consider the circuit shown in the sketch. The current in the 8.0 Ω resistor is 2.0 A, in the direction shown. What is the resistance of the resistor R3?
###### FREE Expert Solution
88% (352 ratings)
###### Problem Details
Consider the circuit shown in the sketch. The current in the 8.0 Ω resistor is 2.0 A, in the direction shown.
What is the resistance of the resistor R3? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9078339338302612, "perplexity": 1328.8382812892569}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039491784.79/warc/CC-MAIN-20210420214346-20210421004346-00001.warc.gz"} |
https://www.mathdoubts.com/pythagorean-identity-secant-tangent/ | # Pythagorean Identity of Secant and Tangent functions
The subtraction of square of tangent function from secant function at an angle equals to one, is called Pythagorean identity for secant and tangent functions.
## Formula
$\sec^2{\theta} \,-\, \tan^2{\theta} = 1$
### Proof
$\Delta CAB$ is a right angled triangle, whose opposite side, adjacent side and hypotenuse are denoted by $\overline{BC}$, $\overline{AB}$ and $\overline{AC}$ respectively. The lengths of the associated sides are $BC$, $AB$ and $AC$ respectively.
#### Expressing Relation between sides
The relation between the sides of the right angled triangle can be written in mathematical form according to the Pythagorean Theorem.
${AC}^2 = {BC}^2 + {AB}^2$
$\implies {BC}^2 + {AB}^2 = {AC}^2$
Try to express the mathematical equation in terms of trigonometric functions. It is possible to express it in terms of secant and tangent functions if the equation is divided by the square of the length of the side $AB$.
$\implies \dfrac{{BC}^2 + {AB}^2}{{AB}^2} = \dfrac{{AC}^2}{{AB}^2}$
$\implies \dfrac{{BC}^2}{{AB}^2} + \dfrac{{AB}^2}{{AB}^2} = \dfrac{{AC}^2}{{AB}^2}$
$\implies \Bigg(\dfrac{BC}{AB}\Bigg)^2 + \Bigg(\dfrac{AB}{AB}\Bigg)^2 = \Bigg(\dfrac{AC}{AB}\Bigg)^2$
$\implies \Bigg(\dfrac{BC}{AB}\Bigg)^2 + 1 = \Bigg(\dfrac{AC}{AB}\Bigg)^2$
#### Expressing the Equation in terms of Secant and Tangent
Now, write the ratios of the sides in terms of secant and tangent functions.
$\dfrac{BC}{AB} = \tan{\theta}$
$\dfrac{AC}{AB} = \sec{\theta}$
Replace the ratios of the sides by the respective trigonometric functions to transform the equation in the form of trigonometric functions.
$\implies (\tan{\theta})^2 + 1 = (\sec \theta)^2$
$\implies \tan^2{\theta} + 1 = \sec^2{\theta}$
$\implies 1 = \sec^2{\theta} \,-\, \tan^2{\theta}$
$\,\,\, \therefore \,\,\,\,\,\, \sec^2{\theta} \,-\, \tan^2{\theta} = 1$
Therefore, it is proved that the subtraction of square of tangent function from square of secant function at an angle is equal to one. The trigonometric identity is actually derived as per Pythagoras Theorem. Thus, it is known as the Pythagorean identity for secant and tangent functions. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.959017813205719, "perplexity": 186.69142927982605}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221213691.59/warc/CC-MAIN-20180818154147-20180818174147-00681.warc.gz"} |
https://math.stackexchange.com/questions/2304855/methods-to-show-an-ideal-in-the-ring-of-integers-mathcalo-k-is-a-proper-ide | # Methods to show an ideal in the ring of integers $\mathcal{O}_K$ is a proper ideal
I was wondering if there were general "tactics" to show if an ideal in the ring of integers $\mathcal{O}_K$, where $K/\mathbb{Q}$ is a number field of degree $n$.
For example, consider $\mathcal{O}_K = \mathbb{Z}[\sqrt{-5}]$ and $P = (2,1+\sqrt{5})$. One method to show this is a proper idea is to argue that if $a + b\sqrt{-5} \in P,$ then $a+b\sqrt{-5} = 2x + y(1+\sqrt{-5})$ for integers $x,y$ and comparing coefficients, we get $a - b \equiv 0 \bmod 2,$ so $P$ cannot be the whole ring. However, this method seems quite laborious, and this may just be a special case when modding out by a prime actually works.
For me, the following isomorphism seems quite intuitive: as $\mathbb{Z}[\sqrt{-5}] \cong \mathbb{Z}[x]/(x^2+5),$ we should have that \begin{align*} \mathbb{Z}[\sqrt{-5}]/(2,1+\sqrt{5}) &\cong \mathbb{Z}[x]/(x^2+5,2,1+x) \\ &\cong \mathbb{Z}/(6,2) \text{ by mapping } x \mapsto -1 \\ &\cong \mathbb{Z}/2\mathbb{Z} \end{align*} so the ideal $P$ is indeed proper. But I can't seem to set up an explicit isomorphism at the moment, so I was wondering if this works generally?
Furthermore, are there any useful techniques one might use to show an ideal is indeed proper?
• Could you please explain what you mean when you say that you can't set up an explicit isomorphism? Doesn't your isomorphism send a coset $a + b\sqrt{-5} + (2, 1+\sqrt{-5})$ in $\mathbb Q [\sqrt{-5}]/(2,1+\sqrt{5})$ to the element $a - b \in \mathbb Z / 2 \mathbb Z$ (i.e. we think of $\sqrt{5}$ as $x$, then map $x \mapsto -1$, like in your second method)? And doesn't this agree perfectly with your first method? – Kenny Wong May 31 '17 at 23:43
• There are other ways to see that $(2, 1+\sqrt{-5})$ is a proper ideal. For example, we can stare at the ideal-theoretical equation $(2)(3) = (1+\sqrt{-5})(1 - \sqrt{-5})$ and deduce that any prime ideal that divides $(2)$ must also divide $(1+\sqrt{-5})$ or $(1-\sqrt{-5})$, and hence, any such prime ideal divides $(2, 1+\sqrt{-5}) = (2, 1- \sqrt{-5})$. Or we can apply Dedekind's factorization theorem to show that $(2) = (2, 1+\sqrt{-5})^2$. But admittedly, these methods don't work for every ideal in every ring of integers. – Kenny Wong May 31 '17 at 23:47
• your isomorphism isn't intuitive at all considering that if you map $x$ to $2$ instead you get something completely different. – mercio Jun 1 '17 at 9:20
• @user1952009 I'm sure you know much more about this than me! For the second method, we're essentially testing whether or not $2$, $1+x$ and $x^2+1$ have g.c.d. $=1$, and it would be natural to use the Euclidean algorithm. I wonder if this is essentially equivalent to using Grobner bases? (I can't remember how the Grobner basis algorithm works off-hand...) – Kenny Wong Jun 1 '17 at 9:23
• @mercio Could you please explain why we would choose to map $x \mapsto 2$ here? The reason we map $x \mapsto -1$ is that $x+ 1$ is in the ideal, so we're quotienting by $x + 1$. (We could just as well map $x \mapsto 85$, because $x + 85 = (x+1)+42\times 2$ is in the ideal, and ultimately we'll get the same answer. But I don't think we can map $x \mapsto 2$...) – Kenny Wong Jun 1 '17 at 9:25
I'm sure there is a much better way, but yes it works generally, assuming you know the multiplication law in $\mathcal{O}_K$ seen as a free $\mathbb{Z}$ module.
Find $r \in I \cap \mathbb{Z}$, list all the elements of the finite ring $$R_0 = \mathcal{O}_K/(r)$$ then write $I = (u_1,\ldots,u_n)$ and list all the elements of the finite quotient rings $$R_1= R_0/(u_1), \quad R_2= R_1/(u_2), \quad R_{m+1} = R_m/(u_{m+1})$$ You'll get $$I = \mathcal{O}_K \qquad \Longleftrightarrow \qquad \mathcal{O}_K / I = R_n = \{0\}$$
The more general setup is $$\mathcal{O}_K = \mathbb{Z}[X_1,\ldots,X_k]/J$$ for some ideal $J$, and we want to know if $$\mathcal{O}_K= I \qquad \Longleftrightarrow \qquad (I,J) = \mathbb{Z}[X_1,\ldots,X_k]$$ there is an algorithm for that using Gröbner basis
A "proper" ideal is any ideal "which is strictly smaller than the whole ring," right?
Then it's enough to show that the given ideal does not contain 1. If the ideal contains a unit, it must also contain 1 and must therefore be the whole ring.
In your example, we have $\mathfrak P = \langle 2, 1 + \sqrt{-5} \rangle$, which is simply the set of all numbers in this domain of the form $2x + y(1 + \sqrt{-5}) = a + b \sqrt{-5}$. The contribution of $2x$ to $a$ and $b$ is even. We can certainly choose $y$ so that $a = 1$, but then $b \neq 0$ like we want, since in fact $b$ must be odd for $a$ to be odd.
I can't set up an isomorphism either, that's a deficiency of mine, but it's kind of overkill if you just want to show that an ideal is proper.
To know is an ideal $\mathfrak{a}$ is proper or not isn't easier if we just compute the size of $\mathcal{O}_K/ \mathfrak{a}$ thinking of $\mathcal{O}_K$ as a lattice?.
As long as you know a $\mathbb{Z}$-basis for $\mathcal{O}_K$ say $\{ \gamma_1, \ldots, \gamma_n \}$ and you're given $\mathfrak{a}$ is terms of $\mathcal{O}_K$-generators say $(\beta_1,\ldots,\beta_m)$ you can esaily find a $\mathbb{Z}$-basis $\{ \alpha_1, \ldots, \alpha_n \}$ for $\mathfrak{a}$ ( reducing the $\mathbb{Z}$-generators of $\mathfrak{a}$ $\{\beta_i \gamma_j \}$ to a $\mathbb{Z}$-basis, for example by hermite form). Then you simply write $\alpha_i=\sum a_{ij} \cdot \gamma_j$ and $| \mathcal{O}_K/ \mathfrak{a} |=|\text{det}(a_{ij})|$ would be $1$ iff $\mathcal{O}_K=\mathfrak{a}$.
In your example $\{1, \sqrt{-5} \}$ is a $\mathbb{Z}$-basis for $\mathcal{O}_K$, $\{2,2\sqrt{-5},1+\sqrt{-5}, (1+\sqrt{-5})\sqrt{-5} \}$ reduces to the $\mathbb{Z}$-basis $\{1+\sqrt{-5},2 \sqrt{-5}\}$ for $P$, because the hermite form of $$\begin{bmatrix} 2 & 0 \\ 0 & 2 \\ 1 & 1\\ -5 & 1 \end{bmatrix}$$ is the matrix
$$\begin{bmatrix} 1 & 1 \\ 0 & 2 \\ 0 & 0\\ 0 & 0 \end{bmatrix}$$
and so $| \mathcal{O}_K/ P |=\text{det}\begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix}$$=2\neq 1$.
• this is what does sagemath (see for example the function __cmp__ where self.pari_rhnf is the Hermite normal form of the ideal) – reuns Jun 2 '17 at 17:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9658228754997253, "perplexity": 143.81026028880427}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256586.62/warc/CC-MAIN-20190521222812-20190522004812-00171.warc.gz"} |
http://math.stackexchange.com/questions/112739/reflexive-transitive-closure-or-transitive-closure | # reflexive transitive closure or transitive closure
This a problem on the definition of reflexive transitive closure in Elements of the Theory of Computation(H.R.Lewis).
Definition 1.6.1: Let $R \subseteq A^2$ be a directed graph defined on a set $A$. The reflexive transitive closure of $R$ is the relation $$R^* = \{ (a,b) : a, b \in A\text{ and there is a path from }a\text{ to }b\text{ in }R\}\;.$$
Also, an example is given as $$R = \{(a_1,a_2), (a_1,a_3), (a_1,a_4), (a_2,a_3), (a_3,a_4)\}$$ and its reflexive transitive closure $$R^* = \{(a_1,a_1), (a_1,a_2), (a_1,a_3), (a_1,a_4), (a_2,a_2), (a_2,a_3), (a_2,a_4), (a_3,a_3), (a_3,a_4), (a_4,a_4) \}\;.$$
My doubt is whether the example goes with the definition and whether the definition is correct itself. By the definition, if $(a, b) \in R^*$ then there is a path from $a$ to $b$ in $R$. However, I can not find a path from $a_1$ to $a_1$ in $R$ but $(a_1,a_1) \in R^*$ as in the example.
I think what the definition wants to say is the transitive closure of $R$.
Edit: here's how the author defines path
A path in a binary relation $R$ is a sequence $(a_1, \ldots, a_n)$ for some $n \geq 1$ such that $(a_i, a_{i+1}) \in R$ for $i = 1, \ldots, n-1$; this path is said to be from $a_1$ to $a_n$. The length of a path $(a_1, \ldots, a_n)$ is $n$.
Although this doesn't seem to clear things up, I find the definition of path in directed graph in Discrete Mathematics and its Applications(Kenneth H.Rosen)
A path from $a$ to $b$ in the directed graph $G$ is a sequence of edges $(x_0,x_1), (x_1,x_2), (x_2,x_3), \ldots, (x_{n-1},x_n)$ in $G$, where $n$ is a nonnegative integer, and $x_0=a$ and $x_n=b$, that is, a sequence of edges where the terminal vertex of an edge is the same as the initial vertex in the next edge in the path. This path is denoted by $x_0, x_1, x_2, \ldots, x_{n-1}, x_n$ and has length $n$. We view the empty set of edges as a path from $a$ to $a$.
Thus, I was wrong about the length of $(a,a)$. It is of length 1 not 0. Moreover, the path denoted by just $x_0$ has length $0$. Since there is no edge of this path we view it from $a$ to $a$. It follows that $(a,a) \in R^*$ no matter whether $(a,a) \in R$ which satisfies the definition of reflexive transitive closure.
-
It seems to me that the Lewis and Rosen definitions disagree. A path $(a,b)$ would have length two by Lewis, one by Rosen. Well, every author is entitled to use his/her own definitions, so long as the definitions are stated clearly and applied consistently. I interpret Lewis as saying $(a)$ is a path of length one from $a$ to $a$, and the last sentence of your edit is in accord with Lewis' definitions. – Gerry Myerson Feb 25 '12 at 2:24
@GerryMyerson, well, so much about the definition stuff. I think I'm gonna stop here since I know which is TC and which is RTC and move on to some practical problems. Thanks for your help!!! – manuzhang Feb 25 '12 at 2:44
Reflexive transitive closure and transitive closure are different. The TC is the smallest transitive relation containing $R$; the RTC is the smallest reflexive transitive relation containing $R$. That's why $(a_1,a_1)$ is in the RTC, but not in the TC.
@manuzhang: The definition is fine if the author’s definition of path allows paths of length $0$. – Brian M. Scott Feb 24 '12 at 11:48
Start at $a$, go to $b$, then go to $c$ - that's a path of length two from $a$ to $c$. Start at $a$, go to $b$ - that's a path of length one from $a$ to $b$. Start at $a$, and don't go anywhere - that's a path of length zero from $a$ to $a$, provided your definitions allow such things. So I think it's time for you to have a look at how the author defines "path". – Gerry Myerson Feb 24 '12 at 22:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9718148708343506, "perplexity": 130.12589952938217}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997894275.63/warc/CC-MAIN-20140722025814-00232-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://iop.figshare.com/articles/_Graphical_representation_of_oscillation_modes_of_the_condensate/1012593/1 | ## Graphical representation of oscillation modes of the condensate
2013-08-19T00:00:00Z (GMT) by
<p><strong>Figure 7.</strong> Graphical representation of oscillation modes of the condensate.</p> <p><strong>Abstract</strong></p> <p>We take into account the higher-order corrections in two-body scattering interactions within a mean-field description, and investigate the stability conditions and collective excitations of a harmonically trapped Bose–Einstein condensate (BEC). Our results show that the presence of higher-order corrections causes drastic changes to the stability condition of a BEC. In particular, we predict that with the help of the higher-order interaction, a BEC can now collapse even for positive scattering lengths; whereas, a usually unstable BEC with a negative scattering length can be stabilized by positive higher-order effects. The low-lying collective excitations are significantly modified as well, compared to those without the higher-order corrections. The conditions for a possible experimental scenario are also proposed.</p> | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8793766498565674, "perplexity": 1105.6939605597463}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202131.54/warc/CC-MAIN-20190319203912-20190319225912-00483.warc.gz"} |
http://mathhelpforum.com/calculus/3145-generalized-mean-value-theorem-print.html | # Generalized Mean Value Theorem
• May 28th 2006, 06:58 AM
TexasGirl
Generalized Mean Value Theorem
Let a<b be elements of R, and let f,g:[a,b]-->R be continuous on [a,b] and differentiable on ]a,b[. Assume that f(a) is less than or equal to g(a) and that f'(x)<g'(x) for x element of ]a,b[. Show that f(x)<g(x) for x element of ]a,b].
Can this be done using Generalized Mean Value Theorem? And either way, could somebody give me some pointers?
Thanks a bunch!
• May 28th 2006, 07:19 AM
CaptainBlack
Quote:
Originally Posted by TexasGirl
Let a<b be elements of R, and let f,g:[a,b]-->R be continuous on [a,b] and differentiable on ]a,b[. Assume that f(a) is less than or equal to g(a) and that f'(x)<g'(x) for x element of ]a,b[. Show that f(x)<g(x) for x element of ]a,b].
Can this be done using Generalized Mean Value Theorem? And either way, could somebody give me some pointers?
Thanks a bunch!
Consider $h:[a,b] \rightarrow \manthbb{R}$ such that
$\forall x \in [a,b]\ h(x)=f(x)-g(x)$.
Then $h$ is strictly decreasing in $[a,b]$, and $h(a)\le 0$.
And the rest should be trivial.
RonL
• May 28th 2006, 07:51 AM
TexasGirl
thanks
thanks a bunch =) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9840098023414612, "perplexity": 1360.0669022947914}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818689411.82/warc/CC-MAIN-20170922235700-20170923015700-00142.warc.gz"} |
http://cvgmt.sns.it/paper/1278/ | # Almost everywhere well-posedness of continuity equations with measure initial data
created by ambrosio on 14 Oct 2009
modified by figalli on 06 Dec 2012
[BibTeX]
Accepted Paper
Inserted: 14 oct 2009
Last Updated: 6 dec 2012
Journal: C. R. Math. Acad. Sci. Paris
Year: 2010
Abstract:
The aim of the note is to present some new results concerning almost everywhere'' well-posedness and stability of continuity equations with measure initial data. The proofs of all such results can be found in \cite{amfifrgi}, together with some application to the semiclassical limit of the Schrödinger equation. } | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8561593890190125, "perplexity": 1433.6962869963736}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039745015.71/warc/CC-MAIN-20181119023120-20181119045120-00416.warc.gz"} |
https://pureportal-staging.strath.ac.uk/en/publications/effect-of-the-damper-property-variability-on-the-seismic-reliabil | # Effect of the damper property variability on the seismic reliability of linear systems equipped with viscous dampers
Andrea Dall’Asta, Fabrizio Scozzese, Laura Ragni, Enrico Tubaldi
Research output: Contribution to journalArticlepeer-review
17 Citations (Scopus)
## Abstract
Viscous dampers are dissipation devices widely employed for seismic structural control. To date, the performance of systems equipped with viscous dampers has been extensively analysed only by employing deterministic approaches. However, these approaches neglect the response dispersion due to the uncertainties in the input as well as the variability of the system properties. Some recent works have highlighted the important role of these seismic input uncertainties in the seismic performance of linear and nonlinear viscous dampers. This study analyses the effect of the variability of damper properties on the probabilistic system response and risk. In particular, the paper aims at evaluating the impact of the tolerance allowed in devices' quality control and production tests in terms of variation of the exceedance probabilities of the Engineering Demand Parameters (EDPs) which are most relevant for the seismic performance. A preliminary study is carried out to relate the variability of the constitutive damper characteristics to the tolerance limit allowed in tests and to evaluate the consequences on the device's dissipation properties. In the subsequent part of the study, the sensitivity of the dynamic response is analysed by harmonic analysis. Finally, the seismic response sensitivity is studied by evaluating the influence of the allowed variability of the constitutive damper characteristics on the response hazard curves, providing the exceedance probability per year of EDPs. A set of linear elastic systems with different dynamic properties, equipped with linear and nonlinear dampers, are considered in the analyses, and subset simulation is employed together with the Markov Chain Monte Carlo method to achieve a confident estimate of small exceedance probabilities.
Original language English 5025-5053 29 Bulletin of Earthquake Engineering 15 11 12 Jun 2017 https://doi.org/10.1007/s10518-017-0169-8 Published - 30 Nov 2017
## Keywords
• damper properties variability
• seismic reliability
• subset simulation
• viscous dampers
## Fingerprint
Dive into the research topics of 'Effect of the damper property variability on the seismic reliability of linear systems equipped with viscous dampers'. Together they form a unique fingerprint. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8884038329124451, "perplexity": 1782.775044460259}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046151699.95/warc/CC-MAIN-20210725143345-20210725173345-00624.warc.gz"} |
https://aas.org/archives/BAAS/v26n2/aas184/abs/S5302.html | Rosat Observations of V1974 Cyg: The Brightest Super Soft X-Ray Source
Session 53 -- Recent Observational Results on Nova Cygni 1992
Oral presentation, Wednesday, 1, 1994, 2:00-5:30
## [53.02] Rosat Observations of V1974 Cyg: The Brightest Super Soft X-Ray Source
S. Starrfield (ASU), J. Krautter (LSW), H. \"{O}gelman (UW), R. Wichmann (LSW), J. Tr\"umper (MPI)
Nova V1974 Cyg was observed by ROSAT on 18 occasions from 22 April 1992 until 3 Dec. 1993. Over this interval it rose from a count rate of 0.3 $\pm$ 0.09 cts s$^{-1}$ to a peak of 76.52 $\pm$ 0.17 cts s$^{-1}$ in the summer of 1993 and then rapidly declined to a value of 0.22 $\pm$ 0.01 cts s$^{-1}$ on the last observation. Its brightness during the summer of 1993 made it the brightest Super Soft Source (SSS) ever observed in X-rays. Our initial observations showed only a hard component with a peak around 1 keV. During the X-ray rise, however, a much softer component appeared that dominated the emitted energy at maximum. It is also this super soft component that decayed the most rapidly. In the same time interval, it declined by a factor of 350 while the harder component declined by less than a factor of 10. We explain the super soft component as the signature of the energy emitted by the underlying white dwarf and its rise caused by the clearing of the ejected nebulae as it expanded and dispersed. The X-ray turn-off was, most likely, caused by the cessation of nuclear burning on the white dwarf as the accreted hydrogen was exhausted. The time to turn-off, about 18 months, implies that the binary system contains a massive ONeMg white dwarf. Although we are still studying the hard component, we suggest that this feature is a signature of mass loss in the system and is caused by the interaction between the differentially moving filaments and the diffuse ejecta. Both of these features were seen and analyzed in high resolution HST ultraviolet spectra (Shore et al. AJ, 106, 2408, 1993). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8645313382148743, "perplexity": 2640.1213819183276}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660706.30/warc/CC-MAIN-20160924173740-00262-ip-10-143-35-109.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/41203/is-physics-very-dependent-on-equipment | # Is physics very dependent on equipment?
I always had the impression that physics depends a lot on particle accelerators and heavy machines for experimentation of new theories, I know there's the field called theoretical physics but until now I have no idea on how TF relies on these machines. Does theoretical physics rely a lot on this machinery?
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There is no simple answer to this question. Physics, and the Natural Science, are basically based on the practice of distinguishing between hypotheses by empirical evidence, which implies experiments, so yes, in that sense we do need our equipment (and we need it well documented).
This is, however, not what you are asking, as I perceive it; what you want to know seems to be to what extent the large scale experiments are necessary.
The answer to this is that, yes, some branches of theoretical physics need large scale experiments if they are to be based in an empirical setting (which is what we desire, since this is how we evaluate scientific content).
The fields in question, cosmology and particle physics for instance, deal with questions of scales far removed from our everyday experience. In order to test the hypothesis of, for instance, the Higgs' boson, access to those scales is necessary, and for that case (and much of particle physics), the key obstacle that needs to be overcome is binding energy: the very small (hypothesised) building blocks of matter are very tightly bound (if they were not, the world we live in would not appear as solid to us, as it does), so in order to study them, we need large scale machines to break them apart.
Likewise, in astrophysics, larger (radio and other) telescopes allow us to probe deeper into the cosmos, which allows us to distinguish between more and more fundamental models of the early universe, for instance. (Incidentally, this brings us back to particle physics, but that is another story.)
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Yes, physics is absolutely dependent on equipment which is needed so an observer can observe and register data so that they can be logically analyzed and the analysis used for predicting future behavior.
Actually there are many levels of instrumentation.
The first one is the observer's brain.
What we observe in nature exists in several scales and physics describes and makes predictions for all scales : from the distances of stars and galaxies and clusters of galaxies to the sizes of atoms and elementary particles.
Observing in human size scale means what our ears hear, what our eyes see, what our hands feel, our nose smells , our mouth tastes. That was the first classification and the level of "proxy", i.e. intermediate between fact and our understanding and classification, which is biological. (the term proxy is widely used in climate researches)
A second level of observing comes when we use proxies, like meters, thermometers, telescopes and microscopes etc. which register on our biological proxies and we accumulate knowledge. These are simple instruments, and with these instruments scientists gathered observations that were used by physicists to construct theories that would describe and predict further observations. At this level we overcome the limits of the human scale and find and study the enormous scales of the galaxies and the tiny scales of the bacteria and microbes. A level of microns and milimeters. We observe waves in liquids with such size wavelengths.
Presently we are studying experimentally dimensions that go from 10^-10 meters and smaller and have hit the basic building blocks of nature, the quantum mechanical level of nature. These experiments need accelerators and the theories describing the results and predicting new ones are at the frontier of physics . Theoretical physicists formulate mathematical models to this effect.
Yes theoretical physics does rely on the data given by large expensive accelerator experiments that need thousands of physicists and engineers to operate and analyze the results. Theoretical physics uses advanced mathematical methods but it is not enough to construct self consistent mathematical models. These models should be validated against the data in order to be called physics theoretical models.
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http://math.stackexchange.com/questions/249614/conditional-probability-question-and-solution | # Conditional probability question and solution
Please could someone review my solutions for the problems below..thanks in advance
An e-mail message can travel through one of three server routes. The probability of transmission of error in each of the servers and the proportion of messages that travel each route are shown in the following table. Assume that the servers are independent.
% messages % errors
Server 1 40 1%
Server 2 25 2%
Server 3 35 1.5%
1) What is the probability of receiving an email containing an error? Solution: this would be .4*.01 + .25*.02 + .35*0.15 = 0.615
2) What is the probability a msg will arrive without error? Solution: .4*.99 + .25*.98 +.35*.95= .9735
3) If a msg arrives without an error, what is the probability that it was sent through server 2? Solution: Let event E = Sent through server 2. Let event F = arrives without an error. We are looking for P(E/F) or the conditional probability. We can use the formula $\frac {P(E \cap F) }{P(F)}$. P(EnF) = .25*.98 = .245. While the P(F) = 1 - (.01) - (.02) - (.15) = .92. Therefore P(E\F) = .245/.92 = .266 or 26.6%
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The procedure used in 1) was correct. Unfortunately, there was a numerical slip. The answer is $$(0.4)(0.01)+(0.25)(0.02)+(0.35)(0.015).$$ The slip was writing that $1.5\%$ is $0.15$.
One should always glance at the answer one gets to check for plausibility: surely these "good" servers cannot produce an error with probability $0.615$!
2) Again, the setup was right, though not optimal. There was a slip: $100\%$ minus $1.5\%$ is not $95\%$.
A more efficient way to solve 2) is to note that the event in 2) is the complement of the event in 1). So to find the answer for 2), it is simplest to find $1-a$, where $a$ is the answer to 1).
3) The procedure used began along correct lines: the conditional probability setup is good. However, $\Pr(F)$ was not computed correctly. The right number is the answer to 2).
There were several problems with the computation of $\Pr(F)$: (i) The wrong idea was used; (ii) There was the error of 1), writing $0.15$ for $1.5\%$; (iii) Even if we assume the expression is right, and $0.15$ is right, the subtraction is incorrect.
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Thanks @andre Nicolas. You mention in 3) that the wrong formula was used. Is this not a conditional probability problem or is it just my interpretation of the formula that is wrong? – bosra Dec 3 '12 at 20:10
I am sorry for possible ambiguity. Have changed "formula" to "idea." Your conditional probability setup was good. It is in the calculation of $\Pr(F)$ that you used, implicitly, an incorrect formula/method. The probability of arrival without error has nothing to do with $1-0.01-0.02-0.015$. – André Nicolas Dec 3 '12 at 20:26
OK, thanks again @Andre Nicolas. I really appreciate your comments and help – bosra Dec 3 '12 at 20:41
You are welcome. Maybe in other questions you ask, numerical stuff could be checked more carefully, so that one can concentrate on mathematical issues. – André Nicolas Dec 3 '12 at 20:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.930978536605835, "perplexity": 328.03820513179494}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257827079.61/warc/CC-MAIN-20160723071027-00050-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/111796/remove-denominators-in-de-rham-cohomology | # Remove denominators in de Rham cohomology
Let $$\omega = \mathrm d \eta$$ be an exact rational $$n$$-form on $$\Bbb P^n$$.
It may happen that the polar locus of $$\eta$$ is not included in the polar locus of $$\omega$$. But is it true that $$\omega = \mathrm d \eta_0$$, where $$\eta_0$$ is an $$(n-1)$$-form which is regular where $$\omega$$ is?
In other words (and considering the affine case), assume that $$F = \partial_1 G_1 + \dotsb + \partial_n G_n$$ with $$F$$ and $$G_i$$'s rational functions in $$x_1,\dotsc,x_n$$. Is it possible to choose the $$G_i$$ such that their denominators are a power of the denominator of $$F$$?
I've been unable to provide a counter example to this question whereas the following formulation seems to indicate that there exists one.
Let $$X$$ be the open set of $$\Bbb P^n$$ where $$\omega$$ is regular, and $$Z$$ the polar locus, in $$X$$, of $$\eta$$, so that $$Z$$ is a hypersurface of $$X$$. If I'm not mistaken, a positive answer to my question is equivalent to the injectivity of the restriction map in the de Rham cohomology $$H^n(X)\to H^n(X\setminus Z).$$
And following Hartshorne's On the De Rham cohomology of algebraic varieties, we have an exact sequence $$\dotsb \to H^{n-1}(X\setminus Z) \to H^{n-2}(Z) \to H^n(X)\to H^n(X\setminus Z),$$ so that the injectivity of the last arrow is equivalent to the nullity of the previous one, or the surjectivity of the one before, the Poincaré residue map. Is there any reason for this residue map to be surjective? What if $$Z$$ is smooth, or is a hyperplane?
For $$n=2$$ we have to check that $$0\to H^1(X) \to H^1(X\setminus Z) \to H^0(Z) \to 0$$ is exact. Which seems to be true … although I don't know why.
Any thought will be much appreciated, including on the specific case $$n=2$$.
I finally found a counter example in a note of Émile Picard written in 1899.
It is an algebraic example but it is easily translated into a rational example with an extra variable. Let $$P(t)$$ be a square-free polynomial of degree at least three. There is a non-zero polynomial $$U(x,y)$$ such that $$\partial_x\left( \frac{\sqrt{P(x)}}{(y-x)\sqrt{P(y)}}\right) - \partial_y\left( \frac{\sqrt{P(y)}}{(x-y)\sqrt{P(x)}}\right) = \frac{U(x,y)}{\sqrt{P(x)}\sqrt{P(y)}}.$$ Let $$F$$ be the right-hand side. Note the pole $$(y-x)$$ which appears inside the derivatives but not in $$F$$.
Picard proved that there is no rational functions $$G_x$$ and $$G_y$$ in $$x$$, $$y$$ and $$\sqrt{P(x)}\sqrt{P(y)}$$ such that:
1. $$F = \partial_x G_x + \partial_y G_y$$
2. $$G_x$$ and $$G_y$$ have no pole outside those of $$F$$
Indeed, this would imply that $$\oint F \mathrm{d}x \mathrm{d}y$$ is zero on every cycle on the Riemann variety of $$\sqrt{P(x)}\sqrt{P(y)}$$. However, let $$a_1$$, $$a_2$$ and $$a_3$$ be three distinct roots of $$P$$, and let $$\gamma_1$$ be a contour in $$\mathbb C$$ which encloses $$a_1$$ and $$a_2$$ but no other root of $$P$$, and let $$\gamma_2$$ be a contour in $$\mathbb C$$ which encloses $$a_2$$ and $$a_3$$ but no other root. $$\gamma_1 \times \gamma_2$$ induces a cycle on the Riemann variety of $$\sqrt{P(x)}\sqrt{P(y)}$$, and we can check that $$\oint_{\gamma_1\times \gamma_2} F(x,y)\mathrm{d}x \mathrm{d}y = 4i\pi.$$
Thanks M. Picard !
Picard, Émile. (1899). Quelques remarques dur les intégrales doubles de seconde espèce dans la théorie des surfaces algébriques, Comptes rendus hebdomadaires des séances de l'Académie des sciences, 129, 539–540, Gallica
Since this question is back on the front page, here is another counterexample. Again, it is algebraic but I think it could be made rational in one more variable; it has the nice feature that all the Hodge structures involved are of Tate type.
Let $$f(z) = \prod_{i=1}^d (z-z_i)$$ with $$z_1$$, $$z_2$$, ..., $$z_d$$ distinct complex numbers. Let $$X$$ be the affine surface $$xy = f(z)$$ and let $$Z = \{ y=0 \}$$. Then $$X \setminus Z$$ projects isomorphically to $$\{ (y,z) : y \neq 0 \}$$, and we see that $$H^1(X \setminus Z) \cong \mathbb{C}$$ and $$H^2(X \setminus Z) = 0$$. However, $$Z$$ is the disjoint union of $$d$$ copies of $$\mathbb{C}$$. So $$H^1(X \setminus Z) \to H^0(Z)$$ is not surjective, and $$H^2(X) \cong \mathbb{C}^{d-1}$$ does not inject into $$H^2(X \setminus Z)$$.
Concretely, let $$\gamma$$ be a path from $$z_i$$ to $$z_j$$ in $$\mathbb{C}$$, not passing through another root of $$f$$. Let $$S = \{ (x,y,z) : |x|=|y|=\sqrt{|f(z)|},\ xy=f(z),\ z \in \gamma \}.$$ So $$S$$ is a $$2$$-sphere.
On $$X$$, we have $$dx/x + dy/y = f'(z)/f(z) dz$$ and thus $$(dx \wedge dz)/x = -(dy \wedge dz)/y$$, wherever these expressions are defined. Now, $$(dx \wedge dz)/x$$ is defined except where $$x=0$$ and $$-(dy \wedge dz)/y$$ is defined except where $$y=0$$, and the locus where $$x=y=0$$ is codimension $$2$$ in the smooth variety $$X$$, so $$(dx \wedge dz)/x$$ extends to a global $$2$$-form on $$X$$. For any polynomial $$g(z)$$, we have $$\int_S g(z) (dx/x) \wedge dz = (2 \pi i) \int_{z_i}^{z_j} g(z) dz$$.
So, for most choices of $$g$$, we have $$\int_S g(z) (dx/x) \wedge dz \neq 0$$, and $$g(z) (dx/x) \wedge dz$$ is a closed, non-exact, $$2$$-form. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 106, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9887552857398987, "perplexity": 90.07939549462665}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371604800.52/warc/CC-MAIN-20200405115129-20200405145629-00438.warc.gz"} |
https://www.physicsforums.com/threads/classical-mechanics.186042/ | # Homework Help: Classical mechanics
1. Sep 21, 2007
### ehrenfest
[SOLVED] classical mechanics
1. The problem statement, all variables and given/known data
is it true that conjugate momentum, canonical momentum, and generalized momentum all mean the same thing (the partial of the Lagrangian with respect to the time derivative of some generalized coordinate)?
2. Relevant equations
3. The attempt at a solution
2. Sep 21, 2007
### siddharth
Yeah.
$$p_j = \frac{\partial L}{\partial \dot{q_j}}$$
is referred to as generalized momentum, canonical momentum or conjugate momentum.
Also, if the potential is velocity dependent, this generalized momentum will be different from the mechanical momentum of the system. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9968039393424988, "perplexity": 1142.776900544546}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823618.14/warc/CC-MAIN-20181211104429-20181211125929-00081.warc.gz"} |
https://www.gradesaver.com/textbooks/science/physics/physics-for-scientists-and-engineers-a-strategic-approach-with-modern-physics-3rd-edition/chapter-25-electric-charges-and-forces-exercises-and-problems-page-746/32 | ## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)
(a) The force on object A is 0.45 N (b) $q_A = 1.0\times 10^{-6}$ $q_B = 5.0\times 10^{-7}~C$
(a) Since both objects experience the same magnitude of electric force, the force on object A is also 0.45 N (b) Let $q$ be the charge on object B. Then the charge on object A is $2q$. We can find the charge $q$. $F = \frac{k~q_A~q_B}{r^2}$ $F = \frac{k~(2q)~q}{r^2}$ $2q^2 = \frac{F~r^2}{k}$ $q^2 = \frac{F~r^2}{2k}$ $q = \sqrt{\frac{F}{2k}}~r$ $q = \sqrt{\frac{0.45~N}{(2)(9.0\times 10^9~N~m^2/C^2)}}~(0.10~m)$ $q = 5.0\times 10^{-7}~C$ Since $q_A = 2q$, then $q_A = 1.0\times 10^{-6}$ and $q_B = 5.0\times 10^{-7}~C$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9682772159576416, "perplexity": 162.00035008646597}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676593051.79/warc/CC-MAIN-20180722061341-20180722081341-00534.warc.gz"} |
http://mathoverflow.net/revisions/88120/list | 2 omitted one of double `the'
You might like to look at my preprint http://arxiv.org/abs/1003.5617 on the homotopy 2-type of a free loop space $LX$. It assumes that $X$ is a 2-type, i.e. the classifying space of a crossed module, and then gives precise formulae for crossed modules representing the the 2-types of the components.
I am aware that the main interest in free loop spaces seems to be their homology, and I can't see how these results help on that.
The paper is planned to be revised with Chris Wensley and to include specific computer calculations, hence the delay.
I might as well quote the theorem.
Let $\mathcal M$ be the crossed module of groups $\delta: M \to P$ and let $X=B\mathcal M$ be the classifying space of $\mathcal M$. Then the components of $LX$, the free loop space on $X$, are determined by equivalence classes of elements $a \in P$ where $a,b$ are equivalent if and only if there are elements $m \in M, p \in P$ such that $$b= p + a + \delta m -p.$$ Further the homotopy $2$-type of a component of $LX$ given by $a \in P$ is determined by the crossed module of groups $L\mathcal M [a]=(\delta_a: M \to P(a))$ where
(i) $P(a)$ is the group of elements $(m,p)\in M \times P$ such that $\delta m= [a,p]$, with composition $(n,q)+(m,p)= (m+n^p,q+p)$;
(ii) $\delta_a(m)= ( -m^a + m,\delta m)$, for $m \in M$;
(iii) the action of $P(a)$ on $M$ is given by $n^{(m,p)}= n^p$ for $n \in M, (m,p) \in P(a)$.
In particular $\pi_1(LX,a)$ is isomorphic to Cok $\delta_a$, and $\pi_2(LX,a) \cong \pi_2(X,*)^{\bar{a}}$, the elements of $\pi_2(X,*)$ fixed under the action of $\bar{a}$, the class of $a$ in $G=\pi_1(X,*)$.
1
You might like to look at my preprint http://arxiv.org/abs/1003.5617 on the homotopy 2-type of a free loop space $LX$. It assumes that $X$ is a 2-type, i.e. the classifying space of a crossed module, and then gives precise formulae for crossed modules representing the the 2-types of the components.
I am aware that the main interest in free loop spaces seems to be their homology, and I can't see how these results help on that.
The paper is planned to be revised with Chris Wensley and to include specific computer calculations, hence the delay.
I might as well quote the theorem.
Let $\mathcal M$ be the crossed module of groups $\delta: M \to P$ and let $X=B\mathcal M$ be the classifying space of $\mathcal M$. Then the components of $LX$, the free loop space on $X$, are determined by equivalence classes of elements $a \in P$ where $a,b$ are equivalent if and only if there are elements $m \in M, p \in P$ such that $$b= p + a + \delta m -p.$$ Further the homotopy $2$-type of a component of $LX$ given by $a \in P$ is determined by the crossed module of groups $L\mathcal M [a]=(\delta_a: M \to P(a))$ where
(i) $P(a)$ is the group of elements $(m,p)\in M \times P$ such that $\delta m= [a,p]$, with composition $(n,q)+(m,p)= (m+n^p,q+p)$;
(ii) $\delta_a(m)= ( -m^a + m,\delta m)$, for $m \in M$;
(iii) the action of $P(a)$ on $M$ is given by $n^{(m,p)}= n^p$ for $n \in M, (m,p) \in P(a)$.
In particular $\pi_1(LX,a)$ is isomorphic to Cok $\delta_a$, and $\pi_2(LX,a) \cong \pi_2(X,*)^{\bar{a}}$, the elements of $\pi_2(X,*)$ fixed under the action of $\bar{a}$, the class of $a$ in $G=\pi_1(X,*)$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8345867395401001, "perplexity": 105.55195027494278}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368705318091/warc/CC-MAIN-20130516115518-00034-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/216176/approximate-diameter-of-polytopes-in-high-dimensions | # approximate diameter of polytopes in high dimensions
I just came across the following problem: Let us consider the unit corner of the n-cube $$\Delta^n = \left\{(t_1,\cdots,t_n)\in\mathbb{R}^n\mid\sum_{i = 1}^{n}{t_i} \leq 1 \mbox{ and } t_i \ge 0 \mbox{ for all } i\right\}.$$ Let $P$ be a polytope in $\Delta_c^n$ generated as the intersection of $m$ half-spaces. Let us equip $\Delta^n$ with the distance defined with a positive-definite matrix $Q$. I am looking to compute the diameter of the set $P$.
From a quick exploration, the diameter is the maximal distance between two extremal points of $P$.
The problem is that the number of these extremal points is typically exponential in $n$.
But if $Q$ has only a few eigenvalues that are not small, is there a smart way to approximate the diameter of $P$ ?
I am interested in any setting, even randomized ones, where the diameter can be approximated correctly numerically. The only hypothesis I do like to keep is the fact that $n$ being large.
This is not a definitive answer.
For the Euclidean distance, it is NP-hard to approximate the diameter within a constant. The best that can be achieved is a factor of $O(\sqrt{n/\log n})$:
Brieden, Andreas. "Geometric optimization problems likely not contained in APX." Discrete and Computational Geometry 28, no. 2 (2002): 201-209. (Springer link.)
See the paper with the discouraging title, "Approximation of diameters: Randomization doesn’t help":
Brieden, Andreas, Peter Gritzman, Ravi Kannan, Victor Klee, László Lovász, and Miklós Simonovits. "Approximation of diameters: Randomization doesn't help." In IEEE Proceedings of the 39th Annual Symposium on Foundations of Computer Science, 1998. pp. 244-251. IEEE, 1998. (IEEE link.)
So if there is any hope for your question, it will depend on the structure of the "positive-definite matrix $Q$," or on restrictions on the shape of the polytope, how it is formed as the intersection of specific halfspaces.
• Thank you for the useful answer and references. I will try to work on s hypothesis on Q and a restriction on the polytopes to make the question more interesting. – rjm Sep 2 '15 at 6:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9530128240585327, "perplexity": 296.955032658364}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496671363.79/warc/CC-MAIN-20191122143547-20191122172547-00307.warc.gz"} |
http://cs.stackexchange.com/questions/9323/where-do-the-length-restrictions-of-the-pumping-lemma-come-from | # Where do the length restrictions of the pumping lemma come from?
For a language $L$ with pumping length $p$, and a string $s\in L$, the pumping lemmas are as follows:
Regular version: If $|s| \geq p$, then $s$ can be written as $xyz$, satisfying the following conditions:
1. $|y|\geq 1$
2. $|xy|\leq p$
3. $\forall i\geq 0: xy^iz\in L$
Context-free version: If $|s| \geq p$, then $s$ can be written as $uvxyz$, satisfying the following conditions:
1. $|vy|\geq 1$
2. $|vxy|\leq p$
3. $\forall i\geq 0: uv^ixy^iz\in L$
My question is this: Can someone give a concise and clear explanation of how regularity (context-freeness) imply the first and second conditions above? The pumping length is determined by (finite) properties (finite number of states or finite properties of production rules, respectively), the third properties guarantee that a state (production rule) can be skipped or repeated arbitrarily many times, but where do the first and second conditions originate? How are they justified?
-
I'm commenting since I might not get your question, but it seems to me obvious that if you have a word of length at least $p \geq 1$, you can partition the word into three parts, $xyz$ such that $|y| \geq 1$ and $|xy| \leq p$. Simply let $x$ be empty, $y$ be the first character and $z$ the rest. Did I completely misunderstand your question? – Pål GD Jan 30 '13 at 19:51
@PålGD, I don't find it obvious in the context of the other conditions of the lemma (non-zero length of the pumpable substring, and arbitrary pumpability). I agree, it's obvious that a string of length at least $x$ can be partitioned into to substrings such that the first has length $\leq x$, but the obviousness (to me) doesn't carry over when you retain the other conditions. Maybe I should clarify the original post? – BlueBomber Jan 30 '13 at 19:55
No, seen in light of the third condition, one should be a bit more careful. As you've noticed, the first two conditions say nothing useful. As you also notice, forcing the $y$ to be nonempty (as is indeed what the first condition says), makes us actually in a position to "pump". – Pål GD Jan 30 '13 at 20:04
Is the context-free PL necessary for your question? It does not make a lot of sense to consider some of these conditions in isolation. – Raphael Jan 30 '13 at 20:27
@BlueBomber, in both case without condition (1) the pumping is pointless (can select the empty substring always). For regular languages, (2) is just that in $n$ symbols the DFA for the language has been in $n + 1$ states, and by the pigeonhole principle if the DFA has $n$ states, one must have repeated (and you can go through the cycle $k$ times, pumping the string). – vonbrand Jan 31 '13 at 20:45
The first condition, i.e. $|y| \geq 1$, is clearly necessary if you want to say something interesting: for $y = \varepsilon$, $xy^iz \in L$ trivially and always holds.
The second condition, i.e. $|xy| \leq p$, is "arbitrary": the lemma still says something interesting if you drop it, and it is still true because the statement becomes weaker.
But remember what we want to use the pumping lemmas for: we want to find a (sufficiently long) word such that all valid decompositions into $x,y,z$ fail to pump. Therefore, it is useful to allow as few such decompositions as possible. Lucky as we are, the proof of the pumping lemma readily yields a strong restriction, namely that there has to be a pumpable decomposition with $|xy| \leq p$ with constant $p$.
Now we have to refute only finitely many prefixes $xy$ (with possibly infinitely many different continuations, of course). If you look at example applications, you will see that they make heavy use of this restriction.
-
condition 2 is "arbitrary" in other ways. E.g., an equivalent lemma is given if we replace (2) by $|yz|<p$. There is even a stronger version of the pumping lemma in which the $y$ part can be anywhere (not necessarily in the beginning or end). The common variant ($|xy|<p$, that is, $y$ is next to the beginning) is due convenience. – Ran G. Jan 31 '13 at 5:10
@RanG. How is a version where the position of $y$ is arbitrary stronger? I think I explained how that would be weaker. Or does that version state other things? – Raphael Jan 31 '13 at 9:37
It is stronger since it gives you more flexibility: maybe the part you wish to pump is not in the beginning of the word. Indeed, it still has an equivalent of $|xy|<p$ for not being "weaker" as you say, but without the need to locate it exactly in the beginning. Think of Ogden's lemma as the most generalized form.. – Ran G. Jan 31 '13 at 17:37
@Raphael, for example the version given above isn't enough to prove that $\{a b^n c^n \colon n \ge 0\} \cup a a^+ b^* c^*$ isn't regular (can pump the $a$), but with $y$ lying anywhere it is easy. – vonbrand Jan 31 '13 at 20:50
@vonbrand: I see now what Ran means, thanks. – Raphael Jan 31 '13 at 23:31
As you move through the states of an automaton, you'll eventually hit a cycle. The cycle is $y$. There is no point in pumping an empty word, hence condition 1. $xy$ is the number of symbols you had to read to find the first cycle. This can't possibly be more than the number of states, hence condition 2. Almost by definition, a cycle can be repeated any number of times, hence condition 3.
The context free variant is very similar. The first condition is we're only looking for useful cycles. The second condition says that a cycle can be found because you'll run out of non-terminals. The third condition says that once you have a cycle you can pump it. Note that to see a cycle in the context free case, you should draw a parse tree.
-
I'll only offer some pointers on the pumping lemma for regular languages; the reasoning is similar enough for the other. Think of $x$ as the part of $s$ generated by everything before starting the first loop for the first time; $y$ as everything generated while in that certain loop for the first loop; and $z$ as everything generated after looping the first time.
1. Since $|s| \geq p$, and $p$ is the number of states in the minimal finite automaton (e.g.) accepting the language, then the automaton must have looped (since at most $p-1$ transitions would have been possible, otherwise). Therefore, $y$ is non-empty; we are stating the fact that since we have a string of a certain length, the automaton accepting it must have looped at some point, so $|y|>1$ is justified.
2. Since $y$ is the first time you go through the first loop, and $x$ is everything before that, you have not visited any state in the automaton more than once. Since there are $|p|$ states in the automaton, and you haven't visited all of them, you have $|xy| < p$.
These aren't so much hypotheses requiring support; they're statements of fact based on how the pumping lemma was dreamed up in terms of finite automata.
-
There are variants of the pumping lemma. I will use yours.
Note that you have really 3 length conditions. The missing one is about the minimum total length of the word. I treat it with the second condition.
In a (big) nutshell :
I call subtree any subpart of the parse tree that has at most one non-terminal at the fringe. The pumping lemma uses recursive subtrees where the non terminal in the fringe is the same as the root of the subtree. The whole parse tree is a subtree.
Subtrees as defined here (and recursive subtrees) are the heart of the matter. Their existence is directly related to the context-freeness.
1st condition: it states simply that if there is an unproductive (fringe with no termina symbol) recursive subtree in the parse tree, it can be short-circuited, so that we are always sure the fringe contains a terminal symbol.
A finiteness issue: It will be used twice. If you have a subtree that contains no recursive subtree, then no path in the subtree has twice the same label (subtree root excepted). The subtree is finitely branching with a limited depth (no more than the number of non-terminals).Hence you have a finite set of such subtrees generating only a finite set of strings at their fringe. Being finite in number, there is an upperbound for the length of the fringes. A contrario, if a fringe exceeds the bound, it is a sure indication it contains a recursive subtree.
"missing condition": The "missing condition" that $\mid s\mid \geq p$ ensure that the string is long enough so that there is at least one recursive subtree in the parse tree for pumping.
2nd condition: you can always get for pumping a recursive subtree that neither dominate nor contains another recursive subtree in the parse tree. If it does, just take the other recursive subtree. Since the parse tree is finite, this terminates. You end up with subtrees (for $vy$ and for $x$) that do not contain recursive subtrees, and the finiteness analysis above garantees the existence of an upperbound.
In the regular grammar case you just have subtrees that do not branch very much. It is really identical to the CF case with some strings replaced by $\epsilon$.
In the CF case, it is often convenient for the proof of the lemma, or its variations, to assume that the grammar is CNF (depending also on the lemma variant)
Much of the formal proof is mathematical presentation, not understanding.
This was an interesting exercise.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8650668859481812, "perplexity": 419.6380563552739}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414637899124.21/warc/CC-MAIN-20141030025819-00168-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/an-integral.116382/ | # An integral
1. Apr 3, 2006
### Ratzinger
Could someone tell me why I'm allowed to change the sign before the second term in (3)? It's certainly baby stuff for you, but could you tell me anyway?
pdf attached
thanks
#### Attached Files:
• ###### an integral.pdf
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2. Apr 4, 2006
### Tom Mattson
Staff Emeritus
As this is from quantum field theory, I'm going to shuffle this over to Advanced Physics to see if it gets more love there.
I don't know the answer off the top of my head, but notice that the second limit of integration changed from p_0=E_p to p_0=-E_p.
3. Apr 4, 2006
### Meir Achuz
It looks like the variable change p_\mu-->-p_\mu has been made.
d3p/E_p won't change sign, but the other signs would.
4. Apr 4, 2006
### Physics Monkey
Hi Ratzinger,
The integration variable in the second part of the integral has been changed from [ itex] \vec{p}[/itex] to [ itex] -\vec{p} [/itex]. The Jacobian of the transformation is unity, and the factor [ itex] E_p = \sqrt{|\vec{p}|^2 + m^2 } [/itex] doesn't change sign. This accounts for the stuff out front. Now what about the stuff in the exponential? Before the change of variable, the argument of the exponential was [ itex] i p_\mu x^\mu = i( - \vec{p}\cdot \vec{r} + p^0 t) = i( - \vec{p}\cdot \vec{r} + E_p t)[/itex] since the original integral specified that [ itex] p^0 = E_p [/itex]. When you change the sign of [ itex] \vec{p} [/itex], [ itex] E_p [/itex] remains fixed, so the exponential now looks like [ itex] i( \vec{p}\cdot \vec{r} + E_p t ) = - i (- \vec{p}\cdot\vec{r} - E_p t )[/itex]. You would like to interpret this as a four vector dot product of the form [ itex] - i (-\vec{p}\cdot \vec{r} + p^0 t ) = - i p_\mu x^\mu [/itex] which clearly means you must now identify [ itex] p^0 = - E_p [/itex].
Edit: Argh, why doesn't the latex work? See the attachment.
#### Attached Files:
• ###### help.pdf
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5. Apr 4, 2006
### Tom Mattson
Staff Emeritus
Well, I was going to say that it isn't working because you put extra spaces in your tex brackets. So I just edited your post to remove the spaces (and to change them to "itex" brackets, so the LaTeX lines up with your text). But then I saw the error messages and concluded that you must have put those spaces in on purpose. :tongue:
I'll stop mucking around here and go approve your attachment...
6. Apr 4, 2006
### Tom Mattson
Staff Emeritus
FYI, I've just learned that the LaTeX problem has been discovered and is being worked on. Sit tight.
7. Apr 4, 2006
### Physics Monkey
Haha! I knew somebody would say something. :rofl:
P.S. Thanks for the info about the itex tag.
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http://mathhelpforum.com/differential-equations/212856-how-solve.html | # Math Help - how to solve this?
1. ## how to solve this?
Use variation of parameter method to solve the differential equation y"+y=secx.
2. ## Re: how to solve this?
First, solve the homogeneous DE. Afterwards you can find the particular solution using 'variation of the parameter'. Show some work!
3. ## Re: how to solve this?
Okay, I did! Turned out to be very easy. But wouldn't it be better for YOU to solve the equation yourself? Or at least try.
Do you know what the "variation of parameters method" is?
Do you know how to find the general solution to the associated homogeneous equation?
4. ## Re: how to solve this?
actually i wanted to check the answer ..... as the answer is not given in the textbook
5. ## Re: how to solve this?
And as has been stated twice above, you should show us what you did and what your answer is, so we can CHECK it, rather than us giving you the solution and answer on the off chance that you actually did do some of the work yourself. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8868962526321411, "perplexity": 454.3373586116214}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246660743.58/warc/CC-MAIN-20150417045740-00247-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.groundai.com/project/scaling-laws-of-turbulence-and-heating-of-fast-solarwind-the-role-of-density-fluctuations/ | Scaling laws of turbulence and heating of fast solar wind: the role of density fluctuations
# Scaling laws of turbulence and heating of fast solar wind: the role of density fluctuations
## Abstract
Incompressible and isotropic magnetohydrodynamic turbulence in plasmas can be described by an exact relation for the energy flux through the scales. This Yaglom-like scaling law has been recently observed in the solar wind above the solar poles observed by the Ulysses spacecraft, where the turbulence is in an Alfvénic state. An analogous phenomenological scaling law, suitably modified to take into account compressible fluctuations, is observed more frequently in the same dataset. Large scale density fluctuations, despite their low amplitude, play thus a crucial role in the basic scaling properties of turbulence. The turbulent cascade rate in the compressive case can moreover supply the energy dissipation needed to account for the local heating of the non-adiabatic solar wind.
###### pacs:
96.50.Ci; 47.27.Gs; 96.50.Tf; 52.35.Ra
The interplanetary space is permeated by the solar wind (1), a magnetized, supersonic flow of charged particles originating in the high solar atmosphere and blowing away from the sun. Low frequency fluctuations of solar wind variables are often described in the framework of fully developed hydromagnetic (MHD) turbulence (2); (3). The large range of scales involved, spanning from 1 AU ( km) down to a few kilometers, make the solar wind the largest “laboratory” where MHD turbulence can be investigated using measurements collected in situ by instruments onboard spacecraft (3). MHD turbulence is often investigated through the Elsässer variables , computed from the local plasma velocity and magnetic field , being the plasma mass density. In terms of such variables, MHD equations can be rewritten as , where is the total hydromagnetic pressure, and diss indicates dissipative terms involving the viscosity and the magnetic diffusivity. As in the Navier-Stokes equations for neutral fluids, the nonlinear terms cause the turbulent energy transfer between different scales, at high Reynolds numbers where dissipative terms can be neglected. However, in the MHD case, they couple the two Elsässer variables, so that the Alfvénic MHD fluctuations , propagating along the background magnetic field, are advected by fluctuations propagating in the opposite direction. The presence of strong correlations (or anti-correlations) between velocity and magnetic fluctuations, along with a nearly constant magnetic intensity and low amplitude density fluctuations, is usually referred to as Alfvénic state of turbulence, and implies that one of the two modes should be negligible, making the nonlinear term of MHD equations vanish for pure Alfvénic fluctuations. In that case, the turbulent energy transfer should also disappear (4). Alfvénic turbulence is observed almost ubiquitously in fast wind. This holds both in the ecliptic fast streams, and in the high latitude wind blowing directly from the sun coronal holes (5); (6); (3). As pointed out in (4), the observation of Alfvénic state turbulence in the solar wind represents therefore a paradoxical “contradiction in terms”.
MHD turbulence however satisfies an important analytical relation, which is the equivalent for magnetized fluids of the Kolmogorov or the Yaglom relations. Under suitable hypotheses, it has been shown (7); (8) that the pseudo-energy fluxes through the scale of the increments of the Elsässer fields follow a linear scaling relation
Y±(ℓ)≡⟨|Δz±|2Δz∓∥⟩=−43ϵ±ℓ . (1)
Here, represents the component of the increment along the direction , and are the dissipation rates per unit mass of the pseudo-energies ( indicates space averages). The scaling law (1) has been recently observed experimentally in polar wind (8) and in the ecliptic plane (9); (10). The confirmation of the scaling law (1) is an important step towards the solution of the apparent paradox of the Alfvénic turbulent state, because it unambiguously shows that an MHD cascade is present, maybe with a weak transfer rate, despite the strong velocity-magnetic fields correlations.
Relation (1), which is of general validity within MHD turbulence, is not always realized in the solar wind observations (8). Indeed, it is possible that local characteristics of the solar wind plasma do not always satisfy the assumptions required for (1) to be valid, namely large-scale homogeneity, isotropy, and incompressibility. The role of anisotropy of solar wind turbulence, which is expected to be important, is not considered in this letter. Density fluctuations in solar wind have a low amplitude, so that nearly incompressible MHD framework is usually considered (11); (12). However, compressible fluctuations are observed, typically convected structures characterized by anti-correlation between kinetic pressure and magnetic pressure (13). Properties and interaction of the basic MHD modes in the compressive case have also been considered in the past (14); (15). In the present paper, we show that density fluctuations, despite their very low amplitude, play a central role in the turbulent energy transfer, and that a phenomenological scaling law obtained by taking into account density fluctuations is observed in a much larger proportion of fast solar wind. We also show that the turbulent dissipation can account for a large fraction of the local heating causing a slower than expected decrease of temperature with distance. A first attempt to include density fluctuations in the framework of fluid turbulence was due to Lighthill (16). He pointed out that in a compressible energy cascade, the mean energy transfer rate per unit volume should be constant in a statistical sense ( being the characteristic velocity fluctuations at the scale ), obtaining . Fluctuations of a density-weighted velocity field should thus follow the usual Kolmogorov scaling . The same phenomenological conjecture can be introduced in MHD turbulence by considering the pseudo-energy dissipation rates per unit volume , and introducing density-weighted Elsässer fields, defined as . The equivalent of the Yaglom-type relation
W±(ℓ)≡⟨|Δw±|2Δw∓∥⟩ ⟨ρ⟩−1=−43ϵ±ℓ (2)
should then hold for the density-weighted increments . Note that we have defined the flux so that it reduces to in the case of constant density, allowing for comparisons between the compressible scaling (2) and the purely incompressible one (1). Despite its simple phenomenological derivation, the introduction of the density fluctuations in the Yaglom-type scaling (2) seems to describe correctly the turbulent cascade for compressible fluid (or magnetofluid) turbulence. The law for the velocity field has been observed in recent numerical simulations (17); (18).
We will now study the cascade properties of compressive MHD turbulence from solar wind data collected by spacecraft Ulysses. In order to avoid as far as possible variations due to solar activity, or other ecliptic disturbances such as slow wind sources, coronal mass ejection, current sheets, we concentrate our analysis on pure Alfvénic state turbulence observed in high speed polar wind. We use here measurements from Ulysses spacecraft in the first six months of 1996. Such period was characterized by low solar activity, so that solar origin disturbances werebv almost absent. Moreover, the spacecraft orbit was at high and slowly decreasing heliolatitude, from about to , and presented small variations of the heliocentric distance , from 3 AU to 4 AU. Since the mean wind speed in the spacecraft frame is much larger than the typical velocity fluctuations, and is nearly aligned with the radial direction , space scales can be viewed as time scales , related through the Taylor hypothesis by . We then used 8 minutes averaged time series of both Elsässer variables and density (obtained as the sum of proton density and 4 times He density), to compute the density-weighted time series . From this time series we calculate the increments for different time lags , and the third-order mixed structure functions by time averaging over windows of fixed duration . The same procedure has been also been used to calculate the quantities using the time series of the Elsässer fields . In order to eliminate instationarities, heliolatitude and heliocentric distance changes, and to explore the wind properties locally, averages are computed over a moving window of about 11 days, consisting of 2048 data points. Accuracy of third order moments estimate (19) was tested with such sample size (20). We found that the third-order structure functions computed from the Ulysses data show a linear scaling
W±(τ)∼43ϵ±⟨vR⟩τ (3)
during a considerable fraction of the period under study. In particular, we observed linear scaling of in about half of the signal, while displays scaling on about a quarter of the sample. As comparison, the corresponding incompressive scaling law for was only observed in a third of the whole period, considerably smaller than the compressible case (8). The portions of wind where the scaling is present are distributed in the whole period, and their extensions span from 6 hours up to 10 days. The linear scaling law generally extends on about 2 decades, from a few minutes to one day or more. For the compressible scaling, the two fluxes coexist in a large number of cases. This does not hold for the incompressive scaling, where in general the scaling periods for the two fluxes are disjoint.
An interesting open question is the problem of the solar wind heating. The first models of solar wind assumed an adiabatic cooling due to spherical expansion of plasma blowing out of the sun. This would result in a radial decrease of the proton temperature with . On the contrary, spacecraft measurements (27) have shown that the temperature decay is slower than the adiabatic prescription, with . This implies that some local heating mechanism is present. One standing hypothesis is that the heating could be provided by energy dissipation occurring at the small scales of a turbulent cascade (23); (9); (24). Using equation (1), the rates at which the incompressible turbulent pseudo-energy is transported down the scales, and eventually dissipated at small scale, can be measured directly from data. This has recently be used to investigate whether or not a turbulent cascade can heat the solar wind. Results were however not conclusive. In fact, the measured incompressible dissipation rate of pseudo-energies can only account for up to 50 % of the solar wind heating (10); (21).
Figure 3 shows the radial profiles of the pseudo-energy transfer rates for both the compressive and incompressive cascades. In the same figure, we show the profiles of the heating rates needed to obtain the observed temperatures, as estimated from heating models (9); (24); (21) and from the measured temperatures (the two different values refer to the different estimates of the temperature obtained from Ulysses instruments). It is evident that, while the incompressive cascade cannot provide all the energy needed to heat the wind, the density fluctuations coupled with magnetohydrodynamic turbulence can supply the amount of energy required. This evidence shows the importance of the density fluctuations in polar, fast solar wind turbulence, confirming that it should be considered as an example of compressive fully developed MHD turbulence. Note that, since in a few samples we measured both and in the same period, the values of the energy and cross-helicity transfer rates can be disentangled. From the values obtained, it is clear that the cross-helicity contribution, indicating the importance of the Alfvénic state of turbulence, can vary from a negligible fraction (less than 1 %) to a considerable 25 % of the energy contribution. Since its amplitude does not appear to be correlated with the observation of the cascade, Alfvénicity seems not to play a crucial role in the cascade at the observed scales. This would be in agreement with previous analysis of solar wind turbulence anisotropy, where the Alfvénic contribution to the field fluctuations is small (25); (26).
To summarize, we used the density-weighted Elsässer fields to show for the first time that a phenomenological compressive Yaglom-like relation is verified to a large extent within the solar wind turbulence. This implies that low amplitude density fluctuations play a crucial role for scaling laws of solar wind turbulence (22). This observation also confirm the recent results for the Kolmogorov -law from numerical simulations of compressible turbulence (17), while no experimental evidences from real fluids had been found so far. This could be attributed to the incompressible nature of flows in ordinary fluids accessible to laboratory experiments. Here in fact, we present the first experimental observation of relation (2) in real systems. Using solar wind data, we have access to a sample of weakly compressible MHD turbulence in nature. Scaling law is found to be quite common and extends on a large range of scales, indicating not only that a nonlinear MHD cascade for pseudo-energies is active in the solar wind turbulence, but also that compressible effects are an important ingredient of the cascade. We point out that the observed departures from the scaling law could be due to presence of inhomogeneity and anisotropy in the solar wind (28). The compressive corrections to the cascade also cause the transfer of a considerably larger amount of energy toward the small scales, where it can be dissipated to heat the plasma locally. The role of anisotropy in the solar wind turbulent cascade still remains an open question.
### References
1. A.J. Hundhausen, Coronal Expansion and Solar Wind, Springer, New York (1972)
2. C.-Y. Tu and E. Marsch, Space Sci. Rev. 73, 1 (1995)
3. R. Bruno and V. Carbone, Living Rev. Solar Phys. 2, 4 (2005)
4. M. Dobrowolny, A. Mangeney and P. Veltri, Phys. Rev. Lett. 45, 144 (1980)
5. J.M. Belcher and L. Davis Jr, J. Geophys. Res. 76, 3534 (1971)
6. R. Bruno, B. Bavassano and U. Villante, J. Geophys. Res. 90, 4373 (1985)
7. H. Politano, and A. Pouquet, J. Geophys. Lett. 25, 273 (1998)
8. L. Sorriso-Valvo, et al., Phys. Rev. Lett. 99, 115001 (2007)
9. B.J. Vasquez, et al. J. Geophys. Res. 112, A07101 (2007)
10. B.T. MacBride, C.W. Smith, and M.A. Forman, Astrophys J. 679, 1644 (2008)
11. D. Montgomery, M.R. Brown, and W.H. Matthaeus, J. Geophys. Res. 92, 282 (1987); W.H. Matthaeus, and M.R. Brown, Phys. Fluids 31, 3634 (1988); G.P. Zank, and W.H. Matthaeus, Phys. Fluids 3, 69 (1991); G.P. Zank, and W.H. Matthaeus, Phys. Fluids 5, 257 (1993)
12. W.H. Matthaeus, et al., J. Geophys. Res. 96, 5421 (1991); B. Bavassano, and R.Bruno, J. Geophys. Res. 100, 9475 (1995)
13. C.-Y. Tu, and E. Marsch, J. Geophys. Res. 99, 21481 (1994)
14. P. Goldreich, and S. Sridhar, Astrophys. J. 438, 763 (1995)
15. J. Cho, and A. Lazarian, Phys. Rev. Lett. 88, 245001 (2002)
16. M.J. Lighthill, in IAU Symp. 2, Gas Dynamics of Cosmic Clouds (Amsterdam: North Holland), 121 (1955)
17. A.G. Kritsuk, et al., Astrophys. J. 665, 416 (2007)
18. G. Kowal, and A. Lazarian, Astrophys. J. 666, L69 (2007)
19. J. J. Podesta, et al., Nonlin. Process. in Geophys. 16, 99 (2009)
20. T. Dudok de Wit, Phys. Rev. E 70, 055302R (2004)
21. R. Marino, et al., Astrophys J. 677, L71 (2008)
22. M.-M. Mac Low, Astrophys. J. 524, 169 (1999)
23. C. W. Smith, et al., J. Geophys. Res. 106, 8253 (2001)
24. M.K. Verma, D.A. Roberts, and M.L. Goldstein, J. Geophys. Res. 100, 19839 (1995)
25. J.W. Bieber, W. Wanner, and W.H. Matthaeus, J. Geophys. Res. 101, 2511 (1996)
26. T.S. Horbury, M.A. Forman, and S. Oughton, Plasma phys. Control. Fusion 47, B703 (2005)
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28. J. J. Podesta, M. A. Forman and C. W. Smith, Phys. Plasmas 14, 092305 (2007)
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The feedback must be of minumum 40 characters | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9572536945343018, "perplexity": 1494.0605033970971}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376826968.71/warc/CC-MAIN-20181215174802-20181215200802-00206.warc.gz"} |
https://homework.zookal.com/questions-and-answers/a-parallelplate-capacitor-made-of-two-disc-shaped-plates-of-284131604 | 1. Science
2. Physics
3. a parallelplate capacitor made of two disc shaped plates of...
# Question: a parallelplate capacitor made of two disc shaped plates of...
###### Question details
A parallel-plate capacitor made of two disc shaped plates of diameter R has a small spacing d between the plates
where R >> d. It is charged by a current I. Derive an expression for the magnetic field as a function of r the radial
distance from the axis of the capacitor. Recall that the electric field inside a parallel plate capacitor is E = σ/(ε0) where σ is the surface charge Q/A (Q is the total charge in the capacitor and A is the area). | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9185934662818909, "perplexity": 1183.1064031435144}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488286726.71/warc/CC-MAIN-20210621151134-20210621181134-00459.warc.gz"} |
http://www.cbsemaster.org/2012/03/cbse-class-6th-vi-mathematics-chapter-7_29.html | ## Thursday, 29 March 2012
### CBSE Class 6th ( VI) Mathematics Chapter 7. Fractions: Exercise 7.2 Solutions
1. Proper Fraction :
1. A proper fraction is a number representing part of a whole.
2. In a proper fraction the denominator shows the number of parts into which the whole is divided and the numerator shows the
number of parts we have taken out.
3. In a proper fraction the numeratoris always less than the denominator.
4. On a number line, all these fractions lie to the left of 1as they are less than 1.For Examples
1 2 3 5 7 , , , , 4 6 9 10 12
are all proper fractions
2. Improper Fraction :
1. The fractions, where the numerator is bigger than the denominator are called
improper Factions. For Examples
3 7 12 18 27 , , , , 2 3 7 5 8
are all improper fractions
3. Mixed Fraction :
1. A mixed fraction has a combination of a whole and a part.For example :
2 3 1 2 5 , 4 , 7 , 3 3 7 6 7
are all mixed fractions
1. Draw number lines and locate the points on them :
a.
1 1 3 4 2 4 4 4
b.
1 2 3 7 8 8 8 8
4. c.
2 3 8 4 5 5 5 5
0 1 1 3 4 | | | | | 0 4 2 4 4
b.
0 1 2 3 7 | | | | | | | | 0 8 8 8 8
c.
0 2 3 4 8 | | | | | | | | | 0 5 5 5 5
2. Express the following as mixed fractions :
20 a. 3
11 b. 5
17 c. 17
28 d. 6
19 e. 6
35 f. 9
Answer: We know that where the numerator is bigger than the denominator are called improper Factions and a mixed fraction has a combination of a whole and a part A improper fraction can be expressed as mixed fraction.
20 2 a. = 6 3 3
11 1 b. = 2 5 5
17 3 c. = 2 7 7
28 3 d. = 5 5 5
19 1 e. = 3 6 6
35 8 f. = 3 9 9
3. Express the following as improper fractions :
3 a. 7 4
6 b. 5 7
5 c. 2 6
3 d. 10 5
3 e. 9 7
4 f. 8 9
Answer: A mixed fraction has a combination of a whole and a part and where the numerator is bigger than the denominator are called improper Factions. A mixed fraction can be expressed as improper fraction.
3 31 a. 7 = 4 4
6 41 b. 5 = 7 7
5 17 c. 2 = 6 6
3 53 d. 10 = 5 5
3 66 e. 9 = 7 7
4 76 f. 8 = 9 9 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8810206651687622, "perplexity": 296.4497760169647}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645348533.67/warc/CC-MAIN-20150827031548-00297-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/69774/invariance-of-states-under-local-unitary-transformations | # Invariance of states under local unitary transformations [closed]
How can I show explicitly that the bell state $$|\psi^{-}>=\frac{1}{\sqrt{2}}(|0>|1>-|1>|0>)$$ is invariant under local unitary transformations $U_{1}\otimes U_{2}$ ?
-
## closed as off-topic by jinawee, Frederic Brünner, Brandon Enright, Waffle's Crazy Peanut, DanDec 18 '13 at 21:04
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It is not invariant: Try $U_{1}\otimes U_{2} = \sigma_x \otimes \mathbb{Id}$ – Trimok Jul 2 '13 at 7:38
I think that you meant that it is invariant (up to a phase) under $U\otimes U$ in which the same unitary transformation is performed on each qubit separately, right? That's true because $U\otimes U$ is the action of an $SU(2)$ rotation and the singlet state is invariant under the total $SU(2)$, the group generated by the total angular momentum. But it is not invariant under the $SU(2)_1$ and $SU(2)_2$ groups separately. For example, a rotation of the first qubit only, a la Trimok, gives you states like $00+11$ etc. – Luboš Motl Jul 2 '13 at 7:42
For demonstration by "brute force", you may use that a unitary operator $U$ of $U(2)$has the form: $U = e^{i\alpha} S$, where $\alpha$ is a real. and $S$ belongs to $SU(2)$ $S$ can be written $S = e^{i~\theta~ \vec n.\vec \sigma} = (cos \theta ~\mathbb{Id} + i sin\theta~\vec n.\vec \sigma)$, where $\theta$ is a real, $\vec n$ is a normed vector ($(\vec n)^2 = 1$), and $\vec \sigma$ are the Pauli matrices. And you have : $S|0> = (cos \theta + i sin\theta~n^z)|0> + i sin\theta(~n^x + in^y)|1>$, $S|1> = i sin\theta(~n^x - in^y) |0> + (cos \theta - i sin\theta~n^z) |1>$ – Trimok Jul 2 '13 at 10:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8803438544273376, "perplexity": 564.2556581765208}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375093400.45/warc/CC-MAIN-20150627031813-00099-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://web2.0calc.com/questions/help-needed-thank-you-in-advanced | +0
# help needed, thank you in advanced
0
123
2
+20
Let a, b, c be real numbers such that min(a + 24b + 38c, a + 16b + 42c) ≥ 2021. Prove that a^2 + b^2 + c^2 > 2021.
I know that we have to break it into two cases, one when a + 24b + 38c is the minimum, and the other when a + 16b + 42c is the minimum, but I am not quite sure how to proceed from there.
Jan 16, 2021
#1
+186
+1
Can't say that I fully understand this question, but here's some algebra that might help.
Consider
$$\displaystyle (a-1)^{2}+(b-24)^{2}+(c-38)^{2} ,$$
(geater than or equal to zero), for all real values of a, b and c.
Expanding each bracket and rearranging,
$$\displaystyle a^{2}+b^{2}+c^{2}-2(a+24b+38c)+1^{2}+24^{2}+38^{2} \geq0, \\ \text{so} \\ a^{2}+b^{2}+c^{2}+2021 \geq2(a+24b+38c), \\ \text{and since } \\ a+24b+38c \geq 2021, \\a^{2}+b^{2}+c^{2} \geq 2(2021)-2021=2021.$$
The other part, a + 16b + 42c, leads to exactly the same result, but how that's to be presented to answer the question, I don't know.
Jan 17, 2021
#2
+112523
0
Thanks Tiggsy.
I have been playing, unsuccessfully with this one.
Never would have thought of that. :)
Melody Jan 17, 2021 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8896154165267944, "perplexity": 962.5805174495594}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178367183.21/warc/CC-MAIN-20210303165500-20210303195500-00308.warc.gz"} |
https://www.physicsforums.com/threads/need-help-with-physics-angles.16540/ | # Need help with physics angles
1. Mar 18, 2004
### mustang
The rear-view mirror of a car is so placed that its upper and lower edges are horizontal and its cener is at the same level as the center of the rear window. The driver's eye is also at this level, and the line of sight from his eye on the center of the mirror makes an angle of 30 degrees with the line joining the centers of the mirror and the window. The distance from his eye to the mirror is 2 feet, and that from the mirror to the window is 8 feet. What is the least width of the mirror is needed if the entire width (3 feet) of the rear window is to be seen?
Would I do 3/8=x/2?
2. Mar 19, 2004
### Chen
A drawing might help here.
Since you know the red angle, you can also know where the driver seats in relation to the center of the mirror/window line (green line). Now you can find the length of the orange line (half the mirror length minus the green line), and you should alreayd know the length of the purple line (half the window length minus half the mirror length).
Now you have two similar triangles, with the purple and orange lines as their bases. You know the heights of the two triangles, so you know what the ratio between the purple line and orange line is. From there you find the length of the orange line, and once you know that you can find the length of the mirror itself.
We ignore the other side of the mirror, because if the driver can see the left side of the window, there's no doubt that he can see the right side of it as well.
The answer, if my rapid calculations are correct (and I wouldn't count on that), is 2.45 feet.
3. Mar 19, 2004
### Chen
Forgot the drawing:
Sorry I used colors instead of variable names, by the way... it's just easier to paint with colors.
#### Attached Files:
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Last edited: Mar 19, 2004 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8305479884147644, "perplexity": 441.03519583188256}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171281.53/warc/CC-MAIN-20170219104611-00089-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://www.helpwithphysics.com/2013/02/finding-number-of-moles.html | Finding number of moles
Calculate how many moles of hydronium ion are in $1.50*10^2$ milliliters of solution, and report your answer using correct rules for significant figures.
solution $4.10*10^{-6}$ M
Therefore the number of moles of hydronium in a $4.1*10^-6 M, 1.5*10^2$ ml solution is:
$N = 4.1*10^(-6)*1.5*10^2*10^(-3) = 6.15*10^(-7)$ moles | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8226109743118286, "perplexity": 2848.688250085993}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038088264.43/warc/CC-MAIN-20210415222106-20210416012106-00568.warc.gz"} |
https://mathoverflow.net/questions/238816/classification-of-fake-quaternionic-octonionic-projective-spaces | # Classification of fake (quaternionic, octonionic) projective spaces
If $X$ is a closed $n$-manifold, a fake $X$ is another closed manifold homotopy equivalent to $X$. There is some interest in classifying manifolds (up to, say, homeomorphism) homotopy equivalent to a given manifold; the Poincare conjecture is the special case of $X = S^n$ (there are no topologically fake $S^n$s). I'm interested in this in the case of $X = \Bbb{KP}^n$, where $\Bbb K = \Bbb {R,C,H,O}$ (where in the $\Bbb O$ case we fix $n=2$; there are no other objects which deserve to be named "$\Bbb{OP}^n$".) One reason to find this interesting is that fake projective spaces are precisely the manifolds that have a sphere bundle over them with total space a sphere, with as usual the unfortunate exception that this is not true for $\Bbb{OP}^2$. (I would note that this is different than the algebro-geometric usage of "fake projective plane", where to my understanding what was sought was a classification of compact complex manifolds with the same Hodge diamond as complex projective space.)
Both real and complex fake projective spaces have been classified: both computations are carried out in Wall's book on surgery theory (see sections 14C and 14D for the classifications of fake complex and real projective spaces, respectively), and there are nice, briefer descriptions of the real and complex cases on the Manifold Atlas.
I've had some difficulty in finding a classification of fake quaternionic projective spaces or fake octonionic projective planes. Has this been carried out, and if so, what is a reference?
• $\mathbb{OP}^1=S^8$ May 14 '16 at 5:43
• Montgomery and Yang wrote a series of papers about the case of $\mathbb{K}=\mathbb{C}$, but I don't know if they achieved a classification. May 16 '16 at 9:13
• @BenMcKay Wall's book provides a PL classification in that case. It seems Montgomery-Yang were interested in the smooth case (and later the case of not-quite-free actions). Wall cites Brumfiel, “Differentiable $S^1$-actions on homotopy spheres", as what seems to be the authoritative reference; I don't know how up-to-date this is. In general, I'm satisfied with a PL classification, or otherwise I would back up to being unsatisfied with the case of $S^n$...
– mme
May 17 '16 at 2:22
• @MikeMiller: Do you know of a reference for the statement that "fake projective spaces are precisely the manifolds that have a sphere bundle over them with total space a sphere"? Jun 19 '17 at 1:34
• @DanRamras I don't know a reference, but I can give you a relatively short proof (assuming 'sphere bundle' is assumed to mean unit sphere bundle of some vector bundle). Feel free to email me.
– mme
Jun 19 '17 at 1:48
It seems that fake projective planes (the case $n=2$) have been completely classified, following work of Eells-Kuiper and Kramer-Stolz:
James Eells, Jr. and Nicolaas H. Kuiper, MR 145544 Manifolds which are like projective planes, Inst. Hautes \'Etudes Sci. Publ. Math. (1962), no. 14, 5--46.
Linus Kramer and Stephan Stolz, MR 2355782 A diffeomorphism classification of manifolds which are like projective planes, J. Differential Geom. 77 (2007), no. 2, 177--188.
• Great! (That second paper is lovely; I haven't had a chance to look in detail at the first.) So we're reduced to the case of the quaternionic projective spaces.
– mme
May 14 '16 at 14:33
I would like to add some more references concerning the (smooth) fake quaternionic projective spaces (FQPS). There is a beautiful paper of Hsiang:
Hsiang, Wu-chung A note on free differentiable actions of S1 and S3 on homotopy spheres. Ann. of Math. (2) 83 1966 266–272
where he proves that there are infinitely many FQPS which can be distinguished by their rational Pontryagin classes. Moreover he shows that every FQPS can be obtained by a smooth and free action of $S^3$ on an exotic sphere.
Finally there is a recent preprint where the authors compute some interesting groups concerning smooth structures on $\mathbb H\mathbb P^ n$ for low $n$.
But in general I believe there is not much known and I think it is interesting subject to think about. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9247725605964661, "perplexity": 258.6052899645766}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585653.49/warc/CC-MAIN-20211023064718-20211023094718-00318.warc.gz"} |
https://crypto.stackexchange.com/tags/trapdoor/hot | # Tag Info
## Hot answers tagged trapdoor
25
A trapdoor function is a function that is easy to perform one way, but has a secret that is required to perform the inverse calculation efficiently. That is, if $f$ is a trapdoor function, then $y = f(x)$ is easy to compute, but $x = f^{-1}(y)$ is hard to compute without some special knowledge $k$. Given $k$, then it is easy to compute $y = f^{-1}(x, k)$. ...
11
It might be a viable example for explaining the concept of a trapdoor function to a layman without a computer science or math background. But from a CS standpoint this is only half-true. For a low number of n where n is the number of rotate-operations performed on the cube, it is true that finding a solution for the cube is more computationally expensive ...
8
Typically, trapdoor functions are used in asymmetric cryptography. For instance, if you have a RSA public key $(n,e)$, it is easy to compute a cipher text $c = Pad(m)^e\mod n$, but hard to get back to $Pad(m)$ from $c$ even if you have the public key. In order to easily reverse the operation you need the corresponding private key $d = e^{-1}\mod\phi(n)$. The ...
6
A trapdoor function doesn't have the security requirements of an encryption function. If you directly use a trapdoor for encryption you open yourself up to related key attacks, known plaintext attacks, malleability attacks and a whole host of other attacks. Trapdoor functions are used as a single piece of a larger encryption function that handles all these ...
6
No: the parameter $m$ and the size $B$ of the coefficients of $(s,e)$ would not be the same in both scenarios. In the first (LWE decryption) the matrix $A$ is high and narrow, which makes the function `strongly injective', and decoding is possible. Most images do not have preimages, i.e., only a small fraction of $y$ are decodable, those corresponding to ...
5
Is it possible for someone to examine $X$ and $Y$ to determine if there is a common factor, as long as $a$ and $b$ are randomly chosen numbers between 0 and the size of the finite field? (Actually, you mean the size of the group) Technically, the answer is "yes, it is easy to determine that". The easiest subcase is if $G$ (and hence $X$ and $Y$) belong to ...
5
A rubik's cube is an NP problem. It is difficult to find a solution to the problem but it is very easy to verify correctness of the solution, by just rotating the cube and checking that all sides have matching colours. Yes, there is a widely known solution for the 3 * 3 * 3 rubik's cube. You can solve it in 20 moves because you have memorised the solution. ...
5
Definition of trapdoor function from Wikipedia; A trapdoor function is a function that is easy to compute in one direction, yet difficult to compute in the opposite direction without special information, called the "trapdoor". The reverse trapdoor function is just the reverse usage of it. Normally, for encryption, we want the encryption easy but ...
4
"$BW_N$ is a permutation over the squares $\mod N$". Does someone know what that means? You define your map $BW_N:\mathbb{QR}_N\rightarrow \mathbb{QR}_N$. Note that $$\mathbb{QR}_N:=\{r\in Z_N: r\equiv y^2 \pmod{N}, y\in Z_N\}$$ and a permutation is a one-to-one mapping (bijection) from a set into the same set. Basically, this map is a permutation if ...
4
The most direct use of a trapdoor one-way function is to create asymmetric cryptosystems such as Public Key Encryption (PKE), Key Encapsulation Mechanisms (KEM), and Digital Signatures. An analogy A trapdoor one-way function is like a lock box that is supplied to the user in an opened configuration. Any user may place an item inside the box, then close the ...
4
If the adversary is a classical algorithm, then the answer to your question is not known. But if the adversary is a quantum algorithm that can query the oracle in superposition, then the answer is yes: by making queries to the oracle on certain (efficiently produceable) quantum states, it can recover a Type-I trapdoor for $A$. For classical algorithms, the ...
4
It is a major open research question whether such a scheme exists, and how to construct one (see, for example, Open Problem 9.10). Of course, we do have schemes like (hashed) ElGamal, which are based on the conjectured hardness of the (computational or decisional) Diffie-Hellman problem. But it is unknown whether either of these problems is equivalent to the ...
3
No, neither of the two blocks of 6 equalities in the current version of the question are correctly describing either a trapdoor one-way function (first citation); or its use for public-key encryption; or signature of a message using a one-way hash and RSA (second citation). The two citations are only distantly related. In particular, the "one-way hash" in ...
3
No, the mathematical problem that is used for Diffie-Hellman is the Diffie-Hellman problem or DHP which is different from the trapdoor function that is behind RSA. One way functions are for instance cryptographically secure hash functions.
3
It is computationally infeasible to find $R$ (or any other short matrix that satisfies the relation) because solving $A R = V \pmod{q}$ for uniformly random $A, V$ is the SIS problem (in its inhomogeneous version). SIS is provably as hard as solving worst-case approximation problems on lattices. (Also, for the parameters considered in the paper, $[\bar{A} \... 3 Your reasoning is flawed. Producing a cipher text for a given key from plaintext without the key is no easier than the reverse for symmetric ciphers. Without the key k, producing ciphertext$Enc_k(x)$is every bit as hard as$Dec_k(x)$for many ciphers it is the same operation entirely (e.g anything in CTR mode). And even for common block ciphers there is ... 3 If you have a (standard) commitment scheme$C$and zero-knowledge proofs of knowledge (both exist from LWE -- in fact, both exist from one-way functions), then you necessarily have equivocal commitments. Here is a straightforward equivocal commitment$C_e$: committing with$C_e$is just committing with$C$, and opening a commitment is done by revealing the ... 3 For the context of this answer I will assume that that the author meant to say doubly enhanced TDP, since that's the only path I can currently think of. (See Definition 7) Constructing CCA secure encryption The relevant construction is due to Dolev, Dwork, and Naor. It is based on a CPA secure public key encryption scheme, a one-time signature scheme and a ... 2 It means that it maps quadratic residues$\mathbb{QR}_{N} \mapsto \mathbb{QR}_{N}$to quadratic residues. A quadratic residue is a number$x$such that$x = y^2 \pmod N$where$N=pq$. A trapdoor means that once you know the factorization of$N$it is easy to break quadratic residuocity problem.$p=q=3 \pmod 4$because you choose 'safe' primes$p,q$such that ... 2 As far as we know, both one-way and pseudorandom permutations do not help us to get public key encryption schemes. The way we obtain these is by using trapdoor functions (also known as trapdoor permutations). These are keyed collections with the following property: there are two keys for each function: one to compute it in the forward direction and one to ... 2 From the paper,$f$is a pseudo-random function that produces a result of$s$bits. You can think of the master key as a vector of$r$values each$s$bits long and for pedagogical purposes I'm sure the trapdoor for$w$could be thought of as the result of hashing the concatenation of each key value$k$with the word$w$. 2 My own answer would be: 2048, 2048, and still 2048 bits. Why ? Because: 2048-bit is the current "standard recommendation"; it has been so for quite some time, and is likely to remain so for quite some time (decades). See this site for pointers. There are plans for removing support for keys shorter than 2048 bits in some widespread software, e.g. Firefox. ... 2 Trapdoor functions only provide one-wayness. This means, that if one uses a trapdoor function to encrypt this may leak large parts of the plaintext. Suppose I have a trapdoor function$F(pk,m)$for say n-bit messages$m$. I can now define an adapted trapdoor function working on$2n$bit messages as$F'(pk,m_1||m_2) = m_1 || F(pk,m_2)$This is still a ... 2 As suggested, here is a hopefully entry-level precis of the paper linked in the comment above. A Discrete Logarithm based asymmetric key system lacks a true trapdoor function - you can't compute a pre-image for an arbitrary image. Instead, a Schnorr signature relies on a slightly weaker condition. A generalized Schnorr signature can be considered to have ... 2 All generality, no—that's pretty much guaranteed to be too broad of a question here. Here's a specific example. Let$S$be a finite set$F$a public trapdoor permutation of$S$with secret inverse$F^{-1}$, and$H\colon S \to \{0,1\}^{256}$a uniform random function. For a member of the public to send a message$m$to the possessor of$F^{-1}$, they can: ... 2 How difficult is it to find$b$? If the matrices within$R$are of dimension$n \times n$, then we can express the equation$u = a \cdot b + b \cdot a$as$n^2$linear equations over the finite field the matrices are over; a simple minded Gaussian Elimination would recover$b$in$O(n^6)$steps. If$u\$ is specified explicitly in the public key or the ...
2
Trapdoor functions are a special kind of one-way function, where it is possible to calculate the reverse function with the knowledge of some secret value, which is called the trapdoor. If a trapdoor function is also a permutation (bijective, from a set to itself), then it's called a trapdoor permutation. When considering encryption, we need an encryption ...
Only top voted, non community-wiki answers of a minimum length are eligible | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8352708220481873, "perplexity": 429.8890717746374}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038088731.42/warc/CC-MAIN-20210416065116-20210416095116-00097.warc.gz"} |
https://mathoverflow.net/questions/291580/equi-h%C3%B6lder-embeddings-of-compact-metric-spaces-of-finite-packing-dimension-into | # Equi-Hölder embeddings of compact metric spaces of finite packing dimension into $\ell_2$
Problem. Does a compact metric space of finite packing dimension admit an equi-Hölder embedding into a Hilbert space?
A map $f:X\to Y$ between metric spaces $(X,d_X)$, $(Y,d_Y)$ is called equi-Hölder embedding if there are positive real constants $c,C,\alpha$ such that $$c\cdot d_X(x,y)^\alpha\le d_Y(f(x),f(y))\le C\cdot d_X(x,y)^\alpha$$ for all $x,y\in X$.
The packing dimension of a compact metric space $(X,d)$ in the (finite or infinite) number $$Dim(X)=\limsup_{\varepsilon\to 0}\frac{\ln N_\varepsilon(X)}{\ln(1/\varepsilon)},$$ where $N_\varepsilon(X)$ is the cardinality of the smallest cover of $X$ by subsets of diameter $\le\varepsilon$.
Remark. By the Assouad Embedding Theorem, a metric space $X$ admits an equi-Hölder embedding in a finite-dimensional Hilbert space if and only if $X$ is doubling. It can be shown that doubling metric spaces have finite packing dimension.
If $X$ has packing dimension (in your sense; they call it "fractal dimension") less than $m/2$, then $X$ has an embedding $f$ into $\mathbb{R}^m$ such that $$L d(x,y)^\gamma \leq |f(x)-f(y)| \leq d(x,y)$$ for all $x,y\in X$, where $L>0$ and $\gamma\geq 1$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9942812919616699, "perplexity": 174.59142190530793}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103329963.19/warc/CC-MAIN-20220627073417-20220627103417-00650.warc.gz"} |
https://www.deepdyve.com/lp/springer_journal/a-natural-approach-to-the-asymptotic-mean-value-property-for-the-p-Yu6htSXDpf | # A natural approach to the asymptotic mean value property for the p-Laplacian
A natural approach to the asymptotic mean value property for the p-Laplacian Let $$1\le p\le \infty$$ 1 ≤ p ≤ ∞ . We show that a function $$u\in C(\mathbb R^N)$$ u ∈ C ( R N ) is a viscosity solution to the normalized p-Laplace equation $$\Delta _p^n u(x)=0$$ Δ p n u ( x ) = 0 if and only if the asymptotic formula \begin{aligned} u(x)=\mu _p(\varepsilon ,u)(x)+o(\varepsilon ^2) \end{aligned} u ( x ) = μ p ( ε , u ) ( x ) + o ( ε 2 ) holds as $$\varepsilon \rightarrow 0$$ ε → 0 in the viscosity sense. Here, $$\mu _p(\varepsilon ,u)(x)$$ μ p ( ε , u ) ( x ) is the p-mean value of u on $$B_\varepsilon (x)$$ B ε ( x ) characterized as a unique minimizer of \begin{aligned} \Vert u-\lambda \Vert _{L^p(B_\varepsilon (x))} \end{aligned} ‖ u - λ ‖ L p ( B ε ( x ) ) with respect to $$\lambda \in {\mathbb {R}}$$ λ ∈ R . This kind of asymptotic mean value property (AMVP) extends to the case $$p=1$$ p = 1 previous (AMVP)’s obtained when $$\mu _p(\varepsilon ,u)(x)$$ μ p ( ε , u ) ( x ) is replaced by other kinds of mean values. The natural definition of $$\mu _p(\varepsilon ,u)(x)$$ μ p ( ε , u ) ( x ) makes sure that this is a monotonic and continuous (in the appropriate topology) functional of u. These two properties help to establish a fairly general proof of (AMVP), that can also be extended to the (normalized) parabolic p-Laplace equation. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Calculus of Variations and Partial Differential Equations Springer Journals
# A natural approach to the asymptotic mean value property for the p-Laplacian
, Volume 56 (4) – Jun 10, 2017
22 pages
/lp/springer_journal/a-natural-approach-to-the-asymptotic-mean-value-property-for-the-p-Yu6htSXDpf
Publisher
Springer Berlin Heidelberg
Subject
Mathematics; Analysis; Systems Theory, Control; Calculus of Variations and Optimal Control; Optimization; Theoretical, Mathematical and Computational Physics
ISSN
0944-2669
eISSN
1432-0835
D.O.I.
10.1007/s00526-017-1188-7
Publisher site
See Article on Publisher Site
### Abstract
Let $$1\le p\le \infty$$ 1 ≤ p ≤ ∞ . We show that a function $$u\in C(\mathbb R^N)$$ u ∈ C ( R N ) is a viscosity solution to the normalized p-Laplace equation $$\Delta _p^n u(x)=0$$ Δ p n u ( x ) = 0 if and only if the asymptotic formula \begin{aligned} u(x)=\mu _p(\varepsilon ,u)(x)+o(\varepsilon ^2) \end{aligned} u ( x ) = μ p ( ε , u ) ( x ) + o ( ε 2 ) holds as $$\varepsilon \rightarrow 0$$ ε → 0 in the viscosity sense. Here, $$\mu _p(\varepsilon ,u)(x)$$ μ p ( ε , u ) ( x ) is the p-mean value of u on $$B_\varepsilon (x)$$ B ε ( x ) characterized as a unique minimizer of \begin{aligned} \Vert u-\lambda \Vert _{L^p(B_\varepsilon (x))} \end{aligned} ‖ u - λ ‖ L p ( B ε ( x ) ) with respect to $$\lambda \in {\mathbb {R}}$$ λ ∈ R . This kind of asymptotic mean value property (AMVP) extends to the case $$p=1$$ p = 1 previous (AMVP)’s obtained when $$\mu _p(\varepsilon ,u)(x)$$ μ p ( ε , u ) ( x ) is replaced by other kinds of mean values. The natural definition of $$\mu _p(\varepsilon ,u)(x)$$ μ p ( ε , u ) ( x ) makes sure that this is a monotonic and continuous (in the appropriate topology) functional of u. These two properties help to establish a fairly general proof of (AMVP), that can also be extended to the (normalized) parabolic p-Laplace equation.
### Journal
Calculus of Variations and Partial Differential EquationsSpringer Journals
Published: Jun 10, 2017
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Export lists, citations | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9577810764312744, "perplexity": 1047.5246154742854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156314.26/warc/CC-MAIN-20180919235858-20180920015858-00404.warc.gz"} |
https://www.clutchprep.com/chemistry/practice-problems/143567/what-is-the-chemical-formula-for-quartz-express-your-answer-as-a-chemical-formul | Law of Definite Proportions Video Lessons
Concept:
# Problem: What is the chemical formula for quartz?Express your answer as a chemical formula.Matter is made up atoms, but in a real chemistry lab, we measure substances in grams. Therefore you need to be able to convert from grams to numbers of atoms, and vice versa.Quartz, which contains one silicon atom and two oxygen atoms per formula unit, is the second-most-common mineral on Earth after feldspar. It is used as a gemstone, as well as in pressure gauges, oscillators, resonators, wave stabilizers, heat-ray lamps, and prismastic lenses, and in the manufacture of glass, paints, abrasives, and precision instruments such as watches.
98% (57 ratings)
###### Problem Details
What is the chemical formula for quartz?
Matter is made up atoms, but in a real chemistry lab, we measure substances in grams. Therefore you need to be able to convert from grams to numbers of atoms, and vice versa.
Quartz, which contains one silicon atom and two oxygen atoms per formula unit, is the second-most-common mineral on Earth after feldspar. It is used as a gemstone, as well as in pressure gauges, oscillators, resonators, wave stabilizers, heat-ray lamps, and prismastic lenses, and in the manufacture of glass, paints, abrasives, and precision instruments such as watches. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8169592618942261, "perplexity": 2604.0380134209045}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362879.45/warc/CC-MAIN-20211203121459-20211203151459-00445.warc.gz"} |
https://math.stackexchange.com/questions/1850875/approximation-to-unsolvable-system-of-equations | # Approximation to unsolvable system of equations
I am working on a project and need to find the "closest" numerical values that satisfy the following equations:
$$\left\{ \begin{array}{} A \cdot C = \frac{1}{2} \\ A \cdot D = \frac{5}{6} \\ B \cdot C = \frac{1}{8} \\ B \cdot D = \frac{1}{2} \end{array} \right.$$
Where $A$ and $B$ are under the constraint that they are non-negative integers and that $C$ and $D$ must be greater than zero but less than or equal to $1$.
From what I have tried so far, it seems no analytical solution exists. For my purposes an approximate solution will suffice (matches the left-hand side of the equation to several decimal points). Having obtained my BS in engineering, my guilty pleasure is Excel's Solver add-in. Using this tool to the best of my ability, I have yet to obtain a satisfactory approximate solution.
I am not asking anyone to "solve" this for me (though I wouldn't object), but would appreciate being pointed in the right direction. To summarize:
• I need to find values for $A$, $B$, $C$, and $D$ which fall under the constraints mentioned above and provide the closest values to the actual values on the left-hand side of the above equations.
• If the closest values are still unsatisfactory, is there someway of proving that they are indeed the "best" set of values (assuming the method used to find them does not inherently identify the "best" set).
• If there is a matter of tradeoff in accuracy vs integer size, I would prefer the smallest value integers possible that still provide reasonably close solutions to the equations (several decimal points).
• You've got a problem: what is $A\times B\times C\times D$? You get one value if you calculate it by multiplying $(A\times C)\times(B\times D)$, but another value if you get it from $(A\times D)\times(B\times C)$. So the equations are inconsistent. How should that be handled? – David Z Jul 6 '16 at 12:23
• @BattleWalrus Yes, but there is no solution for $A,B,C,D$ that will even satisfy such parameters. – Simply Beautiful Art Jul 6 '16 at 12:34
• @BattleWalrus I will warn you that many users on this Q/A site take math very seriously, and to simply say that something found directly from the given information is "meaningless" when others users come to the same conclusion may put you on bad sides. Just a fair warning. – Simply Beautiful Art Jul 6 '16 at 12:36
• If I understand you correctly, you probably should replace your equations by inequalities, for example: $$|A \cdot C-\frac{1}{2}|<\alpha \\ |A \cdot D-\frac{5}{6}|<\beta$$ and so on. Then you work on minimization of all these parameters. So it actually turns into a very complicated optimization problem – Yuriy S Jul 6 '16 at 12:57
• It therefore probably matters quite a lot exactly how the errors matter to you. E.g., one of the answers below tries to minimize the sum (AC-1/2)^2 + etc.; another tries to minimize the sum (log(AC)-log(1/2))^2 + etc. The "best" solution will depend on what "best" means, which will depend on what these numbers are for and what significance your equations actually have. – Gareth McCaughan Jul 6 '16 at 16:29
Let
$$x_1 := \log_2 (a) \qquad \qquad x_2 := \log_2 (b) \qquad \qquad x_3 := \log_2 (c) \qquad \qquad x_4 := \log_2 (d)$$
Thus, the original system of bilinear equations can be transformed into a system of linear equations
$$\begin{bmatrix} 1 & 0 & 1 & 0\\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0\\ 0 & 1 & 0 & 1\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix} = \begin{bmatrix} -1\\ \log_2 \left(\frac 5 6\right)\\ -3\\ -1\end{bmatrix}$$
Find the least-squares solution $(\hat x_1, \hat x_2, \hat x_3, \hat x_4)$ and then compute the corresponding $(\hat a,\hat b,\hat c,\hat d)$.
In MATLAB:
>> A = [1 0 1 0; 1 0 0 1; 0 1 1 0; 0 1 0 1]
A =
1 0 1 0
1 0 0 1
0 1 1 0
0 1 0 1
>> b = [-1; log2(5/6); -3; -1]
b =
-1.0000
-0.2630
-3.0000
-1.0000
Unfortunately, the matrix is singular:
>> det(A)
ans =
0
We then find one solution to the "normal equations":
>> x = (A' * A)\(A' * b)
x =
1.7194
0.3509
-3.0351
-1.6667
The corresponding $(\hat a,\hat b,\hat c,\hat d)$ is, thus:
>> y = 2.^x
y =
3.2930
1.2754
0.1220
0.3150
We introduce a new matrix:
>> Y = y * y'
Y =
10.8437 4.1997 0.4017 1.0372
4.1997 1.6266 0.1556 0.4017
0.4017 0.1556 0.0149 0.0384
1.0372 0.4017 0.0384 0.0992
The absolute error is:
>> Y(1:2,3:4) - [1/2,5/6;1/8,1/2]
ans =
-0.0983 0.2039
0.0306 -0.0983
which is quite large.
• +1: Didn't realize that I was suggesting a similar approach :D. – MrYouMath Jul 6 '16 at 13:14
• @MrYouMath Your approach is nonlinear least-squares, though. – Rodrigo de Azevedo Jul 6 '16 at 13:14
• @MrYouMath and Rodrigo, both your approaches are useful: an NLLS method usually needs a good starting point, so one can use Rodrigo's method as a starting guess, and then MrYouMath's method can then be used for polishing. Also, since we're talking about stuff like this: this optimizes with respect to $\|\cdot\|_2$; one might also consider solutions that optimize with respect to e.g. the Manhattan or Chebyshev norms. – J. M. isn't a mathematician Jul 6 '16 at 14:11
• @J.M. Very nice addition with the other norms :). – MrYouMath Jul 6 '16 at 17:00
Hint: If you stuff the variables into vectors, your problem becomes:
$$M = \left[\begin{array}{c}A\\B\end{array}\right]\left[\begin{array}{cc}C&D\end{array}\right] = \left[\begin{array}{rr} \frac{1}{2} & \frac 56\\\frac 1 8&\frac 1 2 \end{array}\right]$$
This means we want to find the best rank 1 match to the matrix.
Thanks for @Ian s comment which clarified nicely that you can use for example the Singular Value Decomposition ( SVD ) to find the best match.
The SVD says $M = U\Sigma V^*$ where the values in the diagonal $\Sigma$ are the singular values. If we pick the column in U and the row in $V^*$ which correspond to the largest singular value, we can identify the $A,B,C$ and $D$ above.
Some Matlab/Octave code to do the SVD approximation:
M = [1/2,5/6;1/8,1/2];
[U,S,V]=svd(M);
U(:,1)*S(1,1)*V(:,1)' % rank 1 SVD appr. as index 1 contains largest s.v.
$$\left[\begin{array}{cc} 0.445505865263938&0.861513345600555\\ 0.230380036476643&0.445505865263938 \end{array}\right]$$
abs(M-U(:,1)*S(1,1)*V(:,1)') %and the absolute error
$$\left[\begin{array}{cc} 0.0544941347360618&0.028180012267222\\ 0.105380036476643&0.0544941347360618 \end{array}\right]$$
We can see the error distributed so that each element has a rather different error. It differs $10.5/2.8 \approx 3.74$ times smallest compared to largest. As mentioned in comments, if we are not OK with this distribution of the error, we may want to try minimize the error according to some other norm.
• I fail to see the significance of this hint...? – Frenzy Li Jul 6 '16 at 13:41
• It becomes a problem of finding the closest rank-1 match which is a very well studied problem in linear algebra. – mathreadler Jul 6 '16 at 14:35
• What is the closest rank-1 match? – Frenzy Li Jul 6 '16 at 14:40
• @FrenzyLi You have the outer product of two vectors $u v^T$. This is a rank 1 matrix ($uv^Tx$ is always a multiple of $u$). You can identify the rank 1 matrix which is closest to $\begin{bmatrix} 1/2 & 5/6 \\ 1/8 & 1/2 \end{bmatrix}$ using the SVD, and then solve the consistent problem where you put that rank 1 matrix on the right side instead. – Ian Jul 6 '16 at 14:43
• The only thing you might think is bad about this approach is that the $2 \times 2$ matrix 2-norm is quite different from a more familiar norm like the $4 \times 1$ vector 2-norm. Still, you could use this as an initial guess for a method for optimizing in some other norm. – Ian Jul 6 '16 at 15:43
Another way you could try to approach this problem by using a least squares estimator.
Assume $$F(A,B,C,D)=(AC-1/2)^2+(AD-5/6)^2+(BC-1/5)^2+(BD-1/2)^2$$
Now calculate the gradient of $F$ $$\nabla F=\left[\dfrac{\partial F}{\partial A},\cdots,\dfrac{\partial F}{\partial D}\right]$$
Then set the gradient to zero. Solve the resulting nonlinear system using Newton-Raphson algorithm (you should also check the positiveness of the Hessian).
None of the answers thus far have dealt with the integer constraint on $(A,B)$. If we define two vectors, one integer and one real, \eqalign{ a &= [\, A \,\, B \,]^T \cr x &= [\, C \,\, D \,]^T \cr } Then the problem is to minimize the function $$f=\|M-ax^T\|^2_F$$ where $M = [\, 1/2 \,\,\, 5/6;\, 1/8 \,\,\,1/2 \,]$.
Assume that the integer vector, $a$, is known or given. Then by setting the gradient zero, we can solve explicitly for the optimum real vector \eqalign{ x &= \frac{M^Ta}{a^Ta} } Now all we need to do is write a program to search through lots of integer vectors.
#!/usr/bin/env julia
M = [ 1/2 5/6; 1/8 1/2 ];
n=999; fmin=5.0;
for i=1:n, j=1:n
a = [i j]';
x = M'*a/(a'*a);
f = vecnorm(M-a*x');
if f < fmin
@printf("(%d,%d), ", i,j)
fmin = f
end
end
(1,1), (2,1), (15,8), (17,9), (19,10), (21,11), (23,12), (25,13),
(27,14), (29,15), (147,76), (176,91), (205,106), (234,121),
(263,136), (292,151), (555,287), (847,438)
Note that the ratio (i/j) is approximately $1.9$ for all these solutions. To speed things up, we can use this ratio to limit the range on the second index, and extend the search range to even larger integers.
#!/usr/bin/env julia
M = [ 1/2 5/6; 1/8 1/2 ];
i,j = (847,438);
a=[i j]'; x=M'*a/(a'*a);
n=99999; fmin=vecnorm(M-a*x');
for i=1:n, j=round(Int,i/2.1):round(Int,i/1.7)
a = [i j]'
x = M'*a/(a'*a)
f = vecnorm(x*a'-M')
if f < fmin
@printf("(%d,%d), ", i,j)
fmin = f
end
end
(1139,589), (3125,1616), (4264,2205),
(5403,2794), (9667,4999), (52599,27200)
The solution corresponding to the penultimate index pair is \eqalign{ a &= [\, 9667 \,\,\, 4999 \,]^T \cr x &= [\, 4.6085224452467385 \,\,\, 8.91189971076149 \,]^T \times 10^{-5} \cr f &= 0.017838286620463398 \cr } The best of the small ($\le 100$) index pairs is \eqalign{ a &= [\, 29 \,\,\, 15 \,]^T \cr x &= [\, 0.015361163227016885 \,\,\, 0.029706066291432145 \,]^T \cr f &= 0.01783829737335834 \cr }
The OP's last bullet point led me to a different attack on this problem.
First, note that if $(a,b,c,d)$ is a suitable approximate solution, then so is $(k a, k b, c/k, d/k)$ for any positive integer $k$. The last bullet prefers smaller integers, so we will want $\mathrm{gcd}(a,b) = 1$.
Suppose we relax the equalities to memberships in intervals of a given radius: \begin{align} a c &\in [1/2 - \varepsilon, 1/2 + \varepsilon] = 1/2 + \varepsilon[-1,1]\\ a d &\in [5/6 - \varepsilon, 5/6 + \varepsilon] = 5/6 + \varepsilon[-1,1]\\ b c &\in [1/8 - \varepsilon, 1/8 + \varepsilon] = 1/8 + \varepsilon[-1,1]\\ b d &\in [1/2 - \varepsilon, 1/2 + \varepsilon] = 1/2 + \varepsilon[-1,1] \text{,} \end{align} where the first equality uses usual interval notation and the second form uses usual interval arithmetic notation.
Then we can control the number of digits of agreement by setting $\varepsilon$ to suitable negative powers of $10$.
This suggests a process:
• Pick a $(c,d) \in (0,1] \times (0,1]$.
• Find the smallest $\varepsilon$ so that there is still an integer point, $(a,b)$, satisfying the four membership relations.
• Find the largest region on the $c$-$d$ plane where this $(a,b)$ has minimal epsilon of all integer points.
This allows us to partition the square $(0,1] \times (0,1]$ into regions where a particular integer point is optimal and search for the point in that region with minimal $\varepsilon$. In fact, once $a$ and $b$ are fixed, this is a search with linear constraints, so is pretty easy.
By computer algebra system (Mathematica 10.4.1), for $(c,d) = (1/6,1/2)$, the minimum $\varepsilon$ still admitting an integer point satisfying the membership relations is $\varepsilon = 1/6$ for $(a,b) = (2,1)$. For this $(a,b)$, the largest region in the $c$-$d$ plane where each point has minimal $\varepsilon$ at this integer point is $(c,d) \in [1/5,1/4]\times[1/3,5/9]$. In that region, the minimum value of $\varepsilon$ is $1/12$, attained at $(c,d) = (5/24, 5/12)$. This is not so impressive (since $\varepsilon = 1/12$ means that we only have a trifle more than one decimal agreement with the original equalities). (There is another minimum in this region, at $(c,d) = (5/24,7/16)$ yielding the same extreme value of $\varepsilon = 1/12$. However, there do not appear to be criteria for preferring points on the $c$-$d$ plane.)
(It's late/early here, so I'll continue hacking on this analytically some time in the next few days. Note to self: Use the $(ka, kb, c/k, d/k)$ relation to rewrite $c$ and $d$ ($\in [1,\infty)$) to give solutions $(ka, kb, kc, kd)$, then projectively set $b=1$ (or $a=1$, if there are also solutions with $b>a$).)
$$\begin{bmatrix} a\\ b\\ c\\ d\end{bmatrix} \begin{bmatrix} a\\ b\\ c\\ d\end{bmatrix}^\top = \begin{bmatrix} a^2 & a b & a c & a d\\ a b & b^2 & b c & b d\\ a c & b c & c^2 & c d\\ a d & b d & c d & d^2\end{bmatrix} = \begin{bmatrix} a^2 & a b & \frac 12 & \frac 56\\ a b & b^2 & \frac 18 & \frac 12\\ \frac 12 & \frac 18 & c^2 & c d\\ \frac 56 & \frac 12 & c d & d^2\end{bmatrix}$$
Let us look for a symmetric, positive semidefinite, rank-$1$ matrix $\mathrm X \in \mathbb R^{4 \times 4}$ whose northeast and southwest $2 \times 2$ blocks are equal to the northeast and southwest $2 \times 2$ blocks in the matrix above.
Hence, we have a rank-minimization problem
$$\begin{array}{ll} \text{minimize} & \mbox{rank} (\mathrm X) \\ \text{subject to} & x_{13} = \frac 12, \quad x_{14} = \frac 56\\ & x_{23} = \frac 18, \quad x_{24} = \frac 12\\ & \mathrm X = \mathrm X^\top\\ & \mathrm X \succeq \mathrm O_4\end{array}$$
Unfortunately, minimizing the rank is computationally hard [0]. Thus, let us minimize the nuclear norm of $\mathrm X$ instead. Since $\mathrm X$ is symmetric and positive semidefinite, its nuclear norm is given by
$$\|\mathrm X\|_* = \mbox{tr} (\mathrm X)$$
Hence, we obtain the following semidefinite program (SDP)
$$\begin{array}{ll} \text{minimize} & \mbox{tr} (\mathrm X) \\ \text{subject to} & x_{13} = \frac 12, \quad x_{14} = \frac 56\\ & x_{23} = \frac 18, \quad x_{24} = \frac 12\\ & \mathrm X = \mathrm X^\top\\ & \mathrm X \succeq \mathrm O_4\end{array}$$
Using MATLAB + CVX,
clear all; clc;
cvx_begin sdp
variable X(4,4) symmetric
minimize( trace(X) )
subject to
X(1,3)==1/2
X(1,4)==5/6
X(2,3)==1/8
X(2,4)==1/2
X >= 0
cvx_end
% rank of X
disp('rank(X) ='); disp(rank(X))
% build rank-1 matrix
[U,S,V] = svd(X);
X_tilde = U(:,1) * S(1,1) * (V(:,1)');
disp('rank-1 approximation of X ='); disp(X_tilde)
% approximation error
disp('approximation error = '); disp(X_tilde - X)
% error matrix
disp('error matrix = '); disp(X_tilde(1:2,3:4) - [1/2,5/6;1/8,1/2])
Running this MATLAB script, we obtain the following output
Calling sedumi: 10 variables, 4 equality constraints
------------------------------------------------------------
SeDuMi 1.21 by AdvOL, 2005-2008 and Jos F. Sturm, 1998-2003.
Alg = 2: xz-corrector, Adaptive Step-Differentiation, theta = 0.250, beta = 0.500
eqs m = 4, order n = 5, dim = 17, blocks = 2
nnz(A) = 4 + 0, nnz(ADA) = 16, nnz(L) = 10
it : b*y gap delta rate t/tP* t/tD* feas cg cg prec
0 : 9.09E+000 0.000
1 : 2.28E+000 2.11E+000 0.000 0.2322 0.9000 0.9000 1.21 1 1 7.9E-001
2 : 2.30E+000 4.47E-001 0.000 0.2117 0.9000 0.9000 1.18 1 1 1.8E-001
3 : 2.45E+000 1.33E-002 0.000 0.0297 0.9900 0.9900 1.01 1 1 5.5E-003
4 : 2.45E+000 3.56E-007 0.000 0.0000 1.0000 1.0000 1.00 1 1 1.4E-007
5 : 2.45E+000 5.10E-014 0.000 0.0000 1.0000 1.0000 1.00 1 1 2.1E-014
iter seconds digits c*x b*y
5 0.2 13.8 2.4509068616e+000 2.4509068616e+000
|Ax-b| = 4.5e-015, [Ay-c]_+ = 0.0E+000, |x|= 2.2e+000, |y|= 2.8e+000
Detailed timing (sec)
Pre IPM Post
1.719E-001 2.344E-001 7.813E-002
Max-norms: ||b||=8.333333e-001, ||c|| = 1,
Cholesky |add|=0, |skip| = 0, ||L.L|| = 1.41176.
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): +2.45091
rank(X) =
4
rank-1 approximation of X =
0.8615 0.4455 0.4455 0.8615
0.4455 0.2304 0.2304 0.4455
0.4455 0.2304 0.2304 0.4455
0.8615 0.4455 0.4455 0.8615
approximation error =
-0.0282 0.0545 -0.0545 0.0282
0.0545 -0.1054 0.1054 -0.0545
-0.0545 0.1054 -0.1054 0.0545
0.0282 -0.0545 0.0545 -0.0282
error matrix =
-0.0545 0.0282
0.1054 -0.0545
Note that the maximum error is smaller than when using least-squares.
[0] Emmanuel Candès, The Effectiveness of Convex Programming in the Information and Physical Sciences [PDF], Simons Institute Open Lecture, UC Berkeley, October 2013. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.919513463973999, "perplexity": 986.2167589021888}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989526.42/warc/CC-MAIN-20210514121902-20210514151902-00563.warc.gz"} |
https://www.physicsforums.com/threads/finding-original-carbon-nuclei-from-given-sample.600930/ | # Finding Original Carbon nuclei from given sample.
1. Apr 27, 2012
### rcubed
1. The problem statement, all variables and given/known data
A small animal bone fragment found in an archaelogical site has a carbon mass of 155g. When the animal was alive, the ratio of radioactive 146C to the stable 126C was 1.31×10-12. What was the number of 146C nuclei found in the sample when the animal was alive?
2. Relevant equations
None given, but I would assume:
N=N0e-λt
3. The attempt at a solution
Not too sure where to start so I got the decay constant, λ by using half life of Carbon14, 5730 Years
0.5=e-λ(5730)
λ=1.21×10-4
Then I solved the for the number of years since the animal was alive by plugging everything back into the original equation, assuming N/N0 = 1.31×10-12
t=-ln(1.31×10-12)/-1.21×10-4 = 226180 Years
Up until here I don't think I did anything wrong, but here is where I am unsure of what to do.
I tried using the same formula to solve for N0, but this time using the given 155g.
155=N0e-(1.2110-4)(226180)
N0=1.18×1014g
I don't think in doing the right thing here. Can anyone give me some guidance?
Thanks!
2. Apr 27, 2012
### collinsmark
Hello rcubed,
Welcome to Physics Forums!
Okay, you've found that
$$\frac{N}{N_0} = e^{-1.21 \times 10^4 \ t}$$
Although I don't think that helps for this problem. (Maybe it does later in a different part of the problem not listed in the above statement).
No, wait. You're using the 1.31×10-12 out of context. It is *not* the ratio of the final amount of 14C to the original amount of 14C.
As a matter of fact, you don't even know what the final amount of 14C is. That information is not given in the problem statement. And since the age of the animal is not given either, you can't even calculate it (at least not without additional information).
All you're trying to find is the original amount of 14C (when the animal was alive). For this particular problem, don't worry about how much of the sample is 14C today, or even how old the sample is. There's not enough information given anyway.
So here is what we know. There is 155 g sample of carbon. When the animal was alive, the fraction of carbon (in terms of the ratio of the number of nuclei) that was 14C was 1.31×10-12 (Technically that number is the ratio of 14C to 12C, but it's also approximately the same ratio as the number of 14C to total). So when the animal was alive, how much of that sample was 14C?
[Hint: You might want to start by determining how many carbon nuclei (primarily 12C) are in 155 g of carbon. Then, since you know what fraction of that that was 14C, determine the number of 14C nuclei].
-------------
Edit:
By the way, I am presently interpreting the 1.31×10-12 ratio as the ratio of number of nuclei. If instead it is a ratio of masses or weights, then my advice needs to be modified to take that into account (i.e. 14C is heavier than 12C, which needs consideration).
Last edited: Apr 27, 2012
3. May 2, 2012
### rcubed
So its just simply finding the number of 12C Molecules first, then using the ratio to find the 14C?
155g/12gmol-1*NA = 7.78e24
7.78e24*1.32e-12 = 1.026e13 ?
Is that it?
4. May 2, 2012
### collinsmark
It looks like the right idea to me. But I think the original problem statement said that the ratio was 1.31 x 10-12 (you used 1.32 x 10-12).
Similar Discussions: Finding Original Carbon nuclei from given sample. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8922156095504761, "perplexity": 744.8476522182222}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934809419.96/warc/CC-MAIN-20171125051513-20171125071513-00573.warc.gz"} |
https://quizlet.com/subject/term%3Alongitudinal%20wave%20%3D%20a%20wave%20in%20which%20the%20particles%20of%20the%20medium%20vibrate%20parallel%20to%20the%20direction%20of%20wave%20motion/ | Study sets
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medium
a substance solid liquid or gas that a wave can travel through
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https://www.physicsforums.com/threads/chaos-attractor.711381/ | # Chaos attractor
1. ### Shobhit Gupta
2
Can we consider +inf as an attractor for a map for which trajectories emanating from any point in the state space tends to +inf.
2. ### Khashishi
Certainly. The points outside the Mandelbrot set are "attracted to infinity". | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9625322818756104, "perplexity": 2818.8776062661973}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443738009849.87/warc/CC-MAIN-20151001222009-00215-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://www.gamedev.net/forums/topic/310039-continuous-and-differentiable/ | # Continuous and differentiable?
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I'm writing a simple symbolic math library as a way to hone my skills (and hopefully to integrate some statistics-based functionality into my game AI). One of my methods determines if a piecewise function is continuous and differentiable. But I've run into a problem here. Suppose that f(x) = {0, x <= 0; e-1/x, x > 0}. Is f(x) continuous? If so, is it differentiable? If you looked at a graph, you'd probably guess that it is indeed continuous and differentiable. But while it's defined at zero, there will be an infinitesmal break at f(0 + dx) = e-1/dx. If I let dx go to zero, the difference also shrinks to zero. This suggests both continuity (since the function has equal limits from both sides) and differentiability (since the derivative is equal from both sides), but am I missing something here? For now, my algorithm assumes a function is "numerically continuous" if the difference between f(x) and f(x + epsilon) (where epsilon is the smallest possible value that can be represented in an IEEE floating-point number) is less than a specially determined constant. [Edited by - kSquared on March 29, 2005 1:39:07 PM]
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I remember that function as some classical example...
Let's take derivative:
e-1/x ' = e-1/x / x2
and find limit at 0:
lim(e-1/x / x2 , x-->0) = lim(e-y*y2, y-->infinity) = 0
As you can see, limit of derivative is 0. And limit of function is 0.
So we have:
lim((f(0)-f(-dx))/dx, dx-->0) = lim((f(dx)-f(0))/dx, dx-->0) = 0 .
edit: actually it should be written clearer as
lim((f(dx)-f(0))/dx, dx-->0 +) = lim((f(dx)-f(0))/dx, dx-->0 -) = 0
And, IIRC that means it is differenciable in point x=0 by definition.
edit: mathworld , look right under 6 . (Wording is kinda unclear, but as i understand, existence and equality of both limits is both necessary and sufficient , and continuity alone is not.)
As about your algorithm... with many functions in most points you will have
f(x+epsilon)=f(x) because of roundoff errors. Also, with such "bad" functions involving divide by zero, you'll have computational problems...
edit: or take for example sum of sine waves
f(x)=sum(n=0..infinity, sin(x*(e^n))/(e^n))
f is finite and lim(f(x+dx)-f(x))=0 , but lim((f(x+dx)-f(x))/dx) does not exist and therefore function is not differentiable...
[Edited by - Dmytry on March 29, 2005 2:51:38 PM] | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8822228312492371, "perplexity": 1534.4536551505805}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814833.62/warc/CC-MAIN-20180223194145-20180223214145-00065.warc.gz"} |
http://math.stackexchange.com/questions/81821/double-exponential-distribution-is-not-an-exponential-family?answertab=oldest | # Double exponential distribution is not an exponential family
Define a one-parameter exponential family as a family of densities of the form $$f_\theta(x)=\exp(\eta(\theta)T(x) + \xi(\theta))h(x)$$ where $T(x)$ and $h(x)$ are Borel functions, $\theta\in\Theta\subset\mathbb R$ and $\eta$ and $\xi$ are real-valued functions defined on $\Theta$.
Double exponential distribution is a distribution having the density $$p_\theta(x)= \frac{1}{2}\exp(-|x - \theta|)$$ for $\theta\in\mathbb R$.
I am looking for a simple proof of the theorem in the title. I found a proof in the book of Shao "Mathematical Statistics. Exercises and Solutions." but it uses a more general definition of exponential families and doesn't show why the classes are not compatible. What is the special feature of $p_\theta(x)$ that makes the representation as exponential family impossible?
-
The only additional generality assumed in Shao is that the distribution could be from a multiparameter exponential family. In order to prove the statement in your title, you have to show that the double exponential is not in the exponential family for all possible (finite) choices of the dimension of the parameter space. If you want a proof that it's not from a one-parameter family, then just collapse Shao's proof down to the $p = 1$ case. It still holds. :) – cardinal Nov 14 '11 at 0:09
The main feature is that you can't "factor" $|x-\theta|$ into a product of functions, one that depends only on $x$ and the other that depends only on $\theta$ (plus perhaps an additional function of only $\theta$. – cardinal Nov 14 '11 at 0:10
@cardinal Thanks for your comments! I should have mentioned I'm interested in a special case with one-dimensional $\xi$ and $T$. – Julian Wergieluk Nov 14 '11 at 2:28
Assume that $\Theta$ has at least three distinct points $\theta_1$, $\theta_2,$ and $\theta_3$. Suppose that $$\exp(\eta(\theta)T(x)+\xi(\theta)) h(x)={1\over 2}\exp(-|x-\theta|).$$ Since $h(x)$ never takes the value zero, we can write it as $h(x)={1\over 2}\exp(w(x))$, and deduce that, for all $x\in\mathbb{R}$ and $\theta\in\Theta$, $$\eta(\theta)T(x)+\xi(\theta) +w(x)= -|x-\theta|.$$
Substitute $\theta_1, \theta_2$ and subtract the two equations to get $$[\eta(\theta_1)-\eta(\theta_2)]\ T(x)+\xi(\theta_1)- \xi(\theta_2) = |x-\theta_2|-|x-\theta_1|.$$ Since the right hand side is not a constant function of $x$, we find that $\eta(\theta_1)\neq\eta(\theta_2)$ and hence that $T$ is differentiable in $x$, except possibly at $\theta_1$ and $\theta_2$. The same argument using the pairs $\{\theta_1 ,\theta_3\}$ and $\{\theta_2 ,\theta_3\}$ shows that $T$ is, in fact, differentiable everywhere.
We conclude that $|x-\theta_2|-|x-\theta_1|$ is everywhere differentiable in $x$, which is a contradiction.
-
(+1) I think this works, though it still only allows one to conclude that the double-exponential is not from a one-parameter exponential family. – cardinal Nov 14 '11 at 0:54
@cardinal Thanks. I'm not sure what $|x-\theta|$ would mean if $\theta$ were multidimensional. Or are you thinking of parametrizing by something other than $\theta$? – Byron Schmuland Nov 14 '11 at 0:58
I mean that, conceivably, there could be functions $\eta : \mathbb R \to \mathbb R^p$ and $T: \mathbb R \to \mathbb R^p$ such that $\exp( \eta(\theta) \cdot T(x) + \xi(\theta) ) h(x) = \frac{1}{2} \exp(-|x - \theta|)$, in which case at least a curved exponential family would result. – cardinal Nov 14 '11 at 1:02
@cardinal Oh, I see. Thanks for the clarification! – Byron Schmuland Nov 14 '11 at 1:05
Actually my intention was to find a simple proof in case of one-parameter exponential family with restriction $\eta:\mathbb R \to \mathbb R$. Thanks!! – Julian Wergieluk Nov 14 '11 at 2:08
Meanwhile I figured out another proof. But the one of Byron is clearly more elegant.
Following the idea of Shao we consider the quotients $\frac{p_\theta(x)}{p_{-\theta}(x)}= \frac{f_\theta(x)}{f_{-\theta}(x)}$. This allows us to get rid of $h(x)$ and yields $$|x+\theta| - |x-\theta| = \left( \eta(\theta) - \eta(-\theta) \right)T(x) - \left( \xi(\theta) - \xi(-\theta) \right)$$ Since $\eta(\theta) - \eta(-\theta)$ must be non-zero for some $\theta$ we define $$A = \eta(\theta) - \eta(-\theta)$$ $$B =\xi(\theta) - \xi(-\theta)$$ and get $|x+\theta| - |x-\theta| = A T(x) + B$. Moreover $|x+\theta| - |x-\theta|$ is an antisymmetric function and therefore $A T(x) + B = -(A T(-x) + B)$. This yields $T(x) + \frac{B}{A} = - T(-x) - \frac{B}{A}$ and implies that, $T(x)$ shifted by $\frac{B}{A}$ is antisymmetric.
On the other hand, if we consider $\frac{p_\theta(x)}{p_{0}(x)}= \frac{f_\theta(x)}{f_{0}(x)}$ we get $$|x| - |x-\theta| = \left( \eta(\theta) - \eta(0) \right)T(x) - \left( \xi(\theta) - \xi(0) \right)$$ For positive $\theta$ this function is constant in $x$ for $x<0$. But $T(x)$ is also antisymmetric. This implies $T(x) \equiv T$ is constant for all $x\in\mathbb R$. A contradiction.
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http://scholarpedia.org/article/Coherent_activity_in_excitatory_pulse-coupled_networks | # Coherent activity in excitatory pulse-coupled networks
Post-publication activity
Curator: Simona Olmi
An excitatory pulse-coupled neural network is a network composed of neurons coupled via excitatory synapses, where the coupling among the neurons is mediated by the transmission of Excitatory Post-Synaptic Potentials (EPSPs). The coherent activity of a neuronal population usually indicates that some form of correlation is present in the firing of the considered neurons. The article focuses on the influence of dilution on the collective dynamics of these networks: a diluted network is a network where connections have been randomly pruned. Two kind of dilution are examined: massively connected versus sparse networks. A massively (sparse) connected network is characterized by an average connectivity which grows proportionally to (does not depend on) the system size.
Neural collective oscillations have been observed in many contexts in brain circuits, ranging from ubiquitous $\gamma$-oscillations to $\theta$-rhythm in the hippocampus. The origin of these oscillations is commonly associated to the balance between excitation and inhibition in the network, while purely excitatory circuits are believed to lead to “unstructured population bursts” (Buzsàki, 2006). However, coherent activity patterns have been observed also in “in vivo” measurements of the developing rodent neocortex and hippocampus for a short period after birth, despite the fact that at this early stage the nature of the involved synapses is essentially excitatory, while inhibitory synapses will develop only later (Allene et al., 2008). Of particular interest are the so-called Giant Depolarizing Potentials (GDPs), recurrent oscillations which repeatedly synchronizes a relatively small assembly of neurons and whose degree of synchrony is orchestrate by hub neurons (Bonifazi et al., 2009). These experimental results suggest that the macroscopic dynamics of excitatory networks can reveal unexpected behaviors.
On the other hand, numerical and analytical studies of collective motions in networks made of simple spiking neurons have been mainly devoted to balanced excitatory-inhibitory configurations (Brunel, 2000), while few studies focused on the emergence of coherent activity in purely excitatory networks. Pioneering studies of two pulse coupled neurons have revealed that excitatory coupling can have desynchronizing effect, while in general synchronization can be achieved only for sufficiently fast synapses (van Vreeswijk et al., 1994; Hansel et al., 1995). Van Vreeswijk in 1996 has extended these analysis to globally (or fully) coupled excitatory networks of Leaky Integrate-and-Fire (LIF) neurons, where each neuron is connected to all the others. This analysis has confirmed that for slow synapses the collective dynamics is asynchronous ( Splay States ) while for sufficiently fast synaptic responses a quite peculiar coherent regime emerges, characterized by partial synchronization at the population level, while single neurons perform quasi-periodic motions (van Vreeswijk, 1996).
## Introduction
In the recent years, following the seminal study by van Vreeswijk, the robustness of the partially synchronized regime has been examined by considering the influence of external noise and the level of dilution in networks of different topologies. Partial synchronization survives to the introduction of a moderate level of noise (Mohanty and Politi, 2006) and it appears to be quite robust also to dilution.
In particular, for neurons connected as in a directed Erdös-Renyi graph (Albert and Barabàsi, 2002) it has been shown that the coherent activity always emerge for (sufficiently) high connectivities. However, while for massively connected networks, composed by a large number of neurons, the dynamics of the collective state (apart some trivial rescaling) essentially coincide with that observed in the fully coupled system (Olmi et al., 2010; Tattini et al., 2012), for sparse networks this is not the case (Luccioli et al., 2012). This is due to the fact that, for sufficiently large networks, the synaptic currents, driving the dynamics of the single neurons, become essentially identical for massively connected networks, while the differences among them do not vanish for sparse networks.
Sparse and massively connected networks reveal even more striking differences at the microscopic level associated to the membrane potentials' dynamics. As a matter of fact, for finite networks chaotic evolution has been observed in both cases. However, this chaos is weak in the massively connected networks, vanishing for sufficiently large system sizes, while sparse networks remain chaotic for any large number of neurons and the chaotic dynamics is extensive.
## Model and Indicators
In a fully coupled network of $N$ neurons, the membrane potential $$u_i(t)$$ of the $$i−$$th neuron evolves according to the following ordinary differential equation
$\dot{u}_i(t) = a − u_i(t) + g E(t) \qquad i = 1, · · · ,N$
where all variables and parameters are expressed in adimensional rescaled units. According to the above equation, the membrane potential $$u_i$$ relaxes towards the value $$a + gE(t)$$, but as soon as it reaches the threshold value $$u_i = 1$$, it is reset to $$u_i = 0$$ and a spike is simultaneously sent to all neurons. This resetting procedure is an approximate way to describe the discharge mechanism operating in real neurons. The parameter $$a > 1$$ is the supra-threshold input DC current and $$g > 0$$ gauges the synaptic coupling strength of the excitatory interaction with the neural field $$E(t)$$. This field represents the synaptic current injected in each neuron and is given by the superposition of all the pulses emitted by the network in the past. Following (Abbott and van Vreeswijk, 1993), it is assumed that the shape of a pulse emitted at time $$t=0$$ is given by an $\alpha$-function $$s(t)= \frac{\alpha^2 t}{N} {\rm e}^{-\alpha t}$$, where $$1/\alpha$$ is the pulse-width. For this choice of the pulse shape it is easy to show that the field evolution is ruled by the following second order differential equation $\ddot E(t) +2\alpha\dot E(t)+\alpha^2 E(t)= \frac{\alpha^2}{N}\sum_{n|t_n<t} \delta(t-t_n) \ .$ In other words, $$s(t)$$ represents an EPSP emitted at time $$t_n$$ by a neuron reaching the threshold value. The solution $$E(t)$$ for a generic time $$t_n<t<t_{n+1}$$ between two spike emissions is the linear combination of such EPSPs and represents a macroscopic variable reproducing the network activity.
At variance with the fully-coupled network, where all neurons depend on the same "mean field" $$E(t)$$, in a random diluted network neurons have different connectivities. As a result, it is necessary to introduce an explicit dependence of the neural field on the index $$i$$. The field $$E_i(t)$$ represents the linear superposition of the pulses $$s(t)$$ received by neuron $$i$$ at previous spike times $$t_n < t$$ (the integer index $$n$$ orders the sequence of the pulses emitted in the network), namely $E_i(t)= \frac{1}{k_i} \sum_{n|t_n<t} C_{j(n),i} \theta(t-t_n) s(t-t_n),$ where $$\theta(x)$$ is the Heaviside function, $k_i$ is the number of afferent synapses (in-degree connectivity) of neuron $i$ and the pulse shape is still an $\alpha$-function. Furthermore, each pulse $$s(t)$$ is weighted according to the strength of the connection $$C_{j,i}$$ between the emitting ($$j(n)$$) and the receiving ($$i$$) neuron. The matrix entries are chosen randomly with a constant probability: namely, $$C_{j,i}=1$$ (resp. $$C_{j,i}=0$$) with a probability $$p$$ (resp. $$(1-p)$$). In general, the connectivity matrix $$C$$ is non-symmetric. The random network associated to such connectivity matrix is termed directed Erdös-Renyi network and it is characterized by an average (in-degree) connectivity $<k> = p \times N$. An undirected network has a symmetric connectivity matrix.
In order to characterize the evolution of the random neural network at a macroscopic level, it is convenient to introduce the following averaged fields $\bar E(t) = \frac{1}{N} \sum_{i=1}^N E_i(t) \qquad ; \qquad \bar P(t) = \frac{1}{N} \sum_{i=1}^N P_i(t)$ where $$P_i=\alpha E_i + \dot{E}_i$$. Notice that in the fully coupled case $\bar E(t) \equiv E_i(t)$ and $\bar P(t) \equiv P_i(t)$ for any index $i$.
The level of homogeneity in the network can be measured at a "macroscopic level" in terms of the instantaneous standard deviation $$\sigma(t)$$ among the local fields $$E_i$$, namely $\sigma(t) = \left( \frac{\sum_{i=1}^{N}E_{i}^{2}({t})}{N}-\bar{E}^{2}({t})\right)^{1/2}$
Finally, in order to quantify the degree of synchronization among the neurons, the modulus $R$ of the following order parameter (Kuramoto, 1984) is employed: $r(t) = \frac{1}{N} \sum_{j=1}^N {\rm e}^{i \theta_j(t)}=R(t)e^{i\psi(t)}, \qquad \theta_j(t) = 2\pi \frac{(t-t_{j,n})}{t_{j,n+1}-t_{j,n}} \qquad j=1,\ldots,N.$ Here $$\theta_j$$ is the phase of the $$j$$-th neuron at time $t \in [t_{j,n}:t_{j,n+1}]$, where $$t_{j,n}$$ ($$t_{j,n+1}$$) refers to the $$n$$-th ($$n+1$$-th) spiking time of neuron $$j$$. For asynchronous dynamics $R$ is vanishingly small, $R \sim 1/\sqrt{N}$, while for a fully synchronized case $$R=1$$.
## Globally Coupled Networks
In excitatory pulse-coupled LIF networks two distinct collective states can be identified: the splay state and the partial synchronization. Both states can be characterized at two levels: the microscopic one, corresponding to the membrane potential dynamics, and the macroscopic one, associated to the behavior of the field $$E$$.
Splay states have been found in many different contexts such as Josephson devices (Hadley and Beasley, 1987), multi-mode lasers (Wiesenfeld et al., 1990) and electronic circuits (Ashwin et al., 1990). In computational neuroscience, splay states have been mainly investigated for LIF neurons (Abbott and van Vreeswijk, 1993; van Vreeswijk, 1996; Bressloff, 1999; Chow and Kopell, 2000; Zillmer et al., 2007; Olmi et al., 2012), but some studies have been also devoted to the $$\theta$$-neurons (Dipoppa et al., 2012) and to more realistic neuronal models (Brunel and Hansel, 2006). On the other hand partial synchronization (PS) has been discovered in pulse coupled LIF networks (van Vreeswijk, 1996) and more recently observed also for phase oscillators (Rosenblum and Pikovsky, 2007) and electronic devices (Temirbayev et al., 2012) with global nonlinear coupling.
### Splay State
Figure 1: Raster plots of a network with $${N}=200$$ neurons and $$a=1.3$$, $$g=0.4$$ for (a) $$\alpha=3$$ and (b) $$\alpha=9$$.
Figure 2: (a) Minima and maxima of the mean field $${\bar E}(t)$$ as a function of $$\alpha$$ for $$g = 0.4$$ and $$a = 1.3$$.(b) Critical curve $$\alpha_{c}$$ in the parameter space $$({g},\alpha)$$ for $${a}=1.3$$ in the $$N\rightarrow\infty$$ limit.
The splay state is a collective mode emerging in fully coupled oscillator networks. In this state the evolution of all oscillators is periodic of period $T$, and it can be described by the same functional form, as follows ${x}_{j}({t})={X}(t+\frac{j T}{N}) \qquad{j}=1,...,{N}$ where each oscillator $j$ can be characterized by a different phase $\frac{2\pi j}{N}$. The peculiar characteristic of the splay state is that the phases are equally distributed in the interval $$[0,2\pi]$$.
As shown in (Jin, 2002), in fully coupled neural networks neurons reach the threshold in an ordered manner and this order never changes in time. Therefore, to visualize the neuron dynamics it is convenient to order the neurons according to their potential values and then plot the index of the firing neuron as a function of the spike time emission (see Fig. 1(a)). This raster plot clearly shows that in the splay state, the interspike interval between two consecutive spikes in the network is constant and equal to $T/N$.
At a macroscopic level, the field $$E(t)$$ remains constant in time, thus indicating a constant average network activity. In addition to this, the network dynamics is asynchronous, since the modulus of the order parameter $$R$$ is exactly zero, as it can be demonstrated by noticing that the phases of the neurons are given by the following expression $$\{\theta_j(t)\}=\{2\pi [1-(j-1)/N]\}$$.
Therefore splay states are important in that they provide the simplest instance of asynchronous behavior and can be thereby used as a testing ground for the stability of a more general class of dynamical regimes. In addition to this it has been shown in (Zillmer et al., 2007) that, for an excitatory neural network, there exist a critical line $$\alpha_c(a,g)$$ in the parameter space $$(\alpha,g)$$ which defines the region where the splay state is stable (as shown in Fig. 2b).
### Partial Synchronization
Figure 3: (a) Averaged modulus of the order parameter $R$ as a function of $$\alpha$$ for $$g = 0.4$$ and $$a = 1.3$$. (b) Macroscopic attractors as a function of $$\alpha$$.
Above the critical line $$\alpha_c$$ a new stable collective state (the Partial Synchronization) emerges via a super-critical Hopf bifurcation. The transition can be well appreciated by reporting the maximal and minimal value of $$E$$ versus the pulse width, as shown Fig. 2a. Since the field $E$ is constant for splay states and periodically oscillating in the PS regime. This corresponds in the $(\bar E, \bar P)$-plane to point-like attractors for the splay state and closed curves for partially synchronous regimes, see Fig. 3b.
In the partially synchronized regime the dynamics of the neurons' membrane potentials is a quasi-periodic motion. This can be seen by analyzing the raster plot displayed in Fig. 1(b): a group of neuron reaches the threshold almost simultaneously; however the neurons participating to this almost synchronized group change in time. The recombination of the individual quasi-periodic microscopic motions into a macroscopic periodic oscillation is absolutely not trivial and it is still matter of study (Mohanty and Politi, 2006; Rosenblum and Pikovsky, 2007; Popovych and Tass, 2011). The period of the collective periodic oscillations, which arises in this state, does not coincide with (it is longer than) the average interspike-interval of the single neurons and the two quantities are irrationally related. This phenomenon is also called self-organized quasi periodicity.
Furthermore, PS can be characterized in terms of the modulus of the order parameter $$R$$, which in this case is finite and oscillates periodically in time with the same period as the macroscopic field $$\bar E$$. As shown in Fig. 3a the average $R$ value grows with $\alpha$ and tends towards the fully synchronized state. This will be reached only in the limit $\alpha \to \infty$. Indeed it is known that for infinitely rapid synaptic responses, as those associated to exponential- or $\delta$-pulses, the stable state for excitatory synapses is the fully synchronized one (Van Vreeswijk et al. 1994; Van Vreeswijk, 1996; Tsodyks et al., 1993).
## Massively Connected Networks
Figure 4: Characterization of the partially synchronized state (PS) in terms of macroscopic fields in a massively connected network with $z=1$ and for different sizes. Panel a: macroscopic attractors in the $$(\bar E, \bar P)$$ plane. The black curve corresponds to the attractor of a fully coupled networks with properly rescaled coupling constant. Panel b: Enlargement of the Figure (a). The curve at size $$N=100000$$ (not reported for clarity) almost coincides with the fully coupled one. The parameters of the model are $g=0.4$, $a=1.3$ and $\alpha=9$.
The influence of the network properties on the macroscopic neural dynamics has been recently examined in this context in (Tattini et al. 2012). In particular, the authors considered random Erdös-Renyi networks with an average connectivity growing (sub)-linearly with the network size $$N$$. Namely, the average connectivity scales as $$<k> \propto N^z \qquad 0 < z \le 1 \quad;$$ thus exhibiting the same system size dependence as for a truncated power-law distribution of the connectivities, namely $P(k) \propto 1/k^{2-z}$. The authors limited the analysis to $z \in ]0;1]$, since in a recent study of the developing hippocampal networks it has been shown that the functional connectivity is characterized by a truncated power-law distribution with exponent $z \sim 0.7-0.9$ (Bonifazi et al., 2009). In the limit $$z \to 1$$ the massively connected network, with connectivity proportional to $$N$$, is recovered; while for $$z \to 0$$ a sparse network, where the average probability to have a link between two neurons vanishes in the thermodynamic limit is retrieved (Golomb et al., 2001). The topology of Erdös-Renyi networks is modified by varying the parameter $$z$$ in the interval $$]0:1]$$, in particular as far as $$z \geq 0$$ trees and cycles of any order are present in the network, while for $$z \to 1$$ complete subgraphs of increasing order appear in the system (Albert and Barabàsi, 2002).
Figure 5: Phase diagram for the macroscopic activity of the network in the $$(N,\langle k \rangle)$$ plane. The (black) asterisks connected by the solid (black) line correspond to the transition values $$\langle k \rangle_c$$ from asynchronous (AS) to partially synchronized (PS) regime estimated for Erdös-Renyi networks with constant probability. The other symbols refer to Erdös-Renyi with $$z > 0$$: solid (resp. empty) symbols individuate asynchronous (resp. partially synchronized) states. Parameters as in the previous figure. (Modified from Tattini et al., 2012)
Similarly to what observed for fully coupled networks, two distinct dynamical phases are still present: an asynchronous state (AS) corresponding to a desynchronized dynamics of the neurons (which in the fully coupled networks correspond to the splay state) and a regime of partial synchronization (PS) associated with a coherent periodic activity of the network. A peculiar point to stress is that in the limit $N \to \infty$ the macroscopic dynamics of the fully coupled networks will be recovered for random networks for any exponent $z > 0$, as clearly shown in Fig. 4 for in the case $z=1$. Thus a random network is completely equivalent to a fully connected one for sufficiently large system sizes whenever the connectivity grows with the system size, the situation is different for sparse networks where the connectivity stays constant (Olmi et al., 2010; Tattini et al. 2012).
Once the model parameters are fixed, namely the pulse width $\alpha$, the coupling $g$ and the DC current $a$, the transition from AS to PS is now driven by the average connectivity value. In particular, by considering parameter values for which the PS is present in the fully coupled limit, one can observe that at low connectivity the system is in an asynchronous state, while PS emerges only above a certain critical average connectivity $$\langle k \rangle_c$$. Furthermore, for sufficiently large networks, $$\langle k \rangle_c$$ saturates to a constant value (see Fig. 5) suggesting that a minimal average connectivity is sufficient to observe coherent activity in systems of any size irrespectively of the kind of considered network: sparse or massively connected.
Figure 6: Maximal Lyapunov exponents as a function of the system size $$N$$ for various $$z$$-values. Parameters as in the previous figure.(Modified from Tattini et al., 2012)
The average in-degree $$\langle k \rangle$$ also controls the fluctuations in the input synaptic current (or analogously among the different field $E_i$). These can be measured by considering the standard deviation $\sigma(t)$ (defined in Sect. "Model and Indicators"), which due to the central limit theorem scales as $${\bar \sigma} \propto \frac{1}{\sqrt{<k>}} \propto N^{-z/2} \qquad ;$$ where the bar indicates a time average. Therefore, for Erdös-Renyi networks with average in-degree proportional to any positive power of $$N$$, the fluctuations will vanish in the limit $$N \to \infty$$, leading to a homogeneous collective behavior analogous to that of fully connected networks (see Fig. 4). However, the introduction of disorder in the network leads to a chaotic dynamics at the microscopic level of the single neurons. The chaotic motion can be characterized in terms of the maximal Lyapunov exponent $$\lambda_1$$: regular orbits have non positive exponents, while chaotic dynamics is associated with $$\lambda_1 > 0$$. For finite size networks, the dynamics is always chaotic for the considered model, however $$\lambda_1$$ tends to zero for increasing network size whenever $$z > 0$$, as shown in Fig. 6. This kind of deterministic irregular behavior vanishing in the large system size limit has been identified as weak chaos for coupled phase oscillators (Popovych et al., 2005).
## Sparse Networks
Sparse networks represent a peculiar exception, since they remain intrinsically inhomogeneous and chaotic for any system size. In order to examine the influence of this kind of topology it is sufficient to examine a random network with constant connectivity $K$, which is independent of the network size $N$. At a macroscopic level, also in this case a transition from AS to PS can be observed. In particular, the collective dynamics can be characterized in terms of the standard deviation of the average field $\bar E$, namely $$\sigma_E= \sqrt{<\bar{E}^2>- <\bar{E}>^2}$$. For an AS the standard deviation vanishes as $\sigma_E \propto 1/\sqrt{N}$, while in the presence of collective motions it stays finite, as shown in Fig. 7a. Similarly to what observed for massively connected networks, above a finite critical connectivity $$K_c$$ a coherent collective dynamics emerges even in sparse networks, as shown in Fig. 7a.
Figure 7: Standard deviation of the mean field, $$\sigma_E$$, versus $$K$$ for $$N=1,000$$ (black) circles, $$N=5,000$$ (red) squares, $$N=10,000$$ (green) triangles. The inset shows the macroscopic attractors for $$N=5,000$$ and $K=3$ and $$K=200$$. (b) Lyapunov exponent spectra (in the lower inset a zoom of the largest values) for $$K=20$$ and $$N=240-480-960$$. (c) Maximum Lyapunov exponent, $$\lambda_{1}$$, versus N is shown, the (red) line represents the nonlinear fit $$\lambda_{1}=0.0894-2.3562/N$$ and the (green) dashed line marks the asymptotic value.The parameters of the model are $g=0.2$, $a=1.3$ and $\alpha=9$. (Modified from Luccioli et al., 2012)
The most striking difference with respect to massively connected networks concerns the microscopic dynamics, as shown in Fig. 7c the maximal Lyapunov exponent converges to an asymptotic limit for increasing system sizes, therefore these networks will remain chaotic irrespectively of the network size. Furthermore, the dynamics is characterized by extensive high-dimensional chaos (Ruelle, 1982; Grassberger, 1989), i.e. the number of active degrees of freedom, measured by the fractal dimension, increases proportionally to the system size. Extensive chaos has been usually observed in diffusively coupled systems (Livi et al., 1986; Grassberger, 1989; Paul et al., 2007), where the system can be easily decomposed in weakly interacting sub-systems. Whenever the system is chaotically extensive the associated spectra of the Lyapunov exponents $$\{\lambda_i\}$$ collapse onto one another, when they are plotted versus the rescaled index $$i/N$$, as shown in Fig. 7b (Livi et al., 1986). Fully extensive behavior in sparse neural networks has been observed for the Theta neuron model in (Monteforte and Wolf, 2010) and the LIF model in (Luccioli et al., 2012). The previous results are obtained by assuming that all nodes are characterized by the same connectivity $$K$$, but the same scenario holds assuming a Poisson degree distribution with average connectivity $$K$$, as in Erdös-Renyi graphs.
The extensivity property is highly non-trivial in sparse networks, since in this case the dynamics is not additive. Contrary to what happens in spatially extended systems with diffusive coupling, where the dynamical evolution of the whole system can be approximated by the juxtaposition of almost independent sub-structures (Grassberger, 1989; Paul et al., 2007). Extensive chaos has not been observed in globally coupled networks, which exhibit a non-extensive component in the Lyapunov spectrum (Takeuchi et al., 2011).
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• Brunel N. (2000). Dynamics of sparsely connected networks of excitatory and inhibitory spiking neurons. J. Comput. Neurosci. 8: 183.
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### Internal references
• Arkady Pikovsky and Michael Rosenblum (2007) Synchronization Scholarpedia, 2(12):1459. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.902061402797699, "perplexity": 1249.541947241713}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084887729.45/warc/CC-MAIN-20180119030106-20180119050106-00420.warc.gz"} |
https://www.varsitytutors.com/mcat_physical-help/henderson-hasselbalch-equation | # MCAT Physical : Henderson-Hasselbalch Equation
## Example Questions
### Example Question #1 : Henderson Hasselbalch Equation
A solution of acetic acid (pK= 4.75) has a pH of 6.75. The ratio of acid to conjugate base is __________.
0.01 CH3COOH to 100 CH3COO
CH3COOH to 100 CH3COO
100 CH3COO to 1 CH3COOH
100 CH3COOH to 1 CH3COO
1 CH3COO to 100 CH3COOH
CH3COOH to 100 CH3COO
Explanation:
Use the Henderson-Hasselbalch equation:
We want the ratio of acid to conjugate base, which would be the reciprocal,
### Example Question #1 : Henderson Hasselbalch Equation
NaOH is added to a 500mL of 2M acetic acid. If the pKa value of acetic acid is approximately 4.8, what volume of 2M NaOH must be added so that the pH of the solution is 4.8?
250mL
1L
500mL
2L
250mL
Explanation:
To solve this question you need to think about the chemical reaction occurring.
We can ignore water and sodium ions for the sake of this question. The reactants exist in a 1:1 ratio, so that for every mol of NaOH we add, we lose one mol of acetic acid and gain one mol of acetate. We can determine the moles of acetic acid by using M = mol/L, which gives us mol = ML = (2M) * (0.5L) = 1mol acetic acid. If we use the Hendersen Hasselbach equation we can see that the pH equals the pKa when the concentration of conjugate base (acetate) equals the concentration of acid.
If we have 1mol of acetic acid and add 0.5mol of NaOH, we will lose 0.5mol of acetic acid and gain 0.5mol of acetate. We will then be at a point where acetic acid equals acetate. This is summarized in the ICE table below. Now we know the moles of NaOH (0.5 moles) and the concentration (2M) so we can find the volume by doing M = mol/L.
L = mol/M = (0.5mol)/(2M) = 0.25L
Acetic acid NaOH Acetate I 1 mol 0.5 mol 0 mol C -0.5 mol -0.5 mol +0.5 mol E 0.5 mol 0 mol 0.5 mol
### Example Question #1 : Henderson Hasselbalch Equation
The Ka for HCN is .
If there is a solution of 2M HCN, what concentration of NaCN is needed in order for the pH to be 9.2?
2M
No NaCN needs to be added
1M
3M
2M
Explanation:
To answer this question, we need to be able to compare the concentrations of acid and conjugate base in the solution with the pH. The Henderson-Hasselbach equation is used to compare these values, and is written as:. In this question, this equation can be written with the given values of Ka and pH.
Since we know the Ka of HCN, we can derive the pKa, which turns out to be 9.2.
As a result, we want to see to it that the amount of conjugate base is equal to the concentration of acid, so that . Because log(1) = 0, we want to see to it that the concentrations of the acid and the conjugate base are equal to one another. We know from the question that [HCN] = 2M.
As a result, a concentration of 2M NaCN will allow the pH of the solution to be 9.2.
### Example Question #31 : Acid Base Chemistry
You need to produce a buffer with pH of . You have a solution with of acetic acid (). How many moles of sodium acetate must you add to achieve the desired pH?
Explanation:
Use the Henderson-Hasselbalch equation:
Assuming that these ions occupy the same volume,
We know we have 30g of acetic acid, which is equal to 0.5mol (you should memorize the formula for acetic acid).
Plugging in our values gives us
Solving for A gives us 5mol.
### Example Question #2 : Henderson Hasselbalch Equation
A solution of hydrofluoric acid has a concentration of .
The for is .
If sodium hydroxide is slowly added to this solution, what will the pH be at the half equivalence point?
Explanation:
If we use the Henderson-Hasselbalch equation, we do not need to worry about using the molar amounts of both the acid and the base. At the half equivalence point, the conjugate base concentration is equal to that of the weak acid. This means that the equation can be simplified.
Henderson-Hasselbalch equation:
Simplified equation for half equivalence point:
Because we know the acid dissociation constant for hydrofluoric acid, the pH is calculated as:
### Example Question #1 : Henderson Hasselbalch Equation
Calculate the concentration of hydrogen ions in the following acetic acid solution.
Explanation:
To answer this question you need to use the Henderson-Hasselbalch equation:
The ratio given in the question is , or .
To use the correct ratio for the Henderson-Hasselbalch equation, we need to convert this ratio to its reciprocal:
Plugging the given values into the equation gives us:
The question is asking for the concentration of hydrogen ions. To solve for this we have to use the definition of pH.
Solving for the concentration of hydrogen ions gives us:
### Example Question #1 : Henderson Hasselbalch Equation
Which of the following is true regarding the Henderson-Hasselbalch equation?
I. The pH of the solution is always greater than the pKa of the solution
II. As the ratio of conjugate base to acid increases, the pH increases
III. The hydrogen ion concentration can never equal the acid dissociation constant
II and III
I only
II only
I and II
II only
Explanation:
The Henderson-Hasselbalch equation is a tool that allows us to calculate the pH of an acid solution using the pKa of the acid and the relative concentrations of the acid and its conjugate base. It is defined as:
By looking at the equation we can determine that if the ratio inside the logarithm is greater than 1, then the pH of the solution will be greater than the pKa; however, if the ratio is less than 1 (meaning, if the concentration of the acid is greater than the concentration of conjugate base), then the pH will be less than the pKa. Statement I is false.
Increasing the ratio of to will increase the logarithm, and subsequently the pH of the solution. This makes sense because you will have more conjugate base than acid, thereby making the solution more alkaline and increasing the pH. Statement II is true.
pH and pKa are defined as follows:
If we have the same concentration of hydrogen ions as the acid dissociation constant (), then the pH will equal the pKa. According to the Henderson-Hasselbalch equation, the pH equals the pKa if the concentration of the conjugate base equals the concentration of acid; therefore, it is possible for the hydrogen ion concentration to equal the acid dissociation constant. Statement III is false.
### Example Question #1 : Henderson Hasselbalch Equation
Increasing the volume of an acid solution __________ the pH of the solution and __________ the pKa of the acid.
will increase . . . will decrease
will not alter . . . will decrease
will increase . . . will not alter
will not alter . . . will not alter
will increase . . . will not alter
Explanation:
The definition of pH is as follows:
The pH of a solution heavily depends on the concentration of hydrogen ions. Recall that the concentration is in molarity (M), which is defined as:
Increasing the volume of the solution will decrease the concentration (molarity) of the hydrogen ions which will, subsequently, increase the pH; therefore, increasing volume will increase the pH.
Recall that pKa of an acid can never be altered. pKa is a reflection of the strength of the acid, which stays constant under all circumstances.
### Example Question #1 : Acid Base Analysis
A researcher prepares two solutions. Solution A contains an unknown acid, HA, and solution B contains an unknown acid, HB. The researcher performs several tests and collects the following data.
1. Both solutions contain weak acids
2.
3.
4.
5.
What can you conclude about these two solutions?
The acid dissociation constant for HA is greater than that of HB
The acids, HA and HB, are identical
Acid HB is more acidic than acid HA
The hydrogen ion concentration of solution A is greater than that of solution B
The acids, HA and HB, are identical
Explanation:
The Henderson-Hasselbalch equation states that:
The question gives us information regarding the ratio of conjugate base to acid AND the pH for each acidic solution. Using this information, we can solve for the pKa values of both solutions.
The pKa values of both solutions are the same. This means that both solution contains the same acid; therefore, the identity of HA is the same as the identity of HB.
The hydrogen ion concentration of solution A is lower than that of solution B because the pH of solution A is greater. Acidity, or strength, of an acid is determined by the pKa. Since we have the same pKa for both acids, HA and HB will have the same acidity. Acid dissociation constant, Ka, is defined as:
Acid dissociation constant only depends on the pKa; therefore, the Ka for both acids is the same. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9534850716590881, "perplexity": 1986.8630493781368}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267159470.54/warc/CC-MAIN-20180923134314-20180923154714-00092.warc.gz"} |
https://dml.cz/handle/10338.dmlcz/127397 | # Article
Full entry | PDF (0.3 MB)
Summary:
Let $G$ be a graph with order $p$, size $q$ and component number $\omega$. For each $i$ between $p - \omega$ and $q$, let ${\mathcal C}_{i}(G)$ be the family of spanning $i$-edge subgraphs of $G$ with exactly $\omega$ components. For an integer-valued graphical invariant $\varphi$, if $H \rightarrow H^{\prime }$ is an adjacent edge transformation (AET) implies $|\varphi (H) - \varphi (H^{\prime })| \le 1$, then $\varphi$ is said to be continuous with respect to AET. Similarly define the continuity of $\varphi$ with respect to simple edge transformation (SET). Let $M_{j}(\varphi )$ and $m_{j}(\varphi )$ be the invariants defined by $M_{j}(\varphi )(H) = \max _{T \in {\mathcal C}_{j}(H)} \varphi (T)$, $m_{j}(\varphi )(H) = \min _{T \in {\mathcal C}_{j}(H)} \varphi (T)$. It is proved that both $M_{p - \omega }(\varphi )$ and $m_{p - \omega }(\varphi )$ interpolate over $\mathbf{{\mathcal C}_{i}(G)}$, $p - \omega \le i \le q$, if $\varphi$ is continuous with respect to AET, and that $M_{j}(\varphi )$ and $m_{j}(\varphi )$ interpolate over $\mathbf{{\mathcal C}_{i}(G)}$, $p - \omega \le j \le i \le q$, if $\varphi$ is continuous with respect to SET. In this way a lot of known interpolation results, including a theorem due to Schuster etc., are generalized.
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[Zhou2] S.M. Zhou: Several interolation theorems for graphs. Graph Theory Notes of New York XXIX (1995), 18–20.
[Zhou3] S.M. Zhou: Conditional invariants and interpolation theorems for graphs. Submitted. Zbl 0943.05082
Partner of | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9144030809402466, "perplexity": 1182.4645757287096}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948612570.86/warc/CC-MAIN-20171218083356-20171218105356-00303.warc.gz"} |
http://www.aerm.fr/a36-pipe/poland_horizontal_cy_792.html | • # poland horizontal cylindrical tank building volume
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### Tank Volume Calculator
Total volume of a cylinder shaped tank is the area, A, of the circular end times the length, l. A = r 2 where r is the radius which is equal to 1/2 the diameter or d/2. Therefore: V(tank) = r 2 l Calculate the filled volume of a horizontal cylinder tank by first finding the area, A, of a circular segment and multiplying it by the length, l.Horizontal Cylindrical Tank Volume and Level CalculatorVolume calculation on a partially filled cylindrical tank Some Theory. Using the theory. Use this calculator for computing the volume of partially-filled horizontal cylinder-shaped tanks.With horizontal cylinders, volume changes are not linear and in fact are rather complex as the theory above shows. Fortunately you have this tool to do the work for you.Horizontal Cylindrical Tank Volume Calculator - ImperialHorizontal Cylindrical Tank Volume Calculator, Dip Chart and Fill Times - Inch
### Tank Volume Calculator - Oil Tanks
Mar 26, 2015 · Lets look at how to calculate the volume of both of these tanks using tank capacity calculators or a cylindrical tank calculator. In the case of the horizontal cylindrical tank, you need to calculate the area of a cross-section of the tank and then multiply this figure by the total length of the tank.HOW TO CALCULATE THE VOLUMES OF PARTIALLY FULL cylindrical tanks, either in horizontal or vertical configuration. Consider, for example, a cylindrical tank with length L and radius R, filling up to a height H. If you want to obtain the volume of the liquid that partially fills the tank, you should indicate if the tank is in horizontal or vertical position.Horizontal Tank Volume Calculations - HagraHorizontal Cylindrical Tank Volume Calculator. Horizontal Oval Tank Volume Calculator. Disclaimer. The calculations on these pages are a purely theoretical exercise! Therefore the outcomes of the calculations on these pages can only be used for indicative purposes. It might help you to estimate the content of a tank.
### Horizontal Cylindrical Tank Volume Calculator - Metric
Calculate Horizontal Cylindrical Tank Volumes, Dip Charts and Fill Times - Metric. blocklayer, poland horizontal cylindrical tank building volume Directory , poland horizontal cylindrical tank building volume Horizontal cylindrical tank volume end view diagram Fill Rate Fill Times @ Litres / Minute. Total Tank Fill Time Current Time to Fill Current Time to Empty , poland horizontal cylindrical tank building volumecalculus - Volume of a horizontal cylinder using height of , poland horizontal cylindrical tank building volumeVolume of a horizontal cylinder using height of liquid. Ask Question , poland horizontal cylindrical tank building volume length, and height of liquid in the tank to find the volume of the liquid. $\endgroup$ Billybob686 Oct 13 , poland horizontal cylindrical tank building volume then the correction factor is 1/2 -- if the height measured was half the diameter, then you would have half the volume of a horizontal cylinder. Simple. shareCalculation of Liquid Volume in a Horizontal Cylindrical , poland horizontal cylindrical tank building volumeCalculation of Liquid Volume in a Horizontal Cylindrical Container: This calculator calculates the volume of liquid inside a horizontal cylindrical container at any given height of liquid. The other required dimensions are the diameter and length of the tank. Values to be Entered Values to be Calculated; Diameter of Cylinder
### Sloped Cylindrical Tank Level - Chemical plant design , poland horizontal cylindrical tank building volume
Dec 07, 2012 · I am putting together a process design package for a batch plant expansion. We need to install a multipurpose cylindrical 40,000 gallon horizontal storage tank. To minimize holdup we are planning to slope the tank. We would like to measure the liquid level in the tank so we know if there is insufficient material left for a full batch.Content of Horizontal - or Sloped - Cylindrical Tank and PipeVolume of partly filled horizontal or sloped cylindrical tanks and pipes - an online calculator Sponsored Links The online calculator below can be used to calculate the volume and mass of liquid in a partly filled horizontal or sloped cylindrical tank if you know the inside diameter and the level of the liquid the tank.Volume of a Partially Filled Spherical Tank - Calibration , poland horizontal cylindrical tank building volumeThis calculator will tell you the volume of a paritally filled spherical tank. This will also generate a dip chart/table for the tank with the given dimensions you specify. Output is in gallons or liters. If you have a complex tank, or need something special created please don't hesitate contact us for more info.
### Greer Tank Calculator | Greer Tank, Welding & Steel
1. Click the tab below that represents your tank type 2. Enter your tank measurements length, width etc 3. View your result in the bottom bar of the calculator. If you need assistance with our tank volume calculator or would like to inquire about any of our services, give us a call today on 1-800-725-8108, send us an email or request a , poland horizontal cylindrical tank building volumeVolume in horizontal round tanks??? - Excel Help ForumJan 16, 2013 · The following formula will calculate the volume of a fluid in a partially full tank (cylindrical and flat on sides, and set up horizontally) based on the tank dimensions and the fluid depth. Enter the length of tank in A2, diameter of the tank in B2, and fluid depth in C2, and the following formula in D2.calculus - Volume of a horizontal cylinder using height of , poland horizontal cylindrical tank building volumeVolume of a horizontal cylinder using height of liquid. Ask Question , poland horizontal cylindrical tank building volume length, and height of liquid in the tank to find the volume of the liquid. $\endgroup$ Billybob686 Oct 13 , poland horizontal cylindrical tank building volume then the correction factor is 1/2 -- if the height measured was half the diameter, then you would have half the volume of a horizontal cylinder. Simple. share
### Cylindrical Tank Calculator
Cylindrical Tank Calculator, Tank Volume Calculator, Horizontal Cylinder Volume, Dipstick Chart, Dipstick Calculator, Cubic Inches, Cubic Feet, Cubic Meters, Cubic Centimeters, Gallons, Liters, Milliters , poland horizontal cylindrical tank building volume Use this calculator for computing the volume of partially-filled horizontal cylinders. If you need a vertical cylinder calculator, click here.Tank Charts - Hall Tank Company - Hall Tank CompanyUse this form to generate a chart of tank capacities. Hall Tank does not guarantee the capacity charts accuracy and in no way takes liability for loss due to its content. Calculating a chart will be considered acceptance of this agreement.Tank Calibration Chart Calculator - ODay EquipmentFrom fiberglass to steel tanks, we have a tank calculator to help you with your petroleum needs. , poland horizontal cylindrical tank building volume Tank Calibration Chart Calculator; Steel Tanks. Entering information in the following forms create a tank chart for metal rectangular and flat-end cylindrical liquid storage tanks. Horizontal Liquid Storage Tank Charts. Ft. In. Length: Diameter:
### Horizontal tank with hemispherical ends depth to capacity , poland horizontal cylindrical tank building volume
Horizontal tank with hemispherical ends depth to capacity calculation. Ask Question , poland horizontal cylindrical tank building volume I would like to find an equation for calculating the volume of liquid in the tank based on the liquid depth level. , poland horizontal cylindrical tank building volume There is an excel template to calculate horizontal cylindrical tank, with hemispherical both ends, and usually called capsule tank. , poland horizontal cylindrical tank building volumeTank Volume Calculator - ibec language instituteTry Fusion's free tank volume calculator for industrial mixing. Please fill all required fields Tank Volume Calculator , poland horizontal cylindrical tank building volume *Program does not calculate liquid volume into upper head on vertical tanks (mixing with liquid in head space is not good practice).HORIZONTAL CYLINDRICAL TANK VOLUMES - Fuel Storage Tanks HORIZONTAL CYLINDRICAL TANK VOLUMES All dimensions are in inches, volume is U.S. gallons. -Figures must be entered & calculated first in order to create a corresponding Dipstick Chart.
### Horizontal Tank - an overview | ScienceDirect Topics
Horizontal cylindrical tanks are frequently used for water and fuel storage, and in many cases it is important to be able to gauge these vessels to determine the volume of liquid contained in them. However, it is normally much more difficult to establish a volume-per-inch scale for a horizontal tank than for one in a vertical position.Online calculator: Tank Volume CalculatorsTank Volume Calculators. Use one of these to determine your tank's volume. person_outlinePete Mazzschedule 2016-01-16 10:53:10. Cylindrical Tank Volume Use this calculator to determine your cylindrical tank volume in cubic inches and gallons even if one or both ends are rounded. Especially useful if you've cut the tank in length.Calculating Tank VolumeCalculating Tank Volume Saving time, increasing accuracy By Dan Jones, Ph.D., P.E. alculating fluid volume in a horizontal or vertical cylindrical or elliptical tank can be complicated, depending on fluid height and the shape of the heads (ends) of a horizontal tank or the bottom of a vertical tank.
### Volume of horizontal cylindrical tank - OnlineConversion , poland horizontal cylindrical tank building volume
Re: Volume of horizontal cylindrical tank Folks, Re the solution below- I actually need the inverse function, whereby I can calculate the depth measured along a vertical diameter as a function of the occupied volue of the horizontal tank. (I am pumping fluid into the tank at a known rate, and I need to measure how fast the level rises)* Sloped Bottom TankThe easy part the cylindrical section above the slope, which has a volume of: (1) $\displaystyle v = \pi r^2 h$ v = volume; r = tank radius; h = cylindrical section height; More difficult the tank's sloped section, which lies between the tank's bottom and the top of the slope where the tank HORIZONTAL CYLINDRICAL TANK VOLUMES - Fuel Storage Tanks HORIZONTAL CYLINDRICAL TANK VOLUMES All dimensions are in inches, volume is U.S. gallons. -Figures must be entered & calculated first in order to create a corresponding Dipstick Chart.
### (PDF) HOW TO CALCULATE THE VOLUMES OF PARTIALLY FULL TANKS
HOW TO CALCULATE THE VOLUMES OF PARTIALLY FULL TANKS. , poland horizontal cylindrical tank building volume To calculate the fluid volume in a vertical or horizontal tank can be complicated depending of the fluid height and the caps , poland horizontal cylindrical tank building volumeTank Volume Calculator - ibec language instituteTry Fusion's free tank volume calculator for industrial mixing. Please fill all required fields Tank Volume Calculator , poland horizontal cylindrical tank building volume *Program does not calculate liquid volume into upper head on vertical tanks (mixing with liquid in head space is not good practice).Horizontal tank with hemispherical ends depth to capacity , poland horizontal cylindrical tank building volumeHorizontal tank with hemispherical ends depth to capacity calculation. Ask Question , poland horizontal cylindrical tank building volume I would like to find an equation for calculating the volume of liquid in the tank based on the liquid depth level. , poland horizontal cylindrical tank building volume There is an excel template to calculate horizontal cylindrical tank, with hemispherical both ends, and usually called capsule tank. , poland horizontal cylindrical tank building volume
### Tank calculations - MrExcel
Dec 19, 2014 · We use what are called tank strappings for some of our additive tanks. They were built from formulas for cylindrical horizontal or vertical tanks. I pugged in the size of the tanks and a strapping was produced. From there I set up a form using vlookups to the strappings. Give me some tank measurements and I'll see if what I have at work will , poland horizontal cylindrical tank building volumeSpill Prevention Control and Countermeasure (SPCC) Spill Prevention Control and Countermeasure (SPCC) Plan Multiple Horizontal Cylindrical Tanks Inside a Rectangular or Square Dike or Berm EXAMPLE This worksheet calculates the secondary containment volume of a rectangular or square dike or berm for three horizontal cylindrical tanks.Level of liquid in the cylindrical tank, AKA Quarter-Tank , poland horizontal cylindrical tank building volumeLevel of liquid in the cylindrical tank, AKA Quarter-Tank Problem. This online calculator finds the height above the bottom of a horizontal cylinder (such as a cylindrical gas tank) to which the it must be filled for it to be full at specified percentage (for example, one quarter full amounts)
### Calculating diesel volume in a horizontally-cylindrical tank
This is about several diesel tanks that need to be monitored in terms of diesel volume. I want to measure and monitor the volume of Diesel liquid inside horizontally cylindrical tanks with standard dimensions, however the safety rules are strict, where no electrical devices or wires has to exist in the tank zone, which limits the solution to few methods or devices.tank volume formula | Automation & Control Engineering Oct 21, 2006 · It is assumed by reading your description that this is a vertical cylindrical tank. The formula does not account for flanged and dished or cone bottom vessels. Leaving all units in inches - diameter squared x .7854 x height-----231 cubic inches/gallon To calculate a horizontal cylindrical tank's volume is more involved. 120" x 120" x .7854 x 180"
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http://mathhelpforum.com/algebra/202604-inequalities-variable-denominator-both-sides-print.html | # Inequalities: Variable in Denominator both sides
• Aug 27th 2012, 10:49 AM
bajpaiapurva
Inequalities: Variable in Denominator both sides
Please tell me how this question will be solved
1/(x+2) > 3/(x-3)
Though i know the answer but i dont how how will get it
Thanx
• Aug 27th 2012, 11:20 AM
earboth
Re: Inequalities: Variable in Denominator both sides
Quote:
Originally Posted by bajpaiapurva
Please tell me how this question will be solved
1/(x+2) > 3/(x-3)
Though i know the answer but i dont how how will get it
Thanx
$\frac1{x+2}>\frac3{x-3}~\implies~\frac{x-3-3(x+2)}{(x+2)(x-3)}>0$
Examine when a fraction is greater than zero.
If + means positive (> 0) and - means negative (<0) you'll get the following cases:
$\frac{+}{+ \cdot +}~\vee~\frac{+}{- \cdot -} ~\vee~\frac{-}{+ \cdot -}~\vee~\frac{-}{- \cdot +}$
Each fraction describes 3 inequalities.
• Aug 27th 2012, 11:30 AM
Plato
Re: Inequalities: Variable in Denominator both sides
Quote:
Originally Posted by bajpaiapurva
Please tell me how this question will be solved
$\frac{1}{(x+2)} > \frac{3}{(x-3)}$
answer: $(-9/2, -2) \cup (3,\infty)$
Though i know the answer but i dont how how will get it
Thanx
That answer is incorrect. It includes $x=5$ which does not work.
The given answer works for $\frac{1}{(x+2)} < \frac{3}{(x-3)}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9311047792434692, "perplexity": 2371.497305569498}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280872.69/warc/CC-MAIN-20170116095120-00222-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://www.allmathtricks.com/lines-line-segment-ray/ | # Lines In Geometry | line segment math definition | Ray along with their types
In this article explained about some basic concepts in geometry like definitions of lines, line segment and ray. Also explained properties and differences for the same.
## Lines | Line segments | Rays | All Math Tricks
### Line
A line is breadthless length. Line is a set of infinite points which extend indefinitely in both directions without width or thickness.
A line can be defined as the shortest distance between any two points.
From the above figure line has only one dimension of length.
A line may be straight line or curved line. But generally the word “line” usually refers to a straight line.
#### Properties of Lines:
1 . Line is the shortest distance path between two points extending indefinitely in both the directions
2. The length of the line not defined
3. A line has no end points and has infinite number of points on it.
4. We are unable to draw a line on a paper just we represent it on paper.
5. A line denoted by ” ” . Here the arrow heads in opposite directions of line suggests that it extends in definitely in both the directions.
6. We also denoted by small English alphabets as l, m, n . . . . etc ( , , , . . . . etc)
Generally the basic idea of a line is its straightness
Line PQ , Line QR, Line PR , Line PQ , Line QR, Line RP . . . . . . . etc , ( , , . . . . . . . etc )
Line AB , Line AC, Line AD, Line BC , Line CD, Line CB, . . . . . . . . etc ( , , , . . . . . . etc )
### Line-segment
A line-segment is a part or portion of a line with two end points. It is the set of all points between the end points and also including endpoints.
For example P and Q be two points in a plane and “l” be the line passing through P and Q. The portion of the line “l” between points P and Q is called the line segment PQ. It is denoted by
In the above figure P and Q are two points and the shortest distance between them is called a line segment.
Line segments in the above figure – AB, AC, AD, BC, BD & CD
So total line segments – 6 nos.
#### Properties of line segment
1. Line segment is a path between two points
2. It is named using its two end points
3. Line segment has definite length
4. The distance between two points P and Q is called length of denoted as PQ.
##### Congruent line segments:
Two lines segments are said to be congruent, if their lengths are equal.
### Ray
A ray can be defined as, it is a part of a line with one end point. A line segment extending endlessly in one direction is called Ray.
Here PQ is a ray with initial point P and it is denoted as ”
#### Properties of Ray:
1. A ray has only one endpoint also called initial point or starting point.
2. A ray has no definite length.
3. There is one and only one ray having given point as initial point and passing through another given point.
In the above figure is one and only one ray having initial point P and passing through another point Q
4. In the below figure is a ray with starting point P and is a ray with starting point Q. Here and are different.
5. Infinite number of rays can be drawn with the the same initial point.
Here , , , are rays with same initial point O.
6. Since a ray is endless in one direction. We are unable to draw it on paper but just we represent it.
Opposite Rays: Two rays with same initial point and extending indefinitely in opposite directions along the same line, then such rays are called opposite rays.
Here , are called opposite rays.
Related Topics:
Point in Geometry | Collinear Points and non-collinear points Examples
Types of Triangles With examples | Properties of Triangle
Quadrilateral Properties | Trapezium, parallelogram, Rhombus
Two dimensional shapes formulas.
Properties of circle in math | Arc, Perimeter, Segment of circle
Thanks for reading. I Hope you liked this article “Definitions and Properties of Line, Line segment and Ray in math”. Give feed back, comments and please don’t forget to share it. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8530235886573792, "perplexity": 997.4526479103868}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550249490870.89/warc/CC-MAIN-20190223061816-20190223083816-00399.warc.gz"} |
http://pacechallenge.org/2019/vc/ | # PACE 2019 (Track Vertex Cover Exact)
## Vertex Cover
Given: A graph.
Task: Output a vertex cover of smallest size.
#### What is a vertex cover?
An undirected graph or simply a graph is a pair $G=(V,E)$ where $V\neq \emptyset$ is a set of vertices and $E \subseteq \{ \{u, v\} \subseteq V : u \neq v \}$ is a set of edges.
A vertex cover of a graph $G=(V,E)$ is a set $S\subseteq V$ such that for every edge $\{u,v\} \in E$ we have $\{u,v\} \cap S \neq \emptyset$.
See Details
1a. Exact Track
### Literature
[ChenKanjXia06]: Chen J., Kanj I.A., Xia G. Improved Parameterized Upper Bounds for Vertex Cover. In: Královič R., Urzyczyn P. (eds) Mathematical Foundations of Computer Science (MFCS’06). Lecture Notes in Computer Science, vol 4162. Springer. 2006.
[CyganEtAl15]: Marek Cygan, Fedor V. Fomin, Łukasz Kowalik, Daniel Lokshtanov, Daniel Marx, Marcin Pilipczuk, Michał Pilipczuk, Saket Saurabh: Parameterized Algorithms. ISBN 978-3-319-35702-7. Theoretical Computer Science. Springer. 2015.
[DonweyFellows13]: Rod Downey, Michael R. Fellows. Fundamentals of Parameterized Complexity. Texts in Computer Science. Springer. 2013.
[Niedermeier06]: Rolf Niedermeier.
Invitation to Fixed Parameter Algorithms. Oxford Lecture Series in Mathematics And Its Applications. Oxford University Press. 2006.
[Wikipedia]: Wikipedia contributors, “Vertex cover,” Wikipedia, The Free Encyclopedia. (accessed November 28, 2018). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8805904388427734, "perplexity": 4482.993202556886}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371675859.64/warc/CC-MAIN-20200407054138-20200407084638-00483.warc.gz"} |
http://physics.stackexchange.com/questions/74569/how-long-until-we-fall-into-the-sun | # How long until we fall into the Sun?
As a planet moves through the solar system, a bow shock is formed as the solar wind is decelerated by the magnetic field of the planet. Presumably the creation of this shock wave would cause drag on the planet, certainly in the direction of orbit but possibly rotation as well.
Is there an estimate for the amount of drag on the Earth as it orbits the Sun? Based on the drag, how long would it take before the orbital velocity slows to the point that we spiral slowly into the Sun? Would any planets fall into the Sun prior to the Sun expanding into a Red Giant, gobbling them up?
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Assuming the sun is far enough away that we can basically treat it as a point source of solar wind, then doesn't the solar wind slow the Earth down whenever it is moving towards the sun and speed us up as we travel away? From symmetry can we argue that the net effect is basically zero? – Kevin Driscoll Aug 19 '13 at 4:47
I think the bow shock is in the direction of the orbit, so it is perpendicular to the sun. My understanding is the solar wind in the radial direction from the sun is relatively slow moving, particularly in relation to the orbital velocity of the planets. So the bow shock should be perpendicular, or slightly towards the Sun away from perpendicular. Which means there could be a radial force outward which might just offset the slowing factor. But that's all speculation, I don't know enough about astrophysics which is why I asked! – tpg2114 Aug 19 '13 at 4:52
Worth mentioning that this is basically a not-too-relativistic version of Poynting-Robertson drag. – Chris White Aug 19 '13 at 17:45
@ChrisWhite Good thought. I forgot to include radiation pressure in my calculation... wonder how much that affects it? – Michael Brown Aug 19 '13 at 23:11
Hmm...how does solar mass loss through out-gassing factor in here? – dmckee Sep 28 '13 at 1:58
This is a really rough calculation that doesn't take into account the realistic direction of the bow shock, or calculation of the drag force. I just take the net momentum flow in the solar wind and direct it so as to produce the maximum decceleration and see what happens.
Apparently the solar wind pressure is of the order of a nanoPascal. As I write this it's about $0.5\ \mathrm{nPa}$. You can get real time data from NASA's ACE satellite or spaceweather.com (click through "More data" under "Solar wind"). During periods of intense solar activity it can get up to an order or magnitude or so more than this. Let's take this worst case and assume, unrealistically, that all of the pressure is directed retrograde along the Earth's orbit. This will give the maximum deccelerating effect. I get a net force of $\sim 10^6\ \mathrm{N}$. Dividing by the Earth's mass gives a net acceleration $2\times 10^{-19}\ \mathrm{m/s^2}$. Let's fudge up again and call it $10^{-18}\ \mathrm{m/s^2}$. The time it would take for this to make a significant dint the the Earth's orbital velocity ($30\ \mathrm{km/s}$) is of the order of $10^{15}\ \mathrm{yr}$. I think we're safe.
For the other planets there is a $1/r^2$ scaling of the solar wind with the distance from the sun (assuming the solar wind is uniformly distributed) and an $R^2$ scaling with the size of the planet. So for Mercury the former effect gives an order of magnitude increase in drag and the latter effect takes most of that increase away again. There is an additional $R^{-3}$ increase in effect due to the decreased mass of a smaller body (assuming density is similar to the Earth). Then there is the $r^{-1/2}$ increase in orbit velocity due to being closer to the sun. So the total scaling factor for the time is $R r^{3/2}$, which for Mercury is about 0.1. So the end result is not much different for Mercury.
This site always causes me to learn new Mathematica features. It made really quick work of this since it has all sorts of astronomical data built in:
Note that the number of digits displayed in the final column is ludicrous. :)
-
+1 Was just about to try this kind of order of magnitude estimate myself. – Kevin Driscoll Aug 19 '13 at 5:02
I think the highlight of the answer is the Wolfram Alpha link because now I know the Sun exerts the same force on the Earth as 140 American Alligators biting down. And that the space shuttle launching directly in the orbital direction actually applies 10 times more force. – tpg2114 Aug 19 '13 at 5:09
What about tides of the earth on the sun? shouldn't that work in the opposite direction, increasing the distance as the moon is receding from earth due to the tides? – anna v Aug 19 '13 at 5:35
@JackAidley The sun will turn into a red giant in ~5bn years. The most detailed model of what will happen then created so far has the Earth being tidally dragged into the sun just before the end of the red giant phase. However since the Earth is right on the cusp of being destroyed/surviving the result shouldn't be considered definitive. – Dan Neely Aug 19 '13 at 13:10
Using Poynting-Robertson force of $F=vL_\odot R^2/4r^2c^2$ and a timescale of $\tau=v/a=vM_\oplus/F$, I get a timescale for this alternative effect of $10^{17}\ \mathrm{yr}$, scaling as $r^2R\rho$. Assuming my arithmetic is right, this means we can neglect PR for Earth. Note that PR is not the same as raw outward-directed radiation pressure, the latter being an even smaller effect in most cases. – Chris White Aug 19 '13 at 23:58
This question is different from, but related to another question How is it that the Earth's atmosphere is not “blown away”?.
In answering that question with respect to solar wind, I remarked that the orbital speed of Earth is 30 km/s while the speed of the solar wind varies between 300 km/s and 800 km/s in a nearly orthogonal directions (fully orthogonal if the orbit is considered circular). Hence the apparent wind is mostly a side wind, slightly in front (a slightly close reach in sailing terms). As a first small angle approximation, the dragging effect of the solar wind on the planet orbital speed does not come from the solar wind speed, but only from the planet own speed, which is at best a tenth of the solar wind speed.
Hence the actual effect of the solar wind on braking down Earth orbital speed is at best one tenth of the effect computed by Michael Brown, which makes it even less significant.
Another point is that the pressure due to the speed of the solar wind itself is pushing the planet outwards, away from the Sun. I am not sure how this should be analyzed, I mean to give the best insight. One way to do it is to consider that it reduces the centripetal force towards the Sun due to gravity. Furthermore, its effect must also decrase like the solar wind density in proportion to the square of the distance from the Sun, as does gravity. However the effect is proportional to the cross-section surface of the Earth, rather than its mass.
The energy output of the Sun is thought to have increased by about 30% since it formation (some 4.6 billion years ago). So the pressure from the solar wind should have increased in proportion, being equivalent to a minute reduction of the centripetal force that keeps the Earth in orbit. But it also increases the orbital drag in the same proportion. This energy output should continue to increase slowly.
Note that I assumed in these last remarks that the increased output is due to a greater amount of particles being output at the same speed. Some of the energy could be due to a greater speed of the solar wind which would increase the outward push, but not the orbital drag. I do not know which actually occurs.
More detailed calculations, which I have not done, should tell which of the two effects dominates, though they are probably both very negligible.
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https://www.asapovenrepairs.com.au/finding-mrs-lkg/orbital-period-formula-8497a1 | # orbital period formula
Out of these cookies, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Consequently, astronomers expect to be making refinements in calculating its orbital size and shape well into the 21st century. The mass of the Earth is about $$3 \times 10^{-6} M_{\odot}$$, so the approximate formula above gives an orbital period which is … /* astrof003x468x60px */ This website uses cookies to improve your experience while you navigate through the website. 3. What is the orbital period in days? $a\,$ is the orbit's semi-major axis, in meters 2. p = SQRT [ (4*pi*r^3)/G*(M) ] Where p is the orbital period; r is the distance between objects; G is the gravitational constant; M is the mass of the central object The formula is dimensionless, ... = (7.5% of the orbital period in a circular orbit) The fact that the formulas only differ by a constant factor is a priori clear from dimensional analysis. ), $It is important to understand that the following equations are valid for elliptical orbits (i.e., not just circular), and for arbitrary masses (i.e., not just for the case for one mass much less than the other). (T is known) F c = Mo×w 2 ×r= Mo×r×4×π 2 /T 2. [citation needed] Energy. /* astrof004x468x15 */ You will see an orbital period close to the familiar 1 year. We can use the formula for orbital time period: T² = (4π²/GM)a³; where T is in Earth years, a is distance from sun in AU, M is the solar mass (1 for the sun), G is the gravitational constant. T = \ orbital period, Solving for satellite orbit period. The central body could be a planet, the sun or some other large mass capable of causing sufficient acceleration on a less massive nearby object. The full equation looks like the following: where P is the orbital period of the comet, is the mathematical constant pi, a is the semi-major axis of the comet’s orbit, G is the gravitational constant, M is the mass of the Sun, and m is the mass of the comet. * * * * * * * Without Using The Calculator * * * * * * * t 2 = (4 • π 2 • r 3) / (G • m) t 2 = (4 • π 2 • 386,000,000 3) / (6.674x10 -11 • 6.0471x10 24) t 2 = 2.27x10 27 / 4.04 14. t 2 = 5,626,000,000,000. Enter the radius and mass data. When the given parameters are substituted in the orbital velocity formula, we get. Quick and easy wordpress installation. It means that if you know the period of a planet's orbit (P = how long it takes the planet to go around the Sun), then you can determine that planet's distance from the Sun (a = the semimajor axis of the planet's orbit). T = 2\pi\sqrt{a^3/\mu} where: 1. Using the information in the chart, convert the orbital periods of Earth and Mars from days to seconds.$. $a = \$ semimajor axis of the elliptical orbit (i.e., half of the largest symmetry axis of the ellipse), V orbit = √ GM / R = √6.67408 × 10-11 × 1.5 × 10 27 / 70.5×10 6 = √ 10.0095 x 10 16 / … F g = F c. G×M×Mo/r 2 = Mo×r×4×π 2 /T 2. Orbital Period Formula. google_ad_client = "ca-pub-5205698000600672"; Do this by multiplying the number of days by 86,400. According to Kepler's Third Law, the orbital period $T\,$ (in seconds) of two bodies orbiting each other in a circular or elliptic orbitis: 1. Kepler's third law states: The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Solution: Plug into the formula P 2 = k a 3 P 2 = 6.9 x 10-9 x (75,000) 3 P 2 = 2.9 x 10 6 Take the square root of both sides P = 1700 days. The orbital velocity of the International Space Station is 7672 m/s. Since we know that for the Sun-Earth system $T$ is 1 year for $a=1 \ {\rm AU}$ (AU $=$ astronomical units), and for $M+m$ in units of solar mass, \[ That time is simply the orbital period P, which is generally easy to observe. Use astrophysicsformulas for physics, astrophysics assignment and homework help, test prep, exam prep, and as a study aid or memory jogger. Orbital Period Equation In general, two masses, $m$ and $M$ will orbit around the center of mass of the system and the system can be replaced by the motion of the reduced mass, $\mu \equiv mM/(m+M)$. Circumference = C = 2 (pi) A 2) A satellite is orbiting the Earth with an orbital velocity of 3200 m/s. The period of a satellite is the time it takes it to make one full orbit around an object. , . google_ad_client = "ca-pub-5205698000600672"; Thus to maintain the orbital path the gravitational force acted by the planet and centripetal force acted by the moon should be equal. Science Physics Kepler's Third Law. The orbital period is the period of a satellite, the time taken to make one full orbit around an object. where $a_{\rm AU}$ is the semimajor axis in units of AU. Orbital velocity is the velocity of this orbit depends on the distance from the object to the centre of the Earth. The website $\mu = GM \,$ is the velocity of m/s... A mass of mass MCentral orbit depends on the orbital period close to the centre of the Space. Velocity is the velocity of this orbit depends on the distance from the.. / 2.7×106 's semi-major axis, in meters 2 speed of a … orbital period of the axis days! V orbit = √GM / R. v orbit = √GM / R. v orbit = √6.673×10−11 ×8.35×1022 2.7×106., velocity and period based on the orbital period formula from the sun is one year BYJU s. /T 2 in the orbital period is the standard gravitational parameter, in meters.... Gravitational force acted by the moon should be equal s for more such interesting articles Policy Site Map P which! Axis is equal to the radius at which it orbits, the time taken to make one full around! By the visible partner, called the circumference of the Earth is a measure a! Based on the orbital velocity is the standard gravitational parameter, in 2! Small in practice, but not always totally negligible stay tuned with BYJU ’ s encounter with resulted... Input, so if you show your work we may find the problem substituted in the orbital period is orbit... 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https://www.physicsforums.com/threads/what-would-happen-to-the-slope-of-the-graph.549287/ | # What would happen to the slope of the graph?
1. Nov 10, 2011
### physikamagika
1. The problem statement, all variables and given/known data
What would happen to the slope of the graph- Current Vs. Voltage if a higher value resistor was used?
2. Relevant equations
Slope is the Resistance R. This experiement was intended to verify Ohm's Law; however, can you verify Ohm's Law if the resistor overheated in the experiment? How would you rearrange Ohm's Law equation to make it useful to graph y=mx?
3. The attempt at a solution
Maybe there would be a downward curve in the slope because a higher value resistor would decrease the flow of current?
Can you offer guidance or do you also need help?
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https://www.physicsforums.com/threads/stationary-waves-in-composite-strings.729553/ | # Stationary waves in composite strings
1. Dec 22, 2013
### arvindsharma
Dear All,
I was reading the concept of stationary waves in composite string ABC made up of joining two strings AB AND BC with different mass per unit length and a joint at 'B'.the two ends of the composite string are clamped at 'A' and 'C' and a transverse wave is set up by an external source at one of the clamp say clamp 'A'.it was written in the book that stationary wave will set up in the composite string ABC at a particular frequency of external source with joint 'B' as a node.following are my doubts
1)the stationary wave in AB must be due to incident wave and reflected wave at junction B but at junction 'B' there is some reflection and some transmission(so that stationary wave in B can also be set up).due to partial reflection the amplitude of incident wave and reflected wave must be different so a perfect stationary wave is not possible in string AB.however wave in string BC is perfectly stationary(assuming no energy loss at clamp C).is my reasoning correct in this sense?
2)Why there is a node at junction?does this always happen in composite string that joint is a node?
3)the stationary wave in string AB and BC will have different amplitude because the amplitude of reflected and transmitted wave is not same?
2. Dec 22, 2013
### Simon Bridge
1. because of the partial refection at B you won't get opposite travelling waves to interfere to produce a non-zero amplitude there and get standing waves. You can check by solving the wave equation for standing waves or just do the experiment.
2. see above. yes - for standing waves.
3. yes
Merry Xmas.
3. Dec 22, 2013
### AlephZero
THinking about a mode of vibration as a "stationary wave" made from "traveling waves" in opposite directions is useful for a simple structure like a uniform string, but less useful for more complicated structures.
You will certainly get "stationary" modes of vibration of the whole structure, but the traveling waves are complicated because the speed will be different in the two parts of the string, and there will be partial reflection and partial transmission at B, for the traveling waves going in both directions. The "shape" of the traveling waves will not be the same as the shape of the stationary vibration mode, and the "shape" of the stationary wave willl not be a simple sine wave like a uniform string
In general, point B will not be a node. If it is a node in your example, there must be something special about the problem, for example some relation between the mass per unit length of the two strings, or the lengths of parts AB and BC. Or, the book is wrong!
4. Dec 22, 2013
### arvindsharma
not yet satisfied.please clear my doubts
5. Dec 22, 2013
### AlephZero
You need to tell us EXACTLY what the book says about the system.
it should be obvious that if AB and BC were the same length, and the mass of BC was 1.01 times the mass of AB, the vibration modes of the system would be very close to a uniform string, and there would be standing waves that did NOT have a node at B.
If the system in the book has a node at B, it is because of the particular masses, lengths, etc, of that system, not because every possible set of two strings has a node there.
6. Dec 22, 2013
### sophiecentaur
Of course. To get a standing wave, you must have solutions to the wave equations on both sections that are satisfied by the boundary conditions. I have a feeling that the bandwidth of the system (i.e. the losses) may allow some latitude but I can't be sure.
7. Dec 22, 2013
### AlephZero
You are right, the existence of "exact" standing waves with "perfect" nodes (zero motion at all times) is only an approximation, and depends on the way the system is damped.
Making models of damping based on physics is notoriously difficult, especially for lightly damped systems like vibrating strings etc. The usual technique is to make a model that is mathematically convenient, and fits reasonably well to measurements close to the resonant frequency of each mode. What happens away the resonances often isn't important from an engineering point of view, because the amplitudes are small compared with the resonance peaks.
In real life, different positions on a vibrating object don't necessarily vibrate in phase with each other, and don't necessarily even vibrate back and forth through some equilibrium position. For example it a rotating object is vibrating, the coriolis forces make every point move in an ellipse around the equilibrium point (as viewed by somebody rotating at the same speed as the object), not back and forth along a straight line. And the orientations of the principal axes of the ellipses vary at different positions on the object.
But all this is probably irrelevant to the OP's question.
8. Dec 23, 2013
### arvindsharma
i got the point. thank you all for your contribution
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https://www.studypug.com/ie/ie-sixth-year/graph-linear-inequalities-in-two-variables | # Graphing linear inequalities in two variables
### Graphing linear inequalities in two variables
In this lesson, we will talk about how to solve linear inequalities as well as how to graph the solutions on number lines and the xy plane.
#### Lessons
• Introduction
Introduction to Inequalities
a)
Graph the solution on a number line:
$x = 3$
$x \ge 3$
$x$ > $3$
$x \le 3$
$x$ < $3$
b)
Graph the solution on the xy-plane:
$x = 3$
$x \ge 3$
$x$ > $3$
$x \le 3$
$x$ < $3$
c)
Graph the solution on the xy-plane:
$y = -2$
$y \ge -2$
$y$ > $-2$
$y \le -2$
$y$ < $-2$
• 1.
Graphing Linear Inequalities in Two Variables
Graph:
i)
$y = x - 1$
ii)
$y \ge x - 1$
iii)
$y$ < $x - 1$
• 2.
Graphing Linear Inequalities in Two Variables
Graph:
i)
$6x + 5y = 15$
ii)
$6x + 5y \ge 15$
iii)
$6x$ $+ 5y$ < $15$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 30, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8182980418205261, "perplexity": 4506.734767440378}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999620.99/warc/CC-MAIN-20190624171058-20190624193058-00326.warc.gz"} |
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Page 1 of 4 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8276580572128296, "perplexity": 4526.7430181664695}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394678702159/warc/CC-MAIN-20140313024502-00069-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://ask.sagemath.org/answers/12132/revisions/ | # Revision history [back]
I don't think there is any current work. Here is something you might try if you use LaTeX.
• Get sws2tex.
• Use that to make your worksheet into a decent approximation as a LaTeX file.
• Modify that worksheet by changing the document class to Beamer, and then insert slide beginnings and ends.
I agree that this is not ideal for industrial use, but depending on your needs it might be a good workaround.
I don't think there is any current work. Here is something you might try if you use LaTeX. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8427156805992126, "perplexity": 814.5049017063197}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107874026.22/warc/CC-MAIN-20201020162922-20201020192922-00097.warc.gz"} |
https://stats.stackexchange.com/questions/126351/reasons-for-data-to-be-normally-distributed | # Reasons for data to be normally distributed
What are some theorems which might explain (i.e., generatively) why real-world data might be expected to be normally distributed?
There are two that I know of:
1. The Central Limit Theorem (of course), which tells us that the sum of several independent random variables with mean and variance (even when they are not identically distributed) tends towards being normally distributed
2. Let X and Y be independent continuous RV's with differentiable densities such that their joint density only depends on $x^2$ + $y^2$. Then X and Y are normal.
(cross-post from mathexchange)
Edit: To clarify, I am not making any claims about how much real world data is normally distributed. I am just asking about theorems that can give insight into what sort of processes might lead to normally distributed data.
• You might find interesting related material in our thread at stats.stackexchange.com/questions/4364. To avoid potential confusion among some readers, I would like to add (and I hope this was your intention) that your question should not be read as suggesting that all or even most actual datasets can be adequately approximated by a normal distribution. Rather, in certain cases when certain conditions hold, it could be useful to employ a normal distribution as a frame of reference for understanding or interpreting the data: so what might those conditions be? – whuber Dec 2 '14 at 23:19
• Thank you for the link! And that is exactly right, thank you for the clarification. I will edit it to the original post. – anonymous Dec 2 '14 at 23:25
• @user43228, "There are, of course, tons of other distributions that arise in real world problems that don’t look normal at all." askamathematician.com/2010/02/… – Pacerier Jun 3 '15 at 12:43
Many limiting distributions of discrete RVs (poisson, binomial, etc) are approximately normal. Think of plinko. In almost all instances when approximate normality holds, normality kicks in only for large samples.
Most real-world data are NOT normally distributed. A paper by Micceri (1989) called "The unicorn, the normal curve, and other improbable creatures" examined 440 large-scale achievement and psychometric measures. He found a lot of variability in distributions w.r.t. their moments and not much evidence for (even approximate) normality.
In a 1977 paper by Steven Stigler called "Do Robust Estimators Work with Real Data" he used 24 data sets collected from famous 18th century attempts to measure the distance from the earth to the sun and 19th century attempts to measure the speed of light. He reported sample skewness and kurtosis in Table 3. The data are heavy-tailed.
In statistics, we assume normality oftentimes because it makes maximum likelihood (or some other method) convenient. What the two papers cited above show, however, is that the assumption is often tenuous. This is why robustness studies are useful.
• Most of this post is great, but the introductory paragraph bothers me because it could so easily be misinterpreted. It seems to say--rather explicitly--that in general, a "large sample" will look normally distributed. In light of your subsequent remarks I don't believe you really meant to say that. – whuber Dec 2 '14 at 23:25
• I should have been more clear - I'm not suggesting that most real world data is normally distributed. But that is a great point to raise. And I'm assuming what you mean is that binomial distribution with large n is normal, and that poisson distribution with large mean is normal. What other distributions tend towards normality? – anonymous Dec 2 '14 at 23:27
• Thanks, I edited the first paragraph. See Wald and Wolfowitz (1944) for a theorem on linear forms under permutation, for example. I.e., they showed the two sample t statistic under permutation is asymptotically normal. – bsbk Dec 2 '14 at 23:31
• A sampling distribution is not a "real world dataset"! Perhaps the difficulty I am having with apparent inconsistencies in your post stems from this confusion between distribution and data. Perhaps it stems from a lack of clarity about what "limiting" process you actually have in mind. – whuber Dec 2 '14 at 23:33
• The original question was about explaining "generatively" how normal real-world data might come about. It is conceivable that real data might be generated from a binomial or poisson process, both of which can be approximated by the normal distribution. The op asked for other examples and the one that came to mind was the permutation distribution, which is asymptotically normal (in the absence of ties). I can't think of a way off-hand that real data would be generated from that distribution so maybe that one is a stretch. – bsbk Dec 2 '14 at 23:43
There is also an information theoretic justification for use of the normal distribution. Given mean and variance, the normal distribution has maximum entropy among all real-valued probability distributions. There are plenty of sources discussing this property. A brief one can be found here. A more general discussion of the motivation for using Gaussian distribution involving most of the arguments mentioned so far can be found in this article from Signal Processing magazine.
• This is backwards, as I understand it. It's about how making the assumption of normality is in a strictly defined sense a weak assumption. I don't see what that implies about real-world data. You might as well argue that curves are typically straight because that's the simplest assumption you can make about curvature. Epistemology does not limit ontology! If the reference you cite goes beyond that, please spell out the arguments. – Nick Cox Dec 9 '14 at 19:58
In physics it is CLT which is usually cited as a reason for having normally distributed errors in many measurements.
The two most common errors distributions in experimental physics are normal and Poisson. The latter is usually encountered in count measurements, such as radioactive decay.
Another interesting feature of these two distributions is that a sum of random variables from Gaussian and Poisson belongs to Gaussian and Poisson.
There are several books on statistics in experimental sciences such as this one:Gerhard Bohm, Günter Zech, Introduction to Statistics and Data Analysis for Physicists, ISBN 978-3-935702-41-6
The CLT is extremely useful when making inferences about things like the population mean because we get there by computing some sort of linear combination of a bunch of individual measurements. However, when we try to make inferences about individual observations, especially future ones (eg, prediction intervals), deviations from normality are much more important if we are interested in the tails of the distribution. For example, if we have 50 observations, we're making a very big extrapolation (and leap of faith) when we say something about the probability of a future observation being at least 3 standard deviations from the mean. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.831709623336792, "perplexity": 490.14568317036714}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027316783.70/warc/CC-MAIN-20190822042502-20190822064502-00432.warc.gz"} |
https://www.physicsforums.com/threads/confusion-with-regards-to-power-lines-and-the-power-lost.907296/ | # B Confusion with regards to power lines and the power lost
1. Mar 11, 2017
### Coolamebe
Ok, so the title was pretty vague, I'm not sure how to succinctly describe the confusion. Anyway, so I've learnt that the power lost is P=I2R, and so by increasing the voltage, as P=VI and is constant, the current will be lowered, and thus the power lost will decrease.
I'm confused about a couple things. While my physics teacher was specifically talking about P=I2R, should not P=V2/R also give the value, and so by increasing the voltage we increase the power lost? Is this not a contradiction?
I feel like it could be remedied if the wires in power lines are not ohmic conductors and so half the math I did above is invalid.
Anyway, any help would be greatly appreciated, thank you!
2. Mar 11, 2017
### lychette
The calculation relates to the power dissipated (lost) in the transmission cables so the voltage you need is the voltage across the cables. I think you are confusing this with the voltage across the load (at the end of the cables)
3. Mar 11, 2017
### Coolamebe
Can you explain this a little more in depth? My teacher explained this very superficially so I'm not too sure on this. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9634357690811157, "perplexity": 529.3324930083674}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676589404.50/warc/CC-MAIN-20180716154548-20180716174548-00367.warc.gz"} |
https://www.physicsforums.com/threads/finding-length-in-regular-pyramid.567398/ | # Finding length in regular pyramid
1. Jan 13, 2012
### songoku
1. The problem statement, all variables and given/known data
Regular pyramid T.ABCD has square base with side 1 cm. N is the mid-point of AB. Find the length of TN
2. Relevant equations
not sure
3. The attempt at a solution
No information given about the height of the pyramid and the length of the slant-edge, such as TA, so how can we find TN
2. Jan 13, 2012
### Mentallic
Without being given the height of the pyramid (or something equivalent, such as the slant height) then you can't place a value on the length of TN. The only way I can see going about this is to leave the answer in terms of an unknown constant, such as AT, or maybe regular pyramid means something more than just the apex of the pyramid is normal to the centre of the base.
3. Jan 13, 2012
### songoku
Not sure, but I think regular pyramid means that the apex of the pyramid is normal to the centre of the base and the base has equal side, i.e the base is square.
If I remember correctly, the answer is 1/2 √3. But I can't find it. I have provided the complete question
4. Jan 14, 2012
### Mentallic
Ahh ok then, with the answer of $$\frac{\sqrt{3}}{2}$$ that means the edges connecting to the apex are a length of 1 each, which I guess is what was also implied by the "regular" pyramid.
5. Jan 14, 2012
### songoku
Ahhh you are correct. Thanks | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8410371541976929, "perplexity": 600.4220611400748}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257646213.26/warc/CC-MAIN-20180319023123-20180319043123-00143.warc.gz"} |
http://tex.stackexchange.com/questions/58310/top-aligning-text-contents-of-two-parboxes-or-minipages | Top aligning text contents of two parboxes or minipages
I'd appreciate help top aligning the contents of \parboxes containing texts of different sizes as shown in the following graphic. Because one of the texts has a different size it appears to hover above the other:
Source (MWE):
\documentclass[english]{article}
\begin{document}
\mbox{\Huge \textsc{Text}}
\hfill
\parbox[t]{4.25cm}{
I am a fish
I am a fish
I am a fish
I am a fish
}
\end{document}
-
You can also use a tabular environment. – Marco Daniel Jun 2 '12 at 18:46
Sometimes you need to adjust by hand with \raisebox etc but an automated solution to align at the top of the box rather than the first baseline produces:
\documentclass[english]{article}
\begin{document}
\parbox[t]{3cm}{\hrule height 0pt width 0pt
\Huge \textsc{Text}
}
\hfill
\parbox[t]{4.25cm}{\hrule height 0pt width 0pt
I am a fish
I am a fish
I am a fish
I am a fish
}
\end{document}
As noted in the comments, by placing rules in the first line the \parbox[t] code aligns on the rule (so effectively at the top of the first real line rather than its baseline).
-
\vspace{0pt} is the shorter version ... – Herbert Jun 2 '12 at 18:56
@Herbert true, although I tend to use rules myself, partly I think as you can fiddle with visible rules to get special effects and then finally make them 0width (but odd height or depth) – David Carlisle Jun 2 '12 at 19:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.923129141330719, "perplexity": 2516.532040227514}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368697681504/warc/CC-MAIN-20130516094801-00001-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://web2.0calc.com/questions/can-someone-help-me-with-this-calculus-question-please | +0
Can someone help me with this calculus question please?
+1
78
5
+423
I need help w this qustion
Apr 13, 2020
#1
+7826
+2
a)
$$f'(x) \text{ is }\left\{\begin{matrix}\text{positive}\\\text{negative}\end{matrix}\right\}\text{ when }f(x)\text{ is }\left\{\begin{matrix}\text{strictly increasing}\\\text{strictly decreasing}\end{matrix}\right\}.$$
$$\implies f(x) \text{ is strictly increasing on the interval }(-\infty, 1)\cup (3, \infty)\text{ and strictly decreasing on the interval } (1, 3).$$
What about when x = 1 or x = 3?
Let ε be an arbitrary constant, infinitesmally small.
$$f'(1 -\varepsilon) > 0\text{ and }f'(1 + \varepsilon) < 0$$
At x = 1, local maximum of f(x) occurs.
$$f'(3 -\varepsilon) < 0\text{ and }f'(3 + \varepsilon) > 0$$
At x = 3, local minimum of f(x) occurs.
$$f'(x) \text{ is }\left\{\begin{matrix}\text{increasing}\\\text{decreasing}\\\text{constant}\end{matrix}\right\}\text{ when }f(x)\text{ }\left\{\begin{matrix}\text{is convex}\\\text{is concave}\\\text{has a point of inflexion}\end{matrix}\right\}.$$
$$\implies f(x) \text{ is concave on the interval } (-\infty, 2)\text{ and is convex on the interval }(2, \infty).$$
$$\text{Point of inflexion of }f(x)\text{ occurs at } x =2.$$
b)
$$f'(x) \text{ is }\left\{\begin{matrix}\text{increasing}\\\text{decreasing}\\\text{constant}\end{matrix}\right\}\text{ when }f''(x)\text{ is }\left\{\begin{matrix}\text{positive}\\\text{negative}\\\text{0}\end{matrix}\right\}.$$
$$\implies f''(x) \text{ is negative on the interval } (-\infty, 2)\text{ and is positive on the interval }(2, \infty).$$
$$f''(2) = 0$$
.
Apr 14, 2020
#2
+1
what do you mean by f' is constant when f has a point of inflexion?
Guest Apr 14, 2020
edited by Guest Apr 14, 2020
#3
+109524
+1
If the gradient is not increasing and not decreasing then it must be a constant.
At a point of inflection
$$\frac{d^2y}{dx^2}=0\\~ \\and\\~\\ \frac{dy}{dx}=\int \frac{d^2y}{dx^2}dx=\int 0dx=k$$
I am not sure that I have presented this properly.
Feel free to comment Max.
Melody Apr 14, 2020
edited by Melody Apr 14, 2020
#4
+1
a constant where? on the entire line?
Guest Apr 14, 2020
#5
+109524
+1
No just at that one point. At the point of inflextion the gradient is neither increasing nor decreasing.
If ithe rate of change is not + or - then it must be 0. Which means that the gradient must be a constant at that point.
It is not something I would usually add. (I can see the confusion)
I would normally have just left the statement out altogether.
It is true though.
Melody Apr 14, 2020 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9647525548934937, "perplexity": 1859.498183076535}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348502097.77/warc/CC-MAIN-20200605143036-20200605173036-00424.warc.gz"} |
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## Life Group Title How is choosing a boy and a girl from 12 boys and 12 girls to represent a club different from choosing two girls from 12 girls to be president and treasurer of the club? one year ago one year ago Edit Question Delete Cancel Submit
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the first involves 24 choices. the other involve 12 choices.
• one year ago
2. Life
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thats what i thought, thanks for the clarity
• one year ago
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lol you're welcome :)
• one year ago
4. kropot72
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When considering choosing a boy and a girl from 12 boys and 12 girls there are 12 choices of girl for each boy selected. So the total number of ways of choosing is:$12\times 12$ When choosing 2 girls from 12 to be president and treasurer of the club there are 2 ways of assigning the offices for each pair chosen. So in this case the number of ways of choosing is: $12P2=\frac{12!}{(12-2)!}=12\times 11$
• one year ago
5. Life
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What does the exclamation point mean?
• one year ago
6. kropot72
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The exclamation point indicates a factorial. A factorial is the product of all the positive integers from 1 up to and including a given integer. The symbol is n! where n is the given integer. So 5 factorial is written 5! and means $5\times 4\times 3\times 2\times 1(=120)$
• one year ago
7. kropot72
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Also 12P2 means the number of permutations of 12 different things taken 2 at a time.
• one year ago
8. Life
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Thanks :)
• one year ago
9. kropot72
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You're welcome :) You can look at the choice of president and treasurer from the 12 girls very simply as follows: If one of the 12 girls is chosen as president there are 11 choices for treasurer. So for each one of the girls chosen as president there are 11 choices for treasurer. Going through each in sequence it will be seen that each one if the girls has a possibility of being either president or treasurer and the total number of ways of choosing is $12\times 11$
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## anonymous 3 years ago . Delete Cancel Submit
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Solving for t simply means that we want to get t by itself on one side. Lets keep it on the left side since it's already there. Then we would want to get rid of that 9
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• 3 years ago
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To get rid of that 9 we would subtract both sides by 9 Can you tell me what we then get?
3. anonymous
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1. Square it. t + 9^2 = 16^2 = t + 81 = 256 2. Take the 81 from both sides. t = 175 DONE! XD
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• 3 years ago
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Sorry Ryan D: Can you confirm my method and answer though
5. anonymous
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great now we have $\sqrt{t}=7$ now we have to square both sides $\sqrt{t}^2=7^2$ $t=7^2$ Can you tell me what that is?
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Yes you are correct.
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https://www.physicsforums.com/threads/implict-partial-derivative.660199/ | # Homework Help: Implict partial derivative
1. Dec 21, 2012
### unscientific
1. The problem statement, all variables and given/known data
x3 + y3 + z3 - 3xyz = 6
Find (∂y/∂x)z.
2. Relevant equations
3. The attempt at a solution[/
can i simply take the partial derivative of both sides treating z as constant?
x3 + y3 + z3 - 3xyz - 6 = 0
f(x,y,z) = 0
(∂f/∂x)z = 0
2. Dec 21, 2012
### haruspex
Yes.
3. Dec 21, 2012
### HallsofIvy
That z is to be held constant and y thought of as a function of x only is precisely what that subscript "z" means. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.869408905506134, "perplexity": 2572.4718656005116}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267867095.70/warc/CC-MAIN-20180624215228-20180624235228-00278.warc.gz"} |
https://friedfoo.wordpress.com/fearless/fearless-symmetry/3-quadratic-reciprocity-galois-theory-elliptic-curves/ | # Quadratic Reciprocity, Galois Theory, Elliptic Curves
Motivation
Where ${p}$ and ${q}$ are distinct odd primes, there exists a mysterious relationship between solutions of ${x^2=q}$ in ${\mathbb{F}_p}$ and ${x^2=p}$ in ${\mathbb{F}_q}$. This quadratic reciprocity is but the tip of the giant iceberg of reciprocity laws. Quadratic reciprocity is related closely to Galois representations.
The variety defined by equation ${x^2-a=0}$ usually will have either two solutions or none, depending on whether ${a}$ is a square in the number system of ${x}$. For instance, since ${1}$ is a square in all number systems, the variety for ${x^2-1=0}$ is
$\displaystyle S(\mathbb{Z})=S(\mathbb{Q})=S(\mathbb{R})=S(\mathbb{C})=\ldots =\{1, -1\}$
The variety for equation ${x^2+1=0}$ is
$\displaystyle S(\mathbb{Z})=S(\mathbb{Q})=S(\mathbb{R})=S(\mathbb{F}_3)=S(\mathbb{F}_7)=S(\mathbb{F}_{11})=\emptyset$
$\displaystyle S(\mathbb{C})=\{i, -i\} \ \ \ \ S(\mathbb{F}_5)=\{2, 3\} \ \ \ \ S(\mathbb{F}_{13})=\{5, 8\}$
The interesting property here is that ${-1}$ is considered a square in ${\mathbb{C}}$, ${\mathbb{F}_5}$ and ${\mathbb{F}_{13}}$, but not in the other systems. Also note that when ${-1}$ is a square in ${\mathbb{F}_p}$, then the sum of the two solutions equals ${p}$. This is because the two solutions are ${n}$ and ${p-n}$.
In general, one observes that ${\mathbb{F}_p}$ has solutions to ${x^2+a=0}$ when ${p\equiv 1\pmod 4}$ and no solutions when ${p\equiv 3\pmod 4}$.
Returning to the equation ${x^2-a=0}$, recall it has no solutions, one solution ${0}$ (when ${a}$ is a multiple of ${p}$, or two solutions ${\pm a}$. The Legendre Symbol codifies this situation over ${\mathbb{F}_p}$:
$\displaystyle \left (\frac{a}{p}\right ) = \begin{cases} -1 \ \ \text{no solutions}\\ \ \ 0\ \ \text{one solution}\\ \ \ 1\ \ \text{two solutions} \end{cases}$
where this convention provides the following niceties (for integer ${k}$):
$\displaystyle \left (\frac{0}{p}\right ) = 0, \ \ \ \left (\frac{1}{p}\right ) = 1, \ \ \ \left (\frac{a}{p}\right ) \left (\frac{b}{p}\right ) = \left (\frac{ab}{p}\right ), \ \ \ \left (\frac{a}{p}\right ) = \left (\frac{a+kp}{p}\right )$
.
Returning to the equation ${x^2+1=0}$, the solution result can be restated:
$\displaystyle \left (\frac{-1}{p}\right ) = \begin{cases}\ \ 1\ \ p\equiv 1\pmod 4\\ \ \ 0\ \ \text{not possible because} \ p\not| \ a\\ -1\ \ p\equiv 3\pmod 4 \end{cases}$
Similarly,
$\displaystyle \left (\frac{2}{p}\right ) = \begin{cases}\ \ 1\ \ p\equiv 1\pmod 8\\ -1\ \ p\equiv 3\pmod 8\\ -1\ \ p\equiv 5\pmod 8\\ \ \ 1\ \ p\equiv 7\pmod 8 \end{cases} \left (\frac{3}{p}\right ) = \begin{cases}\ \ 1\ \ p\equiv 1\pmod {12}\\ -1\ \ p\equiv 5\pmod {12}\\ -1\ \ p\equiv 7\pmod {12}\\ \ \ 1\ \ p\equiv 11\pmod {12} \end{cases}$
$\displaystyle \left (\frac{5}{p}\right ) = \begin{cases}\ \ 1\ \ p\equiv 1\pmod {20}\\ -1\ \ p\equiv 3\pmod {20}\\ -1\ \ p\equiv 7\pmod {20}\\ \ \ 1\ \ p\equiv 9\pmod {20}\\ \ \ 1\ \ p\equiv 11\pmod {20}\\ -1\ \ p\equiv 13\pmod {20}\\ -1\ \ p\equiv 17\pmod {20}\\ \ \ 1\ \ p\equiv 19\pmod {20} \end{cases} \left (\frac{7}{p}\right ) = \begin{cases}\ \ 1\ \ p\equiv 1\pmod {28}\\ \ \ 1\ \ p\equiv 3\pmod {28}\\ -1\ \ p\equiv 5\pmod {28}\\ \ \ 1\ \ p\equiv 9\pmod {28}\\ -1\ \ p\equiv 11\pmod {28}\\ -1\ \ p\equiv 13\pmod {28}\\ -1\ \ p\equiv 15\pmod {28}\\ -1\ \ p\equiv 17\pmod {28}\\ \ \ 1\ \ p\equiv 19\pmod {28}\\ -1\ \ p\equiv 23\pmod {28}\\ \ \ 1\ \ p\equiv 25\pmod {28}\\ \ \ 1\ \ p\equiv 27\pmod {28} \end{cases}$
Studying the patterns above and additional examples, two facts emerge that are surprisingly difficult to prove, but are very useful for subsequent proofs. Let ${a}$ be a positive integer.
If ${p}$ and ${q}$ are odd primes and ${p\equiv q\pmod {4a}}$, then
$\displaystyle \left (\frac{a}{p}\right ) = \left (\frac{a}{q}\right )$
If ${p}$ and ${q}$ are odd primes and ${p+q\equiv 0\pmod {4a}}$, then
$\displaystyle \left (\frac{a}{p}\right ) = \left (\frac{a}{q}\right )$
This brings us to the Quadratic Reciprocity Theorem.
Given ${p}$ and ${q}$ odd primes:
$\displaystyle \left (\frac{-1}{p}\right ) = \begin{cases}\ \ 1\ \ p\equiv 1\pmod 4\\ -1\ \ p\equiv 3\pmod 4 \end{cases} \left (\frac{2}{p}\right ) = \begin{cases}\ \ 1\ \ p\equiv 1\ \rm{ or }\ 7\pmod 8\\ -1\ \ p\equiv 3\ \rm{ or }\ 5\pmod 8 \end{cases}$
If ${p\equiv q \equiv 3\pmod 4 }$ then
$\displaystyle \left (\frac{p}{q}\right ) = -\left (\frac{q}{p}\right )$
If ${p}$ and/or ${q \equiv 1\pmod 4 }$ then
$\displaystyle \left (\frac{p}{q}\right ) = \left (\frac{q}{p}\right )$
### 8. Galois Theory
Motivation
The absolute Galois group of ${\mathbb{Q}}$, call it ${G}$, is a group of symmetries with deep structure, most of it as yet unknown. This structure is being probed, using representations of ${G}$ by standard objects, namely permutation and matrix groups. ${G}$‘s first applications here will be for studying the structure of solution sets of ${\mathbb{Z}}$-equations, and of torsion points of elliptic curves. But ultimately, it is ${G}$‘s shadowy powers themselves that are attracting the attention of number theorists.
${G}$ consists of permutations ${g}$ on ${\overline{\mathbb{Q}}}$ that preserve the four arithmetic operations, and specifically:
$\displaystyle g(a+b)=g(a)+g(b)\rm,\ g(ab)=g(a)g(b)$
Subtraction and division follow from these properties. It is not hard to show that ${G}$ forms a group.
Exactly two elements of ${G}$ are known explicitly in their entirety: the identity permutation ${g_e : g_e(a)=a}$, and the complex conjugation permutation ${g_c: g_c(a+bi)=a-bi}$. But a few partial characterizations of elements of ${G}$ have been discovered.
Magic 1: For ${g \in G}$
$\displaystyle a \in \mathbb{Q} \implies g(a)=a$
Magic 2: For $\displaystyle f(x)=0$ a ${\mathbb{Z}}$-equation
$\displaystyle f(a)=0 \implies f(g(a))=0 \ \ \ [f(g(a))=g(f(a)]$
Magic 2.1: For ${x^2-2=0}$ (an example),
$\displaystyle g(\sqrt 2)= \begin{cases} -\sqrt 2\ \text{for half of}\ g \in G\\ \sqrt 2\ \text{for half of}\ g \in G \end{cases}$
.
The elements of ${G}$ thus permute the roots of each ${\mathbb{Z}}$-polynomial ${p(x)}$. Depending on ${p(x)}$, there are various degrees of freedom that elements of ${G}$ have for permuting the roots.
Magic 3: The existence proof for an element ${g \in G}$ involves selecting all ${p(x)}$ and determining the compatible root permutations (by application of Zorn’s Lemma).
### 9. Elliptic Curves
Motivation
An elliptic curve is a variety ${E}$ for a type of ${\mathbb{Z}}$-equation. For each number field ${A}$, ${E(A)}$ forms an Abelian group. Because of their group structure, more can be understood about elliptic curves than about varieties with unknown structure. Galois groups can be used to permute the roots of the ${\mathbb{Z}}$-equation to further assist study.
The name elliptic curve relates to using ${E}$ to study the arc-length of certain ellipses. They are called curves because ${E(\mathbb{R})}$ can be plotted and geometrically visualized.
The general ${\mathbb{Z}}$-equation related to ${E}$ is
$\displaystyle y^2=x^3+Ax+B$
where ${A}$ and ${B}$ are integers such that the discriminant ${2(4A^3+27B^2)\ne 0}$. The last condition rules out some number systems from the domain of ${E}$, as seen in the example of equation ${y^2=x^3+1}$. For this equation, the discriminant condition becomes ${6 \ne 0}$ or ${6 \not \equiv 0\pmod p}$, ruling out ${\mathbb{F}_2}$ and ${\mathbb{F}_3}$ from the domain of ${E}$. There are two non-degenerate cases. If the discriminant is ${>0}$, the curve has two components, else the curve has one component.
Note that ${E(\mathbb{Z})}$ is just a set; lack of an inverse prevents it from being a group. To make ${E(A)}$ into a group, ${A}$ must be a field and the solution set in ${A}$ must be augmented by a special identity (neutral) element called the point at infinity, ${\mathcal{O}}$, that is considered in some sense to represent the solution ${x=\infty, y=\infty}$.
Two examples for ${y^2=x^3+1}$:
$\displaystyle E(\mathbb{F}_5)=\{(0,1),(0,4), (2,3),(2,2),(4,0),\mathcal{O}\}$
$\displaystyle E(\mathbb{F}_7)=\{(0,1),(0,6), (1,3),(1,4),(2,3),(2,4),(3.0),(4,3),(4,4),(5,0),(6,0),\mathcal{O}\}$
The group operator has complex operation based on the geometry of the curve. Let ${P}$ and ${Q}$ be points on elliptic curve ${E(A)}$. Then draw a line through ${P}$ and ${Q}$. The line will intersect the curve in just three places, ${P}$, ${Q}$, and a third point ${R}$. Negating the y-coordinate of ${R}$ produces the point ${P+Q}$. This and similar geometric constructions allow ${P+Q}$ to be computed for all possible cases:
If ${P=\mathcal{O}}$, then ${P+Q=Q}$.
If ${Q=\mathcal{O}}$, then ${P+Q=P}$.
If ${x_P=x_Q}$ and ${y_P+y_Q=0}$, then ${P+Q=\mathcal{O}}$.
If ${x_P\neq x_Q}$, compute ${\lambda= (y_Q-y_P)/(x_Q-x_P)}$. If ${x_P=x_Q, y_P=y_Q\ne 0}$, compute ${\lambda=(3x_P^2+A)/2y_P}$. Let ${V=y_P-\lambda x_P}$, and compute ${x_{P+Q}=\lambda^2-x_P-x_Q}$, ${y_{P+Q}=-(\lambda x_{P+Q}+v)}$, where ${P+Q=(x_{P+Q},y_{P+Q})}$.
That there is a group lurking in this geometry is amazing and non-trivial to prove, or even to verify in the case of associativity. Inverses are simple enough; the inverse of solution ${(x,y)}$ is ${(x,-y)}$.
The equation for the elliptic curve arises naturally in a 2500 year old problem, called the congruent number problem: find all right triangles with rational sides and integer areas. Letting the area = 1, by some algebra this can be shown to be equivalent to finding ${E(\mathbb{Q})}$ where ${E}$ corresponds to ${y^2=u^3-u}$. In this geometry, the point at infinity ${\mathcal{O}}$ is given geometric perspective as a degenerate solution of a triangle with one side of length ${0}$, the other side of length ${\infty}$, and area = ${1}$.
Define an element ${P}$ of ${E(\mathbb{C})}$ as part of an ${n}$-torsion if
$\displaystyle \stackrel{n \ \rm{times}}{\overbrace{P+P+P+{ \ldots}+P}}=\mathcal{O}$
The above can be abbreviated as ${nP=\mathcal{O}}$. The set of all ${n}$-torsion points is written ${E[n]}$ and is a subgroup of ${E(A)}$. The number of elements in ${E(\overline{\mathbb{Q}})}$ that are ${n}$-torsion is ${n^2}$. The ${n}$-torsion elements of a variety operate in the same manner as the ${n}$-torsion elements of a general group, so that ${nP=\mathcal{O}}$ is analagous to ${g^n=\rm{E}_{\rm{G}}}$, where ${g}$ is an element of a group G and ${\rm{E}_{\rm{G}}}$ is the identity element in group G.
Proceed to Matrices (Also Basic Notions Of Vector Spaces) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 169, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9476955533027649, "perplexity": 179.0019296178074}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823630.63/warc/CC-MAIN-20171020025810-20171020045810-00741.warc.gz"} |
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• Författare:
## J.C. Taylor
• Förlag:
### Springer-Verlag New York Inc.
• Språk: Engelska
• Utgiven: 199807
• Antal sidor: 324
• Upplaga: 1st ed. 1997. Corr. 2nd printing 1998,
Vikt i gram: 1010
• ISBN10: 0387948309
• ISBN13: 9780387948300
# An Introduction to Measure and Probability
av
## J.C. Taylor
##### Beskrivning:
Assuming only calculus and linear algebra, Professor Taylor introduces readers to measure theory and probability, discrete martingales, and weak convergence. This is a technically complete, self-contained and rigorous approach that helps the reader to develop basic skills in analysis and probability. Students of pure mathematics and statistics can thus expect to acquire a sound introduction to basic measure theory and probability, while readers with a background in finance, business, or engineering will gain a technical understanding of discrete martingales in the equivalent of one semester. J. C. Taylor is the author of numerous articles on potential theory, both probabilistic and analytic, and is particularly interested in the potential theory of symmetric spaces. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8187130093574524, "perplexity": 3936.9875761742037}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107902745.75/warc/CC-MAIN-20201029040021-20201029070021-00370.warc.gz"} |
https://www.positronx.io/solving-node-is-not-recognized-as-an-internal-or-external-command/ | # Solving node is not recognized as an internal or external command
Last updated on: by Editorial Team
Have you ever got this node error “node is not recognized as an internal or external command”? I got into the problem when i found this node error in my command line. This error message can be daunting. I searched the Internet and came to find out that the problem is this node executable was not in my PATH. It was missing for sure.
So I added the node to the path and then restarted the visual studio only to see that the application is working once again. Here are the steps that you will need to follow to get rid of the “the node executable was not in my PATH” error.
• Open control panel => system & security => system => advanced system settings => environment variables
• In “User variables” or “System variables” look for variable PATH and include Node JS folder path. Often it is C:\Program Files\nodejs folder;. If you don’t get variable, then don’t hesitate to create it
Once you have done that, restart the Visual Studio or once again open a new command prompt. In the word bar, enter `'node -v'` to copy the node version which is already installed. Another way to enter the PATH to any other application is directly from the command line. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.875589907169342, "perplexity": 1384.6165057534542}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585671.36/warc/CC-MAIN-20211023095849-20211023125849-00045.warc.gz"} |
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## tbrooks3 Group Title A banner in the shape of an isosceles triangle has a base that is 2 inches shorter than either of the equal sides. If the perimeter of the banner is 43 inches, then what is the length of the equal sides? 2 years ago 2 years ago Edit Question Delete Cancel Submit
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1. FoolForMath
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Let the length of the equal sides be \(x\) inches then: \(2x + x-2 = 43 \implies 3x = 45 \implies x = 15 \) inches
• 2 years ago
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1
We know that the sum of the three sides (the perimeter) is 43 inches. Let x = lenght of one of the equal sides then the base is x-2 inches (given) lets add up the three sides x-2 + x + x which is 3x-2 and equal to 43 3x-2 = 43 3x=45 x=15 in. the equal sides are 15, and the base is 15-2 or 13 inches.
• 2 years ago
3. tbrooks3
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okay i figured u would do it like that but i wasnt sure thanks u 2.
• 2 years ago
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9846197366714478, "perplexity": 3732.1597845536403}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413558066654.17/warc/CC-MAIN-20141017150106-00006-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/8240-probability.html | 1. Probability
Need some help with these please:
The probability that I catch a fish when it is raining is 0.65. The probability that I catch a fish when it is not raining is 0.40. The probability that it will rain tomorrow is 0.70. What is the probability that I catch a fish if I go fishing tomorrow?
In a contest, I get to draw a number from 1-10 from a hat, and roll a die. I win if I get either an odd number from the hat, OR a 3 on the die. What is the probability that I will win with 1 draw and 1 roll?
and
P(E)=0.3, P(F)=0.5, P(E∩F)=0.1
And I have to find:
P(E|F) and P(F|E)
Thanks for any help.
2. For #1. The probability of a catch, C, depends on rain, R, and no rain, R’.
$\begin{array}{rcl}
P(C) & = & P(CR) + P(CR') \\
& = & P(C|R)P(R) + P(C|R')P(R') \\
\end{array}.$
For #2. You have independence so:
$\begin{array}{rcl}
P(O \cup 3) & = & P(O) + P(3) - P(O \cap 3) \\
& = & P(O) + P(3) - P(O)P(3) \\
\end{array}.$
For #3.
$P(A|B) = \frac{{P(A \cap B)}}{{P(B)}}$
3. Originally Posted by Plato
For #1. The probability of a catch, C, depends on rain, R, and no rain, R’.
$\begin{array}{rcl}
P(C) & = & P(CR) + P(CR') \\
& = & P(C|R)P(R) + P(C|R')P(R') \\
\end{array}.$
For #2. You have independence so:
$\begin{array}{rcl}
P(O \cup 3) & = & P(O) + P(3) - P(O \cap 3) \\
& = & P(O) + P(3) - P(O)P(3) \\
\end{array}.$
For #3.
$P(A|B) = \frac{{P(A \cap B)}}{{P(B)}}$
So for #2, I got 0.577.
#3 I got (P|E)=0.2 and (P|F)=0.333...
are those right? I'm still working on #1. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8487562537193298, "perplexity": 974.266207855837}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281263.12/warc/CC-MAIN-20170116095121-00103-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://dantopology.wordpress.com/2009/10/17/lindelof-x-metric-needs-not-be-normal/ | # Lindelof x Metric Needs Not Be Normal
In a previous post, an example for Paracompact x Metric needs not be normal was given. The example is product of the Michael Line and the space of irrationals. In another post, I showed that as long as the set of isolated points in the “Michael Line” construction is not an $F_\sigma-$set, we can obtain a non-normal product. I present here another example using the same technique, but this time the set of isolated points is a Bernstein set. This produces a Lindelof space and the cross product of this Lindelof space with the Bernstein set with the usual topology is not normal.
As in the previous post, let $H \subset \mathbb{R}$. Define a new topology on the real line $\mathbb{R}$ by using open sets of the form $U \cup V$ where $U$ is an open set in the usual topology and $V \subset H$. In this notation, the Michael Line is $\mathbb{R}(\mathbb{P})$ where $\mathbb{P}$ is the space of all irrational numbers. I proved in the previous post that the space $\mathbb{R}(H) \times H$ is normal if and only if the set $H$ is an $F_\sigma-$set in the real line. Thus $\mathbb{R}(\mathbb{P}) \times \mathbb{P}$ is not normal.
I consider the example $\mathbb{R}(S)$ where $S \subset \mathbb{R}$ and both $S$ and its complement $\mathbb{R}-S$ contain no uncountable compact subset of the line. Such a set is called a Bernstein set. The space $\mathbb{R}(S)$ is Lindelof. Let $\mathcal{W}$ be an open cover of $\mathbb{R}-S$ consisting of open sets in the usual topology. Then $\mathcal{W}$ has to cover all of the real line except for countably many points. Otherwise $\mathbb{R}-\cup \mathcal{W}$ is an uncountable closed set, which contains an uncountable compact set that is contained in $S$. Thus any open cover of $\mathbb{R}-S$ made up of usual open sets has to cover the real line except for countably many points. Now let $\mathcal{U}$ be an open cover of $\mathbb{R}(S)$. The usual open sets in $\mathcal{U}$ that covers $\mathbb{R}-S$ has a countable subcover. This countable subcover covers all of the real line except for countably many points. Thus $\mathcal{U}$ has a countable subcover.
Since a Berstein set cannot be an $F_\sigma-$set, $\mathbb{R}(S) \times S$ is not normal. Thus the cross product of a Lindelof space and a separable metric space needs not be normal. Is there a non-nonmal example of hereditarily Lindelof x separable metric space? If there is, it will not be using this “Michael Line” type construction. It is easy to see that $\mathbb{R}(H)$ is hereditarily Lindelof if and only if $H$ is countable. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 32, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9852916598320007, "perplexity": 85.19424708017485}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128321458.47/warc/CC-MAIN-20170627152510-20170627172510-00660.warc.gz"} |
https://www.physicsforums.com/threads/power-of-resistors-in-parallel-vs-series-circuit.916049/ | # Power of resistors in parallel vs series circuit
• Thread starter crostud123
• Start date
#### crostud123
3
0
1. Homework Statement
Two Ohm's resistors in parallel consume power of 76 W and 24 W.
What power will each of them consume if transfer them to the serial circuit with the same voltage of source.
Sorry for my bad english translations I m not use to writing questions in English.. So what confuses me is that on this forum similar question was asked and it stated that Power in serial circuit is 1/P=1/P1+1/P2. (https://www.physicsforums.com/threads/equivalent-power-in-series-and-parallel-combination.827649/page-2 ) Should I use that formula or not? I was thinking that both in parallel and serial power is just P=P1+P1, and power in parallel is therefore 100W, but power of serial is less because total serial resistance is greater than individual resistance of 1 resistor, while in parallel total resistance is less than individual resistance of 1 resistor, so total power would be greater in parallel circuit than in serial circuit?
2. Homework Equations
3. The Attempt at a Solution
In parallel 1/R=1/R1+1/R2 and power is P=V^2/R
IN series R=R+R2 but power here is less since we have the same R1 and R2 but greater R(total) in serial makes it that P(total) in serial is less thatn P(total) in parallel???
Does this make sense or should it be like the previously asked question here where P in series is 1/P=1/P1+1/P2 ?
Last edited by a moderator:
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#### BvU
Homework Helper
12,487
2,799
Hello crostud,
It would be better if you understood the characteristics of parallel and series circuits. The proper formulas to use follow from that understanding.
If the resistors are in parallel, the voltage over each of them is the same.
You already have $P=V^2/R$ They give the power ratio, so $${V^2/R_1 \over 76 } = {V^2/R_2 \over 24 }\ .$$Now what is asked is the power for the serial circuit, with resistance $R_1 + R_2$. The expression for that power is $V^2\over R_1+R_2$. From now on it is math: express e.g. $R_2$ in terms of $R_1$ and calculate the power.
#### crostud123
3
0
Thank you very much but I got the relation between R2 and R1 and I still can't calculate P1 and P2 in serial (power of each resistant like it's asked from me). I don't know where to go from after I got the relation between R1 and R2 since I don't know whats the V, I only know that it's the same in parallel and serial.
#### Carrock
46
10
Thank you very much but I got the relation between R2 and R1 and I still can't calculate P1 and P2 in serial (power of each resistant like it's asked from me). I don't know where to go from after I got the relation between R1 and R2 since I don't know whats the V, I only know that it's the same in parallel and serial.
You know the power with V across each resistor.
With the resistors in series and their ratios known, you can calculate the voltage across each in terms of V e.g. V-x,x.
You can then calculate the power with V-x or x across each resistor compared to V.
You can't calculate and don't need the actual values of V or R1 or R2 or the current.
#### crostud123
3
0
You know the power with V across each resistor.
With the resistors in series and their ratios known, you can calculate the voltage across each in terms of V e.g. V-x,x.
You can then calculate the power with V-x or x across each resistor compared to V.
You can't calculate and don't need the actual values of V or R1 or R2.
Sorry, still quite don't understand. I always get to a dead end :(
#### BvU
Homework Helper
12,487
2,799
R1 and R2 since I don't know whats the V
Show your work. Just shouting I dunno doesn't help you further.
You have $V^2/R_1$; what doe you have for $R_2$ in terms of $R_1$ ? Why is it so difficult to calculate $V^2 \over R_1+R_2$ ?
#### Carrock
46
10
Why is it so difficult to calculate $V^2 \over R_1+R_2$ ?
I would find it difficult to do this; as it is not the question asked I won't try.
This is correct.
In parallel 1/R=1/R1+1/R2 and power is P=V^2/R
IN series R=R1+R2 but power here is less since we have the same R1 and R2 but greater R(total) in serial makes it that P(total) in serial is less than P(total) in parallel???
From #2
$${V^2/R_1 \over 76} = {V^2/R_2 \over 24 }\ .$$
You can work out from this the numeric ratio ${R_1 \over R_2}$.
I don't know whats the V, I only know that it's the same in parallel and serial.
WRONG!!! or English error... the voltage across each resistor in series (not serial) is less than V.
Hint: a related simpler problem:
If you have (say) two resistors in series, one twice as big as the other, connected across any nonzero voltage (say X), can you say anything about the ratio of the voltages across those resistors and express the voltages in terms of X? What must be the sum of those two voltages?
I expect you to also get a similar expression for the main problem; you may then see how to solve it.
Your original attempt was a good start; this is as far as I can presently go without (I hope) breaching forum rules.
#### BvU
Homework Helper
12,487
2,799
What power will each of them consume if transfer them to the serial circuit with the same voltage of source
Once you have the sum of powers, it's easy to find the separate ones.
I would find it difficult to do this; as it is not the question asked I won't try.
You don't have to: it's crostud's exercise.
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15K | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8103967308998108, "perplexity": 806.3499770690439}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670389.25/warc/CC-MAIN-20191120010059-20191120034059-00106.warc.gz"} |
https://mathoverflow.net/questions/322633/gromov-witten-invariants-and-the-mod-2-spectral-flow | # Gromov-Witten invariants and the mod 2 spectral flow
I'm reading Lee-Parker, “A structure theorem for the Gromov-Witten invariants of Kähler surfaces”, which studies Gromov-Witten invariants within symplectic geometry. Lee-Parker write (§9, p. 23) that
In Gromov-Witten theory, the GW invariant associated with a zero-dimensional space of stable maps is the signed count of the maps in that space with the sign of each map $$f$$ specified by the mod 2 spectral flow of the linearization $$D_f$$ (provided each $$D_f$$ is an isomorphism).
I would like to understand why this is, but they don't provide a reference, and I wasn't able to find anything discussing this relationship by searching online. I would guess the mod 2 spectral flow appears somehow in the construction of the virtual fundamental class, but I don't know enough about that construction to fill in the details.
Is there a reference that explains why this fact is true? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8328011631965637, "perplexity": 295.9847597327117}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703515075.32/warc/CC-MAIN-20210118154332-20210118184332-00638.warc.gz"} |
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